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can you prove it… combinatorially?  http://www.mathematicalgemstones.com/gemstones/canyouproveitcombinatorially/ 
phinished!  http://www.mathematicalgemstones.com/gemstones/amber/phinished/ 
what do schubert curves, young tableaux, and ktheory have in common? (part iii)  http://www.mathematicalgemstones.com/gemstones/diamond/whatdoschubertcurvesyoungtableauxandktheoryhaveincommonpartiii/ 
what do schubert curves, young tableaux, and ktheory have in common? (part ii)  http://www.mathematicalgemstones.com/gemstones/diamond/whatdoschubertcurvesyoungtableauxandktheoryhaveincommonpartii/ 
can you prove it… combinatorially?  http://www.mathematicalgemstones.com/gemstones/canyouproveitcombinatorially/#comment696 
can you prove it… combinatorially?  http://www.mathematicalgemstones.com/gemstones/canyouproveitcombinatorially/#comment695 
can you prove it… combinatorially?  http://www.mathematicalgemstones.com/gemstones/canyouproveitcombinatorially/#comment694 
can you prove it… combinatorially?  http://www.mathematicalgemstones.com/gemstones/canyouproveitcombinatorially/#comment693 
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mathematical gemstones on the quest for mathematical beauty and truth search main menu skip to primary content skip to secondary content homeabout this blogcontribute!all posts post navigation ← older posts the structure of the garsiaprocesi modules $r_\mu$ posted on september 23, 2016 by maria gillespie reply somehow, in all the time i’ve posted here, i’ve not yet described the structure of my favorite graded $s_n$modules. i mentioned them briefly at the end of the springer correspondence series, and talked in depth about a particular one of them – the ring of coinvariants – in this post, but it’s about time for… the garsiaprocesi modules! also known as the cohomology rings of the springer fibers in type $a$, or as the coordinate ring of the intersection of the closure of a nilpotent conjugacy class of $n\times n$ matrices with a torus, with a link between these two interpretations given in a paper of deconcini and procesi. but the work of tanisaki, and garsia and procesi, allows us to work with these modules in an entirely elementary way. using tanisaki’s approach, we can define $$r_{\mu}=\mathbb{c}[x_1,\ldots,x_n]/i_{\mu},$$ where $i_{\mu}$ is the ideal generated by the partial elementary symmetric functions defined as follows. recall that the elementary symmetric function $e_d(z_1,\ldots,z_k)$ is the sum of all squarefree monomials of degree $d$ in the set of variables $z_i$. let $s\subset\{x_1,\ldots,x_n\}$ with $s=k$. then the elementary symmetric function $e_r(s)$ in this subset of the variables is called a partial elementary symmetric function, and we have $$i_{\mu}=(e_r(s) : kd_k(\mu) \lt r \le k, s=k).$$ here, $d_k(\mu)=\mu’_n+\mu’_{n1}+\cdots+ \mu’_{nk+1}$ is the number of boxes in the last $k$ columns of $\mu$, where we pad the transpose partition $\mu’$ with $0$’s so that it has $n$ parts. this ring $r_\mu$ inherits the natural action of $s_n$ on $\mathbb{c}[x_1,\ldots,x_n]$ by permuting the variables, since $i_\mu$ is fixed under this action. since $i_\mu$ is also a homogeneous ideal, $r_\mu$ is a graded $s_n$module, graded by degree. to illustrate the construction, suppose $n=4$ and $\mu=(3,1)$. then to compute $i_{\mu}$, first consider subsets $s$ of $\{x_1,\ldots,x_4\}$ of size $k=1$. we have $d_1(\mu)=0$ since the fourth column of the young diagram of $\mu$ is empty (see image below), and so in order for $e_r(s)$ to be in $i_\mu$ we must have $10\lt r \le 1$, which is impossible. so there are no partial elementary symmetric functions in $1$ variable in $i_\mu$. for subsets $s$ of size $k=2$, we have $d_2(\mu)=1$ since there is one box among the last two columns (columns $3$ and $4$) of $\mu$, and we must have $21\lt r\le 2$. so $r$ can only be $2$, and we have the partial elementary symmetric functions $e_2(s)$ for all subsets $s$ of size $2$. this gives us the six polynomials $$x_1x_2,\hspace{0.3cm} x_1x_3,\hspace{0.3cm} x_1x_4,\hspace{0.3cm} x_2x_3,\hspace{0.3cm} x_2x_4,\hspace{0.3cm} x_3x_4.$$ for subsets $s$ of size $k=3$, we have $d_3(\mu)=2$, and so $32 \lt r\le 3$. we therefore have $e_2(s)$ and $e_3(s)$ for each such subset $s$ in $i_\mu$, and this gives us the eight additional polynomials $$x_1x_2+x_1x_3+x_2x_3, \hspace{0.5cm}x_1x_2+x_1x_4+x_2x_4,$$ $$x_1x_3+x_1x_4+x_3x_4,\hspace{0.5cm} x_2x_3+x_2x_4+x_3x_4,$$ $$x_1x_2x_3, \hspace{0.4cm} x_1x_2x_4, \hspace{0.4cm} x_1x_3x_4,\hspace{0.4cm} x_2x_3x_4$$ finally, for $s=\{x_1,x_2,x_3,x_4\}$, we have $d_4(\mu)=4$ and so $44\lt r\le 4$. thus all of the full elementary symmetric functions $e_1,\ldots,e_4$ in the four variables are also relations in $i_{\mu}$. all in all we have $$\begin{align*} i_{(3,1)}= &(e_2(x_1,x_2), e_2(x_1,x_3),\ldots, e_2(x_3,x_4), \\ & e_2(x_1,x_2,x_3), \ldots, e_2(x_2,x_3,x_4), \\ & e_3(x_1,x_2,x_3), \ldots, e_3(x_2,x_3,x_4), \\ & e_1(x_1,\ldots,x_4), e_2(x_1,\ldots,x_4), e_3(x_1,\ldots,x_4), e_4(x_1,\ldots,x_4)) \end{align*}$$ as two more examples, it’s clear that $r_{(1^n)}=\mathbb{c}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$ is the ring of coninvariants under the $s_n$ action, and $r_{(n)}=\mathbb{c}$ is the trivial representation. so $r_\mu$ is a generalization of the coinvariant ring, and in fact the graded frobenius characteristic of $r_\mu$ is the halllittlewood polynomial $\widetilde{h}_\mu(x;q)$. where do these relations come from? the rings $r_\mu$ were originally defined as follows. let $a$ be a nilpotent $n\times n$ matrix over $\mathbb{c}$. then $a$ has all $0$ eigenvalues, and so it is conjugate to a matrix in jordan normal form whose jordan blocks have all $0$’s on the diagonal. the sizes of the jordan blocks, written in nonincreasing order form a partition $\mu’$, and this partition uniquely determines the conjugacy class of $a$. in other words: there is exactly one nilpotent conjugacy class $c_{\mu’}$ in the space of $n\times n$ matrices for each partition $\mu’$ of $n$. the closures of these conjugacy classes $\overline{c_{\mu’}}$ form closed matrix varieties, and their coordinate rings were studied here. however, they are easier to get a handle on after intersecting with the set $t$ of diagonal matrices, leading to an interesting natural question: what is the subvariety of diagonal matrices in the closure of the nilpotent conjugacy class $c_{\mu’}$? defining $r_\mu=\mathcal{o}(\overline{c_{\mu’}}\cap t)$, we obtain the same modules as above. tanisaki found the presentation for $r_\mu$ given above using roughly the following argument. consider the matrix $ati$, where $a\in c_{\mu’}$. then one can show (see, for instance, the discussion of invariant factors and elementary divisors in the article on smith normal form on wikipedia) that the largest power of $t$ dividing all of the $k\times k$ minors of $ati$, say $t^{d_k}$, is fixed under conjugation, so we can assume $a$ is in jordan normal form. then it’s not hard to see, by analyzing the jordan blocks, that this power of $t$ is $t^{d_k(\mu)}$ where $\mu$ is the transpose partition of $\mu’$ and $d_k(\mu)$ is defined as above – the sums of the ending columns of $\mu$. it follows that any element of the closure of $c_\mu$ must also have this property, and so if $x=\mathrm{diag}(x_1,\ldots,x_n)\in \overline{c_\mu}\cap t$ then we have $$t^{d_k(\mu)}  (x_{i_1}t)(x_{i_2}t)\cdots (x_{i_k}t)$$ for any subset $s=\{x_{i_1},\ldots,x_{i_k}\}$ of $\{x_1,\ldots,x_n\}$. expanding the right hand side as a polynomial in $t$ using vieta’s formulas, we see that the elementary symmetric functions $e_r(s)$ vanish on $x$ as soon as $r \gt kd_k(\mu)$, which is exactly the relations that describe $i_\mu$ above. it takes somewhat more work to prove that these relations generate the entire ideal, but this can be shown by showing that $r_\mu$ has the right dimension, namely the multinomial coefficient $\binom{n}{\mu}$. and for that, we’ll discuss on page 2 the monomial basis of garsia and procesi. pages: 1 2 posted in diamond, gemstones  leave a reply can you prove it… combinatorially? posted on august 12, 2016 by maria gillespie 4 this year’s prove it! math academy was a big success, and it was an enormous pleasure to teach the seventeen talented high school students that attended this year. some of the students mentioned that they felt even more inspired to study math further after our twoweek program, but the inspiration went both ways – they inspired me with new ideas as well! one of the many, many things we investigated at the camp was the fibonacci sequence, formed by starting with the two numbers $0$ and $1$ and then at each step, adding the previous two numbers to form the next: $$0,1,1,2,3,5,8,13,21,\ldots$$ if $f_n$ denotes the $(n+1)$st term of this sequence (where $f_0=0$ and $f_1=1$), then there is a remarkable formula for the $n$th term, known as binet’s formula: $$f_n=\frac{1}{\sqrt{5}}\left( \left(\frac{1+\sqrt{5}}{2}\right)^n – \left(\frac{1\sqrt{5}}{2}\right)^n \right)$$ looks crazy, right? why would there be $\sqrt 5$’s showing up in a sequence of integers? at prove it!, we first derived the formula using generating functions. i mentioned during class that it can also be proven by induction, and later, one of our students was trying to work out the induction proof on a white board outside the classroom. she was amazed how many different proofs there could be of the same fact, and it got me thinking: what if we expand each of the terms using the binomial theorem? is there a combinatorial proof of the resulting identity? in particular, suppose we use the binomial theorem to expand $(1+\sqrt{5})^n$ and $(1\sqrt{5})^n$ in binet’s formula. the resulting expression is: $$f_n=\frac{1}{\sqrt{5}\cdot 2^n}\left( \left(\sum_{i=0}^n \binom{n}{i}(\sqrt{5})^i \right) – \left(\sum_{i=0}^n (1)^i\binom{n}{i}(\sqrt{5})^i \right) \right)$$ note that the even terms in the two summations cancel, and combining the odd terms gives us: $$f_n=\frac{1}{\sqrt{5}\cdot 2^n}\left( \sum_{k=0}^{\lfloor n/2\rfloor} 2 \binom{n}{2k+1}(\sqrt{5})^{2k+1} \right)$$ since $(\sqrt{5})^{2k+1}=\sqrt{5}\cdot 5^k$, we can cancel the factors of $\sqrt{5}$ and multiply both sides by $2^{n1}$ to obtain: $$2^{n1}\cdot f_n=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k+1}\cdot 5^k.$$ now, the left hand and right hand side are clearly nonnegative integers, and one handy fact about nonnegative integers is that they count the number of elements in some collection. the proof method of counting in two ways is the simple principle that if by some method one can show that a collection $a$ has $n$ elements, and by another method one can show that $a$ has $m$ elements, then it follows that $m=n$. such a “combinatorial proof” may be able to be used to prove the identity above, with $m$ being the left hand side of the equation and $n$ being the right. i started thinking about this after prove it! ended, and remembered that the $(n+1)$st fibonacci number $f_n$ counts the number of ways to color a row of $n2$ fenceposts either black or white such that no two adjacent ones are black. (can you see why this combinatorial construction would satisfy the fibonacci recurrence?) for instance, we have $f_5=5$, because there are five such colorings of a row of $3$ fenceposts: $$\begin{array}{ccc} \square & \square & \square \\ \\ \square & \square & \blacksquare \\ \\ \square & \blacksquare & \square \\ \\ \blacksquare & \square & \square \\ \\ \blacksquare & \square & \blacksquare \end{array}$$ note also that $2^{n1}$ counts the number of length$(n1)$ sequences of $0$’s and $1$’s. thus, the left hand side of our identity, $2^{n1}\cdot f_n$, counts the number of ways of choosing a binary sequence of length $n1$ and also a fence post coloring of length $n2$. because of their lengths, given such a pair we can interlace their entries, forming an alternating sequence of digits and fence posts such as: $$1\, \square\, 0\, \square\, 1\, \blacksquare\, 1$$ we will call such sequences interlaced sequences. we now need only to show that the right hand side also counts these interlaced sequences. see the next page for my solution, or post your own solution in the comments below! pages: 1 2 posted in gemstones, opal  4 replies phinished! posted on may 24, 2016 by maria gillespie reply sometimes it’s the missteps in life that lead to the greatest adventures down the road. for me, my pursuit of a ph.d. in mathematics, specifically in algebraic combinatorics, might be traced back to my freshman year as an undergraduate at mit. coming off of a series of successes in high school math competitions and other sciencerelated endeavors (thanks to my loving and very mathematical family!), i was a confident and excited 18year old whose dream was to become a physicist and use my mathematical skills to, i don’t know, come up with a unified field theory or something. me at the age of 18ish. but i loved pure math too, and a number of my friends were signed up for the undergraduate algebraic combinatorics class in the spring, so my young ambitious self added it to my already packed course load. i had no idea what “algebraic combinatorics” even meant, but i did hear that it was being taught by richard stanley, a world expert in the area. how could i pass up that chance? what if he didn’t teach it again before i left mit? on the first day of the class, stanley started with a simple combinatorial question. it was something like the following: in a complete graph with $n$ vertices, how many walks of length $k$ starting at vertex $v$ end up back at vertex $v$ on the last step? for instance, if $n=5$ and $k=2$, the graph looks like: and there are four closed walks of length two, from $v$ to any other vertex and back again: there is an elementary (though messy) way to solve this, but stanley went forth with an algebraic proof. he considered the adjacency matrix $a$ of the complete graph, and showed that the total number of loops of length $k$ starting from any vertex is the trace of $a^k$. one can then compute this trace using eigenvalues and divide by $n$ to get the number of loops starting at $v$. beautiful! i remember sitting in my seat, wideeyed, watching richard stanley quietly but authoritatively discuss the technique. it was incredible to me that advanced tools from linear algebra could be used to so elegantly solve such a simple, concrete problem. to use a term from another area of algebraic combinatorics, i was hooked. but i was also a freshman, and didn’t yet have a strong grasp of some of the other algebraic concepts being used in the course. i studied hard but wound up with a b+ in the class. me, get a b+ in a math class? i was horrified, my 18yearold littlemissperfect confidence shattered. now, not only was i fascinated with the subject, i gained respect for it. it was a worthy challenge, and i couldn’t help but come back for more. in the years that followed, i took more courses on similar subjects and wrote several undergraduate research papers. i dabbled in other areas as well, but was always drawn back to the interplay between combinatorics and algebra. i now find myself, as of friday, may 20, 2016, having completed my ph.d. at uc berkeley on a topic in algebraic combinatorics… …and i often wonder how much that silly little b+ motivated me throughout the years. (see page 2 for a summary of my thesis. my full thesis can be found here.) pages: 1 2 3 posted in amber, diamond  leave a reply what do schubert curves, young tableaux, and ktheory have in common? (part iii) posted on april 5, 2016 by maria gillespie 3 this is the third and final post in our expository series of posts (see part i and part ii) on the recent paper coauthored by jake levinson and myself. last time, we discussed the fact that the operator $\omega$ on certain young tableaux is actually the monodromy operator of a certain covering map from the real locus of the schubert curve $s$ to $\mathbb{rp}^1$. now, we’ll show how our improved algorithm for computing $\omega$ can be used to approach some natural questions about the geometry of the curve $s$. for instance, how many (complex) connected components does $s$ have? what is its genus? is $s$ a smooth (complex) curve? the genus of $s$ the arithmetic genus of a connected curve $s$ can be defined as $g=1\chi(\mathcal{o}_s)$ where $$\chi(\mathcal{o}_s)=\dim h^0(\mathcal{o}_s)\dim h^1(\mathcal{o}_s)$$ is the euler characteristic and $\mathcal{o}_s$ is the structure sheaf. so, to compute the genus it suffices to compute the euler characteristic, which can alternatively be defined in terms of the $k$theory of the grassmannian. the $k$theory ring $k(\mathrm{gr}(n,k))$ the $k$theory ring $k(x)$ of a scheme $x$ is defined as follows. first, consider the free abelian group $g$ generated by isomorphism classes of locally free coherent sheaves (a.k.a. vector bundles) on $x$. then define $k(x)$, as a group, to be the quotient of $g$ by “short exact sequences”, that is, the quotient $g/h$ where $h$ is the subgroup generated by expressions of the form $[\mathcal{e}_1]+[\mathcal{e}_2][\mathcal{e}]$ where $0\to \mathcal{e}_1 \to \mathcal{e} \to \mathcal{e}_2 \to 0$ is a short exact sequence of vector bundles on $x$. this gives the additive structure on $k(x)$, and the tensor product operation on vector bundles makes it into a ring. it turns out that, in the case that $x$ is smooth (such as a grassmannian!) then we get the exact same ring if we remove the “locally free” restriction and consider coherent sheaves modulo short exact sequences. where does this construction come from? well, a simpler example of $k$theory is the construction of the grothendieck group of an abelian monoid. consider an abelian monoid m (recall that a monoid is a set with an associative binary operation and an identity element, like a group without inverses). we can construct an associated group $k(m)$ by taking the quotient free abelian group generated by elements $[m]$ for $m\in m$ by the subgroup generated by expressions of the form $[m]+[n][m+n]$. so, for instance, $k(\mathbb{n})=\mathbb{z}$. in a sense we are groupifying monoids. the natural monoidal operation on vector spaces is $\oplus$, so if $x$ is a point, then all short exact sequences split and the $k$theory ring $k(x)$ is the grothendieck ring of this monoid. a good exposition on the basics of $k$theory can be found here, and for the $k$theory of grassmannians, see buch’s paper. for now, we’ll just give a brief description of how the $k$theory of the grassmannian works, and how it gives us a handle on the euler characteristic of schubert curves. recall from this post that the cw complex structure given by the schubert varieties shows that the classes $[\omega_\lambda]$, where $\lambda$ is a partition fitting inside a $k\times (nk)$ rectangle, generate the cohomology ring $h^\ast(\mathrm{gr}(n,k))$. similarly, the $k$theory ring is a filtered ring generated by the classes of the coherent sheaves $[\mathcal{o}_{\lambda}]$ where if $\iota$ is the inclusion map $\iota:\omega_\lambda\to \mathrm{gr}(n,k)$, then $\mathcal{o}_\lambda=\iota_\ast \mathcal{o}_{\omega_\lambda}$. multiplication of these basic classes is given by a variant of the littlewoodrichardson rule: $$[\mathcal{o}_\lambda]\cdot [\mathcal{o}_\nu]=\sum_\nu (1)^{\nu\lambda\mu}c^\nu_{\lambda\mu}[\mathcal{o}_\nu]$$ where if $\nu=\lambda+\mu$ then $c^{\nu}_{\lambda\mu}$ is the usual littlewoodrichardson coefficient. if $\nu\lambda+\mu$ then $c^{\nu}_{\lambda\mu}$ is a nonnegative integer. we will refer to these nonnegative values as $k$theory coefficients. $k$theory and the euler characteristic the $k$theory ring is especially useful in computing euler characteristics. it turns out that the euler characteristic gives an (additive) group homomorphism $\chi:k(x)\to \mathbb{z}$. to show this, it suffices to show that if $0\to \mathcal{a}\to \mathcal{b}\to \mathcal{c}\to 0$ is a short exact sequence of coherent sheaves on $x$, then $\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})=0$. indeed, such a short exact sequence gives rise to a long exact sequence in cohomology: $$ \begin{array}{cccccc} & h^0(\mathcal{a}) & \to & h^0(\mathcal{b}) & \to & h^0(\mathcal{c}) \\ \to &h^1(\mathcal{a}) & \to & h^1(\mathcal{b}) & \to & h^1(\mathcal{c}) \\ \to &h^2(\mathcal{a}) & \to & h^2(\mathcal{b}) & \to & h^2(\mathcal{c}) \\ \cdots & & & & & \end{array} $$ and the alternating sum of the dimensions of any exact sequence must be zero. thus we have $$\begin{eqnarray*}0&=&\sum_i (1)^i\dim h^i(\mathcal{a})\sum_i (1)^i\dim h^i(\mathcal{b})+\sum_i (1)^i\dim h^i(\mathcal{c}) \\ &=&\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})\end{eqnarray*}$$ as desired. therefore, it makes sense to talk about the euler characteristic of a class of coherent sheaves in $k(x)$. in fact, in our situation, we have a closed subset $s$ of $x=\mathrm{gr}(n,k)$, say with inclusion map $j:s\to x$, and so the euler characteristic of the pushforward $j_\ast\mathcal{o}_s$ is equal to $\chi(\mathcal{o}_s)$ itself. we can now compute the euler characteristic $\chi(j_\ast\mathcal{o}_s)$ using the structure of the $k$theory ring of the grassmannian. indeed, $s$ is the intersection of schubert varieties indexed by the three partitions $\alpha$, $\beta$, and $\gamma$ (see part i). so in the $k$theory ring, if we identify structure sheaves of closed subvarieties with their pushforwards under inclusion maps, we have $$[\mathcal{o}_s]=[\mathcal{o}_\alpha]\cdot [\mathcal{o}_\beta]\cdot [\mathcal{o}_\gamma].$$ by the $k$theoretic littlewoodrichardson rule described above, this product expands as a sum of integer multiples of classes $[\mathcal{o}_\nu]$ where $\nu\ge \alpha+\beta+\gamma$. but in our setup we have $\alpha+\beta+\gamma=k(nk)1$, so $\nu$ is either the entire $k\times (nk)$ rectangle (call this $\rho$) or it is the rectangle minus a single box (call this $\rho’$). in other words, we have: $$[\mathcal{o}_s]=c^{\rho’}_{\alpha,\beta,\gamma}[\mathcal{o}_{\rho’}]k[\mathcal{o}_{\rho}]$$ where $k$ is an integer determined by the $k$theory coefficients. notice that $c^{\rho’}_{\alpha,\beta,\gamma}$ is the usual littlewoodrichardson coefficient, and counts exactly the size of the fibers (the set $\omega$ acts on) in our map from part ii. let’s call this number $n$. finally, notice that $\omega_\rho$ and $\omega_{\rho’}$ are a point and a copy of $\mathbb{p}^1$ respectively, and so both have euler characteristic $1$. it follows that $$\chi(\mathcal{o}_s)=nk.$$ going back to the genus, we see that if $s$ is connected, we have $g=1\chi(\mathcal{o}_s)=k(n1)$. computing $k$ in terms of $\omega$ the fascinating thing about our algorithm for $\omega$ is that certain steps of the algorithm combinatorially correspond to certain tableaux that enumerate the $k$theory coefficients, giving us information about the genus of $s$. these tableaux are called “genomic tableaux”, and were first introduced by pechenik and yong. in our case, the genomic tableaux that enumerate $k$ can be defined as follows. the data of a tableau $t$ and two marked squares $\square_1$ and $\square_2$ in $t$ is a genomic tableau if: the marked squares are nonadjacent and contain the same entry $i$, there are no $i$’s between $\square_1$ and $\square_2$ in the reading word of $t$, if we delete either $\square_1$ or $\square_2$, every suffix of the resulting reading word is ballot (has more $j$’s than $j+1$’s for all $j$). for instance, consider the following tableaux with two marked (shaded) squares: property 1 means that the fourth tableau is not genomic: the marked squares, while they do contain the same entry, are adjacent squares. the first tableau above violates property 2, because there is a $2$ between the two marked $2$’s in reading order. finally, the second tableau above violates property 3, because if we delete the top marked $1$ then the suffix $221$ is not ballot. the third tableau above satisfies all three properties, and so it is genomic. finally, consider the algorithm for $\omega$ described in part i. jake and i discovered that the steps in phase 1 in which the special box does not move to an adjacent square are in bijective correspondence with the $k$theory tableau of the total skew shape $\gamma^c/\alpha$ and content $\beta$ (where the marked squares only add $1$ to the total number of $i$’s). the correspondence is formed by simply filling the starting and ending positions of the special box with the entry $i$ that it moved past, and making these the marked squares of a genomic tableau. in other words: the $k$theory coefficient $k$ is equal to the number of nonadjacent moves in all phase 1’s of the local algorithm for $\omega$. geometric consequences this connection allows us to get a handle on the geometry of the schubert curves $s$ using our new algorithm. as one illuminating example, let’s consider the case when $\omega$ is the identity permutation. it turns out that the only way for $\omega$ to map a tableau back to itself is if phase 1 consists of all vertical slides and phase 2 is all horizontal slides; then the final shuffle step simply reverses these moves. this means that we have no nonadjacent moves, and so $k=0$ in this case. since $\omega$, the monodromy operator on the real locus, is the identity, we also know that the number of real connected components is equal to $n$, which is an upper bound on the number of complex connected components (see here), which in turn is an upper bound on the euler characteristic $\chi(\mathcal{o}_s)=\dim h^0(\mathcal{o}_s)\dim h^1(\mathcal{o}_s)$. but the euler characteristic is equal to $n$ in this case, and so there must be $n$ complex connected components, one for each of the real connected components. it follows that $\dim h^1(\mathcal{o}_s)=0$, so the arithmetic genus of each of these components is zero. we also know each of these components is integral, and so they must each be isomorphic to $\mathbb{cp}^1$ (see hartshorne, section 4.1 exercise 1.8). we have therefore determined the entire structure of the complex schubert curve $s$ in the case that $\omega$ is the identity map, using the connection with $k$theory described above. similar analyses lead to other geometric results: we have also shown that the schubert curves $s$ can have arbitrarily high genus, and can arbitrarily many complex connected components, for various values of $\alpha$, $\beta$, and $\gamma$. so, what do schubert curves, young tableaux, and $k$theory have in common? a little monodromy operator called $\omega$. posted in diamond  3 replies what do schubert curves, young tableaux, and ktheory have in common? (part ii) posted on march 18, 2016 by maria gillespie reply in the last post, we discussed an operation $\newcommand{\box}{\square} \omega$ on skew littlewoodrichardson young tableaux with one marked inner corner, defined as a commutator of rectification and shuffling. as a continuation, we’ll now discuss where this operator arises in geometry. schubert curves: relaxing a restriction recall from our post on schubert calculus that we can use schubert varieties to answer the question: given four lines $\ell_1,\ell_2,\ell_3,\ell_4\in \mathbb{c}\mathbb{p}^3$, how many lines intersect all four of these lines in a point? in particular, given a complete flag, i.e. a chain of subspaces $$0=f_0\subset f_1\subset\cdots \subset f_m=\mathbb{c}^m$$ where each $f_i$ has dimension $i$, the schubert variety of a partition $\lambda$ with respect to this flag is a subvariety of the grassmannian $\mathrm{gr}^n(\mathbb{c}^m)$ defined as $$\omega_{\lambda}(f_\bullet)=\{v\in \mathrm{gr}^n(\mathbb{c}^m)\mid \dim v\cap f_{n+i\lambda_i}\ge i.