Condensed Mathematics: Solid Modules¶
Job: define solid abelian groups / solid \(\mathbb{Z}\)-modules, the solid tensor product \(\otimes_{\blacksquare}\), and the free solid module \(\mathbb{Z}[S]^{\blacksquare}\); prove the solid category is a stable \(\infty\)-category / abelian subcategory closed under \(\otimes_{\blacksquare}\) and all limits/colimits and AB5-exact (Clausen–Scholze), and record the dictionary to complete topological abelian groups via the analytic ring \(\overline{\mathbb{Z}}\).
Core¶
Solid completion of the free module¶
Let \(S=\varprojlim_n S_n\) be a profinite set written as a cofiltered limit of finite sets (e.g. \(S_n=S/\mathcal{U}_n\) for a fundamental system of open equivalence relations). The free condensed abelian group is \(\mathbb{Z}[S]=\varinjlim_m\mathbb{Z}[S_m]\) (colimit over finite quotients), a condensed abelian group that is not in general complete. Its solid completion is
a condensed abelian group that is \(M\)-complete in the sense that for every profinite \(T\),
Equivalently, \(\mathbb{Z}[S]^{\blacksquare}\) is the internal \(\mathrm{Hom}\)-dual of the "solid structure sheaf" evaluated on \(S\); this is the value of the analytic-ring structure on \(\overline{\mathbb{Z}}\) at \(S\).
Solid modules¶
A condensed abelian group \(M\) is solid (over \(\mathbb{Z}\)) if for every profinite set \(S\) the natural map
is a bijection. Equivalently, \(M\) is a module over the analytic ring \(\overline{\mathbb{Z}}:=(\mathbb{Z},\mathbb{Z}[-]^{\blacksquare})\). The full subcategory
is the heart of the construction. Write \(\mathrm{Solid}_{\mathbb{Z}}^{(1)}\subset D(\mathrm{Cond}(\mathrm{Ab}))\) for the full subcategory of complexes whose cohomology sheaves are solid — the derived solid category.
| Theorem (Clausen–Scholze, Condensed / Analytic) | Statement |
|---|---|
| Solid = completion | \(\mathrm{Solid}_{\mathbb{Z}}\) is the left completion of \(\mathbb{Z}\)-linear condensed complexes: \(M\) solid iff \(M\simeq M\otimes^{\mathbb{L}}_{\mathbb{Z}}\overline{\mathbb{Z}}\) |
| Abelian | \(\mathrm{Solid}_{\mathbb{Z}}\) is an abelian subcategory of \(\mathrm{Cond}(\mathrm{Ab})\); kernels/cokernels agree |
| AB5 / exact | filtered colimits in \(\mathrm{Solid}_{\mathbb{Z}}\) are exact — the obstruction of \(\mathrm{Cond}(\mathrm{Ab})\) is resolved |
| Stable \(\infty\)-cat | \(\mathrm{Solid}_{\mathbb{Z}}^{(1)}\) is presentable, stable, generated by \(\{\mathbb{Z}[S]^{\blacksquare}\mid S\text{ profinite}\}\) |
| Closed under \(\otimes_{\blacksquare}\) | the solid tensor product \(-\otimes_{\blacksquare}-\) is the left derived functor of \(\otimes\) over \(\overline{\mathbb{Z}}\); monoidal, closed |
| Enough projectives | \(\mathbb{Z}[S]^{\blacksquare}\) for \(S\) ED are projective generators (the ED restriction persists) |
| Full faithful embedding | \(\mathrm{Cond}(\mathrm{Ab})\to\mathrm{Solid}_{\mathbb{Z}},\,M\mapsto M^{\blacksquare}:=M\otimes_{\blacksquare}\overline{\mathbb{Z}}\), is idempotent; restricts to identity on solid \(M\) |
The analytic ring \(\overline{\mathbb{Z}}\)¶
The pair \(\overline{\mathbb{Z}}=(\mathbb{Z},\,\mathcal{A}[-])\) with \(\mathcal{A}[-]=\mathbb{Z}[-]^{\blacksquare}\) is the prototypical analytic ring (Clausen–Scholze, Analytic Rings): an (animated) ring \(\mathcal{A}\) together with a functor \(\mathcal{A}[-]:\mathrm{Cond}(\mathrm{Set})\to D(\mathrm{Cond}(\mathrm{Ab}))\) satisfying the analytic-ring axioms (idempotent completion of a localization, preservation of covers, etc.). Modules over \(\overline{\mathbb{Z}}\) are exactly solid \(\mathbb{Z}\)-modules, and the analytic-ring tensor product recovers \(\otimes_{\blacksquare}\).
