Simple Homotopy, Whitehead Groups, and Torsion

Let $X$ be a finite CW complex and let $n\ge 1$. Write $D^n=\{x\in {\mathbb{R}}^n:\mathrm{\mid }x\mathrm{\mid }\le 1\}$ for the closed unit ball, $S^{n-1}=\mathrm{\partial }{D}^n$ for its boundary sphere, and $D_{+}^{n-1}=\{(x,t)\in S^{n-1}:t\ge 0\}$ for a closed upper hemisphere of $S^{n-1}$.

An elementary expansion of $X$ of dimension $n$ is an inclusion $X\hookrightarrow Y$ of CW complexes together with a homeomorphism $\mathrm{\Phi }:(D^n,D_{+}^{n-1})\to (Q^n,Q^{n-1})$ of ball pairs and a continuous map $\phi :Q^n\to Y$ such that

• $\phi$ is a characteristic map for a new $n$-cell $e^n=\phi (Q^n\setminus \mathrm{\partial }{Q}^n)$,
• $\phi {\mathrm{\mid }}_{Q^{n-1}}$ is a characteristic map for a new $(n-1)$-cell $e^{n-1}=\phi (Q^{n-1}\setminus \mathrm{\partial }{Q}^{n-1})$,
• the remaining boundary is old: $\phi (\mathrm{\partial }{Q}^n\setminus \mathrm{int}{Q}^{n-1})\subseteq X$, and
• $Y=X\cup e^{n-1}\cup e^n$ as a CW complex, with $X$ a subcomplex.

The new $(n-1)$-cell is called the free face of the new $n$-cell. The restriction of $\phi$ to $Q^{n-1}$ maps its interior homeomorphically onto $e^{n-1}$ and maps its boundary into $X$; it need not be a homeomorphism onto the closed cell, whose attaching map may identify boundary points. The complementary boundary of $Q^n$ maps into $X$. In particular, an attachment of only one $n$-cell to a complex already containing the alleged free face is not an elementary expansion under this definition.

We say that $Y$ collapses to $X$ by an elementary collapse and write $Y\searrow X$ when $X\hookrightarrow Y$ is an elementary expansion; the elementary collapse is the inverse formal operation removing the pair $(e^{n-1},e^n)$. A finite sequence of elementary expansions and elementary collapses, each performed relative to the cells retained by the previous steps, is a formal deformation; when every cell of a subcomplex $X_0$ is retained throughout, the deformation is written relative to $X_0$, and the operations are then said to fix the retained subcomplex. The one-cell case $n=1$ attaches a new vertex and a new edge joining it to an old vertex, the new vertex being the free face.

Let $X$ and $Y$ be finite CW complexes. A map $f:X\to Y$ is a simple homotopy equivalence if it is homotopic to a finite composite $X=X_0\stackrel{f_1}{\to }{X}_1\stackrel{f_2}{\to }\cdots \stackrel{f_k}{\to }{X}_k=Y$ of maps between finite CW complexes in which each $f_i$ is either an elementary expansion inclusion, an elementary collapse map (Elementary expansions and collapses of finite CW complexes), or a cellular isomorphism. A cellular isomorphism here means a homeomorphism that carries the cell structure of its source isomorphically onto the cell structure of its target, so that it restricts to a homeomorphism between the interiors of corresponding cells; a general homeomorphism is not included by definition.

For a formal collapse $Y\searrow X$, an elementary collapse map means any retraction $r:Y\to X$ of its expansion inclusion $j:X\hookrightarrow Y$. Such maps exist: the characteristic ball strongly deformation retracts onto the complementary boundary disk, fixing that disk pointwise. This deformation descends through the attaching identifications to a strong deformation retraction of $Y$ onto $X$. If $r_0$ is its endpoint and $r$ is any retraction, composing this deformation with $r$ gives $r\simeq r_0$ relative to $X$. Thus $rj={\mathrm{i}\mathrm{d}}_X$ and $jr\simeq {\mathrm{i}\mathrm{d}}_Y$.

Every such composite is a homotopy equivalence (Homotopy equivalences, homotopy inverses and spaces of the same homotopy type): an expansion inclusion and its collapse map are homotopy inverses, and a cellular isomorphism is a homeomorphism. Consequently every simple homotopy equivalence is a homotopy equivalence.

For disconnected complexes the sequence respects the induced bijection on components: each operation is performed componentwise, and the composite maps the components of $X$ bijectively onto those of $Y$. The empty complex is allowed; the empty sequence exhibits the identity of a finite CW complex as a simple homotopy equivalence.

Homotopy of maps is transitive, so any map homotopic to a simple homotopy equivalence is again a simple homotopy equivalence. Concatenating two finite composites exhibits a composite of two simple homotopy equivalences as a simple homotopy equivalence as well.

