Mixing
Property of smooth functions on the adeles
$begingroup$
Let $k$ be a number field, $mathbb A$ the ring of adeles of $k$, $mathbb A_f$ the finite adeles, and $mathbb A_{infty}$ the infinite adeles.
Let $phi: mathbb A = mathbb A_{infty} times mathbb A_f rightarrow mathbb C$ be a continuous function which is smooth in the first variable and locally constant in the second variable. Some authors call such a function smooth.
Is it the case that for every $x in mathbb A$, there exists an open neighborhood $U$ of $x$, and an open compact subgroup $H$ of $mathbb A_f$, such that
$$phi(x'+h) = phi(x')$$
for all $x' in U, h in H$?
This seems to be the claim in an answer to one of my questions on Mathoverflow, which I am trying to understand.
My attempt:
Lemma 1: For every $x in mathbb A$, there exists an open compact subgroup $H$ of $mathbb A_f$ such that $phi(x+h) = phi(x)$ for all $h in H$.
The problem becomes whether in the Lemma we can choose $H$ uniformly for $x'$ in a sufficiently small neighborhood of $x$.
Proof of Lemma 1: Let $x = (x_1,x_2) in mathbb A = mathbb A_{infty} times mathbb A_f$. There exists an open neighborhood $V subset mathbb A_f$ of $x_2$ such that $$phi(x_1,y') = phi(x_1,x_2)$$ for all $y' in V$. Let $H$ be an open compact subgroup of $mathbb A_f$ which is contained in the neighborhood $V - x_2$ of the identity. Then for all $h in H$, we have $h + x_2 in V$, and so
$$phi(x + h) = phi((x_1,x_2) + (0,h)) = phi(x_1,x_2+h) = phi(x_1,x_2) = phi(x)$$
$blacksquare$
To prove what I want, I will definitely need to make use of the fact that $phi$ is continuous. So far I haven't been successful.
number-theory algebraic-number-theory topological-groups adeles
asked Jan 22 at 4:24
D_SD_S
13.7k61552
$endgroup$
add a comment |
$begingroup$
Let $k$ be a number field, $mathbb A$ the ring of adeles of $k$, $mathbb A_f$ the finite adeles, and $mathbb A_{infty}$ the infinite adeles.
Let $phi: mathbb A = mathbb A_{infty} times mathbb A_f rightarrow mathbb C$ be a continuous function which is smooth in the first variable and locally constant in the second variable. Some authors call such a function smooth.
Is it the case that for every $x in mathbb A$, there exists an open neighborhood $U$ of $x$, and an open compact subgroup $H$ of $mathbb A_f$, such that
$$phi(x'+h) = phi(x')$$
for all $x' in U, h in H$?
This seems to be the claim in an answer to one of my questions on Mathoverflow, which I am trying to understand.
My attempt:
Lemma 1: For every $x in mathbb A$, there exists an open compact subgroup $H$ of $mathbb A_f$ such that $phi(x+h) = phi(x)$ for all $h in H$.
The problem becomes whether in the Lemma we can choose $H$ uniformly for $x'$ in a sufficiently small neighborhood of $x$.
Proof of Lemma 1: Let $x = (x_1,x_2) in mathbb A = mathbb A_{infty} times mathbb A_f$. There exists an open neighborhood $V subset mathbb A_f$ of $x_2$ such that $$phi(x_1,y') = phi(x_1,x_2)$$ for all $y' in V$. Let $H$ be an open compact subgroup of $mathbb A_f$ which is contained in the neighborhood $V - x_2$ of the identity. Then for all $h in H$, we have $h + x_2 in V$, and so
$$phi(x + h) = phi((x_1,x_2) + (0,h)) = phi(x_1,x_2+h) = phi(x_1,x_2) = phi(x)$$
$blacksquare$
To prove what I want, I will definitely need to make use of the fact that $phi$ is continuous. So far I haven't been successful.
number-theory algebraic-number-theory topological-groups adeles
asked Jan 22 at 4:24
D_SD_S
13.7k61552
$endgroup$
add a comment |
$begingroup$
Let $k$ be a number field, $mathbb A$ the ring of adeles of $k$, $mathbb A_f$ the finite adeles, and $mathbb A_{infty}$ the infinite adeles.
Let $phi: mathbb A = mathbb A_{infty} times mathbb A_f rightarrow mathbb C$ be a continuous function which is smooth in the first variable and locally constant in the second variable. Some authors call such a function smooth.
