# How much do homological knot invariants improve the classification problem of knots?

The mutation operation in knots appears to be detected by the Floer homological invariants. See the papers by Ozsvath, Szabo and by Baldwin, Gillam. In addition, the Khovanov homology turns out to be able to detect the unknot. See the papers by Elisenda Grigsby, Wehrli and by Kronheimer, Mrowka.

My question is the following.

How much do homological knot invariants improve the classification problem of knots? Is there something homological invariants cannot distinguish?

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## 1 Answer

I don't think that quantum topology or Heegaard Floer homology are useful as knot tabulation tools. Geometric invariants (e.g. hyperbolic volume) and classical invariants (e.g. signature) have been the weapon of choice for knot tabulators. I don't know of any pair of knots which have been separated using quantum invariants or Floer homology which we did not know how to separate using classical or geometric invariants.

There are several invariants which separate knots at least conjecturally, including the knot group (plus peripheral structure), the collection of all twisted Alexander polynomials of a knot (whatever that means), and the knot quandle. Surely HFH separates knots as well, although I don't think that this has been proven, so it detects "everything" in some sense. The issue is computational efficiency. But I think that the strength of HFH is in revealing structure to the space of knots and in analyzing properties of individual knots as opposed to as a tool to distinguish knots.

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Don't think you're right that Heegaard Floer theory gives complete knot invariants. For $\widehat{HFK}$ at least, I believe that there are counterexamples in the literature; possibly also for $CFK^\infty$. For Khovanov homology, Liam Watson gave counterexamples. –  Tim Perutz Sep 10 '12 at 4:23
gt.postech.ac.kr/~jccha/2009kmsams/tanaka.pdf gives some interesting examples (although I couldn't find a paper to back the slides up). In particular, they say that $\widehat{HFK}(8_{20})$ is isomorphic to $\widehat{HFK}(3_1\#3_1)$ (and this is easy to check). –  Marco Golla Sep 10 '12 at 6:41
For (quasi-)alternating knots, $\widehat{HFK}$ is determined by the Alexander polynomial. –  Tim Perutz Sep 10 '12 at 14:32
Apropos "at least conjecturally", there is of course the Gordon-Luecke theorem MR0965210. –  Lee Mosher Sep 11 '12 at 0:06
By HFH I meant the full theory, not just the hat version. –  Daniel Moskovich Sep 12 '12 at 12:41