The definition of a coroot for non-split reductive groups?
$begingroup$
Let $G$ be a connected, reductive group over a field $k$, and $A_0$ a maximal split torus of $G$. Let $Phi = Phi(G,A_0)$ be the set of roots of $A_0$ in $G$. Then the $mathbb R$-linear span $mathfrak a_0^{G ast}$ of $Phi$ in $mathfrak a_0^{ast} = X(A_0) otimes mathbb R$ is a root system. Then in the dual space $mathfrak a_0^G$ of $mathfrak a_0^{G ast}$ we have a set $Phi^{vee}$ of coroots. By definition, the coroot $alpha^{vee}$ associated to a given root $alpha$ is the unique linear functional on $mathfrak a_0^{G ast}$ such that
$$s_{alpha}(x) := x - langle x, alpha^{vee} rangle alpha$$
is a reflection on $mathfrak a_0^{G ast}$ which sends $alpha$ to $-alpha$ and permutes $Phi$.
If $A_G$ is the split component of $G$, there is a canonical injection of $mathfrak a_G^{ast} = X(A_G) otimes mathbb R$ into $mathfrak a_0^{ast}$, such that $mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus mathfrak a_0^{G ast}$. We then extend each element of $mathfrak a_0^G$ to a linear functional on all of $mathfrak a_0^{ast}$, by setting it to be zero on $mathfrak a_G^{ast}$. Thus the coroots are elements of $$mathfrak a_0 = operatorname{Hom}_{mathbb R}(X(A_0) otimes mathbb R, mathbb R) = operatorname{Hom}(X(A_0),mathbb R) = X^{ast}(A_0)$$
When $G$ is split, there is a very nice way to describe coroots, without reference to reflections or root systems. Namely, if $alpha in Phi$, and $T_{alpha} = (operatorname{Ker} alpha)^0$, then $alpha^{vee}$ is the unique cocharacter of $A_0$ satisfying:
(i): $langle alpha, alpha^{vee} rangle = 2$
(ii): The group generated by $T_{alpha}$ and the image of $alpha^{vee}$ is all of $A_0$.
Suppose that $G$ is not split. Is there any nice way to characterize the coroot $alpha^{vee}$ as in the split case?
representation-theory lie-algebras root-systems reductive-groups
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $G$ be a connected, reductive group over a field $k$, and $A_0$ a maximal split torus of $G$. Let $Phi = Phi(G,A_0)$ be the set of roots of $A_0$ in $G$. Then the $mathbb R$-linear span $mathfrak a_0^{G ast}$ of $Phi$ in $mathfrak a_0^{ast} = X(A_0) otimes mathbb R$ is a root system. Then in the dual space $mathfrak a_0^G$ of $mathfrak a_0^{G ast}$ we have a set $Phi^{vee}$ of coroots. By definition, the coroot $alpha^{vee}$ associated to a given root $alpha$ is the unique linear functional on $mathfrak a_0^{G ast}$ such that
$$s_{alpha}(x) := x - langle x, alpha^{vee} rangle alpha$$
is a reflection on $mathfrak a_0^{G ast}$ which sends $alpha$ to $-alpha$ and permutes $Phi$.
If $A_G$ is the split component of $G$, there is a canonical injection of $mathfrak a_G^{ast} = X(A_G) otimes mathbb R$ into $mathfrak a_0^{ast}$, such that $mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus mathfrak a_0^{G ast}$. We then extend each element of $mathfrak a_0^G$ to a linear functional on all of $mathfrak a_0^{ast}$, by setting it to be zero on $mathfrak a_G^{ast}$. Thus the coroots are elements of $$mathfrak a_0 = operatorname{Hom}_{mathbb R}(X(A_0) otimes mathbb R, mathbb R) = operatorname{Hom}(X(A_0),mathbb R) = X^{ast}(A_0)$$
When $G$ is split, there is a very nice way to describe coroots, without reference to reflections or root systems. Namely, if $alpha in Phi$, and $T_{alpha} = (operatorname{Ker} alpha)^0$, then $alpha^{vee}$ is the unique cocharacter of $A_0$ satisfying:
(i): $langle alpha, alpha^{vee} rangle = 2$
(ii): The group generated by $T_{alpha}$ and the image of $alpha^{vee}$ is all of $A_0$.
Suppose that $G$ is not split. Is there any nice way to characterize the coroot $alpha^{vee}$ as in the split case?
representation-theory lie-algebras root-systems reductive-groups
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
$endgroup$
1
$begingroup$
In the second sentence, do you mean $A_0$ instead of $T$? Also I do not understand the sentence after that, where the notation $mathfrak{a}_0^{G_*}$ is unclear to me; I am quite sure that an $Bbb R$-linear span of whatever cannot be a root system though. Some thing seems to be in a wrong order there.
$endgroup$
– Torsten Schoeneberg
Dec 10 '18 at 0:43
$begingroup$
(1) Yes I meant $A_0$ instead of $T$ (2) The notation $(mathfrak a_0^G)^{ast}$ is standard for the linear span of the roots in $mathfrak a_0^{ast}$, in Arthur-Selberg trace formula stuff they use the fact that there is a canonical direct sum decomposition $$mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus (mathfrak a_0^{G})^{ast}$$where $mathfrak a_G^{ast}$ is the dual real Lie algebra of the split component of $G$. Yes, the pair $(mathfrak a_0^G)^{ast}, Phi$ is definitely a root system.
