In order to generalize Grassmannians to other Lie types, we first need to understand in what sense the ordinary Grassmannian is type A. Recall from this post that the complete flag variety can be written as a quotient of $G=\mathrm{GL}_n$ by a Borel subgroup $B$, such as the group of upper-triangular matrices. It turns out that all *partial flag varieties*, the varieties of partial flags of certain degrees, can be similarly defined as a quotient $$G/P$$ for a parabolic subgroup $P$, namely a closed intermediate subgroup $B\subset P\subset G$.

The (ordinary) Grassmannian $\mathrm{Gr}(n,k)$, then, can be thought of as the quotient of $\mathrm{GL}_n$ by the parabolic subgroup $S=\mathrm{Stab}(V)$ where $V$ is any fixed $k$-dimensional subspace of $\mathbb{C}^n$. Similarly, we can start with a different reductive group, say the special orthogonal group $\mathrm{SO}_{2n+1}$, and quotient by parabolic subgroups to get partial flag varieties in other Lie types.

In particular, the **orthogonal Grassmannian** $\mathrm{OG}(2n+1,k)$ is the quotient $\mathrm{SO}_{2n+1}/P$ where $P$ is the stabilizer of a fixed **isotropic** $k$-dimensional subspace $V$. The term *isotropic* means that $V$ satisfies $\langle v,w\rangle=0$ for all $v,w\in V$ with respect to a chosen symmetric bilinear form $\langle,\rangle$.

The isotropic condition, at first glance, seems very unnatural. After all, how could a nonzero subspace possibly be so orthogonal to itself? Well, it is first important to note that we are working over $\mathbb{C}$, not $\mathbb{R}$, and the bilinear form is symmetric, not conjugate-symmetric. So for instance, if we choose a basis of $\mathbb{C}^{2n+1}$ and define the bilinear form to be the usual dot product $$\langle (a_1,\ldots,a_{2n+1}),(b_1,\ldots,b_{2n+1})\rangle=a_1b_1+a_2b_2+\cdots+a_{2n+1}b_{2n+1},$$ then the vector $(3,5i,4)$ is orthogonal to itself: $3\cdot 3+5i\cdot 5i+4\cdot 4=0$.

While the choice of symmetric bilinear form does not change the fundamental geometry of the orthogonal Grassmannian, one choice in particular makes things easier to work with in practice: the “reverse dot product” given by

$$\langle (a_1,\ldots,a_{2n+1}),(b_1,\ldots,b_{2n+1})\rangle=\sum_{i=1}^{2n+1} a_ib_{2n+1-i}.$$ In particular, with respect to this symmetric form, the “standard” complete flag $\mathcal{F}$, in which $\mathcal{F}_i$ is the span of the first $i$ rows of the identity matrix $I_{2n+1}$, is an **orthogonal flag**, with $\mathcal{F}_i^\perp=\mathcal{F}_{2n+1-i}$ for all $i$. Orthogonal flags are precisely the type of flags that are used to define Schubert varieties in the orthogonal grassmannian.

Other useful variants of the reverse dot product involve certain factorial coefficients, but for this post this simpler version will do.

Going back to the main point, note that isotropic subspaces are sent to other isotropic subspaces under the action of the orthorgonal group: if $\langle v,w\rangle=0$ then $\langle Av,Aw\rangle=\langle v,w\rangle=0$ for any $A\in \mathrm{SO}_{2n+1}$. Thus orthogonal Grassmannian $\mathrm{OG}(2n+1,k)$, which is the quotient $\mathrm{SO}_{2n+1}/\mathrm{Stab}(V)$, can be interpreted as the variety of all $k$-dimensional isotropic subspaces of $\mathbb{C}^{2n+1}$.

Just as in the ordinary Grassmannian, there is a Schubert cell decomposition for the orthogonal Grassmannian, and the combinatorics of Schubert varieties is particularly nice in the case of $\mathrm{OG}(2n+1,n)$ in which the orthogonal subspaces are “half dimension” $n$. (In particular, this corresponds to the “cominuscule” type in which the simple root associated to our maximal parabolic subgroup is the special root in type $B$. See the introduction here or this book for more details.)

Recall that in $\mathrm{Gr}(2n+1,n)$, the Schubert varieties are indexed by partitions $\lambda$ whose Young diagram fit inside an $n\times (n+1)$ rectangle. Suppose we divide this rectangle into two staircases as shown below using the red cut, and only consider the partitions $\lambda$ that are symmetric with respect to the reflective map taking the upper staircase to the lower.