\}$$ here $\lambda$ must fit in a $k\times (nk)$ box in order for the schubert variety to be nonempty. in the case of the question above, we can translate the question into an intersection problem in $\mathrm{gr}^2(\mathbb{c}^4)$ with four general twodimensional subspaces $p_1,p_2,p_3,p_4\subset \mathbb{c}^4$, and construct complete flags $f_\bullet^{(1)},f_\bullet^{(2)},f_\bullet^{(3)},f_\bullet^{(4)}$ such that their second subspace $f^{(i)}_2$ is $p_i$ for each $i=1,2,3,4$. then the intersection condition is the problem of finding a plane that intersects all $p_i$’s in a line. the variety of all planes intersecting $p_i$ in a line is $\omega_\box(f_\bullet^{(i)})$ for each $i$, and so the set of all solutions is the intersection $$\omega_\box(f_\bullet^{(1)})\cap \omega_\box(f_\bullet^{(2)})\cap \omega_\box(f_\bullet^{(3)})\cap \omega_\box(f_\bullet^{(4)}).$$ and, as discussed in our post on schubert calculus, since the $k\times(nk)$ box has size $4$ in $\mathrm{gr}^2(\mathbb{c}^4)$ and the four partitions involved have sizes summing to $4$, this intersection has dimension $0$. the littlewoodrichardson rule then tells us that the number of points in this zerodimensional intersection is the littlewoodrichardson coefficient $c_{\box,\box,\box,\box}^{(2,2)}$. what happens if we relax one of the conditions in the problem, so that we are only intersecting three of the schubert varieties above? in this case we get a oneparameter family of solutions, which we call a schubert curve. to define a schubert curve in general, we require a sufficiently “generic” choice of $r$ flags $f_\bullet^{(1)},\ldots, f_\bullet^{(r)}$ and a list of $r$ partitions $\lambda^{(1)},\ldots,\lambda^{(r)}$ (fitting inside the $k\times (nk)$ box) whose sizes sum to $k(nk)1$. it turns out that one can choose any flags $f_\bullet$ defined by the iterated derivatives at chosen points on the rational normal curve, defined by the locus of points of the form $$(1:t:t^2:t^3:\cdots:t^{n1})\in \mathbb{cp}^n$$ (along with the limiting point $(0:0:\cdots:1)$.) in particular, consider the flag whose $k$th subspace $f_k$ is the span of the first $k$ rows of the matrix of iterated derivatives at the point on this curve parameterized by $t$: $$\left(\begin{array}{cccccc} 1 & t & t^2 & t^3 & \cdots & t^{n1} \\ 0 & 1 & 2t & 3t^2 & \cdots & (n1) t^{n2} \\ 0 & 0 & 2 & 6t & \cdots & \frac{(n1)!}{(n3)!} t^{n3} \\ 0 & 0 & 0 & 6 & \cdots & \frac{(n1)!}{(n4)!} t^{n3} \\ \vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & (n1)! \end{array}\right)$$ this is called the osculating flag at the point $t$, and if we pick a number of points on the curve and choose their osculating flag, it turns out that they are in sufficiently general position in order for the schubert intersections to have the expected dimension. so, we pick some number $r$ of these osculating flags, and choose exactly $r$ partitions $\lambda^{(1)},\ldots,\lambda^{(r)}$ with sizes summing to $k(nk)1$. intersecting the resulting schubert varieties defines a schubert curve $s$. a covering map in order to get a handle on these curves, we consider the intersection of $s$ with another schubert variety of a single box, $\omega_\box(f_\bullet)$. in particular, after choosing our $r$ osculating flags, choose an $(r+1)$st point $t$ on the rational normal curve and choose the singlebox partition $\lambda=(1)$. intersecting the resulting schubert variety $\omega_\box(f_\bullet)$ with our schubert curve $s$ gives us a zerodimensional intersection, with the number of points given by the littlewoodrichardson coefficient $c:=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$ where $b=((nk)^k)$ is the $k\times (nk)$ box partition. by varying the choice of $t$, we obtain a partition of an open subset of the schubert curve into sets of $c$ points. we can then define a map from this open subset of $s$ to the points of $\mathbb{cp}^1$ for which the preimage of $(1:t)$ consists of the $c$ points in the intersection given by choosing the $(r+1)$st point to be $(1:t:t^2:t^3:\cdots:t^{n1})$. in this paper written by my coauthor jake levinson, it is shown that if we choose all $r+1$ points to be real, then this can be extended to a map $s\to \mathbb{cp}^1$ for which the real locus $s(\mathbb{r})$ is a smooth, finite covering of $\mathbb{rp}^1$. the fibers of this map have size $c=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$. note that this littlewoodrichardson coefficient counts the number of ways of filling the box $b$ with skew littlewoodrichardson tableaux with contents $\lambda^{(1)},\ldots,\lambda^{(r)},\box$ such that each skew shape extends the previous shape outwards. in addition, by the symmetry of littlewoodrichardson coefficients, it doesn’t matter what order in which we use the partitions. it turns out that there is a canonical way of labeling the fibers by these chains of tableaux, for which the monodromy of the covering is described by shuffling operators. let’s be more precise, and consider the simplest interesting case, $r=3$. we have three partitions $\alpha$, $\beta$, and $\gamma$ such that $$\alpha+\beta+\gamma=k(nk)1=b1.$$ let’s choose, for simplicity, the three points $0$, $1$, and $\infty$ to define the respective osculating flags. then we can label the points in the fiber $f^{1}(0)$ of the map $f:s\to \mathbb{rp}^1$ by the fillings of the box with a chain of littlewoodrichardson tableaux of contents $\alpha$, $\box$, $\beta$, and $\gamma$ in that order. note that there is only one littlewoodrichardson tableau of straight shape and content $\alpha$, and similarly for the antistraight shape $\gamma$, so the data here consists of a skew littlewoodrichardson tableau of content $\beta$ with an inner corner chosen to be the marked box. this is the same object that we were considering in part i. now, we can also label the fiber $f^{1}(1)$ by the chains of contents $\box$, $\alpha$, $\beta$, and $\gamma$ in that order, and finally label the fiber $f^{1}(\infty)$ by the chains of contents $\alpha$, $\beta$, $\box$, and $\gamma$ in that order, in such a way that if we follow the curve $s$ along an arc from a point in $f^{1}(0)$ to $f^{1}(1)$ or similarly from $1$ to $\infty$ or $\infty$ to $0$, then then the map between the fibers is given by the shuffling operations described in the last post! in particular if we follow the arc from $0$ to $\infty$ that passes through $1$, the corresponding operation on the fibers is given by the “evacuationshuffle”, or the first three steps of the operator $\omega$ described in the previous post. the arc from $\infty$ back to $0$ on the other side is given by the “shuffle” of the $\box$ past $\beta$, which is the last step of $\omega$. all in all, $\omega$ gives us the monodromy operator on the zero fiber $f^{1}(0)$. the following picture sums this up: so, the real geometry of the schubert curve boils down to an understanding of the permutation $\omega$. our local algorithm allows us to get a better handle on the orbits of $\omega$, and hence tell us things about the number of real connected components, the lengths of these orbits, and even in some cases the geometry of the complex curve $s$. next time, i’ll discuss some of these consequences, as well as some fascinating connections to the $k$theory of the grassmannian. stay tuned! posted in diamond  leave a reply what do schubert curves, young tableaux, and ktheory have in common? (part i) posted on january 18, 2016 by maria gillespie reply in a recent and fantastic collaboration between jake levinson and myself, we discovered new links between several different geometric and combinatorial constructions. we’ve weaved them together into a beautiful mathematical story, a story filled with drama and intrigue. so let’s start in the middle. slider puzzles for mathematicians those of you who played with little puzzle toys growing up may remember the “15 puzzle”, a $4\times 4$ grid of squares with 15 physical squares and one square missing. a move consisted of sliding a square into the empty square. the french name for this game is “jeu de taquin”, which translates to “the teasing game”. we can play a similar jeu de taquin game with semistandard young tableaux. to set up the board, we need a slightly more general definition: a skew shape $\lambda/\mu$ is a diagram of squares formed by subtracting the young diagram (see this post) of a partition $\mu$ from the (strictly larger) young diagram of a partition $\lambda$. for instance, if $\lambda=(5,3,3,1)$ and $\mu=(2,1)$, then the skew shape $\lambda/\mu$ consists of the white squares shown below. a semistandard young tableau is then a way of filling the squares in such a skew shape with positive integers in such a way that the entries are weakly increasing across rows and strictly increasing down columns: now, an inner jeu de taquin slide consists of choosing an empty square adjacent to two of the numbers, and successively sliding entries inward into the empty square in such a way that the tableau remains semistandard at each step. this is an important rule, and it implies that, once we choose our inner corner, there is a unique choice between the squares east and south of the empty square at each step; only one can be slid to preserve the semistandard property. an example of an inwards jeu de taquin slide is shown (on repeat) in the animation below: here’s the game: perform a sequence of successive jeu de taquin slides until there are no empty inner corners left. what tableaux can you end up with? it turns out that, in fact, it doesn’t matter how you play this game! no matter which inner corner you pick to start the jeu de taquin slide at each step, you will end up with the same tableau in the end, called the rectification of the original tableau. since we always end up at the same result, it is sometimes more interesting to ask the question in reverse: can we categorize all skew tableaux that rectify to a given fixed tableau? this question has a nice answer in the case that we fix the rectification to be superstandard, that is, the tableau whose $i$th row is filled with all $i$’s: it turns out that a semistandard tableau rectifies to a superstandard tableau if and only if it is littlewoodrichardson, defined as follows. read the rows from bottom to top, and left to right within a row, to form the reading word. then the tableau is littlewoodrichardson if every suffix (i.e. consecutive subword that reaches the end) of the reading word is ballot, which means that it has at least as many $i$’s as $i+1$’s for each $i\ge 1$. for instance, the littlewoodrichardson tableau below has reading word 352344123322111, and the suffix 123322111, for instance, has at least as many $1$’s as $2$’s, $2$’s as $3$’s, etc. littlewoodrichardson tableaux are key to the littlewoodrichardson rule, which allows us to efficiently compute products of schur functions. a convoluted commutator the operation that jake and i studied is a sort of commutator of rectification with another operation called “shuffling”. the process is as follows. start with a littlewoodrichardson tableau $t$, with one of the corners adjacent to $t$ on the inside marked with an “$\times$”. call this extra square the “special box”. then we define $\omega(t)$ to be the result of the following four operations applied to $t$. rectification: treat $\times$ as having value $0$ and rectify the entire skew tableau. shuffling: treat the $\times$ as the empty square to perform an inward jeu de taquin slide. the resulting empty square on the outer corner is the new location of $\times$. unrectification: treat $\times$ as having value $\infty$ and unrectify, using the sequence of moves from the rectification step in reverse. shuffling back: treat the $\times$ as the empty square to perform a reverse jeu de taquin slide, to move the $\times$ back to an inner corner. we can iterate $\omega$ to get a permutation on all pairs $(\times,t)$ of a littlewoodrichardson tableau $t$ with a special box marked on a chosen inner corner, with total shape $\lambda/\mu$ for some fixed $\lambda$ and $\mu$. as we’ll discuss in the next post, this permutation is related to the monodromy of a certain covering space of $\mathbb{rp}^1$ arising from the study of schubert curves. but i digress. one of the main results in our paper provides a new, more efficient algorithm for computing $\omega(t)$. in particular, the first three steps of the algorithm are what we call the “evacuationshuffle”, and our local rule for evacuation shuffling is as follows: phase 1. if the special box does not precede all of the $i$’s in reading order, switch the special box with the nearest $i$ prior to it in reading order. then increment $i$ by $1$ and repeat this step. if, instead, the special box precedes all of the $i$’s in reading order, go to phase 2. phase 2. if the suffix of the reading word starting at the special box has more $i$’s than $i+1$’s, switch the special box with the nearest $i$ after it in reading order whose suffix has the same number of $i$’s as $i+1$’s. either way, increment $i$ by $1$ and repeat this step until $i$ is larger than any entry of $t$. so, to get $\omega(t)$, we first follow the phase 1 and phase 2 steps, and then we slide the special box back with a simple jeu de taquin slide. we can then iterate $\omega$, and compute an entire $\omega$orbit, a cycle of its permutation. an example of this is shown below. that’s all for now! in the next post i’ll discuss the beginning of the story: where this operator $\omega$ arises in geometry and why this algorithm is exactly what we need to understand it. posted in pearl  leave a reply what is a $q$analog? (part 2) posted on december 10, 2015 by maria gillespie reply this is a continuation of part 1 of this series of posts on $q$analogs. counting by $q$’s another important area in which $q$analogs come up is in combinatorics. in this context, $q$ is a formal variable, and the $q$analog is a generating function in $q$, but viewed in a different light than usual generating functions. we think of the $q$analog of as “$q$counting” a set of weighted objects, where the weights are given by powers of $q$. say you’re trying to count permutations of $1,\ldots,n$, that is, ways of rearranging the numbers $1,\ldots,n$ in a row. there are $n$ ways to choose the first number, and once we choose that there are $n1$ remaining choices for the second, then $n2$ for the third, and so on. so there are $n!=n\cdot (n1)\cdot \cdots \cdot 2\cdot 1$ ways to rearrange the entries. for instance, the $3!=6$ permutations of $1,2,3$ are $123$, $132$, $213$, $231$, $312$, $321$. now, say we weight the permutations according to how “mixed up” they are, in the sense of how many pairs of numbers are out of order. an inversion is a pair of entries in the permutation in which the bigger number is to the left of the smaller, and $\mathrm{inv}(\pi)$ denotes the number of inversions of the permutation $\pi$. the table below shows the permutations of 3 along with the number of inversions they contain. $$\begin{array}{ccc} p & \mathrm{inv}(p) & q^{\mathrm{inv}(p)}\\\hline 123 & 0 & 1 \\ 132 & 1 & q\\ 213 & 1 & q\\ 231 & 2 & q^2\\ 312 & 2 & q^2 \\ 321 & 3 & q^3 \end{array} $$ we weight each permutation $p$ by $q^{\mathrm{inv}(p)}$, and $q$count by summing these $q$powers, to form the sum $$\sum_{p\in s_n}q^{\mathrm{inv}(p)}$$ where $s_n$ is the set of all permutations of $1,\ldots,n$. so for $n=3$, the sum is $1+2q+2q^2+q^3$ by our table above. we now come to an important philosophical distinction between $q$analogs and generating functions. as a generating function, the sum $1+2q+2q^2+q^3$ is thought of in terms of the sequence of coefficients, $1,2,2,1$. generatingfunctionologically, we might instead write the sum as $\sum_{i=0}^\infty c_i q^i$ where $c_i$ is the number of permutations of length $n$ with $i$ inversions. but in $q$analog notation, $\sum_{p\in s_n}q^{\mathrm{inv}(p)}$, we understand that it is not the coefficients but rather the exponents of our summation that we are interested in.. in general, a combinatorial $q$analog can be defined as a summation of $q$powers $q^{\mathrm{stat}(p)}$ where $p$ ranges over a certain set of combinatorial objects and $\mathrm{stat}$ is a statistic on these objects. recall that we defined an “interesting $q$analog” of an expression $p$ to be an expression $p_q$ such that setting $q=1$ or taking the limit as $q\to 1$ results in $p$, $p_q$ can be expressed in terms of (possibly infinite) sums or products of rational functions of $q$ over some field, $p_q$ gives us more refined information about something that $p$ describes, and $p_q$ has $p$like properties. certainly setting $q=1$ in a combinatorial $q$analog results in the total number of objects, and the $q$analog gives us more information about the objects than just their total number. it’s also a polynomial in $q$, so it satisfies properties 1, 2, and 3 above. let’s now see how our $q$analog $\sum_{p\in s_n}q^{\mathrm{inv}(p)}$, which is a $q$analog of $n!$, also satisfies property 4. notice that $1+2q+2q^2+q^3$ factors as $(1)(1+q)(1+q+q^2)$. indeed, in general this turns out to be the same $q$factorial we saw in the last post! that is, $$\sum_{p\in s_n}q^{\mathrm{inv}(p)}=(1)(1+q)(1+q+q^2)\cdots(1+q+\cdots+q^n)=(n)_q!.$$ so it satisfies property 4 by exhibiting a product formula like $n!$ itself. i posted a proof of this fact in this post, but let’s instead prove it by building up the theory of $q$counting from the ground up. the multiplication principle in combinatorics is the basic fact that the number of ways of choosing one thing from a set of $m$ things and another from a set of $n$ things is the product $m\cdot n$. but what if the things are weighted? $q$multiplication principle: given two weighted sets $a$ and $b$ with $q$counts $m(q)$ and $n(q)$, the $q$count of the ways of choosing one element from $a$ and another from $b$ is the product $m(q)n(q)$, where the weight of a pair is the sum of the weights of the elements. let’s see how this plays out in the case of $(n)_q!$. if each entry in the permutation is weighted by the number of inversions it forms with smaller entries (to its right), then the first entry can be any of $1,2,\ldots,n$, which contributes a factor of $1+q+q^2+\cdots+q^{n1}$ to the total. the next entry then can be any of the $n1$ remaining entries, and since the first entry cannot be the smaller entry of an inversion with the second, this choice contributes a factor of $1+q+q^2+\cdots+q^{n2}$ to the total by the same argument. continuing in this fashion we get the $q$factorial as our $q$count. notice that even the proof was a $q$analog of the proof that $n!$ is the number of permutations of $1,\ldots,n$, now that we have the $q$multiplication principle. that’s all for now! in the next post we’ll talk about how to use the $q$multiplication principle to derive a combinatorial interpretation of the $q$binomial coefficient, and discuss $q$catalan numbers and other fun $q$analogs. stay tuned! posted in opal  leave a reply what is a $q$analog? (part 1) posted on november 22, 2015 by maria gillespie 4 hi, i’m maria and i’m a $q$analog addict. the theory of $q$analogs is a littleknown gem, and in this series of posts i’ll explain why they’re so awesome and addictive! so what is a $q$analog? it is one of those rare mathematical terms whose definition doesn’t really capture what it is about, but let’s start with the definition anyway: definition: a $q$analog of a statement or expression $p$ is a statement or expression $p_q$, depending on $q$, such that setting $q=1$ in $p_q$ results in $p$. so, for instance, $2q+3q^2$ is a $q$analog of $5$, because if we plug in $q=1$ we get $5$. sometimes, if $p_q$ is not defined at $q=1$, we also say it’s a $q$analog if $p$ can be recovered by taking the limit as $q$ approaches $1$. for instance, the expression $$\frac{q^51}{q1}$$ is another $q$analog of $5$ – even though we get division by zero if we plug in $q=1$, we do have a well defined limit that we can calculate, for instance using l’hospital’s rule: $$\lim_{q\to 1} \frac{q^51}{q1}=\lim_{q\to 1} \frac{5q^4}{1} = 5.$$ now of course, there are an unlimited supply of $q$analogs of the number $5$, but certain $q$analogs are more important than others. when mathematicians talk about $q$analogs, they are usually referring to “good” or “useful” $q$analogs, which doesn’t have a widely accepted standard definition, but which i’ll attempt to define here: more accurate definition: an interesting $q$analog of a statement or expression $p$ is a statement or expression $p_q$ depending on $q$ such that: setting $q=1$ or taking the limit as $q\to 1$ results in $p$, $p_q$ can be expressed in terms of (possibly infinite) sums or products of rational functions of $q$ over some field, $p_q$ gives us more refined information about something that $p$ describes, and $p_q$ has $p$like properties. because of property 2, most people would agree that $5^q$ is not an interesting $q$analog of $5$, because usually we’re looking for polynomiallike things in $q$. on the other hand, $\frac{q^51}{q1}$, is an excellent $q$analog of $5$ for a number of reasons. it certainly satisfies property 2. it can also be easily generalized to give a $q$analog of any real number: we can define $$(a)_q=\frac{q^a1}{q1},$$ a $q$analog of the number $a$. in addition, for positive integers $n$, the expression simplifies: $$(n)_q=\frac{q^n1}{q1}=1+q+q^2+\cdots+q^{n1}.$$ so for instance, $(5)_q=1+q+q^2+q^3+q^4$, which is a natural $q$analog of the basic fact that $5=1+1+1+1+1$. the powers of $q$ are just distinguishing each of our “counts” as we count to $5$. this polynomial also captures the fact that $5$ is prime, in a $q$analogy way: the polynomial $1+q+q^2+q^3+q^4$ cannot be factored into two smallerdegree polynomials with integer coefficients. so the $q$number $(5)_q$ also satisfies properties 3 and 4 above: it gives us more refined information about $5$ness, by keeping track of the way we count to $5$, and behaves like $5$ in the sense that it can’t be factored into smaller $q$analogs of integers. but it doesn’t stop there. properties 3 and 4 can be satisfied in all sorts of ways, and this $q$number is even interesting than we might expect. it comes up in finite geometry, analytic number theory, representation theory, and combinatorics. so much awesome mathematics is involved in the study of $q$analogs that i’ll only cover one aspect of it today: $q$analogs that appear in geometry over a finite field $\mathbb{f}_q$. turn to the next page to see them! pages: 1 2 posted in opal  4 replies glencoe/mcgrawhill doesn’t believe this bijection exists posted on september 14, 2015 by maria gillespie 6 education is a difficult task, it really is. teaching takes a few tries to get the hang of. writing textbooks is even harder. and math is one of those technical fields in which human error is hard to avoid. so usually, when i see a mistake in a math text, it doesn’t bother me much. but some things just hurt my soul. no correspondence between the integers and rationals? yes there is, example 2! yes there is! this horrifying falsehood was stated in the supplementary “study guide and intervention” worksheet for the glencoe/mcgrawhill algebra 2 textbook, and recently pointed out on reddit. or at least it was stated in some version of this worksheet. the original file can be found online at various websites, including one download link from glencoe’s website that shows up on a google search. there are other versions of the document that don’t contain this example, but this version was almost certainly used in some high schools, as the reddit thread claims. luckily, mathematicians are here to set the record straight. the wolfram blog published a fantastic post about this error already, with several proofs of the countability of the rationals. there are also several excellent older expositions on this topic, including on the math less traveled and the division by zero blogs. i’ll discuss two of my favorites here as well. but first, let’s talk about what is wrong with the argument in example 2. the author is correct in stating that listing all the rationals in order would make a onetoone and onto correspondence between the rationals and integers, and so they try to do so in a random way and failed. at that point, instead of trying a different ordering, they gave up and figured it couldn’t be done! that’s not a proof, or even logically sound (as my students at this year’s prove it! math academy would certainly recognize.) if one were going to try to prove that a certain set couldn’t be organized into a list, a common tactic would be to use proof by contradiction: assume there was a way to list them, and then show that something goes wrong and you get a contradiction. of course, this wouldn’t work either in the case of the rationals, because they can be listed. so let’s discuss a correct solution. getting our definitions straight first, let’s state the precise meaning of a onetoone and onto correspondence. a function $f$ from a set $a$ to a set $b$, written $f:a\to b$, is an assignment of each element of $a\in a$ to an element $f(a)\in b$. to clear up another misuse of notation in the glencoe algebra textbook, the set $a$ is called the domain and $b$ is called the codomain (not the range, as glencoe would have you think – the range refers to the set of elements of $b$ that are assigned to by the function.) a function is: onetoone, or injective, if no two elements of $a$ are assigned to the same element of $b$, i.e., if $f(x)=f(y)$ then $x=y$. onto, or surjective, if every element of $b$ is mapped to, i.e., for all $b\in b$, there exists $a\in a$ such that $f(a)=b$. for instance, if $\mathbb{z}$ denotes the set of integers, the function $f:\mathbb{z}\to \mathbb{z}$ defined by $f(x)=2x$ is injective, since if $2x=2y$ then $x=y$. however, it is not surjective, since an odd number like $3$ is not equal to $2x$ for any integer $x$. a function which is both injective and surjective is said to be bijective, and is called a bijection. this is just a shorter way of saying “onetoone and onto correspondence,” which is wordy and cumbersome. so, we want to find a bijection $f:\mathbb{z}\to \mathbb{q}$, where $\mathbb{z}$ denotes the integers and $\mathbb{q}$ the rationals. notice that we can list all the integers in order: $$0,1,1,2,2,3,3,\ldots$$ and so if we list all the rationals in order, $r_0,r_1,r_2,\ldots$, we can define the function $f$ accordingly by $f(0)=r_0$, $f(1)=r_1$, $f(1)=r_2$, and so on. the function will be bijective if and only if every rational number appears in the list exactly once. next, let’s be precise about the rationals. recall that the rational numbers are those numbers which can be written as fractions $a/b$ where $a$ and $b$ are integers with $b\neq 0$. in order to assign every rational number a unique representation, let us restrict to the case where $b>0$ and $a$ is any integer such that $\mathrm{gcd}(a,b)=1$. this condition makes $a/b$ into a reduced fraction. so the number $2/4$ should be written as $1/2$ in this convention. it follows that we can think of the set of rational numbers as the set $$\mathbb{q}=\{(a,b)  b>0\text{ and }\mathrm{gcd}(a,b)=1\text{ and }a,b\in \mathbb{z}\}.