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Let $V$ be a $2n$-dimensional vector space endowed with a nondegenerate skew-symmetric form $q:V \to V^\vee$. We define the isotropic Grassmannian to be $$ X:=G_q(k,V)=\left\{[W] \in \mathbb P \left( \bigwedge^k V\right): W \subset V \text{ and } W \text{ is }q-\text{isotropic}\right\}. $$ We denote the rank $k$ tautological bundle over $X$ by $\mathcal S \hookrightarrow V \otimes \mathcal O_X$. The cokernel will be denoted by $\mathcal Q$. Thanks to the quadratic form, we also have the inclusion $\mathcal S \hookrightarrow \mathcal Q^\vee$ and we denote $\mathcal K$ to be the cokernel $\mathcal Q^\vee/\mathcal S$.

One can prove that the tangent bundle $T_X$ fits in the short exact sequence $$ 0 \to \mathcal S^\vee \otimes \mathcal K \longrightarrow T_X \longrightarrow S^2 \mathcal S^\vee \to 0. $$ Question. Is it true that $T_X$ is the unique non-trivial extension for the short exact sequence? How can we prove it?

I tried to compute $H^1\left(X,S^2\mathcal S \otimes\mathcal S^\vee \otimes \mathcal K\right)$ without significant results.

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  • $\begingroup$ How did you try to compute $H^1$? $\endgroup$
    – Sasha
    Commented Aug 2, 2022 at 14:39
  • $\begingroup$ I tensor the short exact sequence by $\operatorname{Hom}(S^2 \mathcal S^\vee,\cdot)$ and then induce the long exact sequence in cohomology. Since $S^2 \mathcal S^\vee$ is globally generated, its dual is not. Hence $H^0(X, \operatorname{Hom}(S^2\mathcal S^\vee,S^2 \mathcal S^\vee))$ is zero. So I can conclude that $$ H^1(X,\operatorname{Hom}(S^2\mathcal S^\vee,\mathcal S^\vee \otimes \mathcal K)) \hookrightarrow H^1(X,\operatorname{Hom}(S^2\mathcal S^\vee,T_X)). $$ But I cannot say something more for the moment being $\endgroup$
    – Bobech
    Commented Aug 2, 2022 at 14:46
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    $\begingroup$ Why don't you apply Borel--Bott--Weil instead? $\endgroup$
    – Sasha
    Commented Aug 2, 2022 at 18:23
  • $\begingroup$ Because I don't know who are the representations associated with the vector bundles appearing in the sequence $\endgroup$
    – Bobech
    Commented Aug 3, 2022 at 8:01
  • $\begingroup$ Do you know how the equivalence between equivariant bundles on a homogeneous variety and representations of the stabilizer of a point works? $\endgroup$
    – Sasha
    Commented Aug 3, 2022 at 8:39

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