Linear Lie group. [closed]
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Suppose that G is a connected Lie group such that the center of G is trivial.
Question:
Is it true that G is isomorphic (as Lie group) to a closed subgroup of a Linear group GL(n,R) for some natural number n. Or maybe there is an evident counterexample ?
lie-groups
asked Nov 24 at 17:42
Ofra
1206
closed as off-topic by amWhy, Leucippus, Rebellos, KReiser, user10354138 Nov 25 at 3:02
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Rebellos, KReiser, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
Suppose that G is a connected Lie group such that the center of G is trivial.
Question:
Is it true that G is isomorphic (as Lie group) to a closed subgroup of a Linear group GL(n,R) for some natural number n. Or maybe there is an evident counterexample ?
lie-groups
asked Nov 24 at 17:42
Ofra
1206
closed as off-topic by amWhy, Leucippus, Rebellos, KReiser, user10354138 Nov 25 at 3:02
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Rebellos, KReiser, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 24 at 17:43
@mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation.
– Ofra
Nov 24 at 18:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that G is a connected Lie group such that the center of G is trivial.
Question:
Is it true that G is isomorphic (as Lie group) to a closed subgroup of a Linear group GL(n,R) for some natural number n. Or maybe there is an evident counterexample ?
lie-groups
asked Nov 24 at 17:42
Ofra
1206
Suppose that G is a connected Lie group such that the center of G is trivial.
Question:
Is it true that G is isomorphic (as Lie group) to a closed subgroup of a Linear group GL(n,R) for some natural number n. Or maybe there is an evident counterexample ?
lie-groups
lie-groups
asked Nov 24 at 17:42
Ofra
1206
asked Nov 24 at 17:42
Ofra
1206
edited Nov 24 at 18:41
asked Nov 24 at 17:42
Ofra
1206
asked Nov 24 at 17:42
Ofra
1206
asked Nov 24 at 17:42
Ofra
1206
1206
closed as off-topic by amWhy, Leucippus, Rebellos, KReiser, user10354138 Nov 25 at 3:02
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Rebellos, KReiser, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Leucippus, Rebellos, KReiser, user10354138 Nov 25 at 3:02
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Rebellos, KReiser, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 24 at 17:43
@mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation.
– Ofra
Nov 24 at 18:45
add a comment |
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 24 at 17:43
@mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation.
– Ofra
Nov 24 at 18:45
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 24 at 17:43
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 24 at 17:43
@mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation.
– Ofra
Nov 24 at 18:45
@mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation.
– Ofra
Nov 24 at 18:45
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The adjoint representation of $G$ realizes it as a closed subgroup of $mathrm{GL}(mathfrak{g})$, where $mathfrak{g}$ is the Lie algebra of $G$ (in general, the kernel of the adjoint representation consists of elements of $G$ that commute with the connected component of $G$ that contains the identity).
answered Nov 24 at 18:46
Stephen
10.4k12237
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
1
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The adjoint representation of $G$ realizes it as a closed subgroup of $mathrm{GL}(mathfrak{g})$, where $mathfrak{g}$ is the Lie algebra of $G$ (in general, the kernel of the adjoint representation consists of elements of $G$ that commute with the connected component of $G$ that contains the identity).
answered Nov 24 at 18:46
Stephen
10.4k12237
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
1
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
add a comment |
up vote
2
down vote
accepted
The adjoint representation of $G$ realizes it as a closed subgroup of $mathrm{GL}(mathfrak{g})$, where $mathfrak{g}$ is the Lie algebra of $G$ (in general, the kernel of the adjoint representation consists of elements of $G$ that commute with the connected component of $G$ that contains the identity).
answered Nov 24 at 18:46
Stephen
10.4k12237
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
1
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The adjoint representation of $G$ realizes it as a closed subgroup of $mathrm{GL}(mathfrak{g})$, where $mathfrak{g}$ is the Lie algebra of $G$ (in general, the kernel of the adjoint representation consists of elements of $G$ that commute with the connected component of $G$ that contains the identity).
answered Nov 24 at 18:46
Stephen
10.4k12237
The adjoint representation of $G$ realizes it as a closed subgroup of $mathrm{GL}(mathfrak{g})$, where $mathfrak{g}$ is the Lie algebra of $G$ (in general, the kernel of the adjoint representation consists of elements of $G$ that commute with the connected component of $G$ that contains the identity).
answered Nov 24 at 18:46
Stephen
10.4k12237
answered Nov 24 at 18:46
Stephen
10.4k12237
answered Nov 24 at 18:46
Stephen
10.4k12237
answered Nov 24 at 18:46
Stephen
10.4k12237
10.4k12237
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
1
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
add a comment |
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
1
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
Naive question: why the image of G by the adjoint representation is closed in GL(g)?
– Ofra
Nov 24 at 19:01
1
1
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
@Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed.
– Stephen
Nov 24 at 19:15
add a comment |
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 24 at 17:43
@mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation.
– Ofra
Nov 24 at 18:45