Polynomial invariants for unoriented links

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.

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