Mixing
Technical question on adelic quotient of centralizers
$begingroup$
Let $G$ be a connected reductive group over $k =mathbb Q$ with split component $A_G$. Let $M$ be a $k$-Levi subgroup of $G$ with split component $A_M$. Let $gamma$ be a semisimple element of $M(k)$, and suppose that the centralizer $M(mathbb A)_{gamma} = { m in M(mathbb A) : mgamma m^{-1} = gamma}$ is equal to the centralizer $G(mathbb A)_{gamma}$.
Let $G(mathbb A)^1$ be the kernel of the Harish-Chandra map $H_G$. By definition, this is the group of $g in G(mathbb A)$ such that $chi(g) in mathbb A^{ast}$ has norm one for all $k$-rational characters $chi in X(G)_k$. Then $G(mathbb A) = G(mathbb A)^1 A_G(mathbb R)^{0}$ (direct product of groups), where $A_G(mathbb R)^0$ is the connected component of the identity in $A_G(mathbb R)$, and similarly $M(mathbb A) = M(mathbb A)^1 A_M(mathbb R)^0$.
Let $G(mathbb R)^1 = G(mathbb R) cap G(mathbb A)^1$. I want to understand why we have
$$M(mathbb A)_{gamma}^1 backslash G(mathbb A)^1 cong (A_M(mathbb R)^0 cap G(mathbb R)^1) times (G(mathbb A)_{gamma} backslash G(mathbb A)) tag{1}$$
This is claimed in James Arthur's book, Introduction to the Trace Formula, section 11, page 61.
Attempt: It is easy to see that $$G(mathbb A)_{gamma} = G(mathbb A)_{gamma}^1 A_G(mathbb R)^0 tag{2}$$ as a direct product, and therefore $G(mathbb A)_{gamma} backslash G(mathbb A) = G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1$. We also clearly have $A_M(mathbb R)^0 cap G(mathbb R)^1 = A_M(mathbb R)^0 cap G(mathbb A)^1$. Thus the right hand side of (1) is equal to
$$ (A_M(mathbb R)^0 cap G(mathbb A)^1) times (G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1)$$
Now replacing $G$ by $M$ in (2), we also have $M(mathbb A)_{gamma} = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0$, and therefore (using our hypothesis about the centralizer $M(mathbb A)_{gamma}$)
$$G(mathbb A)_{gamma}^1 = G(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 tag{3} $$
Next, I claim that
$$M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 (A_M(mathbb R) cap G(mathbb A)^1) tag{4}$$
To verify this claim, we use the fact that
$$M(mathbb A)^1 subset G(mathbb A)^1 cap M(mathbb A)$$
even though this is not in general an equality. It is clear that the right hand side is contained in the left hand side in equation (4). Conversely, suppose $ma$ is in the left hand side, for $m in M(mathbb A)_{gamma}^1$ and $a in A_M(mathbb R)^0$. Since $ma in G(mathbb A)^1$, and $m in M(mathbb A)_{gamma}^1 subseteq G(mathbb A)^1$, we also have $a in G(mathbb A)^1$. Therefore, $ma in M(mathbb A)_{gamma}^1 (A_M(mathbb R)^0 cap G(mathbb A)^1)$.
reductive-groups adeles
asked Jan 14 at 18:43
D_SD_S
13.6k51552
$endgroup$
add a comment |
$begingroup$
Let $G$ be a connected reductive group over $k =mathbb Q$ with split component $A_G$. Let $M$ be a $k$-Levi subgroup of $G$ with split component $A_M$. Let $gamma$ be a semisimple element of $M(k)$, and suppose that the centralizer $M(mathbb A)_{gamma} = { m in M(mathbb A) : mgamma m^{-1} = gamma}$ is equal to the centralizer $G(mathbb A)_{gamma}$.
Let $G(mathbb A)^1$ be the kernel of the Harish-Chandra map $H_G$. By definition, this is the group of $g in G(mathbb A)$ such that $chi(g) in mathbb A^{ast}$ has norm one for all $k$-rational characters $chi in X(G)_k$. Then $G(mathbb A) = G(mathbb A)^1 A_G(mathbb R)^{0}$ (direct product of groups), where $A_G(mathbb R)^0$ is the connected component of the identity in $A_G(mathbb R)$, and similarly $M(mathbb A) = M(mathbb A)^1 A_M(mathbb R)^0$.
