# Blueprint for the Liquid Tensor Experiment

## 2.4 Condensed abelian groups

Remark 2.4.1

For the time being, the following facts will be used without proof in this text. (They have or will be formalized in Lean though.)

• There is a natural functor $$\operatorname{Top}\to \operatorname{Cond}(\operatorname{Sets})$$.

• The category of condensed abelian groups (resp. condensed $$R$$-modules) is an abelian category with enough projectives. For $$S$$ an extremally disconnected set, the objects $$\mathbb Z[S]$$ (resp. $$R[S]$$) is projective.

• We write $$H^i(S, M)$$ for $$\text{Ext}^i(\mathbb Z[S], M)$$.

Definition 2.4.2
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Consider an exact sequence of abelian groups

$X'\overset {f}{\longrightarrow } X\overset {g}{\longrightarrow } X''$

such that all of $$X'$$, $$X$$ and $$X''$$ carry the structure of compact-Hausdorffly-filtered-pseudonormed abelian groups. Assume that the maps are strict, i.e., $$f(X'_{\leq c})\subset X_{\leq c}$$ and $$g(X_{\leq c})\subset X''_{\leq c}$$. Let $$r \colon \mathbb {R}_{\ge 0} \to \mathbb {R}_{\ge 0}$$ be a function satisfying $$c \le r(c)$$ for all $$c$$. We say that the sequence is strongly exact with respect to $$r$$ if $$\mathrm{ker}(g)\cap X_{\leq c}\subset f(X'_{\leq r(c)})$$.

Proposition 2.4.3
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Consider an inverse system

$(X_i'\overset {f}{\longrightarrow } X_i\overset {g}{\longrightarrow } X_i'')_i$

of exact sequences that are strongly exact with respect to $$r$$ (independent of $$i$$). Moreover, assume that the transition maps $$X'_i\to X'_j$$, $$X_i\to X_j$$ and $$X''_i\to X''_j$$ are strict, and let $$X'$$, $$X$$ and $$X''$$ be their limits. Then

$X'\overset {f}{\longrightarrow } X\overset {g}{\longrightarrow } X''$

is strongly exact with respect to $$r$$.

Proof

Pass to cofiltered limits of compact Hausdorff spaces in the statements $$\mathrm{ker}(g)\cap X_{\leq c}\subset f(X'_{\leq r(c)})$$, noting that cofiltered limits of surjections of compact Hausdorff spaces are still surjective (by an application of Tychonoff).

Definition 2.4.4
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There is a natural functor

\begin{align*} \text{CHPNG} & \longrightarrow \operatorname{Cond}(\operatorname{Ab}) \\ M & \longmapsto \underline{M} \end{align*}

where $$\underline{M}(S)$$ is defined to be collection of functions $$f \colon S \to M$$ that factor as continuous through $$M_c$$, for some $$c$$. In symbols:

$M(S) = \{ f \colon S \to M \mid \exists c, f(S) \subset M_c \text{ and f \colon S \to M_c is continuous}\} .$

Proposition 2.4.5
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Consider an exact sequence of abelian groups

$X'\overset {f}{\longrightarrow } X\overset {g}{\longrightarrow } X''$

such that all of $$X'$$, $$X$$ and $$X''$$ carry the structure of compact-Hausdorffly-filtered-pseudonormed abelian groups. Assume that $$f(X'_{\leq c})\subset X_{\leq c}$$ and $$g(X_{\leq c})\subset X''_{\leq c}$$. If the sequence is strongly exact with respect to $$r$$, then the sequence

$\underline{X'}\longrightarrow \underline{X}\longrightarrow \underline{X''}$

of condensed abelian groups is exact.

Proof

We evaluate at $$S\in \mathrm{ExtrDisc}$$. Since the sequence is exact with respect to $$r$$, we know that for all $$c$$ the natural map

$\phi \colon X_{\le c} \times _{X_{\le r(c)}} X'_{\le r(c)} \to X_{\le c} \times _{X''_c} \{ *\}$

is surjective. Therefore, any continuous map from $$S$$ to the codomain of $$\phi$$ can be lifted; as $$S$$ is extremally disconnected. Since every map $$S \to X$$ factors over some $$X_{\le c}$$, this shows that the kernel of $$g \colon X(S)\to X''(S)$$ is in the image of $$f \colon X'(S)\to X(S)$$.