Is it the case that for every $x in mathbb A$, there exists an open neighborhood $U$ of $x$, and an open compact subgroup $H$ of $mathbb A_f$, such that
$$phi(x'+h) = phi(x')$$
for all $x' in U, h in H$?
This seems to be the claim in an answer to one of my questions on Mathoverflow, which I am trying to understand.
My attempt:
Lemma 1: For every $x in mathbb A$, there exists an open compact subgroup $H$ of $mathbb A_f$ such that $phi(x+h) = phi(x)$ for all $h in H$.
The problem becomes whether in the Lemma we can choose $H$ uniformly for $x'$ in a sufficiently small neighborhood of $x$.
Proof of Lemma 1: Let $x = (x_1,x_2) in mathbb A = mathbb A_{infty} times mathbb A_f$. There exists an open neighborhood $V subset mathbb A_f$ of $x_2$ such that $$phi(x_1,y') = phi(x_1,x_2)$$ for all $y' in V$. Let $H$ be an open compact subgroup of $mathbb A_f$ which is contained in the neighborhood $V - x_2$ of the identity. Then for all $h in H$, we have $h + x_2 in V$, and so
$$phi(x + h) = phi((x_1,x_2) + (0,h)) = phi(x_1,x_2+h) = phi(x_1,x_2) = phi(x)$$
$blacksquare$
To prove what I want, I will definitely need to make use of the fact that $phi$ is continuous. So far I haven't been successful.
number-theory algebraic-number-theory topological-groups adeles
asked Jan 22 at 4:24
D_SD_S
13.7k61552
$endgroup$
Let $k$ be a number field, $mathbb A$ the ring of adeles of $k$, $mathbb A_f$ the finite adeles, and $mathbb A_{infty}$ the infinite adeles.
Let $phi: mathbb A = mathbb A_{infty} times mathbb A_f rightarrow mathbb C$ be a continuous function which is smooth in the first variable and locally constant in the second variable. Some authors call such a function smooth.
Is it the case that for every $x in mathbb A$, there exists an open neighborhood $U$ of $x$, and an open compact subgroup $H$ of $mathbb A_f$, such that
$$phi(x'+h) = phi(x')$$
for all $x' in U, h in H$?
This seems to be the claim in an answer to one of my questions on Mathoverflow, which I am trying to understand.
My attempt:
Lemma 1: For every $x in mathbb A$, there exists an open compact subgroup $H$ of $mathbb A_f$ such that $phi(x+h) = phi(x)$ for all $h in H$.
The problem becomes whether in the Lemma we can choose $H$ uniformly for $x'$ in a sufficiently small neighborhood of $x$.
Proof of Lemma 1: Let $x = (x_1,x_2) in mathbb A = mathbb A_{infty} times mathbb A_f$. There exists an open neighborhood $V subset mathbb A_f$ of $x_2$ such that $$phi(x_1,y') = phi(x_1,x_2)$$ for all $y' in V$. Let $H$ be an open compact subgroup of $mathbb A_f$ which is contained in the neighborhood $V - x_2$ of the identity. Then for all $h in H$, we have $h + x_2 in V$, and so
$$phi(x + h) = phi((x_1,x_2) + (0,h)) = phi(x_1,x_2+h) = phi(x_1,x_2) = phi(x)$$
$blacksquare$
To prove what I want, I will definitely need to make use of the fact that $phi$ is continuous. So far I haven't been successful.
number-theory algebraic-number-theory topological-groups adeles
number-theory algebraic-number-theory topological-groups adeles
asked Jan 22 at 4:24
D_SD_S
13.7k61552
asked Jan 22 at 4:24
D_SD_S
13.7k61552
asked Jan 22 at 4:24
D_SD_S
13.7k61552
asked Jan 22 at 4:24
D_SD_S
13.7k61552
asked Jan 22 at 4:24
D_SD_S
13.7k61552
13.7k61552
add a comment |
add a comment |
1 Answer
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$begingroup$
Say a function $f:Xtimes Yto Z$ is [locally] [constant in the second variable] if each $(x,y)$ has a neighborhood $Utimes Vsubset Xtimes Y$ of $(x,y)$ such that $f(u,v)=f(u,v')$ for all $uin U$ and $v,v'$ in $V.$
Say a function $f:Xtimes Yto Z$ is [locally constant] [in the second variable] if for each $(x,y)$ there is a neighborhood $Vsubset Y$ of $y$ such that $f(x,v)=f(x,v')$ for all $v,v'$ in $V.$
It sounds like you're using the second meaning to define smooth functions. I believe the first definition is right and the second definition is wrong. At least from a quick Google of "smooth adelic function". ☺
Here's a counterexample that I hope answers your question.