Let $G(mathbb R)^1 = G(mathbb R) cap G(mathbb A)^1$. I want to understand why we have
$$M(mathbb A)_{gamma}^1 backslash G(mathbb A)^1 cong (A_M(mathbb R)^0 cap G(mathbb R)^1) times (G(mathbb A)_{gamma} backslash G(mathbb A)) tag{1}$$
This is claimed in James Arthur's book, Introduction to the Trace Formula, section 11, page 61.
Attempt: It is easy to see that $$G(mathbb A)_{gamma} = G(mathbb A)_{gamma}^1 A_G(mathbb R)^0 tag{2}$$ as a direct product, and therefore $G(mathbb A)_{gamma} backslash G(mathbb A) = G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1$. We also clearly have $A_M(mathbb R)^0 cap G(mathbb R)^1 = A_M(mathbb R)^0 cap G(mathbb A)^1$. Thus the right hand side of (1) is equal to
$$ (A_M(mathbb R)^0 cap G(mathbb A)^1) times (G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1)$$
Now replacing $G$ by $M$ in (2), we also have $M(mathbb A)_{gamma} = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0$, and therefore (using our hypothesis about the centralizer $M(mathbb A)_{gamma}$)
$$G(mathbb A)_{gamma}^1 = G(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 tag{3} $$
Next, I claim that
$$M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 (A_M(mathbb R) cap G(mathbb A)^1) tag{4}$$
To verify this claim, we use the fact that
$$M(mathbb A)^1 subset G(mathbb A)^1 cap M(mathbb A)$$
even though this is not in general an equality. It is clear that the right hand side is contained in the left hand side in equation (4). Conversely, suppose $ma$ is in the left hand side, for $m in M(mathbb A)_{gamma}^1$ and $a in A_M(mathbb R)^0$. Since $ma in G(mathbb A)^1$, and $m in M(mathbb A)_{gamma}^1 subseteq G(mathbb A)^1$, we also have $a in G(mathbb A)^1$. Therefore, $ma in M(mathbb A)_{gamma}^1 (A_M(mathbb R)^0 cap G(mathbb A)^1)$.
reductive-groups adeles
asked Jan 14 at 18:43
D_SD_S
13.6k51552
$endgroup$
add a comment |
$begingroup$
Let $G$ be a connected reductive group over $k =mathbb Q$ with split component $A_G$. Let $M$ be a $k$-Levi subgroup of $G$ with split component $A_M$. Let $gamma$ be a semisimple element of $M(k)$, and suppose that the centralizer $M(mathbb A)_{gamma} = { m in M(mathbb A) : mgamma m^{-1} = gamma}$ is equal to the centralizer $G(mathbb A)_{gamma}$.
Let $G(mathbb A)^1$ be the kernel of the Harish-Chandra map $H_G$. By definition, this is the group of $g in G(mathbb A)$ such that $chi(g) in mathbb A^{ast}$ has norm one for all $k$-rational characters $chi in X(G)_k$. Then $G(mathbb A) = G(mathbb A)^1 A_G(mathbb R)^{0}$ (direct product of groups), where $A_G(mathbb R)^0$ is the connected component of the identity in $A_G(mathbb R)$, and similarly $M(mathbb A) = M(mathbb A)^1 A_M(mathbb R)^0$.
Let $G(mathbb R)^1 = G(mathbb R) cap G(mathbb A)^1$. I want to understand why we have
$$M(mathbb A)_{gamma}^1 backslash G(mathbb A)^1 cong (A_M(mathbb R)^0 cap G(mathbb R)^1) times (G(mathbb A)_{gamma} backslash G(mathbb A)) tag{1}$$
This is claimed in James Arthur's book, Introduction to the Trace Formula, section 11, page 61.
Attempt: It is easy to see that $$G(mathbb A)_{gamma} = G(mathbb A)_{gamma}^1 A_G(mathbb R)^0 tag{2}$$ as a direct product, and therefore $G(mathbb A)_{gamma} backslash G(mathbb A) = G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1$. We also clearly have $A_M(mathbb R)^0 cap G(mathbb R)^1 = A_M(mathbb R)^0 cap G(mathbb A)^1$. Thus the right hand side of (1) is equal to
$$ (A_M(mathbb R)^0 cap G(mathbb A)^1) times (G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1)$$
Now replacing $G$ by $M$ in (2), we also have $M(mathbb A)_{gamma} = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0$, and therefore (using our hypothesis about the centralizer $M(mathbb A)_{gamma}$)
$$G(mathbb A)_{gamma}^1 = G(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 tag{3} $$
Next, I claim that
$$M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 (A_M(mathbb R) cap G(mathbb A)^1) tag{4}$$
To verify this claim, we use the fact that
$$M(mathbb A)^1 subset G(mathbb A)^1 cap M(mathbb A)$$
even though this is not in general an equality. It is clear that the right hand side is contained in the left hand side in equation (4). Conversely, suppose $ma$ is in the left hand side, for $m in M(mathbb A)_{gamma}^1$ and $a in A_M(mathbb R)^0$. Since $ma in G(mathbb A)^1$, and $m in M(mathbb A)_{gamma}^1 subseteq G(mathbb A)^1$, we also have $a in G(mathbb A)^1$. Therefore, $ma in M(mathbb A)_{gamma}^1 (A_M(mathbb R)^0 cap G(mathbb A)^1)$.
