Flatness equivalence

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?

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