Lifting of operator under free action
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Suppose a Lie group $K$ acts freely, properly and isometrically on a Riemannian manifold $(M,g)$, which is equipped with a $K$-invariant measure. Then $(M/K,check{g})$ is a Riemannian manifold, where the metric $check{g}$ and the measure descend from the metric and measure on $M$ in the natural way.
Question: Can one interpret a bounded linear operator $T$ on $L^2(M/K)$ as a bounded linear operator $tilde{T}$ on $L^2(M)$ that is invariant under the action of $K$?
Thoughts: What I mean by being invariant under the action of $K$ is that the action of $T$ on $L^2(M)$ commutes with action of $k$ on $L^2(M)$ (defined by $kf(x):=f(k^{-1}x)$) for every $kin K$. One can check that a $K$-invariant operator on $L^2(M)$ always descends to an operator on $L^2(M/K)$. I wonder if there is a reference that addresses the issue of lifting an operator the other way.
functional-analysis differential-geometry lie-groups riemannian-geometry group-actions
asked Nov 14 at 15:01
ougoah
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Suppose a Lie group $K$ acts freely, properly and isometrically on a Riemannian manifold $(M,g)$, which is equipped with a $K$-invariant measure. Then $(M/K,check{g})$ is a Riemannian manifold, where the metric $check{g}$ and the measure descend from the metric and measure on $M$ in the natural way.
Question: Can one interpret a bounded linear operator $T$ on $L^2(M/K)$ as a bounded linear operator $tilde{T}$ on $L^2(M)$ that is invariant under the action of $K$?
Thoughts: What I mean by being invariant under the action of $K$ is that the action of $T$ on $L^2(M)$ commutes with action of $k$ on $L^2(M)$ (defined by $kf(x):=f(k^{-1}x)$) for every $kin K$. One can check that a $K$-invariant operator on $L^2(M)$ always descends to an operator on $L^2(M/K)$. I wonder if there is a reference that addresses the issue of lifting an operator the other way.
functional-analysis differential-geometry lie-groups riemannian-geometry group-actions
asked Nov 14 at 15:01
ougoah
1,134710
No, there are many more of the latter than the former. This issue already arises if $M$ is taken to be a finite set and $K$ a finite group.
– Qiaochu Yuan
Nov 14 at 20:09
I take it you mean there are many more of the former than the latter? In that case, if $M$ is a finite set and $K$ is a finite group, then $Mcong Ntimes K$, for some finite set $N$ and $l^2(M)cong l^2(K)otimes l^2(N)$. I feel like every bounded operator $T$ on $l^2(N)$ lifts to the bounded operator $1otimes Tin B(l^2(K)otimes l^2(N))$?
– ougoah
Nov 14 at 22:00
To keep it really simple let's take $N = 1$. Then $L^2(N)$ is $1$-dimensional so the space of linear operators on it is also $1$-dimensional. But the space of $K$-invariant linear operators $L^2(K) to L^2(K)$ is $|K|$-dimensional (exercise). So there are many more of the latter than the former. I recognize I've slightly misread your question: what I'm claiming is that lifts can't be unique.
– Qiaochu Yuan
Nov 14 at 22:04
By "can one interpret" I meant, is there a $tilde{T}$ that descends to your given $T$. In the case of $N=1$ I think this is true.
– ougoah
Nov 14 at 22:12
I originally read your question as asking whether the two were equivalent / whether there's a natural isomorphism between them.
– Qiaochu Yuan
Nov 14 at 22:13
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose a Lie group $K$ acts freely, properly and isometrically on a Riemannian manifold $(M,g)$, which is equipped with a $K$-invariant measure. Then $(M/K,check{g})$ is a Riemannian manifold, where the metric $check{g}$ and the measure descend from the metric and measure on $M$ in the natural way.
Question: Can one interpret a bounded linear operator $T$ on $L^2(M/K)$ as a bounded linear operator $tilde{T}$ on $L^2(M)$ that is invariant under the action of $K$?
