Ergodic action of dense subgroup
Let $G$ be a group acting ercodically on a probability measure space $(X, mu)$. Let $Gamma$ be a countable dense subgroup of $G$. Is the action of $Gamma$ also ergodic?
The case I am interested in is actually when $G$ is a Lie group acting smoothly on a compact submanifold of $mathbb{R}^n$ with the normalized Lebesgue measure.
Here by ergodic action I mean that if $mu(gA Delta A) = 0$ for all $g in G$, then $mu(A) = 0$ or 1.
lie-groups lebesgue-measure group-actions ergodic-theory
asked Nov 27 '18 at 15:23
frafour
878212
add a comment |
Let $G$ be a group acting ercodically on a probability measure space $(X, mu)$. Let $Gamma$ be a countable dense subgroup of $G$. Is the action of $Gamma$ also ergodic?
The case I am interested in is actually when $G$ is a Lie group acting smoothly on a compact submanifold of $mathbb{R}^n$ with the normalized Lebesgue measure.
Here by ergodic action I mean that if $mu(gA Delta A) = 0$ for all $g in G$, then $mu(A) = 0$ or 1.
lie-groups lebesgue-measure group-actions ergodic-theory
asked Nov 27 '18 at 15:23
frafour
878212
add a comment |
Let $G$ be a group acting ercodically on a probability measure space $(X, mu)$. Let $Gamma$ be a countable dense subgroup of $G$. Is the action of $Gamma$ also ergodic?
The case I am interested in is actually when $G$ is a Lie group acting smoothly on a compact submanifold of $mathbb{R}^n$ with the normalized Lebesgue measure.
Here by ergodic action I mean that if $mu(gA Delta A) = 0$ for all $g in G$, then $mu(A) = 0$ or 1.
lie-groups lebesgue-measure group-actions ergodic-theory
asked Nov 27 '18 at 15:23
frafour
878212
Let $G$ be a group acting ercodically on a probability measure space $(X, mu)$. Let $Gamma$ be a countable dense subgroup of $G$. Is the action of $Gamma$ also ergodic?
The case I am interested in is actually when $G$ is a Lie group acting smoothly on a compact submanifold of $mathbb{R}^n$ with the normalized Lebesgue measure.
Here by ergodic action I mean that if $mu(gA Delta A) = 0$ for all $g in G$, then $mu(A) = 0$ or 1.
lie-groups lebesgue-measure group-actions ergodic-theory
lie-groups lebesgue-measure group-actions ergodic-theory
asked Nov 27 '18 at 15:23
frafour
878212
asked Nov 27 '18 at 15:23
frafour
878212
asked Nov 27 '18 at 15:23
frafour
878212
asked Nov 27 '18 at 15:23
frafour
878212
asked Nov 27 '18 at 15:23
frafour
878212
878212
add a comment |
add a comment |
1 Answer
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No in general, yes for the case your are interested in.
For the "no" part: In the general case, this is lacking a assumption giving a "link" between the topology of $G$ and the action; here is an example of such a discontinuous action:
Let $G=mathbb{Z}oplusmathbb{Z}sqrt{2}oplus mathbb{Z}sqrt{3}$, $Gamma=mathbb{Z}oplus mathbb{Z}sqrt{2}$. We endow $G$ with the induced topology from $mathbb{R}$. In this case, $Gamma$ is a dense subgroup. Now consider an ergodic action of $G/Gammasimeq mathbb{Z}sqrt{3}$ on a measure space (for example, an irrational rotation on the circle). This gives an action of $G$ itself, which is ergodic, but the action of $Gamma$ is trivial.
For the "yes" part: the action of $G$ on $X$ induces a unitary representation on $L^2(X,mu)$ (the Koopman representation)
$$pi:Gto mathcal{U}(L^2(X)),$$
$$pi(f)(x)=f(g^{-1}x).$$
A useful property that may (or may not) have this representation is to be strongly continuous, i.e.
$$g_nto gimplies forall f in L^2(X), ;| pi(g_n)f-pi(g)f|to 0.$$
The relevance of this property to ergodicity of dense subgroup is that in this case, the stabilizer of a characteristic function $f=1_A$
$$Stab(f)={ g in G , : , pi(g)f=f },$$
is then a closed subgroup; so if a set $A$ is $Gamma$-invariant, $f=1_A$ is $Gamma$-invariant, so is also $bar{Gamma}=G$-invariant, so $mu(A)in {0,1}$.
