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The radical of an algebraic group is a torus [closed]
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How to show that the radical of a reductive linear algebraic group is a torus $(mathbb C^*)^n$?
lie-groups algebraic-groups
asked Jul 25 at 7:33
Ronald
1,5841821
closed as off-topic by José Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy Jul 25 at 11:14
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." РJos̩ Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy
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How to show that the radical of a reductive linear algebraic group is a torus $(mathbb C^*)^n$?
lie-groups algebraic-groups
asked Jul 25 at 7:33
Ronald
1,5841821
closed as off-topic by José Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy Jul 25 at 11:14
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." РJos̩ Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy
How do you define reductive? (I define it as the radical being a torus).
– Tobias Kildetoft
Jul 25 at 7:39
the unipotent radical is trivial
– Ronald
Jul 25 at 7:40
Ok, so you need to show that if a connected solvable group has trivial unipotent radical, then it is a torus. For this you can use that it will be isomorphic to a group of upper triangular matrices.
– Tobias Kildetoft
Jul 25 at 7:44
Can't we assume more since the radical actually is abelian?
– Ronald
Jul 25 at 7:48
Sure, if you already know that it is abelian you can use that (though I don't see an obvious way to show this directly).
– Tobias Kildetoft
Jul 25 at 7:50
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to show that the radical of a reductive linear algebraic group is a torus $(mathbb C^*)^n$?
lie-groups algebraic-groups
asked Jul 25 at 7:33
Ronald
1,5841821
How to show that the radical of a reductive linear algebraic group is a torus $(mathbb C^*)^n$?
lie-groups algebraic-groups
asked Jul 25 at 7:33
Ronald
1,5841821
asked Jul 25 at 7:33
Ronald
1,5841821
asked Jul 25 at 7:33
Ronald
1,5841821
asked Jul 25 at 7:33
Ronald
1,5841821
1,5841821
closed as off-topic by José Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy Jul 25 at 11:14
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." РJos̩ Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy
closed as off-topic by José Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy Jul 25 at 11:14
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." РJos̩ Carlos Santos, Parcly Taxel, Shailesh, Arnaud D., amWhy
How do you define reductive? (I define it as the radical being a torus).
– Tobias Kildetoft
Jul 25 at 7:39
the unipotent radical is trivial
– Ronald
Jul 25 at 7:40
Ok, so you need to show that if a connected solvable group has trivial unipotent radical, then it is a torus. For this you can use that it will be isomorphic to a group of upper triangular matrices.
– Tobias Kildetoft
Jul 25 at 7:44
Can't we assume more since the radical actually is abelian?
– Ronald
Jul 25 at 7:48
Sure, if you already know that it is abelian you can use that (though I don't see an obvious way to show this directly).
– Tobias Kildetoft
Jul 25 at 7:50
 |Â
show 2 more comments
How do you define reductive? (I define it as the radical being a torus).
– Tobias Kildetoft
Jul 25 at 7:39
the unipotent radical is trivial
– Ronald
Jul 25 at 7:40
Ok, so you need to show that if a connected solvable group has trivial unipotent radical, then it is a torus. For this you can use that it will be isomorphic to a group of upper triangular matrices.
– Tobias Kildetoft
Jul 25 at 7:44
Can't we assume more since the radical actually is abelian?
– Ronald
Jul 25 at 7:48
Sure, if you already know that it is abelian you can use that (though I don't see an obvious way to show this directly).
– Tobias Kildetoft
Jul 25 at 7:50
How do you define reductive? (I define it as the radical being a torus).
– Tobias Kildetoft
Jul 25 at 7:39
How do you define reductive? (I define it as the radical being a torus).
– Tobias Kildetoft
Jul 25 at 7:39
the unipotent radical is trivial
– Ronald
Jul 25 at 7:40
the unipotent radical is trivial
– Ronald
Jul 25 at 7:40
Ok, so you need to show that if a connected solvable group has trivial unipotent radical, then it is a torus. For this you can use that it will be isomorphic to a group of upper triangular matrices.
