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Please read the edit below.

Is my understanding of this correct? Fix a sufficiently nice and connected topological space $B$ and a topological group $G$. A principal bundle $E\to B$ with structure group $G$ is (modulo equivalence) the same as a collection of transition functions $g_{\alpha,\beta}:V_{\alpha}\cap V_{\beta}\to G$. A fiber bundle $b:Y\to B$ with fiber $F$ defines a collection of transition functions $g_{\alpha,\beta}:V_{\alpha}\cap V_{\beta}\to \mathrm{Aut}(F)$ but you cannot get the fiber bundle back from the transition functions in general. You get only a principal bundle $p:E\to B$ with structure group $G=\mathrm{Aut}(F)$ by the above mentioned equivalence. How does this principal bundle look like (in relation to the fiber bundle $b$) intuitively?

The additional information you need to get a more general fiber bundle from a principal bundle $p:E\to B$ with structure group $G$ is a space $F$ with an action of $G$. Then you define a fiber bundle $$ q:E\times F/\sim\to B $$ where the relation is generated by $(x,y)\sim(xg,gy)$ and the equivalence class of $(x,y)$ is maped to $p(x)$. This bundle has fiber $F$. Applying this construction to the bundle $b$ from above, I suppose that with the obvious action of $G=\mathrm{Aut}(F)$ on $F$ you get the bundle $q\cong b$ back, right?

If the fiber bundle $b$ is a vector bundle (which is generally not a principal bundle) one does not need the second step: They are equivalent to their associated transition functions. Why is this (intuitively) true?

Can someone help me to clarify my picture? Is it really important to take care of the direction of the actions (left action on F, right action on E)? Intuitively, I think that a principal bundle does somehow encode global information where the second construction brings a local action on the fiber in.

Edit: Thank you all for the clarification. Are the following two statements correct?

  1. Fix a topological group $G$ and a topological space $B$. Does this mean that there is an isomorphism of categories $$ X\times Y\to Z $$ where $X$ is the category of topological spaces with a left $G$-action (and $G$-equivariant morphisms), $Y$ is the category of principal bundles over $B$ with structure group $G$ modulo isomorphism (and only identity morphisms) and $Z$ is the category of fiber bundles over $B$ with structure group $G$ (and morphisms over $B$)?

  2. Now fix a topological space $F$ and a topological space $B$. Does this mean that there is an isomorphism of categories $$ X'\times Y'\to Z' $$ where $X'$ is the category of left $\mathrm{Aut}(F)$-actions on $F$ (and $\mathrm{Aut}(F)$-equivariant isomorphisms (?)), $Y'$ is the category of principal bundles over $B$ with structure group $\mathrm{Aut(F)}$ modulo isomorphism (and only identity morphisms) and $Z'$ is the category of fiber bundles over $B$ with fiber $F$ (and isomorphisms over $B$)?

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Since no one has answered the two additional questions, I'll just point out that one needs to be careful in defining the "category of fiber bundles over B with structure group G." For example, I suppose an object in this category includes the data of a mapping `G\to$' Homeo(F), where F is the fiber of the bundle? Even this sounds a bit suspicious to me, since we do not have canonical homeomorphisms among the fibers of the bundle. I'd be interested to see either a precise formulation of these questions (making the answers yes), or a clear explanation of why we shouldn't expect this to work. –  Dan Ramras Mar 2 '10 at 4:00

3 Answers 3

Even in the case of vector bundles, you need the second step: in that case, a representation of $G$ on $GL(V)$, where $V$ is a model fiber.

In all of the cases, the transition functions encode the way that you are supposed to glue fibers together on overlaps, but they don't make any claims about what it is that is being glued. If your transition functions are $G$-valued, you still need to choose a space with a $G$-action to get a bundle.

You have plenty of options here. The easiest is to let $G$ act on itself (or more accurately, a $G$-torsor) on one side or the other. The bundle corresponding to your transition functions is then the associated principal bundle.