We claim that the Schubert varieties of the orthogonal Grassmannian are indexed by the “shifted partitions” formed by ignoring the lower half of these symmetric partition diagrams. In fact, the Schubert varieties consist of the isotropic elements of the ordinary Schubert varieties, giving a natural embedding $\mathrm{OG}(2n+1,n)\to \mathrm{Gr}(2n+1,n)$ that respects the Schubert decompositions.

To get a sense of how this works, let’s look at the example of the partition $(4,3,1)$ shown above, in the case $n=4$. As described in this post, in the ordinary Grassmannian, the Schubert cell $\Omega_\lambda^{\circ}$ with respect to the standard flag is given by the set of vector spaces spanned by the rows of matrices whose reduced row echelon form looks like:

Combinatorially, this is formed by drawing the staircase pattern shown at the left, then adding the partition parts $(4,3,1)$ to it, and placing a $1$ at the end of each of the resulting rows for the reduced row echelon form.

Now, which of these spaces are isotropic? In other words, what is $\Omega_\lambda^{\circ}\cap \mathrm{OG}(2n+1,n)$? Well, suppose we label the starred entries as shown, where we have ommited the $0$’s to make the entries more readable:

Then I claim that the entries $l,j,k,h,i,e$ are all uniquely determined by the values of the remaining variables $a,b,c,d,f,g$. Thus there is one isotropic subspace in this cell for each choice of values $a,b,c,d,f,g$, corresponding to the “lower half” of the partition diagram we started with:

Indeed, let the rows of the matrix be labeled $\mathbf{1},\mathbf{2},\mathbf{3},\mathbf{4}$ from top to bottom as shown, and suppose its row span is isotropic. Since row $\mathbf{1}$ and $\mathbf{4}$ are orthogonal with respect to the reverse dot product, we get the relation $$l+a=0,$$ which expresses $l=-a$ in terms of $a$.

Now, rows $\mathbf{2}$ and $\mathbf{4}$ are also orthogonal, which means that $$b+k=0,$$ so we can similarly eliminate $k$. From rows $\mathbf{2}$ and $\mathbf{3}$, we obtain $f+j=0$, which expresses $j$ in terms of the lower variables. We then pair row $\mathbf{3}$ with itself to see that $h+g^2=0$, eliminating $h$, and finally pairing $\mathbf{3}$ with $\mathbf{4}$ we have $i+gc+d=0$, so $i$ is now expressed in terms of lower variables as well.

Moreover, these are the only relations we get from the isotropic condition – any other pairings of rows give the trivial relation $0=0$. So in this case the Schubert variety restricted to the orthogonal Grassmannian has half the dimension of the original, generated by the possible values for $a,b,c,d,f,g$.

Why does this elimination process work in general, for a symmetric shape $\lambda$? Label the steps of the boundary path of $\lambda$ by $1,2,3,\ldots$ from SW to NE in the lower left half, and label them from NE to SW in the upper right half, as shown:

Then the labels on the vertical steps in the lower left half give the column indices of the $1$’s in the corresponding rows of the matrix. The labels on the horizontal steps in the upper half, which match these labels by symmetry, give the column indices *from the right* of the corresponding starred columns from right to left.

This means that the $1$’s in the lower left of the matrix correspond to the opposite columns of those containing letters in the upper right half. It follows that we can use the orthogonality relations to pair a $1$ (which is leftmost in its row) with a column entry in a higher or equal row so as to express that entry in terms of other letters to its lower left. The $1$ is in a lower or equal row in these pairings precisely for the entries whose corresponding square lies above the staircase cut. Thus we can always express the upper right variables in terms of the lower left, as in our example above.

We can now conversely show that if a point of the Grassmannian is isotropic, then its corresponding partition is symmetric about the staircase cut. Indeed, we claim that if the partition is not symmetric, then two of the $1$’s of the matrix will be in opposite columns, resulting in a nonzero reverse dot product between these two rows.

To show this, first note that we cannot have a $1$ in the middle column, for otherwise its row would not be orthogonal to itself. Therefore, of the remaining $2n$ columns, some $n$ of them have a $1$ in them, and the complementary columns must be the other $n$, the ones without a $1$. But this is the same combinatorial data as choosing some $k$ columns from the first $n$ columns, and the remaining columns being forced to be a pivot column if their opposite is not and vice versa. This implies that the partition is symmetric.

From the above analysis, it follows that the orthogonal Grassmannian is partitioned into Schubert cells given by the data of a **shifted partition**, the upper half of a symmetric partition diagram:

Such a partition is denoted by its distinct “parts”, the number of squares in each row, thought of as the parts of a partition with distinct parts shifted over by the staircase. For instance, the shifted partition above is denoted $(3,1)$.

