# Does the super Temperley-Lieb algebra have a Z-form?

Background Let V denote the standard (2-dimensional) module for the Lie algebra sl2(C), or equivalently for the universal envelope U = U(sl2(C)). The Temperley-Lieb algebra TLd is the algebra of intertwiners of the d-fold tensor power of V.

TLd = EndU(V⊗…⊗V)

Now, let the symmetric group, and hence its group algebra CSd, act on the right of V⊗…⊗V by permuting tensor factors. According to Schur-Weyl duality, V⊗…⊗V is a (U,CSd)-bimodule, with the image of each algebra inside EndC(V⊗…⊗V) being the centralizer of the other.

In other words, TLd is a quotient of CSd. The kernel is easy to describe. First decompose the group algebra into its Wedderburn components, one matrix algebra for each irrep of Sd. These are in bijection with partitions of d, which we should picture as Young diagrams. The representation is faithful on any component indexed by a diagram with at most 2 rows and it annihilates all other components.

So far, I have deliberately avoided the description of the Temperley-Lieb algebra as a diagram algebra in the sense that Kauffman describes it. Here's the rub: by changing variables in Sd to ui = si + 1, where si = (i i+1), the structure coefficients in TLd are all integers so that one can define a ℤ-form TLd(ℤ) by these formulas.

TLd = C ⊗ TLd(ℤ)

As product of matrix algebras (as in the Wedderburn decomposition), TLd has a ℤ-form, as well: namely, matrices of the same dimensions over ℤ. These two rings are very different, the latter being rather trivial from the point of view of knot theory. They only become isomorphic after a base change to C.

There is a super-analog of this whole story. Let U = U(gl1|1(C)), let V be the standard (1|1)-dimensional module, and let the symmetric group act by signed permutations (when two odd vectors cross, a sign pops up). An analogous Schur-Weyl duality statement holds, and so, by analogy, I call the algebra of intertwiners the super-Temperley-Lieb algebra, or STLd.

Over the complex numbers, STLd is a product of matrix algebras corresponding to the irreps of Sd indexed by hook partitions. Young diagrams are confined to one row and one column (super-row!). In that sense, STLd is understood. However, idempotents involved in projecting onto these Wedderburn components are nasty things that cannot be defined over ℤ

Question 1: Does STLd have a ℤ-form that is compatible with the standard basis for CSd?

Question 2: I am pessimistic about Q1; hence, the follow up: why not? I suspect that this has something to do with cellularity.

Question 3: I care about q-deformations of everything mentioned: Uq and the Hecke algebra, respectively. What about here? I am looking for a presentation of STLd,q defined over ℤ[q,q-1].

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Wow, great questions! –  Scott Morrison Oct 29 '09 at 19:41
Though boy did it ever confuse me that you're using d here to denote the number of strings instead of the value of the circle! –  Noah Snyder Oct 30 '09 at 1:28
An important difference between these two cases. For sl_2 the defining representation is self-dual, so the TL algebra captures the full representation theory. For gl(1|1), on the other hand, you're only seeing some of the representation theory. In particular, you only get to see the semisimple part of the representation theory. –  Noah Snyder Oct 30 '09 at 1:32
The question is great --- if I could vote it up more than once, I would. I have no useful answers, though. But your question suggests related ones, which I've posted at [mathoverflow.net/questions/3366/]. –  Theo Johnson-Freyd Oct 30 '09 at 1:36
@Noah: Sorry for the confusion. I think I choose a different letter to index these things every time I talk about them. I like to use m and n for things like gl(m|n). Perhaps, d is poor choice, since $\delta$ is so often the loop value, as you say... –  Sammy Black Oct 30 '09 at 4:31

It depends what you mean by "compatible." For any Z-form of a finite-dimensional C-algebra, there's a canonical Z-form for any quotient just given by the image (the image is a finitely generated abelian subgroup, and thus a lattice). I'll note that the integral form Bruce suggests below is precisely the one induced this way by the Kazhdan-Lusztig basis, since his presentation is the presentation of the Hecke algebra via the K-L basis vectors for reflections, with the additional relations.