Let $psi:mathbb Rtomathbb R$ be a bump function such that $psi(x)=1$ for $|x|leq 1$ and $psi(x)=0$ for $|x|geq 2.$ Define a function $phi:mathbb A_mathbb{Q}tomathbb R$ by
$$phi(x_infty,x_2,dots)=begin{cases}
x_inftycdot psi(x_infty/|x_2|_2)&(x_2neq 0)\
x_infty&(x_2=0)
end{cases}$$
where $|x_2|_2$ is the usual $2$-adic norm.
To see this is continuous, define $N={0}cup{2^{-v}mid vinmathbb Z}.$
If you believe that the map $mathbb A_mathbb{Q}tomathbb Rtimes N$ given by $(x_infty,x_2,dots)mapsto(x_infty,|x_2|_2)$ is continuous, and that the map $mathbb Rtimes Ntomathbb R$ defined by $(x,y)mapsto begin{cases}xcdot psi(x/y)&(yneq 0)\x&(y=0)end{cases}$ is continuous, then you should believe that their composition $phi$ is also continuous.
This function is [locally constant] [in the second variable]: this is hopefully clear for $x_2neq 0,$ while to check the property at $x_2=0$ we need that $phi(0,x_2,dots)=0$ for all $x_2,$ and for each $x_inftyneq 0$ we have $phi(x_infty,x_2,dots)=0$ whenever $|x_2|_2leq tfrac12 |x_infty|.$
But there are no $U,H$ with $U$ an open neighborhood of $0$ and $H$ an open subgroup of $mathbb Q_2$ such that $phi(u,h,dots)=phi(u,0,dots)$ for all $uin U$ and $hin H.$ Just pick a non-zero $hin H$ and take $0<uleq |h|_2$ to get $phi(u,h,dots)=uneq 0=phi(u,0,dots).$
answered Jan 26 at 18:56
DapDap
17.9k841
$endgroup$
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
add a comment |
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$begingroup$
Say a function $f:Xtimes Yto Z$ is [locally] [constant in the second variable] if each $(x,y)$ has a neighborhood $Utimes Vsubset Xtimes Y$ of $(x,y)$ such that $f(u,v)=f(u,v')$ for all $uin U$ and $v,v'$ in $V.$
Say a function $f:Xtimes Yto Z$ is [locally constant] [in the second variable] if for each $(x,y)$ there is a neighborhood $Vsubset Y$ of $y$ such that $f(x,v)=f(x,v')$ for all $v,v'$ in $V.$
It sounds like you're using the second meaning to define smooth functions. I believe the first definition is right and the second definition is wrong. At least from a quick Google of "smooth adelic function". ☺
Here's a counterexample that I hope answers your question.
Let $psi:mathbb Rtomathbb R$ be a bump function such that $psi(x)=1$ for $|x|leq 1$ and $psi(x)=0$ for $|x|geq 2.$ Define a function $phi:mathbb A_mathbb{Q}tomathbb R$ by
$$phi(x_infty,x_2,dots)=begin{cases}
x_inftycdot psi(x_infty/|x_2|_2)&(x_2neq 0)\
x_infty&(x_2=0)
end{cases}$$
where $|x_2|_2$ is the usual $2$-adic norm.
To see this is continuous, define $N={0}cup{2^{-v}mid vinmathbb Z}.$
If you believe that the map $mathbb A_mathbb{Q}tomathbb Rtimes N$ given by $(x_infty,x_2,dots)mapsto(x_infty,|x_2|_2)$ is continuous, and that the map $mathbb Rtimes Ntomathbb R$ defined by $(x,y)mapsto begin{cases}xcdot psi(x/y)&(yneq 0)\x&(y=0)end{cases}$ is continuous, then you should believe that their composition $phi$ is also continuous.