The beauty of shifted partitions is that so much of the original tableaux combinatorics that goes into ordinary Schubert calculus works almost the same way for shifted tableaux and the orthogonal Grassmannian. We can define jeu de taquin, Knuth equivalence, and dual equivalence on shifted tableaux, there is a natural notion of a Littlewood-Richardson shifted tableau, and these notions give rise to formulas for both intersections of Schubert varieties in the orthogonal Grassmannian and to products of Schur $P$-functions or Schur $Q$-functions, both of which are specializations of the Hall-Littlewood polynomials.

I’ll likely go further into the combinatorics of shifted tableaux in future posts, but for now, I hope you enjoyed the breakdown of why shifted partitions are the natural indexing objects for the Schubert decomposition of the orthogonal Grassmannian.

]]>While tutoring a high school student recently, we worked through a problem that was essentially showing just that in a special case. Remarkably:

An ellipse can be defined as the locus of points $P$ for which the distance $|PF|$ to a focus $F$ is $\alpha$ times the distance from $P$ to a fixed line $\ell$ for some positive real number $\alpha\lt 1$.

For instance, if we make $\alpha=1/2$ we would get an ellipse, whereas if $\alpha=1$ then it’s such a large ellipse that it’s a parabola!

Let’s check the math here. Recall that an ellipse is usually defined, synthetically, as the locus of points $P$ in the plane for which the sum of the distances $|PF_1|+|PF_2|$ for two fixed foci $F_1$ and $F_2$ is a fixed constant $s$. By translating this condition into coordinates, one can show that if we place an ellipse with its two lines of symmetry aligned with the $x$- and $y$-axes (with the wider part on the $x$-axis), centered at the origin, then the ellipse will have equation $$(x/b)^2+(y/a)^2=1$$ for some positive real numbers $a$ and $b$ with $a\le b$.

For an ellipse $E$ with this equation, the focii $F_1$ and $F_2$ have coordinates $(-c,0)$ and $(c,0)$ for some $c$. To find $c$ in terms of $a$ and $b$, we have that the sum of the distances of the point $(b,0)$ to the foci is $b-c+b-c=2b$, and the sum of the distances of the point $(0,a)$ to the foci is $2\sqrt{a^2+c^2}$ by the Pythagorean theorem. Therefore $2b=2\sqrt{a^2+c^2}$, and solving we find $c=\sqrt{b^2-a^2}$.

Now, I claim that if we set $\alpha=\frac{\sqrt{b^2-a^2}}{b}$ and let $\ell$ be the vertical line $x=b/\alpha$, then $E$ is the locus of points $P$ for which $|PF_2|$ is $\alpha$ times the distance from $P$ to $\ell$. Indeed, let $P=(x,y)$ be a point that has this property. Then the distance $|PF_2|$ is $\sqrt{(x-\sqrt{b^2-a^2})^2+y^2}$ and the distance from $P$ to $\ell$ is $\frac{b}{\alpha}-x$, so we have

$$\begin{align*}

\sqrt{(x-\sqrt{b^2-a^2})^2+y^2} &= \alpha\left(\frac{b}{\alpha}-x\right) \\

(x-\sqrt{b^2-a^2})^2+y^2 &= (b-\alpha x)^2 \\

x^2-2x\sqrt{b^2-a^2}+b^2-a^2+y^2 &= b^2-2x \sqrt{b^2-a^2} + (\alpha x)^2 \\

x^2-a^2+y^2 &= \frac{b^2-a^2}{b^2}x^2 \\

\frac{a^2}{b^2}x^2 + y^2 &= a^2 \\

\frac{x^2}{b^2} + \frac{y^2}{a^2} &= 1

\end{align*}$$

which is indeed the equation of $E$.

A few noteworthy observations: first, it’s remarkable that the focus of the ellipse as defined in terms of the constant-sum-from-two-focii definition coincides with the focus that appears in the focus-and-directrix version. This makes one wonder if there is some natural relationship between the focii and this new directrix $\ell$, perhaps in terms of *reciprocation* (see this post for the definition of reciprocation in a circle.) And indeed, if we apply the transformation $(x,y)\mapsto (x/b,y/a)$, which maps our ellipse $E$ to the unit circle, the point $F_2$ maps to the point $(\alpha,0)$ and $\ell$ becomes the line $x=1/\alpha$, so indeed $F_2$ and $\ell$ form a reciprocal pair!

