L²-dimension flatness for amenable groups

First let $K\subseteq R^n$ be generated by finitely many columns, and let $B:R^s\to R^n$ be the corresponding matrix map onto $K$. Right exactness of tensor products identifies $K{\otimes }_{R}A$ with $A^s\mathrm{/}\mathrm{im}(({\ker }_{R}B){\otimes }_{R}A)$. The kernel of its natural map to $A^n$ is therefore ${\ker }_{A}B\mathrm{/}\mathrm{im}(({\ker }_{R}B){\otimes }_{R}A)$. This quotient has dimension zero by [F3]. The tensor-product identifications follow directly from the presentation ${\ker }_{R}B\to R^s\to K\to 0$: relations among the chosen generators are exactly the elements of ${\ker }_{R}B$.

For an arbitrary submodule $K\subseteq R^n$, take its directed union of finitely generated submodules $K_j$. A tensor is a finite sum, so every element of $K{\otimes }_{R}A$ comes from some $K_j{\otimes }_{R}A$. If its image in $A^n$ is zero, its representative already lies in the kernel of the natural map $K_j{\otimes }_{R}A\to A^n$. Thus the kernel for $K$ is the directed union of images of the zero-dimensional kernels from step 1.1; it has dimension zero by [F2]. This also proves the assertion if relations identify different representatives, since images are quotients of those kernels.

Let $F={⨁}_{i\in I}R$ be any free module and $K\subseteq F$. Every finite tensor in $K{\otimes }_{R}A$ involves finitely many vectors of $K$, each of finite coordinate support in $F$. Their supports lie in one finite free summand $R^n\subseteq F$. A kernel element is therefore the image of a kernel for a submodule of that finite free summand, handled by step 2.1. Taking the directed union over those submodules and finite supports, and applying [F2], proves that $\ker (K{\otimes }_{R}A\to F{\otimes }_{R}A)$ has dimension zero for every free $F$.

Choose a free presentation $0\to K\to F\to M\to 0$. In the resolution-dependent definition of Tor, first use a free resolution beginning with this presentation. For any other chosen projective resolution, [F5] gives chain maps inverse up to chain homotopy; this use of projective-resolution comparison relies on [A1]. Additive tensoring preserves the homotopy equation $f-g=dh+hd$, since tensoring respects sums and composition. Thus the tensor complexes are homotopy equivalent and their homology modules are isomorphic; the calculation below therefore applies to every retained resolution datum. A free resolution beginning with this presentation computes ${\mathrm{Tor}}_1^R(M,A)$ as $\ker (K{\otimes }_{R}A\to F{\otimes }_{R}A)$: resolving $K$ and splicing onto $F\to M$ gives this equality by right exactness at the last differential. Thus step 3.1 proves the assertion for degree one. For $q>1$, successively apply the same construction to the kernels in a free resolution of $M$; the positive-degree homology after splicing shifts by one, giving ${\mathrm{Tor}}_q^R(M,A)\cong {\mathrm{Tor}}_1^R(K_{q-1},A)$ for the appropriate syzygy $K_{q-1}$. The degree-one argument applies to every such module, so every asserted higher Tor group has dimension zero. No claim that these algebraic quotient modules are zero is used or obtained.