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 $H^1(\Omega_X \otimes \operatorname{End} E)$?


1 Answer 1


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 $\mathbb{P}^1$.

  • $\begingroup$ 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? $\endgroup$ Mar 30, 2014 at 19:08
  • $\begingroup$ 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). $\endgroup$ Mar 30, 2014 at 19:11
  • $\begingroup$ @PavelSafronov Thanks. Also I just realized since $\pi_1(\mathbb P^1) = 0$ the only flat rank 2 vector bundle is trivial. $\endgroup$ Mar 30, 2014 at 19:18
  • $\begingroup$ @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. $\endgroup$ Mar 30, 2014 at 19:36
  • 1
    $\begingroup$ By the way, it was a bit stupid of me to quote Weil's theorem: the Atiyah class of a sum $E\oplus F$ is obviously the sum of the Atiyah classes of $E$ and $F$, via the embedding $\mathcal{E}nd(E)\times \mathcal{E}nd(F)\subset \mathcal{E}nd(E\oplus F)$. $\endgroup$
    – abx
    Mar 31, 2014 at 6:34

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.