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I've heard asserted in talks quite a few times that Lusztig's canonical basis for irreducible representations is known to not always have positive structure coefficents for the action of $E_i$ and $F_i$. There are good geometric reasons the coefficents have to be positive in simply-laced situations, but no such arguments can work for non-simply laced examples. However, this is quite a bit weaker than knowing the result is false.

Does anyone have a good example or reference for a situation where this positivity fails?

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5  
I do not know any reference in the literature nor who found a counter-example at the first time. But, here in Kyoto, everybody believes there are counter-examples. Don't ask me the detail. –  Hiraku Nakajima Sep 29 '10 at 11:31
    
In the related world of Hecke algebras with unequal parameters the "Kazhdan-Lusztig" basis can have structure constants that aren't nonnegative, which happens already for B_2 where you have parameters $q$ and $q^2$. I think this was noticed by Lusztig in the early 80s. (And I'd be stunned if he didn't have a similar example in the case of quantum groups, but I never asked him). –  Kevin McGerty Sep 30 '10 at 17:31
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2 Answers

up vote 19 down vote accepted

Hi,

The following formulas are examples of non-positive structure coefficients for non-symmetric cases which are easily verified by the algorithm presented in Leclerc's paper "Dual Canonical Bases, Quantum Shuffles, and q-characters" or quagroup package in GAP4.

Professor Masaki Kashiwara told me that he has known such non-positive structure coefficient for $G_2$ since Shigenori Yamane found it in 1994 as treated in his master thesis at Osaka University (written in Japanese). You can see similar negative coefficients in at least case $A_{2n}^{(2)}, D_{n+1}^{(2)}$. Anyway, conjecture 52 in Leclerc's paper is false (I already told Professor Leclerc about it).

Shunsuke Tsuchioka

Notation: $G(i_1,\cdots,i_n)$ stands for the canonical basis element corresponds to a crystal element $b(i_1,\cdots,i_n)=\tilde{f}_{i_n}b(i_1,\cdots,i_{n-1})=\cdots$.

$G_2$ (1 is the short root) : $f_2 G(121112211) = G(1211122211) + [2]G(1111222211) + G(2111112221)$

$ + [2]G(1211112221) + G(1111122221) - G(1112211122) + [2]G(1122111122)$

$C_3$ (1,2 are short roots) : $f_3 G(23122312) = [2]G(222333121) + [2]G(312222331) + [2]G(231222331)$

$ + [2]G(122223331) + G(231223312) + [2]G(122233312) - G(223112233) + [2]G(231122233)$

$B_4$ (1,2,3 are long roots) : $f_1G(4342341234) = [2]G(43344423211) + [2]G(43423443211) - G(44233443211)$

$ + [2]G(43423344211) + [2]G(43423442311) + [2]G(34234442311)$

$ + [2]G(43422334411) + G(43423412341)$

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I hope you don't mind that I've added some line breaks in the interest of clarity. Thanks for the answer, by the way. –  Ben Webster Sep 30 '10 at 8:36
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Ben,

I have heard the same thing, but I have never seen an example. After thinking about it a bit, I came up with the following 'heuristic' reason why the structure constants should be positive for half the Chevalley generators.

Assume that it is know that simple modules for affine quiver Hecke algebras have characters given by the dual canonical basis of $U_q({n})$ (many people (not including Nakajima) expected to be true, but in light of Tsuchioka's answer shouldn't!). References for what follows are Leclerc's paper "Dual Canonical Bases, Quantum Shuffles, and $q$-characters" (EDIT (BW) This is also on the arXiv) as well as the paper of Kleshchev and Ram, and my paper with Melvin and Mondragon.

We denote the canonical and dual canonical bases $b_g$ and $b_g^*$, respectively, where $g$ runs over an appropriate index set. These bases are related by the Kashiwara form $(\cdot,\cdot)_K:U_A(n)\times U_A^*(n)\to A$ via

$$ (b_g,b_h^*)_K=\delta_{gh} $$

(above, $A=\mathbb{Z}[q,q^{-1}]$ as usual). This form is defined so that $(1,1)_K=1$ and $(f_iu,v)_K=(u,f_i'v)_K$ and $f_i'$ is Kashiwara's $q$-derivation.

Now, on the level of modules, the $q$-derivation $f_i'$ corresponds to $i$-restriction. As we have assumed $b^*_g$ is the character of a simple module, we have the $f_i'\mathcal{b}^*_g$ is the character of some module, and hence a nonnegative linear combination of dual canonical basis vectors. So now we calculate \begin{align*} f_i\mathcal{b}_g=\sum_h(f_ib_g,b^*_h)_Kb_h =\sum_h(b_g,f_i'b^*_h)_Kb_h. \end{align*} But, as we have explained, $(b_g,f_i'b^*_h)$ is nonnegative.

As I said above, this argument only works for half the generators. I haven't internalized the results of your recent paper, so I'm not sure if you've defined the biadjoint functor $e_i$ or not. If you have, then probably there should be some more information to be teased out of this line of reasoning.

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3  
Could you exclude Nakajima from `everyone' in the 6th line, please ? –  Hiraku Nakajima Sep 30 '10 at 3:35
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I got it for you. –  Harry Gindi Sep 30 '10 at 7:54
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David- You, of course, sussed out my reason for asking the question in the first place; I've seen Aaron Lauda assert in talks that outside symmetric type the basis coming from categorifications can't be Lusztig's because it has positive structure coefficients and Lusztig's doesn't. In symmetric type, it's known that you get the right thing for the enveloping algebra by Vasserot and Varagnolo's paper (arXiv:0901.3992). For reducing to cyclotomic quotients, you might have to use the stuff in my papers (which does include biadjointness of the E_i and F_i functors). –  Ben Webster Sep 30 '10 at 8:32
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Harry - Thank you. After Tsuchioka's answer, it is probably better to change everyone' to nobody'. –  Hiraku Nakajima Sep 30 '10 at 11:43
2  
Okay, I changed it. –  David Hill Sep 30 '10 at 16:18
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