In the case you are interested in, this property of strong continuity is satisfied; see for example Lemma 1.6 in those notes
http://u.math.biu.ac.il/~solomyb/TEACH/17/GrAct/Lec7.pdf
for a statement giving sufficient conditions and a sketch of proof.
edited Nov 28 '18 at 9:42
user39082
1,235513
answered Nov 28 '18 at 9:28
user120527
1,616215
add a comment |
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1 Answer
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1 Answer
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No in general, yes for the case your are interested in.
For the "no" part: In the general case, this is lacking a assumption giving a "link" between the topology of $G$ and the action; here is an example of such a discontinuous action:
Let $G=mathbb{Z}oplusmathbb{Z}sqrt{2}oplus mathbb{Z}sqrt{3}$, $Gamma=mathbb{Z}oplus mathbb{Z}sqrt{2}$. We endow $G$ with the induced topology from $mathbb{R}$. In this case, $Gamma$ is a dense subgroup. Now consider an ergodic action of $G/Gammasimeq mathbb{Z}sqrt{3}$ on a measure space (for example, an irrational rotation on the circle). This gives an action of $G$ itself, which is ergodic, but the action of $Gamma$ is trivial.
For the "yes" part: the action of $G$ on $X$ induces a unitary representation on $L^2(X,mu)$ (the Koopman representation)
$$pi:Gto mathcal{U}(L^2(X)),$$
$$pi(f)(x)=f(g^{-1}x).$$
A useful property that may (or may not) have this representation is to be strongly continuous, i.e.
$$g_nto gimplies forall f in L^2(X), ;| pi(g_n)f-pi(g)f|to 0.$$
The relevance of this property to ergodicity of dense subgroup is that in this case, the stabilizer of a characteristic function $f=1_A$
$$Stab(f)={ g in G , : , pi(g)f=f },$$
is then a closed subgroup; so if a set $A$ is $Gamma$-invariant, $f=1_A$ is $Gamma$-invariant, so is also $bar{Gamma}=G$-invariant, so $mu(A)in {0,1}$.
In the case you are interested in, this property of strong continuity is satisfied; see for example Lemma 1.6 in those notes
http://u.math.biu.ac.il/~solomyb/TEACH/17/GrAct/Lec7.pdf
for a statement giving sufficient conditions and a sketch of proof.
edited Nov 28 '18 at 9:42
user39082
1,235513
answered Nov 28 '18 at 9:28
user120527
1,616215
add a comment |
No in general, yes for the case your are interested in.
For the "no" part: In the general case, this is lacking a assumption giving a "link" between the topology of $G$ and the action; here is an example of such a discontinuous action:
Let $G=mathbb{Z}oplusmathbb{Z}sqrt{2}oplus mathbb{Z}sqrt{3}$, $Gamma=mathbb{Z}oplus mathbb{Z}sqrt{2}$. We endow $G$ with the induced topology from $mathbb{R}$. In this case, $Gamma$ is a dense subgroup. Now consider an ergodic action of $G/Gammasimeq mathbb{Z}sqrt{3}$ on a measure space (for example, an irrational rotation on the circle). This gives an action of $G$ itself, which is ergodic, but the action of $Gamma$ is trivial.
For the "yes" part: the action of $G$ on $X$ induces a unitary representation on $L^2(X,mu)$ (the Koopman representation)
$$pi:Gto mathcal{U}(L^2(X)),$$
$$pi(f)(x)=f(g^{-1}x).$$
A useful property that may (or may not) have this representation is to be strongly continuous, i.e.
$$g_nto gimplies forall f in L^2(X), ;| pi(g_n)f-pi(g)f|to 0.$$
The relevance of this property to ergodicity of dense subgroup is that in this case, the stabilizer of a characteristic function $f=1_A$
$$Stab(f)={ g in G , : , pi(g)f=f },$$
is then a closed subgroup; so if a set $A$ is $Gamma$-invariant, $f=1_A$ is $Gamma$-invariant, so is also $bar{Gamma}=G$-invariant, so $mu(A)in {0,1}$.
In the case you are interested in, this property of strong continuity is satisfied; see for example Lemma 1.6 in those notes
http://u.math.biu.ac.il/~solomyb/TEACH/17/GrAct/Lec7.pdf
for a statement giving sufficient conditions and a sketch of proof.
edited Nov 28 '18 at 9:42
user39082
1,235513
answered Nov 28 '18 at 9:28
user120527
1,616215
add a comment |
No in general, yes for the case your are interested in.
