Let $k^{s}$ be a separable closure of a field $k$. Is $Gal(k^s/k)$ a condensed group in the sense of condensed mathematics? If condensed, is it always solid?
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6$\begingroup$ What is "condensed group"? $\endgroup$– markvsCommented Mar 18, 2021 at 4:47
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$\begingroup$ See Condensed Mathematics $\endgroup$– Peter ScholzeCommented Mar 18, 2021 at 10:07
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1$\begingroup$ The text is 77 page long. Is there a definiton of condensed group there? $\endgroup$– markvsCommented Mar 18, 2021 at 12:28
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1$\begingroup$ @dodd page 7 defines what the category $\operatorname{Cond}(\mathsf{C})$ is. It in particular literally states: "[...] a condensed set/ring/group/... is a functor ..." spelling out in detail what a condensed group is. $\endgroup$– lushCommented Mar 18, 2021 at 14:09
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3$\begingroup$ A condensed group "is" a group just as much as a topological group "is" a group: It's not a property of an abstract group, but extra structure. $\endgroup$– Peter ScholzeCommented Mar 18, 2021 at 14:16
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Like any profinite group (or much more general types of topological groups, such as compactly generated ones), you can consider it as a condensed group in the sense of condensed mathematics. In fact, that's the perspective I am taking, and it is useful for thinking about say continuous group cohomology of $\mathrm{Gal}(k^s/k)$.
On the other hand, "solid" is an adjective that pertains only to condensed abelian groups. But all profinite abelian groups are solid -- in fact, the category of solid abelian groups is a full subcategory of condensed abelian groups stable under all limits and colimits, and containing all discrete objects -- so if $\mathrm{Gal}(k^s/k)$ happens to be abelian, it is solid.
answered Mar 18, 2021 at 8:13