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Feb 2, 2010 at 9:13 comment added Daniel Moskovich In a different direction, I offer the Brandt-Lickorish-Millet-Ho Q-polynomial as an example of an invariant which is defined via a Skein relation but is not a quantum invariant (and is conjectured to contain no quantum invariant which is not a polynomial in the Casson invariant).
Dec 27, 2009 at 14:36 comment added Daniel Moskovich That is certainly one "proof". The skein relation is actually at the end of Alexander's original 1923 paper, although nobody thought of it that way at the time. You can calculate the R-matrix from the Burau representation (or from the Dehn presentation of the knot group as Alexander did), and then reproduce the same matrix from a representation of U_q(U(1|1)) or something. But that gives no conceptual insight into why the Alexander polynomial is a quantum invariant- why is it more than a lucky coincidence? Why that specific quantum group?
Dec 27, 2009 at 14:15 history answered Ben Webster CC BY-SA 2.5