# Infinite family of different prime knots with trivial Alexander polynomial

I am looking for infinite families of prime knots that have all Alexander polynomial equals to 1. I wrote "families" (and not "family") since perhaps there are different constructions out there.

Moreover, is there such a family for which all knots are of genus one (just as the Whitehead doubles)?

## 3 Answers

The P(p,q,r) pretzel knots with p, q, r odd integers have a genus 1 Seifert surface that just consists of two disks connected with three twisted ribbons, with p, q, r twists, respectively. So they have a Seifert matrix $$A = \left(\begin{matrix}p+q & q+1 \\ q-1 & q+r\end{matrix}\right)/2$$.

From this one computes that the determinant of P(p,q,r) is $$|\det(A + A^{\top})| =pq+qr+rp$$. For genus 1 knots, the Alexander polynomial is 1 if and only if the determinant is 1. This gives you plenty of examples of genus 1 knots with Alexander polynomial 1, e.g. P(-3,5,7).

More generally, given any knot $$K$$ with Alexander polynomial 1 and a Seifert surface $$\Sigma$$, you can construct further knots of the same genus as $$K$$ simply by tying a knot into a band of $$\Sigma$$ without changing the framing of the core curve of the band (this is often called "infection").

• Does this "infection" preserve Alexander as well? Dec 17, 2021 at 10:01
• It preserves Seifert matrices, which determine the Alexander polynomial - so yes. Dec 17, 2021 at 18:37

Whitehead doubles are one such family of genus one knots, see this MO answer

• You are right. Unluckily I am really looking for families different than the Whitehead doubles. The properties mentioned in the question are indeed those that the Whitehead doubles satisfy. Dec 8, 2021 at 11:00

Another family of examples is given by the "generalised Kinoshita-Terasaka" knots, here is a picture from Lickorish' "An introduction to Knot Theory".

Here $$d$$ is assumed to be even. Of course, this is not a family of genus 1 knots.