Recently I became through this mathoverflow question aware of the article Codensity and the ultrafilter monad by Tom Leinster. There he shows that the ultrafilter monad on the category $\mathrm{Set}$ arises from the adjunction $$ \mathrm{Set} \rightleftarrows \mathrm{Fun}(\mathrm{FinSet}, \mathrm{Set})^{\mathrm{op}},$$ where the left adjoint is given by the coYoneda-embedding (that it has a right adjoint follows either by a construction or the adjoint functor theorem). Moreover it is known that the category of compact Hausdorff spaces is monadic over $\mathrm{Set}$ and that the corresponding monad is the ultrafilter monad as well, exhibiting the category of compact Hausdorff spaces as algebras over this monad.

Moving to $\infty$-categories, it is natural to replace $\mathrm{Set}$ by the $\infty$-category $\mathcal{S}$ of spaces (or animae, as some call it). This has the sub-$\infty$-category $\mathcal{S}^{\mathrm{fin}}$ of finite spaces (i.e. the smallest finitely cocomplete subcategory containing the point). Using the coYoneda embedding and the adjoint functor theorem, we obtain again an adjunction $$\mathcal{S}\rightleftarrows \mathrm{Fun}(\mathcal{S}^{\mathrm{fin}}, \mathcal{S})^{\mathrm{op}}.$$ Can one describe the resulting monad and algebras over it? Is it a known $\infty$-category? Moreover, one might ask about its relation to to other $\infty$-categories, like profinite spaces or condensed spaces.

Edit: As Denis and Dustin pointed out, it is much more natural to replace $\mathrm{FinSet}$ by the $\infty$-category of $\pi$-finite spaces (instead of $\mathcal{S}^{\mathrm{fin}}$), i.e. spaces whose homotopy groups are concentrated in finitely many degrees and are finite there.

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    $\begingroup$ Is there a reason to choose finite spaces instead of $\pi$-finite spaces? The latter seems to play more the role of "finite sets" (in particular pro-$\pi$-finite spaces embed into condensed spaces, which seems relevant for this question) $\endgroup$ – Denis Nardin Dec 17 '20 at 20:58
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    $\begingroup$ I guess the point is that the ultafilter monad applied to a set $S$ is the same thing as the profinite completion of $S$, the inverse limit of all maps from $S$ to finite sets. It's probably hard to understand maps in $\mathcal{S}$ to objects in $\mathcal{S}^{fin}$ so I'm not sure what one can say about your question as phrased. But as Denis says replacing with $\pi$-finite spaces might be more fruitful. Then it could be that algebras over the monad are the same thing as the coherent condensed anima, which indeed are a kind of notion of "animated compact Hausdorff space". $\endgroup$ – Dustin Clausen Dec 17 '20 at 21:17
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    $\begingroup$ But the phrase "condensed space" without context is not so clear at all, which is the main reason we came up with this "anima" terminology... $\endgroup$ – Dustin Clausen Dec 18 '20 at 13:52
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    $\begingroup$ I don't see how 'anima' out of context is any clearer than 'condensed spaces'. $\endgroup$ – Robert Bruner Dec 18 '20 at 19:40
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    $\begingroup$ It's clearer because the word "space" already means a lot of things and evokes a lot of different images, while "anima" only means one thing: basically, "homotopy type", but more properly an object of the $\infty$-category $\mathcal{S}$. $\endgroup$ – Dustin Clausen Dec 18 '20 at 20:14

That's a good question! I think Barwick and Haine have thought much more about this, and maybe they already know the answer? What I say below is definitely known to them. Also beware that I've written the below in a stream of consciousness, not quite knowing where it will go when I started.

I'll write "anima" for what is variously called homotopy types/spaces/$\infty$-groupoids/..., and denote their $\infty$-category $\mathrm{An}$($=\mathcal S$). We can also consider the $\infty$-category $\mathrm{CondAn}=\mathrm{Cond}(\mathrm{An})$ of condensed anima (this is, by the way, also the animation of the category of condensed sets). If $X\in \mathrm{CondAn}$ is a condensed anima, then $\pi_0 X$ is a condensed set, and for any point $x\in X$, one can define homotopy groups $\pi_i(X,x)$ for $i\geq 1$, which are condensed groups (abelian for $i\geq 2$). Slightly more generally, if $S$ is any profinite set and $g: S\to X$ is any map, one can define a group object $\pi_i(X,g)\to S$ in condensed sets over $S$, whose fibre over any $s\in S$ is $\pi_i(X,g(s))$. Then a map of condensed anima is an equivalence if and only if it induces an equivalence on $\pi_0$ and all $\pi_i$ for $i\geq 1$ (at all base points, including profinite families of basepoints).

