Let π:E→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=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 Au∈u(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)=0I 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*