Quantum groups are useful for making knot/link invariants: for example, $U_q(\mathfrak{sl}_2$) you get the Jones polynomial. This boils down to the fact that $\mathcal C = \operatorname{rep }U_q(\mathfrak{sl}_2)$ is a braided monoidal category, which is not symmetric, hence gives us interesting knot invariants.

A deformation theoretic path to the quantum group $U_q(\mathfrak g)$ from a Lie algebra $\mathfrak g$ is as follows. Fixing a Casimir element $\Omega\in\mathfrak g\otimes \mathfrak g$ endows the category $\mathcal C=\operatorname{rep }\mathfrak g$ with the structure of an infinitesimally braided symmetric monoidal category. We can infinitesimally deform $\mathcal C$ to an honest braided monoidal category $\mathcal C[\hbar]/(\hbar^2)$ with braiding given by $\sigma\circ(1+\hbar\Omega/2)$. The next step is usually to use a Drinfeld associator to integrate this infinitesimal deformation to a formal one $\mathcal C[[\hbar]]$ with braiding given by $\sigma\circ\exp(\hbar\Omega/2)$. After some twisting/scrootching/variable replacement we get $\mathcal C[[\hbar]]=\operatorname{ rep}U_q(\mathfrak g)$.

Question From the vantage point of knot invariants, is there any benefit to integrating the infinitesimal deformation $\mathcal C[\hbar]/(\hbar^2)$ to the formal one? The infinitesimal deformation is already a non-symmetric braided monoidal category.


1 Answer 1


Let $\mathcal{C}$ be the category of finite-dimensional representations of a semisimple Lie algebra $\mathfrak{g}$ and $\mathcal{C}[\![\hbar]\!]$ the ribbon category you mention (which depends on the choice of a Drinfeld associator).

Given an irreducible $\mathfrak{g}$-representation $V$, the corresponding knot invariant obtained from $\mathcal{C}[\![\hbar]\!]$ is the Kontsevich integral evaluated using the weight system coming from $\mathfrak{g}$, $\Omega$ and $V$; this is Theorem 10 in Lê--Murakami's ``The universal Vassiliev-Kontsevich invariant for framed oriented links'' (https://arxiv.org/abs/hep-th/9401016).

The $n$-th Taylor coefficient (with respect to the $\hbar$ expansion) of the Kontsevich integral is a finite type (Vassiliev) invariant of degree less than $n + 1$. But the first nontrivial finite type invariant of oriented knots has degree 2 and so requires working at least with $\mathcal{C}[\hbar]/\hbar^3$.

So, you could consider oriented knot invariants working modulo $\hbar^2$, but they are all independent of the knot.

  • $\begingroup$ Great answer! If you're going to work modulo h^3 you might as well go "all the way" since the deformation theory is unobstructed. $\endgroup$
    – Steve
    Mar 25, 2022 at 14:24
  • 2
    $\begingroup$ You actually don't even need to rely on this thm by Le-Murakami. It's fairly easy to show directly that, after setting $q=e^h$, any knot invariant associated with a representation of a quantum group (or more generally, with any object in a ribbon category obtaind by any deformation of an infinitesimal symmetric monoidal category) has the property that its degree $n$ coefficient is a finite type invariant of degree $n$. This is an invariant of framed knot, but there's only one degree 1 invariant, the self-linking number. $\endgroup$
    – Adrien
    Mar 25, 2022 at 16:43

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