# Associated bundle

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.

-
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 $G$. – Mariano Suárez-Alvarez Sep 9 '11 at 20:49
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! – Mariano Suárez-Alvarez Sep 9 '11 at 20:50
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 $P$. – Manuel Sep 9 '11 at 21:11
What do you mean by "the fibers are not always the group $G$"? – Mariano Suárez-Alvarez Sep 9 '11 at 21:17
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. – Mariano Suárez-Alvarez Sep 9 '11 at 21:18

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.

-
@Tychonoff: he used it to show his section is well-defined. If you try to repeat his argument with the multiplication action, you will see it does not work. Jason is using the fact I observed above, that $1_G$ is a somewhat special fixed point for the conjugation action. – Mariano Suárez-Alvarez Sep 10 '11 at 1:27
@Tychonoff: Exactly what Mariano said. – Jason DeVito Sep 10 '11 at 2:27
@Mariano: I didn't intend to ignore your observation and not give you credit for it. I had actually typed up everything but the continuity argument before you and Tychonoff had your minidiscussion. If you'd prefer, I can edit the post to reference your observation directly. – Jason DeVito Sep 10 '11 at 2:30
@Jason: of course not! Thank you for writing it in detail :) – Mariano Suárez-Alvarez Sep 10 '11 at 2:41