Cellular basis ambiguities vanish in the Whitehead group

Let $n\ge 1$ and let $P\in G{L}_n(R)$ be a product of elementary matrices $e_{ij}(r)$ and their inverses. By [F1] each factor lies in $E(R)$, so $P\in E(R)$; as $K_1(R)=GL(R)\mathrm{/}E(R)$, its class satisfies $[P]=0$, and by additivity of the class in [F2] every finite product of elementary matrices and inverses has class $0$ as well.

In $M_2(R)$ put $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $U=e_{12}(1)e_{21}(-1)e_{12}(1)$. Multiplying the three matrices gives $U=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, hence $S=\mathrm{diag}(1,-1)U$ with $U\in E(R)$; since $\mathrm{diag}(1,-1)$ is the stabilization of the $1\times 1$ matrix $-1$, [F2] gives $[S]=[\mathrm{diag}(1,-1)]+[U]=[-1]$.

Let $R=\mathbb{Z}[\pi ]$ and let a based cellular complex over the universal cover be given as in [F4]. Replacing the chosen lift $\tilde{e}$ of a cell $e$ by the lift $T_h\tilde{e}$ replaces one basis vector by $T_h\tilde{e}=\tilde{e}\cdot h^{-1}$, and reversing the orientation replaces one basis vector by its negative; the change-of-basis matrix is therefore diagonal with one entry $h^{-1}$ or $-1$ and all other entries $1$. Its class in $K_1(R)$ is $[h^{-1}]$ or $[-1]$, both of which are $0$ in $\mathrm{W}\mathrm{h}(\pi )$ by the definition of $\mathrm{W}\mathrm{h}$ in [F2].

For a path $\gamma :x\to x^{\mathrm{\prime }}$, define ${\phi }_{\gamma }([\alpha ])=[\bar{\gamma }\ast \alpha \ast \gamma ]$ on loops at $x$. Concatenating a homotopy rel endpoints with the fixed outer paths gives a homotopy rel endpoints, so the map is well defined and preserves products after inserting the cancellable middle path $\gamma \ast \bar{\gamma }$. Explicitly, for any path $\lambda$, the two formulas $H(s,t)=\lambda (2s(1-t))$ for $s\le \frac{1}{2}$ and $H(s,t)=\lambda (2(1-s)(1-t))$ for $s\ge \frac{1}{2}$ agree at $s=\frac{1}{2}$; by pasting they contract $\lambda \ast \bar{\lambda }$ rel endpoints. Reparametrising by $\lambda ((1-t)r(s)+ts)$ likewise identifies bracketings and deletes constant paths. Thus ${\phi }_{\bar{\gamma }}$ is inverse to ${\phi }_{\gamma }$. For a second path ${\gamma }^{\mathrm{\prime }}:x\to x^{\mathrm{\prime }}$, the composite ${\phi }_{{\gamma }^{\mathrm{\prime }}}^{-1}\circ {\phi }_{\gamma }$ carries $[\alpha ]$ to $[{\gamma }^{\mathrm{\prime }}\ast \bar{\gamma }\ast \alpha \ast \gamma \ast {\bar{\gamma }}^{\mathrm{\prime }}]$, conjugation by the loop ${\gamma }^{\mathrm{\prime }}\ast \bar{\gamma }$ at $x$. By [F2] inner automorphisms act trivially on $\mathrm{W}\mathrm{h}$, so the two paths induce the same Whitehead-group map.

Let $\pi =C_5=\{1,t,t^2,t^3,t^4\}$ and $R=\mathbb{Z}[\pi ]$. In $R$ distribute and use $[t^a][t^b]=[t^{a+b}]$ with $t^5=1$: $(1-t^2-t^3)(1-t-t^4)=1-t-t^4-t^2+t^3+t^6-t^3+t^4+t^7=1-t-t^2+t^6+t^7=1,$ the last step because $t^6=t$ and $t^7=t^2$. Hence $u=1-t^2-t^3$ is a unit of $R$ with inverse $1-t-t^4$.

