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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.

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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.

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  • $\begingroup$ 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 $P$. $\endgroup$
    – Bruce Westbury
    Mar 16, 2010 at 21:01
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    $\begingroup$ 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. $\endgroup$
    – David Jordan
    Mar 16, 2010 at 23:12
  • $\begingroup$ 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. $\endgroup$
    – David Jordan
    Mar 16, 2010 at 23:14
  • $\begingroup$ 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 $\sigma_2 = \left( \begin{array}{cc} -\frac{1}{q^3 \left(q^2+1\right)} & \frac{q^4+q^2+1}{q^2 \left(q^2+1\right)^2} \\ \frac{1}{q^2} & \frac{q}{q^2+1} \end{array} \right)$. $\endgroup$
    – Scott Morrison
    Mar 17, 2010 at 1:25
  • $\begingroup$ 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". $\endgroup$
    – Scott Morrison
    Mar 17, 2010 at 1:26
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Bruce Westbury
    Mar 16, 2010 at 20:51
  • $\begingroup$ Indeed, my comment about bases was just a proviso that my program which produces these uses an ad-hoc basis. $\endgroup$
    – Scott Morrison
    Mar 16, 2010 at 21:04

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