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For a 2-bridge torus knot of the form $T(2,r^2)$, where $r^2$ is an odd number, we have that its determinant is $r^2$, that is a condition that we need for a knot to be slice. Also its Alexander polynomial is $ \ \Delta_{T(2,r^2)}(t)\doteq \frac{t^{r^2}+1}{t+1}=t^{r^2-1}-t^{r^2-2} \ \pm \ ...-t+1$, where $\doteq$ means "up to an integer power of $t$". I would like to know if such a polynomial could factorize into $f(t)f(t^{-1})$, where $f$ is a polynomial with integer coefficients, this is also a condition for the knot to be slice. Has anyone a tip? Is it possible to see it?

Thanks!

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Quickly looking at wikipedia's list of slice knots, which I do not trust blindly, and neither should you (so check on this!), the knot $9_1 = T(2,9)$ is not slice. Thus, assuming they are right, your conjecture in general is not true. Now, maybe with some more conditions, but the next knot where $r^2$ is odd would be $T(2,25)$, which I doubt anyone has really spent time thinking about. But I thought this quick check might help.

Factoring polynomials in my opinion is hard, but good luck! It would be cool if you could show that they were not slice, i.e., there is no $f$ such that $f(t)f(t^{-1})$ is the Alexander polynomial of a $T(2,r^2)$. That would be my guess anyway.

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  • $\begingroup$ Thanks a lot, I didn't think to check the wiki page, I was looking for something called "list of slices knots" :)... Here are the first results from wolfram: $\Delta_{T(2,25)}=(t^4 - t^3 + t^2 - t + 1) (t^{20} - t^{15} + t^{10} - t^5 + 1)$, $\Delta_{T(2,49)}=(t^6 - t^5 + t^4 - t^3 + t^2 - t + 1) (t^{42} - t^{35} + t^{28} - t^{21} + t^{14} - t^7 + 1)$ and $\Delta_{T(2,81)}=(t^2-t+1) (t^6 - t^3 + 1) (t^{18} - t^9 + 1) (t^{54} - t^{27} + 1)$.The factors always have a $1$ and non have the same degree, so I guess in those cases it is not possible... $\endgroup$
    – user2795243
    Commented Sep 3, 2019 at 16:57

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