Dictionary to complete topological abelian groups¶
The solid category is the non-archimedean side of condensed mathematics — no prime is inverted. The dictionary to classical complete topological abelian groups is:
| Topological abelian group \(G\) | Solid object | Solid description |
|---|---|---|
| \(\mathbb{Z}_p\) (profinite) | \(\mathbb{Z}[\mathbb{Z}_p]^{\blacksquare}\cong C(\mathbb{Z}_p,\mathbb{Z}_p)\) | continuous maps \(\mathbb{Z}_p\to\mathbb{Z}_p\) (See it) |
| \(\mathbb{Z}[[T]]\) (formal) | \(\mathbb{Z}[\{0\}\cup\{T^{-1/n}\}]^{\blacksquare}\) | solid power-series |
| \(\ell^2\) (Hilbert, \(\mathbb{R}\)-vector) | not solid — requires liquid \(\mathbb{R}^{\triangleright}\) | see Analytic Rings |
| Discrete \(\mathbb{Z}\) | \(\underline{\mathbb{Z}}=\mathbb{Z}[\ast]^{\blacksquare}\) | already solid |
| \(\prod_I\mathbb{Z}\) (product, \(I\) any) | \(\mathbb{Z}[\prod_I\ast]^{\blacksquare}\)-module | solid products are exact |
| \(\mathbb{R}\) (usual topology) | not solid; \(\mathbb{R}^{\blacksquare}=0\) after \(\mathbb{Z}\)-completion? no — \(\mathbb{R}\) is liquid | archimedean side |
The crucial boundary: archimedean / divisible objects (\(\mathbb{R}\), \(\mathbb{C}\), \(\ell^2\), Banach spaces over \(\mathbb{R}\)) are not solid over \(\overline{\mathbb{Z}}\) — they live over the liquid analytic rings \(\mathbb{R}^{\triangleright}\), \(\mathbb{Z}[\tfrac{1}{p}]^{\triangleright}\). See Analytic Rings.
See it¶
\(\mathbb{Z}[\mathbb{Z}_p]^{\blacksquare}\cong C(\mathbb{Z}_p,\mathbb{Z}_p)\), the solid resolution of a \(p\)-adic group. Let \(S=\mathbb{Z}_p=\varprojlim_n \mathbb{Z}/p^n\). Then
For each \(n\), \(\mathbb{Z}[\mathbb{Z}/p^n]=C(\mathbb{Z}/p^n,\mathbb{Z})\) (continuous maps from a finite discrete set), and the inverse limit is the condensed abelian group of continuous maps \(C(\mathbb{Z}_p,\mathbb{Z}_p)\), where \(\mathbb{Z}_p\) carries the profinite topology and \(\mathbb{Z}\) the discrete topology in the target's \(\mathbb{Z}\)-module structure. Concretely: a section over a profinite \(T\) is a continuous map \(T\times\mathbb{Z}_p\to\mathbb{Z}_p\) that is \(\mathbb{Z}\)-linear in the second variable, which by \(p\)-adic continuity is the same as a compatible system of maps \(T\times\mathbb{Z}/p^n\to\mathbb{Z}/p^n\). Hence \(\mathrm{Hom}_{\blacksquare}(\mathbb{Z}[\mathbb{Z}_p]^{\blacksquare},\mathbb{Z}_p)\cong C(\mathbb{Z}_p,\mathbb{Z}_p)\), and the solid module structure on \(\mathbb{Z}_p\) is exactly the classical \(p\)-adic one. This is the sense in which "solid = complete topological abelian group, made algebraic."
Solid resolution of \(\mathbb{R}\). In the solid category over \(\overline{\mathbb{Z}}\), \(\mathbb{R}\) (with the condensed topology) is not solid — \(\mathbb{R}\otimes_{\blacksquare}\overline{\mathbb{Z}}\) kills the divisible part. The correct resolution of \(\mathbb{R}\) by solid modules is the liquid resolution, over \(\mathbb{R}^{\triangleright}\): one writes \(\mathbb{R}\) as a quotient of an \(\ell^2\)-type liquid module and uses the liquid tensor product theorem (Clausen–Scholze, Analytic) to obtain an exact resolution. This is the archimedean analog of the \(p\)-adic computation above; see Analytic Rings.
Check¶
Show that \(\mathrm{Solid}_{\mathbb{Z}}\) is AB5-exact, and pin down where the AB5-failure of \(\mathrm{Cond}(\mathrm{Ab})\) has gone.
Filtered colimits in \(\mathrm{Solid}_{\mathbb{Z}}\) are computed as \(\varinjlim_{\blacksquare}=\varinjlim\circ(-\otimes_{\blacksquare}\overline{\mathbb{Z}})\): first form the condensed colimit, then solid-complete. The solid-completion functor \((-)^{\blacksquare}\) is left adjoint to the inclusion and is exact (it is a Bousfield localization on \(D(\mathrm{Cond}(\mathrm{Ab}))\) whose kernel is acyclic by the analytic-ring axioms), so exactness of \(\varinjlim\) in \(\mathrm{Cond}(\mathrm{Ab})\) at the sheaf level, followed by exact completion, gives exactness in \(\mathrm{Solid}_{\mathbb{Z}}\). The \(\varprojlim^1\)-obstruction that gave \(\mathrm{Ext}^1_{\mathrm{Cond}}(\mathbb{R},\mathbb{Z})\neq 0\) is killed because the solid completion factors out exactly the non-complete towers — the analytic ring \(\overline{\mathbb{Z}}\) is defined so that its modules see only the completed, AB5-exact part.
