Kostant’s theorem on the ideals defining homogeneous spaces

Posted by: Steven Sam | February 9, 2011

Kostant’s theorem on the ideals defining homogeneous spaces

The goal of this post is to prove Kostant’s theorem on the quadratic generation of the homogeneous ideals of partial flag varieties. This theorem is true in arbitrary characteristic, but I will stick to the complex numbers to give a simpler proof.

Let ${G}$ be a complex semisimple group and let ${G/B}$ be its associated flag variety. Choose a dominant weight ${\lambda}$ and let ${\mathcal{L}(\lambda)}$ be the associated line bundle on ${G/B}$ . Let ${V_\lambda}$ denote the irreducible representation with highest weight ${\lambda}$ . Then ${\mathcal{L}(\lambda)}$ defines an embedding of ${G/P}$ into ${\mathbf{P}(V_\lambda)}$ (the projective space of 1-quotients of ${V_\lambda}$ ) where ${P}$ is the stabilizer subgroup of the lowest weight vector in ${V_\lambda}$ . The homogeneous coordinate ring is ${\bigoplus_{n \ge 0} V_{n\lambda}}$ , which is a quotient of ${\mathrm{Sym}(V_\lambda)}$ . Let ${J}$ be the homogeneous ideal.

Theorem (Kostant). ${J}$ is generated in degree 2.

Note that ${J_d}$ is a direct sum of the representations in ${\mathrm{Sym}^d(V_\lambda)}$ minus the summand ${V_{n\lambda}}$ . We don’t need to know what these representations are, we just need a way to distinguish them. For that, we’ll use the Casimir element. Let ${\mathfrak{g}}$ be the Lie algebra of ${G}$ , and let ${U(\mathfrak{g})}$ be the universal enveloping algebra of ${\mathfrak{g}}$ , and let ${\beta}$ be a choice of Killing form for ${\mathfrak{g}}$ . Let ${\{x_i\}}$ and ${\{y_i\}}$ be dual bases of ${\mathfrak{g}}$ with respect to ${\beta}$ , and set

$\displaystyle \Omega = \sum_i x_iy_i$

to be the Casimir operator. It can be seen that this element does not depend on the choice of bases made above once ${\beta}$ is fixed. Let ${\rho}$ be the sum of the fundamental weights of ${\mathfrak{g}}$ . We need the following fact: the action of ${\Omega}$ on ${V_\lambda}$ is via the scalar

$c_\lambda = (\lambda + 2\rho, \lambda) = (\lambda + \rho, \lambda + \rho)-(\rho, \rho),$

where ${(,)}$ is the symmetric bilinear form on the weight lattice of ${\mathfrak{g}}$ . We’ll use the following lemma:

Lemma. Let ${\lambda}$ and ${\mu}$ be dominant weights such that $\lambda-\mu$ is a sum of positive roots. Then ${(\lambda, \lambda) \ge (\mu, \mu)}$ with equality if and only if ${\lambda = \mu}$ .

Proof. Since ${\lambda-\mu}$ is a sum of positive roots and both ${\lambda}$ and ${\mu}$ are dominant, we get ${(\lambda, \lambda-\mu) \ge 0}$ and ${(\mu, \lambda-\mu) \ge 0}$ . Adding these two inequalities gives ${(\lambda, \lambda) \ge (\mu, \mu)}$ . If both of these inequalities are in fact equalities, then ${(\lambda, \lambda-\mu) = 0}$ , which we can write as ${(\lambda-\mu, \lambda-\mu) + (\mu, \lambda-\mu) = 0}$ . The second summand is nonnegative and the first summand is positive if and only if ${\lambda-\mu \ne 0}$ by positive definiteness, so we get the second statement. QED

Corollary. Let ${\mu_1, \dots, \mu_r}$ be dominant weights and write ${\mu = \mu_1 + \cdots + \mu_r}$ . Then ${c_\mu}$ is the largest eigenvalue of ${\Omega}$ on ${V_{\mu_1} \otimes \cdots \otimes V_{\mu_r}}$ , and ${V_\mu}$ is a subrepresentation of the tensor product with multiplicity 1.

