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Nov 3, 2017 at 1:07 comment added Tom Goodwillie In general in such an "associated bundle with fiber $F$" the action of $\pi_1 BG=\pi_0G$ on homology of the fiber $F$ is given by the given action of $G$ on $F$.
Nov 3, 2017 at 1:06 comment added Tom Goodwillie You can think of $G/H\to BH\to BG$ as a special case of the associated bundle for an action of $G$ on a space $F$, $F\to EG\times_G F\to BG$. Let $F$ be $G/H$ with the action I mentioned. Then the bundle is $G/H\to EG\times_G(G/H)\to BG$, where $G\times_G(G/H)=(EG)/H\sim BH$.
Nov 3, 2017 at 0:46 comment added C. Zhihao Would you mind to illustrate a little bit why the action becomes what you say? it isn't that clear for me
Nov 2, 2017 at 22:04 answer added Nicholas Kuhn timeline score: 4
Nov 2, 2017 at 21:56 comment added Tom Goodwillie The element of $\pi_1(BG)=\pi_0(G)$ represented by $g\in G$ acts on the homology of $G/H$ by the map $xH\mapsto gxH$ from $G/H$ to $G/H$.This can be nontrivial.
Nov 2, 2017 at 20:45 review First posts
Nov 2, 2017 at 21:06
Nov 2, 2017 at 20:43 history asked C. Zhihao CC BY-SA 3.0