Lemma 2.4.6
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Let $$S = \varprojlim S_i$$ be a profinite set. Then $$\mathbb Z[S]$$ is naturally a profinitely filtered pseudo-normed group, via $$\mathbb Z[S]_{\le c} = \varprojlim \mathbb Z[S_i]_{\le c}$$, where $$\mathbb Z[S_i]_{\le c}$$ is the set $$\{ \sum _{s \in S_i} n_s[s] \mid \sum _s |n_s| \le c\}$$.

There is a natural isomorphism between the free condensed abelian group $$\mathbb Z[S]$$ and the colimit $$\varinjlim _c \mathbb Z[S]_{\le c}$$ of condensed sets.

Proof

For now, see Lemma 2.1 of [ Sch20 ] .

Proposition 2.4.7
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Let

$M \colon \mathrm{ProFin}^{\mathrm{op}}\longrightarrow \mathrm{Ab}$

be a functor, i.e. a presheaf of abelian groups on $$\mathrm{ProFin}$$. Assume that $$M$$ preserves finite products, and that for any surjective map $$f: T\to S$$, the complex

$0\longrightarrow M(S)\longrightarrow M(T)\longrightarrow M(T\times _S T)\longrightarrow M(T\times _S T\times _S T)\longrightarrow \ldots$

is exact.

Then $$M$$ is a condensed abelian group, and for all profinite sets $$S$$ and $$i{\gt}0$$, one has $$H^i(S,M)=0$$ for $$i{\gt}0$$.

Proof

We prove by induction on $$i{\gt}0$$ that $$H^i(S,M)=0$$ for all profinite sets $$S$$, so assume the vanishing of $$\mathrm{Ext}^1,\ldots ,\mathrm{Ext}^i$$ for some $$i\geq 0$$. (This is vacuous for $$i=0$$.) We aim to prove that $$H^{i+1}(S,M)=0$$ for all profinite sets $$S$$. Pick any profinite set $$S$$ and a cover $$T\to S$$ with $$T\in \mathrm{ExtrDisc}$$. We get a long exact sequence of condensed abelian groups

$\ldots \longrightarrow \mathbb Z[T\times _S T\times _S T]\longrightarrow \mathbb Z[T\times _S T]\longrightarrow \mathbb Z[T]\longrightarrow \mathbb Z[S]\longrightarrow 0:$

Indeed, taken as presheaves on $$\mathrm{ExtrDisc}$$, this is already true on the level of presheaves, where it reduces to the case of surjections of sets in which case one can write down a contracting homotopy. (Actually, the similar result is true in any topos, where one has to maybe argue a bit more carefully.)

The following argument is making explicit something usually seen through a spectral sequence. Define inductively

$K_1=\mathrm{ker}(\mathbb Z[T]\longrightarrow \mathbb Z[S]),$
$K_2=\mathrm{ker}(\mathbb Z[T\times _S T]\longrightarrow \mathbb Z[T])$

etc. One gets exact sequences

$0\longrightarrow K_n\longrightarrow \mathbb Z[T^{n/S}]\longrightarrow K_{n-1}\longrightarrow 0$

for $$n\geq 2$$. From the long exact sequence

$\ldots \longrightarrow H^i(T,M)\longrightarrow \mathrm{Ext}^i(K_1,M)\longrightarrow H^{i+1}(S,M)\longrightarrow H^{i+1}(T,M)=0$

we see that we have to prove that $$\mathrm{Ext}^i(K_1,M)=0$$ (if $$i{\gt}0$$, otherwise that $$M(T)$$ surjects onto $$\mathrm{Hom}(K_1,M)$$). Assuming $$i{\gt}0$$, we can go on, and using the inductive hypothesis applied to the fibre products $$T^{\ast /S}$$, we inductively see that

