Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

So, if you talk to the right people, they will tell you that the braiding of the category of representations of a quantum group are not unitary and that one can fix this by taking a different commutor (the cactus commutor). But I like the R-matrix better, and I think I know an inner product where it's unitary. I'm throwing this up here in hopes that it will look familiar to someone.

For those of you wondering why I would do such a thing, I should mention that I like categorification. Categorification is a funny thing, and one funny thing about it is that you can't really categorify just a vector space; you can only categorify a vector space with (maybe non-symmetric) inner product, and any map you categorify to an equivalence of categories must be unitary for this inner product. So if one is to categorify a linear operator, it had better be unitary with respect to something.

So, this inner product on V (x) W for any irreducibles V and W is the unique one (up to scalar) such that

$$(\Delta(F_i)v,w)=(v,\Delta(q^{-d}_iK_iE_i)w)$$ $$(\Delta(q^{-d}_iK_iE_i)v,w)= (v,\bar \Delta(F_i)w)$$

(Here, $\Delta(F)=F \otimes K^{-1} + 1\otimes F$ and $\bar \Delta(F)=F \otimes K + 1 \otimes F$).

Ring any bells with anyone?

EDIT: Anyone curious how this story actually works out should look at my papers Knot invariants and higher representation theory I and II. The upshot is that there is a bilinear form on a tensor product for which the R-matrix is an isometry.

share|improve this question
add comment

4 Answers

You're being thrown off by a poor choice of names that Drinfel'd made. (Hey, he had to name a lot of stuff, so I can't really blame him for getting one wrong.) By "unitarization" he means "modifying it so that it's square is 1" which has nothing whatsoever to do with being unitary. The usual R-matrix should be unitary wrt to the usual *-structures on quantum groups if q is a number of the right form (either size 1 or real depending on the *-structure).

Edit: As pointed out in the comments I seem to be wrong here. I'm going to leave up the answer unmodified, however, because the comments clarify the situation and editing it would make the comments more confusing.

share|improve this answer
Actually, Drinfeld is innocent - the physicists are to blame! –  Pavel Etingof Feb 3 '10 at 4:25
At least when $q$ is real, the R-matrix itself won't be unitary with respect to any inner product, since the eigenvalues are $\pm q^t$ for $t$ in some specified set of rational numbers; if $q$ is real and not equal to $\pm 1$, then those eigenvalues will never have modulus 1. On the other hand, if $q$ is real, then one can choose an inner product such that the R-matrix is self-adjoint, and then the cactus commutor is self-adjoint as well. But a self-adjoint operator which squares to the identity is unitary, so in that setting the cactus commutor is indeed unitary. –  MTS May 17 '11 at 23:21
add comment

In the beginning of $SO(3)$-TQFT (Communications in Geometry and Analysis) by F and Kania-Bartoszynska we construct a pairing with respect to which the $R$-matrix for $U_q(sl_2)$ is unitary. The trick is to use pairings on the representations that are not positive, but the induced pairing on intertwiners is. There is a later paper by Kirilov where he does the same thing in a much more general setting, but I forget the citation.

My take-away from struggling to construct TQFT with corners that is Hermitian is, it's a mistake. TQFT as framed in low dimensional topology depends on linearity in an essential way. If some of your basic data is conjugate linear, it messes up the small level structure of the theory.

As a kid who got sucked into mathematics by his desire to understand the quantum mechanics, the fact that TQFT is conceived with bilinear pairings messed me up, because, of course you take the complex conjugate when you are computing the bracket.

share|improve this answer
add comment

Actually if you take the usual R-matrix for type $A$ it is only unitary for the choice $q=1$! There seems to be some conflict between "local" braidings and unitarity (as Z. Wang and I will explain in a paper currently being written).

share|improve this answer
add comment

I'm pretty late to the party here, and Ben, it seems that you already have a satisfactory answer to your question, but I thought for the sake of completeness I would just post this in case anybody stumbles across this question and is still curious.

The key observation is that the eigenvalues of the braiding induced by the $R$-matrix are of the form $\pm q^{t}$, where $2t$ is a rational number in the image of the bilinear pairing of the weight lattice with itself. This follows from a result of Drinfeld which says that the square of the braiding acts as $\Delta(C^{-1})C \otimes C$, where $C$ is the quantum Casimir element. $C$ acts as the scalar $q^{(\lambda, \lambda + 2\rho)}$ in the irrep $V_{\lambda}$, and the statement about the eigenvalues of the braiding follows from this.

If you want the braiding to be unitary, and you want that statement to have some meaning in the category of representations of $U_q(\mathfrak{g})$, then you need a $*$-structure on the quantum group. There are three cases to consider: $q \in \mathbb{R}, q \in i \mathbb{R}$ and $\mathfrak{g}=\mathfrak{sp}(2n,\mathbb{C})$, and $|q|=1$.

The first two cases are ruled out because the eigenvalues of the braiding will not have modulus one, and hence the braiding cannot be unitary. This leaves only the case $|q|=1$.

If you want this inner product to be natural/compatible with the quantum group infrastructure, then it should probably be invariant under the action of $U_q(\mathfrak{g})$ in the sense that $\langle av, w \rangle = \langle v, a^*w\rangle$ for $a \in U_q(\mathfrak{g})$ and $v,w \in V \otimes V$. Of course a $*$-structure on $U_q(\mathfrak{g})$ needs to be specified in order to make sense of this. For $|q|=1$ the standard choice is $K_i^* = K_i, E_i^*=-E_i, F_i^*=-F_i$, although this can also be modified by a diagram automorphism of $\mathfrak{g}$.

To find an invariant inner product on $V \otimes V$, one way is to take the tensor product of an invariant inner product on $V$ with itself (if $V$ is an irrep then this is unique up to a scalar multiple if it exists). However, after a brief glance in Chari-Pressley, I'm not sure if the finite-dimensional irreps have an invariant inner product for the case $|q|=1$. I didn't do any further checking, but I would be interested to know if there are invariant inner products for irreps in that case.

Also, this doesn't rule out the possibility of inner products on $V \otimes V$ that do not come from inner products on $V$.

Summary of answer: the $R$-matrix cannot be made unitary in any inner product for $q \in \mathbb{R}$ or $q \in i \mathbb{R}$. For $|q|=1$ the situation is unclear to me.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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