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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 $G/H$?

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    $\begingroup$ 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. $\endgroup$
    – Andy Putman
    Commented May 12, 2023 at 20:35
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    $\begingroup$ 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. $\endgroup$
    – archipelago
    Commented May 13, 2023 at 14:24

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