Mixing
Fourier Expansion of a function on $mathbb A_k/k$
$begingroup$
Let $k$ be a number field, and let $mathbb A_k$ be the ring adeles of $k$. The quotient group $mathbb A_k/k$ is compact, and the choice of a nontrivial character $psi$ of $mathbb A_k/k$ gives an isomorphism $a mapsto psi_a$ of the additive group $k$ with the Pontryagin dual $(mathbb A_k/k)^{ast}$ of $mathbb A_k/k$, where $psi_a(x) = psi(ax)$.
Question: If $f: mathbb A_k/k rightarrow mathbb C$ is a function, under what circumstances do we have a "Fourier expansion" of $f$, as
$$f(x) = sumlimits_{a in k} c_a psi(ax) tag{$x in mathbb A_k$}$$
The situation of $mathbb A_k$ and $k$ is of course an analogy with $mathbb R$ and $mathbb Z$. The quotient group $mathbb R/mathbb Z$ is compact, and the choice of the character $psi(x) = e^{2 pi i x}$ of $mathbb R/mathbb Z$ gives an isomorphism $n mapsto e^{2pi i nx}$ of $mathbb Z$ with $(mathbb R/mathbb Z)^{ast}$.
The complex Hilbert space $L^2(mathbb R/mathbb Z)$ of square integrable complex valued functions has $e^{2pi i nx} : n in mathbb Z$ as an orthonormal basis (with a suitably normalized Haar measure on $mathbb R/mathbb Z$), so every measurable function $f: mathbb R rightarrow mathbb C$ satisfying $f(x+1) = f(x)$ for almost all $x$, and $int_0^1 |f(x)|^2 dx < infty$ can be written as
$$f(x) = sumlimits_{n in mathbb Z} c_n e^{2pi i nx}$$
for almost all $x$ and for uniquely determined $c_n in mathbb C$.
number-theory fourier-analysis algebraic-number-theory adeles
asked Jan 1 at 20:18
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $k$ be a number field, and let $mathbb A_k$ be the ring adeles of $k$. The quotient group $mathbb A_k/k$ is compact, and the choice of a nontrivial character $psi$ of $mathbb A_k/k$ gives an isomorphism $a mapsto psi_a$ of the additive group $k$ with the Pontryagin dual $(mathbb A_k/k)^{ast}$ of $mathbb A_k/k$, where $psi_a(x) = psi(ax)$.
Question: If $f: mathbb A_k/k rightarrow mathbb C$ is a function, under what circumstances do we have a "Fourier expansion" of $f$, as
$$f(x) = sumlimits_{a in k} c_a psi(ax) tag{$x in mathbb A_k$}$$
The situation of $mathbb A_k$ and $k$ is of course an analogy with $mathbb R$ and $mathbb Z$. The quotient group $mathbb R/mathbb Z$ is compact, and the choice of the character $psi(x) = e^{2 pi i x}$ of $mathbb R/mathbb Z$ gives an isomorphism $n mapsto e^{2pi i nx}$ of $mathbb Z$ with $(mathbb R/mathbb Z)^{ast}$.
The complex Hilbert space $L^2(mathbb R/mathbb Z)$ of square integrable complex valued functions has $e^{2pi i nx} : n in mathbb Z$ as an orthonormal basis (with a suitably normalized Haar measure on $mathbb R/mathbb Z$), so every measurable function $f: mathbb R rightarrow mathbb C$ satisfying $f(x+1) = f(x)$ for almost all $x$, and $int_0^1 |f(x)|^2 dx < infty$ can be written as
$$f(x) = sumlimits_{n in mathbb Z} c_n e^{2pi i nx}$$
for almost all $x$ and for uniquely determined $c_n in mathbb C$.
number-theory fourier-analysis algebraic-number-theory adeles
asked Jan 1 at 20:18
D_SD_S
13.5k51551
$endgroup$
$begingroup$
Okay, I think I know what to do: this should follow from the Peter Weyl theorem
$endgroup$
– D_S
Jan 1 at 22:06
add a comment |
$begingroup$
Let $k$ be a number field, and let $mathbb A_k$ be the ring adeles of $k$. The quotient group $mathbb A_k/k$ is compact, and the choice of a nontrivial character $psi$ of $mathbb A_k/k$ gives an isomorphism $a mapsto psi_a$ of the additive group $k$ with the Pontryagin dual $(mathbb A_k/k)^{ast}$ of $mathbb A_k/k$, where $psi_a(x) = psi(ax)$.