$H^{i+1}(S,M)=\mathrm{Ext}^i(K_1,M)=\mathrm{Ext}^{i-1}(K_2,M)=\ldots =\mathrm{Ext}^1(K_i,M)$

and eventually that this is the same as the cokernel of

$M(T^{i/S})\longrightarrow \mathrm{Hom}(K_{i+1},M).$

But there is an exact sequence

$0\longrightarrow \mathrm{Hom}(K_{i+1},M)\longrightarrow M(T^{(i+1)/S})\longrightarrow \mathrm{Hom}(K_{i+2},M)$

and $$\mathrm{Hom}(K_{i+2},M)$$ injects into $$M(T^{(i+2)/S})$$. We see that

$\mathrm{Hom}(K_{i+1},M)=\mathrm{ker}(M(T^{(i+1)/S})\longrightarrow M(T^{(i+2)/S}))$

and we need to see that

$M(T^{i/S})\longrightarrow \mathrm{Hom}(K_{i+1},M)=\mathrm{ker}(M(T^{(i+1)/S})\longrightarrow M(T^{(i+2)/S}))$

is surjective, which is precisely the exactness of the Čech complex.

Proposition 2.4.8
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Let $$(M,\| \cdot \| )$$ be a complete normed group, regarded as a topological group. Then the corresponding condensed abelian group $$\underline{M}$$ sends any profinite set $$S$$ to the completion of normed group of locally constant maps $$S\to M$$ (with the supremum norm).

Proof

This is a standard result. We omit the proof here, but it is formalized in Lean.

Proposition 2.4.9
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Let $$(M,\| \cdot \| )$$ be a complete normed group, regarded as a topological group. Then for any profinite set $$S$$, one has $$H^i(S,\underline{M})=0$$ for $$i{\gt}0$$.

Proof

This follows Proposition 2.4.7 and the part of [ Sch20 , Proposition 8.19 ] that is already formalized.

Lemma 2.4.10
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Let $$0 {\lt} r {\lt} r' {\lt} 1$$ be real numbers. Let $$S$$ be a profinite set, and let $$V$$ be a $$r$$-normed (Banach?) $$\mathbb Z[T^{\pm 1}]$$-module. Then $$\operatorname{Ext}_{\mathrm{Mod}^{\mathrm{cond}}_{\mathbb Z[T^{-1}]}}^i(\overline{\mathcal L}_{r'}(S), V) = 0$$ for all $$i \ge 0$$. In other words,

$\operatorname{Ext}_{\mathbb Z}^i(\overline{\mathcal L}_{r'}(S), V) \xrightarrow {[T⁻¹]_L - [T⁻¹]_V} \operatorname{Ext}_{\mathbb Z}^i(\overline{\mathcal L}_{r'}(S), V)$

is a bijection for all $$i$$.

Proof

With Proposition 2.3.3, it suffices to prove the following assertion. Pick $$1{\gt}r'{\gt}r{\gt}0$$, a profinite $$S$$, and some $$r$$-Banach $$\mathbb Z[T^{\pm 1}]$$-module $$V$$ as before. Then we want to prove that

$\mathrm{Ext}^i_{\mathbb Z[T^{-1}]}(Q'(\overline{\mathcal M}_{r'}(S)),V)=0$

for all $$i\geq 0$$.

At this point, it is profitable to rewrite this again as the bijectivity of

$(T^{-1})_V - (T^{-1})_{\mathcal M}: \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S)),V)\longrightarrow \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S)),V).$

Now these Ext-groups can be computed! More precisely, recall that $$Q'(\overline{\mathcal M}_{r'}(S))$$ is a complex of the form

$\ldots \longrightarrow \mathbb Z[\overline{\mathcal M}_{r'}(S)^2]\longrightarrow \mathbb Z[\overline{\mathcal M}_{r'}(S)]\longrightarrow 0.$

Termwise, the Ext-groups turn into cohomology groups

$H^i(\overline{\mathcal M}_{r'}(S)^{2^j},V).$

Unfortunately, $$\overline{\mathcal M}_{r'}(S)$$ itself is not profinite, so we cannot directly apply Proposition 2.4.9. To get around this last cliff, we write $$Q'(\overline{\mathcal M}_{r'}(S))$$ as a filtered colimit of complexes

