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


2 Answers 2


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.


$H^1(G,T_G)$=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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.