Question: If $f: mathbb A_k/k rightarrow mathbb C$ is a function, under what circumstances do we have a "Fourier expansion" of $f$, as
$$f(x) = sumlimits_{a in k} c_a psi(ax) tag{$x in mathbb A_k$}$$
The situation of $mathbb A_k$ and $k$ is of course an analogy with $mathbb R$ and $mathbb Z$. The quotient group $mathbb R/mathbb Z$ is compact, and the choice of the character $psi(x) = e^{2 pi i x}$ of $mathbb R/mathbb Z$ gives an isomorphism $n mapsto e^{2pi i nx}$ of $mathbb Z$ with $(mathbb R/mathbb Z)^{ast}$.
The complex Hilbert space $L^2(mathbb R/mathbb Z)$ of square integrable complex valued functions has $e^{2pi i nx} : n in mathbb Z$ as an orthonormal basis (with a suitably normalized Haar measure on $mathbb R/mathbb Z$), so every measurable function $f: mathbb R rightarrow mathbb C$ satisfying $f(x+1) = f(x)$ for almost all $x$, and $int_0^1 |f(x)|^2 dx < infty$ can be written as
$$f(x) = sumlimits_{n in mathbb Z} c_n e^{2pi i nx}$$
for almost all $x$ and for uniquely determined $c_n in mathbb C$.
number-theory fourier-analysis algebraic-number-theory adeles
asked Jan 1 at 20:18
D_SD_S
13.5k51551
$endgroup$
Let $k$ be a number field, and let $mathbb A_k$ be the ring adeles of $k$. The quotient group $mathbb A_k/k$ is compact, and the choice of a nontrivial character $psi$ of $mathbb A_k/k$ gives an isomorphism $a mapsto psi_a$ of the additive group $k$ with the Pontryagin dual $(mathbb A_k/k)^{ast}$ of $mathbb A_k/k$, where $psi_a(x) = psi(ax)$.
Question: If $f: mathbb A_k/k rightarrow mathbb C$ is a function, under what circumstances do we have a "Fourier expansion" of $f$, as
$$f(x) = sumlimits_{a in k} c_a psi(ax) tag{$x in mathbb A_k$}$$
The situation of $mathbb A_k$ and $k$ is of course an analogy with $mathbb R$ and $mathbb Z$. The quotient group $mathbb R/mathbb Z$ is compact, and the choice of the character $psi(x) = e^{2 pi i x}$ of $mathbb R/mathbb Z$ gives an isomorphism $n mapsto e^{2pi i nx}$ of $mathbb Z$ with $(mathbb R/mathbb Z)^{ast}$.
The complex Hilbert space $L^2(mathbb R/mathbb Z)$ of square integrable complex valued functions has $e^{2pi i nx} : n in mathbb Z$ as an orthonormal basis (with a suitably normalized Haar measure on $mathbb R/mathbb Z$), so every measurable function $f: mathbb R rightarrow mathbb C$ satisfying $f(x+1) = f(x)$ for almost all $x$, and $int_0^1 |f(x)|^2 dx < infty$ can be written as
$$f(x) = sumlimits_{n in mathbb Z} c_n e^{2pi i nx}$$
for almost all $x$ and for uniquely determined $c_n in mathbb C$.
