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

It is well-known that representations of quantised enveloping algebras give representations of braid groups. For the examples that I know explicitly the representations of the three string braid group take a specific form. Is there an explanation of this? The examples I know are the simplest examples so what can I expect in general?

More specifically: Fix a quantised enveloping algebra $U$. Let $V$ and $W$ be highest weight finite dimensional representations. Then the three string braid group acts on $Hom_U(\otimes^3V,W)$.

The specific form that appears is the following. Let $P$ be the $n\times n$ matrix with $P_{ij}=1$ if $i+j=n+1$ and $P_{ij}=0$ otherwise. Then we can write $\sigma_1$ and $\sigma_2$ with the following properties

  • $\sigma_1$ is lower triangular
  • $\sigma_2=P\sigma_1P$
  • $\sigma_i^{-1}=\overline{\sigma}_i$ which means apply the involution $q\mapsto q^{-1}$ to each entry

The simplest example is $$\sigma_1=\left(\begin{array}{cc} q & 0 \\\ 1 & -q^{-1}\end{array}\right)$$

I get the feeling this has something to do with canonical bases.

A specific question is: Take $V$ to be the spin representation of $Spin(2n+1)$. Then do these representations have this form and if so how do I find it?

[In fact, I have representations of this specific form which I conjecture are these representations]

Further comment Assume the eigenvalues of $\sigma_i$ are distinct. This condition holds for the spin representation. Then if this basis exists it is unique. Consider a change of basis matrix $A$ which preserves this structure. Then $A$ commutes with $\sigma_1$ so is lower triangular. Then $A$ also commutes with $P$ so is diagonal. Then the final condition requires $A$ to be a scalar matrix.

The problem is existence. The Tuba-Wenzl paper shows such a basis exists in small examples.

share|improve this question
1  
A paper of Tuba and Wenzl may be relevant: pjm.math.berkeley.edu/pjm/2001/197-2/p11.xhtml –  Ian Agol Mar 17 '10 at 3:10

2 Answers 2

The R-matrix is always upper triangular, for any basis of the tensor product which is compatible with the weight spaces on the two factors; depending on your convention, the R-matrix only decreases weight in the first factor and increases it in the second.

The second element in your list is essentially by definition: $\sigma_1$ and $\sigma_2$ involve doing the exact same thing to the first two or last two factors, so they are conjugate by, say, the map that cyclically permutes the factors.

I believe the last condition is what's usually called "unitarity" of the R-matrix, but I should probably just wait for Noah to show up and give a better answer for that one with references and such.

share|improve this answer
    
Ben, Thanks for responding. I am not clear how to go from upper triangular on $\otimes^3 V$ to the irreducible representations. Scott does not claim his matrices will be triangular (does he?). The usual way of saying $\sigma_1$ is conjugate to $\sigma_2$ is to conjugate $\sigma_1$ by $\sigma_1\sigma_2$. Here we have $P=\sigma_1\sigma_2\sigma_1$. Moreover the matrix $P$ is specified. The last condition is not a big deal but is needed for uniqueness. If this is dropped we could conjugate by a diagonal matrix which commutes with $P$. –  Bruce Westbury Mar 16 '10 at 21:01
2  
If a matrix acts in an upper triangular fashion on the entire space V ot V ot V, does it not follow immediately that any operator acts in an upper triangular fashion on a subspace which it preserves? If V ot V ot V = S0 > S1> ... > Sn=0 are the subspaces defining the filtration by which R acts, it seems that letting X = Hom(W,V ot V ot V), we get X = X\cap S0 > .... X\cap Sn, which filtration is preserved by R. Here i blithely identified Hom(W,V ot V ot V) with Hom(V ot V ot V,W) which you wrote, but these are identified since we have a semisimple category. –  David Jordan Mar 16 '10 at 23:12
    
Also, for W irreducible Im regarding Hom(W,V) as a subspace of V, namely its image in V, the W-isotypic component. Also, i edit midstream. "any operator" should be replaced with "that matrix" above. –  David Jordan Mar 16 '10 at 23:14
    
Curiously, looking at my matrices (mentioned in my answer), $\sigma_1$ is actually diagonal for each irrep $W$. e.g., when I type in BraidingData[A1][Irrep[A1][{1}], 3], amongst other things it tell me that with $W$ also the $2$-dimensional representation, $\sigma_1 = \left( \begin{array}{cc} \frac{1}{q} & 0 \\ 0 & -\frac{1}{q^3} \end{array} \right)$ and $\sigma_2 = \left( \begin{array}{cc} -\frac{1}{q^3 \left(q^2+1\right)} & \frac{q^4+q^2+1}{q^2 \left(q^2+1\right)^2} \\ \frac{1}{q^2} & \frac{q}{q^2+1} \end{array} \right)$. –  Scott Morrison Mar 17 '10 at 1:25
    
It's been long enough since I wrote these programs that I forget what basis it's using, but presumably there's a good explanation along the lines of Ben's comment that "the R-matrix only decreases weight in the first factor and increases it in the second". –  Scott Morrison Mar 17 '10 at 1:26

If it would be useful for you, I can show you explicit matrices for the 3-strand braid group with V the spin representation of $Spin(5)$ and $Spin(7)$, and probably $Spin(9)$ and $Spin(11)$ as well. These won't be in any particular basis of $Hom(\otimes^3 V, W)$, however.

share|improve this answer
    
Scott, Thanks for your interest. The spin representation of $Spin(5)$ is also the vector representation of $Sp(4)$ which is understood. The case $Spin(7)$ I worked out and is in arxiv.org/abs/math/0411428 I know $Spin(9)$ will work because all representations of dimension at most 5 have this form. This can essentially be found in arxiv.org/abs/math/9912013 I am now trying to prove a general result so it is not clear what I might learn from the next case. Your wording is unfortunate. You must have a particular basis in order to have matrices - I know what you mean though. –  Bruce Westbury Mar 16 '10 at 20:51
    
Indeed, my comment about bases was just a proviso that my program which produces these uses an ad-hoc basis. –  Scott Morrison Mar 16 '10 at 21:04

Your Answer

 
discard

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.