This function is [locally constant] [in the second variable]: this is hopefully clear for $x_2neq 0,$ while to check the property at $x_2=0$ we need that $phi(0,x_2,dots)=0$ for all $x_2,$ and for each $x_inftyneq 0$ we have $phi(x_infty,x_2,dots)=0$ whenever $|x_2|_2leq tfrac12 |x_infty|.$
But there are no $U,H$ with $U$ an open neighborhood of $0$ and $H$ an open subgroup of $mathbb Q_2$ such that $phi(u,h,dots)=phi(u,0,dots)$ for all $uin U$ and $hin H.$ Just pick a non-zero $hin H$ and take $0<uleq |h|_2$ to get $phi(u,h,dots)=uneq 0=phi(u,0,dots).$
answered Jan 26 at 18:56
DapDap
17.9k841
$endgroup$
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
add a comment |
$begingroup$
Say a function $f:Xtimes Yto Z$ is [locally] [constant in the second variable] if each $(x,y)$ has a neighborhood $Utimes Vsubset Xtimes Y$ of $(x,y)$ such that $f(u,v)=f(u,v')$ for all $uin U$ and $v,v'$ in $V.$
Say a function $f:Xtimes Yto Z$ is [locally constant] [in the second variable] if for each $(x,y)$ there is a neighborhood $Vsubset Y$ of $y$ such that $f(x,v)=f(x,v')$ for all $v,v'$ in $V.$
It sounds like you're using the second meaning to define smooth functions. I believe the first definition is right and the second definition is wrong. At least from a quick Google of "smooth adelic function". ☺
Here's a counterexample that I hope answers your question.
Let $psi:mathbb Rtomathbb R$ be a bump function such that $psi(x)=1$ for $|x|leq 1$ and $psi(x)=0$ for $|x|geq 2.$ Define a function $phi:mathbb A_mathbb{Q}tomathbb R$ by
$$phi(x_infty,x_2,dots)=begin{cases}
x_inftycdot psi(x_infty/|x_2|_2)&(x_2neq 0)\
x_infty&(x_2=0)
end{cases}$$
where $|x_2|_2$ is the usual $2$-adic norm.
To see this is continuous, define $N={0}cup{2^{-v}mid vinmathbb Z}.$
If you believe that the map $mathbb A_mathbb{Q}tomathbb Rtimes N$ given by $(x_infty,x_2,dots)mapsto(x_infty,|x_2|_2)$ is continuous, and that the map $mathbb Rtimes Ntomathbb R$ defined by $(x,y)mapsto begin{cases}xcdot psi(x/y)&(yneq 0)\x&(y=0)end{cases}$ is continuous, then you should believe that their composition $phi$ is also continuous.
This function is [locally constant] [in the second variable]: this is hopefully clear for $x_2neq 0,$ while to check the property at $x_2=0$ we need that $phi(0,x_2,dots)=0$ for all $x_2,$ and for each $x_inftyneq 0$ we have $phi(x_infty,x_2,dots)=0$ whenever $|x_2|_2leq tfrac12 |x_infty|.$
But there are no $U,H$ with $U$ an open neighborhood of $0$ and $H$ an open subgroup of $mathbb Q_2$ such that $phi(u,h,dots)=phi(u,0,dots)$ for all $uin U$ and $hin H.$ Just pick a non-zero $hin H$ and take $0<uleq |h|_2$ to get $phi(u,h,dots)=uneq 0=phi(u,0,dots).$
answered Jan 26 at 18:56
DapDap
17.9k841
$endgroup$
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
add a comment |
$begingroup$
Say a function $f:Xtimes Yto Z$ is [locally] [constant in the second variable] if each $(x,y)$ has a neighborhood $Utimes Vsubset Xtimes Y$ of $(x,y)$ such that $f(u,v)=f(u,v')$ for all $uin U$ and $v,v'$ in $V.$
Say a function $f:Xtimes Yto Z$ is [locally constant] [in the second variable] if for each $(x,y)$ there is a neighborhood $Vsubset Y$ of $y$ such that $f(x,v)=f(x,v')$ for all $v,v'$ in $V.$
It sounds like you're using the second meaning to define smooth functions. I believe the first definition is right and the second definition is wrong. At least from a quick Google of "smooth adelic function". ☺
Here's a counterexample that I hope answers your question.
Let $psi:mathbb Rtomathbb R$ be a bump function such that $psi(x)=1$ for $|x|leq 1$ and $psi(x)=0$ for $|x|geq 2.$ Define a function $phi:mathbb A_mathbb{Q}tomathbb R$ by
$$phi(x_infty,x_2,dots)=begin{cases}
x_inftycdot psi(x_infty/|x_2|_2)&(x_2neq 0)\
x_infty&(x_2=0)
end{cases}$$
where $|x_2|_2$ is the usual $2$-adic norm.
