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I am confused about something regarding hermitian metrics. I understand that since we can define a complex vector bundle over a smooth manifold, it makes sense to consider a hermitian metric that only varies smoothly.

However, in the case that we are defining a hermitian metric on the holomorphic tangent bundle of a complex manifold, why don't we insist that the hermitian metric vary holomorphically?

I know the answer is something along the lines of "why not?" since even if we did insist that hermitian metrics only exist on holomorphic vector bundles and varied holomorphically, we could always generalize the notion to complex vector bundles ask that they only vary smoothly (in this parallel universe, maybe I'd be asking why we bothered to do this). But I was hoping that someone could give me an example of a hermitian metric which does not vary holomorphically but is still a reasonable object to study, and whose exclusion would maybe make some class of examples or problems less natural.

Thanks.

trystero
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    It’s possible to view the Poincaré metric on the unit disk as the curvature form of a Hermitian metric on the trivial line bundle over the disk. In this trivialization the metric is identified with a smooth function on the disk. This function must be real-valued to be a Hermitian metric, so it cannot be holomorphic. More generally, if a metric is actually holomorphic I think it follows that it is flat, and not every bundle admits a flat metric. – Gunnar Þór Magnússon Dec 08 '21 at 08:43
  • @GunnarÞórMagnússon That seems to be an answer (especially the part that $h$ is locally "real-valued"). Will you post that as an answer? – Arctic Char Dec 08 '21 at 15:10
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    Here's a hint: The space of hermitian matrices is a smooth manifold, but not a complex manifold. The defining equation involves conjugation. – Ted Shifrin Dec 08 '21 at 21:28

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It’s possible to view the Poincaré metric on the unit disk as the curvature form of a Hermitian metric on the trivial line bundle over the disk. In this trivialization the metric is identified with a smooth function on the disk. This function must be real-valued to be a Hermitian metric (because a Hermitian inner product on a one-dimensional complex vector space is just a positive real number), so it cannot be holomorphic.

More generally, if a metric $h$ on a holomorphic vector bundle is holomorphic it follows that it is flat: Pick a local holomorphic frame for the bundle and let $H$ be the matrix of $h$ in that frame. Then the entries of $H$ are holomorphic functions. However, as $H$ is Hermitian we have $H = \overline H^t$, so the conjugates of the entries of $H$ are holomorphic functions as well. It follows that the entries of $H$ are constant, so its Chern curvature tensor vanishes. Now, there are holomorphic vector bundles that do not admit flat metrics (like the tangent bundle of the projective space), so those bundles do not admit holomorphic Hermitian metrics.

Gunnar Þór Magnússon
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