number-theory fourier-analysis algebraic-number-theory adeles
number-theory fourier-analysis algebraic-number-theory adeles
asked Jan 1 at 20:18
D_SD_S
13.5k51551
asked Jan 1 at 20:18
D_SD_S
13.5k51551
asked Jan 1 at 20:18
D_SD_S
13.5k51551
asked Jan 1 at 20:18
D_SD_S
13.5k51551
asked Jan 1 at 20:18
D_SD_S
13.5k51551
13.5k51551
$begingroup$
Okay, I think I know what to do: this should follow from the Peter Weyl theorem
$endgroup$
– D_S
Jan 1 at 22:06
add a comment |
$begingroup$
Okay, I think I know what to do: this should follow from the Peter Weyl theorem
$endgroup$
– D_S
Jan 1 at 22:06
$begingroup$
Okay, I think I know what to do: this should follow from the Peter Weyl theorem
$endgroup$
– D_S
Jan 1 at 22:06
$begingroup$
Okay, I think I know what to do: this should follow from the Peter Weyl theorem
$endgroup$
– D_S
Jan 1 at 22:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $G$ be a compact Hausdorff topological group, and let $hat{G}$ be the set of isomorphism classes of irreducible unitary representations of $G$. It is a consequence of the Peter-Weyl theorem that each $pi in hat{G}$ is finite dimensional, and the right regular representation $L^2(G)$ is isomorphic to a Hilbert space direct sum of subrepresentations
$$bigopluslimits_{pi in hat{G}} bigoplus_{i=1}^{operatorname{Dim}pi} pi$$
Explicitly, the direct sum $V = bigoplus_{i=1}^{operatorname{Dim}pi} pi$ occurs in $L^2(G)$ as the span of the matrix coefficients of $pi$. If $e_1, ... , e_n$ is an orthonormal basis of $G$, then $c_{ij}(g) = langle e_j, pi(g)e_irangle$, possibly with some scalar modification, form an orthonormal basis of $V$.
In particular, assume $G$ is abelian, for example $G = mathbb R/mathbb Z$ or $G = mathbb A_k/k$. Then each irreducible unitary representation is one dimensional, i.e. a character. If $chi$ is a character of $G$, the corresponding matrix coefficient is just the function $g mapsto chi(g)$. Thus the characters of $G$ form an orthonormal basis of $L^2(G)$. So if $f in L^2(G)$, we can uniquely write
$$f(g) = sumlimits_{chi in hat{G}} c_{chi} chi(g)$$
for uniquely determined $c_{chi} in mathbb C$. If we want to isolate $c_{chi}$, we use the orthogonality relations
$$intlimits_G chi_1(g) overline{chi_2(g)}dg = begin{cases} operatorname{meas}(G) & textrm{if } chi_1 = chi_2 \ 0 & textrm{if } chi_1 neq chi_2 end{cases}$$
which implies $c_{chi} = frac{1}{operatorname{meas}(G)} intlimits_G f(g) overline{chi(g)} dg$.
EDIT: There is a problem here. The series $F(g) = sumlimits c_{chi} chi(g)$ only converges in the $L^2$ norm to $f$, not pointwise. So I still don't know.
answered Jan 1 at 22:43
D_SD_S
13.5k51551
$endgroup$
add a comment |
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$begingroup$
Let $G$ be a compact Hausdorff topological group, and let $hat{G}$ be the set of isomorphism classes of irreducible unitary representations of $G$. It is a consequence of the Peter-Weyl theorem that each $pi in hat{G}$ is finite dimensional, and the right regular representation $L^2(G)$ is isomorphic to a Hilbert space direct sum of subrepresentations
$$bigopluslimits_{pi in hat{G}} bigoplus_{i=1}^{operatorname{Dim}pi} pi$$
Explicitly, the direct sum $V = bigoplus_{i=1}^{operatorname{Dim}pi} pi$ occurs in $L^2(G)$ as the span of the matrix coefficients of $pi$. If $e_1, ... , e_n$ is an orthonormal basis of $G$, then $c_{ij}(g) = langle e_j, pi(g)e_irangle$, possibly with some scalar modification, form an orthonormal basis of $V$.