Why is \(\mathbb{R}\) not solid over \(\overline{\mathbb{Z}}\), and why does this not contradict \(\mathrm{Cond}(\mathrm{Ab})\hookrightarrow\mathrm{Solid}_{\mathbb{Z}}\) being full faithful on solid objects?
Solid \(\mathbb{Z}\)-modules are \(\mathbb{Z}\)-complete in the sense of being built from profinite free modules \(\mathbb{Z}[S]^{\blacksquare}\); these are \(p\)-adically separated and reduced. \(\mathbb{R}\) is a \(\mathbb{Q}\)-vector space, divisible, with no \(p\)-adic separation, and \(\mathrm{Hom}_{\blacksquare}(\mathbb{Z}[S]^{\blacksquare},\mathbb{R})\) vanishes for connected \(S\) (no continuous \(\mathbb{Z}\)-linear map \(\mathbb{Z}_p\to\mathbb{R}\)), so the solid completion \(\mathbb{R}^{\blacksquare}=0\) (or, more carefully, factors through the \(0\)-object in the solid heart). There is no contradiction with full faithfulness: the embedding \(\mathrm{Cond}(\mathrm{Ab})\to\mathrm{Solid}_{\mathbb{Z}}\) is full faithful on the solid subcategory (it is the identity there), and is not full faithful on all of \(\mathrm{Cond}(\mathrm{Ab})\); \(\mathbb{R}\) is sent to \(0\) and the non-trivial Ext classes of \(\mathrm{Cond}(\mathrm{Ab})\) involving \(\mathbb{R}\) are exactly those killed by completion — this is the point of passing to Analytic Rings.
Pitfalls¶
- Solid = non-archimedean: \(\overline{\mathbb{Z}}\) does not invert any prime; divisible / \(\mathbb{Q}\)-vector / archimedean objects (\(\mathbb{R}\), \(\mathbb{C}\), Banach spaces over \(\mathbb{R}\)) are not solid and require the liquid analytic rings. Computing \(\mathrm{Ext}^1_{\blacksquare}(\mathbb{R},\mathbb{Z})\) and getting \(0\) is correct but the question itself is in the wrong category — \(\mathbb{R}\) should be resolved over \(\mathbb{R}^{\triangleright}\).
- ED projectives persist: the projective generators of \(\mathrm{Solid}_{\mathbb{Z}}\) are \(\mathbb{Z}[S]^{\blacksquare}\) for \(S\) extremally disconnected; the ED hypothesis is unchanged by solid completion. Using \(\mathbb{Z}[S]^{\blacksquare}\) with \(S\) non-ED as if projective produces spurious vanishing of Ext.
- Solid completion is a localization, not an embedding: the map \(\mathrm{Cond}(\mathrm{Ab})\to\mathrm{Solid}_{\mathbb{Z}}\), \(M\mapsto M^{\blacksquare}\), is idempotent and kills the non-solid part; it is full faithful on solid objects but collapses the non-solid ones (e.g. \(\mathbb{R}^{\blacksquare}=0\) over \(\overline{\mathbb{Z}}\)). Conflating "\(M\) embeds" with "\(M^{\blacksquare}\neq 0\)" is the standard error.
- \(\otimes_{\blacksquare}\neq\otimes_{\mathbb{Z}}\): the solid tensor product is the derived tensor over the analytic ring \(\overline{\mathbb{Z}}\), not the underived \(\mathbb{Z}\)-tensor of condensed groups. In particular \(\mathbb{Z}[S]^{\blacksquare}\otimes_{\blacksquare}\mathbb{Z}[T]^{\blacksquare}\neq\mathbb{Z}[S\times T]\) in general (the right side is not solid); one has \(\mathbb{Z}[S]^{\blacksquare}\otimes_{\blacksquare}\mathbb{Z}[T]^{\blacksquare}=\mathbb{Z}[S\times T]^{\blacksquare}\).
Refs¶
- Clausen–Scholze, Condensed Mathematics (2019), Lectures IV–VI (solid modules, the analytic ring \(\overline{\mathbb{Z}}\))
- Clausen–Scholze, Analytic (2019/2020), §I–II (analytic rings, the solid tensor product)
- Scholze, Lectures on Analytic Geometry (Tübingen 2019), §2 (solid completion, dictionary)
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