Why are the fibers of principal G-bundles homeomorphic to G?

Question

I'm trying to get a grip on the modern geometric formulation of gauge theory, in particular connections on principal G-bundles. However, I am stuck right after the definition already: Virtually all introductory texts I found mention (without proof) the fact that the fiber of such a bundle is homeomorphic to the structural group. To avoid ambiguity, let's use the Wikipedia definition:

Definition: Let $G$ be a topological group. A principal $G$-bundle is a fiber bundle $F \hookrightarrow P \xrightarrow{\pi} X$ together with a continuous right action $$P \times G \to P$$ $$(p,g) \mapsto p.g$$ of $G$ on $P$ which preserves fibers, i.e. $$\pi(p.g) = \pi(p) $$ $\forall p \in P$, $g \in G$ and is free and transitive on all $F_x = \pi^{-1}(\{x\})$.

It appears to be a standard fact that $F \cong G$; I haven't a clue how to prove this, though. My best guess is to fix an element $p\in P$ and consider the restricted group action $$G\cong\{p\}\times G \to F_{\pi(p)} \cong F$$ By transitivity, freeness and continuity of the group action, this map is a continuous bijection which is not necessarily a homeomorphism unless e.g. $G$ is compact and $F$ is Hausdorff. I can buy that $F$ is Hausdorff if we restrict ourselves to manifolds, but even then $G$ certainly does not have to be compact. What am I missing? Any help is greatly appreciated.

Answer 1

After closer perusal of Wikipedia's definitions, I think there's an omitted condition. Here are the relevant definitions:

Principal bundle

A principal $G$-bundle, where $G$ denotes any topological group, is a fiber bundle $\pi:P \to X$ together with a continuous right action $P × G \to P$ such that $G$ preserves the fibers of $P$ (i.e. if $y \in P_{x}$ then $yg \in P_{x}$ for all $g \in G$) and acts freely and transitively on them. This implies that each fiber of the bundle is homeomorphic to the group $G$ itself.

(Emphasis added.)

Fibre bundle (Notation changed to match the preceding definition)

A fiber bundle is a structure $(P, X, \pi, F)$, where $P$, $X$, and $F$ are topological spaces and $\pi:P \to X$ is a continuous surjection satisfying a local triviality condition outlined below. The space $X$ is called the base space of the bundle, $P$ the total space, and $F$ the fiber. The map $\pi$ is called the projection map (or bundle projection). We shall assume in what follows that the base space $X$ is connected.

We require that for every $x$ in $X$, there is an open neighborhood $U \subset X$ of $\pi(x)$ (which will be called a trivializing neighborhood) such that there is a homeomorphism $\phi:\pi^{-1}(U) \to U × F$ (where $U × F$ is the product space) in such a way that $\pi$ agrees with the projection onto the first factor.

Using these definitions, it appears one can view a topological group $G$ with a non-discrete topology as the total space of a principal $G$-bundle over a point, with $G$ (as structure group) having the discrete topology: Right multiplication defines a continuous, free, transitive action, but the fibre (non-discrete $G$) is not homeomorphic to the structure group (discrete $G$).

Elsewhere, however:

Principal homogeneous space

In mathematics, a principal homogeneous space, or torsor, for a group $G$ is a homogeneous space $X$ for $G$ in which the stabilizer subgroup of every point is trivial. Equivalently, a principal homogeneous space for a group $G$ is a non-empty set $X$ on which $G$ acts freely and transitively, meaning that for any $x$, $y$ in $X$ there exists a unique $g$ in $G$ such that $x·g = y$ where $·$ denotes the (right) action of $G$ on $X$. [...]

$X$ is a $G$-torsor if $X$ is nonempty and is equipped with a map (in the appropriate category) $X × G \to X$ such that $x·1 = x$ and $x·(gh) = (x·g)·h$ for all $x \in X$ and all $g$, $h \in G$ and such that the map $X × G \to X × X$ given by $(x, g) \mapsto (x, x · g)$ is an isomorphism (of sets, or topological spaces or ..., as appropriate). Note that this means that $X$ and $G$ are isomorphic.

(Emphasis added.) Here, if I'm reading correctly, one explicitly assumes the fibre and structure group are homeomorphic.

I don't have a copy of Kobayashi-Nomizu handy, but that would be the first place I'd go for a working definition. If memory serves, K-N take $F = G$ in the definition, but it's been A Long While since I checked.

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All that said, this was my (in retrospect, overly hasty) idea in the comments: A principal $G$-bundle $\pi:P \to X$ is a fibre bundle with fibre $F$, and equipped with a continuous right $G$-action that is free and transitive on fibres. Local triviality means that if $p \in P$ is arbitrary and if $x = \pi(p)$, there exists an open neighborhood $U$ of $x$ and a $G$-equivariant homeomorphism $\phi:\pi^{-1}(U) \to U × F$ under which $\pi$ corresponds to projection to the first factor. That is, up to $G$-equivariant homeomorphism, we may assume $P = U × F$, and that the $G$-action has the form $$ (p, g) \simeq \bigl((x, f), g\bigr) \mapsto (x, fg). $$ Since the action is free and transitive, fixing an $f$ in $F$ defines a continuous bijection $U × G \to U × F$ by $(x, g) \mapsto (x, fg)$. Superficially, it appeared one could go backward continuously from the fibre to the structure group by including a fibre in $U × F$, using the group action to identify $U × F \simeq U × G$, and projecting to the second factor. Contrary to my initial impression, however, continuity of the reverse identification does not follow from the first definition above.

Ultimately, the fact that the structure group and the fibres of a principal bundle are homeomorphic shouldn't be a technical issue, but a routine (possibly explicit) consequence of the definition. :)

Answer 2

If the action of a topological group $G$ on a manifold $E$ is free, then $G \cong \mathcal{O}_p$ for any $p$ in $E$.

Now, recall that the action on fiber bundles also preserves fibers and is transitive. This is implies that for any $x\in M$ and $p\in \pi^{-1}(x)$, $\mathcal{O}_p = F_x$. Since fibers are homeomorphic, $F \cong G$.