This is a cross-post from [math.SE](https://math.stackexchange.com/q/1022034/120628)


Let $E \to M$ be a complex vector bundle with $P$ the associated $GL(n,\mathbb C)$ frame bundle. The group of gauge transformations is the space of sections of $P\times\_{Ad} GL(n,\mathbb C)$ and the space of hermitian metrics is the space of sections of $P\times\_{GL(n,\mathbb C)} GL(n,\mathbb C)/U(n)$.


What is a clean way to show that the group of gauge transformations acts transitively on the space of hermitian metrics? That is, suppose $h\_1, h\_2$ are hermitian metrics, which are identifiable with $GL(n,\mathbb C)$-equivariant maps $P \to GL(n,\mathbb C)/U(n)$, then we need to find an equivariant map $\phi : P \to GL(n,\mathbb C)$ such that $\phi h\_1 = h\_2$. I could see how to get this if we know that every equivariant map $P\to GL(n,\mathbb C)/U(n)$ lifts to an equivariant map $P\to GL(n,\mathbb C)$ but I'm not sure if that's true. Maybe the fact that $GL(n,\mathbb C)/U(n)$ is contractible comes into play.



Transitivity of the action of the group of gauge transformations on the space of hermitian metrics
View $h\_i$ as conjugate linear bundle isomorphisms $h\_i:E\to E^\*$. 
Then $\phi=h\_1^{-1}\circ h\_2$ satisfies $h\_2(x,y)=h\_1(\phi.x,y)$. 
Since $h\_2$ is positive hermitian, the endomorphism $\phi$ is positive hermitian with respect to $h\_1$, thus $\sqrt{\phi}$ exists with respect to $h\_1$ and $h\_2(x,y)=h\_1(\sqrt{\phi}.x,\sqrt{\phi}.y)$.


