Cyclic branched covers and knot determinants — Examples and Counterexamples

For the trefoil, $V_T+V_T^T=\left(\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right)$. Swap its columns, add twice the first row to the second, then add twice the first column to the second; the result is $\mathrm{diag}(1,-3)$. Each operation is unimodular, so [F1]–[F2] give $H_1({\mathrm{\Sigma }}_2(T))\cong \mathbb{Z}\mathrm{/}3\mathbb{Z}$.

For the figure-eight knot, $V_8+V_8^T=\left(\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right)$. Swap its columns, subtract twice the first row from the second, then add twice the first column to the second; the result is $\mathrm{diag}(1,5)$. Thus its cokernel is $\mathbb{Z}\mathrm{/}5\mathbb{Z}$.

For the unknot, the standard spanning disk has genus zero, so its Seifert matrix is empty and [F1] gives trivial first homology. To identify the whole cover, write $S^3=\{(z,w)\in {\mathbb{C}}^2:\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2=1\}$ and define $f(z,w)=(z\sqrt{1+\mathrm{\mid }w{\mathrm{\mid }}^2},w^2)$. The image lies in $S^3$ because $\mathrm{\mid }z{\mathrm{\mid }}^2(1+\mathrm{\mid }w{\mathrm{\mid }}^2)+\mathrm{\mid }w{\mathrm{\mid }}^4=1$. Off $w=0$, each target point has exactly two preimages, corresponding to the two square roots of its second coordinate; on $w=0$ it has one. The involution $(z,w)\mapsto (z,-w)$ is the deck transformation, and in tubular coordinates about $w=0$ the transverse map is $w\mapsto w^2$. Hence $f$ is the connected double branched cover of $S^3$ over the unknot, so the cover is $S^3$.

Direct calculation gives $\det (V_T-t{V}_T^T)=\det \left(\begin{array}{cc}t-1& -t\\ 1& t-1\end{array}\right)=t^2-t+1$. By [F1], this represents the trefoil Alexander polynomial up to $\pm t^k$. Since ${\zeta }_6^2-{\zeta }_6+1=0$, the Alexander polynomial vanishes at this nontrivial sixth root.

Applying [F2] with $q=6$ shows that $H_1({\mathrm{\Sigma }}_6(T);\mathbb{Z})$ is infinite, contradicting the refuted universal claim. For the boundary comparison, [F3] gives ${\mathrm{\Delta }}_T(-1)$ odd and nonzero, so $H_1({\mathrm{\Sigma }}_2(T);\mathbb{Z})$ is finite. Thus the failure occurs at a higher cyclic cover even though the double-cover case is always finite. The two supplier theorems use the Axiom of Choice in [A1].