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There's a non-unital algebra $\dot{U}$ formed from $U_q (sl_2)$ by including a system of mutually orthogonal idempotents $1_n$, indexed by the weight lattice. You can think of this as a category with objects $\mathbb{Z}$ if you prefer.

Lusztig's basis $\mathbb{\dot{B}}$ for $\dot{U}$ has nice positivity properties: structure coefficients are in $\mathbb{Z}[q,q^{-1}]$.

Has anyone tried to write down a similar type of basis for the algebra associated to $U_q (gl_{1|1})$?

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  • $\begingroup$ Hi Sammy (!!). Can you give a reference for the U(sl_2) idempotent algebra you are referring to in the question? $\endgroup$
    – David Jordan
    Commented Nov 9, 2009 at 22:05
  • $\begingroup$ I've removed the \mathfrak's since they weren't rendering properly. I'll look into this problem. $\endgroup$
    – Anton Geraschenko
    Commented Nov 9, 2009 at 23:19
  • $\begingroup$ One nice description is by Aaron Lauda: arXiv:0803.3652 $\endgroup$
    – Sammy Black
    Commented Nov 9, 2009 at 23:27
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    $\begingroup$ I don't feel sure enough to give an answer, but I'm not optimistic. Certainly Lusztig's original construction doesn't work, and I don't think crystal theory does either, and those are the usual "avatars" of the canonical basis. $\endgroup$
    – Ben Webster ♦
    Commented Nov 9, 2009 at 23:43
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    $\begingroup$ Actually, the structure constants of the canonical basis are in $\mathbb{N}[v,v^{-1}]$ which is explained by the fact that there exists an abelian categorification. If one has a categorification of one half of gl(1,1) using triangulated categories (which seems likely after the recent paper of Khovanov linked in David Hill's answer below) then one would get a basis like the one you are asking for. (Moral: abelian implies $\mathbb{N}$, triangulated implies $\mathbb{Z}$!) $\endgroup$
    – Geordie Williamson
    Commented Sep 4, 2010 at 7:24

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Kashiwara has developed some crystal theoretic methods for the Lie superalgebra $\mathfrak{q}(n)$. However, I think you should look at Khovanov, to get an idea of what it should look like.

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