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 $V^{\ast}$.)

  • 3
    $\begingroup$ +1 for detailed exposition alone. If I could, I'd give another +1 for the question. $\endgroup$ Aug 27, 2010 at 5:30
  • 1
    $\begingroup$ Remember that quadratic Lie algebras have a non-degenerate Killing form, so you're not going to have much luck looking for nonsemisimple ones. Not sure about positive characteristic though. $\endgroup$ Aug 27, 2010 at 5:36
  • 2
    $\begingroup$ Noah: Quadratic Lie algebras most definitely do not have a nondegenerate Killing form: only the semisimple do. There are plenty of examples of quadratic Lie algebras (other people, myself included, call them metric Lie algebras) and it is safe to say that most are not semisimple, not even reductive. $\endgroup$ Aug 27, 2010 at 10:09
  • 1
    $\begingroup$ It is also possible to obtain Vassiliev invariants out of (quadratic) Lie superalgebras or more generally Yang-Baxter algebras. So one could generalise your question to include such objects. It's a very good question. $\endgroup$ Aug 27, 2010 at 10:14
  • 1
    $\begingroup$ I'd like to point out that in positive characteristic the fundamental theorem of Vassiliev invariants is open. So even given such a weight system and a candidate chord diagram, you cannot conclude that you have a Vassiliev invariant which distinguishes a knot from its inverse. It would seem to me that the only relevant weight systems would be in characteristic 0, and would not come from Lie algebras. $\endgroup$ Aug 27, 2010 at 12:30

2 Answers 2


This isn't an answer to your question, more of a too-big comment on your statement "it is not easy to tell whether or not a given knot is invertible".

There is an algorithm to determine if a knot is invertible, or even to check if it is strongly invertible -- or any of the basic symmetry properties of knots. In full generality it can be a slow algorithm. In practice with the use of heuristics it's frequently very fast and easy use. In full generality it's not implemented in software. But it exists.

The pencil-sketch of the algorithm is:

  • Triangulate the knot complement (implemented in SnapPea).

  • Find the JSJ-decomposition of the complement. In principle there is a way to do this using the latest version of Regina, although at present I believe it would be a double-exponential run-time algorithm as you have to enumerate normal tori, cut along them, check incompressibility of the boundary, and then determine which of the bits are Seifert-fibred. This would occasionally give you truly awful run-time even for knots with, say, less than 50 crossings. Jaco and Rubinstein have a single-exponential run-time algorithm which may eventually be implemented in Regina.

  • Find the hyperbolic structures and canonical polyhedral decompositions of the hyperbolic bits in the JSJ-decomposition. There is a heuristic implementing this in SnapPea and it seems to be very difficult to break, moreover frequently you can prove SnapPea to be correct. Although I occasionally still try to break SnapPea. There is an algorithm to find the hyperbolic structure on the various hyperbolic bits, I'm not sure what name people will settle on but I like to call it the cusped Manning algorithm (should appear in the JacoFest proceedings, author is Tillman).

Going from the hyperbolic structure to the Epstein-Penner decomposition, off the top of my head I'm not sure what run-time estimates on that are.

  • Once you have all the above, determining chirality, invertibility, strong invertibility, etc, boils down to a fairly straight-forward combinatorial check on how the various symmetry groups of the manifolds you've decomposed the knot complement into interact.

Some processes that "generate" knots tend to generate a lot of hyperbolic knots, so frequently you can get your answer very quickly using only SnapPea. Other processes generate knots that have a lot of connect-summands (things like random walk knot constructions, or lattice constructions) so you need the full algorithm in those cases.

After all that, a small comment on your actual question. Turchin has some reason to be optimistic that Vassiliev invariants can distinguish knots from their inverses. If I recall correctly, he creates a splitting of the homology of various embedding spaces. This does not work for knots in $\mathbb R^3$, but there is an analogy to the 3-dimensional case. Moreover he computes some classes which, if analogous classes existed in the 3-dimensional case they would be orientation-sensitive. He goes so far as to suggest a certain class of Vassiliev invariant would be a productive place to look for inversion-sensitive invariants (see page 35, just before section 17).

  • $\begingroup$ Ryan -- thanks, that's interesting. The way I understand this program, it's the third step that requires a bit of a black magic. $\endgroup$
    – algori
    Aug 27, 2010 at 4:04

Although, as Jose pointed out above, there are a lot of interesting non-reductive metric Lie algebras out there (see Medina and Revoy - Algèbres de Lie et produit scalaire invariant, Figueroa-O'Farrill and Stanciu - On the structure of symmetric self-dual Lie algebras, and A terminology issue with the Killing form), from the point of view of finite type knot invariants you might as well stick with the reductive ones.

Here's why. Given a nice monoidal category (spherical, trivial object is simple) there is always a maximal planar ideal called the ideal of negligible morphisms. This consists of all morphisms which always give you 0 when you close them off in any way to get an endomorphism of the trivial. Equivalently, they are the kernel of the radical of the inner product given by the trace on the spherical category. From the point of view of knot theory, it's always safe to kill off all negligible morphisms because knots are closed diagrams.

But, if you start with an abelian category (like a nice category of representations of a metric Lie algebra), then the quotient by the negligibles is always semisimple (see Are Abelian, non-degenerate tensor categories semisimple?, Deligne - La cat\'egorie des repr\'esentations du groupe sym\'etrique $S_t$, lorsque $t$ n'est pas un entier naturel, and Barrett and Westbury - Spherical categories). Furthermore by Deligne's reconstruction theorem, once you know that all objects have integer dimensions your category must be the category of representations of a Lie superalgebra.

Combining the above, replace your category of $g$-modules with its quotient by the negligibles. Realize this category as the category of representations of a different metric Lie algebra which is reductive but gives the same knot invariants.

There should be some way to make this construction concrete, i.e., any metric Lie algebra should have some sort of reductive "core" which gives the same knot invariants. But my understanding of double extensions is not good enough to understand what's going on. It would be interesting to see this worked out.

  • $\begingroup$ Noah -- this sounds very interesting, but could you give more details on the last phrase of the second paragraph? In particular, can this procedure give the representation category of some $gl_n$ when applied to a a nilpotent algebra? $\endgroup$
    – algori
    Aug 27, 2010 at 20:44
  • $\begingroup$ On further thought, I may have made a mistake. There are two relevant notions of dimension here. One is the usual dimension of the vector spaces. But what I really care about is the so-called "quantum dimension," that is there's a trace (map from End(V)->k) given by the metric structure and the dimension of an object is the trace of the identity. For a general metric Lie algebra this might not be an integer in which case my argument is bogus. $\endgroup$ Aug 27, 2010 at 20:58
  • $\begingroup$ A good example of the "killing negligible" process is going from gl(a|b) to gl(a-b). The knot invariants should be the same. $\endgroup$ Aug 27, 2010 at 21:00
  • $\begingroup$ Noah -- even if this argument doesn't work it would be interesting if a version of it does i.e. if one can prove that general quadratic Lie algebras give nothing new. $\endgroup$
    – algori
    Aug 28, 2010 at 1:42
  • $\begingroup$ I'm not very good at working directly with these double extensions, but my new theory is that if you compute the trace of the identity on the double extension g+h+h* you'll get dim g (with the h and h* parts contributing in a canceling way) and that you can kill some of the negligibles and end up with g (if g itself is reductive then it's the quotient by all negligibles). Thus taking a double extension wouldn't change the knot invariants at all. In other words, I now think my answer is correct, though I can't quite do the computations myself to prove it. $\endgroup$ Aug 28, 2010 at 2:49

Your Answer

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