Throughout, $R$ is an associative unital ring, not assumed commutative (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). For $n\ge 0$ let $R^n$ be the set of column vectors $v=(v_1,\dots ,v_n)$ with $v_i\in R$, made into a right $R$-module by entrywise addition and the right action $(v\cdot r{)}_i:=v_{i}r(r\in R)$ (Unital left and right modules over a ring; unqualified module means left module); for $n=0$ this is the zero module, whose unique element is the empty column. A map $f:R^n\to R^m$ is right $R$-linear when $f(v+v^{\mathrm{\prime }})=f(v)+f(v^{\mathrm{\prime }})$ and $f(v\cdot r)=f(v)\cdot r$.

Matrices. Every right-linear $f:R^n\to R^m$ has a unique matrix $A\in M_{m\times n}(R)$, written $A=(A_{ij})$, such that $(Av{)}_i={\sum }_{j=1}^n{A}_{ij}{v}_j,$ the sum being the finite sum in the additive group of $R$; uniqueness uses that the standard basis vectors $e_1,\dots ,e_n$ of $R^n$ generate it and that columns of $A$ are the coordinate vectors of $f(e_j)$. If $g:R^m\to R^p$ has matrix $B$, then $g\circ f:R^n\to R^p$ has matrix $BA$, with $(BA{)}_{ik}={\sum }_j{B}_{ij}{A}_{jk}$; this is the displayed order and it uses only associativity and distributivity.

General linear group. Let ${\mathrm{G}\mathrm{L}}_n(R)$ be the group of $n\times n$ matrices over $R$ possessing a two-sided inverse, with multiplication of matrices as the group operation and $I_n$ as the identity; by the previous paragraph ${\mathrm{G}\mathrm{L}}_n(R)$ is exactly the group of right-linear automorphisms of $R^n$. The stabilization $A\mapsto \mathrm{diag}(A,1)$ identifies ${\mathrm{G}\mathrm{L}}_n(R)$ with a subgroup of ${\mathrm{G}\mathrm{L}}_{n+1}(R)$, and $\mathrm{G}\mathrm{L}(R):={\bigcup }_{n\ge 0}{\mathrm{G}\mathrm{L}}_n(R)$ is the stable general linear group, in which every element is represented by some $n\times n$ matrix. A matrix $A$ belongs to ${\mathrm{G}\mathrm{L}}_n(R)$ when there is a matrix $B$ satisfying both $AB=I_n$ and $BA=I_n$; either equation alone need not imply the other over an arbitrary unital ring. No determinant, commutativity, or rank function is used anywhere in this definition.

For a concrete one-sided inverse, take $R={\mathrm{End}}_k(V)$ where $V$ has basis $e_0,e_1,\dots$. Let $S(e_i)=e_{i+1}$ and let $L(e_0)=0$, $L(e_{i+1})=e_i$. Then $LS=1_V$, while $SL(e_0)=0\mathrm{\ne }{e}_0$, so the $1\times 1$ matrices $A=L$ and $B=S$ satisfy $AB=I_1$ but not $BA=I_1$.

Elementary matrices. For $n\ge 1$, $1\le i,j\le n$ with $i\mathrm{\ne }j$ and $r\in R$, let $E_{ij}$ be the matrix with entry $1$ in position $(i,j)$ and $0$ elsewhere, and let $e_{ij}(r):=I_n+r{E}_{ij}$ be the corresponding elementary matrix. It is invertible with two-sided inverse $e_{ij}(-r)$, because $E_{ij}^2=0$ and hence $e_{ij}(r)e_{ij}(-r)=I_n$. Let ${\mathrm{E}}_n(R)$ be the subgroup of ${\mathrm{G}\mathrm{L}}_n(R)$ generated by all elementary matrices in ${\mathrm{G}\mathrm{L}}_n(R)$ for $n\ge 1$, and set ${\mathrm{E}}_0(R)=\{I_0\}$. Let $\mathrm{E}(R):={\bigcup }_{n\ge 0}{\mathrm{E}}_n(R)$ be the stable elementary subgroup, the subgroup of $\mathrm{G}\mathrm{L}(R)$ generated by the images of all elementary matrices. Elementary matrices are stable in the same way: $e_{ij}(r)$ in ${\mathrm{G}\mathrm{L}}_n(R)$ is the image of the elementary matrix with the same indices in ${\mathrm{G}\mathrm{L}}_{n+1}(R)$.

