Mixing
A simple question about homogeneous manifolds
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $G$ be a Lie group and $K$ a compact subgroup of $G$. Then since $K$ acts freely and properly on $G$ by translation, we can form the homogeneous manifold $Kbackslash G$.
Question: is the map $Ktimes (Kbackslash G)rightarrow G$ defined by $(k,[g])mapsto kg$ a diffeomorphism?
Actually I can't see why this isn't true, but I find it hard to believe because it would imply that any Lie group $G$ is always a principal $K$-bundle over $G/K$.
Edit: I meant free principal $K$-bundle.
differential-geometry manifolds lie-groups
asked Jul 25 at 9:33
ougoah
1,060610
add a comment |Â
up vote
1
down vote
favorite
Let $G$ be a Lie group and $K$ a compact subgroup of $G$. Then since $K$ acts freely and properly on $G$ by translation, we can form the homogeneous manifold $Kbackslash G$.
Question: is the map $Ktimes (Kbackslash G)rightarrow G$ defined by $(k,[g])mapsto kg$ a diffeomorphism?
Actually I can't see why this isn't true, but I find it hard to believe because it would imply that any Lie group $G$ is always a principal $K$-bundle over $G/K$.
Edit: I meant free principal $K$-bundle.
differential-geometry manifolds lie-groups
asked Jul 25 at 9:33
ougoah
1,060610
If $K$ acts freely and properly on $G$, shouldn't the manifold you form be $G/K$?
– Osama Ghani
Jul 25 at 9:45
Well $K$ can either multiply from the left or from the right.
– ougoah
Jul 25 at 19:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G$ be a Lie group and $K$ a compact subgroup of $G$. Then since $K$ acts freely and properly on $G$ by translation, we can form the homogeneous manifold $Kbackslash G$.
Question: is the map $Ktimes (Kbackslash G)rightarrow G$ defined by $(k,[g])mapsto kg$ a diffeomorphism?
Actually I can't see why this isn't true, but I find it hard to believe because it would imply that any Lie group $G$ is always a principal $K$-bundle over $G/K$.
Edit: I meant free principal $K$-bundle.
differential-geometry manifolds lie-groups
asked Jul 25 at 9:33
ougoah
1,060610
Let $G$ be a Lie group and $K$ a compact subgroup of $G$. Then since $K$ acts freely and properly on $G$ by translation, we can form the homogeneous manifold $Kbackslash G$.
Question: is the map $Ktimes (Kbackslash G)rightarrow G$ defined by $(k,[g])mapsto kg$ a diffeomorphism?
Actually I can't see why this isn't true, but I find it hard to believe because it would imply that any Lie group $G$ is always a principal $K$-bundle over $G/K$.
Edit: I meant free principal $K$-bundle.
differential-geometry manifolds lie-groups
asked Jul 25 at 9:33
ougoah
1,060610
edited Jul 25 at 12:02
asked Jul 25 at 9:33
ougoah
1,060610
asked Jul 25 at 9:33
ougoah
1,060610
asked Jul 25 at 9:33
ougoah
1,060610
1,060610
If $K$ acts freely and properly on $G$, shouldn't the manifold you form be $G/K$?
– Osama Ghani
Jul 25 at 9:45
Well $K$ can either multiply from the left or from the right.
– ougoah
Jul 25 at 19:33
add a comment |Â
If $K$ acts freely and properly on $G$, shouldn't the manifold you form be $G/K$?
– Osama Ghani
Jul 25 at 9:45
Well $K$ can either multiply from the left or from the right.
– ougoah
Jul 25 at 19:33
If $K$ acts freely and properly on $G$, shouldn't the manifold you form be $G/K$?
– Osama Ghani
Jul 25 at 9:45
If $K$ acts freely and properly on $G$, shouldn't the manifold you form be $G/K$?
– Osama Ghani
Jul 25 at 9:45
Well $K$ can either multiply from the left or from the right.
– ougoah
Jul 25 at 19:33
Well $K$ can either multiply from the left or from the right.
– ougoah
Jul 25 at 19:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The map you write out is not well defined (since you choose a representative of a coset and the map is not independent of this choice). What happens is that $G$ is a principal $K$-bundle over $Kbackslash G$, so locally isomorphic to a product, but not globally. The latter statement basically means that locally you can choose representatives of cosets smoothly.
answered Jul 25 at 9:49
Andreas Cap
9,599622
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
1
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The map you write out is not well defined (since you choose a representative of a coset and the map is not independent of this choice). What happens is that $G$ is a principal $K$-bundle over $Kbackslash G$, so locally isomorphic to a product, but not globally. The latter statement basically means that locally you can choose representatives of cosets smoothly.
answered Jul 25 at 9:49
Andreas Cap
9,599622
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
1
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
add a comment |Â
up vote
3
down vote
accepted
The map you write out is not well defined (since you choose a representative of a coset and the map is not independent of this choice). What happens is that $G$ is a principal $K$-bundle over $Kbackslash G$, so locally isomorphic to a product, but not globally. The latter statement basically means that locally you can choose representatives of cosets smoothly.
answered Jul 25 at 9:49
Andreas Cap
9,599622
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
1
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The map you write out is not well defined (since you choose a representative of a coset and the map is not independent of this choice). What happens is that $G$ is a principal $K$-bundle over $Kbackslash G$, so locally isomorphic to a product, but not globally. The latter statement basically means that locally you can choose representatives of cosets smoothly.
answered Jul 25 at 9:49
Andreas Cap
9,599622
The map you write out is not well defined (since you choose a representative of a coset and the map is not independent of this choice). What happens is that $G$ is a principal $K$-bundle over $Kbackslash G$, so locally isomorphic to a product, but not globally. The latter statement basically means that locally you can choose representatives of cosets smoothly.
answered Jul 25 at 9:49
Andreas Cap
9,599622
answered Jul 25 at 9:49
Andreas Cap
9,599622
answered Jul 25 at 9:49
Andreas Cap
9,599622
answered Jul 25 at 9:49
Andreas Cap
9,599622
9,599622
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
1
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
add a comment |Â
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
1
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
Thanks @Andreas Cap, indeed it’s not well-defined! What’s an easy example where there is no smooth section of the principal bundle $Grightarrow G/K$?
– ougoah
Jul 25 at 12:04
1
1
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
Just take $SO(3)to SO(3)/SO(2)cong S^2$ (given by the action of $SO(3)$ on the unit sphere in $mathbb R^3$). The resulting principal bundle is the oriented orthonormal frame bundle of $S^2$. If this were trivial, then also the associated bundle $TS^2$ would be trivial which contradicts the hairy ball theorem.
– Andreas Cap
Jul 25 at 12:20
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
As a follow-up question: would you still have a measure-theoretic relation $L^2(G)cong L^2(K)otimes L^2(Kbackslash G)$, even though there is no diffeomorphism $Gcong Ktimes Kbackslash G$?
– ougoah
Jul 25 at 19:33
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
Since these are all separable Hilbert spaces, there should be no problem on the level of spaces. I am not sure what happens if you view them as representations of $K$ (I would rather guess that they are non-isomorphic, but I am nor sure).
– Andreas Cap
Jul 26 at 9:05
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862221%2fa-simple-question-about-homogeneous-manifolds%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
If $K$ acts freely and properly on $G$, shouldn't the manifold you form be $G/K$?
– Osama Ghani
Jul 25 at 9:45
Well $K$ can either multiply from the left or from the right.
– ougoah
Jul 25 at 19:33