The Clausen-Scholze theory of condensed mathematics offers an abelian category with enough projective objects that embraces the study of arbitrary locally compact (and Hausdorff) groups. The behaviour of the tensor product is managed by restricting to a subcategory of solid abelian groups within the category of condensed abelian groups and there is a solidification functor that is left adjoint to this inclusion. Putting my first toe in the water, I am restricting to F_p vector spaces where p is prime: in other words, condensed abelian groups of prime exponent. The Clausen-Scholze theory already provides sufficient in this special case to be able to solve problems that were inaccessible with classical Galois cohomology. My question: is every condensed vector space over a finite field automatically solid? And if not, exactly what would be the advantage of solidification in this context?
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1$\begingroup$ Does Theorem 2.9 of Lectures in Analytic Geometry answer your question? $\endgroup$– Z. MCommented Jan 14, 2022 at 16:19
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$\begingroup$ If you look for a concrete example, see the example right after Exercise 2.3, and replacing $\mathbb Z$ by $\mathbb F_p$. $\endgroup$– Z. MCommented Jan 14, 2022 at 16:22
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$\begingroup$ @Z.M : you should post this as an answer :) $\endgroup$– Maxime RamziCommented Jan 15, 2022 at 10:52
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$\begingroup$ @Z.M Thanks for this advice, I had overlooked the discussion in math.uni-bonn.de/people/scholze/Analytic.pdf where Theorem 2.9 gives a very clear characterisation of the solid F_p vector spaces. I guess this means that typically, F_p[S] is not solid when S is a compact Hausdorff space. $\endgroup$– Peter KrophollerCommented Jan 15, 2022 at 12:34
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8$\begingroup$ The condensed $\mathbb F_p$-vector space $\mathbb F_p[S]$ sits in degree $0$ for any compact Hausdorff space $S$. It is naturally an increasing union of compact Hausdorff subsets $\mathbb F_p[S]_{\leq n}$ where this is the set of those sums $\sum_{s\in S} n_s [s]$ where one can choose the $n_s\in \mathbb Z$ with $\sum |n_s|\leq n$. See Proposition 2.1 and Exercise 2.3 in Analytic.pdf for the discussion for $\mathbb Z[S]$ (which carries over without much change to $\mathbb F_p[S]$). These things are not solid as soon as $S$ is infinite. $\endgroup$– Peter ScholzeCommented Jan 17, 2022 at 12:21
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Peter Scholze's comment gives a good answer to the main direct question and tells us that the condensed mod p vector space with basis a compact Hausdorff space S is solid if and only if S is finite. The advantages of solidification are going to become more apparent as we use the condensed maths more widely. Solidification is particularly important when using tensor product. One situation where I believe there is no need to invoke solidification is when tensoring a solid vector space (over F_p) with a finite vector space: such a tensor is automatically solid from the get-go.
answered Jan 20, 2022 at 14:03