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Let $X$ be compact Kahler and $E \to X$ a holomorphic vector bundle. Then $E$ has an Atiyah class, $At(E)$, valued in the sheaf cohomology $H^1(\Omega_X \otimes \operatorname{End} E)$. Suppose the topological Chern classes of $E$ vanish rationally. Evaluating $At(E)$ on an invariant polynomial gives a class in $\bigoplus_k H^k(\Omega_X^k)$, which must vanish in non-zero degrees since, by the Hodge decomposition, they correspond to polynomials in the Chern classes of $E$.

My question is:

Is it the case that the full Atiyah class itself, $At(E)$, is necessarily zero in $H^1(\Omega_X \otimes \operatorname{End} E)$?

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1 Answer 1

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No. A counter-example : the vector bundle $\mathcal{O}_{\mathbb{P}^1}(p)\oplus \mathcal{O}_{\mathbb{P}^1}(-p)$ on $\mathbb{P}^1$ has zero Chern class, but does not admit a holomorphic connection if $p>0$ (by a theorem of Weil, a vector bundle on a curve admits a holomorphic connection iff its indecomposable summands have degree zero), hence its Atiyah class is nonzero. If you want a higher-dimensional example just pull back this one by a morphism to $\mathbb{P}^1$.

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Thanks for the answer! Do you happen to have a reference for the theorem of Weil? Also, do you know offhand if $\mathcal{O}_{\mathbb{P}^1}(p)\oplus \mathcal{O}_{\mathbb{P}^1}(-p)$ admits a flat connection? –  Eric O. Korman Mar 30 at 19:08
    
The theorem of Weil is proved in the original paper by Atiyah "Complex analytic connections..." A vector bundle admits a flat connection on a Riemann surface iff it admits a holomorphic connection (which is necessarily flat). –  Pavel Safronov Mar 30 at 19:11
    
@PavelSafronov Thanks. Also I just realized since $\pi_1(\mathbb P^1) = 0$ the only flat rank 2 vector bundle is trivial. –  Eric O. Korman Mar 30 at 19:18
    
@PavelSafronov Err, now I'm confused. Topologically we have $\mathcal (O(-1)\otimes \mathcal O(-1))\oplus \mathbb C = \mathcal O(-1) \oplus \mathcal O(-1)$. So topologically, $\mathcal O(1) \oplus \mathcal O(-1) = \mathbb C^2$ is trivial and so admits a flat connection. –  Eric O. Korman Mar 30 at 19:36
    
Indeed it is topologically trivial, but does not admit a holomorphic connection. –  abx Mar 30 at 19:48

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