How general a connection on $TM$ can be used in the Chern-Gauss-Bonnet theorem? Wikipedia only states the theorem for the Levi-Civita connection, but this is probably needlessly restrictive. (If the theorem can be proven for the LC connection of some metric then it holds for the LC connection for any metric, and so we see already that there is some family of connections for which it holds. This is also asserted in Q. Yuan's answer here).

What I understand of Chern-Weil theory suggests that the theorem should hold for any connection whatsoever on $TM$, but perhaps I am missing something. Chern seems to only consider Levi-Civita connections in A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds.

  • $\begingroup$ Incidentally, it is a famous open problem due to Chern : Does the existence of a flat affine connection on a compact manifold imply zero Euler characteristic. $\endgroup$ Jul 7, 2019 at 9:37

1 Answer 1


If by generalized Chern-Gauss-Bonnet theorem we mean that $\chi(M)=\frac1{(2\pi)^n}\int_M\textrm{Pf}(\Omega)$, where $\chi$ is the Euler characteristic, $\Omega$ is the curvature form, and Pf is the Pfaffian, then it is false for general connections. Beneventano et al. show in Heat trace asymptotics and the Gauss-Bonnet Theorem for general connections that when the connection is not Levi-Civita the correct integrand is not the Pfaffian. One can write a formula that works in general using the $L_2$ trace of the heat semigroup, but it is not nearly as useful.

Zhao proved a positive result in A note on the Gauss-Bonnet-Chern theorem for general connection. The connection need not be the Levi-Civita, but it has to be metric compatible. This means, in particular, that $\nabla g=0$ for some $g$, i.e. its restricted holonomy group is conjugated to a subgroup of $O(n)$, but it is not necessarily torsion-free, see Does the Levi-Civita connection determine the metric?

  • $\begingroup$ Thank you for these helpful references. If I understood Beneventano et al, the local densities that we integrate to compute the heat trace are no longer only functions of the connection's curvature -- is that correct? And so this is why merely finding a flat affine connection is not enough to conclude that $\chi(M)=0$, whereas a flat metric connection does imply zero Euler characteristic? $\endgroup$
    – Todd N
    Jul 8, 2019 at 6:13
  • $\begingroup$ @ToddN Yes, they identified some asymptotic terms that do not come from curvature but contribute to the sum. But now that I thought about Zhao's result some more, I think what he means by "metric compatible" is that the connection does differentiate the metric to $0$, but is not necessarily torsion-free. $\endgroup$
    – Conifold
    Jul 8, 2019 at 18:28

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