## Conjecture: the quantum link invariants from U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) are “the same”

I’m working on the idea that the quantum link invariants from U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) are the same up to a change of variable. I havn’t gone through all the details yet, but the essential ideas are as follows.

(i) there are representations of the Birman-Wenzl-Murakami (BWM) algebra in the centralisers of repeated tensor powers of the (2n+1)-dimensional irreducible representations of both these quantum algebras

(ii) the quantum link invariants from U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) essentially come from applying the trace functional to elements in BWM(-q^{2n},q) and BWM(q^{2n},-q)

(iii) the truncated Bratteli diagrams for both these BWM algebras quotiented out by the ideal J={x where tr(yx)=0 for all elements y} are the same

(iv) say that a braid is represented by an element X \in BWM(r,q) for indeterminates r and q

(v) the quantum link invariants from U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) essentially come from substituting the numbers (-q^{2n},q), and (q^{2n},-q) for the indeterminates r and q in the trace of X \in BWM(r,q)

(vi) X \in BWM(r,q) can be written as
X = \sum_{\lambda} \sum_{S(lambda), T(lambda)} c_{S(lambda), T(lambda)} e_{S(lambda), T(lambda)}
where \lambda is a Young diagram (a vertex) in the Bratteli diagram for BWM(r,q),
S(lambda) and T(lambda) are paths in the Bratteli diagram that end in \lambda,
c_{S(lambda), T(lambda)} \in C(r,q), and
{e_{S(lambda), T(lambda)}} is a full set of matrix units.

(vii) tr(X) = \sum_{\lambda} \sum_{S(lambda), T(lambda)} c_{S(lambda), T(lambda)} tr(e_{S(lambda), T(lambda)})
= \sum_{\lambda} \sum_{S(lambda), S(lambda)} c_{S(lambda), S(lambda)} tr(e_{S(lambda), S(lambda)}) \in C(r,q)

(viii) X is an element of both BWM(-q^{2n},q) and BWM(q^{2n},-q) when you do the appropriate substitutions: note that X is well-defined in both cases

(ix) in both BWM(-q^{2n},q) and BWM(q^{2n},-q), tr(X) is just the expression in C(r,q) with the appropriate substitutions

(x) the sum done in calculating tr(X) is done only over those lambda in the truncated Bratteli diagram, as the traces of all other matrix units are zero (by definition)

(xi) as the Bratteli diagrams for BWM(-q^{2n},q) and BWM(q^{2n},-q) are the same, tr(X) in both cases is the same up to a change of variables.
I’ll write more about this – the key point is (iii). To see that (iii) is true, think about how the Weyl alcoves for U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) are the same, even at roots of unity.

### 10 Responses to Conjecture: the quantum link invariants from U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) are “the same”

1. Steve says:

Interesting… Actually can’t make head nor tail of it.

When are you back in Sydney?

2. Steve says:

Or rather I should say, when are you likley to be regulary on line again. The hit rate on my site has gone down.

3. Sacha says:

*laugh*

I’m in the Qantas Lounge in HK at the moment, back in Sydney tomorrow morning, and back online (at work) Wednesday…

4. Sacha says:

Wednesday’s too soon…

5. Sacha says:

I think that what I’ve written in the post is true – if it is, then the result is really just a trivial consequence of the equality of the truncated Bratteli diagrams of the relevant BWM algebras.

There is another interesting thing for people interested in the BWM algebras:
let Q_{\lambda}(r,q) be the ratio of polynomials in r and q given in Wenzl’s paper (Quantum groups and subfactors of type B, C and D) – a free pdf of this paper can be found via a dedicated search. For each idempotent matrix unit:
e_{lambda} \in BWM_{f}(r,q)
ending at the Young diagram lambda, the trace of e_{lambda} is:
tr(e_{lambda}) = Q_{lambda}(r,q)/x^{f} where
x = (r-r^{-1})/(q-q^{-1}) + 1.

A cool thing is that:
Q_{\lambda}(-q^{2n},q) = Q_{\lambda}(q^{2n},-q), and also that
sdim_{q}(V) = x(-q^{2n},q) = x(q^{2n},-q) = dim_{-q}(V)
where sdim_{q}(V) is the quantum superdimension of the U_{q}(osp(1|2n)) irrep V, which is a (2n+1)-dimensional irrep, and dim_{-q}(V) is the quantum dimension of V taken as a (2n+1)-irrep of U_{-q}(so(2n+1)).

Another cool thing is that Q_{lambda}(-q^{2n},q) is the quantum superdimension of the irreducible representation V_{lambda} of U_{q}(osp(1|2n)) with highest weight labelled by the Young diagram lambda, even if lambda has more than n rows of boxes. If lambda has more than n rows, it doesn’t immediately represent the highest weight of an irrep, instead, take the new Young diagram mu, given below, to be the Young digram representing the highest weight.

Let lambda’ be the number of boxes in the first column of lambda, then fix mu to be the same Young diagram as lambda except that its first column contains
(2n + 1) – lambda’ boxes.

6. Jason Soon says:

welcome back, Sacha. Can’t say I understand your latest post but welcome back anyway!

7. […] I’m writing a paper on my idea about how two quantum link invariants are really ”the same” that I outline here. Like to submit it for publication within the week. […]

8. Tel says:

Hey Sach,

Is there any special relationship between U_{q}(osp(1/2n)) and U_{-q}(so(2n+1)) or is this some sort of fortuitous accident ?

Tel

9. Sacha says:

Hey Tel,

It’s hard to say which – but the two algebras are isomorphic when you only look at the tensorial representations ie those reps whose weight spaces are integers (rather than half-integers) – I’m sure you’ll understand what I mean. ie, you have an isomorphic when you fix the elements of the cartan subalgebra (all the h_{i}’s) to have integer eigenvalues.

It may be a fortuitous accident. Interestingly, the Cartan matrices are the same. The algebras may be quite different when you look at other kinds of representations though, but I don’t know.

10. […] Last week I finished a draft of a small paper the idea for which came to me at ICM in Madrid in August, possibly while talking to Nathan Geer after my talk. Pretty soon afterwards, I wrote down the beginnings of the idea at this post. It’s taken a fair bit of work to get it to this draft stage. I’m happy to e-mail a pdf of the draft paper to whomever requests it. […]