So, just like in a very very crude approximation an anima $X$ is something like the collection $\pi_0 X,\pi_1 X,\pi_2 X,\ldots$ of a set, a group, and abelian groups, a condensed anima is something like a collection of a condensed set, a condensed group, and condensed abelian groups. In particular, already $\pi_0 X$ can be an interesting topological space like a manifold, so a space. This is why we do not say "condensed space", as then it would seem like forgetting to condensed sets should forget the "space" structure, but it rather forgets "abstract homotopy" structure.

Now the following seems like the obvious "$\infty$-categorical compact Hausdorff spaces":

Definition. A condensed anima $X$ is "compact Hausdorff" if $\pi_0 X$ and all $\pi_i X$ for $i\geq 1$ are compact Hausdorff.

Recall here that compact Hausdorff spaces embed fully faithfully into condensed sets. The second statement means more precisely that for all profinite sets $S$ with a map $g: S\to X$, the group object $\pi_i(X,g)\to S$ in condensed sets over $S$ is compact Hausdorff. (This is a little stronger than only asking it at all fibres.)

So in this case $\pi_0 X$ is a compact Hausdorff space, $\pi_1 X$ is a compact Hausdorff group, and $\pi_2 X,...$ are compact Hausdorff abelian groups.

It turns out that there is a nice characterization of "compact Hausdorff" condensed anima. In fact, there is a general topos-theoretic notion of "coherent"="qcqs" objects. This is usually studied for $1$-topoi, but it generalizes easily to $n$-topoi. Basically, an object is quasicompact if any cover admits a finite subcover; it is quasiseparated if the diagonal is quasicompact; it is 2-quasiseparated if the diagonal is quasiseparated; etc.; and coherent = quasicompact and $n$-quasiseparated for all $n\geq 1$. Then coherent condensed sets are exactly compact Hausdorff spaces, and:

Proposition. Coherent condensed anima are exactly the "compact Hausdorff" condensed anima.

Note: In a $1$-topos, coherent objects often agree with the finitely presented objects, but this fails dramatically for $\infty$-topoi, where coherence and finite presentation are two quite different finiteness conditions. In the case of anima, coherence means finite homotopy groups, while finite presentation should mean generated under finite colimits from the point; these are very different notions. As already discussed in the comments, the "finite homotopy groups" condition seems more relevant for the question.

Now we have a good notion of "$\infty$-categorical compact Hausdorff spaces". The question however started from a different angle, namely as trying to describe it via a monad on anima. The good news is:

Proposition. Compact Hausdorff condensed anima are monadic over anima.

This can be deduced from Barr-Beck-Lurie, although it takes some work.

It remains to understand the monad (and see whether it can be described as a codensity monad). The monad takes an anima $X$ to $\lim_{X\to Y} Y$ where the diagram is over all maps from $X$ to a compact Hausdorff condensed anima $Y$: This computes the desired left adjoint. Assume for the moment that the diagram category was small; then this limit is still a compact Hausdorff condensed anima: The compact Hausdorff condensed anima are stable under all small limits, as they are stable under finite limits and all small products. Now the diagram category is not actually small, so one has to argue slightly more carefully to see the existence of the left adjoint.

If $X$ is actually a set, then one can show that the left adjoint is still the same as usual, given by the Stone-Čech compactification. This is the same as $\lim_{X\to Y} Y$ where we restrict $Y$ to be a finite set. Ultimately, the possibility to restrict $Y$ to finite sets here -- coming from the fact that the Stone-Čech compactification is totally disconnected, and totally disconnected compact Hausdorff spaces are pro-finite -- is what makes it possible to describe compact Hausdorff spaces in terms of the codensity monad for $\mathrm{FinSet}\hookrightarrow \mathrm{Set}$.

