Determinant of Alexander Matrix for Torus Links

Question

The core of the problem
Let $q,r\in\mathbb N$ be natural numbers with greatest common divisor $d$. Consider the $(q-1)\times(q-1)$-matrix $$ B:=\begin{pmatrix} -1\\ 1&-1\\ &1&-1\\ &&\ddots&\ddots\\ &&&1&-1 \end{pmatrix} $$ and form an $(r-1)\times(r-1)$-blockmatrix $$ M:=\begin{pmatrix} B\\ -B&B\\ &-B&B\\ &&\ddots&\ddots\\ &&&-B&B \end{pmatrix} $$ In Theorem 7.3.2 of Murasugi's "Knot Theory and its Applications" is stated that $$ \det(M-t\cdot M^T)=\frac{(1-t)(1-t^\frac{qr}d)}{(1-t^q)(1-t^r)} $$ It is indicated that the proof is by calculating this determinant, but unfortunately no details are given. As bockmatrices are somewhat hazardous when calculating determinants, I am unable to come up with a solution.

Further context
The determinant in question determines the Alexander polynomial of a (q, r)-torus link. Many of the other methods for determining Alexander polynomials are unfortunately not applicable here, as I really would like to prove the formula also for torus links. If someone can give an explanation (or literature reference) on how to calculate the Alexander polynomial in question in a different way this would also answer my question.

Update We can additionally consider the $(r-1)\times(r-1)$-matrix $$ A:=\begin{pmatrix} 1\\ -1&1\\ &-1&1\\ &&\ddots&\ddots\\ &&&-1&1 \end{pmatrix} $$ Then $M=A\otimes B$ is the Kronecker product of $A$ and $B$. By the rules outlined in Wikipedia we may hence calculate $$ \det(M-t\cdot M^T)=\det((A\otimes B)-t\cdot(A\otimes B)^T)\\ =\det((A\otimes B))\cdot\det(1-t\cdot(A^{-1}A^T\otimes B^{-1}B^T)) $$ in particular up to a sign, the determinant I'm looking for equals the characteristic polynomial of $A^{-1}A^T\otimes B^{-1}B^T$. Unfortunatly there does not seem to be a general formula for the characteristic polynomial of a Kronecker-product, but herr one can likely circumvent that by calculating eigenvalues instead.

I will turn this update into an answer once I have written up a proper proof.

Answer 1

We can additionally consider for all $n\in\mathbb N$ the $(n-1)\times(n-1)$ matrix $$ A_n:=\begin{pmatrix} 1\\ -1&1\\ &-1&1\\ &&\ddots&\ddots\\ &&&-1&1 \end{pmatrix} $$ By calculating the characteristic polynomial on can see that there exists $S\in\operatorname{GL}_{n-1}(\mathbb C)$ such that $$ S\cdot A_n^{-1}A_n^T\cdot S^{-1}=D_n:=\begin{pmatrix} -\zeta_n\\ &-\zeta_n^2\\ &&-\zeta_n^3\\ &&&\ddots\\ &&&&-\zeta_n^{n-1} \end{pmatrix} $$ where $\zeta_n:=\exp(2\pi i\cdot\frac1n)$ is a primitiv $n$-th root of unity. As indicated in the upadte we can now calculate using the Kronecker product (and its rules described here) that $$ \det(M-t\cdot M^T)=\det\left(1_{(q-1)(r-1)}-t\cdot\left(A_q^{-1}A_q^T\otimes A_r^{-1}A_r^T\right)\right)=\det\left(1_{(q-1)(r-1)}-t\cdot\left(D_q\otimes D_r\right)\right)\\ =\prod_{\substack{i=1,\dots,q-1\\j=1,\dots,r-1}}(1-t\cdot\zeta_p^i\zeta_q^j) $$ Furthermore is $$ \prod_{\substack{i=1,\dots,q-1\\j=1,\dots,r-1}}(1-t\cdot\zeta_p^i\zeta_q^j)\cdot(1-t^q)(1-t^r)=(1-t)\cdot\prod_{\substack{i=1,\dots,q\\j=1,\dots,r}}(1-t\cdot\zeta_p^i\zeta_q^j)\\=(1-t)\cdot\left(\prod_{i=1,\dots,\frac{qr}d}(1-t\cdot\zeta_{\frac{qr}d}^i)\right)=(1-t)(1-t^\frac{qr}d)^d $$ The second to last equality follows by noticing that the group homomorphism $$ \{q\text{-th roots of unity}\}\times\{r\text{-th roots of unity}\}\to\{\frac{qr}d\text{-th roots of unity}\},~(x,y)\mapsto xy $$ is surjective such that every element of $\{\frac{qr}d\text{-th roots of unity}\}$ has precisely $d$ preimages.

Answer 2

For every positive integer $n$, denote by $J_n$ be the $n\times n$ lower triangular Jordan block for the eigenvalue $-1$. Then $B=J_{q-1}$ and $M=-J_{r-1}\otimes B=-J_{r-1}\otimes J_{q-1}$. Thus $$ \begin{align} \det(M-tM^T) &=\det\big(tJ_{r-1}^T\otimes J_{q-1}^T-J_{r-1}\otimes J_{q-1}\big)\\ &=\det\big(J_{r-1}^T\otimes J_{q-1}^T)\det\big(tI-(J_{r-1}^{-T}J_{r-1})\otimes (J_{q-1}^{-T}J_{q-1})\big)\\ &=\det\big(tI-(J_{r-1}^{-T}J_{r-1})\otimes (J_{q-1}^{-T}J_{q-1})\big).\tag{1}\\ \end{align} $$ Note that $$ -J_n^{-T}J_n =(-J_n)^{-T}J_n =\pmatrix{1&1&\cdots&1\\ &\ddots&\ddots&\vdots\\ &&\ddots&1\\ &&&1} \pmatrix{-1\\ 1&\ddots\\ &\ddots&\ddots\\ &&1&-1} =\pmatrix{0&&&-1\\ 1&\ddots&&\vdots\\ &\ddots&0&\vdots\\ &&1&-1} $$ is the companion matrix of $x^n+x^{n-1}+\cdots+x+1$. The eigenvalues of $J_n^{-T}J_n$ are therefore $-\exp\left(\frac{2k\pi i}{n+1}\right)$ for $k=1,2,\ldots,n$. It follows from $(1)$ that $$ \det(M-tM^T) =\prod_{k=1}^{r-1}\prod_{l=1}^{q-1}\left(t-\omega^k\gamma^l\right)\tag{2} $$ where $\omega=\exp\left(\frac{2\pi i}{r}\right)$ and $\gamma=\exp\left(\frac{2\pi i}{q}\right)$. The equality $(2)$ can perhaps be converted into the one in the book by some clever number-theoretic manipulations.