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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)?

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3 Answers 3

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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").

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  • $\begingroup$ Does this "infection" preserve Alexander as well? $\endgroup$
    – Minkowski
    Dec 17, 2021 at 10:01
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    $\begingroup$ It preserves Seifert matrices, which determine the Alexander polynomial - so yes. $\endgroup$ Dec 17, 2021 at 18:37
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Whitehead doubles are one such family of genus one knots, see this MO answer

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  • $\begingroup$ 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. $\endgroup$
    – Minkowski
    Dec 8, 2021 at 11:00
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Another family of examples is given by the "generalised Kinoshita-Terasaka" knots, here is a picture from Lickorish' "An introduction to Knot Theory".

enter image description here

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

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