reductive-groups adeles
asked Jan 14 at 18:43
D_SD_S
13.6k51552
$endgroup$
Let $G$ be a connected reductive group over $k =mathbb Q$ with split component $A_G$. Let $M$ be a $k$-Levi subgroup of $G$ with split component $A_M$. Let $gamma$ be a semisimple element of $M(k)$, and suppose that the centralizer $M(mathbb A)_{gamma} = { m in M(mathbb A) : mgamma m^{-1} = gamma}$ is equal to the centralizer $G(mathbb A)_{gamma}$.
Let $G(mathbb A)^1$ be the kernel of the Harish-Chandra map $H_G$. By definition, this is the group of $g in G(mathbb A)$ such that $chi(g) in mathbb A^{ast}$ has norm one for all $k$-rational characters $chi in X(G)_k$. Then $G(mathbb A) = G(mathbb A)^1 A_G(mathbb R)^{0}$ (direct product of groups), where $A_G(mathbb R)^0$ is the connected component of the identity in $A_G(mathbb R)$, and similarly $M(mathbb A) = M(mathbb A)^1 A_M(mathbb R)^0$.
Let $G(mathbb R)^1 = G(mathbb R) cap G(mathbb A)^1$. I want to understand why we have
$$M(mathbb A)_{gamma}^1 backslash G(mathbb A)^1 cong (A_M(mathbb R)^0 cap G(mathbb R)^1) times (G(mathbb A)_{gamma} backslash G(mathbb A)) tag{1}$$
This is claimed in James Arthur's book, Introduction to the Trace Formula, section 11, page 61.
Attempt: It is easy to see that $$G(mathbb A)_{gamma} = G(mathbb A)_{gamma}^1 A_G(mathbb R)^0 tag{2}$$ as a direct product, and therefore $G(mathbb A)_{gamma} backslash G(mathbb A) = G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1$. We also clearly have $A_M(mathbb R)^0 cap G(mathbb R)^1 = A_M(mathbb R)^0 cap G(mathbb A)^1$. Thus the right hand side of (1) is equal to
$$ (A_M(mathbb R)^0 cap G(mathbb A)^1) times (G(mathbb A)_{gamma}^1 backslash G(mathbb A)^1)$$
Now replacing $G$ by $M$ in (2), we also have $M(mathbb A)_{gamma} = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0$, and therefore (using our hypothesis about the centralizer $M(mathbb A)_{gamma}$)
$$G(mathbb A)_{gamma}^1 = G(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma} cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 tag{3} $$
Next, I claim that
$$M(mathbb A)_{gamma}^1 A_M(mathbb R)^0 cap G(mathbb A)^1 = M(mathbb A)_{gamma}^1 (A_M(mathbb R) cap G(mathbb A)^1) tag{4}$$
To verify this claim, we use the fact that
$$M(mathbb A)^1 subset G(mathbb A)^1 cap M(mathbb A)$$
even though this is not in general an equality. It is clear that the right hand side is contained in the left hand side in equation (4). Conversely, suppose $ma$ is in the left hand side, for $m in M(mathbb A)_{gamma}^1$ and $a in A_M(mathbb R)^0$. Since $ma in G(mathbb A)^1$, and $m in M(mathbb A)_{gamma}^1 subseteq G(mathbb A)^1$, we also have $a in G(mathbb A)^1$. Therefore, $ma in M(mathbb A)_{gamma}^1 (A_M(mathbb R)^0 cap G(mathbb A)^1)$.
reductive-groups adeles
reductive-groups adeles
asked Jan 14 at 18:43
D_SD_S
13.6k51552
asked Jan 14 at 18:43
D_SD_S
13.6k51552
asked Jan 14 at 18:43
D_SD_S
13.6k51552
asked Jan 14 at 18:43
D_SD_S
13.6k51552
asked Jan 14 at 18:43
D_SD_S
13.6k51552
13.6k51552
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