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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?

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1 Answer 1

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.

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Do you also have counterexamples for higher dimensional bundles on compact Kähler manifolds? –  Generic Mar 8 '12 at 18:15
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@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. –  Peter Dalakov Mar 8 '12 at 20:14
    
You're right, I was careless when reading his question. I will edit the answer accordingly, thanks! –  YangMills Mar 9 '12 at 1:53

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