Dear Pan Zhang:
Your question is not completely clear. I guess you do not really mean a hermitian metric on the manifold $P$ being an $SL(n,C)$-principal bundle over a base space $M$. This does not make to much sense ( $P$ need not even to be even-dimensional). I guess what you want to ask is how to define a hermitian metric on any fibre of an associated complex vector bundle $E$ which is compatible with the given structure group, right?
Thus suppose you have a complex vector bundle $E$ over $M$ with fibre dimension $n$ and with a complex determinant on each fibre, depending smoothly on the base. The determinant can be viewed as a global section of the complex-one-dimensional bundle $\Lambda^n(E)$ (highest exterior power).
Take a finite covering (locally finite would also suffice) of the base space $M$ by contractible open subsets $U_\alpha$ and a corresponding decomposition of unity: a family of functions $\phi_\alpha$ having support on $U_\alpha$ and adding up to one, $\sum_\alpha \phi_\alpha(x) = 1$ for all $x\in M$. Choose any hermitian metric $h_\alpha$ on every fibre $E_x$ over $U_\alpha$, depending smoothly on $x\in U_\alpha$. Add up all $h_\alpha$ to a hermitian metric on the whole bundle $E$, using the decomposition of unity: $h = \sum_\alpha h_\alpha$. Any unitary basis $b = (b_1,...,b_n)$ of $E_x$ has a certain determinant $\det b$ whose modulus $|\det b|$ is independent of the choice of the unitary basis. Let $a(x) > 0$ be the $n$-th root of $|\det b|$. Now change your hermitian metric $h$ to $\tilde h = h/a^2$.
Does this construction meet your question?
Best regards
Jost
Dear Professor,
Thanks very much for your explanation. I think it makes sense. In fact, I am reading a paper titled "Flat G-bundles with canonical metrics". In the third page, it goes "Suppose P has been given a Hermitian metric, i.e. a fixed reduction of structure group to SU(n)".
Yes, I think what I wrote refers to this formulation.
Best regards
Jost
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