Extensions for a short exact sequence on Grassmannians

Question

$\DeclareMathOperator\Sym{Sym}\DeclareMathOperator\Ext{Ext}$Let us consider a $n$-dimensional complex vector space $V$ and denote by $G(k,n)$ the Grassmannian of $k$-planes in $V$. We use the convention that the universal exact sequence on $G(k,n)$ is given by $$ 0 \to S \to V \otimes \mathcal O \to Q \to 0, $$ with $S,Q$ of rank $k$ and $n-k$ respectively. I want to study the (possible) extensions of the following short exact sequence on $G(k,n)$: $$\tag{$\star$} 0 \to Q^\vee \otimes S^\vee \to N \to \Sym^2S^\vee \to 0. $$ More precisely, I'm asking if $N$ has to be $(Q \otimes S)^\vee \oplus \Sym^2 S^\vee$.

In order to attack the problem, I try to compute $$ \Ext^1(\Sym^2 S^\vee,Q^\vee \otimes S^\vee)=H^1(G(k,n),Q^\vee \otimes S^\vee \otimes \Sym^2 S), $$ but now I don't have idea on how to compute such a cohomology piece. Any idea on how to go on?

Edit 1: I also know that $N$ fits in another short exact sequence of vector bundles on $G(k,n)$: $$\tag{$\star\star$} 0 \to N \to S^\vee \otimes (V^\vee \otimes \mathcal O) \to {\bigwedge}^2 S^\vee \to 0 $$ that is $N=\ker \phi$ where $\phi: S^\vee \otimes (V^\vee \otimes \mathcal O) \to {\bigwedge}^2 S^\vee$.

Answer 1

$\DeclareMathOperator\Sym{Sym}$I’ll write the weight of $S^\vee$ as $(0,\ldots,0,-1)$; it might be helpful to think of this bundle as $\bigwedge^{k-1}S \otimes \det(S)^{-1}$. For $\Sym^2(S)$ the weight is $(2,0,\ldots,0)$; both of these are vectors with $k$ entries. By the Pieri rule,

$$S^\vee \otimes \Sym^2(S) \cong S \oplus \mathbb{S}_{(2,0,\dotsc,0,-1)}(S),$$ where $\mathbb{S}$ denotes the Schur functor.

Similarly the weight of $Q^\vee$ is $(0,\dotsc,0,-1)$ (with $n-k$ entries). So we concatenate the weights for each of the two summands:

$$w = (0,\dotsc,0,-1,1,0,\dotsc,0) \text{ and } w'= (0,\dotsc,0,-1,2,0,\dotsc, 0,-1).$$

By Borel–Weil–Bott we add $\rho = (n,n-1,\dotsc,1)$ and count inversions in the resulting word; if there is a repeat, all the cohomology vanishes.

In the first case there is exactly one inversion, so the bundle has nonvanishing $H^1$. After sorting, the resulting weight is

$$\operatorname{sort}(w+\rho)-\rho = \vec{0}.$$ So $H^1$ is one-dimensional (the trivial representation). For the other, $w'+\rho$ has a repeat from the $2$ and the $0$ two steps before it. So all the cohomology vanishes.

Edit: Exception: for the second calculation, if $n-k=1$ then the repeat doesn’t occur. In that case I guess there is one inversion and nonvanishing $H^1$ of weight $(1,0,\dotsc,0,-1)$ which has rank $n^2-1$; it is the kernel of $V \otimes V^\vee \to \mathbb{C}$, i.e. the traceless matrices. Also, if $k=1$ then the second summand simply doesn’t occur at all. I think the calculation above is correct otherwise, that is if $n-k$ and $k$ are both $\geq 2$.

Answer 2

To calculate $$ H^1(G(k,n),Q^\vee \otimes S^\vee \otimes Sym^2 S) $$ you can use the Kostant's version of the Bott-Borel-Weil theorem which computes $H^k(G/P, V_\lambda)$ for an associated bundle to a highest weight represetnation $\mathbb{V}_\lambda$. The answer is given in terms of action of certain elements of the Weyl group, which in the case of Grassmannians and $k=1$ shouldn't be too complicated. However, you first have to decompose the inducing representation of your bundle into irreducible pieces, which can become unwieldy in the general case. Nevertheless, decomposition of tensor product is algorithmic, so you can just run the algorithm for several different ranks and see how it behaves.