Given the diagram of an unoriented knot $K \subset S^3$, one can compute its Jones polynomial by picking an orientation for the knot and then simplifying the diagram using the relations drawn in the "properties" section here (the choice of orientation doesn't matter, although presumably it matters for links). Alternatively, one can give the knot a 0 framing and then simplify using the relations drawn here.

The HOMFLY polynomial of $K$ is computed in a way similar to the first method of computing the Jones polynomial - pick an orientation of the knot and simplify the diagram using the relations here.

Is there a way of computing the HOMFLY polynomial of $K$ analogous to the second way of computing the Jones polynomial described above, i.e. give $K$ the 0 framing (but no orientation) and simplify the diagram using some relations?

  • 2
    $\begingroup$ I really doubt there is an unoriented HOMFLY invariant for two reasons. The first is the asymmetry of the accepted diagrams used in the state sums od the su(n) invariants. The other is that unoriented theories tend to correspond to Lie groups where elements are conjugate to their inverses like su(2) or sp(2n). $\endgroup$ – Charlie Frohman Nov 20 '10 at 0:25
  • $\begingroup$ As stated the problem is insolvable as there is no notion of HOMFLYness that transcends diagrams. $\endgroup$ – Charlie Frohman Nov 20 '10 at 0:27

I'm not the best qualified MO-er to answer this question, but since no one else has answered yet I'll give it a shot. This can be thought of as elaboration of Charlie Frohman's comment above.

Modulo a change of variables, the parameters of the HOMFLY polynomial are $N$, for the Lie algebra $sl_N$, and $q$, the deformation parameter of $U_q(sl_N)$. The strands of the link should be thought of as labeled by the basic representation of $sl_N$. For $N=2$ this is the Jones polynomial. For $N>2$ the basic representation is not self-dual, so the orientations of the link components matter a great deal. So the answer to your question is No, there is no unoriented version of the HOMFLY (a.k.a. HOMFLYPT) polynomial.

The basic representation of $sl_2$ is self dual, but importantly it is antisymmetrically self-dual (Frobenius-Schur indicator = -1). Because of the self-duality, the Jones polynomial has an unoriented version (the Kauffman bracket polynomial), but because of the antisymmetry of the self-duality, the unoriented version has some flaws (e.g. loss of positivity and categorifiability).

[And now we see the real reason I'm answering this question. I want to be the first person in the world (so far as I or Google knows) to use the word "categorifiability" in print.]

If you want an unoriented HOMFLY-ish polynomial, check out the Kauffman Dubrovnik polynomial, which is related to BMW algebras and B, C and D type Lie algebras.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.