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

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
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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.

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