– Tobias Kildetoft
Jul 25 at 7:44
Ok, so you need to show that if a connected solvable group has trivial unipotent radical, then it is a torus. For this you can use that it will be isomorphic to a group of upper triangular matrices.
– Tobias Kildetoft
Jul 25 at 7:44
Can't we assume more since the radical actually is abelian?
– Ronald
Jul 25 at 7:48
Can't we assume more since the radical actually is abelian?
– Ronald
Jul 25 at 7:48
Sure, if you already know that it is abelian you can use that (though I don't see an obvious way to show this directly).
– Tobias Kildetoft
Jul 25 at 7:50
Sure, if you already know that it is abelian you can use that (though I don't see an obvious way to show this directly).
– Tobias Kildetoft
Jul 25 at 7:50
 |Â
show 2 more comments
1 Answer
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A standard example of connected solvable group is the group of (upper) triangular non-singular matrices.
This has a nice structure theorem: semi-direct product of torus with strictly upper triangular (i.e., 1's in the diagonal)
Lie-Kolchin theorem states that connected solvable groups are closed subgroups of the above.
Now unipotent radical is trivial hypothesis combined with the above should tell you that it is a torus.
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A standard example of connected solvable group is the group of (upper) triangular non-singular matrices.
This has a nice structure theorem: semi-direct product of torus with strictly upper triangular (i.e., 1's in the diagonal)
Lie-Kolchin theorem states that connected solvable groups are closed subgroups of the above.
Now unipotent radical is trivial hypothesis combined with the above should tell you that it is a torus.
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
add a comment |Â
up vote
2
down vote
accepted
A standard example of connected solvable group is the group of (upper) triangular non-singular matrices.
This has a nice structure theorem: semi-direct product of torus with strictly upper triangular (i.e., 1's in the diagonal)
Lie-Kolchin theorem states that connected solvable groups are closed subgroups of the above.
Now unipotent radical is trivial hypothesis combined with the above should tell you that it is a torus.
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A standard example of connected solvable group is the group of (upper) triangular non-singular matrices.
This has a nice structure theorem: semi-direct product of torus with strictly upper triangular (i.e., 1's in the diagonal)
Lie-Kolchin theorem states that connected solvable groups are closed subgroups of the above.
Now unipotent radical is trivial hypothesis combined with the above should tell you that it is a torus.
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
A standard example of connected solvable group is the group of (upper) triangular non-singular matrices.
This has a nice structure theorem: semi-direct product of torus with strictly upper triangular (i.e., 1's in the diagonal)
Lie-Kolchin theorem states that connected solvable groups are closed subgroups of the above.
Now unipotent radical is trivial hypothesis combined with the above should tell you that it is a torus.
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
answered Jul 25 at 7:58
P Vanchinathan
13.9k12035
13.9k12035
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
add a comment |Â
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
Well, at first that it is a subgroup of a torus. But since it is also connected, that makes it a torus itself.
– Tobias Kildetoft
Jul 25 at 8:01
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
@TobiasKildetoft: Yes, this point needs to be added. Thanks.
– P Vanchinathan
Jul 25 at 8:03
add a comment |Â
How do you define reductive? (I define it as the radical being a torus).
– Tobias Kildetoft
Jul 25 at 7:39
the unipotent radical is trivial
– Ronald
Jul 25 at 7:40
Ok, so you need to show that if a connected solvable group has trivial unipotent radical, then it is a torus. For this you can use that it will be isomorphic to a group of upper triangular matrices.
– Tobias Kildetoft
Jul 25 at 7:44
Can't we assume more since the radical actually is abelian?
– Ronald
Jul 25 at 7:48
Sure, if you already know that it is abelian you can use that (though I don't see an obvious way to show this directly).
– Tobias Kildetoft
Jul 25 at 7:50