Thoughts: What I mean by being invariant under the action of $K$ is that the action of $T$ on $L^2(M)$ commutes with action of $k$ on $L^2(M)$ (defined by $kf(x):=f(k^{-1}x)$) for every $kin K$. One can check that a $K$-invariant operator on $L^2(M)$ always descends to an operator on $L^2(M/K)$. I wonder if there is a reference that addresses the issue of lifting an operator the other way.
functional-analysis differential-geometry lie-groups riemannian-geometry group-actions
asked Nov 14 at 15:01
ougoah
1,134710
Suppose a Lie group $K$ acts freely, properly and isometrically on a Riemannian manifold $(M,g)$, which is equipped with a $K$-invariant measure. Then $(M/K,check{g})$ is a Riemannian manifold, where the metric $check{g}$ and the measure descend from the metric and measure on $M$ in the natural way.
Question: Can one interpret a bounded linear operator $T$ on $L^2(M/K)$ as a bounded linear operator $tilde{T}$ on $L^2(M)$ that is invariant under the action of $K$?
Thoughts: What I mean by being invariant under the action of $K$ is that the action of $T$ on $L^2(M)$ commutes with action of $k$ on $L^2(M)$ (defined by $kf(x):=f(k^{-1}x)$) for every $kin K$. One can check that a $K$-invariant operator on $L^2(M)$ always descends to an operator on $L^2(M/K)$. I wonder if there is a reference that addresses the issue of lifting an operator the other way.
functional-analysis differential-geometry lie-groups riemannian-geometry group-actions
functional-analysis differential-geometry lie-groups riemannian-geometry group-actions
asked Nov 14 at 15:01
ougoah
1,134710
asked Nov 14 at 15:01
ougoah
1,134710
asked Nov 14 at 15:01
ougoah
1,134710
asked Nov 14 at 15:01
ougoah
1,134710
asked Nov 14 at 15:01
ougoah
1,134710
1,134710
No, there are many more of the latter than the former. This issue already arises if $M$ is taken to be a finite set and $K$ a finite group.
– Qiaochu Yuan
Nov 14 at 20:09
I take it you mean there are many more of the former than the latter? In that case, if $M$ is a finite set and $K$ is a finite group, then $Mcong Ntimes K$, for some finite set $N$ and $l^2(M)cong l^2(K)otimes l^2(N)$. I feel like every bounded operator $T$ on $l^2(N)$ lifts to the bounded operator $1otimes Tin B(l^2(K)otimes l^2(N))$?
– ougoah
Nov 14 at 22:00
To keep it really simple let's take $N = 1$. Then $L^2(N)$ is $1$-dimensional so the space of linear operators on it is also $1$-dimensional. But the space of $K$-invariant linear operators $L^2(K) to L^2(K)$ is $|K|$-dimensional (exercise). So there are many more of the latter than the former. I recognize I've slightly misread your question: what I'm claiming is that lifts can't be unique.
– Qiaochu Yuan
Nov 14 at 22:04
By "can one interpret" I meant, is there a $tilde{T}$ that descends to your given $T$. In the case of $N=1$ I think this is true.
– ougoah
Nov 14 at 22:12
I originally read your question as asking whether the two were equivalent / whether there's a natural isomorphism between them.
– Qiaochu Yuan
Nov 14 at 22:13
|
show 1 more comment
No, there are many more of the latter than the former. This issue already arises if $M$ is taken to be a finite set and $K$ a finite group.
– Qiaochu Yuan
Nov 14 at 20:09
I take it you mean there are many more of the former than the latter? In that case, if $M$ is a finite set and $K$ is a finite group, then $Mcong Ntimes K$, for some finite set $N$ and $l^2(M)cong l^2(K)otimes l^2(N)$. I feel like every bounded operator $T$ on $l^2(N)$ lifts to the bounded operator $1otimes Tin B(l^2(K)otimes l^2(N))$?
– ougoah
Nov 14 at 22:00
To keep it really simple let's take $N = 1$. Then $L^2(N)$ is $1$-dimensional so the space of linear operators on it is also $1$-dimensional. But the space of $K$-invariant linear operators $L^2(K) to L^2(K)$ is $|K|$-dimensional (exercise). So there are many more of the latter than the former. I recognize I've slightly misread your question: what I'm claiming is that lifts can't be unique.
– Qiaochu Yuan
Nov 14 at 22:04
By "can one interpret" I meant, is there a $tilde{T}$ that descends to your given $T$. In the case of $N=1$ I think this is true.