Second, consider the degenerate case of a circle, when $a=b$ and so $\alpha=0$. The condition $\alpha=0$ doesn’t really make sense unless we interpret the diagram in the projective plane and allow the directrix to be the line at infinity, which again makes the focus (the center of the circle) be the polar of this line.

Finally, consider the limit as $b$ approaches $\infty$, so that the ellipse is stretching out further and further until it becomes a parabola. (Exercise for the reader: What projective transformation maps an ellipse into a parabola?) In this limit we have $$\lim_{b\to \infty} \alpha = \lim_{b\to \infty} \frac{\sqrt{b^2-a^2}}{b}=\sqrt{\lim_{b\to \infty}1-(a^2/b^2)}=1,$$ and again we recover the case of a parabola. As a bonus, we find that the focus and directrix of a parabola must be reciprocal to each other across the parabola as well.

That’s all for today – a bit of fun with conics to kick off the new year. Happy and indivisible 2017!

]]>If you’re reading this blog, I’m guessing you’re probably already not too afraid of mathematics. But I hope you share this post with people who are somewhat spooked by it but like to face their fears now and then. And let’s face it, even for math lovers, every difficult-sounding math problem is always a little scary at first… until you work it out and realize that there’s only beauty behind the mask.

I recently made up the following problem for a friend teaching a discrete mathematics class:

Five kids, dressed as a Ghost, a Witch, a Monster, a Skeleton, and a Black Cat, knock at your door. You open it and welcome them in, but you realize you only have $3$ Snickers bars and $3$ Kit Kats left in your candy stash!

Since you have $6$ pieces of candy and there are only $5$ kids, you decide to give both a Kit Kat and a Snickers bar to the scariest costume, and then give the remaining four kids one piece each. How many different ways can you choose who to give what candy to?

Eeek! Ghosts and witches! And combinatorics!

Well, let’s consider. There are $5$ ways you can decide on the scariest costume, since there are $5$ kids to choose from. But for each of those choices, you also have to pick which two of the remaining four get the Kit Kat and which two get the Snickers.

To make this a little easier, suppose we picked the Ghost as the scariest, and so we have to choose which two of the Witch, Monster, Skeleton, and Cat get the Snickers bars (and then the other two will get the Kit Kat). Well, we could pick one of the following options:

1. Witch and Monster

2. Witch and Skeleton

3. Witch and Cat

4. Monster and Skeleton

5. Monster and Cat

6. Skeleton and Cat.

(Notice that I ordered these by first choosing all those that can pair with the Witch, and then all those without the Witch, to make sure I didn’t miss any.)

So there are $6$ ways to choose who gets the Snickers and who gets the Kit Kats, assuming we chose Ghost as Scariest. But this is the same computation no matter who of the $5$ we chose as Scariest. If we chose the Witch as scariest there would still be $6$ possibilities, and same for the other three. Therefore, there are $5\cdot 6=30$ total possibilities.

The nice thing here is that there’s a known formula for computing the $6$ possibilities for the Kit Kats and Snickers – the number of ways of choosing $2$ things out of $4$ things is written $\binom{4}{2}$, pronounced “four choose two”. The formula for $\binom{a}{b}$, the number of ways to choose $b$ things from $a$ things, is known to be $\frac{a!}{b!\cdot (a-b)!}$ where $n!$ means the product of all the numbers from $1$ to $n$. So to compute $\binom{4}{2}$, we just compute $$\frac{4!}{2!\cdot 2!}=\frac{1\cdot 2 \cdot 3 \cdot 4}{1\cdot 2 \cdot 1\cdot 2}=6.$$ This is a shortcut for counting the possibilities without listing them all out.

Can you see why the formula for computing $\binom{a}{b}$ is true?

]]>The Garsia-Procesi 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 square-free 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 $$I_{\mu}=(e_r(S) : k-d_k(\mu) \lt r \le k, |S|=k).$$ Here, $d_k(\mu)=\mu’_n+\mu’_{n-1}+\cdots+ \mu’_{n-k+1}$ 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 $1-0\lt r \le 1$, 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 $2-1\lt r\le 2$. 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 $3-2 \lt r\le 3$. 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 $4-4\lt r\le 4$. 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 Hall-Littlewood 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 $A-tI$, 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 $A-tI$, 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 $r \gt k-d_k(\mu)$, 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.

]]>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 $2^{n-1}$ to obtain:

$$2^{n-1}\cdot F_n=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k+1}\cdot 5^k.$$

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 $n-2$ 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 $2^{n-1}$ counts the number of length-$(n-1)$ sequences of $0$’s and $1$’s. Thus, the left hand side of our identity, $2^{n-1}\cdot F_n$, counts the number of ways of choosing a binary sequence of length $n-1$ and also a fence post coloring of length $n-2$. 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!