$$ listing the rationals, naively one way to construct this list is to think of the rationals as ordered pairs $(a,b)$ of integers with $b>0$ and $\mathrm{gcd}(a,b)=1$ as described above. there is an easy way of ordering all pairs of integers – plot them on a coordinate plane, and use a spiral! now, to list the rationals, follow the spiral from $(0,0)$ outwards. each time we reach an ordered pair of integers, say $(a,b)$, write it down if $b>0$ and $\mathrm{gcd}(a,b)=1$, and otherwise skip it and move on. (these are the green dots above.) this process guarantees that we list all the rationals exactly once. more elegant methods there are many other elegant enumerations of the rationals, and one particularly nice one is due to calkin and wilf. they construct a binary tree in which each rational number appears exactly once, as shown below. the tree is constructed as follows: start with $1/1$ at the top, and for each node $a/b$ in the tree, assign it the left and right children $a/(a+b)$ and $(a+b)/b$ respectively. this tree turns out to contain every positive rational number exactly once. it also has an incredible property: if we read the entries of each row of the tree successively from left to right, the denominator of each entry will match the numerator of the next entry, giving us a sequence of numerators/denominators: $$1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,1,\ldots$$ such that the consecutive quotients give us a list of the positive rationals. this makes me wonder is whether there is a way of listing the integers such that (a) every integer occurs exactly once in the sequence, and (b) the consecutive quotients give all rationals exactly once (allowing the consecutive integers to have common factors greater than one). thoughts? posted in amber  6 replies the rmajor index posted on september 6, 2015 by maria gillespie reply in this post and this one too, we discussed the inv and maj statistics on words, and two different proofs of their equidistribution. in fact, there is an even more unifying picture behind these statistics: they are simply two different instances of an entire parameterized family of statistics, called $r\text{maj}$, all of which are equidistributed! rawlings defined an $r$inversion of a permutation $\pi$ to be a pair of entries $(\pi_i,\pi_j)$ with $i\lt j$ and $$0\lt \pi_i\pi_j\lt r.$$ for instance, $21534$ has three total inversions, $(2,1)$, $(5,4)$, and $(5,3)$, but only the first two have $\pi_i\pi_j$r$descent to be an index $i$ for which $$\pi_i\ge \pi_{i+1}+r,$$ so that $21534$ has only position $3$ as a $2$descent. finally, he defines the $r\text{maj}$ of a permutation to be $$r\text{}\mathrm{maj}(\pi)=\left(\sum_{\pi_i\ge \pi_{i+1}+r}i\right)+\#r\text{}\mathrm{inversions}.$$ thus $2\text{}\mathrm{maj}(21534)=3+2=5$. notice that $1\text{maj}$ is the usual major index, and $n\text{maj}$ is the inv statistic! rawling’s result is that these statistics all have the same distribution: for any $r,s\ge 1$, the number of permutations of $\{1,2,\ldots,n\}$ having $r\text{maj}$ value $k$ is the same as the number of them having $s\text{maj}$ value $k$ for any $k$. more succinctly, $$\sum_{\pi\in s_n} q^{r\text{maj}(\pi)}=\sum_{\pi\in s_n} q^{s\text{maj}(\pi)}.$$ a colleague of mine mentioned this result and challenged me to prove it without reading the proof first, so here goes. i challenge the reader to do the same before turning to the next page. good luck! pages: 1 2 posted in opal  leave a reply post navigation ← older posts search recent posts the structure of the garsiaprocesi modules $r_\mu$ can you prove it… combinatorially? phinished! what do schubert curves, young tableaux, and ktheory have in common? 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divide  0.01% (1) $a^k$.  0.01% (1) diagonal.  0.01% (1) years.  0.01% (1) summary  0.01% (1) $44\lt  0.01% (1) ring.  0.01% (1) $e_1,\ldots,e_4$  0.01% (1) tensor  0.01% (1) (such  0.01% (1) grassmannian!)  0.01% (1) modulo  0.01% (1) free”  0.01% (1) “locally  0.01% (1) remove  0.01% (1) $i_{\mu}$.  0.01% (1) $$\begin{align*}  0.01% (1) e_2(x_1,x_3),\ldots,  0.01% (1) “short  0.01% (1) group,  0.01% (1) bundles)  0.01% (1) sequences”,  0.01% (1) &(e_2(x_1,x_2),  0.01% (1) i_{(3,1)}=  0.01% (1) $[\mathcal{e}_1]+[\mathcal{e}_2][\mathcal{e}]$  0.01% (1) $h$  0.01% (1) $g/h$  0.01% (1) simpler  0.01% (1) $d_4(\mu)=4$  0.01% (1) $\oplus$,  0.01% (1) monoidal  0.01% (1) monoids.  0.01% (1) groupifying  0.01% (1) split  0.01% (1) basics  0.01% (1) x_1x_3x_4,\hspace{0.4cm}  0.01% (1) paper.  0.01% (1) buch’s  0.01% (1) grassmannians,  0.01% (1) $k(\mathbb{n})=\mathbb{z}$.  0.01% (1) $[m]+[n][m+n]$.  0.01% (1) element,  0.01% (1) associative  0.01% (1) (recall  0.01% (1) $s=\{x_1,x_2,x_3,x_4\}$,  0.01% (1) inverses).  0.01% (1) x_2x_3x_4$$  0.01% (1) $m\in  0.01% (1) $[m]$  0.01% (1) $k(m)$  0.01% (1) associated  0.01% (1) (a.k.a.  0.01% (1) e_2(x_3,x_4),  0.01% (1) trivial  0.01% (1) representation.  0.01% (1) generalization  0.01% (1) myself.  0.01% (1) actually  0.01% (1) $r_{(n)}=\mathbb{c}$  0.01% (1) improved  0.01% (1) $r_{(1^n)}=\mathbb{c}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$  0.01% (1) coninvariants  0.01% (1) action,  0.01% (1) coauthored  0.01% (1) expository  0.01% (1) amber,  0.01% (1) eigenvalues,  0.01% (1) here.)  0.01% (1) thesis.  0.01% (1) $\mathbb{c}$.  0.01% (1) originally  0.01% (1) frobenius  0.01% (1) halllittlewood  0.01% (1) april  0.01% (1) $\widetilde{h}_\mu(x;q)$.  0.01% (1) questions  0.01% (1) examples,  0.01% (1) characteristic,  0.01% (1) e_3(x_1,x_2,x_3),  0.01% (1) sheaf.  0.01% (1) $\mathcal{o}_s$  0.01% (1) alternatively  0.01% (1) e_2(x_2,x_3,x_4),  0.01% (1) isomorphism  0.01% (1) e_2(x_1,x_2,x_3),  0.01% (1) scheme  0.01% (1) $k(\mathrm{gr}(n,k))$  0.01% (1) e_3(x_2,x_3,x_4),  0.01% (1) h^1(\mathcal{o}_s)$$  0.01% (1) have?  0.01% (1) e_3(x_1,\ldots,x_4),  0.01% (1) e_4(x_1,\ldots,x_4))  0.01% (1) \end{align*}$$  0.01% (1) e_2(x_1,\ldots,x_4),  0.01% (1) genus?  0.01% (1) $$\chi(\mathcal{o}_s)=\dim  0.01% (1) $g=1\chi(\mathcal{o}_s)$  0.01% (1) e_1(x_1,\ldots,x_4),  0.01% (1) curve?  0.01% (1) load.  0.01% (1) packed  0.01% (1) dividing  0.01% (1) 5$’s  0.01% (1) minors  0.01% (1) right?  0.01% (1) integers?  0.01% (1) it!,  0.01% (1) proven  0.01% (1) during  0.01% (1) largest  0.01% (1) derived  0.01% (1) crazy,  0.01% (1) \left(\frac{1\sqrt{5}}{2}\right)^n  0.01% (1) remarkable  0.01% (1) $f_1=1$),  0.01% (1) $f_0=0$  0.01% (1) conjugation,  0.01% (1) $t^{d_k}$,  0.01% (1) $n$th  0.01% (1) \left(\frac{1+\sqrt{5}}{2}\right)^n  0.01% (1) $$f_n=\frac{1}{\sqrt{5}}\left(  0.01% (1) formula:  0.01% (1) term,  0.01% (1) induction,  0.01% (1) later,  0.01% (1) $(1\sqrt{5})^n$  0.01% (1) $(1+\sqrt{5})^n$  0.01% (1) identity?  0.01% (1) c_{\mu’}$.  0.01% (1) formula.  0.01% (1) (1)^i\binom{n}{i}(\sqrt{5})^i  0.01% (1) us:  0.01% (1) combining  0.01% (1) cancel,  0.01% (1) summations  0.01% (1) (see,  0.01% (1) theorem?  0.01% (1) classroom.  0.01% (1) outside  0.01% (1) smith  0.01% (1) wikipedia)  0.01% (1) article  0.01% (1) amazed  0.01% (1) discussion  0.01% (1) thinking:  0.01% (1) got  0.01% (1) divisors  0.01% (1) $$0,1,1,2,3,5,8,13,21,\ldots$$  0.01% (1) next:  0.01% (1) (x_{i_k}t)$$  0.01% (1) $s=\{x_{i_1},\ldots,x_{i_k}\}$  0.01% (1) $\{x_1,\ldots,x_n\}$.  0.01% (1) diamond,  0.01% (1) (x_{i_1}t)(x_{i_2}t)\cdots  0.01% (1) august  0.01% (1) $x=\mathrm{diag}(x_1,\ldots,x_n)\in  0.01% (1) \overline{c_\mu}\cap  0.01% (1) $$t^{d_k(\mu)}  0.01% (1) 12,  0.01% (1) expanding  0.01% (1) basis  0.01% (1) multinomial  0.01% (1) namely  0.01% (1) dimension,  0.01% (1) kd_k(\mu)$,  0.01% (1) \gt  0.01% (1) $\binom{n}{\mu}$.  0.01% (1) vieta’s  0.01% (1) formulas,  0.01% (1) vanish  0.01% (1) soon  0.01% (1) success,  0.01% (1) enormous  0.01% (1) many,  0.01% (1) well!  0.01% (1) ideas  0.01% (1) $d_k(\mu)$  0.01% (1) $t^{d_k(\mu)}$  0.01% (1) analyzing  0.01% (1) adding  0.01% (1) form.  0.01% (1) camp  0.01% (1) investigated  0.01% (1) inspiration  0.01% (1) program,  0.01% (1) attended  0.01% (1) talented  0.01% (1) seventeen  0.01% (1) pleasure  0.01% (1) year.  0.01% (1) property,  0.01% (1) twoweek  0.01% (1) further  0.01% (1) $c_\mu$  0.01% (1) felt  0.01% (1) \binom{n}{2k+1}(\sqrt{5})^{2k+1}  0.01% (1) $(\sqrt{5})^{2k+1}=\sqrt{5}\cdot  0.01% (1) specifically  0.01% (1) mathematics,  0.01% (1) pursuit  0.01% (1) road.  0.01% (1) traced  0.01% (1) closures  0.01% (1) successes  0.01% (1) off  0.01% (1) coming  0.01% (1) mit.  0.01% (1) adventures  0.01% (1) greatest  0.01% (1) your  0.01% (1) solution,  0.01% (1) varieties,  0.01% (1) 1$$  0.01% (1) $\overline{c_{\mu’}}$  0.01% (1) below!  0.01% (1) life  0.01% (1) missteps  0.01% (1) 24,  0.01% (1) gemstones,  0.01% (1) competitions  0.01% (1) sciencerelated  0.01% (1) pure  0.01% (1) loved  0.01% (1) 18ish.  0.01% (1) something.  0.01% (1) friends  0.01% (1) determines  0.01% (1) added  0.01% (1) uniquely  0.01% (1) ambitious  0.01% (1) spring,  0.01% (1) unified  0.01% (1) know,  0.01% (1) family!),  0.01% (1) loving  0.01% (1) (thanks  0.01% (1) endeavors  0.01% (1) confident  0.01% (1) excited  0.01% (1) skills  0.01% (1) physicist  0.01% (1) become  0.01% (1) dream  0.01% (1) \blacksquare\,  0.01% (1) 0\,  0.01% (1) “combinatorial  0.01% (1) $m=n$.  0.01% (1) $r_\mu=\mathcal{o}(\overline{c_{\mu’}}\cap  0.01% (1) collection.  0.01% (1) proof”  0.01% (1) defining  0.01% (1) ended,  0.01% (1) right.  0.01% (1) equation  0.01% (1) $c_{\mu’}$?  0.01% (1) handy  0.01% (1) clearly  0.01% (1) sides  0.01% (1) multiply  0.01% (1) $\sqrt{5}$  0.01% (1) 5^k$,  0.01% (1) obtain:  0.01% (1) $$2^{n1}\cdot  0.01% (1) roughly  0.01% (1) 5^k.$$  0.01% (1) \binom{n}{2k+1}\cdot  0.01% (1) f_n=\sum_{k=0}^{\lfloor  0.01% (1) remembered  0.01% (1) leading  0.01% (1) easier  0.01% (1) f_n$,  0.01% (1) thus,  0.01% (1) $1$’s.  0.01% (1) $n2$.  0.01% (1) lengths,  0.01% (1) $$1\,  0.01% (1) as:  0.01% (1) digits  0.01% (1) forming  0.01% (1) length$(n1)$  0.01% (1) \end{array}$$  0.01% (1) satisfy  0.01% (1) (can  0.01% (1) black.  0.01% (1) matrices,  0.01% (1) recurrence?)  0.01% (1) $f_5=5$,  0.01% (1) fenceposts:  0.01% (1) colorings  0.01% (1) five  0.01% (1) description  0.01% (1) x_1x_2x_4,  0.01% (1) $$\left(\begin{array}{cccccc}  0.01% (1) $t$:  0.01% (1) span  0.01% (1) $f_k$  0.01% (1) pad  0.01% (1) 2t  0.01% (1) \frac{(n1)!}{(n3)!}  0.01% (1) 6t  0.01% (1) t^{n2}  0.01% (1) 3t^2  0.01% (1) $k$th  0.01% (1) $(0:0:\cdots:1)$.)  0.01% (1) box)  0.01% (1) (fitting  0.01% (1) parts.  0.01% (1) f_\bullet^{(r)}$  0.01% (1) $f_\bullet$  0.01% (1) curve,  0.01% (1) limiting  0.01% (1) (along  0.01% (1) \mathbb{cp}^n$$  0.01% (1) $$(1:t:t^2:t^3:\cdots:t^{n1})\in  0.01% (1) \frac{(n1)!}{(n4)!}  0.01% (1) &\ddots  0.01% (1) $(1:t)$  0.01% (1) preimage  0.01% (1) points.  0.01% (1) boxes  0.01% (1) $(1:t:t^2:t^3:\cdots:t^{n1})$.  0.01% (1) levinson,  0.01% (1) $s\to  0.01% (1) extended  0.01% (1) real,  0.01% (1) $r+1$  0.01% (1) varying  0.01% (1) partition.  0.01% (1) dimension.  0.01% (1) expected  0.01% (1) intersections  0.01% (1) \end{array}\right)$$  0.01% (1) $\omega_\box(f_\bullet)$.  0.01% (1) singlebox  0.01% (1) $b=((nk)^k)$  0.01% (1) $c:=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$  0.01% (1) intersection,  0.01% (1) $\lambda=(1)$.  0.01% (1) $f_\bullet^{(1)},\ldots,  0.01% (1) “generic”  0.01% (1) twodimensional  0.01% (1) $\mathbb{c}[x_1,\ldots,x_n]$  0.01% (1) nonempty.  0.01% (1) permuting  0.01% (1) $p_1,p_2,p_3,p_4\subset  0.01% (1) \mathbb{c}^4$,  0.01% (1) finding  0.01% (1) $i=1,2,3,4$.  0.01% (1) $f^{(i)}_2$  0.01% (1) $f_\bullet^{(1)},f_\bullet^{(2)},f_\bullet^{(3)},f_\bullet^{(4)}$  0.01% (1) i.\}$$  0.01% (1) f_{n+i\lambda_i}\ge  0.01% (1) f_m=\mathbb{c}^m$$  0.01% (1) f_1\subset\cdots  0.01% (1) $$0=f_0\subset  0.01% (1) point?  0.01% (1) $f_i$  0.01% (1) $\mathrm{gr}^n(\mathbb{c}^m)$  0.01% (1) v\cap  0.01% (1) variables,  0.01% (1) \mathrm{gr}^n(\mathbb{c}^m)\mid  0.01% (1) $$\omega_{\lambda}(f_\bullet)=\{v\in  0.01% (1) intersects  0.01% (1) $p_i$’s  0.01% (1) conditions  0.01% (1) happens  0.01% (1) $c_{\box,\box,\box,\box}^{(2,2)}$.  0.01% (1) inherits  0.01% (1) problem,  0.01% (1) above?  0.01% (1) require  0.01% (1) somewhat  0.01% (1) solutions,  0.01% (1) oneparameter  0.01% (1) tells  0.01% (1) $0$.  0.01% (1) $$\omega_\box(f_\bullet^{(1)})\cap  0.01% (1) $\omega_\box(f_\bullet^{(i)})$  0.01% (1) planes  0.01% (1) line.  0.01% (1) \omega_\box(f_\bullet^{(2)})\cap  0.01% (1) \omega_\box(f_\bullet^{(3)})\cap  0.01% (1) $4$,  0.01% (1) $k\times(nk)$  0.01% (1) calculus,  0.01% (1) \omega_\box(f_\bullet^{(4)}).$$  0.01% (1) $s(\mathbb{r})$  0.01% (1) smooth,  0.01% (1) growing  0.01% (1) toys  0.01% (1) played  0.01% (1) $e_d(z_1,\ldots,z_k)$  0.01% (1) “15  0.01% (1) puzzle”,  0.01% (1) missing.  0.01% (1) physical  0.01% (1) grid  0.01% (1) $4\times  0.01% (1) puzzles  0.01% (1) slider  0.01% (1) weaved  0.01% (1) we’ve  0.01% (1) constructions.  0.01% (1) links  0.01% (1) together  0.01% (1) story,  0.01% (1) middle.  0.01% (1) squarefree  0.01% (1) intrigue.  0.01% (1) drama  0.01% (1) consisted  0.01% (1) square.  0.01% (1) $\lambda=(5,3,3,1)$  0.01% (1) $\lambda$.  0.01% (1) larger)  0.01% (1) (strictly  0.01% (1) $\mu=(2,1)$,  0.01% (1) weakly  0.01% (1) numbers,  0.01% (1) $$r_{\mu}=\mathbb{c}[x_1,\ldots,x_n]/i_{\mu},$$  0.01% (1) columns:  0.01% (1) across  0.01% (1) post)  0.01% (1) subtracting  0.01% (1) translates  0.01% (1) taquin”,  0.01% (1) “jeu  0.01% (1) french  0.01% (1) “the  0.01% (1) teasing  0.01% (1) slightly  0.01% (1) board,  0.01% (1) tableaux.  0.01% (1) game”.  0.01% (1) collaboration  0.01% (1) january  0.01% (1) choose,  0.01% (1) $$\alpha+\beta+\gamma=k(nk)1=b1.$$  0.01% (1) $r=3$.  0.01% (1) simplest  0.01% (1) simplicity,  0.01% (1) flags.  0.01% (1) $\gamma$,  0.01% (1) antistraight  0.01% (1) fillings  0.01% (1) $f:s\to  0.01% (1) precise,  0.01% (1) operators.  0.01% (1) extends  0.01% (1) $\lambda^{(1)},\ldots,\lambda^{(r)},\box$  0.01% (1) \mu’_{nk+1}$  0.01% (1) $c=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$.  0.01% (1) symmetry  0.01% (1) $d_k(\mu)=\mu’_n+\mu’_{n1}+\cdots+  0.01% (1) s=k).$$  0.01% (1) labeling  0.01% (1) canonical  0.01% (1) partitions.  0.01% (1) $$i_{\mu}=(e_r(s)  0.01% (1) box.  0.01% (1) hence  0.01% (1) better  0.01% (1) understanding  0.01% (1) boils  0.01% (1) orbits,  0.01% (1) cases  0.01% (1) monomials  0.01% (1) connections  0.01% (1) consequences,  0.01% (1) $d$  0.01% (1) up:  0.01% (1) $f^{1}(0)$.  0.01% (1) $s\subset\{x_1,\ldots,x_n\}$  0.01% (1) $s=k$.  0.01% (1) $f^{1}(\infty)$  0.01% (1) considering  0.01% (1) passes  0.01% (1) corresponding  0.01% (1) $z_i$.  0.01% (1) all,  0.01% (1) “shuffle”  0.01% (1) post.  0.01% (1) homogeneous  0.01% (1) \mathbb{c}\mathbb{p}^3$,  0.01% (1) $32  0.01% (1) $e_3(s)$  0.01% (1) $i_\mu$,  0.01% (1) indexed  0.01% (1) i).  0.01% (1) identify  0.01% (1) $$[\mathcal{o}_s]=[\mathcal{o}_\alpha]\cdot  0.01% (1) maps,  0.01% (1) pushforwards  0.01% (1) subvarieties  0.01% (1) $\chi(j_\ast\mathcal{o}_s)$  0.01% (1) $\chi(\mathcal{o}_s)$  0.01% (1) desired.  0.01% (1) &=&\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})\end{eqnarray*}$$  0.01% (1) h^i(\mathcal{c})  0.01% (1) h^i(\mathcal{b})+\sum_i  0.01% (1) therefore,  0.01% (1) $k(x)$.  0.01% (1) $j_\ast\mathcal{o}_s$  0.01% (1) $j:s\to  0.01% (1) $x=\mathrm{gr}(n,k)$,  0.01% (1) situation,  0.01% (1) [\mathcal{o}_\beta]\cdot  0.01% (1) [\mathcal{o}_\gamma].$$  0.01% (1) $$[\mathcal{o}_s]=c^{\rho’}_{\alpha,\beta,\gamma}[\mathcal{o}_{\rho’}]k[\mathcal{o}_{\rho}]$$  0.01% (1) have:  0.01% (1) $\rho’$).  0.01% (1) minus  0.01% (1) x_3x_4.$$  0.01% (1) $c^{\rho’}_{\alpha,\beta,\gamma}$  0.01% (1) x_2x_4,\hspace{0.3cm}  0.01% (1) ii.  0.01% (1) on)  0.01% (1) acts  0.01% (1) $\rho$)  0.01% (1) $\nu$  0.01% (1) $k=3$,  0.01% (1) expands  0.01% (1) $d_3(\mu)=2$,  0.01% (1) $k$theoretic  0.01% (1) multiples  0.01% (1) $[\mathcal{o}_\nu]$  0.01% (1) $\alpha+\beta+\gamma=k(nk)1$,  0.01% (1) setup  0.01% (1) \alpha+\beta+\gamma$.  0.01% (1) $\nu\ge  0.01% (1) h^i(\mathcal{a})\sum_i  0.01% (1) $$\begin{eqnarray*}0&=&\sum_i  0.01% (1) $$[\mathcal{o}_\lambda]\cdot  0.01% (1) $$x_1x_3+x_1x_4+x_3x_4,\hspace{0.5cm}  0.01% (1) \mathcal{o}_{\omega_\lambda}$.  0.01% (1) $\mathcal{o}_\lambda=\iota_\ast  0.01% (1) [\mathcal{o}_\nu]=\sum_\nu  0.01% (1) (1)^{\nu\lambda\mu}c^\nu_{\lambda\mu}[\mathcal{o}_\nu]$$  0.01% (1) $\nu\lambda+\mu$  0.01% (1) coefficient.  0.01% (1) \hspace{0.5cm}x_1x_2+x_1x_4+x_2x_4,$$  0.01% (1) $\nu=\lambda+\mu$  0.01% (1) $\iota:\omega_\lambda\to  0.01% (1) $\iota$  0.01% (1) $[\omega_\lambda]$,  0.01% (1) $$x_1x_2x_3,  0.01% (1) cw  0.01% (1) works,  0.01% (1) x_2x_3+x_2x_4+x_3x_4,$$  0.01% (1) rectangle,  0.01% (1) $[\mathcal{o}_{\lambda}]$  0.01% (1) filtered  0.01% (1) similarly,  0.01% (1) $h^\ast(\mathrm{gr}(n,k))$.  0.01% (1) integer.  0.01% (1) especially  0.01% (1) h^1(\mathcal{b})  0.01% (1) &h^1(\mathcal{a})  0.01% (1) h^0(\mathcal{c})  0.01% (1) h^0(\mathcal{b})  0.01% (1) h^1(\mathcal{c})  0.01% (1) &h^2(\mathcal{a})  0.01% (1) dimensions  0.01% (1) additional  0.01% (1) h^2(\mathcal{c})  0.01% (1) h^2(\mathcal{b})  0.01% (1) h^0(\mathcal{a})  0.01% (1) $$x_1x_2+x_1x_3+x_2x_3,  0.01% (1) $\chi:k(x)\to  0.01% (1) homomorphism  0.01% (1) (additive)  0.01% (1) characteristics.  0.01% (1) \mathcal{a}\to  0.01% (1) \mathcal{b}\to  0.01% (1) cohomology:  0.01% (1) $\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})=0$.  0.01% (1) $x$,  0.01% (1) \mathcal{c}\to  0.01% (1) $\omega_\rho$  0.01% (1) $\omega_{\rho’}$  0.01% (1) moves,  0.01% (1) moves.  0.01% (1) reverses  0.01% (1) $n=4$  0.01% (1) $k=0$  0.01% (1) case.  0.01% (1) h^1(\mathcal{o}_s)$.  0.01% (1) construction,  0.01% (1) here),  0.01% (1) locus,  0.01% (1) slides;  0.01% (1) horizontal  0.01% (1) past,  0.01% (1) moved  0.01% (1) $i_{\mu}$,  0.01% (1) $i$’s).  0.01% (1) making  0.01% (1) 1’s  0.01% (1) vertical  0.01% (1) illuminating  0.01% (1) algorithm.  0.01% (1) $\mu=(3,1)$.  0.01% (1) components.  0.01% (1) $\dim  0.01% (1) $s_n$module,  0.01% (1) degree.  0.01% (1) $\newcommand{\box}{\square}  0.01% (1) march  0.01% (1) shuffling.  0.01% (1) continuation,  0.01% (1) $\ell_1,\ell_2,\ell_3,\ell_4\in  0.01% (1) relaxing  0.01% (1) curves:  0.01% (1) geometry.  0.01% (1) $\gamma$.  0.01% (1) results:  0.01% (1) hartshorne,  0.01% (1) isomorphic  0.01% (1) integral,  0.01% (1) h^1(\mathcal{o}_s)=0$,  0.01% (1) illustrate  0.01% (1) 4.1  0.01% (1) analyses  0.01% (1) map,  0.01% (1) 1.8).  0.01% (1) exercise  0.01% (1) $\{x_1,\ldots,x_4\}$  0.01% (1) $\gamma^c/\alpha$  0.01% (1) six  0.01% (1) yong.  0.01% (1) pechenik  0.01% (1) introduced  0.01% (1) $2$.  0.01% (1) $2$,  0.01% (1) $\square_2$,  0.01% (1) $4$)  0.01% (1) $21\lt  0.01% (1) if:  0.01% (1) tableaux”,  0.01% (1) “genomic  0.01% (1) respectively,  0.01% (1) $\mathbb{p}^1$  0.01% (1) copy  0.01% (1) x_2x_3,\hspace{0.3cm}  0.01% (1) $$\chi(\mathcal{o}_s)=nk.$$  0.01% (1) connected,  0.01% (1) $$x_1x_2,\hspace{0.3cm}  0.01% (1) x_1x_3,\hspace{0.3cm}  0.01% (1) x_1x_4,\hspace{0.3cm}  0.01% (1) $g=1\chi(\mathcal{o}_s)=k(n1)$.  0.01% (1) (columns  0.01% (1) (has  0.01% (1) ballot.  0.01% (1) $221$  0.01% (1) impossible.  0.01% (1) $i_\mu$.  0.01% (1) $10\lt  0.01% (1) properties,  0.01% (1) $k=1$.  0.01% (1) $d_1(\mu)=0$  0.01% (1) below),  0.01% (1) genomic.  0.01% (1) squares.  0.01% (1) while  0.01% (1) $j$).  0.01% (1) $j+1$’s  0.01% (1) among  0.01% (1) $j$’s  0.01% (1) (shaded)  0.01% (1) squares:  0.01% (1) squares,  0.01% (1) genomic:  0.01% (1) $d_2(\mu)=1$  0.01% (1) remains  0.01% (1)  of the  1.24% (114) is a  0.59% (54) in the  0.53% (49) is the  0.48% (44) at the  0.44% (41) of a  0.41% (38) in a  0.37% (34) that the  0.34% (31) as a  0.31% (29) number of  0.3% (28) the number  0.28% (26) on the  0.27% (25) on a  0.26% (24) to a  0.25% (23) by the  0.23% (21) to the  0.22% (20) we have  0.22% (20) schubert curve  0.21% (19) we can  0.21% (19) can be  0.2% (18) and so  0.18% (17) to be  0.17% (16) with a  0.17% (16) for instance  0.17% (16) and i  0.17% (16) with the  0.17% (16) the rational  0.16% (15) the $k$theory  0.16% (15) for instance,  0.16% (15) the same  0.15% (14) and the  0.15% (14) if we  0.15% (14) in this  0.15% (14) set of  0.14% (13) maria gillespie  0.14% (13) there is  0.14% (13) there a  0.14% (13) prove it  0.13% (12) of in  0.13% (12) there are  0.13% (12) that we  0.13% (12) euler characteristic  0.13% (12) a $q$analog  0.13% (12) $q$analog of  0.13% (12) the rationals  0.13% (12) schubert curves  0.12% (11) so the  0.12% (11) such that  0.12% (11) or the  0.12% (11) 0 &  0.12% (11) is an  0.12% (11) special box  0.11% (10) turns out  0.11% (10) it is  0.11% (10) by maria  0.11% (10) such a  0.11% (10) the set  0.11% (10) posted in  0.11% (10) the schubert  0.11% (10) the euler  0.11% (10) gives us  0.11% (10) young tableau  0.11% (10) was a  0.11% (10) defined as  0.11% (10) as the  0.11% (10) posted on  0.11% (10) and a  0.11% (10) jeu de  0.1% (9) for a  0.1% (9) of $q$  0.1% (9) that a  0.1% (9) sequence of  0.1% (9) the first  0.1% (9) then the  0.1% (9) to an  0.1% (9) the next  0.1% (9) young tableaux  0.1% (9) for the  0.1% (9) de taquin  0.1% (9) to get  0.09% (8) in part  0.09% (8) the fiber  0.09% (8) symmetric function  0.09% (8) which is  0.09% (8) exact sequence  0.09% (8) of these  0.09% (8) curve $s$  0.09% (8) given by  0.09% (8) littlewoodrichardson tableau  0.09% (8) but i  0.09% (8) the case  0.09% (8) can you  0.09% (8) \to &  0.09% (8) in order  0.09% (8) of an  0.09% (8) out that  0.09% (8) it turns  0.09% (8) get a  0.09% (8) one of  0.09% (8) the special  0.09% (8) consider the  0.09% (8) elementary symmetric  0.09% (8) connected components  0.08% (7) permutations of  0.08% (7) that it  0.08% (7) $k$theory ring  0.08% (7) 2 &  0.08% (7) the last  0.08% (7) ring of  0.08% (7) of this  0.08% (7) this is  0.08% (7) of length  0.08% (7) have a  0.08% (7) for each  0.08% (7) the result  0.08% (7) in our  0.08% (7) \square &  0.08% (7) the sum  0.08% (7) taquin slide  0.08% (7) be a  0.08% (7) empty square  0.08% (7) & \square  0.08% (7) & 0  0.08% (7) is not  0.08% (7) rational number  0.07% (6) in common?  0.07% (6) from a  0.07% (6) tableaux, and  0.07% (6) 2016 by  0.07% (6) all the  0.07% (6) curves, young  0.07% (6) in particular  0.07% (6) use the  0.07% (6) leave a  0.07% (6) littlewoodrichardson coefficient  0.07% (6) the permutation  0.07% (6) generated by  0.07% (6) 1 &  0.07% (6) \cdots &  0.07% (6) it was  0.07% (6) reading word  0.07% (6) the intersection  0.07% (6) what is  0.07% (6) with an  0.07% (6) 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variety of  0.05% (5) in reading  0.05% (5) we get  0.05% (5) & \cdots  0.05% (5) osculating flag  0.05% (5) it follows  0.05% (5) each of  0.05% (5) phase 1  0.05% (5) the marked  0.05% (5) algorithm for  0.05% (5) the real  0.05% (5) has a  0.05% (5) consists of  0.05% (5) the empty  0.05% (5) the following  0.05% (5) and ktheory  0.05% (5) common? 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0.01% (1) of $1+q+q^2+\cdots+q^{n2}$  0.01% (1) total by  0.01% (1) fashion we  0.01% (1) $q$factorial as  0.01% (1) proof was  0.01% (1) 2. it  0.01% (1) $i\lt j$  0.01% (1) even the  0.01% (1) our $q$count.  0.01% (1) be easily  0.01% (1) injective, since  0.01% (1) choice contributes  0.01% (1) any real  0.01% (1) $(5)_q=1+q+q^2+q^3+q^4$, which  0.01% (1) then can  0.01% (1) entries, and  0.01% (1) total. the  0.01% (1) natural $q$analog  0.01% (1) that $5=1+1+1+1+1$.  0.01% (1) of $1+q+q^2+\cdots+q^{n1}$  0.01% (1) $$(n)_q=\frac{q^n1}{q1}=1+q+q^2+\cdots+q^{n1}.$$ so  0.01% (1) expression simplifies:  0.01% (1) $$(a)_q=\frac{q^a1}{q1},$$ a  0.01% (1) number: we  0.01% (1) second, this  0.01% (1) $f(x)=2x$ is  0.01% (1) \pi_i\pi_j\lt r.$$  0.01% (1) $n$, the  0.01% (1) addition, for  0.01% (1) $q$count of  0.01% (1) $n(q)$, the  0.01% (1) 4 by  0.01% (1) exhibiting a  0.01% (1) product formula  0.01% (1) $$\sum_{p\in s_n}q^{\mathrm{inv}(p)}=(1)(1+q)(1+q+q^2)\cdots(1+q+\cdots+q^n)=(n)_q!.$$  0.01% (1) we saw  0.01% (1) same $q$factorial  0.01% (1) like $5$  0.01% (1) like $n!$  0.01% (1) itself. i  0.01% (1) and behaves  0.01% (1) let’s instead  0.01% (1) $$1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,1,\ldots$$ such  0.01% (1) an index  0.01% (1) this fact  0.01% (1) posted a  0.01% (1) $i$ for  0.01% (1) sense that  0.01% (1) this turns  0.01% (1) that $21534$  0.01% (1) \pi_{i+1}+r,$$ so  0.01% (1) $n!$, also  0.01% (1) s_n}q^{\mathrm{inv}(p)}$, which  0.01% (1) $q$analog $\sum_{p\in  0.01% (1) above. let’s  0.01% (1) now see  0.01% (1) stop there.  0.01% (1) 4. notice  0.01% (1) $(1)(1+q)(1+q+q^2)$. indeed,  0.01% (1) into smaller  0.01% (1) it can’t  0.01% (1) factors as  0.01% (1) which $$\pi_i\ge  0.01% (1) integers. but  0.01% (1) that $1+2q+2q^2+q^3$  0.01% (1) to $5$,  0.01% (1) by building  0.01% (1) into two  0.01% (1) things are  0.01% (1) weighted? $q$multiplication  0.01% (1) but what  0.01% (1) $f(1)=r_1$, $f(1)=r_2$,  0.01% (1) smallerdegree polynomials  0.01% (1) $m\cdot n$.  0.01% (1) principle: given  0.01% (1) two weighted  0.01% (1) with $q$counts  0.01% (1) is prime,  0.01% (1) $m(q)$ and  0.01% (1) $q$analogy way:  0.01% (1) the polynomial  0.01% (1) sets $a$  0.01% (1) $1+q+q^2+q^3+q^4$ cannot  0.01% (1) things is  0.01% (1) with integer  0.01% (1) function will  0.01% (1) track of  0.01% (1) by keeping  0.01% (1) up. the  0.01% (1) the ground  0.01% (1) have $\pi_i\pi_j$r$descent  0.01% (1) $q$counting from  0.01% (1) first two  0.01% (1) about $5$ness,  0.01% (1) of $m$  0.01% (1) things and  0.01% (1) coefficients. so  0.01% (1) the $q$number  0.01% (1) $(5)_q$ also  0.01% (1) thing from  0.01% (1) 4 above:  0.01% (1) proof that  0.01% (1) $n!$ is  0.01% (1) as $q$  0.01% (1) approaches $1$.  0.01% (1) an odd  0.01% (1) be recovered  0.01% (1) $p$ can  0.01% (1) also say  0.01% (1) $q$analog if  0.01% (1) $$\frac{q^51}{q1}$$ is  0.01% (1) another $q$analog  0.01% (1) do have  0.01% (1) a well  0.01% (1) defined limit  0.01% (1) get division  0.01% (1) though we  0.01% (1) integer $x$.  0.01% (1) – even  0.01% (1) defined at  0.01% (1) if $p_q$  0.01% (1) is said  0.01% (1) and surjective  0.01% (1) both injective  0.01% (1) definition anyway:  0.01% (1) about, but  0.01% (1) doesn’t really  0.01% (1) capture what  0.01% (1) $p_q$, depending  0.01% (1) more unifying  0.01% (1) however, it  0.01% (1) surjective, since  0.01% (1) $5$. sometimes,  0.01% (1) $2q+3q^2$ is  0.01% (1) $p$. so,  0.01% (1) on $q$,  0.01% (1) in $p_q$  0.01% (1) can calculate,  0.01% (1) an even  0.01% (1) which doesn’t  0.01% (1) not equal  0.01% (1) widely accepted  0.01% (1) “useful” $q$analogs,  0.01% (1) “good” or  0.01% (1) referring to  0.01% (1) to $2x$  0.01% (1) standard definition,  0.01% (1) but which  0.01% (1) $p_q$ depending  0.01% (1) number like  0.01% (1) such that:  0.01% (1) $3$ is  0.01% (1) accurate definition:  0.01% (1) i’ll attempt  0.01% (1) here: more  0.01% (1) are usually  0.01% (1) $q$analogs, they  0.01% (1) = 5.