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$f_*$ isomorphism $Rightarrow$ $f$ isomorphism?
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Consider a Lie group endomorphism (bijective homomorphism) $f: G rightarrow G$ and its pushforward $f_*: mathcal{A} rightarrow mathcal{A}$, where $mathcal{A}$ is the Lie algebra of $G$. I have seen that if $f$ is an isomorphism (a diffeomorphism) then $f_*$ is an isomorphism.
Is the converse true? That is, given a group endomorphism $f: G rightarrow G$ with pushforward $f_*: mathcal{A} rightarrow mathcal{A}$ an algebra endomorphism, is it true that if $f_*$ is an isomorphism, the $f$ is an isomorphism too?
group-theory lie-groups lie-algebras
asked yesterday
MBolin
787
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Consider a Lie group endomorphism (bijective homomorphism) $f: G rightarrow G$ and its pushforward $f_*: mathcal{A} rightarrow mathcal{A}$, where $mathcal{A}$ is the Lie algebra of $G$. I have seen that if $f$ is an isomorphism (a diffeomorphism) then $f_*$ is an isomorphism.
Is the converse true? That is, given a group endomorphism $f: G rightarrow G$ with pushforward $f_*: mathcal{A} rightarrow mathcal{A}$ an algebra endomorphism, is it true that if $f_*$ is an isomorphism, the $f$ is an isomorphism too?
group-theory lie-groups lie-algebras
asked yesterday
MBolin
787
It is obviously wrong if $G$ is not connected. Also, "endomorphism" doesn't mean that the map is bijective, it merely means that the domain and the target of the map are the same "space" or object.
– Sebastian Schulz
yesterday
add a comment |
up vote
1
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up vote
1
down vote
favorite
Consider a Lie group endomorphism (bijective homomorphism) $f: G rightarrow G$ and its pushforward $f_*: mathcal{A} rightarrow mathcal{A}$, where $mathcal{A}$ is the Lie algebra of $G$. I have seen that if $f$ is an isomorphism (a diffeomorphism) then $f_*$ is an isomorphism.
Is the converse true? That is, given a group endomorphism $f: G rightarrow G$ with pushforward $f_*: mathcal{A} rightarrow mathcal{A}$ an algebra endomorphism, is it true that if $f_*$ is an isomorphism, the $f$ is an isomorphism too?
group-theory lie-groups lie-algebras
asked yesterday
MBolin
787
Consider a Lie group endomorphism (bijective homomorphism) $f: G rightarrow G$ and its pushforward $f_*: mathcal{A} rightarrow mathcal{A}$, where $mathcal{A}$ is the Lie algebra of $G$. I have seen that if $f$ is an isomorphism (a diffeomorphism) then $f_*$ is an isomorphism.
Is the converse true? That is, given a group endomorphism $f: G rightarrow G$ with pushforward $f_*: mathcal{A} rightarrow mathcal{A}$ an algebra endomorphism, is it true that if $f_*$ is an isomorphism, the $f$ is an isomorphism too?
group-theory lie-groups lie-algebras
group-theory lie-groups lie-algebras
asked yesterday
MBolin
787
asked yesterday
MBolin
787
asked yesterday
MBolin
787
asked yesterday
MBolin
787
asked yesterday
MBolin
787
787
It is obviously wrong if $G$ is not connected. Also, "endomorphism" doesn't mean that the map is bijective, it merely means that the domain and the target of the map are the same "space" or object.
– Sebastian Schulz
yesterday
add a comment |
It is obviously wrong if $G$ is not connected. Also, "endomorphism" doesn't mean that the map is bijective, it merely means that the domain and the target of the map are the same "space" or object.
– Sebastian Schulz
yesterday
It is obviously wrong if $G$ is not connected. Also, "endomorphism" doesn't mean that the map is bijective, it merely means that the domain and the target of the map are the same "space" or object.
– Sebastian Schulz
yesterday
It is obviously wrong if $G$ is not connected. Also, "endomorphism" doesn't mean that the map is bijective, it merely means that the domain and the target of the map are the same "space" or object.
– Sebastian Schulz
yesterday
add a comment |
2 Answers
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1
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As Sebastian Schulz noted, the answer is obviously no if $G$ is not connected.
Furthermore, there are other Lie groups that have isomorphic Lie algebras, but which are not isomorphic since their global properties are different. A well-known example is given by $SU(2)$ and $SO(3)$. The Lie algebras of these groups are isomorphic, but the groups themselves are not: in fact, $SU(2)$ is the double cover of $SO(3)$.
answered yesterday
Stijn B.
1114
New contributor
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He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
add a comment |
up vote
1
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OK, here is the minimal example that I could think of for a connected Lie group. Take $G = U(1) = { z in mathbb{C} | |z| = 1 }$, it allows endomorphisms of the form $phi_n : z mapsto z^n$.
