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










0












$begingroup$


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







share|cite|improve this question









$endgroup$



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
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 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















0












$begingroup$


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







share|cite|improve this question









$endgroup$



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
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 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













0












0








0


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
If this question can be reworded to fit the rules in the help center, please edit the question.







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
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 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




    $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 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


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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













  • 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




-

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