View PDF HTML (experimental)

Abstract:Let $Th(4,2n+1)=\widehat{\beta_{2n+1}}$ be the four-strand Turk's head knot, where $\beta_{2n+1}=\sigma_1\sigma_2^{-1}\sigma_3)^{2n+1}$. In our earlier work the Alexander polynomial of this family was reduced, after the substitution $t=-z$, to the factorization \[
A_{2n+1}(z)=(1+z+\cdots+z^{2n})D_n(z)^2,
\qquad
D_n(z)=\prod_{r=1}^n
\left(z^2+4\sin^2\frac{\pi r}{2n+1}\,z+1\right). \] The remaining difficulty was to prove log-concavity of the coefficient sequence of $D_n(z)$. In this work we prove that the coefficient sequence of $D_n(z)$ is log-concave. The main new ingredient is a four-block smoothing theorem for products of reciprocal quartics $(1+az+z^2)(1+bz+z^2), \: 0\le a,b\le 4,\: a+b\ge 4.$ The smoothing theorem is proved by a finite exact positivity certificate using only integer arithmetic. Combining this smoothing theorem with the trigonometric pairing $r\leftrightarrow n+1-r$ proves that $D_n(z)$ is log-concave for all $n\ge1.$ It follows that the absolute values of the coefficients of the Alexander polynomial of $Th(4,2n+1)$ form a trapezoidal sequence. Thus Fox's trapezoidal conjecture holds for the entire family of four-strand Turk's head knots.

Submission history

From: Suman Saurabh [view email]
[v1] Sat, 13 Jun 2026 11:22:51 UTC (10 KB)