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The Alexander polynomial of a knot is of the form $$\Delta(t)=det(V^T-tV),$$ where $V$ is the Seifert matrix, see http://archive.lib.msu.edu/crcmath/math/math/a/a116.htm. What is geometric or some other meaning of the zeros of the polynomial? I've encountered a similar expression in the theory of 2D/planar electrical networks, where the zeros are essentially multipliers of harmonic continuation on the networks.

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The zeros of the Alexander polynomial are the values where the Tristam-Levine signature of the knot can jump. Let $V$ be a Seifert matrix of $K$ and let $\omega\in\mathbb{C}$ be on the unit circle. The Tristam-Levine signature $\sigma_\omega(K)$ is defined as the signature (number of positive eigenvalues minus number of negative eigenvalues) of the matrix $$(1-\omega)V + (1-\overline{\omega})V^T.$$ The Tristam-Levine signature is locally constant away from zeros of the Alexander polynomial, but can jump from one value to another on either side of a zero.

On another note, the zeros of the Alexander polynomial can say something about the growth rates of the first homology of the cyclic branched covers of the knot. Let $\Sigma_n(K)$ be the $n$-fold cyclic branched cover of $S^3$ branched along $K$. Furthermore, let $b_n=|H_1(\Sigma_n(K))|$ be the size of the first homology of the $n$-fold branched cover. Gordon proved that all of the zeros of the Alexander polynomial of $K$ are roots of unity if and only if the sequence $\{b_n\}$ is periodic. Gonzalez-Acuna and Short proved the if the Alexander polynomial has a zero that is not a root of unity, then the sequence $\{b_n\}$ tends to infinity.

Finally, Hoste conjectured that if $K$ is an alternating knot, then the real part of any zero $\alpha$ of the Alexander polynomial of $K$ satisfies $\operatorname{Re}(\alpha)>-1.$ Lyubich and Murasugi proved Hoste's conjecture in some special cases. The general case is still open. Stoimenow showed that Hoste's conjecture and the conjecture that the coefficients of the Alexander polynomial of an alternating knot form a log-concave sequence are essentially independent.

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  • $\begingroup$ The zeros of $\Delta(t)$ come in reciprocal pairs. Is there a knot symmetry that explains that? $\endgroup$ – DVD Mar 15 '18 at 18:41
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    $\begingroup$ The Alexander polynomial is a symmetric polynomial, and the roots of symmetric polynomials come in reciprocal pairs. The Alexander polynomial can be computed from a presentation of the fundamental group of the knot complement using Fox calculus. There is a way to produce two "dual" presentations of the knot complement that when run through the Fox calculus machinery yields that the Alexander polynomial is symmetric. See this note. $\endgroup$ – Adam Lowrance Mar 15 '18 at 23:39
  • $\begingroup$ Thank you! It also follows from the fact that the matrix $V^TV^{-1}$ is similar to its inverse... $\endgroup$ – DVD Mar 16 '18 at 20:15
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    $\begingroup$ @AdamLowrance Are you sure you mean "symmetric" in the sense of the Wikipedia article? (Plus, the Alexander polynomial is only a symmetric Laurent polynomial when it's been normalized, as it is with the determinant definition.) (Avoiding Fox calculus, the symmetry can also be observed from the interpretation of the Alexander polynomial as being strictly a property of the group, since it is the torsion of the first homology of the commutator subgroup, with coefficients in the group ring of the abelianization. The "dual" presentation gives the same group while "reflecting" the abelianization.) $\endgroup$ – Kyle Miller Mar 18 '18 at 22:40
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    $\begingroup$ @KyleMiller I did use an incorrect term. What I should have said is that the Alexander polynomial is a palindomic polynomial (or a self-reciprocal polynomial). Palindromic polynomials do have roots that come in reciprocal pairs. Thanks for pointing this out. $\endgroup$ – Adam Lowrance Mar 19 '18 at 0:26

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