Uniform spaces as condensed sets

Question

$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Unif{Unif}\DeclareMathOperator\CHaus{CHaus}\DeclareMathOperator\Set{Set}\DeclareMathOperator\op{op}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Cond{Cond}\DeclareMathOperator\Top{Top}$In Barwick–Haine - Pyknotic objects, I. Basic notions Example 2.1.10, they showed that the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$ is a pyknotic set, i.e., a sheaf on the site $\CHaus$ of compact Hausdorff spaces equipped with the coherent topology (Lurie - Ultracategories).

I did not check whether the sheaf above is accessible (update: it is just the underlying topological space, see the Update part). However, I am slightly skeptical about this approach. It seems to me that the "correct" realization of a uniform space should be a condensed set which records its underlying topological space, along with an extra structure which records the uniform structure.

Question: I wonder how much is known about any kind of realization of uniform spaces as condensed sets.

I suppose that this extra structure would be described by a certain kind of enriched groupoid. Indeed, the uniform structure on a topological space could be understood as a groupoid enriched in filters. See nLab page for a description of this sort.

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This is motivated by an attempt to eliminate the restriction that the adjective "solid" only applies to condensed abelian groups.

Let $M$ be a topological abelian group. I was about to understand what it means for $M$ that the condensed abelian group $\underline M$ is solid. Following Lecture II of Scholze - Lectures on Analytic Geometry, for any sequence $(m_n)_{n\in\mathbb N}\in M^{\mathbb N}$ convergent to $0$, we associate a (continuous) map from the profinite set $S\mathrel{:=}\mathbb N\cup\{\infty\}$ to $M$ which maps $n$ to $m_n$ and $\infty$ to $0$, or equivalently, a map $S\to\underline M$ of condensed sets by, say, Yoneda's lemma.

Suppose that $\underline M$ is solid, then this map extends uniquely to a map $\mathbb Z[S]^\blacksquare\to\underline M$ of condensed abelian groups. If I am not mistaken, $\mathbb Z[S]^\blacksquare\to M$ further factors through $\underline{\mathbb Z[[t]]}$, the condensed abelian group associated to the topological abelian group $\mathbb Z[[t]]$ with $(t)$-adic topology, and by full faithfulness, we get a factorization $S\to\mathbb Z[[t]]\to M$ where the second map is additive.

Consequently, for every sequence $(a_n)_{n\in\mathbb N}\in{\mathbb Z}^{\mathbb N}$ of integers, the series $\sum_na_nt^n$ converges in $\mathbb Z[[t]]$, therefore the series $\sum_na_nm_n$ converges in $M$, which should imply, if I am not mistaken, that the uniform structure on $M$ is non-archimedean and complete, at least when $M$ is first countable (by the way, I don't understand why it is claimed that “it is not directly as any kind of limit of finite sums”).

So the non-archimedean nature is rooted in the formalism. I suppose that a natural approach is to generalize the uniform structure to condensed sets, and to generalize the classical Cauchy-completeness. I don't know whether it is convinced that this does not work. The current presentation separates non-archimedean and $\mathbb R$-case, which covers neither non-abelian groups nor the general completeness of topological abelian groups.

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Update: When discussing the functor $\Hom_{\Unif}(-,X)\colon\CHaus^{\op}\to\Set$, I missed the simple fact that $\Hom_{\Unif}(-,X)=\Hom_{\Top}(-,X)\mathrel{=:}\underline X$ by the Heine–Cantor theorem. In other words, when the uniform space in question is $T_0$ (therefore $T_1$), $\underline X$ is thus a condensed set.

Update2: It seems to me that I did not phrase the question unambiguously. In fact, I wanted to ask for any realization of unform spaces as condensed sets with some extra data, which should be recorded in the data of a condensed abelian group when it comes from a topological abelian group. At the time that I posed the question, I did not realize that Barwick–Haine's approach only records the data about the underlying space, but just doubted that it is a "correct" approach to record all data.

Update3: We say that a uniform space is non-archimedean if the uniformity is generated by equivalence relations.

Answer 1

I think there are a few things to untangle here.

First, as concerns your highlighted question, it seems that you've answered it yourself: outside the compact Hausdorff case (where the uniform structure is completely equivalent to the topology), it's unreasonable to think that the associated condensed set "knows" the uniform structure. It really only knows the topology.

