# Flatness equivalence

Let $$π:E→M\pi:E\rightarrow M$$ be a complex vector bundle and $$HH$$ a hermitian metric over it. If $$DD$$ is a connection over $$EE$$, using the metric $$HH$$, we can decompose it as:
$$D=DH+ϕ D=D_H+\phi$$
Where $$DHD_H$$ is an unitary connection and $$ϕ\phi$$ is Hermitian $$11$$-form with values in $$End(E)\mathrm{End}(E)$$. This can be done locally by taking the $$11$$-form matrix $$AA$$ of the connection $$DD$$ and writing it as $$A=AU+AmA=A^{U}+A^{m}$$, with $$Au∈u(n)A^{u}\in\mathfrak{u}(n)$$ and $$AmA^{m}$$ is a Hermitian matrix. Then, we define $$DH=d+AuD_H=d+A^{u}$$ and $$ϕ=Am\phi=A^m$$.

What I want to prove is that, if $$FDF_{D}$$ and $$FHF_{H}$$ denotes the curvatures of the connections $$DD$$ and $$DHD_H$$ respectively, then:
$$FD=0⇔{FH+[ϕ,ϕ]=0DH(ϕ)=0 F_{D}=0 \Leftrightarrow \begin{cases} F_H+[\phi,\phi]=0\\ D_{H}(\phi)=0 \end{cases}$$

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