$endgroup$
– D_S
Dec 10 '18 at 3:05
add a comment |
$begingroup$
Let $G$ be a connected, reductive group over a field $k$, and $A_0$ a maximal split torus of $G$. Let $Phi = Phi(G,A_0)$ be the set of roots of $A_0$ in $G$. Then the $mathbb R$-linear span $mathfrak a_0^{G ast}$ of $Phi$ in $mathfrak a_0^{ast} = X(A_0) otimes mathbb R$ is a root system. Then in the dual space $mathfrak a_0^G$ of $mathfrak a_0^{G ast}$ we have a set $Phi^{vee}$ of coroots. By definition, the coroot $alpha^{vee}$ associated to a given root $alpha$ is the unique linear functional on $mathfrak a_0^{G ast}$ such that
$$s_{alpha}(x) := x - langle x, alpha^{vee} rangle alpha$$
is a reflection on $mathfrak a_0^{G ast}$ which sends $alpha$ to $-alpha$ and permutes $Phi$.
If $A_G$ is the split component of $G$, there is a canonical injection of $mathfrak a_G^{ast} = X(A_G) otimes mathbb R$ into $mathfrak a_0^{ast}$, such that $mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus mathfrak a_0^{G ast}$. We then extend each element of $mathfrak a_0^G$ to a linear functional on all of $mathfrak a_0^{ast}$, by setting it to be zero on $mathfrak a_G^{ast}$. Thus the coroots are elements of $$mathfrak a_0 = operatorname{Hom}_{mathbb R}(X(A_0) otimes mathbb R, mathbb R) = operatorname{Hom}(X(A_0),mathbb R) = X^{ast}(A_0)$$
When $G$ is split, there is a very nice way to describe coroots, without reference to reflections or root systems. Namely, if $alpha in Phi$, and $T_{alpha} = (operatorname{Ker} alpha)^0$, then $alpha^{vee}$ is the unique cocharacter of $A_0$ satisfying:
(i): $langle alpha, alpha^{vee} rangle = 2$
(ii): The group generated by $T_{alpha}$ and the image of $alpha^{vee}$ is all of $A_0$.
Suppose that $G$ is not split. Is there any nice way to characterize the coroot $alpha^{vee}$ as in the split case?
representation-theory lie-algebras root-systems reductive-groups
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
$endgroup$
Let $G$ be a connected, reductive group over a field $k$, and $A_0$ a maximal split torus of $G$. Let $Phi = Phi(G,A_0)$ be the set of roots of $A_0$ in $G$. Then the $mathbb R$-linear span $mathfrak a_0^{G ast}$ of $Phi$ in $mathfrak a_0^{ast} = X(A_0) otimes mathbb R$ is a root system. Then in the dual space $mathfrak a_0^G$ of $mathfrak a_0^{G ast}$ we have a set $Phi^{vee}$ of coroots. By definition, the coroot $alpha^{vee}$ associated to a given root $alpha$ is the unique linear functional on $mathfrak a_0^{G ast}$ such that
$$s_{alpha}(x) := x - langle x, alpha^{vee} rangle alpha$$
is a reflection on $mathfrak a_0^{G ast}$ which sends $alpha$ to $-alpha$ and permutes $Phi$.
If $A_G$ is the split component of $G$, there is a canonical injection of $mathfrak a_G^{ast} = X(A_G) otimes mathbb R$ into $mathfrak a_0^{ast}$, such that $mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus mathfrak a_0^{G ast}$. We then extend each element of $mathfrak a_0^G$ to a linear functional on all of $mathfrak a_0^{ast}$, by setting it to be zero on $mathfrak a_G^{ast}$. Thus the coroots are elements of $$mathfrak a_0 = operatorname{Hom}_{mathbb R}(X(A_0) otimes mathbb R, mathbb R) = operatorname{Hom}(X(A_0),mathbb R) = X^{ast}(A_0)$$
When $G$ is split, there is a very nice way to describe coroots, without reference to reflections or root systems. Namely, if $alpha in Phi$, and $T_{alpha} = (operatorname{Ker} alpha)^0$, then $alpha^{vee}$ is the unique cocharacter of $A_0$ satisfying:
(i): $langle alpha, alpha^{vee} rangle = 2$
(ii): The group generated by $T_{alpha}$ and the image of $alpha^{vee}$ is all of $A_0$.
Suppose that $G$ is not split. Is there any nice way to characterize the coroot $alpha^{vee}$ as in the split case?
representation-theory lie-algebras root-systems reductive-groups
representation-theory lie-algebras root-systems reductive-groups
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
edited Dec 10 '18 at 3:02
D_S
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
asked Dec 9 '18 at 17:08
D_SD_S
13.5k51551
13.5k51551
1
$begingroup$
In the second sentence, do you mean $A_0$ instead of $T$? Also I do not understand the sentence after that, where the notation $mathfrak{a}_0^{G_*}$ is unclear to me; I am quite sure that an $Bbb R$-linear span of whatever cannot be a root system though. Some thing seems to be in a wrong order there.