The Lie algebra of $U(1)$ is $mathbb{R}$ and the derivative of $phi_n$ is just multiplication by $n$ (which is an isomorphism unless $n=0$). However, unless $n= pm 1$, the map is not injective on the level of Lie groups, so it is not an isomorphism.
answered yesterday
Sebastian Schulz
66139
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As Sebastian Schulz noted, the answer is obviously no if $G$ is not connected.
Furthermore, there are other Lie groups that have isomorphic Lie algebras, but which are not isomorphic since their global properties are different. A well-known example is given by $SU(2)$ and $SO(3)$. The Lie algebras of these groups are isomorphic, but the groups themselves are not: in fact, $SU(2)$ is the double cover of $SO(3)$.
answered yesterday
Stijn B.
1114
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
add a comment |
up vote
1
down vote
As Sebastian Schulz noted, the answer is obviously no if $G$ is not connected.
Furthermore, there are other Lie groups that have isomorphic Lie algebras, but which are not isomorphic since their global properties are different. A well-known example is given by $SU(2)$ and $SO(3)$. The Lie algebras of these groups are isomorphic, but the groups themselves are not: in fact, $SU(2)$ is the double cover of $SO(3)$.
answered yesterday
Stijn B.
1114
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
As Sebastian Schulz noted, the answer is obviously no if $G$ is not connected.
Furthermore, there are other Lie groups that have isomorphic Lie algebras, but which are not isomorphic since their global properties are different. A well-known example is given by $SU(2)$ and $SO(3)$. The Lie algebras of these groups are isomorphic, but the groups themselves are not: in fact, $SU(2)$ is the double cover of $SO(3)$.
answered yesterday
Stijn B.
1114
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
As Sebastian Schulz noted, the answer is obviously no if $G$ is not connected.
Furthermore, there are other Lie groups that have isomorphic Lie algebras, but which are not isomorphic since their global properties are different. A well-known example is given by $SU(2)$ and $SO(3)$. The Lie algebras of these groups are isomorphic, but the groups themselves are not: in fact, $SU(2)$ is the double cover of $SO(3)$.
answered yesterday
Stijn B.
1114
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Stijn B.
1114
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Stijn B.
1114
answered yesterday
Stijn B.
1114
1114
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Stijn B. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
add a comment |
He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
He did specifically start with an endomorphism of Lie groups though (so a covering map between Lie groups with isomorphic Lie algebras is not ruled out)
– Sebastian Schulz
yesterday
add a comment |
up vote
1
down vote
OK, here is the minimal example that I could think of for a connected Lie group. Take $G = U(1) = { z in mathbb{C} | |z| = 1 }$, it allows endomorphisms of the form $phi_n : z mapsto z^n$.
The Lie algebra of $U(1)$ is $mathbb{R}$ and the derivative of $phi_n$ is just multiplication by $n$ (which is an isomorphism unless $n=0$). However, unless $n= pm 1$, the map is not injective on the level of Lie groups, so it is not an isomorphism.
answered yesterday
Sebastian Schulz
66139
add a comment |
up vote
1
down vote
OK, here is the minimal example that I could think of for a connected Lie group. Take $G = U(1) = { z in mathbb{C} | |z| = 1 }$, it allows endomorphisms of the form $phi_n : z mapsto z^n$.
The Lie algebra of $U(1)$ is $mathbb{R}$ and the derivative of $phi_n$ is just multiplication by $n$ (which is an isomorphism unless $n=0$). However, unless $n= pm 1$, the map is not injective on the level of Lie groups, so it is not an isomorphism.
answered yesterday
Sebastian Schulz
66139
add a comment |
up vote
1
down vote
up vote
1
down vote
OK, here is the minimal example that I could think of for a connected Lie group. Take $G = U(1) = { z in mathbb{C} | |z| = 1 }$, it allows endomorphisms of the form $phi_n : z mapsto z^n$.
The Lie algebra of $U(1)$ is $mathbb{R}$ and the derivative of $phi_n$ is just multiplication by $n$ (which is an isomorphism unless $n=0$). However, unless $n= pm 1$, the map is not injective on the level of Lie groups, so it is not an isomorphism.
answered yesterday
Sebastian Schulz
66139
OK, here is the minimal example that I could think of for a connected Lie group. Take $G = U(1) = { z in mathbb{C} | |z| = 1 }$, it allows endomorphisms of the form $phi_n : z mapsto z^n$.
The Lie algebra of $U(1)$ is $mathbb{R}$ and the derivative of $phi_n$ is just multiplication by $n$ (which is an isomorphism unless $n=0$). However, unless $n= pm 1$, the map is not injective on the level of Lie groups, so it is not an isomorphism.
answered yesterday
Sebastian Schulz
66139
edited yesterday
answered yesterday
Sebastian Schulz
66139
answered yesterday
Sebastian Schulz
66139
answered yesterday
Sebastian Schulz
66139
66139
add a comment |
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It is obviously wrong if $G$ is not connected. Also, "endomorphism" doesn't mean that the map is bijective, it merely means that the domain and the target of the map are the same "space" or object.
– Sebastian Schulz
yesterday