$Q'(\overline{\mathcal M}_{r'}(S))_{\leq c}: \ldots \longrightarrow \mathbb Z[\overline{\mathcal M}_{r'}(S)^2_{\leq \kappa _1 c}]\longrightarrow \mathbb Z[\overline{\mathcal M}_{r'}(S)_{\leq \kappa _0 c}]\longrightarrow 0$

where the constants $$\kappa _0=1,\kappa _1,\ldots$$ are positive and chosen so that all differentials are well-defined. (The possibility of choosing such constants has already been formalized; TODO include pointer.) It suffices to prove that

$(T^{-1})_V - (T^{-1})_{\mathcal M}: \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq c},V)\longrightarrow \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq r' c},V)$

is a pro-isomorphism in $$c$$, as then the final result follows by passing to a derived limit over $$c$$, see Lemma 2.4.11 below. This final pro-isomorphism assertion can finally be written out, and it unravels to the statement of Theorem 1.7.1.

In passing to the derived limit over $$c$$, we use the following lemma.

Lemma 2.4.11
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Assume that in each degree $$i$$, the map

$(T^{-1})_V - (T^{-1})_{\mathcal M}: \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq c},V)\longrightarrow \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq r' c},V)$

is a pro-isomorphism in $$c$$ (i.e., pro-systems of kernels, and of cokernels, are pro-zero). Then

$(T^{-1})_V - (T^{-1})_{\mathcal M}: \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S)),V)\longrightarrow \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S)),V).$

is an isomorphism.

Proof

We have

$Q'(\overline{\mathcal M}_{r'}(S)) = \varinjlim _n Q'(\overline{\mathcal M}_{r'}(S))_{\leq n},$

inducing a resolution

$0\to \bigoplus _n Q'(\overline{\mathcal M}_{r'}(S))_{\leq n}\to \bigoplus _n Q'(\overline{\mathcal M}_{r'}(S))_{\leq n}\to Q'(\overline{\mathcal M}_{r'}(S))\to 0.$

Passing to a corresponding long exact sequence reduces one to checking that the squares

are bicartesian (here, horizontal maps are shift minus identity, and vertical maps are $$(T^{-1})_V - (T^{-1})_{\mathcal M}$$). Equivalently, the horizontal maps become isomorphisms on vertical kernels, and vertical cokernels. But the vertical kernels and vertical cokernels induce pro-zero systems of abelian groups, and then the horizontal kernels and cokernels compute $$\varprojlim _n$$ and $$\varprojlim _n^1$$ of their systems, which vanish.

Proposition 2.4.12
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Let $$0 {\lt} r {\lt} 1$$ be a real number, and let $$S$$ be a profinite set. Decomposing $$\mathbb Z((T))_r$$ into positive and nonpositive coefficients yields a direct sum decomposition

$\mathbb Z((T))_r = T\mathbb Z[[T]]_r \oplus \mathbb Z[T^{-1}].$

This extends to a decomposition of spaces of measures

$\mathcal L_r(S) = \mathcal L(S,T\mathbb Z((T))_r) = \mathcal L(S,T\mathbb Z[[T]]_r) \oplus \mathcal L(S,\mathbb Z[T^{-1}])$

where $$\mathcal M(S,\mathbb Z[T^{-1}]) = \mathbb Z[T^{-1}][S]$$ is the free condensed $$\mathbb Z[T^{-1}]$$-module on $$S$$. Letting $$\overline{\mathcal L}_r(S)= \mathcal L(S,T\mathbb Z[[T]]_r)$$, we get a short exact sequence of condensed $$\mathbb Z[T^{-1}]$$-modules

$0 \longrightarrow \mathbb Z[T^{-1}][S] \longrightarrow \mathcal L_r(S) \longrightarrow \overline{\mathcal L}_r(S) \longrightarrow 0.$

Proof

On $$\mathbb Z((T))_{r,\leq c}$$, only finitely many nonpositive coefficients can possibly be nonzero, and each of them is bounded. This shows that the nonpositive summand of $$\mathbb Z((T))_r$$ is given by $$\mathbb Z[T^{-1}]$$. To pass to profinite $$S$$, use Proposition 2.4.6.

