Mixing
Comparing the relative and absolute Bruhat decompositions for quasi-split reductive groups
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Let $G$ be a connected, reductive group over a field $k$. Assume $G$ is quasi-split. Let $A_0$ be a maximal split torus of $G$ with centralizer $T$, and let $B$ be a minimal parabolic (Borel) subgroup of $G$ containing $T$. Let $W = N_G(T)/T$ be the absolute Weyl group, and $W_0 = N_G(A_0)/T$. Each element of $W_0$ has a $k$-rational representative, and in fact we have $$W_0 = W(k) = N_G(A_0)(k)/A_0(k) = N_{G(k)}(A_0(k))/A_0(k)$$
There are two Bruhat decompositions of $G$. First, we have
$$G(overline{k}) = coprodlimits_{w in W} B(overline{k})wB(overline{k})$$
and second,
$$G(k) = coprodlimits_{w in W_0} B(k)wB(k)$$
where each representative $w in W_0$ is chosen to be $k$-rational. I'm pretty sure that $B(k)$ is Zariski dense in $B(overline{k})$, and so $overline{B(k)wB(k)} = overline{B(overline{k})wB(overline{k})}$. Also, it is generally true that $G(k)$ is Zariski dense in $G(overline{k})$, so we have
$$coprodlimits_{w in W} B(overline{k})wB(overline{k}) = G(overline{k}) = overline{G(k)} = bigcuplimits_{w in W_0} overline{B(overline{k})wB(overline{k})} tag{1}$$
Now, each $overline{B(overline{k})wB(overline{k})}$ is a union of the cells $B(overline{k})w' B(overline{k})$, where $w' in W$ can be obtained from $w$ by taking a reduced decomposition of $w$ (in the Coxeter system corresponding to $B$ and $T$), and removing generators.
So, where does the long element of $W$ appear on the right hand side? In general, I think the long element of $W$ need not lie in $W_0$, and it seems in this case that the length of every double coset in (1) is strictly less than the long element of $W$.
algebraic-groups reductive-groups
asked Jan 17 at 1:14
D_SD_S
13.6k61552
$endgroup$
add a comment |
$begingroup$
Let $G$ be a connected, reductive group over a field $k$. Assume $G$ is quasi-split. Let $A_0$ be a maximal split torus of $G$ with centralizer $T$, and let $B$ be a minimal parabolic (Borel) subgroup of $G$ containing $T$. Let $W = N_G(T)/T$ be the absolute Weyl group, and $W_0 = N_G(A_0)/T$. Each element of $W_0$ has a $k$-rational representative, and in fact we have $$W_0 = W(k) = N_G(A_0)(k)/A_0(k) = N_{G(k)}(A_0(k))/A_0(k)$$
There are two Bruhat decompositions of $G$. First, we have
$$G(overline{k}) = coprodlimits_{w in W} B(overline{k})wB(overline{k})$$
and second,
$$G(k) = coprodlimits_{w in W_0} B(k)wB(k)$$
where each representative $w in W_0$ is chosen to be $k$-rational. I'm pretty sure that $B(k)$ is Zariski dense in $B(overline{k})$, and so $overline{B(k)wB(k)} = overline{B(overline{k})wB(overline{k})}$. Also, it is generally true that $G(k)$ is Zariski dense in $G(overline{k})$, so we have
$$coprodlimits_{w in W} B(overline{k})wB(overline{k}) = G(overline{k}) = overline{G(k)} = bigcuplimits_{w in W_0} overline{B(overline{k})wB(overline{k})} tag{1}$$
Now, each $overline{B(overline{k})wB(overline{k})}$ is a union of the cells $B(overline{k})w' B(overline{k})$, where $w' in W$ can be obtained from $w$ by taking a reduced decomposition of $w$ (in the Coxeter system corresponding to $B$ and $T$), and removing generators.
So, where does the long element of $W$ appear on the right hand side? In general, I think the long element of $W$ need not lie in $W_0$, and it seems in this case that the length of every double coset in (1) is strictly less than the long element of $W$.
algebraic-groups reductive-groups
asked Jan 17 at 1:14
D_SD_S
13.6k61552
$endgroup$
$begingroup$
You have it backwards. $wleq w'$ in Bruhat order, not $w'leq w$.
$endgroup$
– Matt Samuel
Jan 17 at 1:33
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I didn't mention the Bruhat order, what do you mean?
