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Consider the universal bundle $G\hookrightarrow EG\rightarrow BG$. Is it possible to get another bundle $EG|_{B}$ by restricting $EG$ over a smooth singular $k$-chain $B \in H_k(BG)$ ($k\leq n$)? I know it's perfectly fine to restrict $EG$ to a submanifold of $BG$, but what about chains?

edited Apr 23 2011 at 3:51

asked Apr 22 2011 at 22:12

Kevin Wray
811●2●9

	When you say "a smooth singular chain" where does it live? If you just abstractly think of it then you are asking to build $G$-bundles over $n$-simplices with specified gluing maps. For example if you have two standard simplices which you are gluing along a face then a $G$-bundle over this "chain" is a $G$-bundle over each simplex and a transition data $g:\textrm{face}\to G$ which tells you how to glue the $G$-bundle restricted over one face to the other. – Somnath Basu Apr 22 2011 at 22:47
	So, I am looking for a smooth 4-chain in $BG$ which has boundary the image of a 3-manifold $M$ under the classifying map $\gamma:M\rightarrow BG$; that is $\gamma_\ast[M]$. Then I want to restrict $EG$ to this 4-chain, giving a $G$-bundle over $B$ which restricts to a $G$-bundle over $M$ at the boundary. – Kevin Wray Apr 23 2011 at 2:02
	You should perhaps rephrase your question accordingly to the more specific thing you explained above. – Somnath Basu Apr 23 2011 at 2:32
	Ok, I've rephrased the question. – Kevin Wray Apr 23 2011 at 3:55

Somnath Basu · Answer 1 · 2011-04-23 02:44:50Z UTC

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If you have a $G$-bundle over $M$ (without boundary) then this corresponds to the homotopy class of a map $\gamma:M\to BG$. It is known that for $G$ simply connected, any $G$-bundle over $M$ is trivializable. One way to see this is by obstruction theory. The other is to notice that $\pi_i(BG)=\pi_{i-1}(G)$ and for $i=1,2,3$ this is zero. Therefore, one can get a cellular model for $BG$ which has no $3$-cells. Therefore, $\gamma$ is homotopic to a cellular map $\gamma':M\to BG$ which is necessarily constant. Since $\gamma_\ast[M]=0$ in $H_3(BG)$ by Hurewicz and the previous observations, there is a singular $4$-chain $B$ with boundary $\gamma_\ast[M]$. Then using the necessary pullbacks you get what you want.

answered Apr 23 2011 at 2:44

Somnath Basu
1,692●1●9●18

	Yes, I agree that if $\gamma_\ast[M]=0$ in $H_3(BG)$ then there is some 4-chain $B$ with it as its boundary. However, since $B$ is a 4-chain in $BG$ (which has a bundle over it) does this always mean that I can restrict the bundle over $B$? If so, how does one do this? I mean, don't you need to assume base spaces are manifolds? – Kevin Wray Apr 23 2011 at 3:22
	To clarify the above: say I have a bundle $G\hookrightarrow E\rightarrow N$ over some n-manifold $N$. Then can I always restrict the bundle $E$ to a bundle over a k-chain $C$ in $H_k(N)$ (where $k\leq n$), $E\|_{C}$ giving another bundle? Or, must I take $C$ to be a submanifold of $N$. Thanks. – Kevin Wray Apr 23 2011 at 3:26
	Bundles can be defined with any topological space as base. In the geometric context, you require local triviality, i.e., if $\pi:E\to B$ is a bundle with fibre $F$ then locally $\pi$ should look like the projection map. In the homotopy theoretic context, the homotopy lifting property (also called Serre fibration) defines a fibration $\pi:E\to B$ where a fibre is homotopy equivalent to $F$. – Somnath Basu Apr 23 2011 at 3:49
	So, just to be sure, I can always restrict a principal $G$-bundle to some singular smooth chain, and the resulting object will still have the structure of a principal $G$-bundle? Also, would the construction (in the geometric sense) be analogous to the construction when the base is a manifold? Thanks. – Kevin Wray Apr 23 2011 at 3:55
	(1) There is a distinction that you have to make : if you take an embedded simplex $\Delta$ in $X$ which is part of a $G$-bundles $G\hookrightarrow E\to X$ then $E$ restricted to $2\Delta$ is just two copies of the $G$-bundles over $\Delta$. (2) In general when you construct a $G$-bundle over $X$, you take an open cover of $X$ and take $U_j\times G$. Then you need to glue $U_i \times G$ to $U_j \times G$ over $(U_i\cap U_j)\times G$. If you were looking for bundles with fibre $G$ then the gluing maps would be given by $g_{ij}:U_i\cap U_j\to \textrm{Aut}(G)$. – Somnath Basu Apr 23 2011 at 4:05

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