The dihedral product complement admits no finite CW model

If $K$ is a finite CW complex, let $d$ be its number of one-cells. By [F2], $H_1(K;\mathbb{Z})$ is generated by at most $d$ elements. A homotopy equivalence $K\to M$ would give an isomorphism $H_1(K;\mathbb{Z})\cong H_1(M;\mathbb{Z})$, making the latter generated by at most $d$ elements.

We prove by induction on $d\ge 0$ that any $d+1$ vectors $v_0,\dots ,v_d$ in ${\mathbb{Z}}^d$ have a nonzero integral relation. For $d=0$, $v_0=0$ gives one. For $d>0$, write $a_i$ for the first coordinate of $v_i$. If all $a_i$ vanish, apply the induction hypothesis to the last $d-1$ coordinates of $v_0,\dots ,v_{d-1}$ and give $v_d$ coefficient zero. Otherwise reorder so $a_0\mathrm{\ne }0$. For $1\le i\le d$, the vectors $w_i=a_0{v}_i-a_i{v}_0$ have zero first coordinate. Their last $d-1$ coordinates admit an integral relation ${\sum }_{i=1}^d{b}_i{w}_i=0$ with some $b_i\mathrm{\ne }0$, by induction. Thus $a_0{\sum }_{i=1}^d{b}_i{v}_i-({\sum }_{i=1}^d{b}_i{a}_i)v_0=0$ is a nonzero integral relation, since at least one coefficient $a_0{b}_i$ is nonzero.

Let an abelian group $A$ be generated by $g_1,\dots ,g_d$ and let $h_0,\dots ,h_d\in A$. Write each $h_i$ as ${\sum }_{j=1}^d{c}_{ji}{g}_j$ with integers $c_{ji}$, and let $v_i=(c_{1i},\dots ,c_{di})\in {\mathbb{Z}}^d$. Step 1.2 gives integers $n_i$, not all zero, with ${\sum }_i{n}_i{v}_i=0$. Substituting the expressions for $h_i$ gives ${\sum }_i{n}_i{h}_i=0$. Hence a group generated by $d$ elements contains no independent family of $d+1$ elements; the argument includes $d=0$.

If a finite CW complex $K$ were homotopy equivalent to $M$, step 1.1 would make $H_1(M;\mathbb{Z})$ generated by some finite number $d$ of elements. Choose the $d+1$ meridian classes indexed by $0,\dots ,d$ from [F1]. Their integral independence contradicts step 2.1. Therefore $M$ has no finite CW model.