Let $G$ be a profinite group and $H$ be an open subgroup. As a continuous $H$-topological space, we have $G=\coprod_{G/H} H$. Does this also hold as condensed sets, i.e. do we have an identification of $\underline{H}$-condensed sets $\underline{G}= \coprod_{G/H} \underline{H}$ ? If not, what can we say about the structure of $\underline{G}$ as an $\underline{H}$-condensed set ? More generally, can we say something about the structure of a compactly generated topological group as a condensed set with an action of (the condensed group associated to) an open subgroup of finite index ?

  • $\begingroup$ I think that the functor $X\mapsto\underline X$ from T1 topological spaces to condensed sets preserves coproducts. Filtered colimits of condensed sets could be computed pointwisely w.r.t. extremally disconnected sets, thus it suffices to show that this functor preserves finite coproducts. Finite coproducts of condensed sets is the sheafification of finite coproducts of presheaves on the site of extremally disconnected sets, and the result follows from unrolling the definitions (and the compactness, in general topology, of extremally disconnected sets). $\endgroup$
    – Z. M
    Aug 4 at 16:43
  • $\begingroup$ I think I don't understand how one computes the sheafification in this case, can you say something about that ? Also, extremally disconnected sets are not compact objects in the categorical sense so I don't follow your argument of reduction to finite coproducts (it doesn't matter though as I am only interested in the commutation with finite coproducts anyway). $\endgroup$ Aug 5 at 9:28
  • $\begingroup$ Ok, I understood the sheafification argument, the point is to show that the sheafification of a finite coproduct presheaf $\coprod \underline{X_i}$ is $S\mapsto \mathrm{Cont}(S,\coprod X_i)$ by showing that the latter satisfies the universal property. This comes from the fact that for a map $f:S\to \coprod X_i$ we can write $S=\coprod f^{-1}(X_i)$, a finite disjoint union of extremally disconnected sets; $f$ is then uniquely determined by the maps $f_i : f^{-1}(X_i)\to X_i$ $\endgroup$ Aug 5 at 10:39
  • $\begingroup$ The reduction to finite is argued in the same way. I do not claim that they are compact objects. The disjoint union of the target gives rise to an open cover on the extremally disconnected source. I am not commuting it with arbitrary filtered colimits or so. $\endgroup$
    – Z. M
    Aug 5 at 14:41

1 Answer 1


Yes, that is true. I'm not sure I'm explaining the step that's confusing you, but you are asking about a special case of the statement that the functor $S\mapsto\underline{S}$ from profinite sets to condensed sets commutes with finite coproducts. This is a formal consequence of the fact that the Grothendieck topology on profinite sets used in the definition of condensed sets allows such finite disjoint unions as covers.

Even more generally, if $G$ is any condensed group (for example $\underline{G}$ for a topological group $G$) and $H\subset G$ is a subgroup that is open (in the sense that for any profinite set $S$ with a map $S\to G$, the pullback $S\times_H G\subset S$ is an open subset of $S$ -- this is satisfied for $\underline{H}\subset \underline{G}$ for an open subgroup $H$ of a topological group $G$), then as a condensed $H$-set, one has $G\cong \bigsqcup_{G/H} gH$, where the index set $G/H$ is a discrete condensed set (i.e. just a set). Indeed, it is clear that $\bigsqcup_{G/H} gH$ injects into $G$ (on the level of presheaves, and thus after sheafification). But for any map $S\to G$ from a profinite set $S$, one has, set-theoretically, $S=\bigsqcup_{G/H} S\times_G gH$, and each term $S\times_G gH\subset S$ is open. Thus, this gives an open cover over which one gets a map to $\bigsqcup_{G/H} gH$, as desired.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.