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

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

2009-10-29T17:46:09Z

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 TL_d is the algebra of intertwiners of the d-fold tensor power of V.

TL_d = End_U(V⊗…⊗V)

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

In other words, TL_d is a quotient of CS_d. The kernel is easy to describe. First decompose the group algebra into its Wedderburn components, one matrix algebra for each irrep of S_d. 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 S_d to u_i = s_i + 1, where s_i = (i i+1), the structure coefficients in TL_d are all integers so that one can define a ℤ-form TL_d(ℤ) by these formulas.

TL_d = C ⊗ TL_d(ℤ)

As product of matrix algebras (as in the Wedderburn decomposition), TL_d 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(gl_1|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 STL_d.

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

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

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: U_q and the Hecke algebra, respectively. What about here? I am looking for a presentation of STL_d,q defined over ℤ[q,q^-1].

Answer by Ben Webster for Does the super Temperley-Lieb algebra have a Z-form?

2009-10-30T13:59:44Z

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 TL_d, and more generally to End_{U_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 STL_d 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 STL_d 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.

Answer by Bruce Westbury for Does the super Temperley-Lieb algebra have a Z-form?

2010-03-23T13:02:44Z

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.