]]>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 science-related endeavors (thanks to my loving and very mathematical family!), I was a confident and excited 18-year 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.

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, wide-eyed, 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 18-year-old Little-Miss-Perfect 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.)

]]>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 *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** $K(X)$ 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 $k\times (n-k)$ 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 Littlewood-Richardson 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 Littlewood-Richardson coefficient. If $|\nu|<|\lambda|+|\mu|$ then the coefficient is zero, and if it $|\nu|>|\lambda|+|\mu|$ then $c^{\nu}_{\lambda\mu}$ is a nonnegative integer. We will refer to these nonnegative values as *$K$-theory coefficients*.

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 Littlewood-Richardson 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 $|\alpha|+|\beta|+|\gamma|=k(n-k)-1$, so $\nu$ is either the entire $k\times (n-k)$ 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 Littlewood-Richardson 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 $$\chi(\mathcal{O}_S)=N-k.$$

Going back to the genus, we see that if $S$ is connected, we have $g=1-\chi(\mathcal{O}_S)=k-(N-1)$.

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 non-adjacent 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 non-adjacent moves in all Phase 1’s of the local algorithm for $\omega$.

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 non-adjacent 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 $\omega$.

]]>\omega$ on skew Littlewood-Richardson 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.

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 $k\times (n-k)$ 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 *two-dimensional 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 $k\times(n-k)$ 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 Littlewood-Richardson rule then tells us that the number of points in this zero-dimensional intersection is the Littlewood-Richardson 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 one-parameter 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 $k\times (n-k)$ box) whose sizes sum to $k(n-k)-1$. 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 $$(1:t:t^2:t^3:\cdots:t^{n-1})\in \mathbb{CP}^n$$ (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^{n-1} \\

0 & 1 & 2t & 3t^2 & \cdots & (n-1) t^{n-2} \\

0 & 0 & 2 & 6t & \cdots & \frac{(n-1)!}{(n-3)!} t^{n-3} \\

0 & 0 & 0 & 6 & \cdots & \frac{(n-1)!}{(n-4)!} t^{n-3} \\

\vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\

0 & 0 & 0 & 0 & \cdots & (n-1)!

\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 $k(n-k)-1$. Intersecting the resulting Schubert varieties defines a Schubert curve $S$.

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 single-box partition $\lambda=(1)$. Intersecting the resulting Schubert variety $\Omega_\box(F_\bullet)$ with our Schubert curve $S$ gives us a zero-dimensional intersection, with the number of points given by the Littlewood-Richardson coefficient $c:=c^{B}_{\lambda^{(1)},\ldots,\lambda^{(r)},\box}$ where $B=((n-k)^k)$ is the $k\times (n-k)$ 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 $(1:t:t^2:t^3:\cdots:t^{n-1})$.

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 Littlewood-Richardson coefficient counts the number of ways of filling the box $B$ with skew Littlewood-Richardson 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 Littlewood-Richardson 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 $$|\alpha|+|\beta|+|\gamma|=k(n-k)-1=|B|-1.$$ 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 Littlewood-Richardson tableaux of contents $\alpha$, $\box$, $\beta$, and $\gamma$ in that order. Note that there is only one Littlewood-Richardson tableau of straight shape and content $\alpha$, and similarly for the anti-straight shape $\gamma$, so the data here consists of a skew Littlewood-Richardson 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 “evacuation-shuffle”, 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!

]]>So let’s start in the middle.

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 **Littlewood-Richardson**, 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 Littlewood-Richardson 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 Littlewood-Richardson 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. Littlewood-Richardson tableaux are key to the Littlewood-Richardson rule, which allows us to efficiently compute products of Schur functions.

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 Littlewood-Richardson 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 $\times$.**Un-rectification:**Treat $\times$ as having value $\infty$ and un-rectify, 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 Littlewood-Richardson 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 “evacuation-shuffle”, 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.

]]>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 $n-1$ remaining choices for the second, then $n-2$ for the third, and so on. So there are $n!=n\cdot (n-1)\cdot \cdots \cdot 2\cdot 1$ 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 $1+q+q^2+\cdots+q^{n-1}$ to the total. The next entry then can be any of the $n-1$ remaining entries, and since the first entry cannot be the smaller entry of an inversion with the second, this choice contributes a factor of $1+q+q^2+\cdots+q^{n-2}$ 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!

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