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?
-
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 $B$)? -
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 $B$)?
Best Answer
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.