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The Jones polynomial can be computed from the representation theory of $\mathcal{U}_q(\mathfrak{sl}(2))$. The Alexander polynomial has an analogous description in terms of the representation theory of the superalgebra $\mathcal{U}_q(\mathfrak{gl}(1|1))$ (see, for example, Sartori http://arxiv.org/abs/1308.2047).

From this perspective, is there a nice explanation of the fact that $V_K|_{q = i} = \Delta_K|_{t = -1} = \det K$? Here $K$ is a knot, $V_K(q)$ denotes the Jones polynomial of $K$, $\Delta_K(t)$ denotes the Alexander polynomial of $K$ ($t$ and $q$ are related by $t = q^2$), and $\det K$ is the knot determinant.

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  • $\begingroup$ Maybe I should give a bit more motivation for this question. Rasmussen (arxiv.org/abs/math/0504045) conjectures a spectral sequence from $\delta$-graded reduced Khovanov homology to $\widehat{HFK}$. "Decategorifying" by taking the $\delta$-graded Euler characteristic, we get an equality of two numerical invariants: one is the Poincare polynomial $Kh(q,h)$ for reduced Khovanov homology, evaluated at $q = i$ and $h = -1$, and the other is the Poincare polynomial $HFK(t,h)$ for $\widehat{HFK}$, evaluated at $t = -1$ and $h = -1$. (Here $h$ means homological; the $h$'s may not be the same.) $\endgroup$ – Andy Manion Sep 2 '15 at 22:22
  • $\begingroup$ So, if one hoped to prove Rasmussen's conjecture by relating HFK to some categorified quantum-groups invariant, and then finding relationships with Khovanov homology in this framework, a reasonable first step seems to be to understand the decategorified level first. If a nice explanation exists, it seems like something that's probably been known since before HFK or Khovanov homology were invented. $\endgroup$ – Andy Manion Sep 2 '15 at 22:25
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You haven't specified what sorts of explanations you would find "nice," so the following collection of thoughts might not be helpful, but here goes (or it might be helpful, but contain only things you already know). [By the way, this answer only attempts to "explain" why $V_K(i) = \Delta(-1)$ -- it does not address the additional question of why they both yield the knot determinant. That's a great question; I guess one "explanation" is simply that the invariants are determined by their skein relation.]

Firstly, a reference I like, in addition to the one you mention by Sartori, is this paper by Kauffman and Saleur. In that paper, many (though not all) of the theorems and computations are for the general Lie superalgebra $\mathfrak{gl}(m|n)$, where $m,n \in \mathbb{Z} >= 0$ are not necessarily equal. The Jones and Alexander cases are each specializations of this general case, with $(m,n) = (2,0)$ or $(1,1)$, respectively, so therefore a general strategy for a certain kind of "explanation" would be: observe that for certain specializations of the quantum parameter $q$, the generally-defined $\mathfrak{gl}(m|n)$ quantum link invariants, and even much of the structure of the quantum group $U_q(\mathfrak{gl}(m|n))$, depends only on $m+n$.

Here are two specific instantiations of the strategy. For the first, see page 299 (= page 7) of the paper I linked to. There the authors point out that for $q = i$, the quantum $R$ matrices for $U_q(\mathfrak{gl}(m|n))$ and $U_q(\mathfrak{gl}(m+n))$ agree. This is the essential ingredient for showing the relevant polynomial invariants agree at $q=i$, but to be complete one would want to go through the details of how those invariants are defined, in terms of the $R$ matrix.

The second is along those lines: the first link invariant defined in the Kauffman and Saleur paper has this skein relation (see p. 302):

$$ q^{n-m}P^+ - q^{-(n-m)}P^- = (q-q^{-1})P_{||} $$

(hopefully my notation is clear enough). However, one can see, by passing from a given knot to a split (1,1) tangle, say, that this invariant $P$ is forced to be zero for all knots if the quantum dimension $D_q$ of $U_q(\mathfrak{gl}(m|n))$ is zero. However, one can define another invariant $P':= P/D_q,$ which will be non-zero. The quantum dimension is $$ D_q = \frac{q^{n-m} - q^{-(n-m)}}{q-q^{-1}},$$ which is zero for all $q$ if $n=m$. Anyway, I haven't done it carefully myself, but I think it should be fairly straightforward to go from the above skein relation to the skein relation satisfied by $P'$ at $q = i$, and see that this only depends on $m+n$. (A final step would be to relate these normalizations of the invariants to those in the literature).

Let me know if I've said anything confusing or wrong, and if this helps in any way; or if you were imagining something different in terms of an explanation.

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    $\begingroup$ The skein relation satisfied by $P'$ is the same one satisfied by $P$, right? Plugging in $q = i$, I can reduce this skein relation to $(-1)^m(i^{n+m}P^+ - i^{-n-m}P^-) = 2i P_{||}$. This is close to depending only on $n+m$, but there's an annoying factor of $(-1)^m$ outside. $\endgroup$ – Andy Manion Sep 9 '15 at 2:54
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    $\begingroup$ hi Andy - yes, that seems right. It's strange, I just checked and this is true even according to MathWorld - according to the info on mathworld.wolfram.com/AlexanderPolynomial.html and mathworld.wolfram.com/JonesPolynomial.html, the skein relations for V(-1) and $\Delta(-1)$ differ by a sign, but those same pages claim V(-1) and $\Delta(-1)$ are equal. One can check that something is wrong already for the Hopf link, where the signs differ for the two invariants. This is related to the fact, actually, that the knot determinant is generally claimed to be equal to $|\Delta(-1)|$. $\endgroup$ – Sam Lewallen Sep 9 '15 at 5:39
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    $\begingroup$ I don't have time to look through the details at the moment, but if you do and straighten them out, I'd love to know. If the claim from the Kauff+Saleur paper that the R-matrices agree at q=i is true, then the remaining discrepancy would have to come from the "normalization by the ribbon element," which is the only other part of the definition. It would be interesting to track the sign down, and see if we can figure out what's going on. $\endgroup$ – Sam Lewallen Sep 9 '15 at 6:03
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    $\begingroup$ Again this is off the cuff / off of memory, but I think ultimately one of the invariants needs to be normalized by (-1)^L, where L is the number of components. It would be interesting if this can be incorporated as a slightly tweaked definition of P in the K+S paper, and to see if this invariant is more natural somehow $\endgroup$ – Sam Lewallen Sep 9 '15 at 6:07
  • $\begingroup$ whoops, I meant (-1)^(L-1) (at least to get the value 1 for the unknot). $\endgroup$ – Sam Lewallen Sep 9 '15 at 6:12

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