Flatness equivalence

Let π:EM 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=DH+ϕ
Where DH is an unitary connection and ϕ is Hermitian 1-form with values in End(E). This can be done locally by taking the 1-form matrix A of the connection D and writing it as A=AU+Am, with Auu(n) and Am is a Hermitian matrix. Then, we define DH=d+Au and ϕ=Am.

What I want to prove is that, if FD and FH denotes the curvatures of the connections D and DH respectively, then:
FD=0{FH+[ϕ,ϕ]=0DH(ϕ)=0

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

Answer

Attribution
Source : Link , Question Author : Leonardo Schultz , Answer Author : Community

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