For every associative unital ring $R$, the stable elementary subgroup $\mathrm{E}(R)$ is a normal subgroup of $\mathrm{G}\mathrm{L}(R)$ and $\mathrm{E}(R)=[\mathrm{G}\mathrm{L}(R),\mathrm{G}\mathrm{L}(R)]\mathrm{.}$ Consequently $\mathrm{G}\mathrm{L}(R)\mathrm{/}\mathrm{E}(R)$ is abelian. Moreover, for every $n$, an upper unitriangular matrix $I_n+N$ in specified ordered coordinates, with $N_{ij}=0$ unless $i<j$, lies in ${\mathrm{E}}_n(R)$ in those coordinates. The matrix of the same automorphism in any other ordered basis lies in the stable subgroup $\mathrm{E}(R)$ by normality, and hence lies in ${\mathrm{E}}_N(R)$ after some finite stabilization; this need not hold at the original matrix size.

Let $R$ be an associative unital ring and let $\mathrm{G}\mathrm{L}(R)\supseteq \mathrm{E}(R)$ be its stable groups (Stable general linear and elementary groups for right modules). By Stable elementary matrices equal the commutator subgroup, $\mathrm{E}(R)$ is normal in $\mathrm{G}\mathrm{L}(R)$ and equals the commutator subgroup, so the quotient $K_1(R):=\mathrm{G}\mathrm{L}(R)\mathrm{/}\mathrm{E}(R)$ is an abelian group. This group is written additively: for $A\in {\mathrm{G}\mathrm{L}}_n(R)$ the symbol $[A]$ denotes the class of the stabilized matrix $A$ in $K_1(R)$, and the group law is the one induced by matrix multiplication, so $[AB]=[A]+[B],[I_n]=0,[A^{-1}]=-[A]$ for compatible $A,B$; the class $[A]$ is unchanged by stabilization and equals the class of $\mathrm{diag}(A,I_m)$ for every $m$. The reduced group ${\tilde{K}}_1(R):=K_1(R)\mathrm{/}⟨[-1]⟩$ is the quotient of $K_1(R)$ by the subgroup generated by the class of the $1\times 1$ matrix $-1$ (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups); in ${\tilde{K}}_1(R)$ one has $[-A]=[A]$ for every invertible $A$.

Now let $\pi$ be a discrete group, with integral group ring $\mathbb{Z}[\pi ]$, a unital ring with basis the classes $[g]$ of the group elements (The group ring $R[G]$ of finitely supported formal $R$-linear combinations of group elements, The group ring $R[G]$ is a unital $R$-algebra with basis $G$, and each $g\in G$ is a unit of $R[G]$). The Whitehead group of $\pi$ is $\mathrm{W}\mathrm{h}(\pi ):=K_1(\mathbb{Z}[\pi ])/⟨[\pm g]:g\in \pi ⟩,$ the quotient by the subgroup generated by the classes of the $1\times 1$ unit matrices $(\pm g)$. Reading the generators in the order $[-1]$, then $[g]=[-1]+[-g]$, exhibits the same subgroup as $⟨[-1]⟩+⟨[g]:g\in \pi ⟩$, so there is a natural identification $\mathrm{W}\mathrm{h}(\pi )={\tilde{K}}_1(\mathbb{Z}[\pi ])/⟨[g]:g\in \pi ⟩\mathrm{.}$

Functoriality and well-definedness. A unital ring homomorphism $R\to R^{\mathrm{\prime }}$ carries invertible matrices to invertible matrices and elementary matrices to elementary matrices, hence induces $K_1(R)\to K_1(R^{\mathrm{\prime }})$ and ${\tilde{K}}_1(R)\to {\tilde{K}}_1(R^{\mathrm{\prime }})$; a group homomorphism $\phi :\pi \to {\pi }^{\mathrm{\prime }}$ induces the unital ring homomorphism $\mathbb{Z}[\pi ]\to \mathbb{Z}[{\pi }^{\mathrm{\prime }}]$, $[g]\mapsto [\phi (g)]$, which sends $[\pm g]$ to $[\pm \phi (g)]$ and therefore descends to a homomorphism $\mathrm{W}\mathrm{h}(\pi )\to \mathrm{W}\mathrm{h}({\pi }^{\mathrm{\prime }})$. These assignments are compatible with composition and preserve identities. If $\phi$ is an inner automorphism of $\pi$, given by $c_g(h)=gh{g}^{-1}$, then the induced ring automorphism of $\mathbb{Z}[\pi ]$ is conjugation by the unit $[g]$, so on ${\mathrm{G}\mathrm{L}}_n(\mathbb{Z}[\pi ])$ it acts as $A\mapsto D_{g}A{D}_g^{-1}$ with $D_g$ the scalar matrix $g{I}_n$; conjugation by a fixed invertible matrix is the identity on the quotient $K_1$, because $[D_{g}A{D}_g^{-1}]=[D_g]+[A]-[D_g]=[A]$. Hence inner automorphisms of $\pi$ induce the identity on $\mathrm{W}\mathrm{h}(\pi )$.

