# Are Chern connections on flat bundles flat?

More precisely formulated: We are given a hermitian holomorphic vector bundle $(E,\langle\cdot,\cdot\rangle,\bar{\partial})$ on a $\mathbb{C}$-manifold M such that as a topological bundle, it is flat (i.e. there is some connection, not necessarily compatible with $\langle\cdot,\cdot\rangle$ or $\bar{\partial}$, whose curvature vanishes).

Is it true that the curvature of the Chern connection necessarily vanishes?

The answer is yes if $M$ is compact Kähler (EDIT: and you allow a conformal change in the metric) and $L$ is a line bundle and no in general, already in the case of line bundles.

Take for example $M$ to be the standard Hopf surface $(\mathbb{C}^2\backslash \{(0,0)\})/[(z_1,z_2)\sim (2z_1,2z_2)]$. Then $M$ is diffeomorphic to $S^1\times S^3$ so every line bundle on $M$ has first Chern class zero, therefore it is topologically trivial and it admits a flat connection.

Let us now see that the canonical bundle $K_M$ is not Chern flat (for any Hermitian connection). First of all, write down the explicit Hermitian metric on $M$

$$g_{i\overline{j}}=\frac{\delta_{ij}}{r^2}, r^2=|z_1|^2+|z_2|^2.$$

It induces a Hermitian metric on $K_M$ whose curvature is the first Chern form which can be easily calculated in local coordinates

$$\alpha=-i\partial\overline{\partial}\log\det (g_{i\overline{j}})=\frac{2}{r^2}\left(\delta_{kl}-\frac{\overline{z}_k z_l}{r^2}\right)i dz_k\wedge d\overline{z}_\ell,$$

and $\alpha$ is a nonnegative-definite Hermitian form, which is not identically zero. If you had another Hermitian connection on $K_M$ with zero first Chern form, by the transgression formula it would imply that $\alpha=i\partial\overline{\partial}u$ so $u$ would be a global plurisubharmonic function on $M$, which must be constant by the maximum principle, and you'd get that $\alpha=0$ which is false.

However, if $M$ is compact Kähler then the $\partial\overline{\partial}$-lemma shows that topologically trivial holomorphic line bundles admit flat Hermitian connections. Just start with any Hermitian connection, its Chern curvature form will be $\alpha$, say, which is cohomologous to zero, so $\alpha=i\partial\overline{\partial}u$ by the $\partial\overline{\partial}$-lemma. Then conformally scaling the Hermitian metric by $e^u$ gives you a new Hermitian connection with vanishing Chern curvature.

For a higher rank holomorphic bundle $E$ over a compact Kähler manifold $(M,\omega)$ you have that $E$ admits a flat Hermitian metric iff it has a Hermitian-Yang-Mills connection $H$ with $\omega^{n-1}\wedge F_H=0$ and its second Chern class vanishes, iff it is $[\omega]$-polystable and its first and second Chern classes vanish. This is essentially the Donaldson-Uhlenbeck-Yau Theorem.

So if $E$ admits a flat connection its Chern classes indeed vanish, so it will admit a flat Hermitian metric iff it is $[\omega]$-polystable. I think there should be plenty of examples of such bundles which are not polystable.

• Do you also have counterexamples for higher dimensional bundles on compact Kähler manifolds? Commented Mar 8, 2012 at 18:15
• @YangMills: Just a small nitpick. The way Generic's question is phrased currently, the answer is "no" even in the compact Kaehler case, because the hermitian metric is assumed to be fixed. Commented Mar 8, 2012 at 20:14
• You're right, I was careless when reading his question. I will edit the answer accordingly, thanks! Commented Mar 9, 2012 at 1:53