I want to understand the framing dependence of the Khovanov-Rozansky homology, and as its first step, I am trying to understand the framing dependence of the HOMFLY polynomial (i.e. quantum $sl(n)$ invariants). (Of course, both of them can be defined in a framing independent manner, but as I am mostly interested in quantum 3-manifold invariants, studying how they behave under change of framing seems to be important.)

As we allow our (framing dependent) HOMFLY polynomial to vary under the first Reidemeister move, there is a freedom of choice of the "framing factor" (a multiplicative factor $\delta$ that the polynomial gets under a positive twist). In the end, the following should hold : $$P(L) = \delta^{w(L)}P_{\text{fr}}(L)$$ where $w(L)$ is the writhe of the link $L$, $P$ the HOMFLY polynomial, $P_{\text{fr}}$ a framing dependent version of HOMFLY polynomial.

Probably the most natural choice of $\delta$ would be $t_{\square\square}$ where $\square$ is (the weight of) the defining representation of $sl(n)$, and $t$ is the $t$-matrix : $$t_{\mu\nu}=\delta_{\mu\nu}q^{\langle\mu,\mu+2\rho\rangle}$$ (as defined in Theorem 3.3.20. of Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov, Jr.) Here $\langle,\rangle$ is the usual inner product on $\mathfrak{h}^*$ such that $\langle \alpha,\alpha \rangle=2$ for each simple roots $\alpha$ of $sl(n)$, and $\rho$ is the half sum of positive roots.

So my question is :

  1. Is there a known closed form formula for $\langle\square,\square+2\rho\rangle$ for $sl(n)$?

  2. If this is not the most natural choice of $\delta$, then what should $\delta$ be?


It turned out that it was a very simple calculation to compute $\langle \square,\square+2\rho\rangle$ for $sl(n)$ using Cartan matrices. For $sl(n)$, $\langle \square,\square+2\rho\rangle = 2-\frac{1}{n}$, and hence $$t_{\square\square} = q^{2-\frac{1}{n}}.$$ Note, by the way, that the convention used here is $q=e^{\frac{\pi i}{k+h^\vee}}$, where $k$ is the bare level, and $h^\vee$ is the dual Coxeter number, which is just $n$ for $sl(n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.