Finite based free complexes and contraction torsion

Let $R$ be an associative unital ring. A finite based free right $R$-chain complex is a chain complex $C$ of right $R$-modules (Chain complex in an abelian category, Unital left and right modules over a ring; unqualified module means left module) which is bounded, so that $C_n=0$ for all but finitely many $n$, together with a preferred finite basis $B_n$ of the free right $R$-module $C_n$ for every $n$; the union $B={⨆}_n{B}_n$ is the displayed basis and the elements of $B_n$ are the displayed basis vectors of degree $n$. The degree-ordered bases are $B_{\mathrm{o}\mathrm{d}\mathrm{d}}=(b\in B_n:n\text{ odd},n\text{ increasing}),B_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}=(b\in B_n:n\text{ even},n\text{ increasing}),$ each written as a finite list by increasing degree and, within a degree, in the order fixed by $B_n$. A chain contraction $s$ of $C$ is a right-linear family with $ds+sd=\mathrm{i}\mathrm{d}$ (A chain contraction makes the odd-to-even parity map invertible).

Assume now that $\mathrm{\#}{B}_{\mathrm{o}\mathrm{d}\mathrm{d}}=\mathrm{\#}{B}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}\mathrm{.}$ By the parity lemma (A chain contraction makes the odd-to-even parity map invertible) the odd-to-even component ${\mathrm{\Phi }}_s:=(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}:C_{\mathrm{o}\mathrm{d}\mathrm{d}}\to C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ is an isomorphism of right $R$-modules, so its matrix $A_s$ in the displayed bases $B_{\mathrm{o}\mathrm{d}\mathrm{d}}$, $B_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ is an invertible square matrix over $R$, and the contraction torsion of $(C,s)$ is the class ${\tau }_s(C):=[A_s]\in {\tilde{K}}_1(R)$ of K₁ of a ring and the Whitehead group of a discrete group. The displayed bases fix the sign convention: the odd-to-even parity is used, not the even-to-odd one.

Automatic equality of basis sizes. If $R$ has invariant basis number, then $\mathrm{\#}{B}_{\mathrm{o}\mathrm{d}\mathrm{d}}=\mathrm{\#}{B}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ always holds, because ${\mathrm{\Phi }}_s$ is an isomorphism of free right $R$-modules $C_{\mathrm{o}\mathrm{d}\mathrm{d}}\to C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$; in particular this applies to every group ring $R=\mathbb{Z}[\pi ]$ by Integral group rings have invariant basis number. Over a ring without invariant basis number the matrix formula is asserted only when the two displayed finite basis sizes agree, as above; the parity lemma itself needs no such hypothesis.

The two-term case. Let $q\ge 1$, let $u\in R$ be a unit, and let $C$ be the complex $0\to C_q=R\stackrel{u}{\to }{C}_{q-1}=R\to 0$ with the displayed single basis vector in each of the degrees $q$ and $q-1$ and zero elsewhere. The contraction identity forces $u$ to be a unit and $s_{q-1}=u^{-1}$, $s_q=0$, so the complex is contractible with that contraction. If $q$ is odd the map $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}$ contains the component $C_q\to C_{q-1}$ with matrix $(u)$ and no other nonzero component, so $\tau (C)=[u]$; if $q$ is even the same component belongs to the even-to-odd map, $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=u^{-1}$ on the remaining degree, and $\tau (C)=[u^{-1}]=-[u]$. Thus $\tau (C)=(-1{)}^{q+1}[u]\in {\tilde{K}}_1(R),$ which is the parity sign used throughout this page.