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It is well known that, after a change of variables, the quantum knot invariants (Jones, HOMFLY, Kauffman, etc.) can be written as power series whose coefficients are finite type (i.e., Vassiliev) invariants. But what about their categorifications?

Specifically, do the generating polynomials of dimensions of the Heegard-Floer, Khovanov/Khovanov-Rozansky, etc., homology theories admit a change of variable such that the coefficients of the resulting power series are finite type invariants?

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  • $\begingroup$ This is perhaps relevant: arXiv.org/abs/0803.1200 $\endgroup$ Sep 13, 2010 at 16:59
  • $\begingroup$ @Kevin: As far as I can tell, that paper says something like "categorified Jones invariants can be recovered from categorified finite type invariants". The question I was asking is: are the categorified Jones (and other quantum) invariants recoverable from the non-categorified finite type invariants. $\endgroup$ Sep 13, 2010 at 17:12
  • $\begingroup$ Vivek -- excellent question! Although I think the relationship between finite type invariants and the generating functions for Khovanov homology may be trickier than just making a change of variables and taking the Taylor series. $\endgroup$
    – algori
    Sep 13, 2010 at 18:57
  • $\begingroup$ I suspect this is not known one way or the other. I don't think it's out of the question that the same ridiculously easy proof that works in the decategorified case (the R-matrix is congruent to its inverse mod h) works in cases like Khovanov-Rozansky, but I would have to think through the details before being sure. $\endgroup$
    – Ben Webster
    Sep 13, 2010 at 19:10
  • $\begingroup$ I was expecting the answer to be no, because otherwise you could show that finite type invariants distinguish the unknot by using facts about Heegaar-Floer homology. $\endgroup$ Sep 13, 2010 at 20:46

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If the answer were "yes" by just a change of variable, I think that would imply that the generating polynomials satisfy some kind of skein relation. That's surely false.

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  • $\begingroup$ Oops, I meant to just make a comment. $\endgroup$ Sep 18, 2010 at 4:27

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