Proof. Let ${v_i}$ be a highest weight vector in ${V_{\mu_i}}$ . Then ${v_1 \otimes \cdots \otimes v_r}$ is a highest weight vector in ${V_{\mu_1} \otimes \cdots \otimes V_{\mu_r}}$ with weight ${\mu}$ , and it is clear that the ${\mu}$ -weight space in the tensor product has dimension 1. This proves the second statement. For the first statement, if ${V_\nu}$ is in the tensor product, then ${c_\nu \ge c_\mu}$ . If we have equality, then ${(\nu + \rho, \nu + \rho) = (\mu + \rho, \mu + \rho)}$ . Furthermore, ${\mu-\nu}$ must be a sum of positive roots, so we can apply the lemma to ${\mu + \rho}$ and ${\nu + \rho}$ to conclude that ${\mu = \nu}$ . QED

Applying the corollary to the case when ${\mu_i = \lambda}$ and ${r=d}$ and then passing to the symmetric power, we see that ${V_{d\lambda}}$ is precisely the kernel of ${\Omega-c_{d\lambda}}$ acting on ${\mathrm{Sym}^d(V_\lambda)}$ , and hence ${J_d}$ is the image of ${\Omega-c_{d\lambda}}$ acting on ${\mathrm{Sym}^d(V_\lambda)}$ . Let ${v \in V_\lambda}$ . We claim that the following identity holds:

$\displaystyle (*) (\Omega-c_{d\lambda})(v^d) = \binom{d}{2} (\Omega-c_{2\lambda})(v^2)v^{d-2}.$

Since elements of the form ${v^d}$ linearly span ${\mathrm{Sym}^d(V_\lambda)}$ (recall we are in characteristic 0), this will prove the result.

We will need one other definition. Define

$\displaystyle \Omega_2 = \sum_i x_i \otimes y_i.$

Again, this element does not depend on the basis that we chose. Recall that the comultiplication on ${U(\mathfrak{g})}$ is given by ${x \mapsto 1 \otimes x + x \otimes 1}$ when ${x \in \mathfrak{g}}$ . From this, one can calculate that

$\displaystyle \Omega(v \otimes w) = \Omega(v) \otimes w + 2\Omega_2(v \otimes w) + v \otimes \Omega(w),$

and in general, when applying ${\Omega}$ to a ${d}$ -fold tensor product,

$\displaystyle \Omega(v_1 \otimes \cdots \otimes v_d) = \sum_{i=1}^d v_1 \otimes \cdots \Omega(v_i) \cdots \otimes v_d +$
$\displaystyle 2 \sum_{1 \le i < j \le d} v_1 \otimes \cdots \Omega_2(v_i \otimes v_j) \cdots \otimes v_d.$

In particular, we get

$\displaystyle \Omega(v^d) = d\Omega(v)v^{d-1} + 2\binom{d}{2} \Omega_2(v^2)v^{d-2}$
$\displaystyle = dc_\lambda v^d + \binom{d}{2}(\Omega(v^2)-2\Omega(v)v)v^{d-2}$
$\displaystyle = c_\lambda(d- 2\binom{d}{2}) v^d + \binom{d}{2}\Omega(v^2)v^{d-2}.$

So to prove ${(*)}$ , we just need to show that ${(d-2\binom{d}{2})c_\lambda-c_{d\lambda} =-\binom{d}{2}c_{2\lambda}}$ , but this follows from the fact that ${(,)}$ is a symmetric bilinear form and a direct check.

-Steven

Posted in Algebraic Geometry | Tags: homogeneous spaces, Representation Theory

Responses

Kostant’s theorem, eh? Can you give a reference? (This is a trick question. I’m curious if you have a different one than I do.)

I like to think of this theorem as follows. For any projective X, some Veronese will be quadratically defined. But for flag manifolds, you don’t need the Veronese step.

There’s a relative version, too. For any pair X > Y of projective varieties, in some Veronese, X will be quadratic and Y inside X will be further defined by only linear equations. But for flag manifolds and Schubert varieties, you don’t need the Veronese.

This fact (and Kostant’s theorem) is in [Ramanathan ~’84].
By: Allen Knutson on February 9, 2011
at 4:34 AM
I mainly copied the argument from http://arxiv.org/abs/math.AG/0602316 but tried to fill in more details.

When I first heard this theorem, I looked on mathscinet and was quite frustrated when I couldn’t find any paper of Kostant that seemed to prove this result.

And thanks for the other context. That is a nice way of interpreting this result.
By: Steven Sam on February 9, 2011
at 5:11 AM
A (slight) generalization is given in Proposition 28.3 of http://arxiv.org/abs/math/0602228

There too the theorem is attributed to Kostant, but with a reference to [G. Lancaster and J. Towber, J. Algebra 59 (1979), 16–38.]
By: Bart on February 11, 2011
at 3:32 AM
@Bart: that looks a wonderful reference, thanks for pointing it out.
By: Steven Sam on February 11, 2011
at 3:49 AM

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