$endgroup$
– D_S
Jan 17 at 2:38
$begingroup$
Not explicitly. "Take a reduced decomposition and remove generators." That's a characterization of Bruhat order.
$endgroup$
– Matt Samuel
Jan 17 at 2:39
$begingroup$
I think what I said was right. If for example $w$ is the long element, then every other element $w'$ of $W$ can be obtained by taking a reduced decomposition of $w$ and removing generators. And in turn, $Bw'B$ is contained in the closure of $BwB$.
$endgroup$
– D_S
Jan 17 at 2:52
add a comment |
$begingroup$
Let $G$ be a connected, reductive group over a field $k$. Assume $G$ is quasi-split. Let $A_0$ be a maximal split torus of $G$ with centralizer $T$, and let $B$ be a minimal parabolic (Borel) subgroup of $G$ containing $T$. Let $W = N_G(T)/T$ be the absolute Weyl group, and $W_0 = N_G(A_0)/T$. Each element of $W_0$ has a $k$-rational representative, and in fact we have $$W_0 = W(k) = N_G(A_0)(k)/A_0(k) = N_{G(k)}(A_0(k))/A_0(k)$$
There are two Bruhat decompositions of $G$. First, we have
$$G(overline{k}) = coprodlimits_{w in W} B(overline{k})wB(overline{k})$$
and second,
$$G(k) = coprodlimits_{w in W_0} B(k)wB(k)$$
where each representative $w in W_0$ is chosen to be $k$-rational. I'm pretty sure that $B(k)$ is Zariski dense in $B(overline{k})$, and so $overline{B(k)wB(k)} = overline{B(overline{k})wB(overline{k})}$. Also, it is generally true that $G(k)$ is Zariski dense in $G(overline{k})$, so we have
$$coprodlimits_{w in W} B(overline{k})wB(overline{k}) = G(overline{k}) = overline{G(k)} = bigcuplimits_{w in W_0} overline{B(overline{k})wB(overline{k})} tag{1}$$
Now, each $overline{B(overline{k})wB(overline{k})}$ is a union of the cells $B(overline{k})w' B(overline{k})$, where $w' in W$ can be obtained from $w$ by taking a reduced decomposition of $w$ (in the Coxeter system corresponding to $B$ and $T$), and removing generators.
So, where does the long element of $W$ appear on the right hand side? In general, I think the long element of $W$ need not lie in $W_0$, and it seems in this case that the length of every double coset in (1) is strictly less than the long element of $W$.
algebraic-groups reductive-groups
asked Jan 17 at 1:14
D_SD_S
13.6k61552
$endgroup$
Let $G$ be a connected, reductive group over a field $k$. Assume $G$ is quasi-split. Let $A_0$ be a maximal split torus of $G$ with centralizer $T$, and let $B$ be a minimal parabolic (Borel) subgroup of $G$ containing $T$. Let $W = N_G(T)/T$ be the absolute Weyl group, and $W_0 = N_G(A_0)/T$. Each element of $W_0$ has a $k$-rational representative, and in fact we have $$W_0 = W(k) = N_G(A_0)(k)/A_0(k) = N_{G(k)}(A_0(k))/A_0(k)$$
There are two Bruhat decompositions of $G$. First, we have
$$G(overline{k}) = coprodlimits_{w in W} B(overline{k})wB(overline{k})$$
and second,
$$G(k) = coprodlimits_{w in W_0} B(k)wB(k)$$
where each representative $w in W_0$ is chosen to be $k$-rational. I'm pretty sure that $B(k)$ is Zariski dense in $B(overline{k})$, and so $overline{B(k)wB(k)} = overline{B(overline{k})wB(overline{k})}$. Also, it is generally true that $G(k)$ is Zariski dense in $G(overline{k})$, so we have
$$coprodlimits_{w in W} B(overline{k})wB(overline{k}) = G(overline{k}) = overline{G(k)} = bigcuplimits_{w in W_0} overline{B(overline{k})wB(overline{k})} tag{1}$$
Now, each $overline{B(overline{k})wB(overline{k})}$ is a union of the cells $B(overline{k})w' B(overline{k})$, where $w' in W$ can be obtained from $w$ by taking a reduced decomposition of $w$ (in the Coxeter system corresponding to $B$ and $T$), and removing generators.
