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On the diffeomorphism $mu$ in the definition of a weakly symmetric space
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Selberg in some paper introduced the notion of weakly symmetric space:
Definition. A weakly symmetric (Riemannian) space is a triple $(S, G, mu)$ with $S$ is a Riemannian manifold, $G$ a locally compact transitive group of isometries of $S$ and $mu$ an isometry (not necessarily in $G$) such that:
- $mu^2 in G$
- $mu G mu^-1 = G$
- for all $x, y in S$ there exists $sigma in G$ with
$$ sigma x = mu y quad, quad sigma y = mu x $$
The point is that when $(S, G, mu)$ is such a triplet, then the algebra of $G$-invariant differential operators is commutative.
The same proof works when $G$ is not necessarily transitive and $mu$ is only a diffeomorphism satisfying 1. and 3.
(Examples: $mathbb R^n$ with $H$ the translation group and $mu$ the reflection, a unimodular Lie group with $H$ generated by left and right translations, and $mu$ the inversion, a symmetric space with its full isometry group and $mu=1$.)
Question. In the definition of a weakly symmetric space (and more generally, a triplet $(S, H, mu)$ as above) is $mu$ automatically an isometry? Is $mu G mu^-1 = G$ automatic?
riemannian-geometry isometry differential-operators symmetric-spaces
asked Jul 25 at 13:28
barto
12.9k32478
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up vote
0
down vote
favorite
Selberg in some paper introduced the notion of weakly symmetric space:
Definition. A weakly symmetric (Riemannian) space is a triple $(S, G, mu)$ with $S$ is a Riemannian manifold, $G$ a locally compact transitive group of isometries of $S$ and $mu$ an isometry (not necessarily in $G$) such that:
- $mu^2 in G$
- $mu G mu^-1 = G$
- for all $x, y in S$ there exists $sigma in G$ with
$$ sigma x = mu y quad, quad sigma y = mu x $$
The point is that when $(S, G, mu)$ is such a triplet, then the algebra of $G$-invariant differential operators is commutative.
The same proof works when $G$ is not necessarily transitive and $mu$ is only a diffeomorphism satisfying 1. and 3.
(Examples: $mathbb R^n$ with $H$ the translation group and $mu$ the reflection, a unimodular Lie group with $H$ generated by left and right translations, and $mu$ the inversion, a symmetric space with its full isometry group and $mu=1$.)
Question. In the definition of a weakly symmetric space (and more generally, a triplet $(S, H, mu)$ as above) is $mu$ automatically an isometry? Is $mu G mu^-1 = G$ automatic?
riemannian-geometry isometry differential-operators symmetric-spaces
asked Jul 25 at 13:28
barto
12.9k32478
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Selberg in some paper introduced the notion of weakly symmetric space:
Definition. A weakly symmetric (Riemannian) space is a triple $(S, G, mu)$ with $S$ is a Riemannian manifold, $G$ a locally compact transitive group of isometries of $S$ and $mu$ an isometry (not necessarily in $G$) such that:
- $mu^2 in G$
- $mu G mu^-1 = G$
- for all $x, y in S$ there exists $sigma in G$ with
$$ sigma x = mu y quad, quad sigma y = mu x $$
The point is that when $(S, G, mu)$ is such a triplet, then the algebra of $G$-invariant differential operators is commutative.
The same proof works when $G$ is not necessarily transitive and $mu$ is only a diffeomorphism satisfying 1. and 3.
(Examples: $mathbb R^n$ with $H$ the translation group and $mu$ the reflection, a unimodular Lie group with $H$ generated by left and right translations, and $mu$ the inversion, a symmetric space with its full isometry group and $mu=1$.)
Question. In the definition of a weakly symmetric space (and more generally, a triplet $(S, H, mu)$ as above) is $mu$ automatically an isometry? Is $mu G mu^-1 = G$ automatic?
riemannian-geometry isometry differential-operators symmetric-spaces
asked Jul 25 at 13:28
barto
12.9k32478
Selberg in some paper introduced the notion of weakly symmetric space:
Definition. A weakly symmetric (Riemannian) space is a triple $(S, G, mu)$ with $S$ is a Riemannian manifold, $G$ a locally compact transitive group of isometries of $S$ and $mu$ an isometry (not necessarily in $G$) such that:
- $mu^2 in G$
- $mu G mu^-1 = G$
- for all $x, y in S$ there exists $sigma in G$ with
$$ sigma x = mu y quad, quad sigma y = mu x $$
The point is that when $(S, G, mu)$ is such a triplet, then the algebra of $G$-invariant differential operators is commutative.
The same proof works when $G$ is not necessarily transitive and $mu$ is only a diffeomorphism satisfying 1. and 3.
(Examples: $mathbb R^n$ with $H$ the translation group and $mu$ the reflection, a unimodular Lie group with $H$ generated by left and right translations, and $mu$ the inversion, a symmetric space with its full isometry group and $mu=1$.)
Question. In the definition of a weakly symmetric space (and more generally, a triplet $(S, H, mu)$ as above) is $mu$ automatically an isometry? Is $mu G mu^-1 = G$ automatic?
riemannian-geometry isometry differential-operators symmetric-spaces
asked Jul 25 at 13:28
barto
12.9k32478
edited Jul 25 at 13:43
asked Jul 25 at 13:28
barto
12.9k32478
asked Jul 25 at 13:28
barto
12.9k32478
asked Jul 25 at 13:28
barto
12.9k32478
12.9k32478
add a comment |Â
add a comment |Â
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