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I only ask for the subset of Reshetikhine-Turaev invariants for now.

Is the following sum up into cases a) and b) correct?
- Dots in Dynkin diagrams of Lie groups G come as a) symmetry-equivalent pairs (most of $G=A_n$, some of $D_n$ and $E_6$), b) not.
- The irreps $R$ belonging to them are a) not self-adjunct b) self-adjunct.
- $R\otimes{R}$ contains a) not 1 b) 1
- The knot polynomial belonging to $R$ is defined for oriented knots anyway, and a) it doesn't work with unoriented ones b) you can as well drop the arrows.

At least this would explain why I can't find a R matrix for $A_2$(dim 3 irrep). (Or more precise, an R matrix for unoriented knots. I have exactly one R matrix that suggests itself, but it only works for oriented knots.)

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3 Answers 3

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If I'm remembering correctly, the situation here is a bit subtle. On the one hand, morally the invariants do depend on orientation. In particular, my recollection is that for links they actually depend on orientation. But, on knots they can't detect orientation.

So you should expect your individual formulas which you use to compute the knot invariant to depend on orientation, but in the end the answers for knots won't depend on the orientation.

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If you define link invariants via $R$-matrices and braid representations, all links have to be oriented. It so turns out that for knots the orientation does not matter (although for links it does: reversing the orientation of one component may change the polynomial). The fact that knot orientation can't be determined using invariants that come from semi-simple Lie algebras boils down to the fact that such algebras have an automorphism, called the Cartan involution, that interchanges a representation with its dual (see D. Bar Natan, On Vassiliev knot invariants, 7.9). So (a) and (b) in the first two cases are related, but they have nothing to do with (a) and (b) in the 4-th case, because (b) always holds.

A related question: Lie algebra automorphisms and detecting knot orientation by Vassiliev invariants

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  • $\begingroup$ @Algori: Does your first sentence contain the implicite "...and quantum Lie groups..."? I have no problems at all defining an invariant via R matrices with unoriented links. It "costs" an additional isotypy and a lot of writhe normalization cheating, but it works. (The point of my question was: it works, but NOT for A2,dim 3, and I wondered what was the reason.) $\endgroup$
    – Hauke Reddmann
    Commented Sep 16, 2011 at 10:14
  • $\begingroup$ Hauke -- yes to the first question. Re defining invariants for nonoriented links: could you perhaps edit the posting and briefly describe your construction? The standard procedure does require orientations, or else we don't know which way to insert the $R$-matrices. $\endgroup$
    – algori
    Commented Sep 16, 2011 at 15:18
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I think what you are saying is correct for link invariants, bit even here there is some subtlety. For some self dual reps (e.g. even dimensional $SU(2),$ I am sure the condition in Lie Algebra Theory is straightforward but I don't know it) the intertwiner from $V$ to $V^{**}$ is not the natural linear algebra map, but is off by a sign. This means that if you work with the standard quantum group and $R$ matrix you get an oriented invariant (though dependence on orientation is pretty trivial of course. This can be patched up, but you generally pay some sort of price, such as not getting a unitary theory when it should be.

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