In particular, assume $G$ is abelian, for example $G = mathbb R/mathbb Z$ or $G = mathbb A_k/k$. Then each irreducible unitary representation is one dimensional, i.e. a character. If $chi$ is a character of $G$, the corresponding matrix coefficient is just the function $g mapsto chi(g)$. Thus the characters of $G$ form an orthonormal basis of $L^2(G)$. So if $f in L^2(G)$, we can uniquely write
$$f(g) = sumlimits_{chi in hat{G}} c_{chi} chi(g)$$
for uniquely determined $c_{chi} in mathbb C$. If we want to isolate $c_{chi}$, we use the orthogonality relations
$$intlimits_G chi_1(g) overline{chi_2(g)}dg = begin{cases} operatorname{meas}(G) & textrm{if } chi_1 = chi_2 \ 0 & textrm{if } chi_1 neq chi_2 end{cases}$$
which implies $c_{chi} = frac{1}{operatorname{meas}(G)} intlimits_G f(g) overline{chi(g)} dg$.
EDIT: There is a problem here. The series $F(g) = sumlimits c_{chi} chi(g)$ only converges in the $L^2$ norm to $f$, not pointwise. So I still don't know.
answered Jan 1 at 22:43
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $G$ be a compact Hausdorff topological group, and let $hat{G}$ be the set of isomorphism classes of irreducible unitary representations of $G$. It is a consequence of the Peter-Weyl theorem that each $pi in hat{G}$ is finite dimensional, and the right regular representation $L^2(G)$ is isomorphic to a Hilbert space direct sum of subrepresentations
$$bigopluslimits_{pi in hat{G}} bigoplus_{i=1}^{operatorname{Dim}pi} pi$$
Explicitly, the direct sum $V = bigoplus_{i=1}^{operatorname{Dim}pi} pi$ occurs in $L^2(G)$ as the span of the matrix coefficients of $pi$. If $e_1, ... , e_n$ is an orthonormal basis of $G$, then $c_{ij}(g) = langle e_j, pi(g)e_irangle$, possibly with some scalar modification, form an orthonormal basis of $V$.
In particular, assume $G$ is abelian, for example $G = mathbb R/mathbb Z$ or $G = mathbb A_k/k$. Then each irreducible unitary representation is one dimensional, i.e. a character. If $chi$ is a character of $G$, the corresponding matrix coefficient is just the function $g mapsto chi(g)$. Thus the characters of $G$ form an orthonormal basis of $L^2(G)$. So if $f in L^2(G)$, we can uniquely write
$$f(g) = sumlimits_{chi in hat{G}} c_{chi} chi(g)$$
for uniquely determined $c_{chi} in mathbb C$. If we want to isolate $c_{chi}$, we use the orthogonality relations
$$intlimits_G chi_1(g) overline{chi_2(g)}dg = begin{cases} operatorname{meas}(G) & textrm{if } chi_1 = chi_2 \ 0 & textrm{if } chi_1 neq chi_2 end{cases}$$
which implies $c_{chi} = frac{1}{operatorname{meas}(G)} intlimits_G f(g) overline{chi(g)} dg$.
EDIT: There is a problem here. The series $F(g) = sumlimits c_{chi} chi(g)$ only converges in the $L^2$ norm to $f$, not pointwise. So I still don't know.
answered Jan 1 at 22:43
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $G$ be a compact Hausdorff topological group, and let $hat{G}$ be the set of isomorphism classes of irreducible unitary representations of $G$. It is a consequence of the Peter-Weyl theorem that each $pi in hat{G}$ is finite dimensional, and the right regular representation $L^2(G)$ is isomorphic to a Hilbert space direct sum of subrepresentations
$$bigopluslimits_{pi in hat{G}} bigoplus_{i=1}^{operatorname{Dim}pi} pi$$
Explicitly, the direct sum $V = bigoplus_{i=1}^{operatorname{Dim}pi} pi$ occurs in $L^2(G)$ as the span of the matrix coefficients of $pi$. If $e_1, ... , e_n$ is an orthonormal basis of $G$, then $c_{ij}(g) = langle e_j, pi(g)e_irangle$, possibly with some scalar modification, form an orthonormal basis of $V$.
