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I saw in a paper that the only vector bundle of rank $r$ on the projective plane with trivial chern classes is the trivial vector bundle of rank $r$.

I can see that the trivial vector bundle has trivial chern classes. How do we prove the converse?

user349424
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1 Answers1

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This is not true (probably you missed some assumptions). For instance, take any stable vector budnle $F$ with $c_1(F) = 0$ and $c_2(F) = k^2$ for some $k \in \mathbb{Z}$. Take also $G = \mathcal{O}(kh) \oplus \mathcal{O}(-kh)$. Now set $E = F \oplus G$. Then $$ c(E) = c(F)c(G) = (1 + k^2h^2)(1+kh)(1-kh) = 1, $$ so all Chern classes of $E$ are trivial. However, $E$ is not trivial.

Sasha
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    @ Sasha, I think there was an additional assumption that the vector bundle is semistable. Is that enough to say that the vector bundle is trivial? – user349424 Apr 28 '19 at 18:53