What makes a Lie Group a Differentiable Manifold?
I've recently been trying to glance at the definition of a Lie group, but I'm not clear as to why a Lie group is defined the way it is, and why this becomes a smooth manifold. For example, if we have a Lie group $G$, why do we have the requirement that $G times G to G$, $(x,y) mapsto xy^{-1}$ is a smooth continuous mapping between the product manifold and the manifold $G$.
What is it that makes $G$ a topological manifold in the first place? How does it inherit differentiable structure?
differential-geometry differential-topology lie-groups lie-algebras
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
|
show 1 more comment
I've recently been trying to glance at the definition of a Lie group, but I'm not clear as to why a Lie group is defined the way it is, and why this becomes a smooth manifold. For example, if we have a Lie group $G$, why do we have the requirement that $G times G to G$, $(x,y) mapsto xy^{-1}$ is a smooth continuous mapping between the product manifold and the manifold $G$.
What is it that makes $G$ a topological manifold in the first place? How does it inherit differentiable structure?
differential-geometry differential-topology lie-groups lie-algebras
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
13
It's part of the definition!
– Zhen Lin
Jul 16 '14 at 21:22
You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group.
– Daniel Fischer♦
Jul 16 '14 at 21:23
This isn't a theorem, it's one of the Lie group axioms.
– Adam Hughes
Jul 16 '14 at 21:24
@DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why?
– Anthony Peter
Jul 16 '14 at 21:25
3
@AnthonyPeter: I'd recommend doing some reading about topological groups in general.
– Kyle
Jul 16 '14 at 21:27
|
show 1 more comment
I've recently been trying to glance at the definition of a Lie group, but I'm not clear as to why a Lie group is defined the way it is, and why this becomes a smooth manifold. For example, if we have a Lie group $G$, why do we have the requirement that $G times G to G$, $(x,y) mapsto xy^{-1}$ is a smooth continuous mapping between the product manifold and the manifold $G$.
What is it that makes $G$ a topological manifold in the first place? How does it inherit differentiable structure?
differential-geometry differential-topology lie-groups lie-algebras
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
I've recently been trying to glance at the definition of a Lie group, but I'm not clear as to why a Lie group is defined the way it is, and why this becomes a smooth manifold. For example, if we have a Lie group $G$, why do we have the requirement that $G times G to G$, $(x,y) mapsto xy^{-1}$ is a smooth continuous mapping between the product manifold and the manifold $G$.
What is it that makes $G$ a topological manifold in the first place? How does it inherit differentiable structure?
differential-geometry differential-topology lie-groups lie-algebras
differential-geometry differential-topology lie-groups lie-algebras
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
asked Jul 16 '14 at 21:19
Anthony Peter
3,23211451
3,23211451
13
It's part of the definition!
– Zhen Lin
Jul 16 '14 at 21:22
You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group.
– Daniel Fischer♦
Jul 16 '14 at 21:23
This isn't a theorem, it's one of the Lie group axioms.
– Adam Hughes
Jul 16 '14 at 21:24
@DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why?
– Anthony Peter
Jul 16 '14 at 21:25
3
@AnthonyPeter: I'd recommend doing some reading about topological groups in general.
– Kyle
Jul 16 '14 at 21:27
|
show 1 more comment
13
It's part of the definition!
– Zhen Lin
Jul 16 '14 at 21:22
You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group.
– Daniel Fischer♦
Jul 16 '14 at 21:23
This isn't a theorem, it's one of the Lie group axioms.
– Adam Hughes
Jul 16 '14 at 21:24
@DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why?
– Anthony Peter
Jul 16 '14 at 21:25
3
@AnthonyPeter: I'd recommend doing some reading about topological groups in general.
– Kyle
Jul 16 '14 at 21:27
13
13
It's part of the definition!
– Zhen Lin
Jul 16 '14 at 21:22
It's part of the definition!
– Zhen Lin
Jul 16 '14 at 21:22
You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group.
– Daniel Fischer♦
Jul 16 '14 at 21:23
You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group.
– Daniel Fischer♦
Jul 16 '14 at 21:23
This isn't a theorem, it's one of the Lie group axioms.
– Adam Hughes
Jul 16 '14 at 21:24
This isn't a theorem, it's one of the Lie group axioms.
– Adam Hughes
Jul 16 '14 at 21:24
@DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why?
– Anthony Peter
Jul 16 '14 at 21:25
@DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why?
– Anthony Peter
Jul 16 '14 at 21:25
3
3
@AnthonyPeter: I'd recommend doing some reading about topological groups in general.
– Kyle
Jul 16 '14 at 21:27
@AnthonyPeter: I'd recommend doing some reading about topological groups in general.
– Kyle
Jul 16 '14 at 21:27
|
show 1 more comment
2 Answers
2
active
oldest
votes
When you ask that $G times G to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a smooth structure on $G$ and this property you ask of multiplication is extra.
Hope that helps,
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
add a comment |
As you know that lie group is a structure where G×G→G, it that you can get any element in the same structure with other elements and an operation, now it is a manifold so it's chart is in a same structure which is a diffeomorphism of the elements, so the has to be smooth...
answered Nov 23 at 12:32
explorer
2916
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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votes
When you ask that $G times G to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a smooth structure on $G$ and this property you ask of multiplication is extra.
Hope that helps,
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
add a comment |
When you ask that $G times G to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a smooth structure on $G$ and this property you ask of multiplication is extra.
Hope that helps,
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
add a comment |
When you ask that $G times G to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a smooth structure on $G$ and this property you ask of multiplication is extra.
Hope that helps,
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
When you ask that $G times G to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a smooth structure on $G$ and this property you ask of multiplication is extra.
Hope that helps,
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
answered Jul 16 '14 at 21:25
Patrick Da Silva
32k353106
32k353106
add a comment |
add a comment |
As you know that lie group is a structure where G×G→G, it that you can get any element in the same structure with other elements and an operation, now it is a manifold so it's chart is in a same structure which is a diffeomorphism of the elements, so the has to be smooth...
answered Nov 23 at 12:32
explorer
2916
add a comment |
As you know that lie group is a structure where G×G→G, it that you can get any element in the same structure with other elements and an operation, now it is a manifold so it's chart is in a same structure which is a diffeomorphism of the elements, so the has to be smooth...
answered Nov 23 at 12:32
explorer
2916
add a comment |
As you know that lie group is a structure where G×G→G, it that you can get any element in the same structure with other elements and an operation, now it is a manifold so it's chart is in a same structure which is a diffeomorphism of the elements, so the has to be smooth...
answered Nov 23 at 12:32
explorer
2916
As you know that lie group is a structure where G×G→G, it that you can get any element in the same structure with other elements and an operation, now it is a manifold so it's chart is in a same structure which is a diffeomorphism of the elements, so the has to be smooth...
answered Nov 23 at 12:32
explorer
2916
answered Nov 23 at 12:32
explorer
2916
answered Nov 23 at 12:32
explorer
2916
answered Nov 23 at 12:32
explorer
2916
2916
add a comment |
add a comment |
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13
It's part of the definition!
– Zhen Lin
Jul 16 '14 at 21:22
You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group.
– Daniel Fischer♦
Jul 16 '14 at 21:23
This isn't a theorem, it's one of the Lie group axioms.
– Adam Hughes
Jul 16 '14 at 21:24
@DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why?
– Anthony Peter
Jul 16 '14 at 21:25
3
@AnthonyPeter: I'd recommend doing some reading about topological groups in general.
– Kyle
Jul 16 '14 at 21:27