# Solidification of free abelian group on compact Hausdorff space

In the lecture notes on condensed mathematics 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, 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?

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 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!
• Thank you for the quick and detailed answer. For completeness sake, let me add the following comment: Since $\mathcal{M}(S)$ is a Smith space it is isomorphic to its double dual $\mathcal{M}(S) = \underline{\mathrm{Hom}}_{\mathbb{R}}(\underline{\mathrm{Hom}}_{\mathbb{R}}(\mathcal{M}(S), \mathbb{R}), \mathbb{R})$ and moreover, by what you said, $\underline{\mathrm{Hom}}_{\mathbb{R}}(\mathcal{M}(S), \mathbb{R}) = \underline{\mathrm{Hom}}_{\mathbb{R}}(\mathbb{R}[S]^{\mathcal{M}-cpl}, \mathbb{R}) = C(S, \mathbb{R})$. This implies the expression $\mathcal{M}(S)$ that I mentioned in the question. – J. Steinebrunner Feb 26 at 13:59