What you could lose when you take quotients is positivity (which I presume is one of things you are after). The Hecke algebra of S_n has a basis so nice I would call it "canonical" but usually called Kazhdan-Lusztig. This basis has a a very strong positivity property (its structure coefficients are Laurent polynomials with positive integer coefficients). I would argue that this is the structure you are interested in preserving in the quotient.

If you want a basis of an algebra to descend a quotient, you'd better hope that the intersection of the basis with the kernel is a basis of the kernel (so that the image of the basis is a basis and a bunch of 0's). An ideal in the Hecke algebra which has a basis given by a subset of the KL basis is called "cellular."

The kernel of the map to TLd, and more generally to EndU_q(sl_n)(V⊗d) for any n and d, is cellular. Basically, this is because the parititions corresponding to killed representations form an upper order ideal in the dominance poset of partitions.

However, the kernel of the map to STLd is not cellular. In particular, every cellular ideal contains the alternating representation, so any quotient where the alternating representation survives is not cellular. So, while STLd inherits a perfectly good Z-form, it doesn't inherit any particular basis from the Hecke algebra.

I'm genuinely unsure if this is really a problem from your standpoint. I mean, the representation V⊗d still has a basis on which the image of any positive integral linear combination of KL basis vectors acts with positive integral coefficients. However, I don't think this guarantees any kind of positivity of structure coefficients. Also, Stroppel and Mazorchuk have a categorification of the Artin-Wedderburn basis of S_n, so maybe it's not as bad as you thought.

Anyways, if people want to have a real discussion about this, I suggest we retire to the nLab. I've started a relevant page there.

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Minor point: both here and on the nLab page, you refer to the algebra $STL_d$ as the q-Schur algebra, but I believe that they are centralizers of one another. Either way, thanks. Your comments are helpful. –  Sammy Black Mar 23 '10 at 23:08
You're right. I'll fix that. –  Ben Webster Mar 24 '10 at 2:26

I would define this algebra by a presentation. This algebra is over $\mathbb{Z}[\delta]$ but you can specialise to $\mathbb{Z}[q,q^{-1}]$ by taking $\delta\mapsto q+q^{-1}$.

The generators are $u_1,\ldots ,u_{n-1}$ (I have $n$ where OP has $d$). Then defining relations are
$$u_i^2=\delta u_i$$ $$u_iu_j=u_ju_i\qquad\text{if |i-j|>1}$$ $$u_iu_{i+1}u_i-u_i=u_{i+1}u_iu_{i+1}-u_{i+1}$$ $$u_{i-1}u_{i+1}u_i(\delta-u_{i-1})(\delta-u_{i+1})=0$$ $$(\delta-u_{i-1})(\delta-u_{i+1})u_iu_{i-1}u_{i+1}=0$$

Does this count?

If you want to move this to nLab that's fine by me.

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I don't understand your final comment. When $n=4$, the algebra is 20-dimensional, but only 2 transpositions in $S_4$ have the transposition $(i−1, i+2) = (1 \;4)$, namely $(4,2,3,1)$ and $(4,3,2,1)$ in one-line notation. –  Sammy Black Mar 23 '10 at 20:07
I have withdrawn the final comment. (I meant something different but that's irrelevant as it was mistaken). –  Bruce Westbury Mar 23 '10 at 21:06
Bruce- the KL basis vectors corresponding to hook partitions under RS definitely project to a basis. Unfortunately, this basis won't have the kind of positivity one gets from the KL basis. –  Ben Webster Mar 23 '10 at 21:12
It is not clear to me whether other (elementary) bases also have this property. If you insist on positivity (presumably because your ambition is to categorify this algebra) then is there a Kazhdan-Lusztig basis? –  Bruce Westbury Mar 23 '10 at 22:02
I would be very surprised if there were anything positive. After some discussions last week, I have some ideas about how categorifying this story works, and it definitely has differentials in it. –  Ben Webster Mar 23 '10 at 22:16