$$  0.01% (1) now of  0.01% (1) course, there  0.01% (1) 1} \frac{5q^4}{1}  0.01% (1) 1} \frac{q^51}{q1}=\lim_{q\to  0.01% (1) using l’hospital’s  0.01% (1) rule: $$\lim_{q\to  0.01% (1) are an  0.01% (1) unlimited supply  0.01% (1) than others.  0.01% (1) when mathematicians  0.01% (1) this makes  0.01% (1) more important  0.01% (1) $q$analogs are  0.01% (1) number $5$,  0.01% (1) but certain  0.01% (1) bijective, and  0.01% (1) whose definition  0.01% (1) the $q$binomial  0.01% (1) polynomiallike things  0.01% (1) discuss $q$catalan  0.01% (1) positive rationals.  0.01% (1) interpretation of  0.01% (1) to derive  0.01% (1) to find  0.01% (1) looking for  0.01% (1) numbers and  0.01% (1) most people  0.01% (1) we want  0.01% (1) $q$analogs. stay  0.01% (1) would agree  0.01% (1) that $5^q$  0.01% (1) other fun  0.01% (1) usually we’re  0.01% (1) in $q$.  0.01% (1) $f:\mathbb{z}\to \mathbb{q}$,  0.01% (1) of reasons.  0.01% (1) give us  0.01% (1) $\mathbb{q}$ the  0.01% (1) rationals. notice  0.01% (1) can list  0.01% (1) $(\pi_i,\pi_j)$ with  0.01% (1) it certainly  0.01% (1) $5$ for  0.01% (1) if $2x=2y$  0.01% (1) how to  0.01% (1) where $\mathbb{z}$  0.01% (1) other hand,  0.01% (1) post we’ll  0.01% (1) $\frac{q^51}{q1}$, is  0.01% (1) an excellent  0.01% (1) principle. that’s  0.01% (1) an $r$inversion  0.01% (1) cumbersome. so,  0.01% (1) simply two  0.01% (1) these statistics:  0.01% (1) shorter way  0.01% (1) different instances  0.01% (1) entire parameterized  0.01% (1) posts i’ll  0.01% (1) statistics, called  0.01% (1) just a  0.01% (1) picture behind  0.01% (1) addictive! so  0.01% (1) those rare  0.01% (1) mathematical terms  0.01% (1) a bijection.  0.01% (1) awesome and  0.01% (1) explain why  0.01% (1) they’re so  0.01% (1) $r\text{maj}$, all  0.01% (1) of which  0.01% (1) 4 hi,  0.01% (1) onto correspondence,”  0.01% (1) i’m maria  0.01% (1) november 22,  0.01% (1) wordy and  0.01% (1) rawlings defined  0.01% (1) (part 1)  0.01% (1) and i’m  0.01% (1) “onetoone and  0.01% (1) $q$analogs is  0.01% (1) a littleknown  0.01% (1) gem, and  0.01% (1) are equidistributed!  0.01% (1) addict. the  0.01% (1) their equidistribution.  0.01% (1) of saying  0.01% (1) and 3  0.01% (1) has only  0.01% (1) makes $a/b$  0.01% (1) doesn’t bother  0.01% (1) $n!=n\cdot (n1)\cdot  0.01% (1) reduced fraction.  0.01% (1) the third,  0.01% (1) there exists  0.01% (1) me much.  0.01% (1) \cdots \cdot  0.01% (1) 2\cdot 1$  0.01% (1) of $1,2,3$  0.01% (1) are $123$,  0.01% (1) $132$, $213$,  0.01% (1) $3!=6$ permutations  0.01% (1) entries. for  0.01% (1) this condition  0.01% (1) rearrange the  0.01% (1) number $2/4$  0.01% (1) $n2$ for  0.01% (1) numbers $1,\ldots,n$  0.01% (1) things just  0.01% (1) row. there  0.01% (1) rearranging the  0.01% (1) $1/2$ in  0.01% (1) count permutations  0.01% (1) hurt my  0.01% (1) are $n$  0.01% (1) first number,  0.01% (1) remaining choices  0.01% (1) second, then  0.01% (1) should be  0.01% (1) are $n1$  0.01% (1) but some  0.01% (1) statistics all  0.01% (1) and once  0.01% (1) $231$, $312$,  0.01% (1) $321$. now,  0.01% (1) $\mathrm{inv}(\pi)$ denotes  0.01% (1) statistics on  0.01% (1) and $a$  0.01% (1) smaller, and  0.01% (1) is any  0.01% (1) when i  0.01% (1) bigger number  0.01% (1) permutation $\pi$.  0.01% (1) of numerators/denominators:  0.01% (1) so usually,  0.01% (1) restrict to  0.01% (1) they contain.  0.01% (1) of 3  0.01% (1) integer occurs  0.01% (1) where $b>0$  0.01% (1) below shows  0.01% (1) permutation in  0.01% (1) see a  0.01% (1) $f(a)=b$. for  0.01% (1) to how  0.01% (1) “mixed up”  0.01% (1) permutations according  0.01% (1) weight the  0.01% (1) say we  0.01% (1) next entry,  0.01% (1) they are,  0.01% (1) sense of  0.01% (1) order. an  0.01% (1) inversion is  0.01% (1) mistake in  0.01% (1) integer such  0.01% (1) out of  0.01% (1) that $\mathrm{gcd}(a,b)=1$.  0.01% (1) text, it  0.01% (1) say you’re  0.01% (1) of $q$.  0.01% (1) out on  0.01% (1) recently pointed  0.01% (1) textbook, and  0.01% (1) continuation of  0.01% (1) reddit. or  0.01% (1) set $$\mathbb{q}=\{(a,b)  0.01% (1) reply this  0.01% (1) algebra 2  0.01% (1) posts on  0.01% (1) the glencoe/mcgrawhill  0.01% (1) $k$. more  0.01% (1) having $s\text{maj}$  0.01% (1) area in  0.01% (1) one too,  0.01% (1) $q$analogs. counting  0.01% (1) by $q$’s  0.01% (1) december 10,  0.01% (1) (part 2)  0.01% (1) algorithm is  0.01% (1) and }\mathrm{gcd}(a,b)=1\text{  0.01% (1) original file  0.01% (1) found online  0.01% (1) geometry and  0.01% (1) s_n} q^{r\text{maj}(\pi)}=\sum_{\pi\in  0.01% (1) the story:  0.01% (1) exactly what  0.01% (1) worksheet. the  0.01% (1) in pearl  0.01% (1) will match  0.01% (1)  b>0\text{  0.01% (1) succinctly, $$\sum_{\pi\in  0.01% (1) some version  0.01% (1) to understand  0.01% (1) it. posted  0.01% (1) worksheet for  0.01% (1) which $q$analogs  0.01% (1) $b\in b$,  0.01% (1) this convention.  0.01% (1) of $\{1,2,\ldots,n\}$  0.01% (1) and rationals?  0.01% (1) is, example  0.01% (1) 2! yes  0.01% (1) than usual  0.01% (1) we think  0.01% (1) $r,s\ge 1$,  0.01% (1) soul. no  0.01% (1) are given  0.01% (1) by powers  0.01% (1) objects, where  0.01% (1) of weighted  0.01% (1) as “$q$counting”  0.01% (1) same distribution:  0.01% (1) different light  0.01% (1) but viewed  0.01% (1) the supplementary  0.01% (1) combinatorics. in  0.01% (1) this context,  0.01% (1) same as  0.01% (1) the numerator  0.01% (1) and intervention”  0.01% (1) “study guide  0.01% (1) $q$ is  0.01% (1) a formal  0.01% (1) this horrifying  0.01% (1) there is!  0.01% (1) and maj  0.01% (1) falsehood was  0.01% (1) having $r\text{maj}$  0.01% (1) variable, and  0.01% (1) can think  0.01% (1) $$\begin{array}{ccc} p  0.01% (1) & \mathrm{inv}(p)  0.01% (1) objects. recall  0.01% (1) $\mathbb{f}_q$. turn  0.01% (1) “interesting $q$analog”  0.01% (1) them! pages:  0.01% (1) a statistic  0.01% (1) objects and  0.01% (1) $\mathrm{stat}$ is  0.01% (1) finite field  0.01% (1) (a) every  0.01% (1) one aspect  0.01% (1) $a/b$ where  0.01% (1) as fractions  0.01% (1) of listing  0.01% (1) today: $q$analogs  0.01% (1) that appear  0.01% (1) $p_q$ such  0.01% (1) permutation to  0.01% (1) of combinatorial  0.01% (1) $b$ are  0.01% (1) in.. in  0.01% (1) september 14,  0.01% (1) are interested  0.01% (1) our summation  0.01% (1) difficult task,  0.01% (1) 6 education  0.01% (1) general, a  0.01% (1) a summation  0.01% (1) bijection exists  0.01% (1) believe this  0.01% (1) glencoe/mcgrawhill doesn’t  0.01% (1) be $$r\text{}\mathrm{maj}(\pi)=\left(\sum_{\pi_i\ge  0.01% (1) $p$ ranges  0.01% (1) of $q$powers  0.01% (1) $q^{\mathrm{stat}(p)}$ where  0.01% (1) only cover  0.01% (1) the $r\text{maj}$  0.01% (1) all sorts  0.01% (1) once. next,  0.01% (1) position $3$  0.01% (1) words, and  0.01% (1) properties. certainly  0.01% (1) of ways,  0.01% (1) $2$descent. finally,  0.01% (1) objects, and  0.01% (1) more information  0.01% (1) satisfied in  0.01% (1) it’s also  0.01% (1) 1, 2,  0.01% (1) total number.  0.01% (1) just their  0.01% (1) objects than  0.01% (1) list exactly  0.01% (1) $q$number is  0.01% (1) even interesting  0.01% (1) combinatorics. so  0.01% (1) theory, and  0.01% (1) theory, representation  0.01% (1) much awesome  0.01% (1) mathematics is  0.01% (1) that i’ll  0.01% (1) involved in  0.01% (1) analytic number  0.01% (1) finite geometry,  0.01% (1) precise about  0.01% (1) might expect.  0.01% (1) than we  0.01% (1) it comes  0.01% (1) he defines  0.01% (1) those numbers  0.01% (1) $f:\mathbb{z}\to \mathbb{z}$  0.01% (1) exponents of  0.01% (1) rather the  0.01% (1) $q$count by  0.01% (1) summing these  0.01% (1) $q$powers, to  0.01% (1) $q^{\mathrm{inv}(p)}$, and  0.01% (1) $p$ by  0.01% (1) each permutation  0.01% (1) textbooks is  0.01% (1) sum $$\sum_{p\in  0.01% (1) s_n}q^{\mathrm{inv}(p)}$$ where  0.01% (1) is $1+2q+2q^2+q^3$  0.01% (1) by our  0.01% (1) table above.  0.01% (1) for $n=3$,  0.01% (1) $1,\ldots,n$. so  0.01% (1) $s_n$ is  0.01% (1) statistic! rawling’s  0.01% (1) even harder.  0.01% (1) & q^3  0.01% (1) 132 &  0.01% (1) q\\ 213  0.01% (1) let us  0.01% (1) 1 \\  0.01% (1) 123 &  0.01% (1) & q^{\mathrm{inv}(p)}\\\hline  0.01% (1) to avoid.  0.01% (1) unique representation,  0.01% (1) is hard  0.01% (1) technical fields  0.01% (1) \\ 321  0.01% (1) and math  0.01% (1) q^2\\ 312  0.01% (1) human error  0.01% (1) 231 &  0.01% (1) result is  0.01% (1) come to  0.01% (1) of. writing  0.01% (1) as $\sum_{i=0}^\infty  0.01% (1) c_i q^i$  0.01% (1) where $c_i$  0.01% (1) instead write  0.01% (1) it really  0.01% (1) of coefficients,  0.01% (1) $1,2,2,1$. generatingfunctionologically,  0.01% (1) length $n$  0.01% (1) with $i$  0.01% (1) understand that  0.01% (1) the domain  0.01% (1) coefficients but  0.01% (1) s_n}q^{\mathrm{inv}(p)}$, we  0.01% (1) notation, $\sum_{p\in  0.01% (1) $b\neq 0$.  0.01% (1) inversions. but  0.01% (1) to assign  0.01% (1) \pi_{i+1}+r}i\right)+\#r\text{}\mathrm{inversions}.$$ thus  0.01% (1) $n\text{maj}$ is  0.01% (1) the hang  0.01% (1) few tries  0.01% (1) functions. as  0.01% (1) and generating  0.01% (1) philosophical distinction  0.01% (1) between $q$analogs  0.01% (1) takes a  0.01% (1) index, and  0.01% (1) is thought  0.01% (1) that $1\text{maj}$  0.01% (1) $2\text{}\mathrm{maj}(21534)=3+2=5$. notice  0.01% (1) sum $1+2q+2q^2+q^3$  0.01% (1) is. teaching  0.01% (1) function, the  0.01% (1) usual major  0.01% (1) beginning of  0.01% (1) whose sizes  0.01% (1) length$(n1)$ sequences  0.01% (1) of $0$’s  0.01% (1) and $1$’s.  0.01% (1) thus, the  0.01% (1) that $2^{n1}$  0.01% (1) note also  0.01% (1) colorings of  0.01% (1) of $3$  0.01% (1) fenceposts: $$\begin{array}{ccc}  0.01% (1) \blacksquare \end{array}$$  0.01% (1) our identity,  0.01% (1) $2^{n1}\cdot f_n$,  0.01% (1) interlace their  0.01% (1) entries, forming  0.01% (1) an alternating  0.01% (1) digits and  0.01% (1) given such  0.01% (1) their lengths,  0.01% (1) binary sequence  0.01% (1) $n1$ and  0.01% (1) coloring of  0.01% (1) length $n2$.  0.01% (1) five such  0.01% (1) $f_5=5$, because  0.01% (1) and remembered  0.01% (1) $(n+1)$st fibonacci  0.01% (1) number $f_n$  0.01% (1) color a  0.01% (1) it! ended,  0.01% (1) after prove  0.01% (1) the equation  0.01% (1) and $n$  0.01% (1) right. i  0.01% (1) started thinking  0.01% (1) $n2$ fenceposts  0.01% (1) either black  0.01% (1) this combinatorial  0.01% (1) construction would  0.01% (1) satisfy the  0.01% (1) fibonacci recurrence?)  0.01% (1) see why  0.01% (1) (can you  0.01% (1) or white  0.01% (1) adjacent ones  0.01% (1) are black.  0.01% (1) fence posts  0.01% (1) such as:  0.01% (1) specifically in  0.01% (1) might be  0.01% (1) traced back  0.01% (1) freshman year  0.01% (1) in mathematics,  0.01% (1) a ph.d.  0.01% (1) down the  0.01% (1) road. for  0.01% (1) me, my  0.01% (1) pursuit of  0.01% (1) undergraduate at  0.01% (1) mit. coming  0.01% (1) other sciencerelated  0.01% (1) endeavors (thanks  0.01% (1) loving and  0.01% (1) very mathematical  0.01% (1) competitions and  0.01% (1) school math  0.01% (1) off of  0.01% (1) a series  0.01% (1) of successes  0.01% (1) in high  0.01% (1) greatest adventures  0.01% (1) that lead  0.01% (1) need only  0.01% (1) side also  0.01% (1) counts these  0.01% (1) for my  0.01% (1) such sequences  0.01% (1) will call  0.01% (1) $$1\, \square\,  0.01% (1) 0\, \square\,  0.01% (1) 1\, \blacksquare\,  0.01% (1) 1$$ we  0.01% (1) solution, or  0.01% (1) post your  0.01% (1) 24, 2016  0.01% (1) sometimes it’s  0.01% (1) the missteps  0.01% (1) in life  0.01% (1) on may  0.01% (1) phinished! posted  0.01% (1) own solution  0.01% (1) comments below!  0.01% (1) in gemstones,  0.01% (1) with $m$  0.01% (1) identity above,  0.01% (1) got me  0.01% (1) thinking: what  0.01% (1) expand each  0.01% (1) terms using  0.01% (1) same fact,  0.01% (1) proofs there  0.01% (1) the classroom.  0.01% (1) she was  0.01% (1) amazed how  0.01% (1) many different  0.01% (1) theorem? is  0.01% (1) resulting identity?  0.01% (1) \left(\sum_{i=0}^n \binom{n}{i}(\sqrt{5})^i  0.01% (1) \right) –  0.01% (1) \left(\sum_{i=0}^n (1)^i\binom{n}{i}(\sqrt{5})^i  0.01% (1) \right) \right)$$  0.01% (1) expression is:  0.01% (1) binet’s formula.  0.01% (1) suppose we  0.01% (1) to expand  0.01% (1) $(1+\sqrt{5})^n$ and  0.01% (1) $(1\sqrt{5})^n$ in  0.01% (1) board outside  0.01% (1) a white  0.01% (1) at prove  0.01% (1) it!, we  0.01% (1) first derived  0.01% (1) the formula  0.01% (1) of integers?  0.01% (1) 5$’s showing  0.01% (1) looks crazy,  0.01% (1) right? why  0.01% (1) would there  0.01% (1) be $\sqrt  0.01% (1) using generating  0.01% (1) functions. i  0.01% (1) students was  0.01% (1) work out  0.01% (1) the induction  0.01% (1) proof on  0.01% (1) later, one  0.01% (1) induction, and  0.01% (1) mentioned during  0.01% (1) class that  0.01% (1) proven by  0.01% (1) the even  0.01% (1) terms in  0.01% (1) in two  0.01% (1) ways is  0.01% (1) principle that  0.01% (1) if by  0.01% (1) of counting  0.01% (1) proof method  0.01% (1) one handy  0.01% (1) fact about  0.01% (1) they count  0.01% (1) collection. the  0.01% (1) some method  0.01% (1) a collection  0.01% (1) a “combinatorial  0.01% (1) proof” may  0.01% (1) be able  0.01% (1) prove the  0.01% (1) $m=n$. such  0.01% (1) $m$ elements,  0.01% (1) $n$ elements,  0.01% (1) and by  0.01% (1) another method  0.01% (1) clearly nonnegative  0.01% (1) side are  0.01% (1) n/2\rfloor} 2  0.01% (1) \binom{n}{2k+1}(\sqrt{5})^{2k+1} \right)$$  0.01% (1) since $(\sqrt{5})^{2k+1}=\sqrt{5}\cdot  0.01% (1) 5^k$, we  0.01% (1) 2^n}\left( \sum_{k=0}^{\lfloor  0.01% (1) us: $$f_n=\frac{1}{\sqrt{5}\cdot  0.01% (1) summations cancel,  0.01% (1) and combining  0.01% (1) the odd  0.01% (1) terms gives  0.01% (1) can cancel  0.01% (1) the factors  0.01% (1) $$2^{n1}\cdot f_n=\sum_{k=0}^{\lfloor  0.01% (1) n/2\rfloor} \binom{n}{2k+1}\cdot  0.01% (1) 5^k.$$ now,  0.01% (1) hand and  0.01% (1) to obtain:  0.01% (1) by $2^{n1}$  0.01% (1) of $\sqrt{5}$  0.01% (1) and multiply  0.01% (1) both sides  0.01% (1) family!), i  0.01% (1) confident and  0.01% (1) strong grasp  0.01% (1) of some  0.01% (1) other algebraic  0.01% (1) concepts being  0.01% (1) didn’t yet  0.01% (1) freshman, and  0.01% (1) a term  0.01% (1) from another  0.01% (1) area of  0.01% (1) hooked. but  0.01% (1) the course.  0.01% (1) hard but  0.01% (1) confidence shattered.  0.01% (1) now, not  0.01% (1) only was  0.01% (1) i fascinated  0.01% (1) 18yearold littlemissperfect  0.01% (1) horrified, my  0.01% (1) wound up  0.01% (1) the class.  0.01% (1) me, get  0.01% (1) math class?  0.01% (1) concrete problem.  0.01% (1) a simple,  0.01% (1) beautiful! i  0.01% (1) remember sitting  0.01% (1) in my  0.01% (1) seat, wideeyed,  0.01% (1) at $v$.  0.01% (1) loops starting  0.01% (1) trace using  0.01% (1) eigenvalues and  0.01% (1) divide by  0.01% (1) $n$ to  0.01% (1) watching richard  0.01% (1) stanley quietly  0.01% (1) tools from  0.01% (1) linear algebra  0.01% (1) so elegantly  0.01% (1) solve such  0.01% (1) that advanced  0.01% (1) to me  0.01% (1) but authoritatively  0.01% (1) technique. it  0.01% (1) was incredible  0.01% (1) subject, i  0.01% (1) gained respect  0.01% (1) ph.d. at  0.01% (1) uc berkeley  0.01% (1) topic in  0.01% (1) algebraic combinatorics…  0.01% (1) completed my  0.01% (1) 2016, having  0.01% (1) now find  0.01% (1) myself, as  0.01% (1) of friday,  0.01% (1) may 20,  0.01% (1) …and i  0.01% (1) often wonder  0.01% (1) years. (see  0.01% (1) summary of  0.01% (1) my thesis.  0.01% (1) my full  0.01% (1) throughout the  0.01% (1) motivated me  0.01% (1) how much  0.01% (1) that silly  0.01% (1) little b+  0.01% (1) algebra. i  0.01% (1) combinatorics and  0.01% (1) for more.  0.01% (1) years that  0.01% (1) followed, i  0.01% (1) took more  0.01% (1) come back  0.01% (1) help but  0.01% (1) for it.  0.01% (1) a worthy  0.01% (1) challenge, and  0.01% (1) i couldn’t  0.01% (1) courses on  0.01% (1) similar subjects  0.01% (1) well, but  0.01% (1) was always  0.01% (1) drawn back  0.01% (1) interplay between  0.01% (1) areas as  0.01% (1) i dabbled  0.01% (1) and wrote  0.01% (1) several undergraduate  0.01% (1) research papers.  0.01% (1) compute this  0.01% (1) $a^k$. one  0.01% (1) had no  0.01% (1) idea what  0.01% (1) “algebraic combinatorics”  0.01% (1) even meant,  0.01% (1) load. i  0.01% (1) packed course  0.01% (1) so my  0.01% (1) young ambitious  0.01% (1) self added  0.01% (1) my already  0.01% (1) did hear  0.01% (1) was being  0.01% (1) i pass  0.01% (1) up that  0.01% (1) chance? what  0.01% (1) if he  0.01% (1) how could  0.01% (1) the area.  0.01% (1) taught by  0.01% (1) richard stanley,  0.01% (1) a world  0.01% (1) expert in  0.01% (1) the spring,  0.01% (1) class in  0.01% (1) skills to,  0.01% (1) i don’t  0.01% (1) know, come  0.01% (1) a unified  0.01% (1) my mathematical  0.01% (1) a physicist  0.01% (1) excited 18year  0.01% (1) old whose  0.01% (1) dream was  0.01% (1) to become  0.01% (1) field theory  0.01% (1) or something.  0.01% (1) my friends  0.01% (1) were signed  0.01% (1) up for  0.01% (1) the undergraduate  0.01% (1) math too,  0.01% (1) loved pure  0.01% (1) me at  0.01% (1) the age  0.01% (1) of 18ish.  0.01% (1) didn’t teach  0.01% (1) it again  0.01% (1) this, but  0.01% (1) stanley went  0.01% (1) forth with  0.01% (1) an algebraic  0.01% (1) to solve  0.01% (1) messy) way  0.01% (1) other vertex  0.01% (1) and back  0.01% (1) again: there  0.01% (1) elementary (though  0.01% (1) proof. he  0.01% (1) considered the  0.01% (1) loops of  0.01% (1) starting from  0.01% (1) any vertex  0.01% (1) trace of  0.01% (1) showed that  0.01% (1) graph, and  0.01% (1) adjacency matrix  0.01% (1) $a$ of  0.01% (1) the complete  0.01% (1) to any  0.01% (1) from $v$  0.01% (1) something like  0.01% (1) the following:  0.01% (1) with $n$  0.01% (1) vertices, how  0.01% (1) combinatorial question.  0.01% (1) started with  0.01% (1) before i  0.01% (1) left mit?  0.01% (1) first day  0.01% (1) class, stanley  0.01% (1) many walks  0.01% (1) $v$ end  0.01% (1) graph looks  0.01% (1) like: and  0.01% (1) four closed  0.01% (1) length two,  0.01% (1) $k=2$, the  0.01% (1) $n=5$ and  0.01% (1) up back  0.01% (1) $v$ on  0.01% (1) step? for  0.01% (1) \left(\frac{1\sqrt{5}}{2}\right)^n \right)$$  0.01% (1) \left(\frac{1+\sqrt{5}}{2}\right)^n –  0.01% (1) size $k=2$,  0.01% (1) $d_2(\mu)=1$ since  0.01% (1) one box  0.01% (1) among the  0.01% (1) in $i_\mu$.  0.01% (1) $1$ variable  0.01% (1) \le 1$,  0.01% (1) impossible. so  0.01% (1) no partial  0.01% (1) functions in  0.01% (1) last two  0.01% (1) columns (columns  0.01% (1) can only  0.01% (1) be $2$,  0.01% (1) $e_2(s)$ for  0.01% (1) all subsets  0.01% (1) so $r$  0.01% (1) r\le 2$.  0.01% (1) $3$ and  0.01% (1) $4$) of  0.01% (1) $\mu$, and  0.01% (1) have $21\lt  0.01% (1) $10\lt r  0.01% (1) $i_\mu$ we  0.01% (1) the construction,  0.01% (1) suppose $n=4$  0.01% (1) and $\mu=(3,1)$.  0.01% (1) then to  0.01% (1) to illustrate  0.01% (1) by degree.  0.01% (1) action. since  0.01% (1) homogeneous ideal,  0.01% (1) a graded  0.01% (1) $s_n$module, graded  0.01% (1) compute $i_{\mu}$,  0.01% (1) first consider  0.01% (1) (see image  0.01% (1) below), and  0.01% (1) $e_r(s)$ to  0.01% (1) be in  0.01% (1) is empty  0.01% (1) fourth column  0.01% (1) of $\{x_1,\ldots,x_4\}$  0.01% (1) $k=1$. we  0.01% (1) have $d_1(\mu)=0$  0.01% (1) size $2$.  0.01% (1) six polynomials  0.01% (1) &(e_2(x_1,x_2), e_2(x_1,x_3),\ldots,  0.01% (1) e_2(x_3,x_4), \\  0.01% (1) & e_2(x_1,x_2,x_3),  0.01% (1) \ldots, e_2(x_2,x_3,x_4),  0.01% (1) $$\begin{align*} i_{(3,1)}=  0.01% (1) $i_{\mu}$. all  0.01% (1) the full  0.01% (1) functions $e_1,\ldots,e_4$  0.01% (1) four variables  0.01% (1) relations in  0.01% (1) e_3(x_1,x_2,x_3), \ldots,  0.01% (1) e_3(x_2,x_3,x_4), \\  0.01% (1) it’s clear  0.01% (1) that $r_{(1^n)}=\mathbb{c}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$  0.01% (1) coninvariants under  0.01% (1) the $s_n$  0.01% (1) more examples,  0.01% (1) as two  0.01% (1) & e_1(x_1,\ldots,x_4),  0.01% (1) e_2(x_1,\ldots,x_4), e_3(x_1,\ldots,x_4),  0.01% (1) e_4(x_1,\ldots,x_4)) \end{align*}$$  0.01% (1) 4$. thus  0.01% (1) $44\lt r\le  0.01% (1) therefore have  0.01% (1) $e_2(s)$ and  0.01% (1) $e_3(s)$ for  0.01% (1) each such  0.01% (1) so $32  0.01% (1) $d_3(\mu)=2$, and  0.01% (1) $$x_1x_2,\hspace{0.3cm} x_1x_3,\hspace{0.3cm}  0.01% (1) x_1x_4,\hspace{0.3cm} x_2x_3,\hspace{0.3cm}  0.01% (1) x_2x_4,\hspace{0.3cm} x_3x_4.$$  0.01% (1) size $k=3$,  0.01% (1) in $i_\mu$,  0.01% (1) the eight  0.01% (1) x_1x_3x_4,\hspace{0.4cm} x_2x_3x_4$$  0.01% (1) finally, for  0.01% (1) $s=\{x_1,x_2,x_3,x_4\}$, we  0.01% (1) have $d_4(\mu)=4$  0.01% (1) x_1x_2x_4, \hspace{0.4cm}  0.01% (1) $$x_1x_2x_3, \hspace{0.4cm}  0.01% (1) additional polynomials  0.01% (1) $$x_1x_2+x_1x_3+x_2x_3, \hspace{0.5cm}x_1x_2+x_1x_4+x_2x_4,$$  0.01% (1) $$x_1x_3+x_1x_4+x_3x_4,\hspace{0.5cm} x_2x_3+x_2x_4+x_3x_4,$$  0.01% (1) under this  0.01% (1) the variables,  0.01% (1) – in  0.01% (1) but it’s  0.01% (1) about time  0.01% (1) for… the  0.01% (1) of coinvariants  0.01% (1) them –  0.01% (1) series, and  0.01% (1) talked in  0.01% (1) depth about  0.01% (1) a particular  0.01% (1) garsiaprocesi modules!  0.01% (1) also known  0.01% (1) link between  0.01% (1) these two  0.01% (1) interpretations given  0.01% (1) paper of  0.01% (1) a torus,  0.01% (1) matrices with  0.01% (1) cohomology rings  0.01% (1) springer fibers  0.01% (1) in type  0.01% (1) $a$, or  0.01% (1) springer correspondence  0.01% (1) briefly at  0.01% (1) content skip  0.01% (1) to secondary  0.01% (1) content homeabout  0.01% (1) this blogcontribute!all  0.01% (1) to primary  0.01% (1) menu skip  0.01% (1) quest for  0.01% (1) mathematical beauty  0.01% (1) and truth  0.01% (1) search main  0.01% (1) posts post  0.01% (1) $r_\mu$ posted  0.01% (1) yet described  0.01% (1) favorite graded  0.01% (1) $s_n$modules. i  0.01% (1) mentioned them  0.01% (1) i’ve not  0.01% (1) posted here,  0.01% (1) 23, 2016  0.01% (1) somehow, in  0.01% (1) time i’ve  0.01% (1) deconcini and  0.01% (1) procesi. but  0.01% (1) s=k).$$ here,  0.01% (1) $d_k(\mu)=\mu’_n+\mu’_{n1}+\cdots+ \mu’_{nk+1}$  0.01% (1) boxes in  0.01% (1) $k$ columns  0.01% (1) \le k,  0.01% (1) : kd_k(\mu)  0.01% (1) this subset  0.01% (1) variables is  0.01% (1) symmetric function,  0.01% (1) have $$i_{\mu}=(e_r(s)  0.01% (1) where we  0.01% (1) pad the  0.01% (1) natural action  0.01% (1) of $s_n$  0.01% (1) on $\mathbb{c}[x_1,\ldots,x_n]$  0.01% (1) by permuting  0.01% (1) inherits the  0.01% (1) ring $r_\mu$  0.01% (1) $\mu’$ with  0.01% (1) $0$’s so  0.01% (1) parts. this  0.01% (1) $e_r(s)$ in  0.01% (1) $s=k$. then  0.01% (1) an entirely  0.01% (1) elementary way.  0.01% (1) using tanisaki’s  0.01% (1) approach, we  0.01% (1) modules in  0.01% (1) with these  0.01% (1) the work  0.01% (1) of tanisaki,  0.01% (1) and garsia  0.01% (1) and procesi,  0.01% (1) $$r_{\mu}=\mathbb{c}[x_1,\ldots,x_n]/i_{\mu},$$ where  0.01% (1) $i_{\mu}$ is  0.01% (1) $d$ in  0.01% (1) of variables  0.01% (1) $z_i$. let  0.01% (1) $s\subset\{x_1,\ldots,x_n\}$ with  0.01% (1) of degree  0.01% (1) squarefree monomials  0.01% (1) the ideal  0.01% (1) functions defined  0.01% (1) $e_d(z_1,\ldots,z_k)$ is  0.01% (1) action, and  0.01% (1) $r_{(n)}=\mathbb{c}$ is  0.01% (1) work to  0.01% (1) entire ideal,  0.01% (1) shown by  0.01% (1) showing that  0.01% (1) somewhat more  0.01% (1) it takes  0.01% (1) kd_k(\mu)$, which  0.01% (1) the relations  0.01% (1) that describe  0.01% (1) $i_\mu$ above.  0.01% (1) $r_\mu$ has  0.01% (1) dimension, namely  0.01% (1) procesi. pages:  0.01% (1) diamond, gemstones  0.01% (1) combinatorially? posted  0.01% (1) on august  0.01% (1) basis of  0.01% (1) the monomial  0.01% (1) the multinomial  0.01% (1) coefficient $\binom{n}{\mu}$.  0.01% (1) that, we’ll  0.01% (1) discuss on  0.01% (1) $r \gt  0.01% (1) soon as  0.01% (1) also have  0.01% (1) this property,  0.01% (1) if $x=\mathrm{diag}(x_1,\ldots,x_n)\in  0.01% (1) \overline{c_\mu}\cap t$  0.01% (1) $c_\mu$ must  0.01% (1) any element  0.01% (1) and $d_k(\mu)$  0.01% (1) sums of  0.01% (1) the ending  0.01% (1) $\mu$. it  0.01% (1) have $$t^{d_k(\mu)}  0.01% (1)  (x_{i_1}t)(x_{i_2}t)\cdots  0.01% (1) vieta’s formulas,  0.01% (1) functions $e_r(s)$  0.01% (1) vanish on  0.01% (1) $x$ as  0.01% (1) $t$ using  0.01% (1) $\{x_1,\ldots,x_n\}$. expanding  0.01% (1) (x_{i_k}t)$$ for  0.01% (1) any subset  0.01% (1) $s=\{x_{i_1},\ldots,x_{i_k}\}$ of  0.01% (1) 12, 2016  0.01% (1) big success,  0.01% (1) adding the  0.01% (1) previous two  0.01% (1) numbers to  0.01% (1) next: $$0,1,1,2,3,5,8,13,21,\ldots$$  0.01% (1) then at  0.01% (1) by starting  0.01% (1) many things  0.01% (1) we investigated  0.01% (1) camp was  0.01% (1) sequence, formed  0.01% (1) if $f_n$  0.01% (1) $(n+1)$st term  0.01% (1) the $n$th  0.01% (1) term, known  0.01% (1) as binet’s  0.01% (1) formula: $$f_n=\frac{1}{\sqrt{5}}\left(  0.01% (1) formula for  0.01% (1) a remarkable  0.01% (1) sequence (where  0.01% (1) $f_0=0$ and  0.01% (1) $f_1=1$), then  0.01% (1) the many,  0.01% (1) as well!  0.01% (1) that attended  0.01% (1) this year.  0.01% (1) the students  0.01% (1) mentioned that  0.01% (1) school students  0.01% (1) talented high  0.01% (1) was an  0.01% (1) enormous pleasure  0.01% (1) to teach  0.01% (1) the seventeen  0.01% (1) they felt  0.01% (1) inspired to  0.01% (1) went both  0.01% (1) ways –  0.01% (1) they inspired  0.01% (1) new ideas  0.01% (1) the inspiration  0.01% (1) program, but  0.01% (1) study math  0.01% (1) further after  0.01% (1) our twoweek  0.01% (1) of $\mu’$  0.01% (1) where $\mu$  0.01% (1) this partition  0.01% (1) uniquely determines  0.01% (1) the conjugacy  0.01% (1) exactly one  0.01% (1) $\mu’$, and  0.01% (1) order form  0.01% (1) diagonal. the  0.01% (1) sizes of  0.01% (1) blocks, written  0.01% (1) in nonincreasing  0.01% (1) matrices for  0.01% (1) each partition  0.01% (1) matrix varieties,  0.01% (1) and their  0.01% (1) coordinate rings  0.01% (1) were studied  0.01% (1) form closed  0.01% (1) classes $\overline{c_{\mu’}}$  0.01% (1) $\mu’$ of  0.01% (1) $n$. the  0.01% (1) closures of  0.01% (1) these conjugacy  0.01% (1) all $0$’s  0.01% (1) blocks have  0.01% (1) the halllittlewood  0.01% (1) polynomial $\widetilde{h}_\mu(x;q)$.  0.01% (1) the rings  0.01% (1) $r_\mu$ were  0.01% (1) graded frobenius  0.01% (1) fact the  0.01% (1) the trivial  0.01% (1) representation. so  0.01% (1) a generalization  0.01% (1) coinvariant ring,  0.01% (1) originally defined  0.01% (1) let $a$  0.01% (1) $0$ eigenvalues,  0.01% (1) conjugate to  0.01% (1) a matrix  0.01% (1) whose jordan  0.01% (1) has all  0.01% (1) then $a$  0.01% (1) nilpotent $n\times  0.01% (1) n$ matrix  0.01% (1) over $\mathbb{c}$.  