Mixing
Lie group intuition
$begingroup$
Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.
Thank you!
lie-groups smooth-manifolds
asked Jan 23 at 14:28
mipmip
425
$endgroup$
add a comment |
$begingroup$
Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.
Thank you!
lie-groups smooth-manifolds
asked Jan 23 at 14:28
mipmip
425
$endgroup$
add a comment |
$begingroup$
Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.
Thank you!
lie-groups smooth-manifolds
asked Jan 23 at 14:28
mipmip
425
$endgroup$
Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.
Thank you!
lie-groups smooth-manifolds
lie-groups smooth-manifolds
asked Jan 23 at 14:28
mipmip
425
asked Jan 23 at 14:28
mipmip
425
asked Jan 23 at 14:28
mipmip
425
asked Jan 23 at 14:28
mipmip
425
asked Jan 23 at 14:28
mipmip
425
425
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.
Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).
We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.
answered Jan 23 at 14:35
YankoYanko
7,7751830
$endgroup$
add a comment |
$begingroup$
Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.
Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).
We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.
answered Jan 23 at 14:35
YankoYanko
7,7751830
$endgroup$
add a comment |
$begingroup$
I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.
Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).
We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.
answered Jan 23 at 14:35
YankoYanko
7,7751830
$endgroup$
add a comment |
$begingroup$
I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.
Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).
We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.
answered Jan 23 at 14:35
YankoYanko
7,7751830
$endgroup$
I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.
Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).
We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.
answered Jan 23 at 14:35
YankoYanko
7,7751830
answered Jan 23 at 14:35
YankoYanko
7,7751830
answered Jan 23 at 14:35
YankoYanko
7,7751830
answered Jan 23 at 14:35
YankoYanko
7,7751830
7,7751830
add a comment |
add a comment |
$begingroup$
Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 14:35
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
add a comment |
add a comment |
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