Simple Homotopy, Whitehead Groups, and Torsion: Examples

Realize $Y$ as the CW complex with vertex set $\{v,w\}$ and the single $1$-cell $e$ attached by the map $q:S^0=\{-1,+1\}\to Y^0$ with $q(-1)=v$ and $q(+1)=w$. With $Q^1=D^1=[-1,1]$, $Q^0=D_{+}^0=\{+1\}$ and $\phi$ the characteristic map of $e$, the restriction $\phi {\mathrm{\mid }}_{Q^0}$ is a characteristic map for the new $0$-cell $w$ (mapping $+1$ homeomorphically onto $w$), the remaining boundary value $\phi (-1)=v$ lies in $X$, and $Y=X\cup \{w\}\cup e$. Hence $X\hookrightarrow Y$ is an elementary expansion of dimension $1$ by [F1], and the new vertex $w$ is a free face of $e$.

The pair $(Y,X)$ deformation retracts onto $X$ by [F1]; the retraction $r:Y\to X$ therefore satisfies $r\circ i={\mathrm{i}\mathrm{d}}_X$ for the inclusion $i:X=\{v\}\hookrightarrow Y$ and $i\circ r\simeq {\mathrm{i}\mathrm{d}}_Y$. The one-point space $X$ is a nonempty convex subset of ${\mathbb{R}}^1$, hence contractible with ${\pi }_1(X,v)=1$ by [F5], and functoriality together with based-homotopy invariance of the induced maps gives $i_{\ast }\circ r_{\ast }=(i\circ r{)}_{\ast }=\mathrm{i}\mathrm{d}$ and $r_{\ast }\circ i_{\ast }=(r\circ i{)}_{\ast }=\mathrm{i}\mathrm{d}$ on ${\pi }_1$, so $i_{\ast }:{\pi }_1(X,v)\to {\pi }_1(Y,v)$ is an isomorphism and ${\pi }_1(Y,v)=1$; the edge $e$ supplies a path from $v$ to $w$, so basepoint change also gives ${\pi }_1(Y,w)=1$.

Over the universal cover the pair $(Y,X)$ has exactly two relative cells, the lift of $w$ in degree $0$ and the lift of $e$ in degree $1$, so by [F2] the relative based chain complex $T_{\bullet }$ is concentrated in degrees $0$ and $1$ with a single displayed basis vector in each degree and its differential has the $1\times 1$ right-module matrix $\pm 1$: the characteristic map restricts to a homeomorphism on the free face, so the incidence of $w$ in the boundary of the lift of $e$ is $\pm 1$. This is the complex displayed in the statement, where the orientation is chosen so that the entry is $1$.

With the contraction $s(e_0)=e_1$ for the displayed differential $d(e_1)=e_0$, the odd-to-even map $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}:T_1\to T_0$ is $d$ and has the $1\times 1$ matrix $(1)$. Thus $\tau (T_{\bullet })=[1]=0$ in ${\tilde{K}}_1(\mathbb{Z})$. Reversing either cell orientation changes the matrix to $(-1)$, whose class is also $0$ in the reduced group. The resulting image in $\mathrm{W}\mathrm{h}(1)$ vanishes.

By [F3] applied to the elementary expansion $j:X\hookrightarrow Y$ of step 1.1, $\tau (X\hookrightarrow Y)=0$ in $\mathrm{W}\mathrm{h}({\pi }_{1}Y)=\mathrm{W}\mathrm{h}(1)$ by step 1.2; the relative boundary is the unit $\pm g$ of [F3] with $g$ the group element determined by the chosen lifts, and $[\pm g]=0$ in $\mathrm{W}\mathrm{h}({\pi }_{1}Y)$ since here $g=1$.

The identical computation applies to an elementary expansion performed on any finite CW complex: by [F3] the only nonzero relative cellular boundary of the pair is the unit $\pm g$ in two consecutive degrees and the class $[\pm g]$ dies in $\mathrm{W}\mathrm{h}({\pi }_{1}Y)$, so the boundary unit $\pm g$ produces zero Whitehead class over any group $\pi$.

Since $\mathbb{Z}[1]=\mathbb{Z}$, integer determinant sends $G{L}_m(\mathbb{Z})$ to $\{\pm 1\}$ and sends each elementary matrix to $1$. It therefore induces a homomorphism $K_1(\mathbb{Z})\to \{\pm 1\}$, which is onto because the one-by-one matrices $(1)$ and $(-1)$ occur.

