Mixing
Square integrable functions on Lie groups
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If $Gamma$ is a countably infinite discrete group and $F$ a finite subgroup of $Gamma$, then one knows that there is an isomorphism of $F$-representations
$l^2(Gamma)cong l^2(F)otimes l^2(FbackslashGamma)$,
where $F$ acts on $l^2(Gamma)$ and $l^2(F)$ by via the left regular representation and on $l^2(FbackslashGamma)$ trivially.
Now suppose $G$ is a Lie group and $K$ a compact subgroup of $G$ (equipped with suitable Haar measures). Is there an isomorphism analogous to the above, that is, an isomorphism of $K$-representations
$L^2(G)cong L^2(K)otimes L^2(Kbackslash G),$
where $K$ acts on $L^2(G)$ and $L^2(K)$ via the left regular representation and on $L^2(Kbackslash G)$ trivially?
Thanks!
Edit: I guess the first question is whether there is a measure-theoretic isomorphism $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$. If $K$ were a maximal compact subgroup this would be true since $G$ would be a trivial $K$-bundle over $Kbackslash G$, but even for an arbitrary compact subgroup $K$ this may be true at a measure-theoretic level; I'm not entirely sure. Such an isomorphism of $K$-representations seems plausible since $K$ acts only on each fibre of the principal $K$-bundle $Grightarrow Kbackslash G$.
functional-analysis measure-theory differential-geometry lie-groups topological-groups
asked Jul 21 at 9:08
ougoah
1,060610
add a comment |Â
up vote
2
down vote
favorite
If $Gamma$ is a countably infinite discrete group and $F$ a finite subgroup of $Gamma$, then one knows that there is an isomorphism of $F$-representations
$l^2(Gamma)cong l^2(F)otimes l^2(FbackslashGamma)$,
where $F$ acts on $l^2(Gamma)$ and $l^2(F)$ by via the left regular representation and on $l^2(FbackslashGamma)$ trivially.
Now suppose $G$ is a Lie group and $K$ a compact subgroup of $G$ (equipped with suitable Haar measures). Is there an isomorphism analogous to the above, that is, an isomorphism of $K$-representations
$L^2(G)cong L^2(K)otimes L^2(Kbackslash G),$
where $K$ acts on $L^2(G)$ and $L^2(K)$ via the left regular representation and on $L^2(Kbackslash G)$ trivially?
Thanks!
Edit: I guess the first question is whether there is a measure-theoretic isomorphism $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$. If $K$ were a maximal compact subgroup this would be true since $G$ would be a trivial $K$-bundle over $Kbackslash G$, but even for an arbitrary compact subgroup $K$ this may be true at a measure-theoretic level; I'm not entirely sure. Such an isomorphism of $K$-representations seems plausible since $K$ acts only on each fibre of the principal $K$-bundle $Grightarrow Kbackslash G$.
functional-analysis measure-theory differential-geometry lie-groups topological-groups
asked Jul 21 at 9:08
ougoah
1,060610
Well, "there is an isomorphism" is surely weaker than what you want. The mere existence of such an isomorphism of Hilbert spaces is trivial, since Hilbert spaces are classified by their dimension. Presumably you want the isomorphism to have some additional properties.
– Eric Wofsey
Jul 21 at 13:49
Thanks Eric, indeed I wanted a particular isomorphism of representations of $K$, and I've edited my question. I hope it is specific enough now.
– ougoah
Jul 21 at 15:11
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $Gamma$ is a countably infinite discrete group and $F$ a finite subgroup of $Gamma$, then one knows that there is an isomorphism of $F$-representations
$l^2(Gamma)cong l^2(F)otimes l^2(FbackslashGamma)$,
where $F$ acts on $l^2(Gamma)$ and $l^2(F)$ by via the left regular representation and on $l^2(FbackslashGamma)$ trivially.
Now suppose $G$ is a Lie group and $K$ a compact subgroup of $G$ (equipped with suitable Haar measures). Is there an isomorphism analogous to the above, that is, an isomorphism of $K$-representations
$L^2(G)cong L^2(K)otimes L^2(Kbackslash G),$
where $K$ acts on $L^2(G)$ and $L^2(K)$ via the left regular representation and on $L^2(Kbackslash G)$ trivially?
Thanks!
