About gauge transformation

Question

If $E$ is a vector bundle with a bundle metric, so we have ${\rm Aut}\ (E)$ whose fiber at $x\in M$ is the group of orthogonal transformation in $E_x$. Then gauge transformation is a section of ${\rm Aut}\ E$. Here I have a question. ${\rm Aut}\ E$ is principal. Right ? If so, since there exists a global section, ${\rm Aut}\ E$ is trivial. Am I right ? Thank you in anticipation.

Answer 1

No, $\operatorname{Aut}(E)$ is not a principal bundle. In a local (orthonormal) trivialization of $E$ over $U\subseteq M$, we get a local trivialization of $\operatorname{Aut}(E)$ of the form $\pi_{\operatorname{Aut}(E)}^{-1}(U) \leftrightarrow U\times \operatorname{O}(n)$; however, when two such local trivializations overlap, the transition functions act on $\operatorname{O}(n)$ by conjugation. To be a principal bundle would require that the action on $\operatorname{O}(n)$ be by left multiplication.

The principal bundle associated to $E$ is the bundle of orthonormal frames for $E$. This is trivial if and only if $E$ is trivial.

Answer 2

A global section would indeed trivialize the bundle - if it were a principal bundle! I'm going to work more generally here and talk about the gauge group of a principal $G$-bundle $\pi: P \to M$. The gauge group is defined to be the group of $G$-equivariant diffeomorphisms $\varphi: P \to P$ that cover the identity (take fibers to the same fiber). That is, $\varphi(p \cdot g) = \varphi(p) \cdot g$, and $\pi(p \cdot g) = \pi(p)$.

So let's be careful to say what kind of symmetries this bundle $\text{Aut}(P)$ actually has. Fiberwise it's diffeomorphisms $\varphi_x: G \to G$ such that $\varphi_x(g \cdot h) = \varphi_x(g) \cdot h$. It's not hard to see that $\varphi_x(g) = g' \cdot g$ for some $g' \in G$. (Note left-multiplication instead of right!)

One might want to define an isomorphism $\text{Aut}(P) \to M \times G$ by $(x,\varphi_x) \mapsto (x,g')$ where $g'$ is extracted as above. But $g'$ depended on our identification of the fiber with $G$ - pick a different identification and $g'$ becomes $hg'h^{-1}$. (So, as Jack Lee said, under different trivializations the bundle transforms by conjugation.)

One alternate way of describing $\text{Aut}(P)$ is by passing to $P \times_G G \to P/G$, where $G$ acts on itself by conjugation. This is the (not principal!) $G$-bundle $\text{Aut}(P)$. This bundle is trivial iff the structure group of $P$ could be reduced to $Z(G)$. In particular if $G$ is abelian $\text{Aut}(P)$ is the trivial bundle.