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I have seen that usually one finds polynomial invariants for oriented links (for example the Jones polynomial, the Hompfly polynomial). Does anyone know what polynomial invariants exist for non-oriented links? Thank you for the help.

Update I have found out that there exists another polynomial, not mentioned in the other answers/comments. It is the BLM/Ho polynomial. One can read the definition in the original papers (Brandt, R. D.; Lickorish, W. B. R.; and Millett, K. C. "A Polynomial Invariant for Unoriented Knots and Links." Invent. Math. 84, 563-573, 1986; Ho, C. F. "A New Polynomial for Knots and Links--Preliminary Report." Abstracts Amer. Math. Soc. 6, 300, 1985) or http://mathworld.wolfram.com/BLMHoPolynomial.html

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  • $\begingroup$ One option is to take the product (or sum, or whatever symmetric operation) of -say- the Alexander/Jones/HOMFLY polynomial over all possible orientations, but I guess it's not what you're looking for. $\endgroup$
    – Marco Golla
    Mar 28, 2015 at 11:56
  • $\begingroup$ Thank you for the answer. To be honest I thought about this possibility, but it doesn't seem very easy to use in "real life". I was interested in learning what are the available polynomials. $\endgroup$
    – John N.
    Mar 28, 2015 at 12:09
  • $\begingroup$ How about the oldest of all knot polynomials, the Alexander polynomial? $\endgroup$
    – Liviu Nicolaescu
    Oct 26, 2015 at 22:14
  • $\begingroup$ @Liviu: Doesn't the Alexander polynomial of a link depend on the relative orientation of the link components? See for instance Linkinfo links L6n1{0,1} with polynomial -1+t^3 and L6n1{1,1} with polynomial t-t^2. They differ by flipping the orientation of one component. $\endgroup$
    – Danny Ruberman
    Oct 27, 2015 at 2:08
  • $\begingroup$ @Danny I misspoke. I was thinking of the Reidemeister torsion of the complement. That is independent of the link orientation and it is (essentially) a polynomial in as many variables as the number of components. $\endgroup$
    – Liviu Nicolaescu
    Oct 27, 2015 at 10:12

3 Answers 3

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Colored Kauffman polynomials are independent of orientations of links. If you look at the skein relation of Kauffman polynomial, there is no arrow. This is true for colored cases. Kauffman polynomials are related to SO(N) quantum knot invariants by substitution $q^{N-1}=a$. Since the representations of SO(N) are pseudo-real, they are independent of orientation.

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  • $\begingroup$ Thanks for the asnwer. Could you suggest me some references? $\endgroup$
    – John N.
    Mar 28, 2015 at 14:29
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    $\begingroup$ There are lots of books and papers about Kauffman polynomials. Maybe you can start with Lickorish "An introduction to Knot theory" or Kauffman "Knots and Physics". $\endgroup$
    – Satoshi Nawata
    Mar 28, 2015 at 15:43
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Let $\mathfrak g$ be a simple Lie algebra, $U_q(\mathfrak g)$ the corresponding quantum group and $V$ a simple module over it. From this you get an invariant of framed oriented links. Now since $V$ is simple, it depends on the framing only up to scalar multiplication, so there is a standard way to renormalize your invariant so that it becomes an invariant of oriented links.

Now assume that $V$ is self-dual, ie that there exists a $U_q(\mathfrak g)$-module isomorphism $V\rightarrow V^*$. If you fix one of those isomorphism you can compose it with the evaluation and coevaluation to get maps $V\otimes V\rightarrow \mathbb{C}$ and $ \mathbb{C} \rightarrow V\otimes V$ and interpret them as unoriented cap and cup (since they don't involve $V^*$ anymore). So now it seems that you can build an invariant of unoriented link. Thing is, it won't work if you use the standard ribbon element, but it will, I think (I'm not 100% sure), if you make a certain non standard choice: the square of a "half-twist" in the sense of Snyder-Tingley.

For more details :

  • Peter Tingley, A minus sign that used to annoy me but now I know why it is there
  • Noah Snyder, Peter Tingley, The half-twist for U_q(g) representations
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In Kauffmans "abstract tensor" approach, you can translate the Reidemeister moves to tensor equations. (This works equally well for directed and undirected case.) Of course, solving n^6 cubics and whatnot is hell, but I still have tabulated all "nice" ("charge conserved") cases up to n=6 and each gives rise to an undirected invariant (and you can immediately give its skein equation too). For example, with S being the S matrix, $1/q*S^2+q*S^{-2}-(q+1/q)*S^0+(1/q^2-q^2)*S^1+(q^2-1/q^2)*S^{-1}=0$ is my simplest example (n=4) that might be unknown. For knots it's identical to the Jones polynomial, but not for links. Note that these solutions may or may not be related to a Lie algebra (cf. Adriens answer), that's way over my head. (Feel free to PM me for details)

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