It is a simple fact that if $L \to B$ is a complex line bundle endowed with an Hermitian product and a compatible connection $\nabla$, then the curvature $F_\nabla$ is imaginary (and so are the local connection $1$-forms). I am curious, though, if the following converse is true: if $\nabla$ is given such that $F_\nabla$ is imaginary, is it possible to find an Hermitian product such that $\nabla$ be compatible with it?
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Even flat connections don't have to arise from a Hermitian inner product; they can have holonomy not unitary.
answered Apr 21, 2016 at 13:42
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$\begingroup$ For example, on any Hopf surface, the canonical bundle has a unique holomorphic connection, and the holonomy of this connection is not in the unitary group $U(1)$, so there is no Hermitian inner product compatible with the connection. $\endgroup$ Commented Apr 21, 2016 at 14:27
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$\begingroup$ ... and that connection is flat too. $\endgroup$ Commented Apr 21, 2016 at 16:04
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$\begingroup$ The Poincare lemma for $\partial\bar\partial$ ensures that there are local Hermitian inner products with the given curvature, on any contractible open set. $\endgroup$ Commented Apr 21, 2016 at 16:28