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Let $ G \hookrightarrow P \xrightarrow{ \pi } M $ be a principal $ G $-bundle over $M$. Denote by $\mathrm{Ad} (P) = P \times _{ \mathrm{Ad} } G $ the non-linear adjoint bundle.

It seems to me that a bundle automorphism $\tilde f$ induces a section of the non linear adjoint bundle $f:M\rightarrow \mathrm{Ad}(P)$. In fact, given $\tilde f$, for $x\in M $ define \begin{equation} f(x) =[\tilde f(p),e], \end{equation} where $p$ is any point in the fibre over $x$. This is well defined since \begin{equation} [ \tilde f( p ^\prime ),e] =[ \tilde f( p \cdot g ),e] =[ \tilde f( p )\cdot g,e] =[ \tilde f( p ), g ^{-1} e g] =[ \tilde f( p ), e]. \end{equation} It is a section since \begin{equation} (\pi\circ f)(x)=\pi\left ([\tilde f(p),e]\right )=[\pi_P(\tilde f(p)),e]=[\pi_P(p),e]=x. \end{equation}

On the other hand it seems to me that a section $f:M\rightarrow \mathrm{Ad}(P)$ of the nonlinear adjoint bundle does not contain enough information to determine a bundle autmorphism, in fact if we try to define $\tilde f(p)$ via \begin{equation} f(\pi (p)) =[\tilde f(p),e] \end{equation} we find that $f$ satisfies the properties of an automorphism, but is defined only up to a $G$-action, that is we cannot tell the difference between $f(p)$ and $f(p)\cdot g$.

Is what I say correct?

EDIT: I have read on a book that there is a bijective correspondence between principal bundle automorphisms and sections of the adjoint bundle. While I still think that what I have written above is correct, and would like to receive feedback on that, I think that in order to get a bijective correspondence one defines a $G$-equivariant map $g_f: P\rightarrow G$ in terms of a section of $\mathrm{Ad}(P)$ / a section of $\mathrm{Ad}(P)$ in terms of a $G$-equivariant map $g_f$ via the formula \begin{equation} f(x)=[p,g_f(p)]. \end{equation} Then a $G$-equivariant map $g_f: P\rightarrow G$ determines / is determined by a bundle automorphism $\tilde f$ via \begin{equation} \tilde{f} (p)=p\cdot g_f (p). \end{equation} Again, is this correct?

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A section of the adjoint bundle $\mathrm{Ad}(P)$ determines an automorphism of $P$ since the group of automorphisms can be identified with $\{\lambda:P\to G\mid\lambda(pg)=g^{-1}\lambda(p)g\}$, which can be identified itself with the sections of the associated bundle with fiber $G$ under the adjoint action, i.e. $\mathrm{Ad}(P)$. – gofvonx Jun 29 '14 at 10:53
@gofvonx thanks, I was just reading that one is determined by the other. I have edited my post in accord. – GFR Jun 29 '14 at 11:01

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