Suppose $G$ is a Grassmannian variety. Let $T_G$ be the tangent bundle. Then $H^0(G,T_G)$ is non zero. I wondered if one knows $H^0(G,\wedge^iT_G)$, for $i>0$.



1 Answer 1


You need Borel--Bott--Weil Theorem. It gives the following answer: if $G = G(k,n)$ then $H^0(G,\Lambda^m_G)$ is the direct sum over all Young diagrams $\alpha = (a_1,\dots,a_k)$ inscribed in an $(n-k)\times k$ rectangle with $m$ boxes of the highest weight representations of the group $GL_n$ with highest weight $(a_1,\dots,a_k,-b_{n-k},\dots,-b_1)$, where $\beta = (b_1,\dots,b_{n-k})$ is the transposition of $\alpha$.

  • $\begingroup$ Should Bott really be a part of it? I mean computing $H^0$ :) $\endgroup$ Apr 9, 2013 at 19:54

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