Minuscule Schubert Calculus and the Geometric Satake Correspondence

Abstract

We describe a relationship between work of Gatto, Laksov, and their collaborators on realizations of (generalized) Schubert calculus of Grassmannians, and the geometric Satake correspondence of Lusztig, Ginzburg, and Mirković and Vilonen. Along the way we obtain new proofs of equivariant Giambelli formulas for the ordinary and orthogonal Grassmannians, as well as a simple derivation of the “rim-hook” rule for computing in the equivariant quantum cohomology of the Grassmannian.

DA was partially supported by a postdoctoral fellowship from the Instituto Nacional de Matemática Pura e Aplicada (IMPA) and by NSF Grant DMS-1502201.

Appendix: Pfaffians and Spinors

In this appendix, we prove a change-of-basis formula in the spin representation, where Pfaffians play the role analogous to determinants in the exterior algebra. This refines similar formulas of Chevalley and Manivel [10, 37].

First, for any complex vector space V with symmetric bilinear form $\langle \;,\;\rangle$, the associated Clifford algebra is the quotient of the tensor algebra by two-sided ideal forcing the relation $v\cdot w + v\cdot w = \langle v,w\rangle 1$ for all vectors $v,w\in V$:

$$ Cl(V) := T^\bullet (V) / ( v\otimes w + w\otimes v - \langle v,w\rangle 1). $$

(Note that this definition differs slightly from the standard one, where $\langle v,w \rangle 1$ would be replaced by $2\langle v,w \rangle 1$—but it is equivalent, up to rescaling the form $\langle \;,\; \rangle$, since we are not in characteristic 2.) If the bilinear form is zero, this is just the exterior algebra: $Cl(V) \cong \textstyle \bigwedge ^\bullet V$. In general, $\dim Cl(V) = 2^{\dim V}$, with a basis consisting of products $v_I=v_{i_1}\cdots v_{i_k}$ of distinct basis elements of V. (One can prove this by degenerating to the exterior algebra.)

We note the following general formulas in Cl(V), which are immediate from the defining relations. First, if x and y are orthogonal vectors in V (i.e., $\langle x,y \rangle = 0$), then

$$\begin{aligned} x\cdot y&= -y\cdot x \end{aligned}$$

(A.1)

in Cl(V). Next, suppose $x = x_1\cdots x_r \in Cl(V)$ is a product of vectors in V such that $X={{\,\mathrm{span}\,}}\{x_1,\ldots ,x_r\} \subseteq V$ is isotropic. Then for any vector $y\in X$,

$$\begin{aligned} y\cdot x = 0 \end{aligned}$$

(A.2)

in Cl(V).

From now on, we assume $\dim V=2n$ and its bilinear form is nondegenerate. Let $y_{\overline{n-1}},\ldots ,y_{\overline{0}},y_0,\ldots ,y_{{n-1}}$ be an “orthogonal basis”, meaning $\langle y_{\overline{\imath }}, y_j \rangle = \delta _{ij}$. (We interpret the bar as a negative sign, so $\overline{\overline{\imath }}=i$.) Let $y=y_0\cdots y_{n-1}$. The spin representation is the left ideal

$$ \mathbb {S}= Cl(V)\cdot y. $$

A (pure) spinor is an element of the form $z\cdot y \in \mathbb {S}$, where $z = z_1\cdots z_n$ is a product of vectors $z_1,\ldots ,z_n \in V$ which span a maximal isotropic subspace of V.

Let $[m] = \{0,\ldots ,m-1\}$ for any integer $m\ge 1$. For a subset $I=\{i_1<\cdots <i_r\} \subseteq [n]$, let $I' = \{i'_1<\cdots <i'_{n-r}\} = [n]\smallsetminus I$. Given such an I, the corresponding standard spinor is

$$ y_I := y_{\overline{\imath }'_1} \cdots y_{\overline{\imath }'_{n-r}}\cdot y. $$

These form a basis for $\mathbb {S}$, as I ranges over all subsets of [n]. Note that $y_\emptyset = y_{\overline{0}}\cdots y_{\overline{n-1}}\cdot y$, and

$$\begin{aligned} y_{i_1}\cdots y_{i_r}\cdot y_\emptyset = (-1)^{i_1+\cdots +i_r} y_I, \end{aligned}$$

(A.3)

which is sometimes useful.

