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Let $\pi:E\rightarrow M$ be a complex vector bundle and $H$ a hermitian metric over it. If $D$ is a connection over $E$, using the metric $H$, we can decompose it as: $$ D=D_H+\phi $$ Where $D_H$ is an unitary connection and $\phi$ is Hermitian $1$-form with values in $\mathrm{End}(E)$. This can be done locally by taking the $1$-form matrix $A$ of the connection $D$ and writing it as $A=A^{U}+A^{m}$, with $A^{u}\in\mathfrak{u}(n)$ and $A^{m}$ is a Hermitian matrix. Then, we define $D_H=d+A^{u}$ and $\phi=A^m$.

What I want to prove is that, if $F_{D}$ and $F_{H}$ denotes the curvatures of the connections $D$ and $D_H$ respectively, then: $$ F_{D}=0 \Leftrightarrow \begin{cases} F_H+[\phi,\phi]=0\\ D_{H}(\phi)=0 \end{cases} $$

I tried to see this in many different ways but the calculations led me to nowhere. Does anyone have any suggestion?

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    $\begingroup$ $F_D = F_H + D_H(\phi) + \phi \wedge \phi$. You can conclude by taking on the right hand side the hermitian part and the anti-hermitian part. $\endgroup$
    – InfiniteLooper
    Jul 16, 2019 at 10:54
  • $\begingroup$ Nice, thanks for your comment @InfiniteLooper . Do you know how can I see that $D_{H}(\phi)$ is hermitian? $\endgroup$
    – Leonardo Schultz
    Jul 16, 2019 at 16:42
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    $\begingroup$ Write $D_H(\phi)$ as $D_H \circ \phi - \phi \circ D_H$ and you're done as $D_H$ is anti-Hermitian and $\phi$ is Hermitian. $\endgroup$
    – InfiniteLooper
    Jul 16, 2019 at 17:03

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