A question about *A new proof of a theorem of Narasimhan and Seshadri*

Question

I am reading S. K. Donsaldson's paper A new proof of a theorem of Narasimhan and Seshadri. Now $M$ is a compact Riemann surface and $E$ is a holomorphic bundle over $M$ with a Hermitian metric. There is a statement (p.276) : for $g$ a self-adjoint complex gauge transformation $g=g^*$, $A$ a Hermitian metric \begin{equation} F(A)=F(g.A)\implies \partial\bar\partial(g^2)=-\{(\bar\partial g^2)g^{-1}\}\{(\bar\partial g^2)g^{-1}\}^* \end{equation} where $F$ is the curvature form.

I try to prove this.

Here I think a problem happens.

We have \begin{equation} g.A=A-(\bar\partial g)g^{-1}+((\bar\partial g)g^{-1})^*=A-\bar\partial gg^{-1}+g^{-1}\partial g \end{equation} and for a section $\sigma$ \begin{equation} \begin{split} F(g.A)\sigma=&(d_{g.A})(d_{g.A})\sigma\\ =&(d_A-\bar\partial gg^{-1}+g^{-1}\partial g)(d_A-\bar\partial gg^{-1}+g^{-1}\partial g)\sigma \end{split} \end{equation} and just consider the second order derivative of $g$ \begin{equation} \begin{split} F(g.A)\sigma=&F(A)\sigma+d_A(-\bar\partial g g^{-1}\sigma)+d_A(g^{-1}\partial g\sigma)+\cdots\\ =&F(A)\sigma-\partial\bar\partial gg^{-1}\sigma-g^{-1}\bar\partial\partial g\sigma+\cdots \end{split} \end{equation} so from $F(A)=F(g.A)$ we will get something like \begin{equation} \partial\bar\partial gg^{-1}-g^{-1}\bar\partial\partial g+\cdots=0 \end{equation} but I think this is different from $\partial\bar\partial(g^2)=\cdots$, since in which the second derivatives are \begin{equation} \partial\bar\partial g g+g\partial\bar\partial g \end{equation} and the $\partial$ operator is always ahead of $\bar\partial$ operator. If $\partial\bar\partial g+\bar\partial\partial g=0$, then the problem is solved but I do not think this is naturally true.

Answer 1

Indeed $\partial \bar\partial g + \bar\partial\partial g=0$. In fact $\partial \bar\partial + \bar\partial\partial$ is the zero operator. We have that $d= \partial + \bar\partial$ is the induced connection on $End(E)$. Then $$ d^2 = \partial^2 + \partial \bar\partial + \bar\partial\partial + \bar\partial^2 = \partial \bar\partial + \bar\partial\partial $$ is the curvature of the induced connection on $End(E)$, since $\partial^2 = \bar\partial^2 =0$ (by dimension). Then the result follows once we have $End(E)$ flat.

We have that the curvature of $E$ is a multiple of the identity automorphism by hypothesis, $F(A) = \mu 1$ in page 276. Computing the curvature of the induced connection on $E\otimes E^\ast \simeq End(E)$ we see that it vanishes.