0.01% (1) here. however,  0.01% (1) easier to  0.01% (1) $t$ dividing  0.01% (1) k$ minors  0.01% (1) of $ati$,  0.01% (1) say $t^{d_k}$,  0.01% (1) the largest  0.01% (1) wikipedia) that  0.01% (1) elementary divisors  0.01% (1) article on  0.01% (1) smith normal  0.01% (1) form on  0.01% (1) under conjugation,  0.01% (1) so we  0.01% (1) by analyzing  0.01% (1) blocks, that  0.01% (1) this power  0.01% (1) is $t^{d_k(\mu)}$  0.01% (1) to see,  0.01% (1) not hard  0.01% (1) can assume  0.01% (1) normal form.  0.01% (1) then it’s  0.01% (1) factors and  0.01% (1) of invariant  0.01% (1) $c_{\mu’}$? defining  0.01% (1) $r_\mu=\mathcal{o}(\overline{c_{\mu’}}\cap t)$,  0.01% (1) modules as  0.01% (1) above. tanisaki  0.01% (1) the nilpotent  0.01% (1) question: what  0.01% (1) after intersecting  0.01% (1) set $t$  0.01% (1) matrices, leading  0.01% (1) interesting natural  0.01% (1) found the  0.01% (1) presentation for  0.01% (1) where $a\in  0.01% (1) c_{\mu’}$. then  0.01% (1) show (see,  0.01% (1) the discussion  0.01% (1) matrix $ati$,  0.01% (1) following argument.  0.01% (1) $r_\mu$ given  0.01% (1) above using  0.01% (1) roughly the  0.01% (1) thesis can  0.01% (1) here.) pages:  0.01% (1) all $p_i$’s  0.01% (1) line. the  0.01% (1) all planes  0.01% (1) intersecting $p_i$  0.01% (1) that intersects  0.01% (1) a plane  0.01% (1) is $p_i$  0.01% (1) $i=1,2,3,4$. then  0.01% (1) condition is  0.01% (1) of finding  0.01% (1) line is  0.01% (1) $\omega_\box(f_\bullet^{(i)})$ for  0.01% (1) calculus, since  0.01% (1) the $k\times(nk)$  0.01% (1) size $4$  0.01% (1) four partitions  0.01% (1) discussed in  0.01% (1) and, as  0.01% (1) each $i$,  0.01% (1) solutions is  0.01% (1) $$\omega_\box(f_\bullet^{(1)})\cap \omega_\box(f_\bullet^{(2)})\cap  0.01% (1) \omega_\box(f_\bullet^{(3)})\cap \omega_\box(f_\bullet^{(4)}).$$  0.01% (1) subspace $f^{(i)}_2$  0.01% (1) their second  0.01% (1) f_{n+i\lambda_i}\ge i.\}$$  0.01% (1) must fit  0.01% (1) box in  0.01% (1) variety to  0.01% (1) \dim v\cap  0.01% (1) $$\omega_{\lambda}(f_\bullet)=\{v\in \mathrm{gr}^n(\mathbb{c}^m)\mid  0.01% (1) to this  0.01% (1) flag is  0.01% (1) a subvariety  0.01% (1) grassmannian $\mathrm{gr}^n(\mathbb{c}^m)$  0.01% (1) be nonempty.  0.01% (1) above, we  0.01% (1) subspaces $p_1,p_2,p_3,p_4\subset  0.01% (1) \mathbb{c}^4$, and  0.01% (1) construct complete  0.01% (1) flags $f_\bullet^{(1)},f_\bullet^{(2)},f_\bullet^{(3)},f_\bullet^{(4)}$  0.01% (1) general twodimensional  0.01% (1) with four  0.01% (1) can translate  0.01% (1) into an  0.01% (1) intersection problem  0.01% (1) involved have  0.01% (1) to $4$,  0.01% (1) normal curve,  0.01% (1) the locus  0.01% (1) form $$(1:t:t^2:t^3:\cdots:t^{n1})\in  0.01% (1) \mathbb{cp}^n$$ (along  0.01% (1) at chosen  0.01% (1) $f_\bullet$ defined  0.01% (1) $k(nk)1$. it  0.01% (1) that one  0.01% (1) can choose  0.01% (1) any flags  0.01% (1) limiting point  0.01% (1) $(0:0:\cdots:1)$.) in  0.01% (1) matrix of  0.01% (1) point on  0.01% (1) this curve  0.01% (1) parameterized by  0.01% (1) $k$ rows  0.01% (1) span of  0.01% (1) particular, consider  0.01% (1) the flag  0.01% (1) whose $k$th  0.01% (1) subspace $f_k$  0.01% (1) sum to  0.01% (1) mathematical gemstones  0.01% (1) the conditions  0.01% (1) problem, so  0.01% (1) are only  0.01% (1) intersecting three  0.01% (1) we relax  0.01% (1) happens if  0.01% (1) this intersection  0.01% (1) $0$. the  0.01% (1) then tells  0.01% (1) $c_{\box,\box,\box,\box}^{(2,2)}$. what  0.01% (1) above? in  0.01% (1) a oneparameter  0.01% (1) flags $f_\bullet^{(1)},\ldots,  0.01% (1) f_\bullet^{(r)}$ and  0.01% (1) (fitting inside  0.01% (1) (nk)$ box)  0.01% (1) “generic” choice  0.01% (1) a sufficiently  0.01% (1) solutions, which  0.01% (1) curve. to  0.01% (1) we require  0.01% (1) with respect  0.01% (1) each $f_i$  0.01% (1) know each  0.01% (1) integral, and  0.01% (1) must each  0.01% (1) be isomorphic  0.01% (1) is zero.  0.01% (1) h^1(\mathcal{o}_s)=0$, so  0.01% (1) $n$ complex  0.01% (1) one for  0.01% (1) connected components.  0.01% (1) that $\dim  0.01% (1) to $\mathbb{cp}^1$  0.01% (1) (see hartshorne,  0.01% (1) $k$theory described  0.01% (1) above. similar  0.01% (1) analyses lead  0.01% (1) to other  0.01% (1) connection with  0.01% (1) identity map,  0.01% (1) section 4.1  0.01% (1) exercise 1.8).  0.01% (1) therefore determined  0.01% (1) that $\omega$  0.01% (1) case, and  0.01% (1) h^0(\mathcal{o}_s)\dim h^1(\mathcal{o}_s)$.  0.01% (1) these moves.  0.01% (1) this means  0.01% (1) have no  0.01% (1) nonadjacent moves,  0.01% (1) simply reverses  0.01% (1) shuffle step  0.01% (1) 2 is  0.01% (1) all horizontal  0.01% (1) slides; then  0.01% (1) the final  0.01% (1) $k=0$ in  0.01% (1) this case.  0.01% (1) of complex  0.01% (1) (see here),  0.01% (1) turn is  0.01% (1) characteristic $\chi(\mathcal{o}_s)=\dim  0.01% (1) to $n$,  0.01% (1) know that  0.01% (1) since $\omega$,  0.01% (1) locus, is  0.01% (1) the identity,  0.01% (1) geometric results:  0.01% (1) also shown  0.01% (1) calculus that  0.01% (1) use schubert  0.01% (1) varieties to  0.01% (1) answer the  0.01% (1) a restriction  0.01% (1) curves: relaxing  0.01% (1) a continuation,  0.01% (1) we’ll now  0.01% (1) discuss where  0.01% (1) geometry. schubert  0.01% (1) question: given  0.01% (1) four lines  0.01% (1) i.e. a  0.01% (1) subspaces $$0=f_0\subset  0.01% (1) f_1\subset\cdots \subset  0.01% (1) f_m=\mathbb{c}^m$$ where  0.01% (1) complete flag,  0.01% (1) a point?  0.01% (1) $\ell_1,\ell_2,\ell_3,\ell_4\in \mathbb{c}\mathbb{p}^3$,  0.01% (1) lines intersect  0.01% (1) all four  0.01% (1) shuffling. as  0.01% (1) rectification and  0.01% (1) of $\alpha$,  0.01% (1) $\gamma$. so,  0.01% (1) and $k$theory  0.01% (1) common? a  0.01% (1) various values  0.01% (1) components, for  0.01% (1) have arbitrarily  0.01% (1) high genus,  0.01% (1) and can  0.01% (1) arbitrarily many  0.01% (1) little monodromy  0.01% (1) operator called  0.01% (1) discussed an  0.01% (1) operation $\newcommand{\box}{\square}  0.01% (1) marked inner  0.01% (1) corner, defined  0.01% (1) post, we  0.01% (1) march 18,  0.01% (1) $\omega$. posted  0.01% (1)  3  0.01% (1) replies what  0.01% (1) $t$: $$\left(\begin{array}{cccccc}  0.01% (1) t^3 &  0.01% (1) box. this  0.01% (1) same object  0.01% (1) were considering  0.01% (1) i. now,  0.01% (1) corner chosen  0.01% (1) $\beta$ with  0.01% (1) similarly for  0.01% (1) the antistraight  0.01% (1) shape $\gamma$,  0.01% (1) data here  0.01% (1) also label  0.01% (1) $f^{1}(1)$ by  0.01% (1) in $f^{1}(0)$  0.01% (1) to $f^{1}(1)$  0.01% (1) or similarly  0.01% (1) from $1$  0.01% (1) along an  0.01% (1) order, in  0.01% (1) $\box$, $\alpha$,  0.01% (1) and finally  0.01% (1) fiber $f^{1}(\infty)$  0.01% (1) $\beta$, $\box$,  0.01% (1) $\alpha$, and  0.01% (1) one littlewoodrichardson  0.01% (1) choose, for  0.01% (1) simplicity, the  0.01% (1) three points  0.01% (1) $0$, $1$,  0.01% (1) $$\alpha+\beta+\gamma=k(nk)1=b1.$$ let’s  0.01% (1) case, $r=3$.  0.01% (1) operators. let’s  0.01% (1) be more  0.01% (1) precise, and  0.01% (1) simplest interesting  0.01% (1) and $\infty$  0.01% (1) the respective  0.01% (1) tableaux of  0.01% (1) $\box$, $\beta$,  0.01% (1) order. note  0.01% (1) is only  0.01% (1) the fillings  0.01% (1) \mathbb{rp}^1$ by  0.01% (1) osculating flags.  0.01% (1) can label  0.01% (1) map $f:s\to  0.01% (1) or $\infty$  0.01% (1) to $0$,  0.01% (1) and hence  0.01% (1) tell us  0.01% (1) things about  0.01% (1) the lengths  0.01% (1) of $\omega$,  0.01% (1) the orbits  0.01% (1) boils down  0.01% (1) understanding of  0.01% (1) $\omega$. our  0.01% (1) a better  0.01% (1) orbits, and  0.01% (1) some cases  0.01% (1) well as  0.01% (1) some fascinating  0.01% (1) connections to  0.01% (1) grassmannian. stay  0.01% (1) consequences, as  0.01% (1) discuss some  0.01% (1) complex curve  0.01% (1) $s$. next  0.01% (1) time, i’ll  0.01% (1) real geometry  0.01% (1) so, the  0.01% (1) $1$, the  0.01% (1) corresponding operation  0.01% (1) previous post.  0.01% (1) from $\infty$  0.01% (1) passes through  0.01% (1) $\infty$ that  0.01% (1) then then  0.01% (1) the shuffling  0.01% (1) operations described  0.01% (1) $0$ to  0.01% (1) $0$ on  0.01% (1) side is  0.01% (1) the zero  0.01% (1) fiber $f^{1}(0)$.  0.01% (1) picture sums  0.01% (1) this up:  0.01% (1) all, $\omega$  0.01% (1) of $\omega$.  0.01% (1) the “shuffle”  0.01% (1) $\box$ past  0.01% (1) $\beta$, which  0.01% (1) by shuffling  0.01% (1) is described  0.01% (1) map in  0.01% (1) these curves,  0.01% (1) we consider  0.01% (1) single box,  0.01% (1) a covering  0.01% (1) varieties defines  0.01% (1) these osculating  0.01% (1) flags, and  0.01% (1) choose exactly  0.01% (1) $\lambda^{(1)},\ldots,\lambda^{(r)}$ with  0.01% (1) $\omega_\box(f_\bullet)$. in  0.01% (1) particular, after  0.01% (1) $\omega_\box(f_\bullet)$ with  0.01% (1) our schubert  0.01% (1) a zerodimensional  0.01% (1) intersection, with  0.01% (1) $\lambda=(1)$. intersecting  0.01% (1) singlebox partition  0.01% (1) choosing our  0.01% (1) $r$ osculating  0.01% (1) flags, choose  0.01% (1) an $(r+1)$st  0.01% (1) $r$ of  0.01% (1) some number  0.01% (1) 6 &  0.01% (1) \frac{(n1)!}{(n4)!} t^{n3}  0.01% (1) \\ \vdots  0.01% (1) &\ddots &  0.01% (1) & \frac{(n1)!}{(n3)!}  0.01% (1) & 6t  0.01% (1) t^{n1} \\  0.01% (1) 2t &  0.01% (1) 3t^2 &  0.01% (1) (n1) t^{n2}  0.01% (1) \vdots \\  0.01% (1) (n1)! \end{array}\right)$$  0.01% (1) general position  0.01% (1) schubert intersections  0.01% (1) the expected  0.01% (1) dimension. so,  0.01% (1) in sufficiently  0.01% (1) osculating flag,  0.01% (1) point $t$,  0.01% (1) and if  0.01% (1) choose their  0.01% (1) coefficient $c:=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$  0.01% (1) where $b=((nk)^k)$  0.01% (1) coefficient counts  0.01% (1) contents $\lambda^{(1)},\ldots,\lambda^{(r)},\box$  0.01% (1) each skew  0.01% (1) shape extends  0.01% (1) this littlewoodrichardson  0.01% (1) size $c=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$.  0.01% (1) a smooth,  0.01% (1) finite covering  0.01% (1) of $\mathbb{rp}^1$.  0.01% (1) map have  0.01% (1) shape outwards.  0.01% (1) symmetry of  0.01% (1) of labeling  0.01% (1) by these  0.01% (1) tableaux, for  0.01% (1) the covering  0.01% (1) canonical way  0.01% (1) partitions. it  0.01% (1) littlewoodrichardson coefficients,  0.01% (1) matter what  0.01% (1) order in  0.01% (1) $s(\mathbb{r})$ is  0.01% (1) map $s\to  0.01% (1) this open  0.01% (1) of $\mathbb{cp}^1$  0.01% (1) the preimage  0.01% (1) of $(1:t)$  0.01% (1) points. we  0.01% (1) of $c$  0.01% (1) box partition.  0.01% (1) by varying  0.01% (1) the choice  0.01% (1) into sets  0.01% (1) the $c$  0.01% (1) choosing the  0.01% (1) choose all  0.01% (1) $r+1$ points  0.01% (1) real, then  0.01% (1) be extended  0.01% (1) jake levinson,  0.01% (1) my coauthor  0.01% (1) $(1:t:t^2:t^3:\cdots:t^{n1})$. in  0.01% (1) this paper  0.01% (1) written by  0.01% (1) vertical slides  0.01% (1) 1 consists  0.01% (1) a brief  0.01% (1) description of  0.01% (1) how the  0.01% (1) works, and  0.01% (1) just give  0.01% (1) paper. for  0.01% (1) the basics  0.01% (1) found here,  0.01% (1) of grassmannians,  0.01% (1) see buch’s  0.01% (1) how it  0.01% (1) curves. recall  0.01% (1) partition fitting  0.01% (1) inside a  0.01% (1) rectangle, generate  0.01% (1) ring $h^\ast(\mathrm{gr}(n,k))$.  0.01% (1) where $\lambda$  0.01% (1) classes $[\omega_\lambda]$,  0.01% (1) post that  0.01% (1) the cw  0.01% (1) complex structure  0.01% (1) varieties shows  0.01% (1) exposition on  0.01% (1) a good  0.01% (1) $[m]$ for  0.01% (1) $m\in m$  0.01% (1) form $[m]+[n][m+n]$.  0.01% (1) instance, $k(\mathbb{n})=\mathbb{z}$.  0.01% (1) by elements  0.01% (1) group $k(m)$  0.01% (1) group without  0.01% (1) inverses). we  0.01% (1) can construct  0.01% (1) an associated  0.01% (1) sense we  0.01% (1) are groupifying  0.01% (1) then all  0.01% (1) sequences split  0.01% (1) $k(x)$ is  0.01% (1) this monoid.  0.01% (1) a point,  0.01% (1) is $\oplus$,  0.01% (1) monoids. the  0.01% (1) natural monoidal  0.01% (1) vector spaces  0.01% (1) similarly, the  0.01% (1) filtered ring  0.01% (1) sheaves on  0.01% (1) $x$, then  0.01% (1) $\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})=0$. indeed,  0.01% (1) sequence gives  0.01% (1) \mathcal{b}\to \mathcal{c}\to  0.01% (1) $0\to \mathcal{a}\to  0.01% (1) (additive) group  0.01% (1) homomorphism $\chi:k(x)\to  0.01% (1) \mathbb{z}$. to  0.01% (1) show this,  0.01% (1) rise to  0.01% (1) a long  0.01% (1) &h^1(\mathcal{a}) &  0.01% (1) h^1(\mathcal{b}) &  0.01% (1) h^1(\mathcal{c}) \\  0.01% (1) \to &h^2(\mathcal{a})  0.01% (1) & h^0(\mathcal{c})  0.01% (1) & h^0(\mathcal{b})  0.01% (1) in cohomology:  0.01% (1) $$ \begin{array}{cccccc}  0.01% (1) & h^0(\mathcal{a})  0.01% (1) gives an  0.01% (1) characteristics. it  0.01% (1) variant of  0.01% (1) rule: $$[\mathcal{o}_\lambda]\cdot  0.01% (1) [\mathcal{o}_\nu]=\sum_\nu (1)^{\nu\lambda\mu}c^\nu_{\lambda\mu}[\mathcal{o}_\nu]$$  0.01% (1) $\nu=\lambda+\mu$ then  0.01% (1) basic classes  0.01% (1) \mathcal{o}_{\omega_\lambda}$. multiplication  0.01% (1) $[\mathcal{o}_{\lambda}]$ where  0.01% (1) if $\iota$  0.01% (1) $\iota:\omega_\lambda\to \mathrm{gr}(n,k)$,  0.01% (1) then $\mathcal{o}_\lambda=\iota_\ast  0.01% (1) littlewoodrichardson coefficient.  0.01% (1) if $\nu\lambda+\mu$  0.01% (1) characteristic the  0.01% (1) is especially  0.01% (1) useful in  0.01% (1) computing euler  0.01% (1) $k$theory and  0.01% (1) values as  0.01% (1) nonnegative integer.  0.01% (1) refer to  0.01% (1) these nonnegative  0.01% (1) like a  0.01% (1) identity element,  0.01% (1) $$\chi(\mathcal{o}_s)=\dim h^0(\mathcal{o}_s)\dim  0.01% (1) h^1(\mathcal{o}_s)$$ is  0.01% (1) characteristic and  0.01% (1) $\mathcal{o}_s$ is  0.01% (1) $g=1\chi(\mathcal{o}_s)$ where  0.01% (1) a connected  0.01% (1) $s$ have?  0.01% (1) its genus?  0.01% (1) is $s$  0.01% (1) (complex) curve?  0.01% (1) sheaf. so,  0.01% (1) characteristic, which  0.01% (1) by isomorphism  0.01% (1) (a.k.a. vector  0.01% (1) bundles) on  0.01% (1) $x$. then  0.01% (1) $g$ generated  0.01% (1) the free  0.01% (1) can alternatively  0.01% (1) ring $k(\mathrm{gr}(n,k))$  0.01% (1) scheme $x$  0.01% (1) first, consider  0.01% (1) components does  0.01% (1) (complex) connected  0.01% (1) expository series  0.01% (1) recent paper  0.01% (1) coauthored by  0.01% (1) and myself.  0.01% (1) final post  0.01% (1) third and  0.01% (1) 3 posted  0.01% (1) in amber,  0.01% (1) april 5,  0.01% (1) 3 this  0.01% (1) last time,  0.01% (1) on certain  0.01% (1) $\omega$ can  0.01% (1) to approach  0.01% (1) some natural  0.01% (1) questions about  0.01% (1) improved algorithm  0.01% (1) we’ll show  0.01% (1) is actually  0.01% (1) operator of  0.01% (1) $\mathbb{rp}^1$. now,  0.01% (1) define $k(x)$,  0.01% (1) group, to  0.01% (1) modulo short  0.01% (1) exact sequences.  0.01% (1) where does  0.01% (1) this construction  0.01% (1) free” restriction  0.01% (1) the “locally  0.01% (1) a grassmannian!)  0.01% (1) exact same  0.01% (1) ring if  0.01% (1) we remove  0.01% (1) well, a  0.01% (1) simpler example  0.01% (1) a monoid  0.01% (1) set with  0.01% (1) an associative  0.01% (1) binary operation  0.01% (1) (recall that  0.01% (1) monoid m  0.01% (1) construction of  0.01% (1) group of  0.01% (1) monoid. consider  0.01% (1) (such as  0.01% (1) is smooth  0.01% (1) form $[\mathcal{e}_1]+[\mathcal{e}_2][\mathcal{e}]$  0.01% (1) where $0\to  0.01% (1) \mathcal{e}_1 \to  0.01% (1) \mathcal{e} \to  0.01% (1) where $h$  0.01% (1) quotient $g/h$  0.01% (1) quotient of  0.01% (1) $g$ by  0.01% (1) “short exact  0.01% (1) sequences”, that  0.01% (1) \mathcal{e}_2 \to  0.01% (1) of vector  0.01% (1) tensor product  0.01% (1) makes it  0.01% (1) ring. it  0.01% (1) that $x$  0.01% (1) on $k(x)$,  0.01% (1) additive structure  0.01% (1) bundles on  0.01% (1) $x$. this  0.01% (1) gives the  0.01% (1) & h^2(\mathcal{b})  0.01% (1) & h^2(\mathcal{c})  0.01% (1) marked (shaded)  0.01% (1) squares: property  0.01% (1) 1 means  0.01% (1) fourth tableau  0.01% (1) with two  0.01% (1) following tableaux  0.01% (1) (has more  0.01% (1) $j$’s than  0.01% (1) $j+1$’s for  0.01% (1) all $j$).  0.01% (1) genomic: the  0.01% (1) marked squares,  0.01% (1) finally, the  0.01% (1) second tableau  0.01% (1) property 3,  0.01% (1) marked $1$  0.01% (1) $2$’s in  0.01% (1) 2, because  0.01% (1) while they  0.01% (1) do contain  0.01% (1) entry, are  0.01% (1) adjacent squares.  0.01% (1) $\square_2$, every  0.01% (1) $\square_1$ or  0.01% (1) yong. in  0.01% (1) our case,  0.01% (1) the genomic  0.01% (1) enumerate $k$  0.01% (1) pechenik and  0.01% (1) introduced by  0.01% (1) $s$. these  0.01% (1) called “genomic  0.01% (1) tableaux”, and  0.01% (1) were first  0.01% (1) follows. the  0.01% (1) data of  0.01% (1) entry $i$,  0.01% (1) no $i$’s  0.01% (1) between $\square_1$  0.01% (1) delete either  0.01% (1) and contain  0.01% (1) are nonadjacent  0.01% (1) $t$ and  0.01% (1) squares $\square_1$  0.01% (1) if: the  0.01% (1) suffix $221$  0.01% (1) ballot. the  0.01% (1) the local  0.01% (1) $\omega$. geometric  0.01% (1) consequences this  0.01% (1) connection allows  0.01% (1) 1’s of  0.01% (1) all phase  0.01% (1) these the  0.01% (1) genomic tableau.  0.01% (1) of nonadjacent  0.01% (1) moves in  0.01% (1) $s$ using  0.01% (1) our new  0.01% (1) only way  0.01% (1) to map  0.01% (1) itself is  0.01% (1) if phase  0.01% (1) permutation. it  0.01% (1) case when  0.01% (1) algorithm. as  0.01% (1) one illuminating  0.01% (1) example, let’s  0.01% (1) and making  0.01% (1) moved past,  0.01% (1) i discovered  0.01% (1) not move  0.01% (1) bijective correspondence  0.01% (1) $k$theory tableau  0.01% (1) finally, consider  0.01% (1) is genomic.  0.01% (1) third tableau  0.01% (1) above satisfies  0.01% (1) all three  0.01% (1) properties, and  0.01% (1) total skew  0.01% (1) shape $\gamma^c/\alpha$  0.01% (1) is formed  0.01% (1) by simply  0.01% (1) starting and  0.01% (1) ending positions  0.01% (1) the correspondence  0.01% (1) of $i$’s).  0.01% (1) $\beta$ (where  0.01% (1) squares only  0.01% (1) add $1$  0.01% (1) us information  0.01% (1) coefficients, giving  0.01% (1) $\gamma$ (see  0.01% (1) part i).  0.01% (1) ring, if  0.01% (1) we identify  0.01% (1) indexed by  0.01% (1) indeed, $s$  0.01% (1) $j_\ast\mathcal{o}_s$ is  0.01% (1) $\chi(\mathcal{o}_s)$ itself.  0.01% (1) now compute  0.01% (1) characteristic $\chi(j_\ast\mathcal{o}_s)$  0.01% (1) structure sheaves  0.01% (1) of closed  0.01% (1) $k$theoretic littlewoodrichardson  0.01% (1) rule described  0.01% (1) above, this  0.01% (1) product expands  0.01% (1) [\mathcal{o}_\beta]\cdot [\mathcal{o}_\gamma].$$  0.01% (1) have $$[\mathcal{o}_s]=[\mathcal{o}_\alpha]\cdot  0.01% (1) subvarieties with  0.01% (1) their pushforwards  0.01% (1) under inclusion  0.01% (1) maps, we  0.01% (1) the pushforward  0.01% (1) $j:s\to x$,  0.01% (1) h^i(\mathcal{a})\sum_i (1)^i\dim  0.01% (1) h^i(\mathcal{b})+\sum_i (1)^i\dim  0.01% (1) h^i(\mathcal{c}) \\  0.01% (1) &=&\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})\end{eqnarray*}$$ as  0.01% (1) $$\begin{eqnarray*}0&=&\sum_i (1)^i\dim  0.01% (1) zero. thus  0.01% (1) \\ \cdots  0.01% (1) & \end{array}  0.01% (1) the alternating  0.01% (1) the dimensions  0.01% (1) desired. therefore,  0.01% (1) it makes  0.01% (1) our situation,  0.01% (1) a closed  0.01% (1) of $x=\mathrm{gr}(n,k)$,  0.01% (1) say with  0.01% (1) fact, in  0.01% (1) $k(x)$. in  0.01% (1) sense to  0.01% (1) a class  0.01% (1) sheaves in  0.01% (1) integer multiples  0.01% (1) of classes  0.01% (1) $\mathbb{p}^1$ respectively,  0.01% (1) both have  0.01% (1) $1$. it  0.01% (1) $$\chi(\mathcal{o}_s)=nk.$$ going  0.01% (1) copy of  0.01% (1) $\omega_{\rho’}$ are  0.01% (1) let’s call  0.01% (1) this number  0.01% (1) $n$. finally,  0.01% (1) $\omega_\rho$ and  0.01% (1) the genus,  0.01% (1) connected, we  0.01% (1) certain steps  0.01% (1) algorithm combinatorially  0.01% (1) correspond to  0.01% (1) certain tableaux  0.01% (1) our algorithm  0.01% (1) thing about  0.01% (1) have $g=1\chi(\mathcal{o}_s)=k(n1)$.  0.01% (1) computing $k$  0.01% (1) the fascinating  0.01% (1) part ii.  0.01% (1) acts on)  0.01% (1) entire $k\times  0.01% (1) $\rho$) or  0.01% (1) the rectangle  0.01% (1) minus a  0.01% (1) either the  0.01% (1) $\nu$ is  0.01% (1) $[\mathcal{o}_\nu]$ where  0.01% (1) $\nu\ge \alpha+\beta+\gamma$.  0.01% (1) our setup  0.01% (1) $\alpha+\beta+\gamma=k(nk)1$, so  0.01% (1) $\rho’$). in  0.01% (1) other words,  0.01% (1) that $c^{\rho’}_{\alpha,\beta,\gamma}$  0.01% (1) counts exactly  0.01% (1) fibers (the  0.01% (1) set $\omega$  0.01% (1) coefficients. notice  0.01% (1) determined by  0.01% (1) we have:  0.01% (1) $$[\mathcal{o}_s]=c^{\rho’}_{\alpha,\beta,\gamma}[\mathcal{o}_{\rho’}]k[\mathcal{o}_{\rho}]$$ where  0.01% (1) an integer  0.01% (1) on january  0.01% (1)  the number of  0.23% (21) the euler characteristic  0.11% (10) by maria gillespie  0.11% (10) jeu de taquin  0.1% (9) it turns out  0.09% (8) the special box  0.09% (8) elementary symmetric function  0.09% (8) turns out that  0.09% (8) there is a  0.09% (8) short exact sequence  0.07% (6) & \to &  0.07% (6) curves, young tableaux,  0.07% (6)  leave a  0.07% (6) can you prove  0.07% (6) do schubert curves,  0.07% (6) young tableaux, and  0.07% (6) a $q$analog of  0.07% (6) in the case  0.07% (6) leave a reply  0.07% (6) maria gillespie reply  0.07% (6) have in common?  0.07% (6) schubert curves, young  0.07% (6) you prove it…  0.07% (6) what do schubert  0.07% (6) the set of  0.07% (6) 2016 by maria  0.07% (6) of the schubert  0.05% (5) and so the  0.05% (5) in terms of  0.05% (5) the schubert curve  0.05% (5) in reading order  0.05% (5) one of the  0.05% (5) given by the  0.05% (5) to get a  0.05% (5) & 0 &  0.05% (5) it follows that  0.05% (5) & \cdots &  0.05% (5) elementary symmetric functions  0.05% (5) pages: 1 2  0.05% (5) defined as follows.  0.05% (5) is a $q$analog  0.05% (5) ktheory have in  0.05% (5) tableaux, and ktheory  0.05% (5) in common? (part  0.05% (5) and ktheory have  0.05% (5) \\ 0 &  0.04% (4) each of the  0.04% (4) partial elementary symmetric  0.04% (4) the $k$theory of  0.04% (4) that we can  0.04% (4) $q$analog of $5$  0.04% (4) a statement or  0.04% (4) 2 posted in  0.04% (4) counts the number  0.04% (4) number of ways  0.04% (4) & \square &  0.04% (4) schubert curve $s$  0.04% (4) in the last  0.04% (4) a handle on  0.04% (4) out that the  0.04% (4) $\alpha$, $\beta$, and  0.04% (4) structure of the  0.04% (4) gives us more  0.04% (4) for instance, the  0.04% (4) subsets $s$ of  0.04% (4) is equal to  0.04% (4) handle on the  0.04% (4) allows us to  0.04% (4) the structure of  0.04% (4) is given by  0.04% (4) 0 & 0  0.04% (4) in opal   0.03% (3) can be defined  0.03% (3) posted on september  0.03% (3) can be found  0.03% (3) $s$ of size  0.03% (3) the intersection of  0.03% (3) which is a  0.03% (3) right hand side  0.03% (3) the $k$theory coefficient  0.03% (3) what is a  0.03% (3) short exact sequences  0.03% (3) is the identity  0.03% (3) $q$analog of the  0.03% (3) real connected components  0.03% (3) of ways of  0.03% (3) in order for  0.03% (3) taking the limit  0.03% (3) of the number  0.03% (3) variety of a  0.03% (3) we have the  0.03% (3) in this post  0.03% (3) the total number  0.03% (3) the garsiaprocesi modules  0.03% (3) number of inversions  0.03% (3) the limit as  0.03% (3) box with the  0.03% (3) list all the  0.03% (3) gives us the  0.03% (3) there is an  0.03% (3) euler characteristic of  0.03% (3) $\gamma$ in that  0.03% (3) in the next  0.03% (3) & 1 &  0.03% (3) & 2 &  0.03% (3) a number of  0.03% (3) algorithm for $\omega$  0.03% (3) such a way  0.03% (3) we can define  0.03% (3) of permutations of  0.03% (3) steps of the  0.03% (3) is the usual  0.03% (3) the closure of  0.03% (3) us more refined  0.03% (3) think of the  0.03% (3) get a handle  0.03% (3) a short exact  0.03% (3) $k$theory of the  0.03% (3) be defined as  0.03% (3) nilpotent conjugacy class  0.03% (3) number of points  0.03% (3) defined as a  0.03% (3) refined information about  0.03% (3) the case of  0.03% (3) all the rationals  0.03% (3) and $\gamma$ in  0.03% (3) for which the  0.03% (3) is the number  0.03% (3) ways of choosing  0.03% (3) that the number  0.03% (3) a way that  0.03% (3) the geometry of  0.03% (3) a sequence of  0.03% (3) & \square \\  0.03% (3) \square & \blacksquare  0.03% (3) a set of  0.03% (3) gillespie reply in  0.03% (3) from a set  0.03% (3) the real locus  0.03% (3) the littlewoodrichardson rule  0.03% (3) of a partition  0.03% (3) posted in diamond  0.03% (3) the case that  0.03% (3) to be a  0.03% (3) and onto correspondence  0.03% (3) \square & \square  0.03% (3) of $5$, because  0.02% (2) if we plug  0.02% (2) by taking the  0.02% (2) $k$theory ring $k(x)$  0.02% (2) basic fact that  0.02% (2) to get the  0.02% (2) we plug in  0.02% (2) label the fiber  0.02% (2) terms of the  0.02% (2) it was a  0.02% (2) fibers is given  0.02% (2) stay tuned! posted  0.02% (2) in the permutation  0.02% (2) the next entry  0.02% (2) contributes a factor  0.02% (2) that if we  0.02% (2) number of loops  0.02% (2) \mathbb{cp}^1$ for which  0.02% (2) sum of the  0.02% (2) is the product  0.02% (2) chains of contents  0.02% (2) we follow the  0.02% (2) expression $p$ is  0.02% (2) is one of  0.02% (2) maria gillespie 4  0.02% (2) the arithmetic genus  0.02% (2) the last post!  0.02% (2) that setting $q=1$  0.02% (2) field, $p_q$ gives  0.02% (2) \\ \\ \square  0.02% (2) consists of a  0.02% (2) one of those  0.02% (2) the integers and  0.02% (2) was stated in  0.02% (2) partitions $\alpha$, $\beta$,  0.02% (2) \\ \blacksquare &  0.02% (2) \\ \\ \blacksquare  0.02% (2) to the next  0.02% (2) to show that  0.02% (2) opal  4  0.02% (2) the monodromy of  0.02% (2) skew littlewoodrichardson tableau  0.02% (2) left hand side  0.02% (2) method one can  0.02% (2) a certain set  0.02% (2) can also be  0.02% (2) points in the  0.02% (2) of the rationals,  0.02% (2) at each step,  0.02% (2) it! math academy  0.02% (2) this year’s prove  0.02% (2) can show that  0.02% (2) that there is  0.02% (2) proofs of the  0.02% (2) the binomial theorem  0.02% (2) of the resulting  0.02% (2) the right hand  0.02% (2) we use the  0.02% (2) in $p$, $p_q$  0.02% (2) $q\to 1$ results  0.02% (2) can be expressed  0.02% (2) with a simple  0.02% (2) (possibly infinite) sums  0.02% (2) $q=1$ or taking  0.02% (2) of length $k$  0.02% (2) or expression $p$  0.02% (2) of a statement  0.02% (2) is a statement  0.02% (2) or expression $p_q$  0.02% (2) depending on $q$  0.02% (2) or products of  0.02% (2) rational functions of  0.02% (2) powers of $q$  0.02% (2) the basic fact  0.02% (2) be factored into  0.02% (2) properties 3 and  0.02% (2) by the chains  0.02% (2) up with a  0.02% (2) interesting $q$analog of  0.02% (2) the theory of  0.02% (2) $q$ over some  0.02% (2) information about something  0.02% (2) that $p$ describes,  0.02% (2) and $p_q$ has  0.02% (2) an interesting $q$analog  0.02% (2) expressed in terms  0.02% (2) the schubert curves  0.02% (2) $\times$ as having  0.02% (2) us to get  0.02% (2) treat the $\times$  0.02% (2) square to perform  0.02% (2) as the empty  0.02% (2) operator on the  0.02% (2) an upper bound  0.02% (2) in order to  0.02% (2) 1$. for instance,  0.02% (2) at least as  0.02% (2) bound on the  0.02% (2) is an upper  0.02% (2) de taquin slide.  