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Hi,

In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$ where $I={1,2}$ 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 the set 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 $p\geq 2$.

show/hide this revision's text 4 added 20 characters in body; added 13 characters in body

Hi,

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 the set $B(\infty)$ 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 $G^{*}(b)$ 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 ${G^*}(0,0,1,0,0,3)$ 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_2{G^{}}(0,0,1,0,0,3) e_2G^{\ast}(0,0,1,0,0,3) ={G^{}}(1,0,1,0,0,3) G^{\ast}(1,0,1,0,0,3) +q^{-3}{G^{}}(0,3,0,0,0,3) q^{-3}G^{\ast}(0,3,0,0,0,3) -q^{-3}{G^{}}(0,2,0,1,0,2) q^{-3}G^{\ast}(0,2,0,1,0,2) +q^{-6}{G^{}}(0,0,1,0,1,0). q^{-6}G^{\ast}(0,0,1,0,1,0). \end{align}end{align*}

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

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

show/hide this revision's text 3 I'm sorry I have some texnical problem. I hope you can guess their display.; added 22 characters in body

Hi,

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 the set $B(\infty)$ 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 $G^{*}(b)$ 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 $G^*(0,0,1,0,0,3)$ {G^*}(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^e_2{G^{}(0,0,1,0,0,3) }(0,0,1,0,0,3) =G^{{G^{}(1,0,1,0,0,3) }(1,0,1,0,0,3) +q^{-3}G^{q^{-3}{G^{}(0,3,0,0,0,3) }(0,3,0,0,0,3) -q^{-3}G^{q^{-3}{G^{}(0,2,0,1,0,2) }(0,2,0,1,0,2) +q^{-6}G^{q^{-6}{G^{}(0,0,1,0,1,0). }(0,0,1,0,1,0). \end{align}

I checked that

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

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

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