## 1.6 Polyhedral lattices

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 \(c_1, \dots , c_n \in \mathbb Q\) such that \(\lambda = \sum c_i \lambda _i\) and \(\| \lambda \| = c_i\| \lambda _i\| \).

Equivalently (but not verified in Lean): the norm is given by the supremum of finitely many linear functions on \(\Lambda \); or once more, equivalently, the “unit ball” \(\{ \lambda \in \Lambda \otimes \mathbb R\mid ‖\lambda ‖_\Lambda \leq 1\} \) is a polyhedron.

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.

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

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

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

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 \(d\).

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

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.

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

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.

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.

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

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.

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 \(n_j\in \{ 0,\ldots ,N-1\} \).

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 \(\sum _{i \in A_j} x_i \le (\sum _{i=0}^\infty x_i)/N + 1\)

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

is minimal.

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

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

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

Pick \(\lambda _1,\ldots ,\lambda _m\in \Lambda \) generating the norm. We fix a finite subset \(A\subset \Lambda ^\vee \) satisfying the conclusion of the previous lemma. Write

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

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

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

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 \(\sum _{a\in A} |a(\lambda _i)|\).)

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

This is a polyhedral lattice.

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.

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

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

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