A Lie algebra with trivial center and commutative radical
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Let $mathfrak g$ be a complex linear Lie algebra. Assume that the center $mathfrak z$ of $mathfrak g$ is trivial
Let $mathfrak r$ be the radical of $mathfrak g$. If $mathfrak r$ is abelian, then $mathfrak g$ is semisimple?
What if $mathfrak g$ is the Lie algebra of an algebraic complex linear group?
lie-groups lie-algebras algebraic-groups
asked Sep 1 at 0:04
Amrat A
1135
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up vote
3
down vote
favorite
Let $mathfrak g$ be a complex linear Lie algebra. Assume that the center $mathfrak z$ of $mathfrak g$ is trivial
Let $mathfrak r$ be the radical of $mathfrak g$. If $mathfrak r$ is abelian, then $mathfrak g$ is semisimple?
What if $mathfrak g$ is the Lie algebra of an algebraic complex linear group?
lie-groups lie-algebras algebraic-groups
asked Sep 1 at 0:04
Amrat A
1135
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $mathfrak g$ be a complex linear Lie algebra. Assume that the center $mathfrak z$ of $mathfrak g$ is trivial
Let $mathfrak r$ be the radical of $mathfrak g$. If $mathfrak r$ is abelian, then $mathfrak g$ is semisimple?
What if $mathfrak g$ is the Lie algebra of an algebraic complex linear group?
lie-groups lie-algebras algebraic-groups
asked Sep 1 at 0:04
Amrat A
1135
Let $mathfrak g$ be a complex linear Lie algebra. Assume that the center $mathfrak z$ of $mathfrak g$ is trivial
Let $mathfrak r$ be the radical of $mathfrak g$. If $mathfrak r$ is abelian, then $mathfrak g$ is semisimple?
What if $mathfrak g$ is the Lie algebra of an algebraic complex linear group?
lie-groups lie-algebras algebraic-groups
lie-groups lie-algebras algebraic-groups
asked Sep 1 at 0:04
Amrat A
1135
asked Sep 1 at 0:04
Amrat A
1135
edited Sep 1 at 2:06
asked Sep 1 at 0:04
Amrat A
1135
asked Sep 1 at 0:04
Amrat A
1135
asked Sep 1 at 0:04
Amrat A
1135
1135
add a comment |Â
add a comment |Â
1 Answer
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Your question whether $mathfrakg$ is semisimple is equivalent to whether necessarily $mathfrakr = 0$. The answer is no. A counterexample is given by
$$mathfrakg = lbrace pmatrixa & b & d\
c& -a& e\
0&0&0 : a,b,c,d,e in Bbb Crbrace$$
where the radical
$$mathfrakr = lbrace pmatrix0 & 0 & d\
0& 0& e\
0&0&0 : d,e in Bbb Crbrace$$
is two-dimensional.
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your question whether $mathfrakg$ is semisimple is equivalent to whether necessarily $mathfrakr = 0$. The answer is no. A counterexample is given by
$$mathfrakg = lbrace pmatrixa & b & d\
c& -a& e\
0&0&0 : a,b,c,d,e in Bbb Crbrace$$
where the radical
$$mathfrakr = lbrace pmatrix0 & 0 & d\
0& 0& e\
0&0&0 : d,e in Bbb Crbrace$$
is two-dimensional.
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
add a comment |Â
up vote
1
down vote
accepted
Your question whether $mathfrakg$ is semisimple is equivalent to whether necessarily $mathfrakr = 0$. The answer is no. A counterexample is given by
$$mathfrakg = lbrace pmatrixa & b & d\
c& -a& e\
0&0&0 : a,b,c,d,e in Bbb Crbrace$$
where the radical
$$mathfrakr = lbrace pmatrix0 & 0 & d\
0& 0& e\
0&0&0 : d,e in Bbb Crbrace$$
is two-dimensional.
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your question whether $mathfrakg$ is semisimple is equivalent to whether necessarily $mathfrakr = 0$. The answer is no. A counterexample is given by
$$mathfrakg = lbrace pmatrixa & b & d\
c& -a& e\
0&0&0 : a,b,c,d,e in Bbb Crbrace$$
where the radical
$$mathfrakr = lbrace pmatrix0 & 0 & d\
0& 0& e\
0&0&0 : d,e in Bbb Crbrace$$
is two-dimensional.
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
Your question whether $mathfrakg$ is semisimple is equivalent to whether necessarily $mathfrakr = 0$. The answer is no. A counterexample is given by
$$mathfrakg = lbrace pmatrixa & b & d\
c& -a& e\
0&0&0 : a,b,c,d,e in Bbb Crbrace$$
where the radical
$$mathfrakr = lbrace pmatrix0 & 0 & d\
0& 0& e\
0&0&0 : d,e in Bbb Crbrace$$
is two-dimensional.
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
answered Sep 1 at 10:14
Torsten Schoeneberg
2,7551732
2,7551732
add a comment |Â
add a comment |Â
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