MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question implicitly asked in Ben Webster's question is: Does the canonical basis of Uq(n+) agree with the basis coming from categorification via Khovanov-Lauda-Rouqier algebras?

Thanks to Shunsuke Tsuchioka's answer to the very same question, the answer is No.

One can now ask for more, and ask for an explicit example where these two bases disagree. So that's what I am going to do here.

In case someone brings it up, I'm disqualifying Example 3.25 of Khovanov-Lauda I on the grounds that it doesn't count (the algebra isn't quite defined "correctly" in characteristic zero (it doesn't match the geometry), and weird stuff is to be expected outside of finite type for KLR algebras in positive characteristic). I'd prefer to work in finite type please.

Of course Tsuchioka's answer provides us with an upper bound on where to look.

share|cite|improve this question
I'll just note, for the record: I don't know. I have some ideas about how to go about looking, but it's actually not so easy to do the calculations. – Ben Webster Mar 23 '12 at 1:23


In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$ where $I=\{1,2\}$ and $\alpha_2$ is the long root.

We choose a reduced expression for the longest element $w_0$ of the Weyl group $W(G_2)$ as $w_0=s_2s_1s_2s_1s_2s_1$ and thus identify $\mathbb{N}^{6}$ and Kashiwara's crystal $B(\infty)$ via Lusztig's parametrization, \begin{align*} \mathbb{N}^6 \stackrel{\sim}{\longrightarrow} B(\infty)=L(\infty)/q^{-1}L(\infty), (a,b,c,d,e,f)\longmapsto e_2^{(a)}e_{12}^{(b)}e_{11122}^{(c)}e_{112}^{(d)}e_{1112}^{(e)}e_1^{(f)}+{q^{-1}L(\infty)} \end{align*} Here $e_{*}$ are suitable elements defined by Lusztig's braid group symmetry which we don't repeat it here.

For each $b\in B(\infty)$, we denote by $G^{\ast}(b)$ the corresponding dual canonical basis element.

Let $\mathcal{H}_n$ be Khovanov-Lauda-Rouquier algebra of type $G_2$ over $\mathbb{Q}$. You can see that $G^{\ast}(0,0,1,0,0,3)$ corresponds to an irreducible representation of dimension 168 of $\mathcal{H}_8$.

Note that we have the negative occurrence \begin{align*} e_2G^{\ast}(0,0,1,0,0,3) =G^{\ast}(1,0,1,0,0,3) +q^{-3}G^{\ast}(0,3,0,0,0,3) -q^{-3}G^{\ast}(0,2,0,1,0,2) +q^{-6}G^{\ast}(0,0,1,0,1,0). \end{align*}

I checked that \begin{align*} G^{\ast}(1,0,1,0,0,3), G^{\ast}(0,3,0,0,0,3)-G^{\ast}(0,2,0,1,0,2), G^{\ast}(0,2,0,1,0,2), G^{\ast}(0,0,1,0,1,0) \end{align*} correspond to irreducible representations of dimension 168,1176,168,168 of $\mathcal{H}_9$ respectively.

Thus, the irreducible $\mathcal{H}_9$-module $V$ whose character in the quantum Shuffle is given by $G^{\ast}(0,3,0,0,0,3)-G^{\ast}(0,2,0,1,0,2)$ is an example you ask. $V$ is realizable over $\mathbb{Z}$ and its irreducibility is always preserved under the modulo-$p$ reduction for every prime $p\geq 2$.

share|cite|improve this answer
Excellent! Very nice. – Ben Webster Mar 23 '12 at 14:33

Since this question was first asked, nontrivial decomposition numbers for Quiver Hecke algebras (=KLR algebras) have been discovered in finite type. Thus there is an alternative family of examples where the canonical and KLR basis disagree, if one is willing to accept ones algebras living over a field of positive characteristic.

Known examples are not always explicated in the literature right now. The historically first example is due to Geordie Williamson - see or

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.