Every compact semi-simple Lie group has finite center
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I am reading introductory lecture notes on Lie groups and Lie algebras. There it is stated as a fact without proof, that any compact semi-simple Lie group has finite center.
Here, semi-simple means, that the corresponding Lie algebra can be written as a direct sum of simple Lie algebras (having no non-trivial ideals).
Since this is not immediately obvious for me, I wonder if the proof is actually complicated or if I am just too ignorant to see it.
If anyone can give an idea of where this fact comes from, I would be thankful.
lie-groups lie-algebras
asked May 31 '16 at 13:13
TheAbelian
314111
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up vote
2
down vote
favorite
I am reading introductory lecture notes on Lie groups and Lie algebras. There it is stated as a fact without proof, that any compact semi-simple Lie group has finite center.
Here, semi-simple means, that the corresponding Lie algebra can be written as a direct sum of simple Lie algebras (having no non-trivial ideals).
Since this is not immediately obvious for me, I wonder if the proof is actually complicated or if I am just too ignorant to see it.
If anyone can give an idea of where this fact comes from, I would be thankful.
lie-groups lie-algebras
asked May 31 '16 at 13:13
TheAbelian
314111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading introductory lecture notes on Lie groups and Lie algebras. There it is stated as a fact without proof, that any compact semi-simple Lie group has finite center.
Here, semi-simple means, that the corresponding Lie algebra can be written as a direct sum of simple Lie algebras (having no non-trivial ideals).
Since this is not immediately obvious for me, I wonder if the proof is actually complicated or if I am just too ignorant to see it.
If anyone can give an idea of where this fact comes from, I would be thankful.
lie-groups lie-algebras
asked May 31 '16 at 13:13
TheAbelian
314111
I am reading introductory lecture notes on Lie groups and Lie algebras. There it is stated as a fact without proof, that any compact semi-simple Lie group has finite center.
Here, semi-simple means, that the corresponding Lie algebra can be written as a direct sum of simple Lie algebras (having no non-trivial ideals).
Since this is not immediately obvious for me, I wonder if the proof is actually complicated or if I am just too ignorant to see it.
If anyone can give an idea of where this fact comes from, I would be thankful.
lie-groups lie-algebras
asked May 31 '16 at 13:13
TheAbelian
314111
asked May 31 '16 at 13:13
TheAbelian
314111
asked May 31 '16 at 13:13
TheAbelian
314111
asked May 31 '16 at 13:13
TheAbelian
314111
314111
add a comment |Â
add a comment |Â
1 Answer
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If $G$ is a semi-simple Lie group, then its Lie algebra $frak g$ is semi-simple, which implies that it has trivial center. Therefore, the Lie algebra of the center $Z(G)$ of $G$ is trivial, which means that $Z(G)$ is discrete. Since $G$ is compact, $Z(G)$ is finite.
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If $G$ is a semi-simple Lie group, then its Lie algebra $frak g$ is semi-simple, which implies that it has trivial center. Therefore, the Lie algebra of the center $Z(G)$ of $G$ is trivial, which means that $Z(G)$ is discrete. Since $G$ is compact, $Z(G)$ is finite.
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
add a comment |Â
up vote
4
down vote
accepted
If $G$ is a semi-simple Lie group, then its Lie algebra $frak g$ is semi-simple, which implies that it has trivial center. Therefore, the Lie algebra of the center $Z(G)$ of $G$ is trivial, which means that $Z(G)$ is discrete. Since $G$ is compact, $Z(G)$ is finite.
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If $G$ is a semi-simple Lie group, then its Lie algebra $frak g$ is semi-simple, which implies that it has trivial center. Therefore, the Lie algebra of the center $Z(G)$ of $G$ is trivial, which means that $Z(G)$ is discrete. Since $G$ is compact, $Z(G)$ is finite.
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
If $G$ is a semi-simple Lie group, then its Lie algebra $frak g$ is semi-simple, which implies that it has trivial center. Therefore, the Lie algebra of the center $Z(G)$ of $G$ is trivial, which means that $Z(G)$ is discrete. Since $G$ is compact, $Z(G)$ is finite.
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
edited Aug 21 at 8:17
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
answered May 31 '16 at 13:40
Johannes Huisman
3,035516
3,035516
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
add a comment |Â
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
And a semi-simple Lie algebra $frakg$ has trivial center, since if the center was non-trivial then one of the centers of the simple Lie algebras $frakg_i$ forming the decomposition would be non-trivial, which is a contradiction to $frakg_i$ being simple, since the center is an ideal. Is that correct?
– TheAbelian
May 31 '16 at 13:54
add a comment |Â
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