The first interesting new case is $X=K(G,1)$, for some discrete group $G$. Ignoring higher homotopy groups, we are then interested in the universal compact group $H$ with a map $G\to H$. In general, this is known as the "Bohr compactification" of $G$. If $G=\mathbb Z$, then we look for the free compact group on one generator. This is necessarily abelian, and then one can use Pontrjagin duality to actually determine this (I hope I didn't screw this up): Take $\prod_{\mathbb R/\mathbb Z}\mathbb R/\mathbb Z$, the product of $\mathbb R/\mathbb Z$ (as a discrete set) many copies of the circle $\mathbb R/\mathbb Z$, with its tautological "diagonal" element, and take the closed subgroup generated by this element.

What we see from the example is that already for the anima $X=K(\mathbb Z,1)$ (aka the circle), the monad takes an extremely complicated value (note that we were ignoring higher homotopy groups, but the computation of $\pi_1$ is correct), that in particular is not itself totally disconnected, and so cannot be written as a limit of finite anima. So I gather that these "$\infty$-categorical compact Hausdorff spaces" cannot be described in the way the question started.

This, then again, begs the question what algebras for the monad in the question are!

Well, I don't know the precise answer, but one can also consider "totally disconnected compact Hausdorff" condensed anima, asking now that all $\pi_i X$ are totaly disconnected compact Hausdorff. So $\pi_0 X$ is a profinite set, $\pi_1 X$ is a profinite group, and $\pi_2 X,\ldots$ are profinite abelian groups.

Proposition. "Totally disconnected compact Hausdorff condensed $n$-truncated anima" are equivalent to the Pro-category of $n$-truncated anima with finite homotopy groups.

One can also pass to the limit $n\to \infty$ in some sense, but has to be careful as this does not exactly commute with passage to Pro-categories. It is still true that any totally disconnected compact Hausdorff condensed anima $X$ maps isomorphically to the $\lim_{X\to Y} Y$ where $Y$ runs over anima with finite homotopy groups.

Now totally disconnected compact Hausdorff condensed anima are not monadic anymore over anima, but the forgetful functor still detects isomorphisms, and has a left adjoint, so gives rise to a monad on anima, and totally disconnected compact Hausdorff condensed anima embed fully faithfully into algebras over this monad. And this monad, by the last paragraph, can be identified with the codensity monad for the inclusion $\mathrm{An}^{\mathrm{coh}}\hookrightarrow \mathrm{An}$ of coherent anima (=anima with finite homotopy groups) into all anima.

So, if I'm not screwing this up, then the category of algebras over this monad is some kind of hull of totally disconnected compact Hausdorff condensed anima (including all geometric realizations that are split on underlying anima); this hull is contained in compact Hausdorff condensed anima.

In summary, if one takes "finite anima" in the question to mean "finite homotopy groups", then this gives rise to a monad whose algebras lie somewhere between totally disconnected compact Hausdorff condensed anima, and all compact Hausdorff condensed anima. I think they definitely include all those for which $\pi_0 X$ is arbitrary compact Hausdorff, but $\pi_i X$ for $i\geq 1$ is totally disconnected.

Hmm... OK, let me make the following:

Conjecture: Algebras over the codensity monad for $\mathrm{An}^{\mathrm{coh}}\hookrightarrow \mathrm{An}$ are exactly those compact Hausdorff condensed anima $X$ for which all $\pi_i X$ for $i\geq 1$ are totally disconnected.

I'm willing to conjecture this for the following reason: while one can obtain all compact Hausdorff spaces as quotients of profinite sets by closed equivalence relations, nothing like this happens for groups: a quotient of a profinite group by a closed equivalence relation is still a profinite group.

  • $\begingroup$ With apologies for distracting from a nice answer -- is "anima" singular or plural or both? Is the singular "animum" or is the plural "animae", or something else altogether? $\endgroup$ – Tim Campion Jan 5 at 21:45
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    $\begingroup$ We decided to use it both as singular and plural, although that is technically wrong I guess (I believe the correct plural would be "animae", but this sounds awful to me). $\endgroup$ – Peter Scholze Jan 5 at 23:43
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    $\begingroup$ Interesting! I don't mind "animae", except that it's a homophone for "anime" which would potentially lead to even more terminological confusion in a talk :). Your comment also clarifies that in languages with gender, "anima" should be feminine. $\endgroup$ – Tim Campion Jan 5 at 23:58

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