Mixing
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Split component is the identity component of the intersection of the kernels of the roots
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Let $G$ be a connected, reductive group over a field $k$. Let $S$ be a maximal $k$-split torus of $G$, and $T$ a maximal torus of $G$ which contains $S$ and which is defined over $k$. Let $A$ be the split component of $G$, the largest split torus inside the center of $G$.
I recall a result like this:
$$A = (bigcaplimits_alpha in Phi operatornameKer alpha)^circ$$
where $Phi = Phi(S,G)$, the roots of $S$ in $G$. However, I can't seem to remember why this is true. I would appreciate a proof or reference.
When $G$ is split, a stronger result is true:
$$Z_G = bigcaplimits_alpha in PhioperatornameKer alpha$$
which implies the result I want.
algebraic-groups reductive-groups
asked Jul 18 at 3:28
D_S
12.8k51550
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Let $G$ be a connected, reductive group over a field $k$. Let $S$ be a maximal $k$-split torus of $G$, and $T$ a maximal torus of $G$ which contains $S$ and which is defined over $k$. Let $A$ be the split component of $G$, the largest split torus inside the center of $G$.
I recall a result like this:
$$A = (bigcaplimits_alpha in Phi operatornameKer alpha)^circ$$
where $Phi = Phi(S,G)$, the roots of $S$ in $G$. However, I can't seem to remember why this is true. I would appreciate a proof or reference.
When $G$ is split, a stronger result is true:
$$Z_G = bigcaplimits_alpha in PhioperatornameKer alpha$$
which implies the result I want.
algebraic-groups reductive-groups
asked Jul 18 at 3:28
D_S
12.8k51550
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be a connected, reductive group over a field $k$. Let $S$ be a maximal $k$-split torus of $G$, and $T$ a maximal torus of $G$ which contains $S$ and which is defined over $k$. Let $A$ be the split component of $G$, the largest split torus inside the center of $G$.
I recall a result like this:
$$A = (bigcaplimits_alpha in Phi operatornameKer alpha)^circ$$
where $Phi = Phi(S,G)$, the roots of $S$ in $G$. However, I can't seem to remember why this is true. I would appreciate a proof or reference.
When $G$ is split, a stronger result is true:
$$Z_G = bigcaplimits_alpha in PhioperatornameKer alpha$$
which implies the result I want.
algebraic-groups reductive-groups
asked Jul 18 at 3:28
D_S
12.8k51550
Let $G$ be a connected, reductive group over a field $k$. Let $S$ be a maximal $k$-split torus of $G$, and $T$ a maximal torus of $G$ which contains $S$ and which is defined over $k$. Let $A$ be the split component of $G$, the largest split torus inside the center of $G$.
I recall a result like this:
$$A = (bigcaplimits_alpha in Phi operatornameKer alpha)^circ$$
where $Phi = Phi(S,G)$, the roots of $S$ in $G$. However, I can't seem to remember why this is true. I would appreciate a proof or reference.
When $G$ is split, a stronger result is true:
$$Z_G = bigcaplimits_alpha in PhioperatornameKer alpha$$
which implies the result I want.
algebraic-groups reductive-groups
asked Jul 18 at 3:28
D_S
12.8k51550
asked Jul 18 at 3:28
D_S
12.8k51550
asked Jul 18 at 3:28
D_S
12.8k51550
asked Jul 18 at 3:28
D_S
12.8k51550
12.8k51550
add a comment |Â
add a comment |Â
1 Answer
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$(S cap Z_G)^circ$ is a split torus contained in the center of $G$, so $(S cap Z_G)^0 subseteq A$. Let $widetildeDelta$ be a set of simple roots of $T$ in $G$, chosen so that the set $Delta = _S : alpha in widetildeDelta backslash 0$ is a set of simple roots of $S$ in $G$. Since
$$Z_G = bigcaplimits_alpha in Phi(T,G)operatornameKeralpha= bigcaplimits_alpha in widetildeDeltaoperatornameKeralpha$$
we have
$$S cap Z_G = bigcaplimits_alpha in Delta operatornameKeralpha = bigcaplimits_alpha in Phi(S,G) operatornameKeralpha$$
which implies that
$$(bigcaplimits_alpha in Phi(S,G) operatornameKeralpha)^circ subseteq A $$
On the other hand, the right hand side is clearly contained in the left hand side, since $t in A$ implies $operatornameAdt$ is trivial.