$endgroup$
– Torsten Schoeneberg
Dec 10 '18 at 0:43
$begingroup$
(1) Yes I meant $A_0$ instead of $T$ (2) The notation $(mathfrak a_0^G)^{ast}$ is standard for the linear span of the roots in $mathfrak a_0^{ast}$, in Arthur-Selberg trace formula stuff they use the fact that there is a canonical direct sum decomposition $$mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus (mathfrak a_0^{G})^{ast}$$where $mathfrak a_G^{ast}$ is the dual real Lie algebra of the split component of $G$. Yes, the pair $(mathfrak a_0^G)^{ast}, Phi$ is definitely a root system.
$endgroup$
– D_S
Dec 10 '18 at 3:05
add a comment |
1
$begingroup$
In the second sentence, do you mean $A_0$ instead of $T$? Also I do not understand the sentence after that, where the notation $mathfrak{a}_0^{G_*}$ is unclear to me; I am quite sure that an $Bbb R$-linear span of whatever cannot be a root system though. Some thing seems to be in a wrong order there.
$endgroup$
– Torsten Schoeneberg
Dec 10 '18 at 0:43
$begingroup$
(1) Yes I meant $A_0$ instead of $T$ (2) The notation $(mathfrak a_0^G)^{ast}$ is standard for the linear span of the roots in $mathfrak a_0^{ast}$, in Arthur-Selberg trace formula stuff they use the fact that there is a canonical direct sum decomposition $$mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus (mathfrak a_0^{G})^{ast}$$where $mathfrak a_G^{ast}$ is the dual real Lie algebra of the split component of $G$. Yes, the pair $(mathfrak a_0^G)^{ast}, Phi$ is definitely a root system.
$endgroup$
– D_S
Dec 10 '18 at 3:05
1
1
$begingroup$
In the second sentence, do you mean $A_0$ instead of $T$? Also I do not understand the sentence after that, where the notation $mathfrak{a}_0^{G_*}$ is unclear to me; I am quite sure that an $Bbb R$-linear span of whatever cannot be a root system though. Some thing seems to be in a wrong order there.
$endgroup$
– Torsten Schoeneberg
Dec 10 '18 at 0:43
$begingroup$
In the second sentence, do you mean $A_0$ instead of $T$? Also I do not understand the sentence after that, where the notation $mathfrak{a}_0^{G_*}$ is unclear to me; I am quite sure that an $Bbb R$-linear span of whatever cannot be a root system though. Some thing seems to be in a wrong order there.
$endgroup$
– Torsten Schoeneberg
Dec 10 '18 at 0:43
$begingroup$
(1) Yes I meant $A_0$ instead of $T$ (2) The notation $(mathfrak a_0^G)^{ast}$ is standard for the linear span of the roots in $mathfrak a_0^{ast}$, in Arthur-Selberg trace formula stuff they use the fact that there is a canonical direct sum decomposition $$mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus (mathfrak a_0^{G})^{ast}$$where $mathfrak a_G^{ast}$ is the dual real Lie algebra of the split component of $G$. Yes, the pair $(mathfrak a_0^G)^{ast}, Phi$ is definitely a root system.
$endgroup$
– D_S
Dec 10 '18 at 3:05
$begingroup$
(1) Yes I meant $A_0$ instead of $T$ (2) The notation $(mathfrak a_0^G)^{ast}$ is standard for the linear span of the roots in $mathfrak a_0^{ast}$, in Arthur-Selberg trace formula stuff they use the fact that there is a canonical direct sum decomposition $$mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus (mathfrak a_0^{G})^{ast}$$where $mathfrak a_G^{ast}$ is the dual real Lie algebra of the split component of $G$. Yes, the pair $(mathfrak a_0^G)^{ast}, Phi$ is definitely a root system.
$endgroup$
– D_S
Dec 10 '18 at 3:05
add a comment |
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$begingroup$
In the second sentence, do you mean $A_0$ instead of $T$? Also I do not understand the sentence after that, where the notation $mathfrak{a}_0^{G_*}$ is unclear to me; I am quite sure that an $Bbb R$-linear span of whatever cannot be a root system though. Some thing seems to be in a wrong order there.
$endgroup$
– Torsten Schoeneberg
Dec 10 '18 at 0:43
$begingroup$
(1) Yes I meant $A_0$ instead of $T$ (2) The notation $(mathfrak a_0^G)^{ast}$ is standard for the linear span of the roots in $mathfrak a_0^{ast}$, in Arthur-Selberg trace formula stuff they use the fact that there is a canonical direct sum decomposition $$mathfrak a_0^{ast} = mathfrak a_G^{ast} oplus (mathfrak a_0^{G})^{ast}$$where $mathfrak a_G^{ast}$ is the dual real Lie algebra of the split component of $G$. Yes, the pair $(mathfrak a_0^G)^{ast}, Phi$ is definitely a root system.
$endgroup$
– D_S
Dec 10 '18 at 3:05