Let $A\in G{L}_m(\mathbb{Z})$. Its first column is primitive: if an integer $d>1$ divided every entry, it would divide the determinant $\pm 1$, a contradiction. By [F2], elementary integer row additions reduce this column to $(\pm 1,0,\dots ,0{)}^{\mathsf{T}}$; a row swap is a product of elementary matrices and a diagonal $-1$, so its Whitehead class is accounted for by $[-1]$. Clear the remainder of the first row by elementary column additions and repeat on the invertible $(m-1)\times (m-1)$ minor. Induction gives an elementary-equivalent diagonal matrix with entries $\pm 1$. Stabilized diagonal $-1$ entries add to a single $[-1]$ class, since $[-1]+[-1]=[1]=0$. Hence the determinant homomorphism is injective, and $K_1(\mathbb{Z})\cong \{\pm 1\}$.

The quotient defining $\mathrm{Wh}(1)$ kills this entire two-element group, so $\mathrm{Wh}(1)=0$. A finite simply connected homotopy equivalence has torsion in $\mathrm{Wh}(1)$ on each connected component and thus has zero torsion; [F3] makes it simple. ∎

Write $e_q,e_{q-1}$ for the displayed right-module basis vectors, so the matrix convention means $d(e_q)=e_{q-1}\cdot u$. Define the right-linear map $s$ by $s(e_{q-1})=e_q\cdot u^{-1}$ and set its other components to zero. Then $ds(e_{q-1})=d(e_q)\cdot u^{-1}=e_{q-1}\cdot u{u}^{-1}=e_{q-1}$, while $sd(e_q)=s(e_{q-1}\cdot u)=e_q\cdot u^{-1}u=e_q$. Thus $ds+sd=1$ in both nonzero degrees, so $C_{\bullet }$ is contractible with one odd and one even displayed basis vector.

If $q$ is odd then $C_{\mathrm{o}\mathrm{d}\mathrm{d}}=C_q$, $C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}=C_{q-1}$ and $s$ vanishes on $C_q$, so $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=d$ has the $1\times 1$ matrix $u$ in the displayed bases and $\tau (C_{\bullet })=[u]$ by [F1] and [F2]. If $q$ is even then $C_{\mathrm{o}\mathrm{d}\mathrm{d}}=C_{q-1}$, $C_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}=C_q$ and $d$ vanishes on $C_{q-1}$, so $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=s$ has the $1\times 1$ matrix $u^{-1}$ and $\tau (C_{\bullet })=[u^{-1}]=-[u]$ by [F3]. This proves assertions 1 and 2, with the single formula $\tau (C_{\bullet })=(-1{)}^{q+1}[u]$.

For $R=\mathbb{Z}[\pi ]$, apply the quotient homomorphism $K_1(R)\to \mathrm{W}\mathrm{h}(\pi )$ to the torsion class $(-1{)}^{q+1}[u]$ computed in step 1.2. Its image is $(-1{)}^{q+1}$ times the image of $[u]$, which proves the same formula in the Whitehead group.

In $R=\mathbb{Z}[t]\mathrm{/}(t^5-1)$, direct multiplication gives $(1-t^2-t^3)(1-t-t^4)=1$: the coefficient vector in the basis $(1,t,t^2,t^3,t^4)$ is $(1,0,0,0,0)$. Thus $u$ is a unit with the stated inverse. Its coefficient vector is $(1,0,-1,-1,0)$, unlike every $\pm t^k$, so it is not a trivial unit.

Since $C_5$ is abelian, $R$ is commutative. Determinant sends $G{L}_m(R)$ to $R^{\times }$, is multiplicative, and sends elementary matrices to $1$; hence it descends to $K_1(R)$ and then gives a homomorphism $\mathrm{Wh}(C_5)\to R^{\times }\mathrm{/}⟨\pm t^k⟩$. The determinant of the one-by-one matrix $(u)$ is $u$, whose coset is nontrivial by step 1.1. Therefore $[u]\mathrm{\ne }0$ in $\mathrm{Wh}(C_5)$.

The differential $u:R\to R$ is invertible, so the displayed two-term complex has contraction $u^{-1}$ and is acyclic as an $R$-complex and after forgetting to abelian groups. Its degree-one based torsion is $[u]\mathrm{\ne }0$ by [F1] and step 2.1. This exhibits why ordinary homology alone does not retain the chosen group-ring bases and their torsion.

Take the finite presentation complex $X$ of $C_5=⟨t\mid t^5⟩$: one vertex, one loop and one two-cell. Apply [F3] to $X$ and the nonzero class $[u]$ to obtain a finite CW homotopy equivalence $i:X\hookrightarrow Y$ with $\tau (i)=[u]$. By [F4] and step 2.1 it is not simple. Its relative group-ring complex may be chosen as the two-term invertible-matrix complex constructed in [F3], so its ordinary relative homology also vanishes. ∎