Root-product test for cyclic branched-cover homology

Quotienting the module presentation in [F2] by $(t^q-1)A_K$ gives $A_K\mathrm{/}(t^q-1)A_K\cong \mathrm{coker}(V-t{V}^T:R_q^{2g}\to R_q^{2g})$ as abelian groups. The group $R_q^{2g}$ is free abelian of rank $2gq$ on $t^k{a}_i$ for $0\le k<q$ and $1\le i\le 2g$, so this map is represented by a square integer matrix $Q_q$.

After tensoring with $\mathbb{C}$, multiplication by $t$ on $R_q\otimes \mathbb{C}$ is diagonalizable: its eigenvalues are the $q$ distinct roots ${\zeta }_q^j$ for $0\le j<q$, each with one-dimensional eigenspace. On the corresponding $2g$-dimensional summand, $Q_q$ acts by $V-{\zeta }_q^j{V}^T$. Therefore $\det Q_q={\prod }_{j=0}^{q-1}\det (V-{\zeta }_q^j{V}^T)$. The $j=0$ factor is $\det (V-V^T)=\pm 1$ by [F3]. Thus $\det Q_q\mathrm{\ne }0$ exactly when every factor with $1\le j<q$ is nonzero, and its absolute value is the product of the absolute values of those factors.

By [F1], the group in question is $\mathrm{coker}(Q_q)$. Applying [F5] shows it is finite exactly when $\det Q_q\mathrm{\ne }0$, with order $\mathrm{\mid }\det Q_q\mathrm{\mid }$; if any nontrivial-root factor vanishes, the square integer cokernel has positive free rank and is infinite.

By [F4], $D(t)=\pm t^k{\mathrm{\Delta }}_K(t)$ for some integer $k$. At every root of unity, $\mathrm{\mid }\pm ({\zeta }_q^j{)}^k\mathrm{\mid }=1$, so $\mathrm{\mid }D({\zeta }_q^j)\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Delta }}_K({\zeta }_q^j)\mathrm{\mid }$. Hence the determinant-root criterion and product from step 3.1 give exactly the Alexander-polynomial criterion and order in the Statement, including the empty-matrix case $g=0$.