# Blueprint for the Liquid Tensor Experiment

## 1.6 Polyhedral lattices

The following definition deviates slightly from  [ Sch20 ] .

Definition 1.6.1
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A polyhedral lattice is a finite free abelian group $$\Lambda$$ equipped with a norm $$‖\cdot ‖_\Lambda \colon \Lambda \otimes \mathbb R\to \mathbb R$$ such that there exists a finite set $$\{ \lambda _1, \dots , \lambda _n\} \subset \Lambda$$ that generate the norm: that is to say, for every $$\lambda \in \Lambda$$ there exist $$c_1, \dots , c_n \in \mathbb N$$ such that $$\lambda = \sum c_i \lambda _i$$ and $$\| \lambda \| = \sum c_i\| \lambda _i\|$$.

Finally, we can prove the key combinatorial lemma, ensuring that any element of $$\operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))$$ can be decomposed into $$N$$ elements whose norm is roughly $$\tfrac 1N$$ of the original element.

Definition 1.6.2
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Let $$M$$ be a pseudo-normed group, $$N \in \mathbb N$$, and $$d \in \mathbb R_{\ge 0}$$. We say that $$M$$ is $$N$$-splittable with error term $$d$$, if for all $$c$$ and $$x \in M_c$$, there exists a decomposition

$x = x_1 + x_2 + \dots + x_N,$

with $$x_i \in M_{c/N + d}$$.

Proposition 1.6.3
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Let $$\Lambda$$ be a polyhedral lattice, and $$S$$ a profinite set. Then for all positive integers $$N$$ there is a constant $$d$$ such that for all $$c{\gt}0$$ one can write any $$x\in \operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c}$$ as

$x=x_1+\ldots +x_N$

where all $$x_i\in \operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c/N+d}$$.

In other words, for all $$N$$, there exists a $$d$$ such that $$\operatorname{Hom}(\Lambda , \overline{\mathcal L}_{r'}(S))$$ is $$N$$-splittable with error term $$d$$.

Proof

The desired statement is equivalent to the surjectivity of the map of profinite sets

$\operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c/N+d}^N\times _{\operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c+Nd}} \operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c} \longrightarrow \operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c}.$

Note that, as a functor of $$S$$, both sides commute with cofiltered limits, so it is enough to handle finite $$S$$, by Tychonoff. But that is exactly the following Lemma 1.6.4.

Lemma 1.6.4
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Let $$\Lambda$$ be a polyhedral lattice, and $$S$$ a finite set. Then for all positive integers $$N$$ there is a constant $$d$$ such that for all $$c{\gt}0$$ one can write any $$x\in \operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c}$$ as

$x=x_1+\ldots +x_N$

where all $$x_i\in \operatorname{Hom}(\Lambda ,\overline{\mathcal L}_{r'}(S))_{\leq c/N+d}$$.

In other words, for all $$N$$, there exists a $$d$$ such that $$\operatorname{Hom}(\Lambda , \overline{\mathcal L}_{r'}(S))$$ is $$N$$-splittable with error term $$d$$.

As preparation for the proof, we have the following results.

Lemma 1.6.5 Gordan’s lemma
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Let $$\Lambda$$ be a finite free abelian group, and let $$\lambda _1, \ldots , \lambda _m \in \Lambda$$ be elements. Let $$M \subset \operatorname{Hom}(\Lambda , \mathbb Z)$$ be the submonoid $$\{ x \mid x(\lambda _i) \ge 0 \text{ for all i = 1, \dots , m}\}$$. Then $$M$$ is finitely generated as monoid.

Proof

This is a standard result. We omit the proof here. It is done in Lean.

Lemma 1.6.6
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Let $$\Lambda$$ be a finite free abelian group, let $$N$$ be a positive integer, and let $$\lambda _1,\ldots ,\lambda _m\in \Lambda$$ be elements. Then there is a finite subset $$A\subset \Lambda ^\vee$$ such that for all $$x\in \Lambda ^\vee =\operatorname{Hom}(\Lambda ,\mathbb Z)$$ there is some $$x'\in A$$ such that $$x-x'\in N\Lambda ^\vee$$ and for all $$i=1,\ldots ,m$$, the numbers $$x'(\lambda _i)$$ and $$(x-x')(\lambda _i)$$ have the same sign, i.e. are both nonnegative or both nonpositive.

Proof

It suffices to prove the statement for all $$x$$ such that $$\lambda _i(x)\geq 0$$ for all $$i$$; indeed, applying this variant to all $$\pm \lambda _i$$, one gets the full statement.

Thus, consider the submonoid $$\Lambda ^\vee _+\subset \Lambda ^\vee$$ of all $$x$$ that pair nonnegatively with all $$\lambda _i$$. This is a finitely generated monoid by Lemma 1.6.5; let $$y_1,\ldots ,y_M$$ be a set of generators. Then we can take for $$A$$ all sums $$n_1y_1+\ldots +n_My_M$$ where all $$n_j\in \{ 0,\ldots ,N-1\}$$.