The spin representation becomes an $\mathfrak {so}_{2n}$-module via the embedding $\mathfrak {so}_{2n} \cong \textstyle \bigwedge ^2 V \hookrightarrow Cl(V)$ by

$$ v\wedge w \mapsto \frac{1}{2}(v\cdot w - w\cdot v) = v\cdot w - \frac{1}{2}\langle v,w \rangle 1. $$

So $\mathfrak {so}_{2n}$ preserves the parity of basis vectors of Cl(V), and the spin representation breaks into two irreducible half-spin representations

$$ \mathbb {S}= \mathbb {S}^+ \oplus \mathbb {S}^-. $$

Our convention will be that $\mathbb {S}^+$ is spanned by standard spinors $y_I$, with I an even subset of [n], i.e., it has even cardinality.

Now let $x_{\overline{n-1}},\ldots ,x_{\overline{0}},x_0,\ldots ,x_{{n-1}}$ be another orthogonal basis, and assume it is related to the $y_i$’s by a unitriangular matrix:

$$\begin{aligned} x_{\overline{\imath }} = y_{\overline{\imath }} + \sum _{j<i} \overline{c}_{ji} y_{\overline{\jmath }} + \sum _{j=0}^{n-1} c_{ji} y_j \end{aligned}$$

and

$$\begin{aligned} x_i = y_i + \sum _{j>i} b_{ji}y_j. \end{aligned}$$

Thus $x=x_0\cdots x_{n-1}$ is equal to $y=y_0\cdots y_{n-1}$.

We define pure spinors $x_I$ analogously, by

$$ x_I := x_{\overline{\imath }'_1} \cdots x_{\overline{\imath }'_{n-r}}\cdot y. $$

Our goal is to compute the expansion of $x_I$ in the basis $y_K$.

In fact, we will be free to multiply by the inverse of the matrix $\overline{C}=(\overline{c}_{ji})$ and assume that $x_{\overline{\imath }}$ is written

$$\begin{aligned} x_{\overline{\imath }}&= y_{\overline{\imath }} + \sum _{j=0}^{n-1} a_{ji} y_j. \end{aligned}$$

In this case, the fact that the $x_{\overline{\imath }}$ span an isotropic space is equivalent to the fact that the matrix $A=(a_{ji})$ is skew-symmetric. Up to appropriately indexing rows and columns, the x basis is related to the y basis by a matrix of the form

$$ \left( \begin{array}{c|c} A &{} B \\ \hline w_\circ &{} 0\end{array}\right) , $$

where $w_\circ$ is the matrix with 1’s on the antidiagonal and zeroes elsewhere. The top n rows determine a skew-symmetric matrix (by replacing $w_\circ$ with $-B^t$). For a subset $I\subseteq [n]$, let B(I) be the submatrix formed by taking columns I of B, and let A(I) be the skew-symmetric matrix

$$ A(I) = \left( \begin{array}{c|c} A &{} B(I) \\ \hline -B(I)^t &{} 0\end{array}\right) . $$

Only the top n rows are needed to perform operations with such a matrix. For instance, if $n=3$ and $I=\{1\}$, the top rows are

$$ \left( \begin{array}{ccc|ccc} 0 &{} a_{21} &{} a_{20} &{} b_{20} &{} b_{21} &{} 1 \\ &{} 0 &{} a_{10} &{} b_{10} &{} 1 &{} 0 \\ &{} &{} 0 &{} 1 &{} 0 &{} 0\end{array}\right) $$

and

$$ A(I) = \left( \begin{array}{ccc|c}0 &{} a_{21} &{} a_{20} &{} b_{21} \\ -a_{21} &{} 0 &{} a_{10} &{} 1 \\ -a_{20} &{} -a_{10} &{} 0 &{} 0 \\ \hline -b_{21} &{} -1 &{} 0 &{} 0\end{array}\right) $$

For even r, the Pfaffian of any $r\times r$ skew-symmetric matrix A may be computed recursively using the Laplace-type expansion formula

$$ {{\,\mathrm{Pf}\,}}(A) = \sum _{j=1}^{r-1} (-1)^{j-1} a_{jr}{{\,\mathrm{Pf}\,}}_{\widehat{j,r}}(A), $$

where ${{\,\mathrm{Pf}\,}}_{\widehat{j,r}}(A)$ is the Pfaffian of the submatrix of A obtained by removing the jth and rth rows and columns. Let ${{\,\mathrm{Pf}\,}}_K(A)$ denote the Pfaffian of the submatrix on rows and columns K, and we always use the convention ${{\,\mathrm{Pf}\,}}_\emptyset (A)=1$.

