TITLE: Unsplitting sequence of vector bundles
QUESTION [5 upvotes]: Let $V$ be a $n$-dimensional complex vector space. Using Grothendieck's notation, we define the Grassmannian $G(k,V)$ as the space of $k$-quotients of $V$ or, equivalently, as
$$
G(k,V)=\{ \mathbb P W \subset \mathbb P V: \dim W=k\}.
$$
On $G(k,V)$ we have the tautological bundle sequence:
$$
0 \to \mathcal S^\vee \to V \otimes \mathcal O_{G(k,V)} \to \mathcal Q \to 0.
$$
We suppose that the following short exact sequence holds:
$$
0 \to \mathcal Q \otimes \mathcal S^\vee \to N \to \wedge^2 \mathcal Q \to 0,
$$
where $N$ is another vector bundle on $G(k,V)$. I know that in the case $k=2$
$$
Ext^1(\wedge^2 \mathcal Q, \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V), \mathcal Q \otimes \mathcal S^\vee \otimes (\wedge^2 \mathcal Q)^\vee)
$$
become, using $\wedge^2 \mathcal Q=\mathcal O(1)$ and $\mathcal Q(-1)=\mathcal Q^\vee$,
$$
Ext^1(\mathcal O(1), \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V),\mathcal Q^\vee \otimes \mathcal S^\vee)=H^1(G(k,V), \Omega^1)=\mathbb C,
$$
hence the short exact sequence do not split. I expect that this happens also in the general case, but I cannot prove it. How can I proceed?

REPLY [5 votes]: $\def\CC{\mathbb{C}}$A splitting would be a global map $f : G(k,n) \times \CC^n \to \CC^n$ such that $f(L,v) \in L$ for all $L \in G(k,n)$ and $v \in \CC^n$. But, since $G(k,n)$ is projective and connected and the target $\CC^n$ is affine, any map $f : G(k,n) \times \CC^n \to \CC^n$ is constant on each $G(k,n) \times \{ v \}$. So we would have a map $f : \CC^n \to \CC^n$ such that $f(v) \in L$ for each $v \in \CC^n$ and each $L$ in $G(k,n)$. The only such map is the zero map.