0.02% (2) in other words:  0.02% (2) contain the same  0.02% (2) tableau above violates  0.02% (2) algorithm for computing  0.02% (2) special box does  0.02% (2) of the $i$’s  0.02% (2) reading word is  0.02% (2) in reading order.  0.02% (2) the study of  0.02% (2) $t$ on the  0.02% (2) treat $\times$ as  0.02% (2) the sequence of  0.02% (2) box does not  0.02% (2) of a certain  0.02% (2) $i$’s as $i+1$’s  0.02% (2) least as many  0.02% (2) a skew shape  0.02% (2) schubert curve in  0.02% (2) of the complex  0.02% (2) that we are  0.02% (2) sizes summing to  0.02% (2) semistandard young tableau  0.02% (2) the $k\times (nk)$  0.02% (2) reply what do  0.02% (2) at the point  0.02% (2) \cdots & (n1)  0.02% (2) on the rational  0.02% (2) \vdots & \vdots  0.02% (2) into the empty  0.02% (2) $r$ partitions $\lambda^{(1)},\ldots,\lambda^{(r)}$  0.02% (2) order for the  0.02% (2) the resulting schubert  0.02% (2) ii) posted on  0.02% (2) $\omega$ is the  0.02% (2) of the reading  0.02% (2) these components is  0.02% (2) the question in  0.02% (2) commutator of rectification  0.02% (2) schubert variety of  0.02% (2) a partition $\lambda$  0.02% (2) a chain of  0.02% (2) out that, in  0.02% (2) our post on  0.02% (2) in reading order,  0.02% (2) if we delete  0.02% (2) $p_q$ can be  0.02% (2) results in $p$,  0.02% (2) jake levinson and  0.02% (2) of (possibly infinite)  0.02% (2) sums or products  0.02% (2) free abelian group  0.02% (2) as $q\to 1$  0.02% (2) setting $q=1$ or  0.02% (2) the schubert varieties  0.02% (2) a $k\times (nk)$  0.02% (2) on the euler  0.02% (2) a combinatorial $q$analog  0.02% (2) the first three  0.02% (2) of rational functions  0.02% (2) of $q$ over  0.02% (2) turns out to  0.02% (2) turns out that,  0.02% (2) consists of the  0.02% (2) \to 0$ is  0.02% (2) subgroup generated by  0.02% (2) expressions of the  0.02% (2) we get the  0.02% (2) an abelian monoid  0.02% (2) open subset of  0.02% (2) some field, $p_q$  0.02% (2) something that $p$  0.02% (2) describes, and $p_q$  0.02% (2) has $p$like properties.  0.02% (2) subset of the  0.02% (2) 1 & q\\  0.02% (2) and phase 2  0.02% (2) tableaux that enumerate  0.02% (2) back to the  0.02% (2) all for now!  0.02% (2) reply what is  0.02% (2) arises in geometry  0.02% (2) rational normal curve  0.02% (2) with the nearest  0.02% (2) $\square_1$ and $\square_2$  0.02% (2) switch the special  0.02% (2) the same entry  0.02% (2) increment $i$ by  0.02% (2) $1$ and repeat  0.02% (2) a $q$analog? (part  0.02% (2) this series of  0.02% (2) the last step  0.02% (2) of the permutation  0.02% (2) along with the  0.02% (2) exact sequence of  0.02% (2) is a short  0.02% (2) in the sense  0.02% (2) monodromy operator on  0.02% (2) $k\times (nk)$ rectangle  0.02% (2) a single box  0.02% (2) and choose the  0.02% (2) intersecting the resulting  0.02% (2) of coherent sheaves  0.02% (2) that is, the  0.02% (2) genus of $s$  0.02% (2) recall that the  0.02% (2) opal  leave  0.02% (2) navigation ← older  0.02% (2) to prove that  0.02% (2) is called a  0.02% (2) $\mathbb{z}$ denotes the  0.02% (2) the transpose partition  0.02% (2) and so if  0.02% (2) posts the structure  0.02% (2) of the garsiaprocesi  0.02% (2) of my favorite  0.02% (2) prove it! math  0.02% (2) combinatorially?maria gillespie on  0.02% (2) gillespie on can  0.02% (2) of the springer  0.02% (2) in this post,  0.02% (2) common? (part iii)  0.02% (2) the jordan blocks,  0.02% (2) in jordan normal  0.02% (2) the partial elementary  0.02% (2) and we have  0.02% (2) we must have  0.02% (2) a way of  0.02% (2) consecutive quotients give  0.02% (2) so it is  0.02% (2) this gives us  0.02% (2) such that the  0.02% (2) be written as  0.02% (2) rationals exactly once  0.02% (2) rational number appears  0.02% (2) power of $t$  0.02% (2) pair of entries  0.02% (2) the young diagram  0.02% (2) of diagonal matrices  0.02% (2) posted in amber  0.02% (2) so there are  0.02% (2) it… combinatorially?maria gillespie  0.02% (2) since $i_\mu$ is  0.02% (2) ← older posts  0.02% (2) a reverse jeu  0.01% (1) de taquin slide,  0.01% (1) to move the  0.01% (1) reverse. shuffling back:  0.01% (1) only one littlewoodrichardson  0.01% (1) moves from the  0.01% (1) rectification step in  0.01% (1) $\times$ back to  0.01% (1) an inner corner.  0.01% (1) quotients give all  0.01% (1) of a littlewoodrichardson  0.01% (1) tableau $t$ with  0.01% (1) all pairs $(\times,t)$  0.01% (1) a permutation on  0.01% (1) we can iterate  0.01% (1) $\omega$ to get  0.01% (1) (allowing the consecutive  0.01% (1) and unrectify, using  0.01% (1) an inward jeu  0.01% (1)  6 replies  0.01% (1) tableau of straight  0.01% (1) the rmajor index  0.01% (1) 6, 2015 by  0.01% (1) rectify the entire  0.01% (1) skew tableau. shuffling:  0.01% (1) miscellaneous (6) meta  0.01% (1) than one). thoughts?  0.01% (1) the resulting empty  0.01% (1) common factors greater  0.01% (1) integers to have  0.01% (1) having value $\infty$  0.01% (1) of $\times$. unrectification:  0.01% (1) the new location  0.01% (1) square on the  0.01% (1) outer corner is  0.01% (1) log in entries  0.01% (1) a special box  0.01% (1) $\alpha$, $\box$, $\beta$,  0.01% (1) new, more efficient  0.01% (1) every integer occurs  0.01% (1) paper provides a  0.01% (1) results in our  0.01% (1) of the main  0.01% (1) rss comments rss  0.01% (1) $\omega(t)$. in particular,  0.01% (1) algorithm are what  0.01% (1) as follows: phase  0.01% (1) 1. if the  0.01% (1) such that (a)  0.01% (1) evacuation shuffling is  0.01% (1) local rule for  0.01% (1) we call the  0.01% (1) “evacuationshuffle”, and our  0.01% (1) i digress. one  0.01% (1) schubert curves. but  0.01% (1) $\lambda/\mu$ for some  0.01% (1) fixed $\lambda$ and  0.01% (1) $\mu$. as we’ll  0.01% (1) that order. note  0.01% (1) with total shape  0.01% (1) marked on a  0.01% (1) chosen inner corner,  0.01% (1) discuss in the  0.01% (1) next post, this  0.01% (1) $\mathbb{rp}^1$ arising from  0.01% (1) the sequence, and  0.01% (1) exactly once in  0.01% (1) covering space of  0.01% (1) (b) the consecutive  0.01% (1) permutation is related  0.01% (1) to the monodromy  0.01% (1) value $0$ and  0.01% (1) and this one  0.01% (1) the littlewoodrichardson tableau  0.01% (1) below has reading  0.01% (1) word 352344123322111, and  0.01% (1) are simply two  0.01% (1) different instances of  0.01% (1) an entire parameterized  0.01% (1) for each $i\ge  0.01% (1) the suffix 123322111,  0.01% (1) for instance, has  0.01% (1) $2$’s, $2$’s as  0.01% (1) $3$’s, etc. littlewoodrichardson  0.01% (1) tableaux are key  0.01% (1) many $1$’s as  0.01% (1) an even more  0.01% (1) these statistics: they  0.01% (1) unifying picture behind  0.01% (1) family of statistics,  0.01% (1) called $r\text{maj}$, all  0.01% (1) for the antistraight  0.01% (1) row, to form  0.01% (1) the reading word.  0.01% (1) right within a  0.01% (1) equidistributed! rawlings defined  0.01% (1) and left to  0.01% (1) an $r$inversion of  0.01% (1) then the tableau  0.01% (1) is littlewoodrichardson if  0.01% (1) word is ballot,  0.01% (1) which means that  0.01% (1) it has at  0.01% (1) of which are  0.01% (1) reaches the end)  0.01% (1) every suffix (i.e.  0.01% (1) consecutive subword that  0.01% (1) fact, there is  0.01% (1) to the littlewoodrichardson  0.01% (1) then we define  0.01% (1) different proofs of  0.01% (1) $\omega(t)$ to be  0.01% (1) the “special box”.  0.01% (1) this extra square  0.01% (1) inside marked with  0.01% (1) an “$\times$”. call  0.01% (1) words, and two  0.01% (1) the result of  0.01% (1) the inv and  0.01% (1) too, we discussed  0.01% (1) shape and content  0.01% (1) maj statistics on  0.01% (1) $t$. rectification: treat  0.01% (1) the following four  0.01% (1) operations applied to  0.01% (1) corners adjacent to  0.01% (1) tableau $t$, with  0.01% (1) their equidistribution. in  0.01% (1) schur functions. a  0.01% (1) convoluted commutator the  0.01% (1) compute products of  0.01% (1) us to efficiently  0.01% (1) $\alpha$, and similarly  0.01% (1) rule, which allows  0.01% (1) operation that jake  0.01% (1) and i studied  0.01% (1) the process is  0.01% (1) as follows. start  0.01% (1) with a littlewoodrichardson  0.01% (1) operation called “shuffling”.  0.01% (1) rectification with another  0.01% (1) is a sort  0.01% (1) of commutator of  0.01% (1) not precede all  0.01% (1) listing the integers  0.01% (1) say you’re trying  0.01% (1) to count permutations  0.01% (1) number exactly once.  0.01% (1) powers of $q$.  0.01% (1) are given by  0.01% (1) of weighted objects,  0.01% (1) where the weights  0.01% (1) of $1,\ldots,n$, that  0.01% (1) is, ways of  0.01% (1) every positive rational  0.01% (1) choose the first  0.01% (1) number, and once  0.01% (1) $n$ ways to  0.01% (1) row. there are  0.01% (1) rearranging the numbers  0.01% (1) $1,\ldots,n$ in a  0.01% (1) “$q$counting” a set  0.01% (1) $q$analog of as  0.01% (1) come up is  0.01% (1) in combinatorics. in  0.01% (1) this context, $q$  0.01% (1) in which $q$analogs  0.01% (1) an incredible property:  0.01% (1) counting by $q$’s  0.01% (1) another important area  0.01% (1) is a formal  0.01% (1) variable, and the  0.01% (1) light than usual  0.01% (1) generating functions. we  0.01% (1) it also has  0.01% (1) in a different  0.01% (1) $q$, but viewed  0.01% (1) $q$analog is a  0.01% (1) generating function in  0.01% (1) we choose that  0.01% (1) out to contain  0.01% (1) tree, assign it  0.01% (1) of how many  0.01% (1) pairs of numbers  0.01% (1) up” they are,  0.01% (1) to how “mixed  0.01% (1) say we weight  0.01% (1) the permutations according  0.01% (1) are out of  0.01% (1) order. an inversion  0.01% (1) $a/b$ in the  0.01% (1) number is to  0.01% (1) the left of  0.01% (1) which the bigger  0.01% (1) the permutation in  0.01% (1) is a pair  0.01% (1) of entries in  0.01% (1) $312$, $321$. now,  0.01% (1) $132$, $213$, $231$,  0.01% (1) the second, then  0.01% (1) $n2$ for the  0.01% (1) third, and so  0.01% (1) remaining choices for  0.01% (1) and $(a+b)/b$ respectively.  0.01% (1) there are $n1$  0.01% (1) this tree turns  0.01% (1) on. so there  0.01% (1) are $n!=n\cdot (n1)\cdot  0.01% (1) the left and  0.01% (1) $3!=6$ permutations of  0.01% (1) $1,2,3$ are $123$,  0.01% (1) rearrange the entries.  0.01% (1) 1$ ways to  0.01% (1) \cdots \cdot 2\cdot  0.01% (1) right children $a/(a+b)$  0.01% (1) posts on $q$analogs.  0.01% (1) if we read  0.01% (1) $i$ after it  0.01% (1) next entry, giving  0.01% (1) whose suffix has  0.01% (1) us a sequence  0.01% (1) than $i+1$’s, switch  0.01% (1) word starting at  0.01% (1) has more $i$’s  0.01% (1) the same number  0.01% (1) of $i$’s as  0.01% (1) than any entry  0.01% (1) of $t$. so,  0.01% (1) to get $\omega(t)$,  0.01% (1) $i$ is larger  0.01% (1) numerator of the  0.01% (1) $i+1$’s. either way,  0.01% (1) this step until  0.01% (1) if the suffix  0.01% (1) 2. phase 2.  0.01% (1) of the positive  0.01% (1) us a list  0.01% (1) nearest $i$ prior  0.01% (1) rationals. this makes  0.01% (1) me wonder is  0.01% (1) tableaux of contents  0.01% (1) whether there is  0.01% (1) to it in  0.01% (1) chain of littlewoodrichardson  0.01% (1) instead, the special  0.01% (1) box precedes all  0.01% (1) go to phase  0.01% (1) this step. if,  0.01% (1) of numerators/denominators: $$1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,1,\ldots$$  0.01% (1) reading order. then  0.01% (1) box with a  0.01% (1) we first follow  0.01% (1) the phase 1  0.01% (1) algorithm is exactly  0.01% (1) what we need  0.01% (1) to understand it.  0.01% (1) and why this  0.01% (1) from left to  0.01% (1) the story: where  0.01% (1) this operator $\omega$  0.01% (1) posted in pearl  0.01% (1) the tree successively  0.01% (1) a continuation of  0.01% (1) part 1 of  0.01% (1) the entries of  0.01% (1) reply this is  0.01% (1) december 10, 2015  0.01% (1) each row of  0.01% (1) 2) posted on  0.01% (1) the beginning of  0.01% (1) post i’ll discuss  0.01% (1) slide. we can  0.01% (1) then iterate $\omega$,  0.01% (1) will match the  0.01% (1) special box back  0.01% (1) we slide the  0.01% (1) wordpress.org proudly powered  0.01% (1) steps, and then  0.01% (1) and compute an  0.01% (1) entire $\omega$orbit, a  0.01% (1) shown below. that’s  0.01% (1) of each entry  0.01% (1) right, the denominator  0.01% (1) of this is  0.01% (1) fillings of the  0.01% (1) cycle of its  0.01% (1) permutation. an example  0.01% (1) bottom to top,  0.01% (1) the rows from  0.01% (1) (part i) posted  0.01% (1) on january 18,  0.01% (1) the same as  0.01% (1) them having $s\text{maj}$  0.01% (1) value $k$ for  0.01% (1) grassmannian. stay tuned!  0.01% (1) any $k$. more  0.01% (1) value $k$ is  0.01% (1) $\{1,2,\ldots,n\}$ having $r\text{maj}$  0.01% (1) the same distribution:  0.01% (1) myself, we discovered  0.01% (1) box. this is  0.01% (1) for any $r,s\ge  0.01% (1) 1$, the number  0.01% (1) a recent and  0.01% (1) fantastic collaboration between  0.01% (1) connections to the  0.01% (1) succinctly, $$\sum_{\pi\in s_n}  0.01% (1) us things about  0.01% (1) real connected components,  0.01% (1) the lengths of  0.01% (1) a colleague of  0.01% (1) and hence tell  0.01% (1) mine mentioned this  0.01% (1) orbits of $\omega$,  0.01% (1) these orbits, and  0.01% (1) even in some  0.01% (1) some of these  0.01% (1) consequences, as well  0.01% (1) as some fascinating  0.01% (1) time, i’ll discuss  0.01% (1) curve $s$. next  0.01% (1) cases the geometry  0.01% (1) q^{r\text{maj}(\pi)}=\sum_{\pi\in s_n} q^{s\text{maj}(\pi)}.$$  0.01% (1) new links between  0.01% (1) statistics all have  0.01% (1) squares and one  0.01% (1) square missing. a  0.01% (1) move consisted of  0.01% (1) with 15 physical  0.01% (1) grid of squares  0.01% (1) the “15 puzzle”,  0.01% (1) a $4\times 4$  0.01% (1) is the inv  0.01% (1) sliding a square  0.01% (1) the usual major  0.01% (1) de taquin”, which  0.01% (1) translates to “the  0.01% (1) game is “jeu  0.01% (1) name for this  0.01% (1) index, and $n\text{maj}$  0.01% (1) square. the french  0.01% (1) up may remember  0.01% (1) puzzle toys growing  0.01% (1) together into a  0.01% (1) beautiful mathematical story,  0.01% (1) a story filled  0.01% (1) we’ve weaved them  0.01% (1) is that these  0.01% (1) several different geometric  0.01% (1) and combinatorial constructions.  0.01% (1) with drama and  0.01% (1) intrigue. so let’s  0.01% (1) of you who  0.01% (1) be the marked  0.01% (1) played with little  0.01% (1) for mathematicians those  0.01% (1) middle. slider puzzles  0.01% (1) statistic! rawling’s result  0.01% (1) start in the  0.01% (1) result and challenged  0.01% (1) get a better  0.01% (1) in particular if  0.01% (1) the fiber $f^{1}(1)$  0.01% (1) arc from $0$  0.01% (1) ii) recent commentsmaria  0.01% (1) operations described in  0.01% (1) map between the  0.01% (1) by the shuffling  0.01% (1) to $\infty$ that  0.01% (1) passes through $1$,  0.01% (1) modules $r_\mu$ can  0.01% (1) i. now, we  0.01% (1) the “evacuationshuffle”, or  0.01% (1) on the fibers  0.01% (1) combinatorially? phinished! what  0.01% (1) can also label  0.01% (1) the corresponding operation  0.01% (1) then then the  0.01% (1) $\infty$ to $0$,  0.01% (1) it… combinatorially?steven on  0.01% (1) order, and finally  0.01% (1) follow the curve  0.01% (1) that order, in  0.01% (1) in the complex  0.01% (1) on equilateral triangles  0.01% (1) $\alpha$, $\beta$, $\box$,  0.01% (1) $s$ along an  0.01% (1) arc from a  0.01% (1) to $f^{1}(1)$ or  0.01% (1) similarly from $1$  0.01% (1) to $\infty$ or  0.01% (1) combinatorially?carlos on can  0.01% (1) of contents $\box$,  0.01% (1) point in $f^{1}(0)$  0.01% (1) planebrowse by category  0.01% (1) considering in part  0.01% (1) gemstones (49) amber  0.01% (1) picture sums this  0.01% (1) up: so, the  0.01% (1) real geometry of  0.01% (1) $f^{1}(0)$. the following  0.01% (1) reader to do  0.01% (1) turning to the  0.01% (1) the same before  0.01% (1) i challenge the  0.01% (1) so here goes.  0.01% (1) the proof first,  0.01% (1) it without reading  0.01% (1) me to prove  0.01% (1) our local algorithm  0.01% (1) the permutation $\omega$.  0.01% (1) boils down to  0.01% (1) an understanding of  0.01% (1) the zero fiber  0.01% (1) next page. good  0.01% (1) from $\infty$ back  0.01% (1) to $0$ on  0.01% (1) the other side  0.01% (1) post. the arc  0.01% (1) posts search recent  0.01% (1) operator $\omega$ described  0.01% (1) in the previous  0.01% (1) the “shuffle” of  0.01% (1) the $\box$ past  0.01% (1) in all, $\omega$  0.01% (1) the same object  0.01% (1) luck! pages: 1  0.01% (1) of $\omega$. all  0.01% (1) a reply post  0.01% (1) $\beta$, which is  0.01% (1) that we were  0.01% (1) that $1\text{maj}$ is  0.01% (1) teasing game”. we  0.01% (1) (9) diamond (11)  0.01% (1) no matter which  0.01% (1) inner corner you  0.01% (1) play this game!  0.01% (1) matter how you  0.01% (1) fact, it doesn’t  0.01% (1) opal (12) sapphire  0.01% (1) pick to start  0.01% (1) the jeu de  0.01% (1) three total inversions,  0.01% (1) tableau in the  0.01% (1) end, called the  0.01% (1) with the same  0.01% (1) will end up  0.01% (1) taquin slide at  0.01% (1) each step, you  0.01% (1) (10) pearl (8)  0.01% (1) $(2,1)$, $(5,4)$, and  0.01% (1) game: perform a  0.01% (1) with an inner  0.01% (1) sequence of successive  0.01% (1) below: here’s the  0.01% (1) in the animation  0.01% (1) taquin slide is  0.01% (1) shown (on repeat)  0.01% (1) slides until there  0.01% (1) are no empty  0.01% (1) you end up  0.01% (1) with? it turns  0.01% (1) $(5,3)$, but only  0.01% (1) what tableaux can  0.01% (1) inner corners left.  0.01% (1) of content $\beta$  0.01% (1) the first two  0.01% (1) rectification of the  0.01% (1) original tableau. since  0.01% (1) superstandard, that is,  0.01% (1) the tableau whose  0.01% (1) $i$th row is  0.01% (1) a permutation $\pi$  0.01% (1) rectification to be  0.01% (1) the data here  0.01% (1) shape $\gamma$, so  0.01% (1) filled with all  0.01% (1) $i$’s: it turns  0.01% (1) only if it  0.01% (1) is littlewoodrichardson, defined  0.01% (1) as follows. read  0.01% (1) tableau if and  0.01% (1) to a superstandard  0.01% (1) out that a  0.01% (1) semistandard tableau rectifies  0.01% (1) we fix the  0.01% (1) nice answer in  0.01% (1) \pi_i\pi_j\lt r.$$ for  0.01% (1) is sometimes more  0.01% (1) interesting to ask  0.01% (1) instance, $21534$ has  0.01% (1) same result, it  0.01% (1) we always end  0.01% (1) up at the  0.01% (1) j$ and $$0\lt  0.01% (1) reverse: can we  0.01% (1) $(\pi_i,\pi_j)$ with $i\lt  0.01% (1) question has a  0.01% (1) the smaller, and  0.01% (1) fixed tableau? this  0.01% (1) to a given  0.01% (1) categorize all skew  0.01% (1) tableaux that rectify  0.01% (1) inwards jeu de  0.01% (1) example of an  0.01% (1) and $\mu=(2,1)$, then  0.01% (1) the skew shape  0.01% (1) $\lambda/\mu$ consists of  0.01% (1) instance, if $\lambda=(5,3,3,1)$  0.01% (1) partition $\lambda$. for  0.01% (1) (strictly larger) young  0.01% (1) diagram of a  0.01% (1) finally, he defines  0.01% (1) the white squares  0.01% (1) is then a  0.01% (1) way of filling  0.01% (1) the squares in  0.01% (1) that $21534$ has  0.01% (1) only position $3$  0.01% (1) as a $2$descent.  0.01% (1) shown below. a  0.01% (1) $\mu$ from the  0.01% (1) the $r\text{maj}$ of  0.01% (1) to set up  0.01% (1) the board, we  0.01% (1) thus $2\text{}\mathrm{maj}(21534)=3+2=5$. notice  0.01% (1) semistandard young tableaux.  0.01% (1) taquin game with  0.01% (1) can play a  0.01% (1) similar jeu de  0.01% (1) need a slightly  0.01% (1) more general definition:  0.01% (1) (see this post)  0.01% (1) a permutation to  0.01% (1) corner chosen to  0.01% (1) formed by subtracting  0.01% (1) diagram of squares  0.01% (1) be $$r\text{}\mathrm{maj}(\pi)=\left(\sum_{\pi_i\ge \pi_{i+1}+r}i\right)+\#r\text{}\mathrm{inversions}.$$  0.01% (1) $\lambda/\mu$ is a  0.01% (1) such a skew  0.01% (1) shape with positive  0.01% (1) corner, there is  0.01% (1) a unique choice  0.01% (1) between the squares  0.01% (1) choose our inner  0.01% (1) that, once we  0.01% (1) an important rule,  0.01% (1) and it implies  0.01% (1) east and south  0.01% (1) of the empty  0.01% (1) to preserve the  0.01% (1) semistandard property. an  0.01% (1) have $\pi_i\pi_j$r$descent to  0.01% (1) be an index  0.01% (1) can be slid  0.01% (1) square at each  0.01% (1) step; only one  0.01% (1) step. this is  0.01% (1) semistandard at each  0.01% (1) rows and strictly  0.01% (1) increasing down columns:  0.01% (1) now, an inner  0.01% (1) weakly increasing across  0.01% (1) the entries are  0.01% (1) integers in such  0.01% (1) $$\pi_i\ge \pi_{i+1}+r,$$ so  0.01% (1) $i$ for which  0.01% (1) slide consists of  0.01% (1) sliding entries inward  0.01% (1) square in such  0.01% (1) the tableau remains  0.01% (1) numbers, and successively  0.01% (1) two of the  0.01% (1) choosing an empty  0.01% (1) square adjacent to  0.01% (1) $f^{1}(\infty)$ by the  0.01% (1) with $1/1$ at  0.01% (1) that it can’t  0.01% (1) smaller $q$analogs of  0.01% (1) integers. but it  0.01% (1) $x=y$. however, it  0.01% (1) behaves like $5$  0.01% (1) track of the  0.01% (1) way we count  0.01% (1) to $5$, and  0.01% (1) doesn’t stop there.  0.01% (1) if $2x=2y$ then  0.01% (1) is even interesting  0.01% (1) than we might  0.01% (1) expect. it comes  0.01% (1) and this $q$number  0.01% (1) sorts of ways,  0.01% (1) 4 can be  0.01% (1) satisfied in all  0.01% (1) $5$ness, by keeping  0.01% (1) 4 above: it  0.01% (1) this polynomial also  0.01% (1) captures the fact  0.01% (1) that $5$ is  0.01% (1) count to $5$.  0.01% (1) “counts” as we  0.01% (1) number like $3$  0.01% (1) are just distinguishing  0.01% (1) each of our  0.01% (1) prime, in a  0.01% (1) $q$analogy way: the  0.01% (1) so the $q$number  0.01% (1) $(5)_q$ also satisfies  0.01% (1) is not surjective,  0.01% (1) with integer coefficients.  0.01% (1) two smallerdegree polynomials  0.01% (1) polynomial $1+q+q^2+q^3+q^4$ cannot  0.01% (1) since an odd  0.01% (1) up in finite  0.01% (1) is injective, since  0.01% (1) a difficult task,  0.01% (1) it really is.  0.01% (1) teaching takes a  0.01% (1) 6 education is  0.01% (1) september 14, 2015  0.01% (1) exists posted on  0.01% (1) denotes the set  0.01% (1) instance, if $\mathbb{z}$  0.01% (1) few tries to  0.01% (1) get the hang  0.01% (1) of. writing textbooks  0.01% (1) mapped to, i.e.,  0.01% (1) of $b$ is  0.01% (1) for all $b\in  0.01% (1) b$, there exists  0.01% (1) that $f(a)=b$. for  0.01% (1) $a\in a$ such  0.01% (1) believe this bijection  0.01% (1) replies glencoe/mcgrawhill doesn’t  0.01% (1) is involved in  0.01% (1) $q$analogs that i’ll  0.01% (1) only cover one  0.01% (1) much awesome mathematics  0.01% (1) and combinatorics. so  0.01% (1) defined by $f(x)=2x$  0.01% (1) geometry, analytic number  0.01% (1) theory, representation theory,  0.01% (1) aspect of it  0.01% (1) today: $q$analogs that  0.01% (1) page to see  0.01% (1) them! pages: 1  0.01% (1) of integers, the  0.01% (1) function $f:\mathbb{z}\to \mathbb{z}$  0.01% (1) field $\mathbb{f}_q$. turn  0.01% (1) appear in geometry  0.01% (1) over a finite  0.01% (1) that $5=1+1+1+1+1$. the  0.01% (1) is not equal  0.01% (1) in order, $r_0,r_1,r_2,\ldots$,  0.01% (1) if we list  0.01% (1) $$0,1,1,2,2,3,3,\ldots$$ and so  0.01% (1) the function $f$  0.01% (1) more accurate definition:  0.01% (1) standard definition, but  0.01% (1) which i’ll attempt  0.01% (1) to define here:  0.01% (1) integers in order:  0.01% (1) the rationals. notice  0.01% (1) we want to  0.01% (1) and cumbersome. so,  0.01% (1) which is wordy  0.01% (1) find a bijection  0.01% (1) $f:\mathbb{z}\to \mathbb{q}$, where  0.01% (1) such that: setting  0.01% (1) integers and $\mathbb{q}$  0.01% (1) a widely accepted  0.01% (1) which doesn’t have  0.01% (1) there are an  0.01% (1) unlimited supply of  0.01% (1) $q$analogs of the  0.01% (1) now of course,  0.01% (1) $f(1)=r_1$, $f(1)=r_2$, and  0.01% (1) l’hospital’s rule: $$\lim_{q\to  0.01% (1) 1} \frac{q^51}{q1}=\lim_{q\to 1}  0.01% (1) \frac{5q^4}{1} = 5.$$  0.01% (1) number $5$, but  0.01% (1) certain $q$analogs are  0.01% (1) accordingly by $f(0)=r_0$,  0.01% (1) referring to “good”  0.01% (1) or “useful” $q$analogs,  0.01% (1) they are usually  0.01% (1) talk about $q$analogs,  0.01% (1) more important than  0.01% (1) others. when mathematicians  0.01% (1) and onto correspondence,”  0.01% (1) of saying “onetoone  0.01% (1) any real number:  0.01% (1) is both injective  0.01% (1) $$(a)_q=\frac{q^a1}{q1},$$ a $q$analog  0.01% (1) and surjective is  0.01% (1) generalized to give  0.01% (1) certainly satisfies property  0.01% (1) 2. it can  0.01% (1) also be easily  0.01% (1) a function which  0.01% (1) $a$. in addition,  0.01% (1) for instance, $(5)_q=1+q+q^2+q^3+q^4$,  0.01% (1) to $2x$ for  0.01% (1) natural $q$analog of  0.01% (1) simplifies: $$(n)_q=\frac{q^n1}{q1}=1+q+q^2+\cdots+q^{n1}.$$ so  0.01% (1) $n$, the expression  0.01% (1) for positive integers  0.01% (1) any integer $x$.  0.01% (1) of reasons. it  0.01% (1) for a number  0.01% (1) $p$like properties. because  0.01% (1) of property 2,  0.01% (1) most people would  0.01% (1) bijective, and is  0.01% (1) called a bijection.  0.01% (1) a shorter way  0.01% (1) this is just  0.01% (1) agree that $5^q$  0.01% (1) is not an  0.01% (1) $q$. on the  0.01% (1) other hand, $\frac{q^51}{q1}$,  0.01% (1) is an excellent  0.01% (1) polynomiallike things in  0.01% (1) we’re looking for  0.01% (1) said to be  0.01% (1) $5$, because usually  0.01% (1) if every element  0.01% (1) is even harder.  0.01% (1) logically sound (as  0.01% (1) my students at  0.01% (1) of elements of  0.01% (1) proof, or even  0.01% (1) that’s not a  0.01% (1) they gave up  0.01% (1) and figured it  0.01% (1) couldn’t be done!  0.01% (1) would certainly recognize.)  0.01% (1) if one were  0.01% (1) into a list,  0.01% (1) a common tactic  0.01% (1) would be to  0.01% (1) couldn’t be organized  0.01% (1) the range refers  0.01% (1) to the set  0.01% (1) going to try  0.01% (1) a different ordering,  0.01% (1) instead of trying  0.01% (1) in order would  0.01% (1) the function.) a  0.01% (1) make a onetoone  0.01% (1) function is: onetoone,  0.01% (1) stating that listing  0.01% (1) argument in example  0.01% (1) 2. the author  0.01% (1) is correct in  0.01% (1) assigned to by  0.01% (1) $b$ that are  0.01% (1) in a random  0.01% (1) way and failed.  0.01% (1) at that point,  0.01% (1) to do so  0.01% (1) so they try  0.01% (1) between the rationals  0.01% (1) and integers, and  0.01% (1) you think –  0.01% (1) use proof by  0.01% (1) set $b$, written  0.01% (1) $f:a\to b$, is  0.01% (1) the domain and  0.01% (1) $a$ to a  0.01% (1) $b$ is called  0.01% (1) of a onetoone  0.01% (1) and onto correspondence.  0.01% (1) a function $f$  0.01% (1) an assignment of  0.01% (1) each element of  0.01% (1) of notation in  0.01% (1) the glencoe algebra  0.01% (1) textbook, the set  0.01% (1) up another misuse  0.01% (1) b$. to clear  0.01% (1) $a\in a$ to  0.01% (1) an element $f(a)\in  0.01% (1) the precise meaning  0.01% (1) the codomain (not  0.01% (1) wrong and you  0.01% (1) get a contradiction.  