Image of the Harish-Chandra map is of finite index?

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Let $G$ be a connected, reductive group over a global or local field $k$ with absolute value $| cdot |$. Let $X(G)_k$ be the group of rational characters of $G$ which are defined over $k$, and let $H_G: G(k) rightarrow operatornameHom_mathbb Z(X(G)_k,mathbb R)$ be the Harish-Chandra map, defined by

$$H_G(g)(chi) = log |chi(g)|$$

What can be said about the image of $H_G$? I know that it is a discrete subgroup of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$? Does it span all of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$?

I know that this is true when $G$ is a split torus. What about in general?

algebraic-groups reductive-groups

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edited Jul 19 at 3:15

asked Jul 18 at 2:31

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Let $G$ be a connected, reductive group over a global or local field $k$ with absolute value $| cdot |$. Let $X(G)_k$ be the group of rational characters of $G$ which are defined over $k$, and let $H_G: G(k) rightarrow operatornameHom_mathbb Z(X(G)_k,mathbb R)$ be the Harish-Chandra map, defined by

$$H_G(g)(chi) = log |chi(g)|$$

What can be said about the image of $H_G$? I know that it is a discrete subgroup of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$? Does it span all of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$?

I know that this is true when $G$ is a split torus. What about in general?

algebraic-groups reductive-groups

share|cite|improve this question

edited Jul 19 at 3:15

asked Jul 18 at 2:31

D_S

12.8k51550

add a comment | 

up vote
0
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up vote
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down vote

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Let $G$ be a connected, reductive group over a global or local field $k$ with absolute value $| cdot |$. Let $X(G)_k$ be the group of rational characters of $G$ which are defined over $k$, and let $H_G: G(k) rightarrow operatornameHom_mathbb Z(X(G)_k,mathbb R)$ be the Harish-Chandra map, defined by

$$H_G(g)(chi) = log |chi(g)|$$

What can be said about the image of $H_G$? I know that it is a discrete subgroup of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$? Does it span all of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$?

I know that this is true when $G$ is a split torus. What about in general?

algebraic-groups reductive-groups

share|cite|improve this question

edited Jul 19 at 3:15

asked Jul 18 at 2:31

D_S

12.8k51550

Let $G$ be a connected, reductive group over a global or local field $k$ with absolute value $| cdot |$. Let $X(G)_k$ be the group of rational characters of $G$ which are defined over $k$, and let $H_G: G(k) rightarrow operatornameHom_mathbb Z(X(G)_k,mathbb R)$ be the Harish-Chandra map, defined by

$$H_G(g)(chi) = log |chi(g)|$$

What can be said about the image of $H_G$? I know that it is a discrete subgroup of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$? Does it span all of $operatornameHom_mathbb Z(X(G)_k,mathbb R)$?

I know that this is true when $G$ is a split torus. What about in general?

algebraic-groups reductive-groups

share|cite|improve this question

edited Jul 19 at 3:15

asked Jul 18 at 2:31

D_S

12.8k51550

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 3:15

edited Jul 19 at 3:15

edited Jul 19 at 3:15

asked Jul 18 at 2:31

D_S

12.8k51550

asked Jul 18 at 2:31

D_S

12.8k51550

asked Jul 18 at 2:31

D_S

12.8k51550

12.8k51550

add a comment | 

add a comment | 

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The image always spans $operatornameHom_mathbb Z(X(G)_k,mathbb R)$. The proof essentially reduces to that of when $G$ is a split torus. Let $chi_1, ... , chi_n$ be a basis for $X(G)_k$. Then $f_1, ... , f_n$ is a basis for $operatornameHom_mathbb Z(X(G)_k,mathbb R)$, where $f_i(chi_j) = delta_ij$.

Fix $1 leq i leq n$. We need to show that there exists a $g in G(F)$ and a real number $m$ such that $m H_G(g) = f_i$.

Let $A_G$ be the split component of $G$. It is a well known result that the restriction homomorphism $X(G)_k rightarrow X(A_G)$ is injective, and the image is of finite index in $X(A_G)$. So in the beginning, we can choose the basis $chi_i$ of $X(G)_k$ as well as a basis $eta_1, ... , eta_n$ of $X(A_G)$, so that there exist nonzero integers $d_i$ such that $chi_i|_A_G = d_i eta_i$.

