Blueprint for the Liquid Tensor Experiment: Condensed abelian groups

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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

$\begin{tikzcd} \prod_n \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq n},V)\ar[d]\ar[r] & \prod_n \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq n},V)\ar[d]\\ \prod_n \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq n},V)\ar[r] & \prod_n \mathrm{Ext}^i(Q'(\overline{\mathcal M}_{r'}(S))_{\leq n},V) \end{tikzcd}$

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 ✓

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 ✓

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 ✓

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 ✓

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.