answered Jul 19 at 1:02
D_S
12.8k51550
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$(S cap Z_G)^circ$ is a split torus contained in the center of $G$, so $(S cap Z_G)^0 subseteq A$. Let $widetildeDelta$ be a set of simple roots of $T$ in $G$, chosen so that the set $Delta = _S : alpha in widetildeDelta backslash 0$ is a set of simple roots of $S$ in $G$. Since
$$Z_G = bigcaplimits_alpha in Phi(T,G)operatornameKeralpha= bigcaplimits_alpha in widetildeDeltaoperatornameKeralpha$$
we have
$$S cap Z_G = bigcaplimits_alpha in Delta operatornameKeralpha = bigcaplimits_alpha in Phi(S,G) operatornameKeralpha$$
which implies that
$$(bigcaplimits_alpha in Phi(S,G) operatornameKeralpha)^circ subseteq A $$
On the other hand, the right hand side is clearly contained in the left hand side, since $t in A$ implies $operatornameAdt$ is trivial.
answered Jul 19 at 1:02
D_S
12.8k51550
add a comment |Â
up vote
0
down vote
$(S cap Z_G)^circ$ is a split torus contained in the center of $G$, so $(S cap Z_G)^0 subseteq A$. Let $widetildeDelta$ be a set of simple roots of $T$ in $G$, chosen so that the set $Delta = _S : alpha in widetildeDelta backslash 0$ is a set of simple roots of $S$ in $G$. Since
$$Z_G = bigcaplimits_alpha in Phi(T,G)operatornameKeralpha= bigcaplimits_alpha in widetildeDeltaoperatornameKeralpha$$
we have
$$S cap Z_G = bigcaplimits_alpha in Delta operatornameKeralpha = bigcaplimits_alpha in Phi(S,G) operatornameKeralpha$$
which implies that
$$(bigcaplimits_alpha in Phi(S,G) operatornameKeralpha)^circ subseteq A $$
On the other hand, the right hand side is clearly contained in the left hand side, since $t in A$ implies $operatornameAdt$ is trivial.
answered Jul 19 at 1:02
D_S
12.8k51550
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$(S cap Z_G)^circ$ is a split torus contained in the center of $G$, so $(S cap Z_G)^0 subseteq A$. Let $widetildeDelta$ be a set of simple roots of $T$ in $G$, chosen so that the set $Delta = _S : alpha in widetildeDelta backslash 0$ is a set of simple roots of $S$ in $G$. Since
$$Z_G = bigcaplimits_alpha in Phi(T,G)operatornameKeralpha= bigcaplimits_alpha in widetildeDeltaoperatornameKeralpha$$
we have
$$S cap Z_G = bigcaplimits_alpha in Delta operatornameKeralpha = bigcaplimits_alpha in Phi(S,G) operatornameKeralpha$$
which implies that
$$(bigcaplimits_alpha in Phi(S,G) operatornameKeralpha)^circ subseteq A $$
On the other hand, the right hand side is clearly contained in the left hand side, since $t in A$ implies $operatornameAdt$ is trivial.
answered Jul 19 at 1:02
D_S
12.8k51550
$(S cap Z_G)^circ$ is a split torus contained in the center of $G$, so $(S cap Z_G)^0 subseteq A$. Let $widetildeDelta$ be a set of simple roots of $T$ in $G$, chosen so that the set $Delta = _S : alpha in widetildeDelta backslash 0$ is a set of simple roots of $S$ in $G$. Since
$$Z_G = bigcaplimits_alpha in Phi(T,G)operatornameKeralpha= bigcaplimits_alpha in widetildeDeltaoperatornameKeralpha$$
we have
$$S cap Z_G = bigcaplimits_alpha in Delta operatornameKeralpha = bigcaplimits_alpha in Phi(S,G) operatornameKeralpha$$
which implies that
$$(bigcaplimits_alpha in Phi(S,G) operatornameKeralpha)^circ subseteq A $$
On the other hand, the right hand side is clearly contained in the left hand side, since $t in A$ implies $operatornameAdt$ is trivial.
answered Jul 19 at 1:02
D_S
12.8k51550
answered Jul 19 at 1:02
D_S
12.8k51550
answered Jul 19 at 1:02
D_S
12.8k51550
answered Jul 19 at 1:02
D_S
12.8k51550
12.8k51550
add a comment |Â
add a comment |Â
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