non nilpotent Lie algebras
$begingroup$
Let $L$ be a nonperfect Lie algebra (i.e. $L ne [L, L]$) which is
not nilpotent. Is there a ideal of $L$ such as $M$ such that intersection drived subalgebra of $L$ and $Z(M)$ is nonzero?
group-theory lie-algebras
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
$endgroup$
add a comment |
$begingroup$
Let $L$ be a nonperfect Lie algebra (i.e. $L ne [L, L]$) which is
not nilpotent. Is there a ideal of $L$ such as $M$ such that intersection drived subalgebra of $L$ and $Z(M)$ is nonzero?
group-theory lie-algebras
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
$endgroup$
add a comment |
$begingroup$
Let $L$ be a nonperfect Lie algebra (i.e. $L ne [L, L]$) which is
not nilpotent. Is there a ideal of $L$ such as $M$ such that intersection drived subalgebra of $L$ and $Z(M)$ is nonzero?
group-theory lie-algebras
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
$endgroup$
Let $L$ be a nonperfect Lie algebra (i.e. $L ne [L, L]$) which is
not nilpotent. Is there a ideal of $L$ such as $M$ such that intersection drived subalgebra of $L$ and $Z(M)$ is nonzero?
group-theory lie-algebras
group-theory lie-algebras
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
edited Dec 18 '18 at 10:35
Andreas Caranti
56.6k34395
56.6k34395
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
asked Dec 18 '18 at 5:56
AfsanehAfsaneh
83
83
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Yes, there is such an ideal. Take $L$ to be the $2$-dimensional nonabelian Lie algebra with basis $e_1,e_2$ and Lie bracket $[e_1,e_2]=e_1$ and the ideal $M=langle e_1rangle$. Then $L$ is not nilpotent, $Lneq [L,L]$ and $M$ is an ideal with $Z(M)=[L,L]neq 0$.
In general, however, this need not be true. Take $L=mathfrak{gl}_2(Bbb{C})=mathfrak{sl}_2(Bbb{C})oplus Bbb{C}$. Then $L$ is not perfect and not nilpotent, but $Z(M)$ is either zero or $Bbb{C}$ for all ideals $M$, and the intersection with $[L,L]$ is zero.
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
$endgroup$
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
Yes, there is such an ideal. Take $L$ to be the $2$-dimensional nonabelian Lie algebra with basis $e_1,e_2$ and Lie bracket $[e_1,e_2]=e_1$ and the ideal $M=langle e_1rangle$. Then $L$ is not nilpotent, $Lneq [L,L]$ and $M$ is an ideal with $Z(M)=[L,L]neq 0$.
In general, however, this need not be true. Take $L=mathfrak{gl}_2(Bbb{C})=mathfrak{sl}_2(Bbb{C})oplus Bbb{C}$. Then $L$ is not perfect and not nilpotent, but $Z(M)$ is either zero or $Bbb{C}$ for all ideals $M$, and the intersection with $[L,L]$ is zero.
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
$endgroup$
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
add a comment |
$begingroup$
Yes, there is such an ideal. Take $L$ to be the $2$-dimensional nonabelian Lie algebra with basis $e_1,e_2$ and Lie bracket $[e_1,e_2]=e_1$ and the ideal $M=langle e_1rangle$. Then $L$ is not nilpotent, $Lneq [L,L]$ and $M$ is an ideal with $Z(M)=[L,L]neq 0$.
In general, however, this need not be true. Take $L=mathfrak{gl}_2(Bbb{C})=mathfrak{sl}_2(Bbb{C})oplus Bbb{C}$. Then $L$ is not perfect and not nilpotent, but $Z(M)$ is either zero or $Bbb{C}$ for all ideals $M$, and the intersection with $[L,L]$ is zero.
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
$endgroup$
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
add a comment |
$begingroup$
Yes, there is such an ideal. Take $L$ to be the $2$-dimensional nonabelian Lie algebra with basis $e_1,e_2$ and Lie bracket $[e_1,e_2]=e_1$ and the ideal $M=langle e_1rangle$. Then $L$ is not nilpotent, $Lneq [L,L]$ and $M$ is an ideal with $Z(M)=[L,L]neq 0$.
In general, however, this need not be true. Take $L=mathfrak{gl}_2(Bbb{C})=mathfrak{sl}_2(Bbb{C})oplus Bbb{C}$. Then $L$ is not perfect and not nilpotent, but $Z(M)$ is either zero or $Bbb{C}$ for all ideals $M$, and the intersection with $[L,L]$ is zero.
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
$endgroup$
Yes, there is such an ideal. Take $L$ to be the $2$-dimensional nonabelian Lie algebra with basis $e_1,e_2$ and Lie bracket $[e_1,e_2]=e_1$ and the ideal $M=langle e_1rangle$. Then $L$ is not nilpotent, $Lneq [L,L]$ and $M$ is an ideal with $Z(M)=[L,L]neq 0$.
In general, however, this need not be true. Take $L=mathfrak{gl}_2(Bbb{C})=mathfrak{sl}_2(Bbb{C})oplus Bbb{C}$. Then $L$ is not perfect and not nilpotent, but $Z(M)$ is either zero or $Bbb{C}$ for all ideals $M$, and the intersection with $[L,L]$ is zero.
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
edited Dec 18 '18 at 9:25
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
answered Dec 18 '18 at 9:02
Dietrich BurdeDietrich Burde
79.4k647103
79.4k647103
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
add a comment |
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Thank you for your help.
$endgroup$
– Afsaneh
Dec 18 '18 at 11:35
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L?
$endgroup$
– Afsaneh
Dec 18 '18 at 11:38
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
$begingroup$
In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above.
$endgroup$
– Dietrich Burde
Dec 18 '18 at 12:18
add a comment |
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