In a previous post, we discussed Schubert calculus in the Grassmannian and the various intersection problems it can solve. I have recently been thinking about problems involving the type B variant of Schubert calculus, namely, intersections in the orthogonal Grassmanian. So, it’s time for a combinatorial introduction to the orthogonal Grassmannian!

What is the orthogonal Grassmannian?

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 $\mathbb{C}^{2n+1}$.

Schubert varieties and row reduction in $\mathrm{OG}(2n+1,n)$

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 $a,b,c,d,f,g$.

General elimination argument

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.

Isotropic implies symmetric

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.

Coming soon: Combinatorics of shifted partitions and tableaux

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.

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 $k\times (n-k)$ 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.

$K$-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 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 $g=1-\chi(\mathcal{O}_S)=k-(N-1)$.

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:

1. The marked squares are non-adjacent and contain the same entry $i$,
2. There are no $i$’s between $\square_1$ and $\square_2$ in the reading word of $T$,
3. 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 $\omega$.

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 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 $\omega$.

In the last post, we discussed an operation $\newcommand{\box}{\square}
\omega$ 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.

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 $k\times (n-k)$ 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 $S$.

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 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!