4
$\begingroup$

If $H$ is a Lie subgroup of $G$, then there is a fibration sequence $$ G/H\to BH\to BG. $$ By choosing a model for $EG$ we can promote this into a fibre bundle.

My question is about how to understand the monodromy action for this fibration, which should be an action of $\pi_1(BG)\cong\pi_0(G)$ on the homotopy groups of $G/H$. Is it really just given by lifting an element of $\pi_0(G)$ to $G$ and then left-translation on $G/H$?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ I think the fact that it has the description you give is obvious if you use the right models. Here I'm thinking of the model for $EG$ as the simplicial space whose space of $k$-simplices are sequences $(g_0,\ldots,g_k)$ with the $g_i$ in $G$. You have $BH = EG/H$ and $BG=EG/G$, and $BH \rightarrow BG$ is the obvious map. Now $BG$ is the bar resolution, so $\pi_1(BG)$ is generated by the loops corresponding to $1$-simplices $[g]$, and these have obvious lifts to $BH$ (namely, the image of the $1$-simplex $(1,g)$ in $BH$). For these, the monodromy action is obviously what it is supposed to be. $\endgroup$ Commented May 12, 2023 at 20:35
  • 1
    $\begingroup$ The fundamental group of the base of a fibre sequence acts on the fibre over the basepoint in the \emph{unpointed} homotopy category, so in general there is no induced action on the homotopy groups. The fundamental group of the total space however acts on the fibre in the \emph{pointed} homotopy category, and thus in particular on its homotopy groups. In your example, the action of $\pi_1(BG)=\pi_0(G)$ on $G/H$ is indeed given by left-translation, which is not pointed, but can be enhanced to a pointed action once one restricts it to $\pi_0(H)$ since $[1] \in G/H$ is now preserved. $\endgroup$ Commented May 13, 2023 at 14:24

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.