Dimension of image of one parameter subgroup. [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is a basis for the Lie algebra of a Lie group also a set of infinitesimal generators for the Lie group?One parameter subgroupReconstructing Lie group globally from the exponential mapAm I correct in saying that there are no non-commuting connected one-parameter Lie groups?1 parameter subgroups and Lie groupsOne parameter subgroups of group of isometries of planeClosures of one-parameter subgroups of lie groupsOne parameter subgroups of a Lie subgroupFinding all one-parameter subgroups of a matrix Lie groupUnique Lie subgroup structure
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Dimension of image of one parameter subgroup. [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is a basis for the Lie algebra of a Lie group also a set of infinitesimal generators for the Lie group?One parameter subgroupReconstructing Lie group globally from the exponential mapAm I correct in saying that there are no non-commuting connected one-parameter Lie groups?1 parameter subgroups and Lie groupsOne parameter subgroups of group of isometries of planeClosures of one-parameter subgroups of lie groupsOne parameter subgroups of a Lie subgroupFinding all one-parameter subgroups of a matrix Lie groupUnique Lie subgroup structure
$begingroup$
If $G$ is a Lie group then $eta : mathbbRto G$ is called one parameter subgroup if it is a continuous group homomorphism.
I need to show that the images of one-parameter subgroups in a Lie group $G$ are precisely the connected Lie subgroups of dimension less than or equal to $1$.
I am not getting any idea how to start. Any hints are appreciated.
Thank you.
differential-geometry lie-groups lie-algebras
asked Apr 8 at 18:58
SaikatSaikat
369216
$endgroup$
closed as off-topic by Moishe Kohan, Saad, Alexander Gruber♦ Apr 9 at 16:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, Alexander Gruber
add a comment |
$begingroup$
If $G$ is a Lie group then $eta : mathbbRto G$ is called one parameter subgroup if it is a continuous group homomorphism.
I need to show that the images of one-parameter subgroups in a Lie group $G$ are precisely the connected Lie subgroups of dimension less than or equal to $1$.
I am not getting any idea how to start. Any hints are appreciated.
Thank you.
differential-geometry lie-groups lie-algebras
asked Apr 8 at 18:58
SaikatSaikat
369216
$endgroup$
closed as off-topic by Moishe Kohan, Saad, Alexander Gruber♦ Apr 9 at 16:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, Alexander Gruber
1
$begingroup$
The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem.
$endgroup$
– Moishe Kohan
Apr 8 at 21:41
add a comment |
$begingroup$
If $G$ is a Lie group then $eta : mathbbRto G$ is called one parameter subgroup if it is a continuous group homomorphism.
I need to show that the images of one-parameter subgroups in a Lie group $G$ are precisely the connected Lie subgroups of dimension less than or equal to $1$.
I am not getting any idea how to start. Any hints are appreciated.
Thank you.
differential-geometry lie-groups lie-algebras
asked Apr 8 at 18:58
SaikatSaikat
369216
$endgroup$
If $G$ is a Lie group then $eta : mathbbRto G$ is called one parameter subgroup if it is a continuous group homomorphism.
I need to show that the images of one-parameter subgroups in a Lie group $G$ are precisely the connected Lie subgroups of dimension less than or equal to $1$.
I am not getting any idea how to start. Any hints are appreciated.
Thank you.
differential-geometry lie-groups lie-algebras
differential-geometry lie-groups lie-algebras
asked Apr 8 at 18:58
SaikatSaikat
369216
asked Apr 8 at 18:58
SaikatSaikat
369216
asked Apr 8 at 18:58
SaikatSaikat
369216
asked Apr 8 at 18:58
SaikatSaikat
369216
asked Apr 8 at 18:58
SaikatSaikat
369216
369216
closed as off-topic by Moishe Kohan, Saad, Alexander Gruber♦ Apr 9 at 16:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, Alexander Gruber
closed as off-topic by Moishe Kohan, Saad, Alexander Gruber♦ Apr 9 at 16:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Moishe Kohan, Saad, Alexander Gruber
1
$begingroup$
The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem.
$endgroup$
– Moishe Kohan
Apr 8 at 21:41
add a comment |
1
$begingroup$
The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem.
$endgroup$
– Moishe Kohan
Apr 8 at 21:41
1
1
$begingroup$
The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem.
$endgroup$
– Moishe Kohan
Apr 8 at 21:41
$begingroup$
The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem.
$endgroup$
– Moishe Kohan
Apr 8 at 21:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not true, consider the 2-dimensional torus $T^2$, there exists morphisms $mathbbRrightarrow T^2$ whose image are dense, such an image is not a subgroup since it is not closed.
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
1
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
1
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
2
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
1
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
1
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not true, consider the 2-dimensional torus $T^2$, there exists morphisms $mathbbRrightarrow T^2$ whose image are dense, such an image is not a subgroup since it is not closed.
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
1
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
1
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
2
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
1
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
1
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
|
show 6 more comments
$begingroup$
This is not true, consider the 2-dimensional torus $T^2$, there exists morphisms $mathbbRrightarrow T^2$ whose image are dense, such an image is not a subgroup since it is not closed.
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
1
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
1
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
2
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
1
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
1
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
|
show 6 more comments
$begingroup$
This is not true, consider the 2-dimensional torus $T^2$, there exists morphisms $mathbbRrightarrow T^2$ whose image are dense, such an image is not a subgroup since it is not closed.
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
This is not true, consider the 2-dimensional torus $T^2$, there exists morphisms $mathbbRrightarrow T^2$ whose image are dense, such an image is not a subgroup since it is not closed.
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
answered Apr 8 at 19:00
Tsemo AristideTsemo Aristide
60.7k11446
60.7k11446
1
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
1
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
2
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
1
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
1
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
|
show 6 more comments
1
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
1
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
2
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
1
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
1
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
1
1
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
$begingroup$
As far as I know, they are not group homomorphisms, however.
$endgroup$
– Randall
Apr 8 at 19:01
1
1
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
$begingroup$
If they were, the image would be a subgroup.
$endgroup$
– Randall
Apr 8 at 19:02
2
2
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
$begingroup$
Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup
$endgroup$
– Moishe Kohan
Apr 8 at 19:11
1
1
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
$begingroup$
@Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $mathbbR^n^2=Gl(n, mathbbR)$. Now use the matrix exponential.
$endgroup$
– Laz
Apr 8 at 22:06
1
1
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
$begingroup$
@TsemoAristide is right about his counterexample. Just precompose the homomorphism $mathbbS^1rightarrow mathbbT^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:mathbbRrightarrow mathbbS^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited.
$endgroup$
– Laz
Apr 8 at 23:55
|
show 6 more comments
1
$begingroup$
The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem.
$endgroup$
– Moishe Kohan
Apr 8 at 21:41