Edit: I guess the first question is whether there is a measure-theoretic isomorphism $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$. If $K$ were a maximal compact subgroup this would be true since $G$ would be a trivial $K$-bundle over $Kbackslash G$, but even for an arbitrary compact subgroup $K$ this may be true at a measure-theoretic level; I'm not entirely sure. Such an isomorphism of $K$-representations seems plausible since $K$ acts only on each fibre of the principal $K$-bundle $Grightarrow Kbackslash G$.
functional-analysis measure-theory differential-geometry lie-groups topological-groups
asked Jul 21 at 9:08
ougoah
1,060610
If $Gamma$ is a countably infinite discrete group and $F$ a finite subgroup of $Gamma$, then one knows that there is an isomorphism of $F$-representations
$l^2(Gamma)cong l^2(F)otimes l^2(FbackslashGamma)$,
where $F$ acts on $l^2(Gamma)$ and $l^2(F)$ by via the left regular representation and on $l^2(FbackslashGamma)$ trivially.
Now suppose $G$ is a Lie group and $K$ a compact subgroup of $G$ (equipped with suitable Haar measures). Is there an isomorphism analogous to the above, that is, an isomorphism of $K$-representations
$L^2(G)cong L^2(K)otimes L^2(Kbackslash G),$
where $K$ acts on $L^2(G)$ and $L^2(K)$ via the left regular representation and on $L^2(Kbackslash G)$ trivially?
Thanks!
Edit: I guess the first question is whether there is a measure-theoretic isomorphism $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$. If $K$ were a maximal compact subgroup this would be true since $G$ would be a trivial $K$-bundle over $Kbackslash G$, but even for an arbitrary compact subgroup $K$ this may be true at a measure-theoretic level; I'm not entirely sure. Such an isomorphism of $K$-representations seems plausible since $K$ acts only on each fibre of the principal $K$-bundle $Grightarrow Kbackslash G$.
functional-analysis measure-theory differential-geometry lie-groups topological-groups
asked Jul 21 at 9:08
ougoah
1,060610
edited Jul 26 at 1:51
asked Jul 21 at 9:08
ougoah
1,060610
asked Jul 21 at 9:08
ougoah
1,060610
asked Jul 21 at 9:08
ougoah
1,060610
1,060610
Well, "there is an isomorphism" is surely weaker than what you want. The mere existence of such an isomorphism of Hilbert spaces is trivial, since Hilbert spaces are classified by their dimension. Presumably you want the isomorphism to have some additional properties.
– Eric Wofsey
Jul 21 at 13:49
Thanks Eric, indeed I wanted a particular isomorphism of representations of $K$, and I've edited my question. I hope it is specific enough now.
– ougoah
Jul 21 at 15:11
add a comment |Â
Well, "there is an isomorphism" is surely weaker than what you want. The mere existence of such an isomorphism of Hilbert spaces is trivial, since Hilbert spaces are classified by their dimension. Presumably you want the isomorphism to have some additional properties.
– Eric Wofsey
Jul 21 at 13:49
Thanks Eric, indeed I wanted a particular isomorphism of representations of $K$, and I've edited my question. I hope it is specific enough now.
– ougoah
Jul 21 at 15:11
Well, "there is an isomorphism" is surely weaker than what you want. The mere existence of such an isomorphism of Hilbert spaces is trivial, since Hilbert spaces are classified by their dimension. Presumably you want the isomorphism to have some additional properties.
– Eric Wofsey
Jul 21 at 13:49
Well, "there is an isomorphism" is surely weaker than what you want. The mere existence of such an isomorphism of Hilbert spaces is trivial, since Hilbert spaces are classified by their dimension. Presumably you want the isomorphism to have some additional properties.
– Eric Wofsey
Jul 21 at 13:49
Thanks Eric, indeed I wanted a particular isomorphism of representations of $K$, and I've edited my question. I hope it is specific enough now.
– ougoah
Jul 21 at 15:11
Thanks Eric, indeed I wanted a particular isomorphism of representations of $K$, and I've edited my question. I hope it is specific enough now.
– ougoah
Jul 21 at 15:11
add a comment |Â
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Well, "there is an isomorphism" is surely weaker than what you want. The mere existence of such an isomorphism of Hilbert spaces is trivial, since Hilbert spaces are classified by their dimension. Presumably you want the isomorphism to have some additional properties.
– Eric Wofsey
Jul 21 at 13:49
Thanks Eric, indeed I wanted a particular isomorphism of representations of $K$, and I've edited my question. I hope it is specific enough now.
– ougoah
Jul 21 at 15:11