By [F7] the elements $1,t,t^2,t^3,t^4$ form a $\mathbb{Z}$-basis of $R$. Every element of the subgroup $⟨\pm t^k⟩$ is a finite product of monomials $\pm t^k$, hence is itself $\pm t^m$ for some $m$, an element whose coefficient vector in that basis has exactly one nonzero entry; the coefficient vector of $u=1+(-1)t^2+(-1)t^3$ has the three nonzero entries $1,-1,-1$. Therefore $u\mathrm{\notin }⟨\pm t^k⟩$ and $u\mathrm{\ne }0$.

Let $S$ be a commutative unital ring. Since $\det$ is multiplicative with $\det (I_n)=1$ by [F5] and $\det (e_{ij}(r))=1$ because $e_{ij}(r)$ is obtained from $I_n$ by adding $r$ times the $j$-th row to the $i$-th row, the determinant is unchanged under right multiplication by any product of elementary matrices; moreover $\det (\mathrm{diag}(B,1))=\det (B)$ for $B\in M_n(S)$: the function $B\mapsto \det (\mathrm{diag}(B,1))$ is normalized, column-multilinear and alternating in the columns of $B$, so it equals $\det$ by the uniqueness in [F6]. Hence $\det$ is compatible with stabilization and descends to a well-defined homomorphism $\det :K_1(S)\to S^{\times }$ with $\det [u]=u$ for every unit $u\in S^{\times }$.

The matrix of a permutation of a finite basis that is a product of $m$ transpositions is a product of $m$ copies of $S$ padded by identity blocks, because permutation matrices for $0\mathrm{/}1$ entries multiply by the composition rule of [F1] and the padded matrices are the stabilized transpositions. Hence by [F2] its class is $m[-1]$, which is $[I]=0$ for $m$ even and $[-1]$ for $m$ odd, so every reordering has class $0$ or $[-1]$ in $K_1(R)$, and class $0$ in ${\tilde{K}}_1(R)$ and in $\mathrm{W}\mathrm{h}(\pi )$ where $[-1]=0$.

For $S=R=\mathbb{Z}[C_5]$ the homomorphism of step 1.7 sends the class $[\pm t^k]$ of the $1\times 1$ matrix $\pm t^k$ to $\pm t^k$, so it induces a homomorphism $\mathrm{W}\mathrm{h}(C_5)=K_1(R)\mathrm{/}⟨[\pm t^k]⟩\to R^{\times }\mathrm{/}⟨\pm t^k⟩$ carrying the class of the unit $u$ of step 1.5 to the coset $u⟨\pm t^k⟩$.

By the basis-change formula of [F3], a change of the displayed bases whose change-of-basis matrices all have class $0$ in $\mathrm{W}\mathrm{h}(\pi )$ leaves the torsion class in $\mathrm{W}\mathrm{h}(\pi )$ unchanged; combining with steps 2.1 and 1.3, neither a reordering of the cells, nor a reversal of an orientation, nor a change of the chosen lifts alters the torsion class in $\mathrm{W}\mathrm{h}(\pi )$, and an elementary basis change does not alter the class in $K_1(R)$.

Since $u\mathrm{\notin }⟨\pm t^k⟩$ by step 1.6, that coset is not the identity coset, so the image of $[u]$ in $\mathrm{W}\mathrm{h}(C_5)$ is nonzero: the quotient map $K_1(R)\to \mathrm{W}\mathrm{h}(C_5)$ does not kill every unit.

Clauses 1 and 2 are steps 1.1 and 2.1, clause 3 is steps 1.3 and 1.4, and clause 4 is steps 1.5, 1.6, 1.7, 2.2 and 3.2; no step assumed commutativity of $R$ except in clauses about the commutative group ring $\mathbb{Z}[C_5]$, and no step used any choice principle.