Infinitely many two-variable generalisations of the Alexander–Conway polynomial

Abstract

We show that the Alexander-Conway polynomial ∆ is obtainable via a particular one-variable reduction of each two-variable Links-Gould invariant LG m,1 , where m is a positive integer. Thus there exist infinitely many two-variable generalisations of ∆. This result is not obvious since in the reduction, the representation of the braid group generator used to define LG m,1 does not satisfy a second-order characteristic identity unless m = 1. To demonstrate that the one-variable reduction of LG m,1 satisfies the defining skein relation of ∆, we evaluate the kernel of a quantum trace.

Multivariable link invariants arising from sl(2|1) and the Alexander polynomial

“…This generalizes the results of [6] which state that the two-variable Links-Gould invariant dominates the one-variable Alexander polynomial.…”

“…The proof of uniqueness given by Turaev has two steps. The first step uses axioms (1)- (6) to show that the two one-variable specializations are the same: ∇ 1 = ∇ 2 . This part applies in our context without any change.…”

Rozansky-Witten geometry of Coulomb branches and logarithmic knot invariants

“…[14]). The structure on the right-hand side of (3.5) also appears in the literature on generalized Alexander invariants and can be viewed as a TQFT reason why such generalizations often end up related to the Alexander polynomial [62][63][64][65]. (See, however, the discussion in Section 4).…”

On the Links–Gould invariant and the square of the Alexander polynomial

“…For any oriented link L, LG(L, t 0 , t 1 ) is a two-variable Laurent polynomial in variables t 0 and t 1 . It is known to be a generalization of the Alexander-Conway polynomial ∆ L (t 0 ) in at least two different ways (see [2,10,11]):…”

“…LG(L, t 0 , −t −1 0 ) = ∆ L (t 2 0 ) ; LG(L, t 0 , t −1 0 ) = ∆ L (t 0 ) 2 . Therefore, using the Seifert inequality for the Alexander polynomial ∆ K of a knot K, the minimal genus g(K) among all possible Seifert surfaces for K satisfies the following inequality:…”