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## Totally disconnected homogeneous spaces

I was hoping to give the proof of Rodier’s Mackey-type theorem in this post (the one that implies the Stone-von Neumann theorem), but in the interest of keeping this at a reasonable length I’m first going to prove a fact which I think was used implicitly in Rodier’s proof. Even if we don’t directly use this, it seems more or less essential to understanding the quasi-inverse of the equivalence in the theorem.

This result is also pretty interesting on its own. The case where $G$ is compact, hence profinite, is at the beginning of Serre’s Galois Cohomology.

Lemma If $G$ is an $\ell$-group and $H$ is a closed subgroup of $G$, then $\pi : G \to G/H$ is a locally trivial $H$-bundle. This means that any point $x \in G/H$ has an open neighborhood $U$ such that there is an $H$-equivariant homeomorphism $\pi^{-1}(U) \to U \times H$ which commutes with $\pi$ and the natural projection $U \times H \to U$.

Proof. First, we claim that if $G$ is compact, i.e. profinite, then $\pi : G \to G/H$ is a globally trivial $H$-bundle. More generally, if $K \subset H \subset G$ are closed subgroups, then the projection $\pi : G/K \to G/H$ admits a continuous section $s : G/H \to G/K$. To see why this implies the previous statement, take $K = 1$ and consider the map $G/H \times H \to G$ which sends $(x,h) \mapsto s(x)h$, which is clearly a continuous $H$-equivariant bijection and commutes with the projections to $G/H$.  But a continuous bijection between compact Hausdorff spaces is a homeomorphism.

To prove the existence of this continuous section, assume for the moment that $H/K$ is finite, so there is an open subgroup $U$ of $G$ such that $H \cap U \subset K$. This means $\pi|_{UK/K}$ is injective, and a continuous bijection between compact Hausdorff spaces is a homeomorphism. So $\pi$ has a continuous section over $\pi(UK/K)$, and since $G/H$ is covered by the open translates $g \cdot \pi(UK/K)$ we can put these together to obtain a continuous section over all of $G/H$. Now if $H/K$ is not necessarily finite, consider the set of pairs $(L,s)$ where $K \subset L \subset H$ is a closed subgroup and $s : G/H \to G/L$ is a continuous section. There is a natural partial ordering on these pairs: let $(L_1,s_1) \leq (L_2,s_2)$ provided that $L_1 \subset L_2$ and $s_1,s_2$ commute with the natural projection $G/L_1 \to G/L_2$. Obviously this partially ordered set is nonempty, since we can take $L = H$ and $s$ to be the identity. If $\mathcal{C}$ is a chain of such pairs, then it is not hard to see that if $L$ is the intersection of all the subgroups and $s$ is the restriction of the sections to $L$, then $(L,s)$ is a lower bound for $\mathcal{C}$, since the natural map

$G/L \to \varprojlim_{L_i \in \mathcal{C}} G/L_i$

is a homeomorphism. Thus by Zorn’s lemma there is a minimal element $(L,s)$, and we will show that $L = K$. Otherwise, we can find an open subgroup $U$ of $G$ such that $L \cap U$ is strictly contained in $L$, and since $L/L \cap U$ is finite the first argument yields a continuous section $t : G/L \cap U \to G/L$. But then $(L \cap U,t \circ s) < (L,s)$, which contradicts the minimality of $(L,s)$.

Finally, let $G$ be any $\ell$-group and $H$ a closed subgroup. Clearly it suffices to show that $\pi : G \to G/H$ is a trivial $H$-bundle over some neighborhood of $\pi(1)$. For this, choose a compact open subgroup $K \subset G$ and put $U = \pi(K)$. By the compact case there is a continuous section $s : U \to K \subset \pi^{-1}(U)$, so the map $(u,h) \mapsto s(u)h$ is a continuous $H$-equivariant bijection $U \times H \to \pi^{-1}(U)$ which commutes with the projections to $U$. But this map is the disjoint union of the maps $U \times hH \cap K \to Kh(H \cap K)$ for various $h \in H$, and these are all homeomorphisms by the same argument as above. Since the $Kh(H \cap K)$ are open in $KH = \pi^{-1}(U)$, it follows that $U \times H \to \pi^{-1}(U)$ is a homeomorphism.

$\Box$