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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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    In general, the fibers of $P\times_\rho X$ with $X$ a $G$-space, are homeomorphic to $X$. That means the fibers of your $P\times\phi G$ are homeomorphic to $G$.2011-09-09
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    What definition of "principal $G$-bundle" are you using? It is important to know that if we want to prove something is not a principal $G$-bundle!2011-09-09
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    Right, the fibers of $P \times_{\rho} G$ are homeomorphic to $G$ as a space for any action $\rho$ of $G$ on itself, but the fibers are not always the group $G$. Otherwise, we would have a global section (choose the identity element in the fibers) and in the first example I gave, when $\rho$ is multiplication, the principal G-bundle $P \times_{\rho} G$ would be trivial. My definition of a principal bundle is a locally trivial bundle $P\rightarrow X$ with a fiberwise free and transitive $G$ action on $P$.2011-09-09
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    What do you mean by "the fibers are not always the group $G$"?2011-09-09
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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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