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Let me first explain what I mean by the "classical limit". For quantum group invariants of links and webs (such as colored Jones polynomials), the "classical limit" means the limit $k\rightarrow +\infty$, or equivalently $q \rightarrow 1$. In this limit, quantum invariants become insensitive to the embedding of links and webs inside $\mathbb{R}^3$ (or $S^3$), and they count the number of certain edge colorings. So, in this sense, "quantization" (deforming $q$ away from $1$) is what makes those invariants sensitive to embedding.

Categorification deforms another variable $t$ away from $-1$ (if we just focus on the Poincare polynomial). Hence, starting from the "classical invariant" counting certain edge colorings, we have two different deformations (see the diagram below), and it is natural to wonder if we can complete the square by filling in the "???" part. $$\begin{matrix}??? & \xleftarrow{?} & \text{categorified quantum invariant} \\ \quad\quad\downarrow\small{t=-1} & & \quad\downarrow\small{t=-1}\\ \text{classical invariant} &\xleftarrow[]{q=1} &\text{quantum invariant}\end{matrix}$$

That is, I am curious if there is a certain specialization "?" (which I want to call the "classical limit") of the categorified quantum invariant (e.g. Khovanov homology) such that it is independent of the embedding of links and link cobordisms inside $\mathbb{R}^3$ and $\mathbb{R}^3\times \mathbb{R}$, and when specialized to $t=-1$ it becomes the classical invariant. Ideally, "???" must be a certain type of 2d TQFT.

Certainly, the naive guess of $q=1$ doesn't work in this case, as even in such specialization, Khovanov homology is still sensitive to embedding. Therefore, my question can be summarized as : Is there a specialization of Khovanov homology (or any other categorified quantum invariants) of type $f(q,t)=0$ for some polynomial $f$ such that the invariants become insensitive to the embedding under such specialization?

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  • $\begingroup$ Why do you insist that the words "classical limit" mean that the invariant be insensitive to the embedding? $\endgroup$ Commented Aug 5, 2018 at 22:19
  • $\begingroup$ @TheoJohnson-Freyd My choice of the word is probably not the best one, but I call it that way because that's how I understand classical limit in uncategorified cases. $\endgroup$
    – Henry
    Commented Aug 5, 2018 at 22:22
  • $\begingroup$ @TheoJohnson-Freyd To elaborate a little more, in case of abelian Chern-Simons theory, the corresponding quantum invariant is related to the Gauss linking number, and physically, it can be understood as interaction of link with itself. The interaction is determined by the propagator $\langle A(x)A(y)\rangle$, and it is in turn proportional to $1/k$. Hence in this case the classical limit $k\rightarrow \infty$ is exactly the limit where the interaction of the link with itself vanishes. $\endgroup$
    – Henry
    Commented Aug 5, 2018 at 22:32

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