Lemma 1.6.7
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Let $$x_0, x_1, \dots$$ be a sequence of reals, and assume that $$\sum _{i=0}^\infty x_i$$ converges absolutely. For every natural number $$N {\gt} 0$$, there exists a partition $$\mathbb N = A_1 \sqcup A_2 \sqcup \dots \sqcup A_N$$ such that for each $$j = 1,\dots ,N$$ we have $$\sum _{i \in A_j} x_i \le (\sum _{i=0}^\infty x_i)/N + 1$$

Proof

Define the $$A_j$$ recursively: assume that the natural numbers $$0, \dots , n$$ have been placed into the sets $$A_1, \dots , A_N$$. Then add the number $$n+1$$ to the set $$A_j$$ for which

$\sum _{i=0, i\in A_j}^n x_i$

is minimal.

Lemma 1.6.8
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For all natural numbers $$N {\gt} 0$$, and for all $$x\in \overline{\mathcal L}_{r'}(S)_{\leq c}$$ one can decompose $$x$$ as a sum

$x=x_1+\ldots +x_N$

with all $$x_i\in \overline{\mathcal L}_{r'}(S)_{\leq c/N+1}$$.

Proof

Choose a bijection $$S \times \mathbb N \cong \mathbb N$$, and transport the result from Lemma 1.6.7.

Proof of Lemma 1.6.4

Pick $$\lambda _1,\ldots ,\lambda _m\in \Lambda$$ generating the norm. We fix a finite subset $$A\subset \Lambda ^\vee$$ satisfying the conclusion of Lemma 1.6.6. Write

$x=\sum _{n\geq 1, s\in S} x_{n,s} T^n [s]$

with $$x_{n,s}\in \Lambda ^\vee$$. Then we can decompose

$x_{n,s} = N x_{n,s}^0 + x_{n,s}^1$

where $$x_{n,s}^1\in A$$ and we have the same-sign property of the last lemma. Letting $$x^0 = \sum _{n\geq 1, s\in S} x_{n,s}^0 T^n [s]$$, we get a decomposition

$x = Nx^0 + \sum _{a\in A} a x_a$

with $$x_a\in \overline{\mathcal L}_{r'}(S)$$ (with the property that in the basis given by the $$T^n [s]$$, all coefficients are $$0$$ or $$1$$). Crucially, we know that for all $$i=1,\ldots ,m$$, we have

$‖x(\lambda _i)‖ = N ‖x^0(\lambda _i)‖ + \sum _{a\in A} |a(\lambda _i)| ‖x_a‖$

by using the same sign property of the decomposition.

Using this decomposition of $$x$$, we decompose each term into $$N$$ summands. This is trivial for the first term $$Nx^0$$, and each summand of the second term decomposes with $$d = 1$$ by Lemma 1.6.8. (It follows that in general one can take for $$d$$ the supremum over all $$i$$ of $$\sum _{a\in A} |a(\lambda _i)|$$.)

Definition 1.6.9
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Let $$\Lambda$$ be a polyhedral lattice, and let $$N {\gt} 0$$ be a natural number. (We think of $$N$$ as being fixed once and for all, and thus it does not show up in the notation below.)

By $$\Lambda '$$ we denote $$\Lambda ^N$$ endowed with the norm

$‖(\lambda _1,\ldots ,\lambda _N)‖_{\Lambda '} = \tfrac 1N(‖\lambda _1‖_\Lambda +\ldots +‖\lambda _N‖_\Lambda ).$

This is a polyhedral lattice.

Lemma 1.6.10
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For any $$m\geq 1$$, let $$\Lambda '^{(m)}$$ be given by $$\Lambda '^m / \Lambda \otimes (\mathbb Z^m)_{\sum =0}$$; for $$m=0$$, we set $$\Lambda '^{(0)} = \Lambda$$. Then $$\Lambda '^{(m)}$$ is a polyhedral lattice.

Proof

The proof is done in Lean. TODO: write down a proof here.

Definition 1.6.11
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For any $$m\geq 1$$, let $$\Lambda '^{(m)}$$ be given by $$\Lambda '^m / \Lambda \otimes (\mathbb Z^m)_{\sum =0}$$; for $$m=0$$, we set $$\Lambda '^{(0)} = \Lambda$$. Then $$\Lambda '^{(\bullet )}$$ is a cosimplicial polyhedral lattice, the Čech conerve of $$\Lambda \to \Lambda '$$.

In particular, $$\Lambda '^{(0)} = \Lambda \to \Lambda ' = \Lambda '^{(1)}$$ is the diagonal embedding.

Definition 1.6.12
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Let $$\Lambda$$ be a polyhedral lattice, and $$M$$ a profinitely filtered pseudo-normed group.

Endow $$\operatorname{Hom}(\Lambda , M)$$ with the subspaces

$\operatorname{Hom}(\Lambda , M)_{\leq c} = \{ f \colon \Lambda \to M \mid \forall x \in \Lambda , f(x) \in M_{\leq c‖x‖} \} .$

As $$\Lambda$$ is polyhedral, it is enough to check the given condition on $$f$$ for a finite collection of $$x$$ that generate the norm.

These subspaces are profinite subspaces of $$M^\Lambda$$, and thus they make $$\operatorname{Hom}(\Lambda , M)$$ ito a profinitely filtered pseudo-normed group.

If $$M$$ has an action of $$T^{-1}$$, then so does $$\operatorname{Hom}(\Lambda , M)$$.