Let $P:=P(a_1,\dots,a_n)$ be a Pretzel link ( http://en.wikipedia.org/wiki/Pretzel_link ). For every permutation $\sigma\in S_n$ we can consider the link $$\sigma P:=P(a_{\sigma(1)},\dots,a_{\sigma(n)})$$. If $\sigma=(12\dots n)^k$ for some k then $P$ and $\sigma P$ are equivalent. My questions are:

  • Are there any results related to the general case?

  • Which invariants can be used to distinguish pretzel links which differ only by a permutation of their coefficients?

Note that $\sigma P$ is a mutant of $P$ for every $\sigma\in S_n$. So this problem is related to a more general problem i.e. how to distinguish mutant knots (or links).


Richard Bedient has proved in 1984 that two Pretzel knots $P$ and $\sigma P$ are equivalent if and only if $\sigma$ is a cyclic permutation, an order reversing permutation, or a composition of both. He computes a particular group constructed from the fundamental group of the complement to distinguish every non-equivalent pair of knots.

Finding invariants is non-obvious since mutant knots have the same Jones (and HOMFLY) polynomial and hyperbolic volume. These particular mutant knots also have the same branched double covering (a Seifert manifold fibering over $S^2$ with singular fibers of order $a_1, \ldots, a_n$). In fact, Bedient was interested in finding families of non-equivalent knots sharing the same branched double covering.

Edit. I missed some quite restricting hypothesis on the coefficients. The theorem holds if the $a_i$'s are distinct and all strictly greater than $1$. Moreover we must have $\sum_i 1/a_i <1$.

As noted by Ian, such a theorem cannot hold in general. For instance, a consecutive pair $(+1,-1)$ can be moved everywhere.

  • $\begingroup$ Bruno, I think you have to be slightly careful - if there are coefficients $a_i=+1, a_j=-1$, then they may be inserted anywhere. $\endgroup$
    – Ian Agol
    Nov 22 '10 at 19:28
  • $\begingroup$ Ian, you are right: there is a strong hypothesis on the coefficients, they are all greater than 1. $\endgroup$ Nov 22 '10 at 20:02

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