0.01% (1) of course, this  0.01% (1) that something goes  0.01% (1) and then show  0.01% (1) contradiction: assume there  0.01% (1) was a way  0.01% (1) to list them,  0.01% (1) glencoe would have  0.01% (1) wouldn’t work either  0.01% (1) correct solution. getting  0.01% (1) our definitions straight  0.01% (1) first, let’s state  0.01% (1) let’s discuss a  0.01% (1) be listed. so  0.01% (1) the range, as  0.01% (1) because they can  0.01% (1) wrong with the  0.01% (1) about what is  0.01% (1) glencoe/mcgrawhill algebra 2  0.01% (1) textbook, and recently  0.01% (1) assigned to the  0.01% (1) worksheet for the  0.01% (1) guide and intervention”  0.01% (1) this horrifying falsehood  0.01% (1) same element of  0.01% (1) the supplementary “study  0.01% (1) pointed out on  0.01% (1) reddit. or at  0.01% (1) file can be  0.01% (1) found online at  0.01% (1) various websites, including  0.01% (1) worksheet. the original  0.01% (1) version of this  0.01% (1) least it was  0.01% (1) stated in some  0.01% (1) yes there is!  0.01% (1) is, example 2!  0.01% (1) avoid. so usually,  0.01% (1) when i see  0.01% (1) a mistake in  0.01% (1) is hard to  0.01% (1) which human error  0.01% (1) and math is  0.01% (1) onto, or surjective,  0.01% (1) technical fields in  0.01% (1) a math text,  0.01% (1) it doesn’t bother  0.01% (1) $f(x)=f(y)$ then $x=y$.  0.01% (1) rationals? yes there  0.01% (1) $b$, i.e., if  0.01% (1) no correspondence between  0.01% (1) hurt my soul.  0.01% (1) me much. but  0.01% (1) some things just  0.01% (1) one download link  0.01% (1) of $a$ are  0.01% (1) rationals. there are  0.01% (1) also several excellent  0.01% (1) older expositions on  0.01% (1) countability of the  0.01% (1) or injective, if  0.01% (1) a fantastic post  0.01% (1) about this error  0.01% (1) already, with several  0.01% (1) this topic, including  0.01% (1) on the math  0.01% (1) my favorites here  0.01% (1) as well. but  0.01% (1) first, let’s talk  0.01% (1) discuss two of  0.01% (1) zero blogs. i’ll  0.01% (1) less traveled and  0.01% (1) the division by  0.01% (1) wolfram blog published  0.01% (1) record straight. the  0.01% (1) search. there are  0.01% (1) other versions of  0.01% (1) the document that  0.01% (1) no two elements  0.01% (1) on a google  0.01% (1) from glencoe’s website  0.01% (1) that shows up  0.01% (1) don’t contain this  0.01% (1) example, but this  0.01% (1) thread claims. luckily,  0.01% (1) mathematicians are here  0.01% (1) to set the  0.01% (1) as the reddit  0.01% (1) some high schools,  0.01% (1) version was almost  0.01% (1) certainly used in  0.01% (1) for instance using  0.01% (1) we can calculate,  0.01% (1) coordinate plane, and  0.01% (1) them on a  0.01% (1) certainly setting $q=1$  0.01% (1) use a spiral!  0.01% (1) now, to list  0.01% (1) $(0,0)$ outwards. each  0.01% (1) the spiral from  0.01% (1) the rationals, follow  0.01% (1) in a combinatorial  0.01% (1) integers – plot  0.01% (1) about the objects  0.01% (1) than just their  0.01% (1) total number. it’s  0.01% (1) us more information  0.01% (1) the $q$analog gives  0.01% (1) $q$analog results in  0.01% (1) of objects, and  0.01% (1) time we reach  0.01% (1) an ordered pair  0.01% (1) defined an “interesting  0.01% (1) $q$analog” of an  0.01% (1) expression $p$ to  0.01% (1) recall that we  0.01% (1) green dots above.)  0.01% (1) objects and $\mathrm{stat}$  0.01% (1) is a statistic  0.01% (1) on these objects.  0.01% (1) be an expression  0.01% (1) $p_q$ such that  0.01% (1) down if $b>0$  0.01% (1) $(a,b)$, write it  0.01% (1) of integers, say  0.01% (1) and $\mathrm{gcd}(a,b)=1$, and  0.01% (1) otherwise skip it  0.01% (1) (these are the  0.01% (1) and move on.  0.01% (1) also a polynomial  0.01% (1) in $q$, so  0.01% (1) 4 by exhibiting  0.01% (1) list is to  0.01% (1) to construct this  0.01% (1) it satisfies property  0.01% (1) $$\sum_{p\in s_n}q^{\mathrm{inv}(p)}=(1)(1+q)(1+q+q^2)\cdots(1+q+\cdots+q^n)=(n)_q!.$$ so  0.01% (1) post! that is,  0.01% (1) pairs $(a,b)$ of  0.01% (1) rationals as ordered  0.01% (1) a product formula  0.01% (1) like $n!$ itself.  0.01% (1) post, but let’s  0.01% (1) instead prove it  0.01% (1) by building up  0.01% (1) fact in this  0.01% (1) proof of this  0.01% (1) naively one way  0.01% (1) i posted a  0.01% (1) integers with $b>0$  0.01% (1) $q$factorial we saw  0.01% (1) all pairs of  0.01% (1) s_n}q^{\mathrm{inv}(p)}$, which is  0.01% (1) way of ordering  0.01% (1) our $q$analog $\sum_{p\in  0.01% (1) now see how  0.01% (1) it satisfies properties  0.01% (1) 1, 2, and  0.01% (1) 3 above. let’s  0.01% (1) $n!$, also satisfies  0.01% (1) is an easy  0.01% (1) in general this  0.01% (1) and $\mathrm{gcd}(a,b)=1$ as  0.01% (1) be the same  0.01% (1) as $(1)(1+q)(1+q+q^2)$. indeed,  0.01% (1) that $1+2q+2q^2+q^3$ factors  0.01% (1) described above. there  0.01% (1) property 4. notice  0.01% (1) set of combinatorial  0.01% (1) over a certain  0.01% (1) permutation $p$ by  0.01% (1) $q^{\mathrm{inv}(p)}$, and $q$count  0.01% (1) by summing these  0.01% (1) we weight each  0.01% (1) q^3 \end{array} $$  0.01% (1) wilf. they construct  0.01% (1) q^2 \\ 321  0.01% (1) & 3 &  0.01% (1) $q$powers, to form  0.01% (1) the sum $$\sum_{p\in  0.01% (1) of $1,\ldots,n$. so  0.01% (1) for $n=3$, the  0.01% (1) sum is $1+2q+2q^2+q^3$  0.01% (1) of all permutations  0.01% (1) is the set  0.01% (1) to calkin and  0.01% (1) s_n}q^{\mathrm{inv}(p)}$$ where $s_n$  0.01% (1) & q^2\\ 312  0.01% (1) 231 & 2  0.01% (1) permutations of 3  0.01% (1) as follows: start  0.01% (1) they contain. $$\begin{array}{ccc}  0.01% (1) $a$ is called  0.01% (1) below shows the  0.01% (1) for each node  0.01% (1) the top, and  0.01% (1) $\pi$. the table  0.01% (1) tree is constructed  0.01% (1) shown below. the  0.01% (1) q\\ 213 &  0.01% (1) in which each  0.01% (1) a binary tree  0.01% (1) exactly once, as  0.01% (1) 1 \\ 132  0.01% (1) p & \mathrm{inv}(p)  0.01% (1) & q^{\mathrm{inv}(p)}\\\hline 123  0.01% (1) by our table  0.01% (1) above. we now  0.01% (1) coefficients but rather  0.01% (1) the exponents of  0.01% (1) our summation that  0.01% (1) are many other  0.01% (1) is not the  0.01% (1) in $q$analog notation,  0.01% (1) $\sum_{p\in s_n}q^{\mathrm{inv}(p)}$, we  0.01% (1) understand that it  0.01% (1) we are interested  0.01% (1) in.. in general,  0.01% (1) of $q$powers $q^{\mathrm{stat}(p)}$  0.01% (1) this process guarantees  0.01% (1) where $p$ ranges  0.01% (1) that we list  0.01% (1) as a summation  0.01% (1) elegant methods there  0.01% (1) exactly once. more  0.01% (1) $i$ inversions. but  0.01% (1) elegant enumerations of  0.01% (1) a generating function,  0.01% (1) the sum $1+2q+2q^2+q^3$  0.01% (1) is thought of  0.01% (1) generating functions. as  0.01% (1) between $q$analogs and  0.01% (1) come to an  0.01% (1) important philosophical distinction  0.01% (1) one is due  0.01% (1) coefficients, $1,2,2,1$. generatingfunctionologically,  0.01% (1) q^i$ where $c_i$  0.01% (1) the rationals, and  0.01% (1) length $n$ with  0.01% (1) as $\sum_{i=0}^\infty c_i  0.01% (1) one particularly nice  0.01% (1) we might instead  0.01% (1) write the sum  0.01% (1) listing the rationals,  0.01% (1) $q$counting from the  0.01% (1) and i’m a  0.01% (1) $q$analog addict. the  0.01% (1) theory of $q$analogs  0.01% (1) hi, i’m maria  0.01% (1) can be written  0.01% (1) $q$analog? (part 1)  0.01% (1) posted on november  0.01% (1) 22, 2015 by  0.01% (1) is a littleknown  0.01% (1) gem, and in  0.01% (1) so what is  0.01% (1) a $q$analog? it  0.01% (1) rational numbers are  0.01% (1) awesome and addictive!  0.01% (1) those numbers which  0.01% (1) posts i’ll explain  0.01% (1) why they’re so  0.01% (1) as fractions $a/b$  0.01% (1) where $a$ and  0.01% (1) with $b\neq 0$.  0.01% (1) $q$multiplication principle. that’s  0.01% (1) post we’ll talk  0.01% (1) $1,\ldots,n$, now that  0.01% (1) proof that $n!$  0.01% (1) that even the  0.01% (1) proof was a  0.01% (1) assign every rational  0.01% (1) about how to  0.01% (1) use the $q$multiplication  0.01% (1) $q$catalan numbers and  0.01% (1) other fun $q$analogs.  0.01% (1) $b$ are integers  0.01% (1) coefficient, and discuss  0.01% (1) of the $q$binomial  0.01% (1) principle to derive  0.01% (1) a combinatorial interpretation  0.01% (1) those rare mathematical  0.01% (1) terms whose definition  0.01% (1) limit as $q$  0.01% (1) approaches $1$. for  0.01% (1) instance, the expression  0.01% (1) bijective if and  0.01% (1) can be recovered  0.01% (1) $q=1$, we also  0.01% (1) say it’s a  0.01% (1) $q$analog if $p$  0.01% (1) $$\frac{q^51}{q1}$$ is another  0.01% (1) function will be  0.01% (1) $q=1$, we do  0.01% (1) have a well  0.01% (1) defined limit that  0.01% (1) so on. the  0.01% (1) by zero if  0.01% (1) – even though  0.01% (1) we get division  0.01% (1) not defined at  0.01% (1) if $p_q$ is  0.01% (1) about the rationals.  0.01% (1) let’s be precise  0.01% (1) expression $p_q$, depending  0.01% (1) definition anyway: definition:  0.01% (1) start with the  0.01% (1) doesn’t really capture  0.01% (1) what it is  0.01% (1) about, but let’s  0.01% (1) on $q$, such  0.01% (1) exactly once. next,  0.01% (1) only if every  0.01% (1) in $q=1$ we  0.01% (1) get $5$. sometimes,  0.01% (1) in the list  0.01% (1) for instance, $2q+3q^2$  0.01% (1) in $p_q$ results  0.01% (1) in $p$. so,  0.01% (1) our $q$count. notice  0.01% (1) the $q$factorial as  0.01% (1) the ways of  0.01% (1) choosing one element  0.01% (1) \mathbb{rp}^1$ by the  0.01% (1) $1/2$ in this  0.01% (1) the $q$count of  0.01% (1) $b$ with $q$counts  0.01% (1) convention. it follows  0.01% (1) $m(q)$ and $n(q)$,  0.01% (1) from $a$ and  0.01% (1) another from $b$  0.01% (1) fraction. so the  0.01% (1) weights of the  0.01% (1) into a reduced  0.01% (1) pair is the  0.01% (1) weight of a  0.01% (1) number $2/4$ should  0.01% (1) $m(q)n(q)$, where the  0.01% (1) sets $a$ and  0.01% (1) given two weighted  0.01% (1) b>0\text{ and }\mathrm{gcd}(a,b)=1\text{  0.01% (1) $m$ things and  0.01% (1) set $$\mathbb{q}=\{(a,b)   0.01% (1) one thing from  0.01% (1) and }a,b\in \mathbb{z}\}.$$  0.01% (1) ground up. the  0.01% (1) multiplication principle in  0.01% (1) combinatorics is the  0.01% (1) another from a  0.01% (1) set of $n$  0.01% (1) but what if  0.01% (1) the things are  0.01% (1) weighted? $q$multiplication principle:  0.01% (1) product $m\cdot n$.  0.01% (1) things is the  0.01% (1) numbers as the  0.01% (1) set of rational  0.01% (1) elements. let’s see  0.01% (1) how this plays  0.01% (1) inversion with the  0.01% (1) second, this choice  0.01% (1) of $1+q+q^2+\cdots+q^{n2}$ to  0.01% (1) entry of an  0.01% (1) be the smaller  0.01% (1) $n1$ remaining entries,  0.01% (1) and since the  0.01% (1) first entry cannot  0.01% (1) the total by  0.01% (1) and $a$ is  0.01% (1) the same argument.  0.01% (1) continuing in this  0.01% (1) fashion we get  0.01% (1) number a unique  0.01% (1) representation, let us  0.01% (1) case where $b>0$  0.01% (1) restrict to the  0.01% (1) any of the  0.01% (1) then can be  0.01% (1) is weighted by  0.01% (1) that $\mathrm{gcd}(a,b)=1$. this  0.01% (1) inversions it forms  0.01% (1) condition makes $a/b$  0.01% (1) if each entry  0.01% (1) out in the  0.01% (1) case of $(n)_q!$.  0.01% (1) with smaller entries  0.01% (1) (to its right),  0.01% (1) factor of $1+q+q^2+\cdots+q^{n1}$  0.01% (1) to the total.  0.01% (1) any integer such  0.01% (1) which contributes a  0.01% (1) any of $1,2,\ldots,n$,  0.01% (1) then the first  0.01% (1) entry can be  0.01% (1) $\mathrm{inv}(\pi)$ denotes the  0.01% (1) in particular, after  0.01% (1) $2^{n1}\cdot f_n$, counts  0.01% (1) a binary sequence  0.01% (1) of length $n1$  0.01% (1) and also a  0.01% (1) of our identity,  0.01% (1) $1$’s. thus, the  0.01% (1) also that $2^{n1}$  0.01% (1) of length$(n1)$ sequences  0.01% (1) of $0$’s and  0.01% (1) fence post coloring  0.01% (1) of length $n2$.  0.01% (1) alternating sequence of  0.01% (1) digits and fence  0.01% (1) posts such as:  0.01% (1) $$1\, \square\, 0\,  0.01% (1) entries, forming an  0.01% (1) can interlace their  0.01% (1) because of their  0.01% (1) lengths, given such  0.01% (1) a pair we  0.01% (1) \blacksquare \end{array}$$ note  0.01% (1) \blacksquare \\ \\  0.01% (1) or white such  0.01% (1) that no two  0.01% (1) adjacent ones are  0.01% (1) black. (can you  0.01% (1) fenceposts either black  0.01% (1) row of $n2$  0.01% (1) $(n+1)$st fibonacci number  0.01% (1) $f_n$ counts the  0.01% (1) to color a  0.01% (1) see why this  0.01% (1) combinatorial construction would  0.01% (1) of a row  0.01% (1) of $3$ fenceposts:  0.01% (1) $$\begin{array}{ccc} \square &  0.01% (1) five such colorings  0.01% (1) because there are  0.01% (1) satisfy the fibonacci  0.01% (1) recurrence?) for instance,  0.01% (1) we have $f_5=5$,  0.01% (1) \square\, 1\, \blacksquare\,  0.01% (1) 1$$ we will  0.01% (1) my freshman year  0.01% (1) as an undergraduate  0.01% (1) at mit. coming  0.01% (1) off of a  0.01% (1) traced back to  0.01% (1) combinatorics, might be  0.01% (1) pursuit of a  0.01% (1) ph.d. in mathematics,  0.01% (1) specifically in algebraic  0.01% (1) series of successes  0.01% (1) in high school  0.01% (1) was a confident  0.01% (1) and excited 18year  0.01% (1) old whose dream  0.01% (1) was to become  0.01% (1) mathematical family!), i  0.01% (1) loving and very  0.01% (1) math competitions and  0.01% (1) other sciencerelated endeavors  0.01% (1) (thanks to my  0.01% (1) for me, my  0.01% (1) down the road.  0.01% (1) see the next  0.01% (1) page for my  0.01% (1) solution, or post  0.01% (1) your own solution  0.01% (1) these interlaced sequences.  0.01% (1) side also counts  0.01% (1) call such sequences  0.01% (1) interlaced sequences. we  0.01% (1) now need only  0.01% (1) in the comments  0.01% (1) below! pages: 1  0.01% (1) missteps in life  0.01% (1) that lead to  0.01% (1) the greatest adventures  0.01% (1) sometimes it’s the  0.01% (1) 24, 2016 by  0.01% (1) gemstones, opal   0.01% (1) 4 replies phinished!  0.01% (1) posted on may  0.01% (1) remembered that the  0.01% (1) it! ended, and  0.01% (1) suppose we use  0.01% (1) to expand $(1+\sqrt{5})^n$  0.01% (1) and $(1\sqrt{5})^n$ in  0.01% (1) binet’s formula. the  0.01% (1) identity? in particular,  0.01% (1) a combinatorial proof  0.01% (1) of the terms  0.01% (1) using the binomial  0.01% (1) theorem? is there  0.01% (1) resulting expression is:  0.01% (1) $$f_n=\frac{1}{\sqrt{5}\cdot 2^n}\left( \left(\sum_{i=0}^n  0.01% (1) summations cancel, and  0.01% (1) combining the odd  0.01% (1) terms gives us:  0.01% (1) $$f_n=\frac{1}{\sqrt{5}\cdot 2^n}\left( \sum_{k=0}^{\lfloor  0.01% (1) in the two  0.01% (1) the even terms  0.01% (1) \binom{n}{i}(\sqrt{5})^i \right) –  0.01% (1) \left(\sum_{i=0}^n (1)^i\binom{n}{i}(\sqrt{5})^i \right)  0.01% (1) \right)$$ note that  0.01% (1) we expand each  0.01% (1) thinking: what if  0.01% (1) proven by induction,  0.01% (1) and later, one  0.01% (1) of our students  0.01% (1) was trying to  0.01% (1) class that it  0.01% (1) i mentioned during  0.01% (1) it!, we first  0.01% (1) derived the formula  0.01% (1) using generating functions.  0.01% (1) work out the  0.01% (1) induction proof on  0.01% (1) be of the  0.01% (1) same fact, and  0.01% (1) it got me  0.01% (1) proofs there could  0.01% (1) how many different  0.01% (1) a white board  0.01% (1) outside the classroom.  0.01% (1) she was amazed  0.01% (1) n/2\rfloor} 2 \binom{n}{2k+1}(\sqrt{5})^{2k+1}  0.01% (1) \right)$$ since $(\sqrt{5})^{2k+1}=\sqrt{5}\cdot  0.01% (1) show that $a$  0.01% (1) has $m$ elements,  0.01% (1) then it follows  0.01% (1) that $m=n$. such  0.01% (1) and by another  0.01% (1) has $n$ elements,  0.01% (1) that if by  0.01% (1) some method one  0.01% (1) a collection $a$  0.01% (1) a “combinatorial proof”  0.01% (1) may be able  0.01% (1) and $n$ being  0.01% (1) the right. i  0.01% (1) started thinking about  0.01% (1) this after prove  0.01% (1) of the equation  0.01% (1) $m$ being the  0.01% (1) to be used  0.01% (1) to prove the  0.01% (1) identity above, with  0.01% (1) the simple principle  0.01% (1) two ways is  0.01% (1) obtain: $$2^{n1}\cdot f_n=\sum_{k=0}^{\lfloor  0.01% (1) n/2\rfloor} \binom{n}{2k+1}\cdot 5^k.$$  0.01% (1) now, the left  0.01% (1) hand and right  0.01% (1) by $2^{n1}$ to  0.01% (1) multiply both sides  0.01% (1) 5^k$, we can  0.01% (1) cancel the factors  0.01% (1) of $\sqrt{5}$ and  0.01% (1) hand side are  0.01% (1) clearly nonnegative integers,  0.01% (1) in some collection.  0.01% (1) the proof method  0.01% (1) of counting in  0.01% (1) number of elements  0.01% (1) they count the  0.01% (1) and one handy  0.01% (1) fact about nonnegative  0.01% (1) integers is that  0.01% (1) a physicist and  0.01% (1) use my mathematical  0.01% (1) subject, i gained  0.01% (1) respect for it.  0.01% (1) worthy challenge, and  0.01% (1) i couldn’t help  0.01% (1) fascinated with the  0.01% (1) only was i  0.01% (1) was horrified, my  0.01% (1) 18yearold littlemissperfect confidence  0.01% (1) shattered. now, not  0.01% (1) but come back  0.01% (1) for more. in  0.01% (1) research papers. i  0.01% (1) dabbled in other  0.01% (1) areas as well,  0.01% (1) but was always  0.01% (1) wrote several undergraduate  0.01% (1) similar subjects and  0.01% (1) the years that  0.01% (1) followed, i took  0.01% (1) more courses on  0.01% (1) math class? i  0.01% (1) b+ in a  0.01% (1) was hooked. but  0.01% (1) i was also  0.01% (1) a freshman, and  0.01% (1) didn’t yet have  0.01% (1) algebraic combinatorics, i  0.01% (1) another area of  0.01% (1) a simple, concrete  0.01% (1) problem. to use  0.01% (1) a term from  0.01% (1) a strong grasp  0.01% (1) of some of  0.01% (1) with a b+  0.01% (1) in the class.  0.01% (1) me, get a  0.01% (1) but wound up  0.01% (1) i studied hard  0.01% (1) the other algebraic  0.01% (1) concepts being used  0.01% (1) in the course.  0.01% (1) drawn back to  0.01% (1) the interplay between  0.01% (1) our expository series  0.01% (1) of posts (see  0.01% (1) part i and  0.01% (1) part ii) on  0.01% (1) final post in  0.01% (1) the third and  0.01% (1) iii) posted on  0.01% (1) april 5, 2016  0.01% (1) 3 this is  0.01% (1) the recent paper  0.01% (1) coauthored by jake  0.01% (1) young tableaux is  0.01% (1) actually the monodromy  0.01% (1) operator of a  0.01% (1) certain covering map  0.01% (1) $\omega$ on certain  0.01% (1) that the operator  0.01% (1) levinson and myself.  0.01% (1) last time, we  0.01% (1) discussed the fact  0.01% (1) amber, diamond   0.01% (1) 3 posted in  0.01% (1) my ph.d. at  0.01% (1) uc berkeley on  0.01% (1) a topic in  0.01% (1) algebraic combinatorics… …and  0.01% (1) 2016, having completed  0.01% (1) friday, may 20,  0.01% (1) combinatorics and algebra.  0.01% (1) i now find  0.01% (1) myself, as of  0.01% (1) i often wonder  0.01% (1) how much that  0.01% (1) my thesis. my  0.01% (1) full thesis can  0.01% (1) be found here.)  0.01% (1) a summary of  0.01% (1) page 2 for  0.01% (1) silly little b+  0.01% (1) motivated me throughout  0.01% (1) the years. (see  0.01% (1) elegantly solve such  0.01% (1) used to so  0.01% (1) how could i  0.01% (1) pass up that  0.01% (1) chance? what if  0.01% (1) he didn’t teach  0.01% (1) in the area.  0.01% (1) a world expert  0.01% (1) hear that it  0.01% (1) was being taught  0.01% (1) by richard stanley,  0.01% (1) it again before  0.01% (1) i left mit?  0.01% (1) the following: in  0.01% (1) a complete graph  0.01% (1) with $n$ vertices,  0.01% (1) how many walks  0.01% (1) was something like  0.01% (1) combinatorial question. it  0.01% (1) on the first  0.01% (1) day of the  0.01% (1) class, stanley started  0.01% (1) but i did  0.01% (1) combinatorics” even meant,  0.01% (1) of 18ish. but  0.01% (1) i loved pure  0.01% (1) math too, and  0.01% (1) my friends were  0.01% (1) at the age  0.01% (1) or something. me  0.01% (1) skills to, i  0.01% (1) don’t know, come  0.01% (1) unified field theory  0.01% (1) signed up for  0.01% (1) the undergraduate algebraic  0.01% (1) packed course load.  0.01% (1) i had no  0.01% (1) idea what “algebraic  0.01% (1) to my already  0.01% (1) self added it  0.01% (1) combinatorics class in  0.01% (1) the spring, so  0.01% (1) my young ambitious  0.01% (1) starting at vertex  0.01% (1) $v$ end up  0.01% (1) compute this trace  0.01% (1) using eigenvalues and  0.01% (1) divide by $n$  0.01% (1) starting at $v$.  0.01% (1) one can then  0.01% (1) trace of $a^k$.  0.01% (1) that the total  0.01% (1) starting from any  0.01% (1) vertex is the  0.01% (1) beautiful! i remember  0.01% (1) sitting in my  0.01% (1) me that advanced  0.01% (1) tools from linear  0.01% (1) algebra could be  0.01% (1) was incredible to  0.01% (1) the technique. it  0.01% (1) seat, wideeyed, watching  0.01% (1) richard stanley quietly  0.01% (1) but authoritatively discuss  0.01% (1) graph, and showed  0.01% (1) of the complete  0.01% (1) graph looks like:  0.01% (1) and there are  0.01% (1) four closed walks  0.01% (1) of length two,  0.01% (1) and $k=2$, the  0.01% (1) instance, if $n=5$  0.01% (1) back at vertex  0.01% (1) $v$ on the  0.01% (1) last step? for  0.01% (1) from $v$ to  0.01% (1) any other vertex  0.01% (1) an algebraic proof.  0.01% (1) he considered the  0.01% (1) adjacency matrix $a$  0.01% (1) went forth with  0.01% (1) this, but stanley  0.01% (1) and back again:  0.01% (1) elementary (though messy)  0.01% (1) way to solve  0.01% (1) integers? at prove  0.01% (1) showing up in  0.01% (1) two columns (columns  0.01% (1) $3$ and $4$)  0.01% (1) of $\mu$, and  0.01% (1) $21\lt r\le 2$.  0.01% (1) among the last  0.01% (1) is one box  0.01% (1) $i_\mu$. for subsets  0.01% (1) $k=2$, we have  0.01% (1) $d_2(\mu)=1$ since there  0.01% (1) so $r$ can  0.01% (1) only be $2$,  0.01% (1) x_1x_4,\hspace{0.3cm} x_2x_3,\hspace{0.3cm} x_2x_4,\hspace{0.3cm}  0.01% (1) x_3x_4.$$ for subsets  0.01% (1) $k=3$, we have  0.01% (1) $d_3(\mu)=2$, and so  0.01% (1) polynomials $$x_1x_2,\hspace{0.3cm} x_1x_3,\hspace{0.3cm}  0.01% (1) us the six  0.01% (1) symmetric functions $e_2(s)$  0.01% (1) for all subsets  0.01% (1) $2$. this gives  0.01% (1) $1$ variable in  0.01% (1) symmetric functions in  0.01% (1) $i_{\mu}$, first consider  0.01% (1) $\{x_1,\ldots,x_4\}$ of size  0.01% (1) $k=1$. we have  0.01% (1) $d_1(\mu)=0$ since the  0.01% (1) then to compute  0.01% (1) $n=4$ and $\mu=(3,1)$.  0.01% (1) $s_n$module, graded by  0.01% (1) degree. to illustrate  0.01% (1) the construction, suppose  0.01% (1) fourth column of  0.01% (1) of $\mu$ is  0.01% (1) r \le 1$,  0.01% (1) which is impossible.  0.01% (1) no partial elementary  0.01% (1) must have $10\lt  0.01% (1) in $i_\mu$ we  0.01% (1) empty (see image  0.01% (1) below), and so  0.01% (1) $e_r(s)$ to be  0.01% (1) $32 \lt r\le  0.01% (1) 3$. we therefore  0.01% (1) \end{align*}$$ as two  0.01% (1) more examples, it’s  0.01% (1) clear that $r_{(1^n)}=\mathbb{c}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$  0.01% (1) is the ring  0.01% (1) e_2(x_1,\ldots,x_4), e_3(x_1,\ldots,x_4), e_4(x_1,\ldots,x_4))  0.01% (1) \\ & e_1(x_1,\ldots,x_4),  0.01% (1) & e_2(x_1,x_2,x_3), \ldots,  0.01% (1) e_2(x_2,x_3,x_4), \\ &  0.01% (1) e_3(x_1,x_2,x_3), \ldots, e_3(x_2,x_3,x_4),  0.01% (1) of coninvariants under  0.01% (1) the $s_n$ action,  0.01% (1) and in fact  0.01% (1) the graded frobenius  0.01% (1) characteristic of $r_\mu$  0.01% (1) is the halllittlewood  0.01% (1) the coinvariant ring,  0.01% (1) a generalization of  0.01% (1) and $r_{(n)}=\mathbb{c}$ is  0.01% (1) the trivial representation.  0.01% (1) so $r_\mu$ is  0.01% (1) e_2(x_1,x_3),\ldots, e_2(x_3,x_4), \\  0.01% (1) $$\begin{align*} i_{(3,1)}= &(e_2(x_1,x_2),  0.01% (1) polynomials $$x_1x_2+x_1x_3+x_2x_3, \hspace{0.5cm}x_1x_2+x_1x_4+x_2x_4,$$  0.01% (1) $$x_1x_3+x_1x_4+x_3x_4,\hspace{0.5cm} x_2x_3+x_2x_4+x_3x_4,$$ $$x_1x_2x_3,  0.01% (1) \hspace{0.4cm} x_1x_2x_4, \hspace{0.4cm}  0.01% (1) x_1x_3x_4,\hspace{0.4cm} x_2x_3x_4$$ finally,  0.01% (1) the eight additional  0.01% (1) in $i_\mu$, and  0.01% (1) have $e_2(s)$ and  0.01% (1) $e_3(s)$ for each  0.01% (1) such subset $s$  0.01% (1) for $s=\{x_1,x_2,x_3,x_4\}$, we  0.01% (1) have $d_4(\mu)=4$ and  0.01% (1) also relations in  0.01% (1) $i_{\mu}$. all in  0.01% (1) all we have  0.01% (1) four variables are  0.01% (1) $e_1,\ldots,e_4$ in the  0.01% (1) so $44\lt r\le  0.01% (1) 4$. thus all  0.01% (1) of the full  0.01% (1) is a graded  0.01% (1) homogeneous ideal, $r_\mu$  0.01% (1) garsiaprocesi modules! also  0.01% (1) known as the  0.01% (1) cohomology rings of  0.01% (1) the springer fibers  0.01% (1) time for… the  0.01% (1) but it’s about  0.01% (1) one of them  0.01% (1) – the ring  0.01% (1) of coinvariants –  0.01% (1) in type $a$,  0.01% (1) or as the  0.01% (1) a link between  0.01% (1) these two interpretations  0.01% (1) given in a  0.01% (1) paper of deconcini  0.01% (1) a torus, with  0.01% (1) n$ matrices with  0.01% (1) coordinate ring of  0.01% (1) a nilpotent conjugacy  0.01% (1) class of $n\times  0.01% (1) about a particular  0.01% (1) talked in depth  0.01% (1) to secondary content  0.01% (1) homeabout this blogcontribute!all  0.01% (1) posts post navigation  0.01% (1) $r_\mu$ posted on  0.01% (1) primary content skip  0.01% (1) menu skip to  0.01% (1) the quest for  0.01% (1) mathematical beauty and  0.01% (1) truth search main  0.01% (1) september 23, 2016  0.01% (1) reply somehow, in  0.01% (1) mentioned them briefly  0.01% (1) at the end  0.01% (1) correspondence series, and  0.01% (1) graded $s_n$modules. i  0.01% (1) described the structure  0.01% (1) all the time  0.01% (1) i’ve posted here,  0.01% (1) i’ve not yet  0.01% (1) and procesi. but  0.01% (1) the work of  0.01% (1) of boxes in  0.01% (1) the last $k$  0.01% (1) columns of $\mu$,  0.01% (1) where we pad  0.01% (1) here, $d_k(\mu)=\mu’_n+\mu’_{n1}+\cdots+ \mu’_{nk+1}$  0.01% (1) \le k, s=k).$$  0.01% (1) function, and we  0.01% (1) have $$i_{\mu}=(e_r(s) :  0.01% (1) kd_k(\mu) \lt r  0.01% (1) $\mu’$ with $0$’s  0.01% (1) so that it  0.01% (1) permuting the variables,  0.01% (1) fixed under this  0.01% (1) action. since $i_\mu$  0.01% (1) is also a  0.01% (1) on $\mathbb{c}[x_1,\ldots,x_n]$ by  0.01% (1) action of $s_n$  0.01% (1) has $n$ parts.  0.01% (1) this ring $r_\mu$  0.01% (1) inherits the natural  0.01% (1) of the variables  0.01% (1) in this subset  0.01% (1) elementary way. using  0.01% (1) tanisaki’s approach, we  0.01% (1) can define $$r_{\mu}=\mathbb{c}[x_1,\ldots,x_n]/i_{\mu},$$  0.01% (1) where $i_{\mu}$ is  0.01% (1) in an entirely  0.01% (1) with these modules  0.01% (1) tanisaki, and garsia  0.01% (1) and procesi, allows  0.01% (1) us to work  0.01% (1) the ideal generated  0.01% (1) by the partial  0.01% (1) $s\subset\{x_1,\ldots,x_n\}$ with $s=k$.  0.01% (1) then the elementary  0.01% (1) symmetric function $e_r(s)$  0.01% (1) variables $z_i$. let  0.01% (1) degree $d$ in  0.01% (1) $e_d(z_1,\ldots,z_k)$ is the  0.01% (1) sum of all  0.01% (1) squarefree monomials of  0.01% (1) polynomial $\widetilde{h}_\mu(x;q)$. where  0.01% (1) do these relations  0.01% (1) showing that $r_\mu$  0.01% (1) has the right  0.01% (1) dimension, namely the  0.01% (1) multinomial coefficient $\binom{n}{\mu}$.  