– ougoah
Nov 14 at 22:12
I originally read your question as asking whether the two were equivalent / whether there's a natural isomorphism between them.
– Qiaochu Yuan
Nov 14 at 22:13
No, there are many more of the latter than the former. This issue already arises if $M$ is taken to be a finite set and $K$ a finite group.
– Qiaochu Yuan
Nov 14 at 20:09
No, there are many more of the latter than the former. This issue already arises if $M$ is taken to be a finite set and $K$ a finite group.
– Qiaochu Yuan
Nov 14 at 20:09
I take it you mean there are many more of the former than the latter? In that case, if $M$ is a finite set and $K$ is a finite group, then $Mcong Ntimes K$, for some finite set $N$ and $l^2(M)cong l^2(K)otimes l^2(N)$. I feel like every bounded operator $T$ on $l^2(N)$ lifts to the bounded operator $1otimes Tin B(l^2(K)otimes l^2(N))$?
– ougoah
Nov 14 at 22:00
I take it you mean there are many more of the former than the latter? In that case, if $M$ is a finite set and $K$ is a finite group, then $Mcong Ntimes K$, for some finite set $N$ and $l^2(M)cong l^2(K)otimes l^2(N)$. I feel like every bounded operator $T$ on $l^2(N)$ lifts to the bounded operator $1otimes Tin B(l^2(K)otimes l^2(N))$?
– ougoah
Nov 14 at 22:00
To keep it really simple let's take $N = 1$. Then $L^2(N)$ is $1$-dimensional so the space of linear operators on it is also $1$-dimensional. But the space of $K$-invariant linear operators $L^2(K) to L^2(K)$ is $|K|$-dimensional (exercise). So there are many more of the latter than the former. I recognize I've slightly misread your question: what I'm claiming is that lifts can't be unique.
– Qiaochu Yuan
Nov 14 at 22:04
To keep it really simple let's take $N = 1$. Then $L^2(N)$ is $1$-dimensional so the space of linear operators on it is also $1$-dimensional. But the space of $K$-invariant linear operators $L^2(K) to L^2(K)$ is $|K|$-dimensional (exercise). So there are many more of the latter than the former. I recognize I've slightly misread your question: what I'm claiming is that lifts can't be unique.
– Qiaochu Yuan
Nov 14 at 22:04
By "can one interpret" I meant, is there a $tilde{T}$ that descends to your given $T$. In the case of $N=1$ I think this is true.
– ougoah
Nov 14 at 22:12
By "can one interpret" I meant, is there a $tilde{T}$ that descends to your given $T$. In the case of $N=1$ I think this is true.
– ougoah
Nov 14 at 22:12
I originally read your question as asking whether the two were equivalent / whether there's a natural isomorphism between them.
– Qiaochu Yuan
Nov 14 at 22:13
I originally read your question as asking whether the two were equivalent / whether there's a natural isomorphism between them.
– Qiaochu Yuan
Nov 14 at 22:13
|
show 1 more comment
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No, there are many more of the latter than the former. This issue already arises if $M$ is taken to be a finite set and $K$ a finite group.
– Qiaochu Yuan
Nov 14 at 20:09
I take it you mean there are many more of the former than the latter? In that case, if $M$ is a finite set and $K$ is a finite group, then $Mcong Ntimes K$, for some finite set $N$ and $l^2(M)cong l^2(K)otimes l^2(N)$. I feel like every bounded operator $T$ on $l^2(N)$ lifts to the bounded operator $1otimes Tin B(l^2(K)otimes l^2(N))$?
– ougoah
Nov 14 at 22:00
To keep it really simple let's take $N = 1$. Then $L^2(N)$ is $1$-dimensional so the space of linear operators on it is also $1$-dimensional. But the space of $K$-invariant linear operators $L^2(K) to L^2(K)$ is $|K|$-dimensional (exercise). So there are many more of the latter than the former. I recognize I've slightly misread your question: what I'm claiming is that lifts can't be unique.
– Qiaochu Yuan
Nov 14 at 22:04
By "can one interpret" I meant, is there a $tilde{T}$ that descends to your given $T$. In the case of $N=1$ I think this is true.
– ougoah
Nov 14 at 22:12
I originally read your question as asking whether the two were equivalent / whether there's a natural isomorphism between them.
– Qiaochu Yuan
Nov 14 at 22:13