For every discrete group $\pi$, the integral group ring $\mathbb{Z}[\pi ]$ has invariant basis number: an isomorphism of finite free right $\mathbb{Z}[\pi ]$-modules $\mathbb{Z}[\pi {]}^m\cong \mathbb{Z}[\pi {]}^n$ forces $m=n$. More generally, if $R$ is an associative unital ring admitting a unital ring homomorphism $R\to S$ into a nonzero commutative unital ring $S$, then an isomorphism of finite free right $R$-modules $R^m\cong R^n$ forces $m=n$.

Let $R$ be an associative unital ring and let $C$ be a bounded free right $R$-chain complex, so that $C_n=0$ for all but finitely many $n$ and each $C_n$ is a free right $R$-module. A chain contraction of $C$ is a family of right-linear maps $s_n:C_n\to C_{n+1}$ with $d_{n+1}{s}_n+s_{n-1}{d}_n={\mathrm{i}\mathrm{d}}_{C_n}$ for every $n$, that is, the identity of $C$ is null-homotopic and $C$ is contractible; equivalently $ds+sd=\mathrm{i}\mathrm{d}$ as maps of graded modules. Write $C_{\mathrm{o}\mathrm{d}\mathrm{d}}={⨁}_n{C}_{2n+1}$ and $C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}={⨁}_n{C}_{2n}$, and let $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}:C_{\mathrm{o}\mathrm{d}\mathrm{d}}\to C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ and $(d+s{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}:C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}\to C_{\mathrm{o}\mathrm{d}\mathrm{d}}$ be the odd-to-even and even-to-odd components of the degree-one perturbation $d+s$ of the differential.

Then:

For the matrix assertions in clause 2, choose a finite ordered basis $B_n$ of each $C_n$ and suppose that the concatenated bases $B_{\mathrm{o}\mathrm{d}\mathrm{d}}$ and $B_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ have equal size. Use these same bases for every parity map and every contraction below. A bracket on a parity map means the $K_1(R)$ class of its square matrix in these source and target bases. The triangular assertions use decreasing degree order; the equality of classes holds in any fixed ordering of these bases.

1. $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}$ and $(d+s{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ are isomorphisms of right $R$-modules, mutually inverse up to the unipotent correction $\mathrm{i}\mathrm{d}+s^2$: one has $(d+s{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=\mathrm{i}\mathrm{d}+s^2$ on $C_{\mathrm{o}\mathrm{d}\mathrm{d}}$ and $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}(d+s{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}=\mathrm{i}\mathrm{d}+s^2$ on $C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$, where $s^2$ raises degrees by two and is nilpotent. This holds over an arbitrary unital $R$ and uses no rank, freeness, commutativity or invariant-basis-number hypothesis.
2. If $t$ is a second chain contraction, $u=s-t$, ${\mu }_n=(s_{n+1}-t_{n+1})t_n$ and ${\nu }_n=(t_{n+1}-s_{n+1})s_n$, then $(\mathrm{i}\mathrm{d}+\mu {)}_{\mathrm{o}\mathrm{d}\mathrm{d}}$, $(\mathrm{i}\mathrm{d}+\nu {)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ and the composites $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}(\mathrm{i}\mathrm{d}+\mu {)}_{\mathrm{o}\mathrm{d}\mathrm{d}}(d+t{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}},(d+t{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}(\mathrm{i}\mathrm{d}+\nu {)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}$ are the identity plus maps that strictly raise degrees by even positive amounts. In particular, when the two displayed bases are finite and of the same size and ordered by decreasing degree, all four matrices are unipotent upper triangular and hence have class $0$ in $K_1(R)$, and $[(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}]=-[(d+t{)}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}]\in K_1(R)\mathrm{.}$

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.

Let $C$ be a bounded finite based free right $R$-chain complex over an associative unital ring $R$, with displayed bases $B_{\mathrm{o}\mathrm{d}\mathrm{d}},B_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ of equal finite size, and let $s$ and $t$ be two chain contractions of $C$ with $ds+sd=\mathrm{i}\mathrm{d}=dt+td$. Then ${\tau }_s(C)={\tau }_t(C)\in {\tilde{K}}_1(R),$ where ${\tau }_s,{\tau }_t$ are the contraction torsions of Finite based free complexes and contraction torsion. Their common value is written $\tau (C)$ and called the torsion of the based complex $C$. The equality uses no choice and no further hypothesis on the contraction.