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L²-dimension estimates for amenable D(R) groups

Statement

Assume the Axiom of Choice. Let R be Q or F_p, let Γ be a countable discrete amenable group in Strebel's class D(R), and let C_* be a bounded-below chain complex of right ZΓ-modules whose C_i are finitely generated free for i≤n. Give R the trivial Γ-action. If H_i(C_⊗{ZΓ}R)=0 for every i≤n, then H_i(C⊗_{ZΓ}NΓ) has von Neumann dimension zero for every i≤n. This is a dimension-zero conclusion and does not assert that the L²-homology module vanishes.

Statement references

Facts & Assumptions

Given: The Axiom of Choice, the field R∈{Q,Fp}, the amenable group Γ∈D(R), the bounded-below chain complex and its finite-free terms through degree n, and the stated trivial-coefficient homology vanishing.

[F1]

The class D(R) detects injectivity of maps between projective RΓ-modules from injectivity after augmentation to R. Integral half-solutions, Strebel classes, and L²-signatures

[F2]

For every CΓ-module M, amenability gives dim⁡NΓTor⁡1CΓ(M,NΓ)=0. L²-dimension flatness for amenable groups

[F3]

Finite-matrix kernels are projection modules, polar decomposition preserves their trace dimensions, and quotienting a finite projection module by a Hilbert-dense submodule gives dimension zero. Dimension-zero modules have dimension-zero quotients. Finite von Neumann modules, Hilbert closures, and dimension-zero quotients

[A1]

Full AC is used by the Hilbert-space and von Neumann dimension supplier [F3] and by the amenable Tor dimension theorem [F2]. No additional choice is used in the finite matrix reductions below. The Axiom of Choice

Proof

Given: The data and foundational convention above. Write A=NΓ, and let l be a lower bound with Ci=0 for i<l. If n<l, every claimed group is already zero, so assume n≥l.

1.1givenF3constructalgebra

First replace the term above degree n by a finite free module if necessary. The image of dn+1 after tensoring with R is the kernel of dn over R, a finite-dimensional space. Choose finitely many tensors in Cn+1⊗R whose boundaries span it, and expand each tensor as a finite sum c⊗r. The finitely many resulting elements c∈Cn+1 define a finite free module Pn+1 with a map to Cn+1; compose this map with dn+1. Keep Pi=Ci for l≤i≤n and set all other terms to zero. Since dndn+1=0, this defines a bounded finite-free complex P∗ whose R-homology vanishes through degree n. Below n its A-homology is the original one, and in degree n its A-homology surjects onto the original one, since it has the same cycles and possibly fewer boundaries. It suffices by [F3] to prove the result for P∗.

2.1givenstep 1.1constructalgebra

Over the field R, split the exact complex through degree n: choose vector-space complements of the kernels in each of its finitely many finite-dimensional terms, and invert each boundary map onto its image on the chosen complement. This gives maps si:Pi⊗R→Pi+1⊗R, l≤i≤n, with ds+sd=I through degree n. For R=Fp, lift the finitely many matrix entries to integers and then to constant coefficients in ZΓ, obtaining maps Di. For R=Q, choose a single positive integer a clearing the denominators of all these finitely many matrices, and lift asi instead; in the finite-field case put a=1. Set Di=0 outside the indicated range. Then f=dD+Dd is a chain map over ZΓ, chain homotopic to zero, and its augmented matrix in degree i≤n is aI, which is injective over R.

3.1F1step 2.1algebra

Apply [F1] to fi⊗ZR on the free RΓ-module Pi⊗ZR; its augmentation is injective by step 2.1, so this map is injective. Hence fi is injective over ZΓ. For R=Q this follows by viewing a nonzero integral vector as a nonzero rational vector. For R=Fp, an integral kernel vector has finitely many coefficients; divide them all by the largest common power of p. Torsion-freeness of the integral target preserves the kernel equation, and the resulting nonzero vector modulo p contradicts injectivity. Passing to CΓ preserves this injection: any finite complex relation can be decomposed using a basis of the finite-dimensional rational span of its coefficients into rational relations, each of which clears denominators to an integral one.

4.1A1F2F3step 3.1algebra

Resolve the cokernel of the injective square matrix fi over CΓ. The degree-one Tor group is the kernel of fi:Ari→Ari after tensoring, where ri is the rank of Pi; this is the same free-presentation computation used to prove [F2]. Consequently [F2] makes that kernel dimension zero. Its projection has trace zero and is therefore zero by faithfulness in [F3]. Thus fi is injective on the Hilbert completion. Let Zi be the Hilbert kernel of di. The chain identity makes fi preserve Zi. Its restriction to Zi has zero kernel, and polar decomposition in [F3] makes its Hilbert-closed image have the same trace dimension as Zi; since that image lies in Zi, faithfulness makes it dense in Zi.

5.1A1F3step 1.1step 4.1algebra∎

The algebraic A-cycle module is the projection module PZiAri and is Hilbert-dense in Zi by [F3]. On these cycles, fi=di+1Di+Di−1di=di+1Di, so its image is contained in the algebraic boundary module. Step 4.1 shows that these boundaries are Hilbert-dense in Zi. The dense-quotient assertion in [F3] therefore gives dimension-zero A-homology of P∗ through degree n. The identity/surjection comparison of step 1.1 transfers this conclusion to C∗. The proof establishes vanishing of the dimension and does not assert algebraic vanishing of the homology module.

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