Blueprint for the Liquid Tensor Experiment: Polyhedral lattices

1.6 Polyhedral lattices

The following definition deviates slightly from [ Sch20 ] .

Definition 1.6.1 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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 ✓

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

Definition 1.6.12 ✓

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 $\operatorname{Hom}(\Lambda , M)$.