0.01% (1) be shown by  0.01% (1) but this can  0.01% (1) somewhat more work  0.01% (1) these relations generate  0.01% (1) the entire ideal,  0.01% (1) and for that,  0.01% (1) we’ll discuss on  0.01% (1) a reply can  0.01% (1) combinatorially? posted on  0.01% (1) august 12, 2016  0.01% (1) 4 this year’s  0.01% (1) gemstones  leave  0.01% (1) posted in diamond,  0.01% (1) page 2 the  0.01% (1) monomial basis of  0.01% (1) garsia and procesi.  0.01% (1) above. it takes  0.01% (1) that describe $i_\mu$  0.01% (1) then we have  0.01% (1) $$t^{d_k(\mu)}  (x_{i_1}t)(x_{i_2}t)\cdots  0.01% (1) (x_{i_k}t)$$ for any  0.01% (1) subset $s=\{x_{i_1},\ldots,x_{i_k}\}$ of  0.01% (1) $x=\mathrm{diag}(x_1,\ldots,x_n)\in \overline{c_\mu}\cap t$  0.01% (1) have this property,  0.01% (1) columns of $\mu$.  0.01% (1) any element of  0.01% (1) $c_\mu$ must also  0.01% (1) $\{x_1,\ldots,x_n\}$. expanding the  0.01% (1) as a polynomial  0.01% (1) as $r \gt  0.01% (1) kd_k(\mu)$, which is  0.01% (1) exactly the relations  0.01% (1) $x$ as soon  0.01% (1) $e_r(s)$ vanish on  0.01% (1) in $t$ using  0.01% (1) vieta’s formulas, we  0.01% (1) see that the  0.01% (1) academy was a  0.01% (1) big success, and  0.01% (1) two numbers to  0.01% (1) form the next:  0.01% (1) $$0,1,1,2,3,5,8,13,21,\ldots$$ if $f_n$  0.01% (1) denotes the $(n+1)$st  0.01% (1) adding the previous  0.01% (1) $1$ and then  0.01% (1) formed by starting  0.01% (1) with the two  0.01% (1) numbers $0$ and  0.01% (1) term of this  0.01% (1) sequence (where $f_0=0$  0.01% (1) – \left(\frac{1\sqrt{5}}{2}\right)^n \right)$$  0.01% (1) looks crazy, right?  0.01% (1) why would there  0.01% (1) be $\sqrt 5$’s  0.01% (1) formula: $$f_n=\frac{1}{\sqrt{5}}\left( \left(\frac{1+\sqrt{5}}{2}\right)^n  0.01% (1) known as binet’s  0.01% (1) and $f_1=1$), then  0.01% (1) remarkable formula for  0.01% (1) the $n$th term,  0.01% (1) the fibonacci sequence,  0.01% (1) the camp was  0.01% (1) this year. some  0.01% (1) of the students  0.01% (1) mentioned that they  0.01% (1) felt even more  0.01% (1) students that attended  0.01% (1) talented high school  0.01% (1) it was an  0.01% (1) enormous pleasure to  0.01% (1) teach the seventeen  0.01% (1) inspired to study  0.01% (1) math further after  0.01% (1) ideas as well!  0.01% (1) many, many things  0.01% (1) we investigated at  0.01% (1) me with new  0.01% (1) – they inspired  0.01% (1) our twoweek program,  0.01% (1) but the inspiration  0.01% (1) went both ways  0.01% (1) of the ending  0.01% (1) – the sums  0.01% (1) space of $n\times  0.01% (1) n$ matrices for  0.01% (1) each partition $\mu’$  0.01% (1) of $n$. the  0.01% (1) $c_{\mu’}$ in the  0.01% (1) is exactly one  0.01% (1) the conjugacy class  0.01% (1) of $a$. in  0.01% (1) other words: there  0.01% (1) closures of these  0.01% (1) conjugacy classes $\overline{c_{\mu’}}$  0.01% (1) handle on after  0.01% (1) intersecting with the  0.01% (1) set $t$ of  0.01% (1) diagonal matrices, leading  0.01% (1) they are easier  0.01% (1) studied here. however,  0.01% (1) form closed matrix  0.01% (1) varieties, and their  0.01% (1) coordinate rings were  0.01% (1) partition uniquely determines  0.01% (1) $\mu’$, and this  0.01% (1) $n\times n$ matrix  0.01% (1) over $\mathbb{c}$. then  0.01% (1) $a$ has all  0.01% (1) $0$ eigenvalues, and  0.01% (1) be a nilpotent  0.01% (1) follows. let $a$  0.01% (1) come from? the  0.01% (1) rings $r_\mu$ were  0.01% (1) originally defined as  0.01% (1) conjugate to a  0.01% (1) matrix in jordan  0.01% (1) jordan blocks, written  0.01% (1) in nonincreasing order  0.01% (1) form a partition  0.01% (1) sizes of the  0.01% (1) the diagonal. the  0.01% (1) normal form whose  0.01% (1) jordan blocks have  0.01% (1) all $0$’s on  0.01% (1) to an interesting  0.01% (1) natural question: what  0.01% (1) say $t^{d_k}$, is  0.01% (1) fixed under conjugation,  0.01% (1) so we can  0.01% (1) assume $a$ is  0.01% (1) minors of $ati$,  0.01% (1) the $k\times k$  0.01% (1) form on wikipedia)  0.01% (1) that the largest  0.01% (1) dividing all of  0.01% (1) form. then it’s  0.01% (1) not hard to  0.01% (1) partition of $\mu’$  0.01% (1) and $d_k(\mu)$ is  0.01% (1) defined as above  0.01% (1) is the transpose  0.01% (1) $t^{d_k(\mu)}$ where $\mu$  0.01% (1) see, by analyzing  0.01% (1) that this power  0.01% (1) of $t$ is  0.01% (1) on smith normal  0.01% (1) in the article  0.01% (1) we obtain the  0.01% (1) same modules as  0.01% (1) above. tanisaki found  0.01% (1) the presentation for  0.01% (1) defining $r_\mu=\mathcal{o}(\overline{c_{\mu’}}\cap t)$,  0.01% (1) conjugacy class $c_{\mu’}$?  0.01% (1) is the subvariety  0.01% (1) in the closure  0.01% (1) of the nilpotent  0.01% (1) $r_\mu$ given above  0.01% (1) using roughly the  0.01% (1) instance, the discussion  0.01% (1) of invariant factors  0.01% (1) and elementary divisors  0.01% (1) show (see, for  0.01% (1) then one can  0.01% (1) following argument. consider  0.01% (1) the matrix $ati$,  0.01% (1) where $a\in c_{\mu’}$.  0.01% (1) from the real  0.01% (1) locus of the  0.01% (1) these lines in  0.01% (1) a point? in  0.01% (1) particular, given a  0.01% (1) complete flag, i.e.  0.01% (1) all four of  0.01% (1) many lines intersect  0.01% (1) answer the question:  0.01% (1) given four lines  0.01% (1) $\ell_1,\ell_2,\ell_3,\ell_4\in \mathbb{c}\mathbb{p}^3$, how  0.01% (1) subspaces $$0=f_0\subset f_1\subset\cdots  0.01% (1) \subset f_m=\mathbb{c}^m$$ where  0.01% (1) the grassmannian $\mathrm{gr}^n(\mathbb{c}^m)$  0.01% (1) defined as $$\omega_{\lambda}(f_\bullet)=\{v\in  0.01% (1) \mathrm{gr}^n(\mathbb{c}^m)\mid \dim v\cap  0.01% (1) f_{n+i\lambda_i}\ge i.\}$$ here  0.01% (1) a subvariety of  0.01% (1) this flag is  0.01% (1) each $f_i$ has  0.01% (1) dimension $i$, the  0.01% (1) with respect to  0.01% (1) schubert varieties to  0.01% (1) we can use  0.01% (1) last post, we  0.01% (1) discussed an operation  0.01% (1) $\newcommand{\box}{\square} \omega$ on  0.01% (1) skew littlewoodrichardson young  0.01% (1) reply in the  0.01% (1) march 18, 2016  0.01% (1) a little monodromy  0.01% (1) operator called $\omega$.  0.01% (1)  3 replies  0.01% (1) tableaux with one  0.01% (1) marked inner corner,  0.01% (1) curves: relaxing a  0.01% (1) restriction recall from  0.01% (1) schubert calculus that  0.01% (1) in geometry. schubert  0.01% (1) this operator arises  0.01% (1) and shuffling. as  0.01% (1) a continuation, we’ll  0.01% (1) now discuss where  0.01% (1) $\lambda$ must fit  0.01% (1) in a $k\times  0.01% (1) $i$, and so  0.01% (1) all solutions is  0.01% (1) the intersection $$\omega_\box(f_\bullet^{(1)})\cap  0.01% (1) \omega_\box(f_\bullet^{(2)})\cap \omega_\box(f_\bullet^{(3)})\cap \omega_\box(f_\bullet^{(4)}).$$  0.01% (1) $\omega_\box(f_\bullet^{(i)})$ for each  0.01% (1) a line is  0.01% (1) line. the variety  0.01% (1) of all planes  0.01% (1) intersecting $p_i$ in  0.01% (1) and, as discussed  0.01% (1) in our post  0.01% (1) partitions involved have  0.01% (1) $4$, this intersection  0.01% (1) has dimension $0$.  0.01% (1) then tells us  0.01% (1) and the four  0.01% (1) $4$ in $\mathrm{gr}^2(\mathbb{c}^4)$  0.01% (1) on schubert calculus,  0.01% (1) since the $k\times(nk)$  0.01% (1) box has size  0.01% (1) $p_i$’s in a  0.01% (1) that intersects all  0.01% (1) the question into  0.01% (1) an intersection problem  0.01% (1) in $\mathrm{gr}^2(\mathbb{c}^4)$ with  0.01% (1) four general twodimensional  0.01% (1) we can translate  0.01% (1) the question above,  0.01% (1) (nk)$ box in  0.01% (1) schubert variety to  0.01% (1) be nonempty. in  0.01% (1) subspaces $p_1,p_2,p_3,p_4\subset \mathbb{c}^4$,  0.01% (1) and construct complete  0.01% (1) intersection condition is  0.01% (1) the problem of  0.01% (1) finding a plane  0.01% (1) $i=1,2,3,4$. then the  0.01% (1) $p_i$ for each  0.01% (1) flags $f_\bullet^{(1)},f_\bullet^{(2)},f_\bullet^{(3)},f_\bullet^{(4)}$ such  0.01% (1) that their second  0.01% (1) subspace $f^{(i)}_2$ is  0.01% (1) tableaux, and $k$theory  0.01% (1) so, what do  0.01% (1) if phase 1  0.01% (1) consists of all  0.01% (1) vertical slides and  0.01% (1) phase 2 is  0.01% (1) to itself is  0.01% (1) a tableau back  0.01% (1) permutation. it turns  0.01% (1) only way for  0.01% (1) $\omega$ to map  0.01% (1) all horizontal slides;  0.01% (1) then the final  0.01% (1) so $k=0$ in  0.01% (1) this case. since  0.01% (1) $\omega$, the monodromy  0.01% (1) real locus, is  0.01% (1) nonadjacent moves, and  0.01% (1) we have no  0.01% (1) shuffle step simply  0.01% (1) reverses these moves.  0.01% (1) this means that  0.01% (1) case when $\omega$  0.01% (1) let’s consider the  0.01% (1) making these the  0.01% (1) marked squares of  0.01% (1) a genomic tableau.  0.01% (1) $k$ is equal  0.01% (1) moved past, and  0.01% (1) $i$ that it  0.01% (1) the starting and  0.01% (1) ending positions of  0.01% (1) with the entry  0.01% (1) to the number  0.01% (1) of nonadjacent moves  0.01% (1) $s$ using our  0.01% (1) new algorithm. as  0.01% (1) one illuminating example,  0.01% (1) this connection allows  0.01% (1) $\omega$. geometric consequences  0.01% (1) in all phase  0.01% (1) 1’s of the  0.01% (1) local algorithm for  0.01% (1) the identity, we  0.01% (1) also know that  0.01% (1) curve $s$ in  0.01% (1) identity map, using  0.01% (1) the connection with  0.01% (1) $k$theory described above.  0.01% (1) the complex schubert  0.01% (1) entire structure of  0.01% (1) section 4.1 exercise  0.01% (1) 1.8). we have  0.01% (1) therefore determined the  0.01% (1) similar analyses lead  0.01% (1) to other geometric  0.01% (1) many complex connected  0.01% (1) components, for various  0.01% (1) values of $\alpha$,  0.01% (1) $\beta$, and $\gamma$.  0.01% (1) and can arbitrarily  0.01% (1) arbitrarily high genus,  0.01% (1) results: we have  0.01% (1) also shown that  0.01% (1) $s$ can have  0.01% (1) $\mathbb{cp}^1$ (see hartshorne,  0.01% (1) be isomorphic to  0.01% (1) euler characteristic $\chi(\mathcal{o}_s)=\dim  0.01% (1) h^0(\mathcal{o}_s)\dim h^1(\mathcal{o}_s)$. but  0.01% (1) $n$ in this  0.01% (1) case, and so  0.01% (1) which in turn  0.01% (1) components (see here),  0.01% (1) $n$, which is  0.01% (1) on the number  0.01% (1) of complex connected  0.01% (1) there must be  0.01% (1) $n$ complex connected  0.01% (1) know each of  0.01% (1) integral, and so  0.01% (1) they must each  0.01% (1) zero. we also  0.01% (1) of each of  0.01% (1) components, one for  0.01% (1) real connected components.  0.01% (1) $\dim h^1(\mathcal{o}_s)=0$, so  0.01% (1) of points in  0.01% (1) this zerodimensional intersection  0.01% (1) $c$ points in  0.01% (1) the intersection given  0.01% (1) by choosing the  0.01% (1) $(r+1)$st point to  0.01% (1) preimage of $(1:t)$  0.01% (1) points of $\mathbb{cp}^1$  0.01% (1) then define a  0.01% (1) map from this  0.01% (1) $s$ to the  0.01% (1) be $(1:t:t^2:t^3:\cdots:t^{n1})$. in  0.01% (1) this paper written  0.01% (1) to be real,  0.01% (1) then this can  0.01% (1) be extended to  0.01% (1) a map $s\to  0.01% (1) all $r+1$ points  0.01% (1) if we choose  0.01% (1) by my coauthor  0.01% (1) jake levinson, it  0.01% (1) is shown that  0.01% (1) points. we can  0.01% (1) sets of $c$  0.01% (1) singlebox partition $\lambda=(1)$.  0.01% (1) schubert variety $\omega_\box(f_\bullet)$  0.01% (1) with our schubert  0.01% (1) curve $s$ gives  0.01% (1) an $(r+1)$st point  0.01% (1) osculating flags, choose  0.01% (1) single box, $\omega_\box(f_\bullet)$.  0.01% (1) mathematical gemstones on  0.01% (1) choosing our $r$  0.01% (1) us a zerodimensional  0.01% (1) intersection, with the  0.01% (1) obtain a partition  0.01% (1) of an open  0.01% (1) schubert curve into  0.01% (1) of $t$, we  0.01% (1) varying the choice  0.01% (1) littlewoodrichardson coefficient $c:=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$  0.01% (1) where $b=((nk)^k)$ is  0.01% (1) box partition. by  0.01% (1) $s(\mathbb{r})$ is a  0.01% (1) smooth, finite covering  0.01% (1) operators. let’s be  0.01% (1) more precise, and  0.01% (1) consider the simplest  0.01% (1) interesting case, $r=3$.  0.01% (1) described by shuffling  0.01% (1) the covering is  0.01% (1) the fibers by  0.01% (1) these chains of  0.01% (1) tableaux, for which  0.01% (1) we have three  0.01% (1) and $\gamma$ such  0.01% (1) the respective osculating  0.01% (1) flags. then we  0.01% (1) can label the  0.01% (1) fiber $f^{1}(0)$ of  0.01% (1) $\infty$ to define  0.01% (1) $0$, $1$, and  0.01% (1) that $$\alpha+\beta+\gamma=k(nk)1=b1.$$ let’s  0.01% (1) choose, for simplicity,  0.01% (1) the three points  0.01% (1) way of labeling  0.01% (1) is a canonical  0.01% (1) filling the box  0.01% (1) $b$ with skew  0.01% (1) littlewoodrichardson tableaux with  0.01% (1) contents $\lambda^{(1)},\ldots,\lambda^{(r)},\box$ such  0.01% (1) this littlewoodrichardson coefficient  0.01% (1) $c=c^{b}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$. note that  0.01% (1) of $\mathbb{rp}^1$. the  0.01% (1) fibers of this  0.01% (1) map have size  0.01% (1) that each skew  0.01% (1) shape extends the  0.01% (1) order in which  0.01% (1) partitions. it turns  0.01% (1) out that there  0.01% (1) doesn’t matter what  0.01% (1) littlewoodrichardson coefficients, it  0.01% (1) previous shape outwards.  0.01% (1) in addition, by  0.01% (1) the symmetry of  0.01% (1) with another schubert  0.01% (1) intersection of $s$  0.01% (1) the iterated derivatives  0.01% (1) at chosen points  0.01% (1) normal curve, defined  0.01% (1) by the locus  0.01% (1) $f_\bullet$ defined by  0.01% (1) choose any flags  0.01% (1) box) whose sizes  0.01% (1) sum to $k(nk)1$.  0.01% (1) that one can  0.01% (1) of points of  0.01% (1) the form $$(1:t:t^2:t^3:\cdots:t^{n1})\in  0.01% (1) $f_k$ is the  0.01% (1) span of the  0.01% (1) first $k$ rows  0.01% (1) of the matrix  0.01% (1) whose $k$th subspace  0.01% (1) consider the flag  0.01% (1) \mathbb{cp}^n$$ (along with  0.01% (1) the limiting point  0.01% (1) $(0:0:\cdots:1)$.) in particular,  0.01% (1) $\lambda^{(1)},\ldots,\lambda^{(r)}$ (fitting inside  0.01% (1) of $r$ partitions  0.01% (1) the problem, so  0.01% (1) only intersecting three  0.01% (1) varieties above? in  0.01% (1) this case we  0.01% (1) the conditions in  0.01% (1) relax one of  0.01% (1) is the littlewoodrichardson  0.01% (1) coefficient $c_{\box,\box,\box,\box}^{(2,2)}$. what  0.01% (1) happens if we  0.01% (1) get a oneparameter  0.01% (1) family of solutions,  0.01% (1) choice of $r$  0.01% (1) flags $f_\bullet^{(1)},\ldots, f_\bullet^{(r)}$  0.01% (1) and a list  0.01% (1) a sufficiently “generic”  0.01% (1) general, we require  0.01% (1) which we call  0.01% (1) a schubert curve.  0.01% (1) to define a  0.01% (1) of iterated derivatives  0.01% (1) on this curve  0.01% (1) the schubert intersections  0.01% (1) to have the  0.01% (1) expected dimension. so,  0.01% (1) we pick some  0.01% (1) sufficiently general position  0.01% (1) they are in  0.01% (1) on the curve  0.01% (1) and choose their  0.01% (1) osculating flag, it  0.01% (1) number $r$ of  0.01% (1) these osculating flags,  0.01% (1) a covering map  0.01% (1) on these curves,  0.01% (1) we consider the  0.01% (1) schubert curve $s$.  0.01% (1) varieties defines a  0.01% (1) and choose exactly  0.01% (1) with sizes summing  0.01% (1) to $k(nk)1$. intersecting  0.01% (1) we pick a  0.01% (1) $t$, and if  0.01% (1) 1 & 2t  0.01% (1) & 3t^2 &  0.01% (1) t^{n2} \\ 0  0.01% (1) 2 & 6t  0.01% (1) \cdots & t^{n1}  0.01% (1) & t^3 &  0.01% (1) parameterized by $t$:  0.01% (1) $$\left(\begin{array}{cccccc} 1 &  0.01% (1) t & t^2  0.01% (1) \frac{(n1)!}{(n3)!} t^{n3} \\  0.01% (1) 6 & \cdots  0.01% (1) & (n1)! \end{array}\right)$$  0.01% (1) this is called  0.01% (1) the osculating flag  0.01% (1) 0 & \cdots  0.01% (1) & \vdots \\  0.01% (1) & \frac{(n1)!}{(n4)!} t^{n3}  0.01% (1) \\ \vdots &  0.01% (1) & \vdots &\ddots  0.01% (1) by simply filling  0.01% (1) correspondence is formed  0.01% (1) how it gives  0.01% (1) us a handle  0.01% (1) characteristic of schubert  0.01% (1) curves. recall from  0.01% (1) grassmannian works, and  0.01% (1) of how the  0.01% (1) paper. for now,  0.01% (1) we’ll just give  0.01% (1) a brief description  0.01% (1) this post that  0.01% (1) the cw complex  0.01% (1) rectangle, generate the  0.01% (1) cohomology ring $h^\ast(\mathrm{gr}(n,k))$.  0.01% (1) similarly, the $k$theory  0.01% (1) ring is a  0.01% (1) partition fitting inside  0.01% (1) $\lambda$ is a  0.01% (1) structure given by  0.01% (1) shows that the  0.01% (1) classes $[\omega_\lambda]$, where  0.01% (1) grassmannians, see buch’s  0.01% (1) here, and for  0.01% (1) in a sense  0.01% (1) we are groupifying  0.01% (1) monoids. the natural  0.01% (1) monoidal operation on  0.01% (1) for instance, $k(\mathbb{n})=\mathbb{z}$.  0.01% (1) form $[m]+[n][m+n]$. so,  0.01% (1) generated by elements  0.01% (1) $[m]$ for $m\in  0.01% (1) m$ by the  0.01% (1) vector spaces is  0.01% (1) $\oplus$, so if  0.01% (1) monoid. a good  0.01% (1) exposition on the  0.01% (1) basics of $k$theory  0.01% (1) ring of this  0.01% (1) is the grothendieck  0.01% (1) $x$ is a  0.01% (1) point, then all  0.01% (1) split and the  0.01% (1) filtered ring generated  0.01% (1) by the classes  0.01% (1) group homomorphism $\chi:k(x)\to  0.01% (1) \mathbb{z}$. to show  0.01% (1) this, it suffices  0.01% (1) if $0\to \mathcal{a}\to  0.01% (1) gives an (additive)  0.01% (1) euler characteristics. it  0.01% (1) characteristic the $k$theory  0.01% (1) ring is especially  0.01% (1) useful in computing  0.01% (1) \mathcal{b}\to \mathcal{c}\to 0$  0.01% (1) coherent sheaves on  0.01% (1) $$ \begin{array}{cccccc} &  0.01% (1) h^0(\mathcal{a}) & \to  0.01% (1) & h^0(\mathcal{b}) &  0.01% (1) \to & h^0(\mathcal{c})  0.01% (1) sequence in cohomology:  0.01% (1) a long exact  0.01% (1) $x$, then $\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})=0$.  0.01% (1) indeed, such a  0.01% (1) gives rise to  0.01% (1) and the euler  0.01% (1) $k$theory coefficients. $k$theory  0.01% (1) $\mathcal{o}_\lambda=\iota_\ast \mathcal{o}_{\omega_\lambda}$. multiplication  0.01% (1) of these basic  0.01% (1) classes is given  0.01% (1) by a variant  0.01% (1) $\iota:\omega_\lambda\to \mathrm{gr}(n,k)$, then  0.01% (1) the inclusion map  0.01% (1) of the coherent  0.01% (1) sheaves $[\mathcal{o}_{\lambda}]$ where  0.01% (1) if $\iota$ is  0.01% (1) of the littlewoodrichardson  0.01% (1) rule: $$[\mathcal{o}_\lambda]\cdot [\mathcal{o}_\nu]=\sum_\nu  0.01% (1) integer. we will  0.01% (1) refer to these  0.01% (1) nonnegative values as  0.01% (1) is a nonnegative  0.01% (1) $\nu\lambda+\mu$ then $c^{\nu}_{\lambda\mu}$  0.01% (1) (1)^{\nu\lambda\mu}c^\nu_{\lambda\mu}[\mathcal{o}_\nu]$$ where if  0.01% (1) $\nu=\lambda+\mu$ then $c^{\nu}_{\lambda\mu}$  0.01% (1) littlewoodrichardson coefficient. if  0.01% (1) taking the quotient  0.01% (1) group $k(m)$ by  0.01% (1) be defined in  0.01% (1) grassmannian. the $k$theory  0.01% (1) ring $k(\mathrm{gr}(n,k))$ the  0.01% (1) of a scheme  0.01% (1) which can alternatively  0.01% (1) the euler characteristic,  0.01% (1) so, to compute  0.01% (1) the genus it  0.01% (1) suffices to compute  0.01% (1) $x$ is defined  0.01% (1) as follows. first,  0.01% (1) (a.k.a. vector bundles)  0.01% (1) on $x$. then  0.01% (1) define $k(x)$, as  0.01% (1) a group, to  0.01% (1) free coherent sheaves  0.01% (1) classes of locally  0.01% (1) consider the free  0.01% (1) abelian group $g$  0.01% (1) generated by isomorphism  0.01% (1) the structure sheaf.  0.01% (1) and $\mathcal{o}_s$ is  0.01% (1) to approach some  0.01% (1) natural questions about  0.01% (1) the curve $s$.  0.01% (1) for instance, how  0.01% (1) can be used  0.01% (1) for computing $\omega$  0.01% (1) to $\mathbb{rp}^1$. now,  0.01% (1) we’ll show how  0.01% (1) our improved algorithm  0.01% (1) many (complex) connected  0.01% (1) components does $s$  0.01% (1) curve $s$ can  0.01% (1) $g=1\chi(\mathcal{o}_s)$ where $$\chi(\mathcal{o}_s)=\dim  0.01% (1) h^0(\mathcal{o}_s)\dim h^1(\mathcal{o}_s)$$ is  0.01% (1) of a connected  0.01% (1) (complex) curve? the  0.01% (1) have? what is  0.01% (1) its genus? is  0.01% (1) $s$ a smooth  0.01% (1) be the quotient  0.01% (1) of $g$ by  0.01% (1) well, a simpler  0.01% (1) example of $k$theory  0.01% (1) is the construction  0.01% (1) of the grothendieck  0.01% (1) construction come from?  0.01% (1) where does this  0.01% (1) restriction and consider  0.01% (1) coherent sheaves modulo  0.01% (1) short exact sequences.  0.01% (1) group of an  0.01% (1) abelian monoid. consider  0.01% (1) identity element, like  0.01% (1) a group without  0.01% (1) inverses). we can  0.01% (1) construct an associated  0.01% (1) operation and an  0.01% (1) an associative binary  0.01% (1) m (recall that  0.01% (1) a monoid is  0.01% (1) a set with  0.01% (1) the “locally free”  0.01% (1) if we remove  0.01% (1) \mathcal{e} \to \mathcal{e}_2  0.01% (1) sequence of vector  0.01% (1) bundles on $x$.  0.01% (1) this gives the  0.01% (1) $0\to \mathcal{e}_1 \to  0.01% (1) form $[\mathcal{e}_1]+[\mathcal{e}_2][\mathcal{e}]$ where  0.01% (1) “short exact sequences”,  0.01% (1) quotient $g/h$ where  0.01% (1) $h$ is the  0.01% (1) additive structure on  0.01% (1) $k(x)$, and the  0.01% (1) smooth (such as  0.01% (1) a grassmannian!) then  0.01% (1) exact same ring  0.01% (1) that $x$ is  0.01% (1) a ring. it  0.01% (1) tensor product operation  0.01% (1) on vector bundles  0.01% (1) makes it into  0.01% (1) the map $f:s\to  0.01% (1) \\ \to &h^1(\mathcal{a})  0.01% (1) and two marked  0.01% (1) squares $\square_1$ and  0.01% (1) $\square_2$ in $t$  0.01% (1) is a genomic  0.01% (1) a tableau $t$  0.01% (1) the data of  0.01% (1) yong. in our  0.01% (1) case, the genomic  0.01% (1) $k$ can be  0.01% (1) tableau if: the  0.01% (1) marked squares are  0.01% (1) either $\square_1$ or  0.01% (1) $\square_2$, every suffix  0.01% (1) ballot (has more  0.01% (1) $j$’s than $j+1$’s  0.01% (1) word of $t$,  0.01% (1) in the reading  0.01% (1) nonadjacent and contain  0.01% (1) $i$, there are  0.01% (1) no $i$’s between  0.01% (1) by pechenik and  0.01% (1) were first introduced  0.01% (1) $k$ in terms  0.01% (1) of $\omega$ the  0.01% (1) fascinating thing about  0.01% (1) our algorithm for  0.01% (1) have $g=1\chi(\mathcal{o}_s)=k(n1)$. computing  0.01% (1) is connected, we  0.01% (1) that $$\chi(\mathcal{o}_s)=nk.$$ going  0.01% (1) genus, we see  0.01% (1) that if $s$  0.01% (1) $\omega$ is that  0.01% (1) certain steps of  0.01% (1) of $s$. these  0.01% (1) tableaux are called  0.01% (1) “genomic tableaux”, and  0.01% (1) about the genus  0.01% (1) giving us information  0.01% (1) the algorithm combinatorially  0.01% (1) correspond to certain  0.01% (1) the $k$theory coefficients,  0.01% (1) for all $j$).  0.01% (1) for instance, consider  0.01% (1) i discovered that  0.01% (1) the steps in  0.01% (1) phase 1 in  0.01% (1) which the special  0.01% (1) i. jake and  0.01% (1) described in part  0.01% (1) properties, and so  0.01% (1) it is genomic.  0.01% (1) finally, consider the  0.01% (1) move to an  0.01% (1) adjacent square are  0.01% (1) $\beta$ (where the  0.01% (1) marked squares only  0.01% (1) add $1$ to  0.01% (1) of $i$’s). the  0.01% (1) $\gamma^c/\alpha$ and content  0.01% (1) total skew shape  0.01% (1) in bijective correspondence  0.01% (1) with the $k$theory  0.01% (1) tableau of the  0.01% (1) satisfies all three  0.01% (1) third tableau above  0.01% (1) is not genomic:  0.01% (1) the marked squares,  0.01% (1) while they do  0.01% (1) entry, are adjacent  0.01% (1) the fourth tableau  0.01% (1) 1 means that  0.01% (1) the following tableaux  0.01% (1) with two marked  0.01% (1) (shaded) squares: property  0.01% (1) squares. the first  0.01% (1) property 2, because  0.01% (1) $1$ then the  0.01% (1) suffix $221$ is  0.01% (1) not ballot. the  0.01% (1) the top marked  0.01% (1) property 3, because  0.01% (1) $2$ between the  0.01% (1) two marked $2$’s  0.01% (1) finally, the second  0.01% (1) h^1(\mathcal{b}) & \to  0.01% (1) $1$. it follows  0.01% (1) the pushforward $j_\ast\mathcal{o}_s$  0.01% (1) $\chi(\mathcal{o}_s)$ itself. we  0.01% (1) can now compute  0.01% (1) $\chi(j_\ast\mathcal{o}_s)$ using the  0.01% (1) map $j:s\to x$,  0.01% (1) say with inclusion  0.01% (1) situation, we have  0.01% (1) a closed subset  0.01% (1) $s$ of $x=\mathrm{gr}(n,k)$,  0.01% (1) $k$theory ring of  0.01% (1) the grassmannian. indeed,  0.01% (1) i). so in  0.01% (1) the $k$theory ring,  0.01% (1) if we identify  0.01% (1) structure sheaves of  0.01% (1) $\gamma$ (see part  0.01% (1) the three partitions  0.01% (1) $s$ is the  0.01% (1) intersection of schubert  0.01% (1) varieties indexed by  0.01% (1) fact, in our  0.01% (1) in $k(x)$. in  0.01% (1) & & \end{array}  0.01% (1) $$ and the  0.01% (1) alternating sum of  0.01% (1) the dimensions of  0.01% (1) & & &  0.01% (1) h^2(\mathcal{c}) \\ \cdots  0.01% (1) & h^1(\mathcal{c}) \\  0.01% (1) \to &h^2(\mathcal{a}) &  0.01% (1) \to & h^2(\mathcal{b})  0.01% (1) any exact sequence  0.01% (1) must be zero.  0.01% (1) it makes sense  0.01% (1) to talk about  0.01% (1) of a class  0.01% (1) as desired. therefore,  0.01% (1) h^i(\mathcal{c}) \\ &=&\chi(\mathcal{a})+\chi(\mathcal{c})\chi(\mathcal{b})\end{eqnarray*}$$  0.01% (1) thus we have  0.01% (1) $$\begin{eqnarray*}0&=&\sum_i (1)^i\dim h^i(\mathcal{a})\sum_i  0.01% (1) (1)^i\dim h^i(\mathcal{b})+\sum_i (1)^i\dim  0.01% (1) have euler characteristic  0.01% (1) closed subvarieties with  0.01% (1) and counts exactly  0.01% (1) the size of  0.01% (1) the fibers (the  0.01% (1) set $\omega$ acts  0.01% (1) usual littlewoodrichardson coefficient,  0.01% (1) $c^{\rho’}_{\alpha,\beta,\gamma}$ is the  0.01% (1) an integer determined  0.01% (1) by the $k$theory  0.01% (1) coefficients. notice that  0.01% (1) on) in our  0.01% (1) map from part  0.01% (1) and a copy  0.01% (1) of $\mathbb{p}^1$ respectively,  0.01% (1) and so both  0.01% (1) are a point  0.01% (1) $\omega_\rho$ and $\omega_{\rho’}$  0.01% (1) ii. let’s call  0.01% (1) this number $n$.  0.01% (1) finally, notice that  0.01% (1) where $k$ is  0.01% (1) we have: $$[\mathcal{o}_s]=c^{\rho’}_{\alpha,\beta,\gamma}[\mathcal{o}_{\rho’}]k[\mathcal{o}_{\rho}]$$  0.01% (1) product expands as  0.01% (1) a sum of  0.01% (1) integer multiples of  0.01% (1) classes $[\mathcal{o}_\nu]$ where  0.01% (1) described above, this  0.01% (1) $k$theoretic littlewoodrichardson rule  0.01% (1) inclusion maps, we  0.01% (1) have $$[\mathcal{o}_s]=[\mathcal{o}_\alpha]\cdot [\mathcal{o}_\beta]\cdot  0.01% (1) [\mathcal{o}_\gamma].$$ by the  0.01% (1) their pushforwards under  0.01% (1) $\nu\ge \alpha+\beta+\gamma$. but  0.01% (1) the rectangle minus  0.01% (1) (call this $\rho’$).  0.01% (1) in other words,  0.01% (1) or it is  0.01% (1) (call this $\rho$)  0.01% (1) we have $\alpha+\beta+\gamma=k(nk)1$,  0.01% (1) in our setup  0.01% (1) so $\nu$ is  0.01% (1) either the entire  0.01% (1) by wordpress  0% (0) 