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If one has a $\mathbb{C}$-linear representation of the braid algebra into e.g. the Temperley-Lieb algebra i.e. $\rho:\mathbb{C}[B_{n}]\to TL_{n}(\delta)$, we can deduce a skein relation $\mathcal{S}$. Then given some diagram $D$ (for a framed link $L$) of the form

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(i.e. the closure of some $n$-braid $b$) then we can apply $\mathcal{S}$ to $D$ in order to obtain some (multivariate) Laurent polynomial $f$.

It is clear that $f$ is invariant under braid isotopy on $b$. The skein relation for e.g. the Kauffman bracket can be found by considering the appropriate $\rho$.

Question: Is the above enough to claim that $\mathcal{S}$ defines a framed link invariant (i.e. that $f$ is an invariant of $L$)?

I would've thought the answer to be yes (since we can use ambient isotopy to make a type-II or III Reidemeister move on $D$ locally look like a braid isotopy).

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Yes, but you need to check a normalization condition. (Thanks to Student for noting the error in my previous answer.)

An invariant $f$ of braids $b$ is an invariant of links (obtained as braid closures) if it satisfies the Markov moves:

  1. $f(\sigma_i b \sigma_i^{-1}) = f(b)$
  2. $f(\sigma_n b) = f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

The first condition is essentially the same as braid isotopy: I think you can always use an ambient isotopy to reduce it to applying either the invertibility of the braid generators or a braid-type relation. However, the second condition is a bit more subtle: you need something like it to deal with "extra" strands in your braid, but it is not the same as braid istopy and also fails for framed links, which are usually the natural objects to consider in this context.

One way to deal with this is to replace 2 with a weaker condition:

2'. There is a constant $x$ so that $f(\sigma_n b) = x f(b)$ whenever $b$ contains only the generators $\sigma_1, \dots, \sigma_{n-1}$ (i.e., when $b$ is a braid on strands $1, \dots, n$).

This will give you an invariant of framed links that changes by a power of $x$ when you change the framing, which is typically what happens for quantum invariants.

For a more systematic answer you can keep track of the framings as part of the braid instead of using the blackboard framing; this leads to the framed braid groups discussed in [1].

[1] Ko, Ki Hyoung; Smolinsky, Lawrence, The framed braid group and 3-manifolds, Proc. Am. Math. Soc. 115, No. 2, 541-551 (1992). ZBL0760.57007.

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    $\begingroup$ I don't think this answer is correct. If you just take away the second (i.e. (de)stabilization) relation), the result can't be what characterizes the framed links. Indeed, obviously there are braids with different amount of stings whose closures are isotopic framed links. With the first condition along this is missed. A reference I found for the Markov theorem for framed links is lemma 1 in Framed Braid Group and 3-Manifolds-[Ki Hyoung Ko and Lawrence Smolinsky]. Notice that the group is larger, but the relations are similar to Markov. $\endgroup$
    – Student
    Commented Oct 28, 2023 at 12:03
  • $\begingroup$ I think you're right. Upon further thought I don't see a good way to avoid having a stabilization-type relation, although there's a way to deal with this at the level of invariants to avoid passing to the framed braid group. $\endgroup$ Commented Oct 30, 2023 at 11:43

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