There exists an $a in k^ast$ and an $m in mathbb R$ such that $|a|^md_i = e$. Choose $g in A_G$ such that $eta_i(g) = a$, and $eta_j(g) = 1$ for all $j neq i$. Then for all $chi = c_1chi_1 + cdots + c_nchi_n in X(G)_k$ ($c_i in mathbb Z$), we have

$$mH_G(g)(chi) = m log |chi_1(g)^c_1 cdots chi_n(g)^c_n| = m log |eta_1^c_1d_1(g) cdots eta_n^c_nd_n(g)| = m log |a^c_id_i|$$

$$ = c_i log |a|^md_i = c_i = f_i(chi)$$

which shows that $m H_G(g) = f_i$.

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edited Jul 19 at 3:16

answered Jul 18 at 2:31

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1 Answer
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1 Answer
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up vote
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The image always spans $operatornameHom_mathbb Z(X(G)_k,mathbb R)$. The proof essentially reduces to that of when $G$ is a split torus. Let $chi_1, ... , chi_n$ be a basis for $X(G)_k$. Then $f_1, ... , f_n$ is a basis for $operatornameHom_mathbb Z(X(G)_k,mathbb R)$, where $f_i(chi_j) = delta_ij$.

Fix $1 leq i leq n$. We need to show that there exists a $g in G(F)$ and a real number $m$ such that $m H_G(g) = f_i$.

Let $A_G$ be the split component of $G$. It is a well known result that the restriction homomorphism $X(G)_k rightarrow X(A_G)$ is injective, and the image is of finite index in $X(A_G)$. So in the beginning, we can choose the basis $chi_i$ of $X(G)_k$ as well as a basis $eta_1, ... , eta_n$ of $X(A_G)$, so that there exist nonzero integers $d_i$ such that $chi_i|_A_G = d_i eta_i$.

There exists an $a in k^ast$ and an $m in mathbb R$ such that $|a|^md_i = e$. Choose $g in A_G$ such that $eta_i(g) = a$, and $eta_j(g) = 1$ for all $j neq i$. Then for all $chi = c_1chi_1 + cdots + c_nchi_n in X(G)_k$ ($c_i in mathbb Z$), we have

$$mH_G(g)(chi) = m log |chi_1(g)^c_1 cdots chi_n(g)^c_n| = m log |eta_1^c_1d_1(g) cdots eta_n^c_nd_n(g)| = m log |a^c_id_i|$$

$$ = c_i log |a|^md_i = c_i = f_i(chi)$$

which shows that $m H_G(g) = f_i$.

share|cite|improve this answer

edited Jul 19 at 3:16

answered Jul 18 at 2:31

D_S

12.8k51550

add a comment | 

up vote
0
down vote

The image always spans $operatornameHom_mathbb Z(X(G)_k,mathbb R)$. The proof essentially reduces to that of when $G$ is a split torus. Let $chi_1, ... , chi_n$ be a basis for $X(G)_k$. Then $f_1, ... , f_n$ is a basis for $operatornameHom_mathbb Z(X(G)_k,mathbb R)$, where $f_i(chi_j) = delta_ij$.

Fix $1 leq i leq n$. We need to show that there exists a $g in G(F)$ and a real number $m$ such that $m H_G(g) = f_i$.

Let $A_G$ be the split component of $G$. It is a well known result that the restriction homomorphism $X(G)_k rightarrow X(A_G)$ is injective, and the image is of finite index in $X(A_G)$. So in the beginning, we can choose the basis $chi_i$ of $X(G)_k$ as well as a basis $eta_1, ... , eta_n$ of $X(A_G)$, so that there exist nonzero integers $d_i$ such that $chi_i|_A_G = d_i eta_i$.