Lemma 2.4.13
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Let $$0 {\lt} r {\lt} r' {\lt} 1$$ be real numbers. Let $$S$$ be a profinite set, and let $$V$$ be an $$r$$-normed $$\mathbb Z[T^{\pm 1}]$$-module. Then $$\operatorname{Ext}_{\mathrm{Mod}^{\mathrm{cond}}_{\mathbb Z[T^{-1}]}}^i(\mathcal L_{r'}(S), V) = 0$$ for all $$i {\gt} 0$$. In other words,

$\operatorname{Ext}_{\mathbb Z}^i(\mathcal L_{r'}(S), V) \xrightarrow {[T⁻¹]_L - [T⁻¹]_V} \operatorname{Ext}_{\mathbb Z}^i(\mathcal L_{r'}(S), V)$

is a bijection for all $$i {\gt} 0$$ and a surjection for $$i = 0$$.

Proof

Consider the long exact sequence of Ext-groups arising form the short exact sequence (Lemma 2.4.12)

$0 \longrightarrow \mathbb Z[T^{-1}][S] \longrightarrow \mathcal L_{r'}(S) \longrightarrow \overline{\mathcal L}_{r'}(S) \longrightarrow 0$

by applying $$\operatorname{Ext}^*(\_ , V)$$.

By Lemma 2.4.7 all groups $$\operatorname{Ext}_{\operatorname{Cond}(\operatorname{Ab})}^i(\mathbb {Z}[S], V)$$ vanish for $$i {\gt} 0$$. And by Lemma 2.4.10 all groups $$\operatorname{Ext}_{\mathrm{Mod}^{\mathrm{cond}}_{\mathbb Z[T^{-1}]}}^i(\overline{\mathcal L}_{r'}(S), V)$$ vanish for $$i \ge 0$$. The result follows.

The “In other words” version can be proved without mentioning $$\mathbb Z[T^{-1}]$$-linear Ext groups, by using the same ingredients and the five lemma.

Lemma 2.4.14
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Let $$0 {\lt} p' {\lt} 1$$ be a real number, let $$S$$ be a profinite set, and let $$r'$$ denote $$(\tfrac 12)^{p'}$$. There is a short exact sequence of condensed $$\mathbb Z[T^{-1}]$$-modules

$0 \longrightarrow \mathcal L_{r'}(S) \longrightarrow \mathcal L_{r'}(S) \longrightarrow \mathcal M_{p'}(S) \longrightarrow 0$

where the first map is multiplication by $$2T - 1$$, and the second is evaluation at $$T = \tfrac 12$$.

Proof

By Proposition 2.4.5 it suffices to show that the corresponding sequence of pseudonormed groups is short exact, and by Proposition 2.4.3 we may also assume that $$S$$ is finite. With this reductions, the lemma is precisely Theorem 2.2.8.

Theorem 2.4.15 Clausen–Scholze
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Let $$0 {\lt} p' {\lt} p \le 1$$ be real numbers, let $$S$$ be a profinite set, and let $$V$$ be a $$p$$-Banach space. Let $$\mathcal M_{p'}(S)$$ be the space of real $$p'$$-measures on $$S$$. Then

$\operatorname{Ext}^i_{\operatorname{Cond}(\operatorname{Ab})}(\mathcal M_{p'}(S),V) = 0$

for $$i \ge 1$$.

Proof

Recall from Lemma 2.4.14 the short exact sequence

$0 \longrightarrow \mathcal L_{r'}(S) \longrightarrow \mathcal L_{r'}(S) \longrightarrow \mathcal M_{p'}(S) \longrightarrow 0.$

Apply to this $$\text{Ext}^*(\_ , V)$$ to obtain a long exact sequence. Note that $$T$$ acts on $$V$$ via multiplication by $$\tfrac 12$$ (by Lemma 2.1.2). Hence we can use Lemma 2.4.13 to obtain isomorphisms between the Ext-groups involving $$\mathcal L_{r'}(S)$$, for $$i {\gt} 0$$, and a surjection for $$i = 0$$. The result follows.