Theorem

For a subset $I\subseteq [n]$, we have

$$\begin{aligned} x_I = \mathop {\sum _{K\subseteq [n]}}_{K \text { even}} {{\,\mathrm{Pf}\,}}_K(A(I))\, y_K, \end{aligned}$$

(A.4)

where the sum is over subsets K of even cardinality.

See also [43, Corollary 2.4] for a related Pfaffian identity.

Proof

First we establish the formula for $x_\emptyset$. In fact, for any $m\le n$, we have

$$ x_{\overline{0}}\cdots x_{\overline{m-1}}\cdot y = \mathop {\sum _{K\subseteq [m]}}_{K \text { even}} {{\,\mathrm{Pf}\,}}_K(A)\, y_{\overline{k}'_1}\cdots y_{\overline{k}'_s}\cdot y, $$

where $K' = \{k'_1<\cdots <k'_s\} = [m]\smallsetminus K$. This is proved by induction on m, the base case $m=0$ being the tautology $y=y$. For the induction step, we compute using the Laplace expansion formula:

$$\begin{aligned} x_{\overline{0}} \cdots x_{\overline{m}}\cdot y&= (-1)^m x_{\overline{m}} \cdot x_{\overline{0}} \cdots x_{\overline{m-1}}\cdot y \\&= (-1)^m \left( y_{\overline{m}} + \sum _{j=0}^{m-1} a_{jm} y_j \right) \sum _{K \subseteq [m]} {{\,\mathrm{Pf}\,}}_K(A) y_{\overline{k}'_1} \cdots y_{\overline{k}'_r} \cdot y \\&= \sum _{K \subseteq [m]} {{\,\mathrm{Pf}\,}}_K(A) y_{\overline{k}'_1} \cdots y_{\overline{k}'_r}\cdot y_{\overline{m}} \cdot y \\&\quad + \sum _{j=0}^{m-1}\sum _{K \subseteq [m]} (-1)^m a_{jm}{{\,\mathrm{Pf}\,}}_K(A) y_{j}\cdot y_{\overline{k}'_1} \cdots y_{\overline{k}'_r} \cdot y. \end{aligned}$$

Note

$$ y_j\cdot y_{\overline{k}'_1} \cdots y_{\overline{k}'_r} \cdot y = {\left\{ \begin{array}{ll} 0 &{}\text {if }j\in K; \\ (-1)^{a-1} y_{\overline{k}'_1} \cdots \widehat{y_{\overline{k}'_a}}\cdots y_{\overline{k}'_r} \cdot y &{}\text {if }j=k'_a \in K'. \end{array}\right. } $$

So the second sum above becomes

$$ \sum _K \sum _{j\not \in K} (-1)^{m+a-1} a_{jm} {{\,\mathrm{Pf}\,}}_K(A) y_{\overline{k}''_1} \cdots y_{\overline{k}''_{r-1}} \cdot y, $$

where $K'' = K'\smallsetminus j$. Combining the two and using Laplace expansion yields the claimed formula for $x_\emptyset$.

The general case is similar. Using (A.3), it is equivalent to show

$$ x_{i_1}\cdots x_{i_r}\cdot x_\emptyset = (-1)^{i_1+\cdots +i_r} \mathop {\sum _{K\subseteq [n]}}_{K \text { even}} {{\,\mathrm{Pf}\,}}_K(A(I))\, y_K, $$

and we do this by induction on r. Let $I=\{i_1<\cdots <i_r\}$ and consider $i>i_r$. We compute:

$$\begin{aligned} x_{i_1} \cdots x_{i_r} \cdot x_i \cdot x_\emptyset&= (-1)^{r} x_{i} \cdot x_{i_1} \cdots x_{i_r}\cdot x_\emptyset \\&= (-1)^r \left( \sum _{j} b_{ji} y_j \right) (-1)^{i_1+\cdots +i_r} \sum _{K} {{\,\mathrm{Pf}\,}}_K(A(I)) y_K \\&= (-1)^{r+i_1+\cdots +i_r} \sum _{j}\sum _{K \not \ni j} b_{ji}{{\,\mathrm{Pf}\,}}_K(A(I)) y_{j}\cdot y_K \\&= \cdots \\&= (-1)^{i+i_1+\cdots +i_r} \sum _K {{\,\mathrm{Pf}\,}}_K( A(I\cup \{i\}) ) y_K, \end{aligned}$$

as desired.