To see this is continuous, define $N={0}cup{2^{-v}mid vinmathbb Z}.$
If you believe that the map $mathbb A_mathbb{Q}tomathbb Rtimes N$ given by $(x_infty,x_2,dots)mapsto(x_infty,|x_2|_2)$ is continuous, and that the map $mathbb Rtimes Ntomathbb R$ defined by $(x,y)mapsto begin{cases}xcdot psi(x/y)&(yneq 0)\x&(y=0)end{cases}$ is continuous, then you should believe that their composition $phi$ is also continuous.
This function is [locally constant] [in the second variable]: this is hopefully clear for $x_2neq 0,$ while to check the property at $x_2=0$ we need that $phi(0,x_2,dots)=0$ for all $x_2,$ and for each $x_inftyneq 0$ we have $phi(x_infty,x_2,dots)=0$ whenever $|x_2|_2leq tfrac12 |x_infty|.$
But there are no $U,H$ with $U$ an open neighborhood of $0$ and $H$ an open subgroup of $mathbb Q_2$ such that $phi(u,h,dots)=phi(u,0,dots)$ for all $uin U$ and $hin H.$ Just pick a non-zero $hin H$ and take $0<uleq |h|_2$ to get $phi(u,h,dots)=uneq 0=phi(u,0,dots).$
answered Jan 26 at 18:56
DapDap
17.9k841
$endgroup$
Say a function $f:Xtimes Yto Z$ is [locally] [constant in the second variable] if each $(x,y)$ has a neighborhood $Utimes Vsubset Xtimes Y$ of $(x,y)$ such that $f(u,v)=f(u,v')$ for all $uin U$ and $v,v'$ in $V.$
Say a function $f:Xtimes Yto Z$ is [locally constant] [in the second variable] if for each $(x,y)$ there is a neighborhood $Vsubset Y$ of $y$ such that $f(x,v)=f(x,v')$ for all $v,v'$ in $V.$
It sounds like you're using the second meaning to define smooth functions. I believe the first definition is right and the second definition is wrong. At least from a quick Google of "smooth adelic function". ☺
Here's a counterexample that I hope answers your question.
Let $psi:mathbb Rtomathbb R$ be a bump function such that $psi(x)=1$ for $|x|leq 1$ and $psi(x)=0$ for $|x|geq 2.$ Define a function $phi:mathbb A_mathbb{Q}tomathbb R$ by
$$phi(x_infty,x_2,dots)=begin{cases}
x_inftycdot psi(x_infty/|x_2|_2)&(x_2neq 0)\
x_infty&(x_2=0)
end{cases}$$
where $|x_2|_2$ is the usual $2$-adic norm.
To see this is continuous, define $N={0}cup{2^{-v}mid vinmathbb Z}.$
If you believe that the map $mathbb A_mathbb{Q}tomathbb Rtimes N$ given by $(x_infty,x_2,dots)mapsto(x_infty,|x_2|_2)$ is continuous, and that the map $mathbb Rtimes Ntomathbb R$ defined by $(x,y)mapsto begin{cases}xcdot psi(x/y)&(yneq 0)\x&(y=0)end{cases}$ is continuous, then you should believe that their composition $phi$ is also continuous.
This function is [locally constant] [in the second variable]: this is hopefully clear for $x_2neq 0,$ while to check the property at $x_2=0$ we need that $phi(0,x_2,dots)=0$ for all $x_2,$ and for each $x_inftyneq 0$ we have $phi(x_infty,x_2,dots)=0$ whenever $|x_2|_2leq tfrac12 |x_infty|.$
But there are no $U,H$ with $U$ an open neighborhood of $0$ and $H$ an open subgroup of $mathbb Q_2$ such that $phi(u,h,dots)=phi(u,0,dots)$ for all $uin U$ and $hin H.$ Just pick a non-zero $hin H$ and take $0<uleq |h|_2$ to get $phi(u,h,dots)=uneq 0=phi(u,0,dots).$
answered Jan 26 at 18:56
DapDap
17.9k841
answered Jan 26 at 18:56
DapDap
17.9k841
answered Jan 26 at 18:56
DapDap
17.9k841
answered Jan 26 at 18:56
DapDap
17.9k841
17.9k841
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
add a comment |
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
$begingroup$
Thanks very much. So what I wanted to happen is actually built into the definition of smoothness.
$endgroup$
– D_S
Jan 26 at 21:01
add a comment |
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