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?


1 Answer 1


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 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!

  • $\begingroup$ 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. $\endgroup$ Feb 26, 2021 at 13:59

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