So, where does the long element of $W$ appear on the right hand side? In general, I think the long element of $W$ need not lie in $W_0$, and it seems in this case that the length of every double coset in (1) is strictly less than the long element of $W$.
algebraic-groups reductive-groups
algebraic-groups reductive-groups
asked Jan 17 at 1:14
D_SD_S
13.6k61552
asked Jan 17 at 1:14
D_SD_S
13.6k61552
asked Jan 17 at 1:14
D_SD_S
13.6k61552
asked Jan 17 at 1:14
D_SD_S
13.6k61552
asked Jan 17 at 1:14
D_SD_S
13.6k61552
13.6k61552
$begingroup$
You have it backwards. $wleq w'$ in Bruhat order, not $w'leq w$.
$endgroup$
– Matt Samuel
Jan 17 at 1:33
$begingroup$
I didn't mention the Bruhat order, what do you mean?
$endgroup$
– D_S
Jan 17 at 2:38
$begingroup$
Not explicitly. "Take a reduced decomposition and remove generators." That's a characterization of Bruhat order.
$endgroup$
– Matt Samuel
Jan 17 at 2:39
$begingroup$
I think what I said was right. If for example $w$ is the long element, then every other element $w'$ of $W$ can be obtained by taking a reduced decomposition of $w$ and removing generators. And in turn, $Bw'B$ is contained in the closure of $BwB$.
$endgroup$
– D_S
Jan 17 at 2:52
add a comment |
$begingroup$
You have it backwards. $wleq w'$ in Bruhat order, not $w'leq w$.
$endgroup$
– Matt Samuel
Jan 17 at 1:33
$begingroup$
I didn't mention the Bruhat order, what do you mean?
$endgroup$
– D_S
Jan 17 at 2:38
$begingroup$
Not explicitly. "Take a reduced decomposition and remove generators." That's a characterization of Bruhat order.
$endgroup$
– Matt Samuel
Jan 17 at 2:39
$begingroup$
I think what I said was right. If for example $w$ is the long element, then every other element $w'$ of $W$ can be obtained by taking a reduced decomposition of $w$ and removing generators. And in turn, $Bw'B$ is contained in the closure of $BwB$.
$endgroup$
– D_S
Jan 17 at 2:52
$begingroup$
You have it backwards. $wleq w'$ in Bruhat order, not $w'leq w$.
$endgroup$
– Matt Samuel
Jan 17 at 1:33
$begingroup$
You have it backwards. $wleq w'$ in Bruhat order, not $w'leq w$.
$endgroup$
– Matt Samuel
Jan 17 at 1:33
$begingroup$
I didn't mention the Bruhat order, what do you mean?
$endgroup$
– D_S
Jan 17 at 2:38
$begingroup$
I didn't mention the Bruhat order, what do you mean?
$endgroup$
– D_S
Jan 17 at 2:38
$begingroup$
Not explicitly. "Take a reduced decomposition and remove generators." That's a characterization of Bruhat order.
$endgroup$
– Matt Samuel
Jan 17 at 2:39
$begingroup$
Not explicitly. "Take a reduced decomposition and remove generators." That's a characterization of Bruhat order.
$endgroup$
– Matt Samuel
Jan 17 at 2:39
$begingroup$
I think what I said was right. If for example $w$ is the long element, then every other element $w'$ of $W$ can be obtained by taking a reduced decomposition of $w$ and removing generators. And in turn, $Bw'B$ is contained in the closure of $BwB$.
$endgroup$
– D_S
Jan 17 at 2:52
$begingroup$
I think what I said was right. If for example $w$ is the long element, then every other element $w'$ of $W$ can be obtained by taking a reduced decomposition of $w$ and removing generators. And in turn, $Bw'B$ is contained in the closure of $BwB$.
$endgroup$
– D_S
Jan 17 at 2:52
add a comment |
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$begingroup$
You have it backwards. $wleq w'$ in Bruhat order, not $w'leq w$.
$endgroup$
– Matt Samuel
Jan 17 at 1:33
$begingroup$
I didn't mention the Bruhat order, what do you mean?
$endgroup$
– D_S
Jan 17 at 2:38
$begingroup$
Not explicitly. "Take a reduced decomposition and remove generators." That's a characterization of Bruhat order.
$endgroup$
– Matt Samuel
Jan 17 at 2:39
$begingroup$
I think what I said was right. If for example $w$ is the long element, then every other element $w'$ of $W$ can be obtained by taking a reduced decomposition of $w$ and removing generators. And in turn, $Bw'B$ is contained in the closure of $BwB$.
$endgroup$
– D_S
Jan 17 at 2:52