In the [lecture notes on condensed mathematics](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf) the solidification of the free condensed abelian group $\mathbb{Z}[S]$ on a profinite set $S$ is defined as the inverse limit $\lim\_{\leftarrow} \mathbb{Z}[S\_i]$, but it can also alternatively be described as
$$
 \mathbb{Z}[S]^\blacksquare \cong \underline{\mathrm{Hom}}(C(S,\mathbb{Z}), \mathbb{Z}).
$$
My question is whether this also holds for general compact Hausdorff spaces $S$. In the case that $X$ is a CW complex this seems to follow from example 6.5 where $\mathbb{Z}[X]^{\blacksquare} \cong H\_\bullet(X)$ is shown.


This also seems to follow from it's derived analogue, which was stated in part 2 of [the answer here](https://mathoverflow.net/questions/379076/computations-in-condensed-mathematics-ii-page-43/379103#379103),
if we can show that $\mathbb{Z}[X]$ is pseudo-coherent for any compact Hausdorff space $X$.
I think this holds because we can resolve $X$ by extremally disconnected spaces, which should yield a resolution of $\mathbb{Z}[X]$ by compact projectives.


By analogy I also have the same question for the $\mathcal{M}$-completeness of condensed $\mathbb{R}$-vector spaces. Is it true that the condensed vector space of signed Radon measures on a compact Hausdorff space $X$ can be given as:
$$
 \mathcal{M}(X) \cong \underline{\mathrm{Hom}}\_{\mathbb{R}}( C( X, \mathbb{R}) , \mathbb{R})
$$
Again it seems like all the hints are in the lecture notes, but I'm not comfortable enough with the subject to pin down the details.




---


A short note on why I'm asking this: the main reason is that I want to better understand the analogy with measure theory where one says that a condensed $\mathbb{R}$-vector space $V$ is $\mathcal{M}$-complete if for any continuous $f: K \to A$ and measure $\mu$ on $K$ one can form the "integral" $\int f\ d\mu$. I would find this much more convincing if I can let $K$ be a finite CW-complex rather than just a profinite set.


On this note I also have a third, optional question: if the above two descriptions work, is there something similar one can say in the case of $\mathcal{M}\_p(S)$ and for $p$-liquid vector spaces?



Solidification of free abelian group on compact Hausdorff space
Good question! You essentially already give the answer, but let me spell it out.


First, in the solid case, as you say one can compute the derived solidification $\mathbb Z[S]^\blacksquare$ for any compact Hausdorff $S$ as
$$
\mathbb Z[S]^\blacksquare = R\underline{\mathrm{Hom}}(R\Gamma(S,\mathbb Z),\mathbb Z),
$$
where $R\Gamma(S,\mathbb Z)$ is the Cech cohomology of $S$. The answer is nicer for profinite $S$ as then this is concentrated in degree $0$. If you are only interested in the solidification on the abelian level, then you can actually deduce from this that the map $S\to \pi\_0 S$ (a profinite set) induces an isomorphism $H\_0(\mathbb Z[S]^\blacksquare)\to \mathbb Z[\pi\_0 S]^\blacksquare$.


In the case of $\mathcal M$-complete $\mathbb R$-vector spaces, it is actually true that the $\mathcal M$-completion of $\mathbb R[S]$ is $\mathcal M(S,\mathbb R)$, the space of signed Radon measures on $S$ (with its Smith space topology), for any compact Hausdorff $S$. One can actually show that for any simplicial resolution $S\_\bullet\to S$ by profinite sets $S\_i$, the complex
$$
\ldots \to \mathcal M(S\_1,\mathbb R)\to \mathcal M(S\_0,\mathbb R)\to \mathcal M(S,\mathbb R)\to 0
$$
is exact. This follows from the anti-equivalence of Smith spaces with Banach spaces, and the exact sequence
$$
0\to C(S,\mathbb R)\to C(S\_0,\mathbb R)\to C(S\_1,\mathbb R)\to \ldots,
$$
the latter being exact by the computation $H^i\_{\mathrm{cond}}(S,\mathbb R)=0$ for $i>0$ (and $C(S,\mathbb R)$ for $i=0$). [I actually find the definition of signed Radon measures on general $S$ a bit hard to process (while for profinite sets, it is completely transparent, one only has to give measures on open and closed subsets). So I actually prefer to think of the above as the definition of $\mathcal M(S,\mathbb R)$ for general compact Hausdorff $S$. But I guess it's nice that one can describe this space explicitly, intrinsically on $S$.]


Finally, in the $p$-liquid case, one can also show that for any compact Hausdorff $S$, there is a $p$-Smith space $\mathcal M\_p(S,\mathbb R)$ such that for any resolution $S\_\bullet\to S$ as above, the complex
$$
\ldots \to \mathcal M\_p(S\_1,\mathbb R)\to \mathcal M\_p(S\_0,\mathbb R)\to \mathcal M\_p(S,\mathbb R)\to 0
$$
is exact. This follows from Proposition 10.1 (i) in [Analytic Geometry](https://www.math.uni-bonn.de/people/scholze/Analytic.pdf) by passing to the quotient $\mathbb Z((T))\_r\to \mathbb R$ (sending $T$ to $r^{1/p}$) as usual (using also Proposition 7.2). Unfortunately, I don't really know how to describe $\mathcal M\_p(S,\mathbb R)$ intrinsically on $S$, without choosing this surjection $\mathbb Z((T))\_r\to \mathbb R$; I'd be happy to see a description!


