We know that a line bundle $L$ on the complex flag variety $G/P$ is trivial iff $c_1(E) = 0$. But if we have a homogeneous vector bundle $E$ of higher rank, then is it true that $c_i(E) = 0$ $ \forall i$ implies that $E$ is trivial? If not, is there anything analogous we can say about such vector bundles? I apologize if this question is trivial as my Representation theory background isn't that great.


Take for instance $G/P = Gr(k,n)$, and let $$ 0 \to U \to V \otimes \mathcal{O} \to Q \to 0 $$ be the tautological exact sequence. Then $$ c_\bullet(U \oplus Q) = c_\bullet(U) \cdot c_\bullet(Q) = c_\bullet(V \otimes \mathcal{O}) = 1, $$ but $U \oplus Q$ is a nontrivial vector bundle.

A similar example can be constructed on any flag variety.

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  • $\begingroup$ Thanks for your answer. I realized after asking this question that it's actually really easy to produce counterexamples: \mathcal{O}(-n) \bigoplus \mathcal{O}(n) on \mathbf{P}^n being easy counterexamples. In fact more generally, any homogeneous bundle all of whose weights are non zero linear combinations of the weights corresponding to the parabolic $P$ will have zero chern classes, but not be trivial. I think this question should be taken off this site: it's too silly. $\endgroup$ – user69183 Oct 20 '19 at 19:18
  • $\begingroup$ Not quite, in fact $c_2(\mathcal{O}(-n) \oplus \mathcal{O}(n)) \ne 0$. $\endgroup$ – Sasha Oct 21 '19 at 4:32
  • $\begingroup$ Yes, you're right, it should be $-n^2$. But whatever, the Chern class formula for homogeneous bundles sees only the weights which aren't in $P$. So there exist lots of counterexamples. $\endgroup$ – user69183 Oct 21 '19 at 15:33

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