Does the super Temperley-Lieb algebra have a Z-form? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:15:51Z http://mathoverflow.net/feeds/question/3299 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3299/does-the-super-temperley-lieb-algebra-have-a-z-form Does the super Temperley-Lieb algebra have a Z-form? Sammy Black 2009-10-29T17:46:09Z 2010-03-28T18:34:45Z <p><strong>Background</strong> Let V denote the standard (2-dimensional) module for the Lie algebra sl<sub>2</sub>(C), or equivalently for the universal envelope U = U(sl<sub>2</sub>(C)). The Temperley-Lieb algebra TL<sub>d</sub> is the algebra of intertwiners of the d-fold tensor power of V.</p> <blockquote> <p>TL<sub>d</sub> = End<sub>U</sub>(V&otimes;&hellip;&otimes;V)</p> </blockquote> <p>Now, let the symmetric group, and hence its group algebra CS<sub>d</sub>, act on the right of V&otimes;&hellip;&otimes;V by permuting tensor factors. According to Schur-Weyl duality, V&otimes;&hellip;&otimes;V is a (U,CS<sub>d</sub>)-bimodule, with the image of each algebra inside End<sub>C</sub>(V&otimes;&hellip;&otimes;V) being the centralizer of the other.</p> <p>In other words, TL<sub>d</sub> is a quotient of CS<sub>d</sub>. The kernel is easy to describe. First decompose the group algebra into its Wedderburn components, one matrix algebra for each irrep of S<sub>d</sub>. 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.</p> <p>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<sub>d</sub> to u<sub>i</sub> = s<sub>i</sub> + 1, where s<sub>i</sub> = (i i+1), the structure coefficients in TL<sub>d</sub> are all integers so that one can define a &#8484;-form TL<sub>d</sub>(&#8484;) by these formulas.</p> <blockquote> <p>TL<sub>d</sub> = C &otimes; TL<sub>d</sub>(&#8484;)</p> </blockquote> <p>As product of matrix algebras (as in the Wedderburn decomposition), TL<sub>d</sub> has a &#8484;-form, as well: namely, matrices of the same dimensions over &#8484;. 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.</p> <p><hr /></p> <p>There is a super-analog of this whole story. Let U = U(gl<sub>1|1</sub>(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<sub>d</sub>.</p> <p>Over the complex numbers, STL<sub>d</sub> is a product of matrix algebras corresponding to the irreps of S<sub>d</sub> indexed by hook partitions. Young diagrams are confined to one row and one column (super-row!). In that sense, STL<sub>d</sub> is understood. However, idempotents involved in projecting onto these Wedderburn components are nasty things that cannot be defined over &#8484;</p> <p><hr /></p> <p><strong>Question 1:</strong> Does STL<sub>d</sub> have a &#8484;-form that is compatible with the standard basis for CS<sub>d</sub>?</p> <p><strong>Question 2:</strong> I am pessimistic about Q1; hence, the follow up: why not? I suspect that this has something to do with cellularity.</p> <p><strong>Question 3:</strong> I care about q-deformations of everything mentioned: U<sub>q</sub> and the Hecke algebra, respectively. What about here? I am looking for a presentation of STL<sub>d,q</sub> defined over &#8484;[q,q<sup>-1</sup>].</p> http://mathoverflow.net/questions/3299/does-the-super-temperley-lieb-algebra-have-a-z-form/3442#3442 Answer by Ben Webster for Does the super Temperley-Lieb algebra have a Z-form? Ben Webster 2009-10-30T13:59:44Z 2010-03-28T18:34:45Z <p>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.</p> <p>What you could lose when you take quotients is positivity (which I presume is one of things you are after). The Hecke algebra of <code>S_n</code> 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.</p> <p>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."</p> <p>The kernel of the map to TL<sub>d</sub>, and more generally to End<sub><code>U_q(sl_n)</code></sub>(V<sup>&otimes;d</sup>) 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. </p> <p>However, the kernel of the map to STL<sub>d</sub> is <strong>not</strong> cellular. In particular, every cellular ideal contains the alternating representation, so any quotient where the alternating representation survives is not cellular. So, while STL<sub>d</sub> inherits a perfectly good Z-form, it doesn't inherit any particular basis from the Hecke algebra.</p> <p>I'm genuinely unsure if this is really a problem from your standpoint. I mean, the representation V<sup>&otimes;d</sup> 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.</p> <p>Anyways, if people want to have a real discussion about this, I suggest we retire to the nLab. I've started a <a href="http://ncatlab.org/nlab/show/super+q-Schur+algebra" rel="nofollow">relevant page</a> there.</p> http://mathoverflow.net/questions/3299/does-the-super-temperley-lieb-algebra-have-a-z-form/19106#19106 Answer by Bruce Westbury for Does the super Temperley-Lieb algebra have a Z-form? Bruce Westbury 2010-03-23T13:02:44Z 2010-03-23T21:09:26Z <p>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 $\delta\mapsto q+q^{-1}$.</p> <p>The generators are $u_1,\ldots ,u_{n-1}$ (I have $n$ where OP has $d$). Then defining relations are<br> $$u_i^2=\delta u_i$$ $$u_iu_j=u_ju_i\qquad\text{if |i-j|>1}$$ $$u_iu_{i+1}u_i-u_i=u_{i+1}u_iu_{i+1}-u_{i+1}$$ $$u_{i-1}u_{i+1}u_i(\delta-u_{i-1})(\delta-u_{i+1})=0$$ $$(\delta-u_{i-1})(\delta-u_{i+1})u_iu_{i-1}u_{i+1}=0$$</p> <p>Does this count?</p> <p>If you want to move this to nLab that's fine by me.</p>