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5
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edited Mar 23 2012 at 2:17
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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 $p\geq 2$.
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4
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edited Mar 23 2012 at 2:04
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Hi,
In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$
where $I={1,2}$ 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 $p\geq 2$.
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3
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edited Mar 23 2012 at 1:58
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Hi,
In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$
where $I={1,2}$ 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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2
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edited Mar 23 2012 at 1:52
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Hi,
In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$
where $I={1,2}$ 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}.
$\mathbb{Q}$.
You can see that $G^*(0,0,1,0,0,3)$ corresponds to an irreducible representation of dimension 168 of $\mathcal{H}_8$. You can also see 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)
corresponds to an irreducible representations of dimension 168,1176,168,168 of $\mathcal{H}_9$.
Note that we have the negative occurrence
\begin{align*}
e_2G^{}(0,0,1,0,0,3)
=G^{G^{}(1,0,1,0,0,3)
+q^{-3}G^{}(0,3,0,0,0,3)
-q^{-3}G^{}(0,2,0,1,0,2)
+q^{-6}G^{}(0,0,1,0,1,0).
\end{align}
I checked that there exists an irreducible $\mathcal{H}_{10}$-module $V$ which
corresponds to $G^{}(1,3,0,0,0,3)-G^{}(1,2,0,1,0,2)$. Note that
\begin{align*}
e_2G^G^{}(0,2,0,1,0,2) = (1,0,1,0,0,3),
G^{}(1,2,0,1,0,2),
e_2G^(0,3,0,0,0,3)-G^{}(0,2,0,1,0,2),
G^{}(0,3,0,0,0,3) = (0,2,0,1,0,2),
G^{}(1,3,0,0,0,3).
(0,0,1,0,1,0)
\end{align*}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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1
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answered Mar 23 2012 at 1:43
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Hi,
In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$
where $I={1,2}$ 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 $G^*(0,0,1,0,0,3)$ corresponds to an irreducible representation of dimension 168 of $\mathcal{H}_8$. You can also see 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)
corresponds to an irreducible representations of dimension 168,1176,168,168 of $\mathcal{H}_9$.
Note that we have
\begin{align*}
e_2G^{}(0,0,1,0,0,3)
= G^{}(1,0,1,0,0,3)
+q^{-3}G^{}(0,3,0,0,0,3)
-q^{-3}G^{}(0,2,0,1,0,2)
+q^{-6}G^{}(0,0,1,0,1,0).
\end{align}
I checked that there exists an irreducible $\mathcal{H}_{10}$-module $V$ which
corresponds to $G^{}(1,3,0,0,0,3)-G^{}(1,2,0,1,0,2)$. Note that
\begin{align*}
e_2G^{}(0,2,0,1,0,2) = G^{}(1,2,0,1,0,2),
e_2G^{}(0,3,0,0,0,3) = G^{}(1,3,0,0,0,3).
\end{align*}
Thus, $V$ 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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