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Given a principal $G$-Bundle $P\rightarrow X$ and if we let $G$ act on itself by multiplication (denote this action by $\rho$) we obtain an associated bundle $P\times_{\rho} G=(P\times G)/\sim$ where $(p,f)\sim(gp, gf)$ with fibers homeomorphic to $G$. It is easy to check that this bundle is G-principal; the G-action is given by $g[p,f]=[p,fg]$.

If we change the action of $G$ on itself to conjugation (denote this action by $\phi$) we can construct an associated bundle $P \times_{\phi} G$ in the same way. However, I want to show that this is not a principal bundle by showing that the fibers are not only homeomorphic to $G$ but they are groups isomorphic to $G$. (This would imply that the bundle has a section and if it were principal it would be trivial and there are examples of non trivial associated conjugation bundles).

How can I show that each fiber of $P \times_{\phi} G$ is a fiber bundle of groups? I would probably have to use the fact that conjugation by a fixed element is a group isomorphism, since in the case of multiplication, which is not a group isomorphism, the obtained bundle is principal.

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    If that is your definition, then you simply have to check that the action of $G$ on the fibers is not transitive---this is immediate, because $1_G$ is a fixed point.2011-09-09

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I'm not sure exactly what you mean by "fibers are not only homeomorphic to $G$ but they are groups isomorphic to $G$" - what opeartion are you putting on the fibers before asking this question?

On the other hand, I think I can prove there is a section without this part.

Define a map $X\rightarrow P\times_\phi G$ by $x\rightarrow [(p,e)]$ where $p$ is any element of $P$ with $\pi(p) = x$ and $e\in G$ the identity element.

I first claim this map is well defined. For if \pi(p) = \pi(p'), then there is an element $g\in G$ with gp = p'. For this $g$, we then have [(p,e)] = [g*(p,e)] = [(gp, geg^{-1}] = [(p',e)], so the map is well defined.

To see it's continuous, note that this can be checked locally. If $U\subseteq X$ with both the principal and associated bundles trivial over $U$, then, when restricted to $U$, the map from $X$ to $P\times_\phi G$ factors as $X\rightarrow U\times G\times G\rightarrow P\times G\rightarrow P\times_\phi G$ where the first map sends $x$ to $(x,e,e)$ and the second map sends $U\times G$ to $\pi^{-1}(U)\subseteq P$ via a trivialization and the third map is the natural projection. Since all of these maps are continuous, our section factors (locally) as a composition of continuous functions, hence is continuous itself.

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    @Jason: of course not! Thank you for writing it in detail :)2011-09-10