In particular, assume $G$ is abelian, for example $G = mathbb R/mathbb Z$ or $G = mathbb A_k/k$. Then each irreducible unitary representation is one dimensional, i.e. a character. If $chi$ is a character of $G$, the corresponding matrix coefficient is just the function $g mapsto chi(g)$. Thus the characters of $G$ form an orthonormal basis of $L^2(G)$. So if $f in L^2(G)$, we can uniquely write
$$f(g) = sumlimits_{chi in hat{G}} c_{chi} chi(g)$$
for uniquely determined $c_{chi} in mathbb C$. If we want to isolate $c_{chi}$, we use the orthogonality relations
$$intlimits_G chi_1(g) overline{chi_2(g)}dg = begin{cases} operatorname{meas}(G) & textrm{if } chi_1 = chi_2 \ 0 & textrm{if } chi_1 neq chi_2 end{cases}$$
which implies $c_{chi} = frac{1}{operatorname{meas}(G)} intlimits_G f(g) overline{chi(g)} dg$.
EDIT: There is a problem here. The series $F(g) = sumlimits c_{chi} chi(g)$ only converges in the $L^2$ norm to $f$, not pointwise. So I still don't know.
answered Jan 1 at 22:43
D_SD_S
13.5k51551
$endgroup$
Let $G$ be a compact Hausdorff topological group, and let $hat{G}$ be the set of isomorphism classes of irreducible unitary representations of $G$. It is a consequence of the Peter-Weyl theorem that each $pi in hat{G}$ is finite dimensional, and the right regular representation $L^2(G)$ is isomorphic to a Hilbert space direct sum of subrepresentations
$$bigopluslimits_{pi in hat{G}} bigoplus_{i=1}^{operatorname{Dim}pi} pi$$
Explicitly, the direct sum $V = bigoplus_{i=1}^{operatorname{Dim}pi} pi$ occurs in $L^2(G)$ as the span of the matrix coefficients of $pi$. If $e_1, ... , e_n$ is an orthonormal basis of $G$, then $c_{ij}(g) = langle e_j, pi(g)e_irangle$, possibly with some scalar modification, form an orthonormal basis of $V$.
In particular, assume $G$ is abelian, for example $G = mathbb R/mathbb Z$ or $G = mathbb A_k/k$. Then each irreducible unitary representation is one dimensional, i.e. a character. If $chi$ is a character of $G$, the corresponding matrix coefficient is just the function $g mapsto chi(g)$. Thus the characters of $G$ form an orthonormal basis of $L^2(G)$. So if $f in L^2(G)$, we can uniquely write
$$f(g) = sumlimits_{chi in hat{G}} c_{chi} chi(g)$$
for uniquely determined $c_{chi} in mathbb C$. If we want to isolate $c_{chi}$, we use the orthogonality relations
$$intlimits_G chi_1(g) overline{chi_2(g)}dg = begin{cases} operatorname{meas}(G) & textrm{if } chi_1 = chi_2 \ 0 & textrm{if } chi_1 neq chi_2 end{cases}$$
which implies $c_{chi} = frac{1}{operatorname{meas}(G)} intlimits_G f(g) overline{chi(g)} dg$.
EDIT: There is a problem here. The series $F(g) = sumlimits c_{chi} chi(g)$ only converges in the $L^2$ norm to $f$, not pointwise. So I still don't know.
answered Jan 1 at 22:43
D_SD_S
13.5k51551
edited Jan 2 at 3:46
answered Jan 1 at 22:43
D_SD_S
13.5k51551
answered Jan 1 at 22:43
D_SD_S
13.5k51551
answered Jan 1 at 22:43
D_SD_S
13.5k51551
13.5k51551
add a comment |
add a comment |
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$begingroup$
Okay, I think I know what to do: this should follow from the Peter Weyl theorem
$endgroup$
– D_S
Jan 1 at 22:06