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Forgive me for this elementary question.

Let $E$ be a holomorphic vector bundle over a Riemann surface $M$ equipped with a Hermitian metric. Let $\nabla$ be the compatible connection on $E$ amd $g$ is a self adjoint complexified gauge transformation. Denote $\partial=\nabla^{1,0}$ and $\overline{\partial}=\nabla^{0,1}$. Then

$g.\nabla=\nabla-(\overline{\partial}g)g^{-1}+((\overline{\partial}g)g^{-1})^\dagger=\nabla-(\overline{\partial}g)g^{-1}+(g^{-1}\partial g)$

since $g$ is self adjoint.

Let $F_\nabla$ be the curvature of $\nabla$. We have the formula

$F_{\nabla+A}=F_\nabla+\nabla A+\frac{1}{2}[A,A].$

How to show

$F_{g.\nabla}=F_{\nabla}-\partial((\overline{\partial}g)g^{-1})+\overline{\partial}(g^{-1}(\partial g))-(\overline{\partial}g)g^{-2}(\partial g)+g^{-1}(\partial g)(\overline{\partial}g)g^{-1}$?

(Let $A=-(\overline{\partial}g)g^{-1}+(g^{-1}\partial g)$. How to compute $[A,A]$? How to compute even just $[(\overline{\partial}g)g^{-1},(\overline{\partial}g)g^{-1}]$?)

This question is taken from page 8 of Donaldson's proof of Narasimhan-Seshadri theorem.

Thank you.

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$\frac{1}{2} [A,A] = A^2$ is multiplication of $A$ with itself (taking the usual multiplication on the endomorphism factor and the wedge product on the one-form factor). Note that $[(\overline{\partial}g)g^{-1},(\overline{\partial}g)g^{-1}] = 0$ since this involves the wedge product of two (0,1) forms, which necessarily vanishes since $M$ has complex dimension 1. Incidentally, I think there is a sign mistake-- it looks like there should be a minus sign on the term $g^{-1}(\partial g)(\overline{\partial}g)g^{-1}$. (One consistency check is that in the rank 1 case, changing the connection by the 1-form $A$ changes the curvature by $dA$).

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  • $\begingroup$ Is $[g^{-1}\partial g,g^{-1}\partial g]=0$? One more thing, the term $\nabla A$ in the formula $F_{\nabla+A}=F_\nabla +\nabla A+\frac{1}{2}[A,A]$. It should give two extra terms, which are $-\overline{\partial}((\overline{\partial}g)g^{-1})$ and $\partial(g^{-1}\partial g)$, in the final output. Why don't we see these terms? $\endgroup$
    – HLC
    Commented Jul 28, 2016 at 17:46
  • $\begingroup$ Yes, $[g^{-1}\partial g,g^{-1}\partial g]=0$ since it involves the wedge product of two forms of type (1,0) on a one-dimensional complex manifold (i.e. they are multiples of $dz$ where $z$ is a local holomorphic coordinate). Similarly by type considerations those two extra terms are zero: the first is $\bar\partial$ of something that is (0,1) and so is (0,2), which is zero. The second is $\partial$ of something that is (1,0) and so is also zero. $\endgroup$
    – Eric O. Korman
    Commented Jul 28, 2016 at 17:55
  • $\begingroup$ $\partial g$ is of type $(1,0)$ but do we know that $g^{-1}\partial g$ is also of type $(1,0)$? I mean $g$ is just a gauge transformation, not necessarily holomorphic, right? $\endgroup$
    – HLC
    Commented Jul 28, 2016 at 20:27

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