Or maybe $G$ has a natural representation $G \to GL(V)$. The corresponding bundle is a vector bundle with fiber $V$.

Or maybe $G$ acts on a topological space $G \to Aut(X)$. The corresponding bundle is a bundle of $X$s.

In each case, your transition functions / cocycle only describe the gluing information. Information about what is to be glued has to be provided by picking an action of $G$ on $\text{Thing}$, giving you a $\text{Thing}$-bundle glued up by the pattern described by your cocycle.

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Why do you need the second step in the case of vector bundles? I read in Switzer, p.195 that a vector bundle over B is the same as a principal $GL_n(F)$-bundle over B. But a principal $G=GL_n(F)$ bundle has fiber $G$ and a vector bundle has not (but $F^n$)! I am confused. –  veit79 Feb 16 '10 at 8:59
    
As you say, they are certainly not the same. However, they are equivalent structures: given one kind of bundle, one can canonically construct the other kind of bundle, with the same set of transition functions. –  Pete L. Clark Feb 16 '10 at 15:07
    
It is confusing because the group which your transition function takes values in often has several canonical actions on various spaces. This leads authors to sometimes sweep over the fact that these actions are also important data. In your case, $GL_n(F)$ has natural actions on both $F^n$ (giving a vector bundle) and the set of frames in $F^n$ (giving a principal $GL_n(F)$ bundle, as in Dan's answer). The same cocycle can be used to construct either type of bundle; the only difference is which action you provide. –  Matt Noonan Feb 17 '10 at 1:58

Almost all of what you say is correct, so it's not easy for me to see exactly where your confusion lies.

Maybe it's here: the structure of an $(F,G)$ bundle includes a topological space $F$ and a topological action of a topological group $G$ on $F$. In the case of a principal bundle, we take $F = G$ and the action to be the (left- or right-, depending upon how you set it up) regular one.

Once you have the structure of $F$ and the action of $G$, the transition functions completely determine the $(F,G)$-bundle up to equivalence. The existence is handled via a quotient construction as you have indicated above.

In the case of (say real) vector bundles, you still need the structure group $\operatorname{GL}_n(\mathbb{R})$. This acts naturally by homeomorphisms on $\mathbb{R}^n$ but it is by no means the entire self-homeomorphism group of $\mathbb{R}^n$ -- the latter is an enormous, infinite-dimensional object.

The fact that the structure of the fiber bundle is determined by a cocycle with values in $\underline{G}$ -- the sheaf of continuous [or perhaps smooth, analytic,...] $G$-valued functions on the base space -- is extremely convenient and useful in the theory: it gives you natural bijective correspondences between geometric objects with very different looking fibers.

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I started writing my answer before Matt Noonan's answer appeared. I think that mine is essentially the same answer as his, but maybe it will be helpful to hear essentially the same thing in two slightly different ways. –  Pete L. Clark Feb 15 '10 at 20:38

For comparing vector bundles to principal GL$(n)$ bundles, there is a second, more canonical viewpoint that you may find useful. If $E \to B$ is a vector bundle, then the bundle of $n$-frames in E is a principal GL$(n)$ bundle over $B$, denoted by Fr$(E) \to B$. This bundle consists of all $n$-tuples $(v_1, \ldots, v_n)$ such that the $v_i$ form a basis for some fiber of $B$. The general linear group acts on Fr$(E)$ by the following formula, which just mimics matrix multiplication: $$((v_1, \ldots, v_n) \cdot (a_{ij}) = (\sum a_{i1} v_i, \ldots, \sum_{in} v_i),$$ where $A = (a_{ij}) \in \textrm{GL}(n)$.

You can check that Fr$(E)$ is a principal GL$(n)$ bundle, and that for any choice of transition functions for the bundle $E$, the principal GL$(n)$ bundle built from that data is isomorphic to Fr$(E)$.

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