I've been reviewing fiber bundles from the nitty-gritty standpoint of atlases and want an intrinsic definition of the structure group based solely on the topology of the bundle $(E,B,\pi)$. Steenrod includes a top group action in the def of a fiber bundle, thus imposing it externally. Some other authors define a fiber bundle as taking any open cover of charts, but then ignore the question altogether.
If we start with a bundle $(E,B,\pi)$ which admits some atlas (and thus has homeomorphic fibers), there should be an intrinsic structure group that reflects how fibers topologically twist as we move around $B$. I.e. we shouldn't need to be told to confine ourselves to atlases with overlap $g_{ij}$'s in some external $G$. I'll use the term "atlas" in this question to refer to one relative to $Homeo(F)$ as a whole (i.e. no $G$-restriction on the overlaps).
The topology of $E$ dictates the set of open subsets $U\subset B$ which appear in charts, but any given $U$ has a vast number of associated charts. Ex. given any $\psi:\pi^{-1}(U)\rightarrow U\times F$ and any $\lambda\in Homeo(F)$, we can construct another chart with the same $U$ and $\psi'= (Id_U,\lambda)\circ\psi$. In fact, we can compose with any continuous fn $\lambda':U\rightarrow Homeo(F)$. Similarly, given an atlas, we can obtain another by composing $\lambda$ across the board (or composing each chart with a fn $\lambda'_i$ as above).
Given any atlas, the topology of the bundle manifests in the overlap $g_{ij}$'s. Atlas $A$ has associated group $G(A)$ generated by those $g_{ij}$'s (i.e. the homeomorphism on $F$ at each point in $U_i\cap U_j$) induced by the overlap maps.
An atlas $A$ has the property of being "maximal" if no new charts can be added relative to its own $G(A)$. It is maximal only relative to that group, and in fact manifests (in some overlap map) every element of $G(A)$. Every atlas $A$ has a unique maximal atlas $\hat{A}$ which has $G(\hat{A})= G(A)$. An atlas is "minimal" if we can't remove a single chart without ceasing to be a cover of $B$.
There exist atlases (and minimal atlases) which have varying groups $G(A)$. Consider the Mobius strip $M$, with $B=S^1$ and $F=I$. There are atlases with $4$ charts whose $U$'s cover a little over a quarter of the circle each and whose overlap $g_{ij}$'s happen to be constant. If we denote $i$ the flip map $t\rightarrow 1-t$, $e= Id_F$, and $x$ some other homeomorphism on $F$ (ex. $t\rightarrow t^2$), the ops $(e,i,x)$ generate some group $H\subset Homeo(F)$. There are minimal atlases of the type described with the following values (among others) for the overlap $g$'s: $[e,i,e,e]$, $[i,i,i,e]$, $[e,xix^{-1},e,e]$.
The question is how do we intrinsically (i.e. without being told $G$) identify that $(e,i)\approx Z_2$ is the correct structure group of the bundle $(M,S^1,\pi)$?
There are two possibilities which leap out at me, and I don't know which (if either) is the case or how to prove them. The idea is that either there is
- (Possibility 1): A unique (modulo inner-automorphism of $Homeo(F)$) largest subgroup $H\subset Homeo(F)$ s.t. for every minimal atlas $A$, $G(A)$ contains $H$ modulo an inner-automorphism of $Homeo(F)$ (i.e. $H\subseteq \lambda G(A)\lambda^{-1}$ for some $\lambda\in Homeo(F)$).
or
- (Possibility 2): A unique (modulo inner-automorphism of $Homeo(F)$) smallest subgroup $H\subset Homeo(F)$ s.t. every maximal atlas has some minimal atlas with $H=G(A)$ modulo an inner automorphism (i.e. $H= \lambda G(A)\lambda^{-1}$ for some $\lambda\in Homeo(F)$).
My take is that we at least need the "modulo an inner-automorphism" due to the $\lambda$-flexibility above, and that the best we can do is narrow down the structure group to a class of groups related by inner-automorphisms of $Homeo(F)$.
Anyway, I'd appreciate any insight into how we would determine the appropriate structure group intrinsically from the basic bundle topology and whether anyone has a simple proof (or counterexample) for either of these two possibilities.
Cheers, Ken
Let $F$ be a $G$-space. It is a classical that isomorphism classes of $F$-bundles with structure group $G$ are in correspondence with principal $G$-bundles, so one can often restrict their attention to principal $G$-bundles. Principal $G$-bundles are classified by homotopy classes of maps into a space called $BG$ which is characterized by i) being connected and ii) having an equivalence $\Omega BG \simeq G$.
In terms of principal bundles, the question of reducing the structure group from $G$ to $H \leq G$ is equivalent to asking what the image of $[X,BH] \rightarrow [X,BG]$ is. As one might imagine, this is a difficult question in general. One case in which it is approachable is if one can argue that $BH \rightarrow BG$ is a homotopy equivalence. This happens, if and only if, the inclusion $H \leq G$ is a homotopy equivalence of spaces. This happens a lot in practice, notably, for the chain of inclusions $O(n) \leq GL(n) \leq \mathrm{Diff}(n)$, but not for the "inclusion" $\mathrm{Diff}(n) \leq \mathrm{Homeo}(n)$. I write "inclusion" here because there are subtle issues with the topologies involved.
If one is dealing with an inclusion that is not a homotopy equivalence, one could start by understanding $\mathrm{hofiber}(BH \rightarrow BG)$. From ii) it is fairly straightforward to show $\mathrm{hofiber}(BH \rightarrow BG) \simeq G/H$, at least when their is nothing weird going on with the topologies of the groups. An example of where we know this homotopy type is $\mathrm{Top}(n)/ PL(n)$ when $n\geq 4$. It has the homotopy type of $K(\mathbb{Z}/2,3)$. This calculation is the fundamental calculation in Triangulation theory which characterizes which manifolds (above dimension 4) can be made PL in terms of a single cohomology class called the Kirby–Siebenmann invariant in $H^4(M; \mathbb{Z}/2)$.
There is a lot more to be said about this question, particularly in regard to characteristic classes. These are elements in the cohomology of the classifying space $BG$, and these can be used to decide questions of structure group reduction. The most elementary example of this is that a vector bundle is orientable, if and only if, its first Stiefel-Whitney class is trivial. For an introduction to characteristic classes of vector bundles, I highly recommend Milnor-Stasheff.