cohomology of tangent bundle

Question

Suppose G is a Grassmannian variety. What are cohomology groups $H^i(G,T_G)$ $(i\geq 0)$ of the tangent bundle ? thanks.

Answer 1

This answer is in characteristic zero so that I can use Borel-Bott-Weil; I'm not sure if it's still right in finite characteristic. As Serge says, $H^0(G(k,V), T) = \mathrm{End}(V)/\langle \mathrm{Id} \rangle$. All the other cohomology groups are zero.

Proof sketch: Let $Fl(V)$ be the variety of complete flags in $V$; we'll write such a flag as $F_1 \subset F_2 \subset \cdots \subset F_{n-1} \subset V$. Let $L$ be the line bundle $F_1^{\vee}$ over $Fl(V)$ and let $L'$ be the line bundle $V/F_{n-1}$. (I write $E^{\vee}$ for the dual of the vector bundle $E$.)

Let $\pi$ be the projection map $Fl(V) \to G(k,n)$ taking $(F_{\bullet})$ to $F_k$. Let $S$ be the tautological subbundle on $G(k,n)$ and $Q$ the tautological quotient bundle.

Let $[W]$ be a point of $G(k,V)$, with corresponding $k$-dimensional subspace $W$. The fiber $\pi^{-1}([W])$ is $Fl(W) \times Fl(V/W)$. The sections of $L$ and $L'$ on this fiber are naturally $W^{\vee}$ and $V/W$ respectively. The sections of $L \otimes L'$ are $W^{\vee} \otimes (V/W) = \mathrm{Hom}(W,V/W)$. The line bundles $L$ and $L'$ on this fiber have no higher cohomology.

I claim (but have not checked carefully) that the above paragraph works in families, so $\pi_{\ast}(L \otimes L') \cong \mathrm{Hom}(S, Q)$ and $R^i \pi_{\ast}(L \otimes L') =0$. As is well known, $T_G \cong \mathrm{Hom}(S, Q)$.

So the Serre spectral sequence for $Fl(V) \to G(k,V) \to \mathrm{pt}$ collapses and $H^i(G(k,V), T_G) \cong H^i(Fl(V), L \otimes L')$.

By Borel-Bott-Weil (in characteristic zero), $H^0(Fl(V), L \otimes L')$ is the $GL_V$ irrep indexed by $(1,0,0,\ldots,-1)$. This is an $n^2-1$ dimensional vector space, with the explicit description given by Serge. Also by Borel-Bott-Weil, $H^i(Fl(V), L \otimes L')$ vanishes.

Answer 2

$H^1(G,T_G)$=0`. If $G$ is the Grassmannian of $k$-dimensional linear subspaces in a linear space $E$, then $H^0(G,T_G)$ is isomorphic to the quotient $\mathrm{End}(E)$ modulo multiples of identity. Don't know about higher cohomology.