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I was studying about fiber bundles with a $G$-structure, and I arrive to the definition (below all the spaces are smooth manifolds):

Given a topological group $G$ a $G$-structure on fiber bundle $(E,M,F)$ with projection $\pi:E\rightarrow M$ and typical fiber $G$ (or a $G$-bundle) consists (roughly) on:

1) A faithful action of $G$ on the typical fiber $F$.

2) A bundle atlas (collection of open trivialization neighborhoods which cover the base space $M$) $\{U_\alpha , \phi_\alpha \}_{\alpha \in \mathcal{A}}$ with $\phi_\alpha:\pi^{-1}(U_\alpha) \rightarrow U_\alpha \times F$. It is usually called $G$-atlas.

3) For all $\alpha,\beta \in \mathcal{A}$ there exists a smooth map $g_{\alpha,\beta}:U_\alpha \cap U_\beta \rightarrow G$, called the transition function from $\phi_\alpha$ to $\phi_\beta$, such $g_{\alpha,\beta}(p)=\phi_\beta \circ \phi_\alpha^{-1}(p)$.

The first thing that bothers me is that in the above definition is important which is the $G$-atlas you first consider. I used to think that in differential geometry, given any atlas of a manifold $M$, if you change it by a bigger atlas (adding new coordinates charts compatible with de previous one) or by a smaller atlas (keeping a subset of charts which they still covers $M$), all the relevant topological and differential structure should not change (please let me known if this is incorrect). But, this is what not happen in the definition of the G-structure of a fiber bundle.

After all, the same fiber bundle may be equipped with many different G-structures, depending on the $G$-atlas you considered. E.g. I'm thinking a globally trivial vector bundle, with can be equipped by a trivial group structure according to the trivial group $G=\{1\}$ or by the $GL(\mathbb{R^n})$-structure.

Then, the two main questions are:

1) There is an "intrinsic" definition of G-structure on a fiber bundle such not depends on the $G$-atlas, i.e. some kind of "coordinate free" definition? If I'm well understanding, I hope such definition can't exist.

2) Given any fiber bundle $(E,M,F)$ (without a priori G-structure) and consider some bundle atlas of such fiber bundle. Now I can construct all the transition functions $t_{\alpha\ \beta}=\phi_\alpha \circ \phi_\beta^{-1} : F \rightarrow F$. Does the collection of such transition functions ${t_{\alpha\ \beta}}$ define a G-structure on $F$ ? It may not define a topological group, but I expect that the algebraic group properties of the definition of G-structure on a fiber bundle must be satisfied. In other words, every fiber bundle admits a $G$-structure, but it usually depends on the bundle atlas considered.

Thanks,

D

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2 Answers 2

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What you are getting at, I think, is the idea of reduction of structure group. This is a very nice aspect of bundles actually. For example, if we only allowed some minimal group to act, then we could not consider unoriented vector bundles with structure group $O(n)$, and oriented bundles with structure group $SO(n)$ at the same time.

There is however a coordinate free definition of $G$-principal bundles, and then through those bundles a coordinate free definition of $G$ bundles.

One way to define a $G$ principal bundle is as a homotopy class of map $$f:M\to BG$$. Then pull back $EG$ along $f$. To get a general bundle with fiber a $G$-space $F$, simply define $E=f^*(EG)\times_G F$.

As for your second question, yes this gives a fibre bundle with structure group $Aut(F)$.

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If you suppose that $E$ is a principal $H$-bundle and $G\subset H$, a $G$-structure is defined by a global section of $E/H$.

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  • $\begingroup$ Did you mean E/G? E/H isn’t even a bundle. $\endgroup$ Commented Mar 27 at 14:33

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