# Can the ribbon category of f.d. reps of $\mathcal{U}_q(\mathfrak{sl}(2))$ be modified so the twist is trivial on the vector representation?

Consider the ribbon category of finite-dimensional representations of $\mathcal{U}_q(\mathfrak{sl}(2))$, with twist $\theta$. If $V$ is the vector representation, then $\theta_V$ is multiplication by $q^{-3}$. As described e.g. in Snyder-Tingley http://arxiv.org/abs/0810.0084, one could use an alternate twist $\theta'$ with $\theta'_V$ given by multiplication by $-q^{-3}$, but these are the only two options compatible with the braiding.

As a result, the Reshetikhin-Turaev invariant of framed links obtained by labeling all link components with $V$ depends on the framing: different framings change the invariant by $\pm q^{-3}$. This invariant is the Kauffman bracket; the Jones polynomial, which is independent of framing, is obtained from the Kauffman bracket by rescaling by powers of $q^3$ to remove the framing dependence.

I'm wondering whether the Jones polynomial itself, rather than the Kauffman bracket, can be obtained from a ribbon category using the Reshetikhin-Turaev connstruction. A similar question may have been asked at Invariants of unframed, oriented links from Reshetikhin Turaev construction, but the answer isn't quite what I'm looking for: I know the framing-dependent invariants can be rescaled at the end to produce a framing-independent invariant. What I want to know is whether this rescaling can be built in earlier in the Reshetikhin-Turaev construction.

It seems like this would involve rescaling not just the ribbon element $\theta$, but the braiding / $R$-matrix for $\mathcal{U}_q(\mathfrak{sl}(2))$ as well. But I don't see a way to do this that still satisfies the axioms of a braided monoidal category.

Ideally, I'd like a ribbon structure on the monoidal category of finite-dimensional representations of $\mathcal{U}_q(\mathfrak{sl}(2))$, such that the twist is trivial on the vector representation, and such that the Reshetikhin-Turaev invariant associated to the vector representation is the Jones polynomial. Alternatively, I'd take an explanation of why no such thing exists, or some weaker version that satisfies different axioms but is morally what I'm looking for.

Here's some additional motivation for this question: in http://arxiv.org/abs/1308.2047, Sartori presents a ribbon structure on the category of finite-dimensional representations of $\mathcal{U}_q(\mathfrak{gl}(1|1))$ and shows how to obtain the Alexander polynomial from (a slight modification of) the Reshetikhin-Turaev construction. In this setting, the twist acts as the identity on the vector representation and its dual (Lemma 4.2 of Sartori). Thus, the RT invariants for the vector representation are naturally framing-independent. I'm wondering if there's a way to set up the RT construction for $\mathcal{U}_q(\mathfrak{sl}(2))$ such that the same thing happens.

There is a category built from the HOMFLYPT skein relation in the same way that $\mathcal U_q(\mathfrak sl(2))$ is built from the Kauffman skein relation. It is a version of $\mathcal U_q(\mathfrak gl(t))$, where $t$ is a variable which I will use to parameterize the value of the unknot, and where I don't have the determinant representation (which you could add in if you want — it will essentially give you a $\mathbb Z$-graded version). If you normalize the relation as $\alpha L_+ -\alpha^{-1} L_- = (q - q^{-1})L_0$, then $t = \frac{\alpha - \alpha^{-1}}{q - q^{-1}}$. Perhaps the usual conventions have $q^{1/2}$ or something. Because HOMFLYPT is an invariant of oriented unframed links, this category can be made ribbon with the trivial ribbon structure. (You could also put on the ribbon structure which is $-1$ on the defining object.)

This category can be built from Iwahori–Hecke algebras in a way similar to the way that $\mathcal U_q(\mathfrak{sl}(2))$ can be built from Temperley–Lieb algebras.

When $\alpha = q^n$, so that $t = [n] =$ quantum $n$, the category has interesting quotients that look more like $\mathcal U_q(\mathfrak{gl}(n))$. If my memory is correct, you produce these quotients the same way you produce MTCs from root-of-unity categories: you quotient by the ideal of negligible morphisms. Recall the usual (type A) "quantum Schur–Weyl duality", which says that both quantum $GL$ and Iwahori–Hecke algebras have representation theory indexed by Young diagrams. The quotient that gets you down to $\mathcal U_q(\mathfrak{gl}(n))$ corresponds on the Iwahori–Hecke side to the quotient of Iwahori–Hecke algebras whose representations are those just with at most $n$ columns (or rows, depending on your convention). ($SL(n)$ is the sub of $GL(n)$ that doesn't distinguish certain modules. Namely, in $GL(n)$ you are building Young diagrams with at most $n$ columns. In $SL(n)$, you do this but also play "tetris", so that if you ever completely fill a row — build a diagram with a row of length $n$ — then you can remove it.)

In your question, the reason to work with $\mathfrak{gl}$ and not $\mathfrak{sl}$ is that you specifically asked for invariants of oriented knots. In $\mathfrak{gl}(2)$ the vector is not isomorphic to its dual, whereas in $\mathfrak{sl}(2)$ they are.

• I should warn: I'm not sure how much of my representation theory claims are on firm footing. Indeed, my impression is that there are (or at least recently were) open questions about whether you actually produce $\mathcal U_q(\mathfrak{gl}(n))$ from following a procedure like I said, or just some category that maps to it. The question is: "is $\mathrm{Rep}(\mathcal U_q\mathfrak g)$ described by local relations"? The Kuperberg spiders give the answer "yes" for the $\mathfrak g$ of rank-2, and for a long time it was open in higher rank. I know that there's been major recent work on quantum... – Theo Johnson-Freyd Feb 4 '16 at 16:41
• ... Schur–Weyl duality recently, including by regulars here on MO, which perhaps has answered the question, but I admit I have not kept up to date on the results. – Theo Johnson-Freyd Feb 4 '16 at 16:42
• Come to think of it, I said "quotient out by the ideal of negligible morphisms", which I do think always works, but I'm going off of memory, and one regularly simplifies claims in memory that are more complicated in real life. If so, then one version of the "local relations" question is whether that ideal is finitely generated. – Theo Johnson-Freyd Feb 4 '16 at 16:46
• The point here is roughly that central character gives a grading on representations, and so elements of the center of the corresponding Lie group give you the ways of changing ribbon structure. For SL the center is finite and so you only get a few ribbon structures, but for GL the center is the scalars and so you can change ribbon structure at the vector rep however you want. The ribbon structure won't be trivial everywhere, in particular the quantum determinant will no longer have trivial ribbon element (which is why this ribbon structure doesn't descend to SL). – Noah Snyder Feb 4 '16 at 23:49
• @AndyManion Just the vector. Already for the symmetric and exterior squares, the ribbon is something like multiplication by $q$. – Theo Johnson-Freyd Feb 6 '16 at 17:05