Recall that there are knots in $\mathbf{R}^3$ that are not invertible, i.e. not isotopic to themselves with the orientation reversed. However, it is not easy to tell whether or not a given knot is invertible; I believe the easiest known example involves 8 crossings. In particular, all knot invariants that are (more or less) easy to compute (e.g. the Jones or Homfly polynomials) fail to detect knot orientation -- there is a simple Lie algebra trick, the Cartan involution, that prohibits that. But this trick works for complex semi-simple Lie algebras and the question I'd like to ask is: can one perhaps circumvent this by using other kind of Lie algebras?

Here are some more details. A long standing problem is whether or not knot orientation can be detected by finite type (aka Vassiliev) invariants. Recall that these are the elements of the dual of a certain graded vector space $\mathcal{A}=\bigoplus_{i\geq 0} \mathcal{A}^i$; the vector space itself is infinite dimensional, but each graded piece $\mathcal{A}^i$ is finite dimensional and can be identified with the space spanned by all chord diagram with a given number of chords modulo the 1-term and 4-term relations (the 4-term relation is shown e.g. on figure 1, Bar-Natan, On Vassiliev knot invariants, Topology 34, and the 1-term relation says that the chord diagram containing an isolated chord is zero).

The above question on whether or not finite type invariants detect the orientation is equivalent to asking whether there are chord diagrams that are not equal to themselves with the orientation of the circle reversed modulo the 1-term and 4-term relations. (There are several ways to rephrase this using other kinds of diagrams, see e.g. Bar-Natan, ibid.)

However, although $\mathcal{A}^i$'s are finite-dimensional, their dimensions grow very fast as $i\to \infty$ (conjecturally faster that the exponential, I believe). So if we are given two diagrams with 20 or so chords, checking whether or not they are the same modulo the 1-term and 4-term relations by brute force is completely hopeless. Fortunately, there is a way to construct a linear function on $\mathcal{A}$ starting from a representation of a quadratic Lie algebra (i.e. a Lie algebra equipped with an ad-invariant quadratic form); these linear functions can be explicitly evaluated on each diagram and are zero on the relations. So sometimes one can tell whether two diagrams are equivalent using weight functions. But unfortunately, a weight function that comes from a representation of a complex semi-simple Lie algebra always takes the same value on a chord diagram and the same diagram with the orientation reversed.

As explained in Bar-Natan, ibid, hint 7.9, the reason for that is that each complex semi-simple Lie algebra $g$ admits an automorphism $\tau:g\to g$ that interchanges a representation and its dual. (This means that if $\rho:g\to gl_n$ is a representation, then $\rho\tau$ is isomorphic to the dual representation.) Given a system of simple roots and the corresponding Weyl chamber $C$, $\tau$ acts as minus the element of the Weyl group that takes $C$ to the opposite chamber. On the level of the Dynkin diagrams $\tau$ gives the only non-trivial automorphism of the diagram (for $so_{4n+2}, n\geq 1,sl_n,n\geq 3$ and $E_6$) and the identity for other simple algebras. (Recall that the automorphism group of the diagram is the outer automorphism group of the Lie algebra.)

However, so far as I understand, the existence of such an automorphism $\tau$ that interchanges representations with their duals is somewhat of an accident. (I would be interested to know if it isn't.) So I'd like to ask: is there a quadratic Lie algebra $g$ in positive characteristic (or a non semi-simple algebra in characteristic 0) and a $g$-module $V$ such that there is no automorphism of $g$ taking $V$ to its dual?

(More precisely, if $\rho:g\to gl(V)$ is a representation, we require that there is no automorphism $\tau:g\to g$ such that if we equip $V$ with a $g$-module structure via $\rho\tau$, we get a $g$-module isomorphic to $V^{\ast}$.)

nothave a nondegenerate Killing form: only the semisimple do. There are plenty of examples of quadratic Lie algebras (other people, myself included, call themmetricLie algebras) and it is safe to say that most are not semisimple, not even reductive. – José Figueroa-O'Farrill Aug 27 '10 at 10:09superalgebrasor more generallyYang-Baxter algebras. So one could generalise your question to include such objects. It's a very good question. – José Figueroa-O'Farrill Aug 27 '10 at 10:14