For the "no" part: In the general case, this is lacking a assumption giving a "link" between the topology of $G$ and the action; here is an example of such a discontinuous action:
Let $G=mathbb{Z}oplusmathbb{Z}sqrt{2}oplus mathbb{Z}sqrt{3}$, $Gamma=mathbb{Z}oplus mathbb{Z}sqrt{2}$. We endow $G$ with the induced topology from $mathbb{R}$. In this case, $Gamma$ is a dense subgroup. Now consider an ergodic action of $G/Gammasimeq mathbb{Z}sqrt{3}$ on a measure space (for example, an irrational rotation on the circle). This gives an action of $G$ itself, which is ergodic, but the action of $Gamma$ is trivial.
For the "yes" part: the action of $G$ on $X$ induces a unitary representation on $L^2(X,mu)$ (the Koopman representation)
$$pi:Gto mathcal{U}(L^2(X)),$$
$$pi(f)(x)=f(g^{-1}x).$$
A useful property that may (or may not) have this representation is to be strongly continuous, i.e.
$$g_nto gimplies forall f in L^2(X), ;| pi(g_n)f-pi(g)f|to 0.$$
The relevance of this property to ergodicity of dense subgroup is that in this case, the stabilizer of a characteristic function $f=1_A$
$$Stab(f)={ g in G , : , pi(g)f=f },$$
is then a closed subgroup; so if a set $A$ is $Gamma$-invariant, $f=1_A$ is $Gamma$-invariant, so is also $bar{Gamma}=G$-invariant, so $mu(A)in {0,1}$.
In the case you are interested in, this property of strong continuity is satisfied; see for example Lemma 1.6 in those notes
http://u.math.biu.ac.il/~solomyb/TEACH/17/GrAct/Lec7.pdf
for a statement giving sufficient conditions and a sketch of proof.
edited Nov 28 '18 at 9:42
user39082
1,235513
answered Nov 28 '18 at 9:28
user120527
1,616215
No in general, yes for the case your are interested in.
For the "no" part: In the general case, this is lacking a assumption giving a "link" between the topology of $G$ and the action; here is an example of such a discontinuous action:
Let $G=mathbb{Z}oplusmathbb{Z}sqrt{2}oplus mathbb{Z}sqrt{3}$, $Gamma=mathbb{Z}oplus mathbb{Z}sqrt{2}$. We endow $G$ with the induced topology from $mathbb{R}$. In this case, $Gamma$ is a dense subgroup. Now consider an ergodic action of $G/Gammasimeq mathbb{Z}sqrt{3}$ on a measure space (for example, an irrational rotation on the circle). This gives an action of $G$ itself, which is ergodic, but the action of $Gamma$ is trivial.
For the "yes" part: the action of $G$ on $X$ induces a unitary representation on $L^2(X,mu)$ (the Koopman representation)
$$pi:Gto mathcal{U}(L^2(X)),$$
$$pi(f)(x)=f(g^{-1}x).$$
A useful property that may (or may not) have this representation is to be strongly continuous, i.e.
$$g_nto gimplies forall f in L^2(X), ;| pi(g_n)f-pi(g)f|to 0.$$
The relevance of this property to ergodicity of dense subgroup is that in this case, the stabilizer of a characteristic function $f=1_A$
$$Stab(f)={ g in G , : , pi(g)f=f },$$
is then a closed subgroup; so if a set $A$ is $Gamma$-invariant, $f=1_A$ is $Gamma$-invariant, so is also $bar{Gamma}=G$-invariant, so $mu(A)in {0,1}$.
In the case you are interested in, this property of strong continuity is satisfied; see for example Lemma 1.6 in those notes
http://u.math.biu.ac.il/~solomyb/TEACH/17/GrAct/Lec7.pdf
for a statement giving sufficient conditions and a sketch of proof.
edited Nov 28 '18 at 9:42
user39082
1,235513
answered Nov 28 '18 at 9:28
user120527
1,616215
edited Nov 28 '18 at 9:42
user39082
1,235513
edited Nov 28 '18 at 9:42
user39082
1,235513
edited Nov 28 '18 at 9:42
user39082
1,235513
1,235513
answered Nov 28 '18 at 9:28
user120527
1,616215
answered Nov 28 '18 at 9:28
user120527
1,616215
answered Nov 28 '18 at 9:28
user120527
1,616215
1,616215
add a comment |
add a comment |
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