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So, there's a construction of Reshetikhin and Turaev which extracts knot invariants from ribbon monoidal categories, which are (usually) the representation category a Hopf algebra with a choice of ribbon element.

How do these knot invariants change if I pick a different ribbon element in the same Hopf algebra? In particular, will something strange happen with 3-manifold invariants?

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2 Answers 2

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Did you look at prop 5.21 in the paper with Peter? I think that should answer your question.

There are two slightly different questions you could ask. First how does the framing-dependent invariant change. Here it is just (\pm 1)^#L where # is the number of components. Second how does the framing-corrected invariant change? Here it's (\pm 1)^#L (\pm 1)^writhe. In both cases the \pm 1 just measures whether you've changed the FS indicator of your rep V.

If you want to think about things labelled with components labelled by more than one irrep it'll get yuckier.

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  • $\begingroup$ I think you may have reversed "dependent" and "independent" up there. Is the answer for a link with different labels written up somewhere? It's not just "multiply by the ratios of FS-indicators for labels of all components"? $\endgroup$
    – Ben Webster
    Oct 25, 2009 at 15:35
  • $\begingroup$ Also, that formula is extremely confusing when V is not self-dual. You defined the FS indicator to be 0 in that case. I'm pretty sure I know how to interpret the resulting 0/0 (basically by David's answer), but you guys have committed a grave crime against notation. $\endgroup$
    – Ben Webster
    Oct 25, 2009 at 15:38
  • $\begingroup$ Yuck, good point. $\endgroup$ Oct 25, 2009 at 17:01
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Ben,

As I mentioned in response to your previous question about ribbon elements, the element u which is defined from the R-matrix, u=\mu\circ(S\ot \id)(R21) has the property that uS(u)=v^2 (well this is not the formula I gave for u in that post, because the one I gave was incorrect; this one appears to be correct according to wikipedia).

This relation v^2=uS(u) is true in any ribbon Hopf algebra, and in particular it implies that v has to be a square root of uS(u). So I think this means that the ribbon element is almost unique.

More precisely, let v and w be two ribbon elements. Then v/w is a grouplike element of order two. I think this implies that the corresponding invariant applied to a link will be multiplied by the constant v/w applied to each link. Now if you choose irreducible representations to label your link, then this number would have to be +/- 1.

Does this seem correct? -david

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  • $\begingroup$ This seems problematic because it doesn't take into account the ribbon nature of the knot. When I add an extra ribbon twist to my knot, then I'm multiplying by different scalars, but it's not clear to me if there's anywhere I can compare them. Maybe the 0-framings are the same? $\endgroup$
    – Ben Webster
    Oct 24, 2009 at 13:59
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    $\begingroup$ I'm sorry I don't understand the reply. I think the above argument is saying that any two ribbon elements in a ribbon Hopf algebra differ by multiplication by a central group-like element of order two. Unless I made an error, this seems to follow trivially from the axioms. Now this central group-like element acts on an irrep as +/-1, just because it's order 2. So when you evaluate a tangle diagram as a morphism, each time you use the isomorphism V-->V**, your new invariant is +/-1 times your old one. In particular the scalar coming from viewing a link in End(1,1) will be +/-1 of the old. $\endgroup$ Oct 24, 2009 at 23:54
  • $\begingroup$ Right, and I was saying that that answer couldn't possibly right for all ribbon knots. I think (based on Noah's answer) that's it's right for 0-framed ribbon knots, but that means there are some other funny scalars for other framings. $\endgroup$
    – Ben Webster
    Oct 25, 2009 at 15:24

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