Second, you are correct in your unpacking of the definition of "solid" as it applies to (let's say metrizable) toplogical abelian groups $M$. If $\underline{M}$ is solid, then for any nullsequence $(m_n)$ in $M$ and any sequence of integers $(a_n)$ the sum $\sum a_nm_n$ must converge in $M$. I agree that this feels a lot saying that $M$ is nonarchimedean and complete. However, though I gather for you "complete" means "Cauchy-complete" as usual, I'm not sure what general definition of a topological abelian group $M$ being "nonarchimedean" you're thinking of here to say that you think it is actually equivalent.

Here's something I find helpful to keep in mind. There are several differences between the notion of "solid" and the notion of "complete", beyond the fact that solid enforces some kind of non-archimedeanness. (Indeed, the same remarks apply in the liquid setting, which does not enforce nonarchimedeanness.)

1. First, while both the definition of "complete" and the definition of "solid" are of the form "for every [...] there exists a unique {...}", the differences in nature of the ...'s occuring lead to drastic divergence of the general notions. Most importantly, the solid condition in no way implies any kind of Hausdorff behavior. Essentially, the reason is that in the solid condition you already require in [...] a bunch of limits to exist uniquely (because you map in from a compact Hausdorff space). Then the solid condition is only that this generates more limits (and uniquely) by taking certain $\mathbb{Z}$-linear combinations. Whereas in the usual Cauchy completeness every limit that you posit exists, you also posit exists uniquely.

2. This phenomenon, that solid abelian groups incorporate non-Hausdorff behvaior, it absolutely crucial to having a functioning theory. Why? Because non-Hausdorff behavior is inevitable once you require an abelian category. If $M$ is some non-discrete condensed abelian group which you want to call "complete", and $N$ is any discrete dense subgroup mapping into $M$, the cokernel has to be "complete" as well, even though it's a classic example of a "bad quotient" which is non-Hausdorff.

(This is an example of the trade-off between "good categories of (mostly) bad objects" vs. "bad categories of good objects". But for me personally, I've seen enough examples of the solid formalism gracefully handling non-Hausdorff spaces in practice that I no longer think of them as "bad objects".)

3. You might be tempted to then think that solid abelian groups are more like weakened analog of complete topological abelian groups, where you drop the uniqueness requirement in the definition of completeness and only require existence of limits for Cauchy sequences. But no, the fact that solidness is an "exists a unique" condition is also crucial for it being an abelian category. "Exists" is just not stable enough a notion.

4. The other big difference is that a solid abelian group is "only required to be complete as far as compact subsets are concerned". If you have a Cauchy sequence which is not contained in a compact subset, the solid condition says nothing about it. Thus, for example, solid abelian groups are closed under arbitrary direct sums, whereas I don't think there's any reasonable topology on the direct sum $\oplus \mathbb{Z}_p$ for which it's complete [Edit: I thought wrong! See https://mathoverflow.net/questions/387322/countable-sum-bigoplus-n-0-infty-mathbb-z-p-as-a-topological-group]. Again, the fact that solid abelian groups are closed under all colimits is very important for us theoretically: it's part of what makes sure that the category of solid abelian groups "behaves like the category of modules over a ring" (formally, it is an abelian category generated by compact projective objects), and therefore has convenient algebraic properties.

Now, it seems the main thrust of your question as about defining possibile non-abelian analogs of solidness. I don't want to say this can't be done (I doubt it can but I certainly could be wrong), but I hope that the above remarks show that if you want to define such a notion, you shouldn't do it by trying to follow the usual presentation via Cauchy-completeness and uniform structures. Despite the fact that the two notions agree on many familiar objects, there is a huge divergence in general.

Answer 2

Here is an essentially tautological answer. The notion of uniformity makes sense also for condensed sets -- it is a condensed set $X$ together with certain subsets $U\subset X\times X$ termed entourages satisfying all the usual axioms. Moreover, one can also define the completion of $X$ with respect to its uniform structure "by interpreting everything internally in condensed sets".

If $X$ is a condensed abelian group with a nonarchimedean uniformity -- meaning say that the $U$'s above can be chosen to be subgroups, and all $X/U$ are discrete (as condensed sets) -- then the completion of $X$ is the inverse limit of the $X/U$, and hence is solid.