There exists an $a in k^ast$ and an $m in mathbb R$ such that $|a|^md_i = e$. Choose $g in A_G$ such that $eta_i(g) = a$, and $eta_j(g) = 1$ for all $j neq i$. Then for all $chi = c_1chi_1 + cdots + c_nchi_n in X(G)_k$ ($c_i in mathbb Z$), we have

$$mH_G(g)(chi) = m log |chi_1(g)^c_1 cdots chi_n(g)^c_n| = m log |eta_1^c_1d_1(g) cdots eta_n^c_nd_n(g)| = m log |a^c_id_i|$$

$$ = c_i log |a|^md_i = c_i = f_i(chi)$$

which shows that $m H_G(g) = f_i$.

share|cite|improve this answer

edited Jul 19 at 3:16

answered Jul 18 at 2:31

D_S

12.8k51550

add a comment | 

up vote
0
down vote

up vote
0
down vote

The image always spans $operatornameHom_mathbb Z(X(G)_k,mathbb R)$. The proof essentially reduces to that of when $G$ is a split torus. Let $chi_1, ... , chi_n$ be a basis for $X(G)_k$. Then $f_1, ... , f_n$ is a basis for $operatornameHom_mathbb Z(X(G)_k,mathbb R)$, where $f_i(chi_j) = delta_ij$.

Fix $1 leq i leq n$. We need to show that there exists a $g in G(F)$ and a real number $m$ such that $m H_G(g) = f_i$.

Let $A_G$ be the split component of $G$. It is a well known result that the restriction homomorphism $X(G)_k rightarrow X(A_G)$ is injective, and the image is of finite index in $X(A_G)$. So in the beginning, we can choose the basis $chi_i$ of $X(G)_k$ as well as a basis $eta_1, ... , eta_n$ of $X(A_G)$, so that there exist nonzero integers $d_i$ such that $chi_i|_A_G = d_i eta_i$.

There exists an $a in k^ast$ and an $m in mathbb R$ such that $|a|^md_i = e$. Choose $g in A_G$ such that $eta_i(g) = a$, and $eta_j(g) = 1$ for all $j neq i$. Then for all $chi = c_1chi_1 + cdots + c_nchi_n in X(G)_k$ ($c_i in mathbb Z$), we have

$$mH_G(g)(chi) = m log |chi_1(g)^c_1 cdots chi_n(g)^c_n| = m log |eta_1^c_1d_1(g) cdots eta_n^c_nd_n(g)| = m log |a^c_id_i|$$

$$ = c_i log |a|^md_i = c_i = f_i(chi)$$

which shows that $m H_G(g) = f_i$.

share|cite|improve this answer

edited Jul 19 at 3:16

answered Jul 18 at 2:31

D_S

12.8k51550

The image always spans $operatornameHom_mathbb Z(X(G)_k,mathbb R)$. The proof essentially reduces to that of when $G$ is a split torus. Let $chi_1, ... , chi_n$ be a basis for $X(G)_k$. Then $f_1, ... , f_n$ is a basis for $operatornameHom_mathbb Z(X(G)_k,mathbb R)$, where $f_i(chi_j) = delta_ij$.

Fix $1 leq i leq n$. We need to show that there exists a $g in G(F)$ and a real number $m$ such that $m H_G(g) = f_i$.

Let $A_G$ be the split component of $G$. It is a well known result that the restriction homomorphism $X(G)_k rightarrow X(A_G)$ is injective, and the image is of finite index in $X(A_G)$. So in the beginning, we can choose the basis $chi_i$ of $X(G)_k$ as well as a basis $eta_1, ... , eta_n$ of $X(A_G)$, so that there exist nonzero integers $d_i$ such that $chi_i|_A_G = d_i eta_i$.

There exists an $a in k^ast$ and an $m in mathbb R$ such that $|a|^md_i = e$. Choose $g in A_G$ such that $eta_i(g) = a$, and $eta_j(g) = 1$ for all $j neq i$. Then for all $chi = c_1chi_1 + cdots + c_nchi_n in X(G)_k$ ($c_i in mathbb Z$), we have

$$mH_G(g)(chi) = m log |chi_1(g)^c_1 cdots chi_n(g)^c_n| = m log |eta_1^c_1d_1(g) cdots eta_n^c_nd_n(g)| = m log |a^c_id_i|$$

$$ = c_i log |a|^md_i = c_i = f_i(chi)$$

which shows that $m H_G(g) = f_i$.

share|cite|improve this answer

edited Jul 19 at 3:16

answered Jul 18 at 2:31

D_S

12.8k51550

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 3:16

edited Jul 19 at 3:16

edited Jul 19 at 3:16

answered Jul 18 at 2:31

D_S

12.8k51550

answered Jul 18 at 2:31

D_S

12.8k51550

answered Jul 18 at 2:31

D_S

12.8k51550

12.8k51550

add a comment | 

add a comment | 

 

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