maximal subalgebras of nilpotent Lie algebra
I know that if every maximal subalgebra is an ideal, then L is nilpotent. Is every maximal subalgebra of a nilpotent Lie algebra an ideal?
lie-algebras
asked Nov 24 at 19:16
Afsaneh
83
add a comment |
I know that if every maximal subalgebra is an ideal, then L is nilpotent. Is every maximal subalgebra of a nilpotent Lie algebra an ideal?
lie-algebras
asked Nov 24 at 19:16
Afsaneh
83
Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central.
– Torsten Schoeneberg
Nov 25 at 19:44
add a comment |
I know that if every maximal subalgebra is an ideal, then L is nilpotent. Is every maximal subalgebra of a nilpotent Lie algebra an ideal?
lie-algebras
asked Nov 24 at 19:16
Afsaneh
83
I know that if every maximal subalgebra is an ideal, then L is nilpotent. Is every maximal subalgebra of a nilpotent Lie algebra an ideal?
lie-algebras
lie-algebras
asked Nov 24 at 19:16
Afsaneh
83
asked Nov 24 at 19:16
Afsaneh
83
edited Nov 24 at 19:25
asked Nov 24 at 19:16
Afsaneh
83
asked Nov 24 at 19:16
Afsaneh
83
asked Nov 24 at 19:16
Afsaneh
83
83
Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central.
– Torsten Schoeneberg
Nov 25 at 19:44
add a comment |
Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central.
– Torsten Schoeneberg
Nov 25 at 19:44
Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central.
– Torsten Schoeneberg
Nov 25 at 19:44
Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central.
– Torsten Schoeneberg
Nov 25 at 19:44
add a comment |
1 Answer
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For any Lie algebra $L$, the lower central series is defined by $C_0(L) := L$, $C_{n+1}(L):= [L, C_n(L)]$. It is well-known that $L$ is nilpotent if and only if $C_n(L) = 0$ for sufficiently high $n$.
Now let $M$ be a maximal proper subalgebra of a Lie algebra $L$. Then $M +C_1(L)$ is also a subalgebra, so by maximality we have either $M +C_1(L)=M$ or $M +C_1(L)=L$. In the first case, $M$ is an ideal. To see that the second case does not occur in our situation, first show via induction:
Lemma: $L=M +C_1(L) Rightarrow C_n(L)=C_n(M) +C_{n+1}(L)$ for all $n$.
Namely,
$C_{n+1}(L)\ =[L, C_n(L)]= [M+C_1(L), C_n(M) +C_{n+1}(L)] =\ underbrace{[M, C_n(M)]}_{C_{n+1}(M)} + underbrace{[M, C_{n+1}(L)] +[C_1(L), C_n(M)] + [C_1(L), C_{n+1}(L)]}_{C_{n+2}(L)}$
Now since $L$ is nilpotent, there is $k$ with $C_k(L) neq 0 = C_{k+1}(L)$, hence $C_k(L) =C_k(M)$. Working backwards ($kto k-1to...$) with the formula in the lemma gives $C_i(L) =C_i(M) subseteq M$ for all $i$, in particular $L=M$, contradiction to $M$ being a proper subalgebra.
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
add a comment |
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1 Answer
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1 Answer
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For any Lie algebra $L$, the lower central series is defined by $C_0(L) := L$, $C_{n+1}(L):= [L, C_n(L)]$. It is well-known that $L$ is nilpotent if and only if $C_n(L) = 0$ for sufficiently high $n$.
Now let $M$ be a maximal proper subalgebra of a Lie algebra $L$. Then $M +C_1(L)$ is also a subalgebra, so by maximality we have either $M +C_1(L)=M$ or $M +C_1(L)=L$. In the first case, $M$ is an ideal. To see that the second case does not occur in our situation, first show via induction:
Lemma: $L=M +C_1(L) Rightarrow C_n(L)=C_n(M) +C_{n+1}(L)$ for all $n$.
Namely,
$C_{n+1}(L)\ =[L, C_n(L)]= [M+C_1(L), C_n(M) +C_{n+1}(L)] =\ underbrace{[M, C_n(M)]}_{C_{n+1}(M)} + underbrace{[M, C_{n+1}(L)] +[C_1(L), C_n(M)] + [C_1(L), C_{n+1}(L)]}_{C_{n+2}(L)}$
Now since $L$ is nilpotent, there is $k$ with $C_k(L) neq 0 = C_{k+1}(L)$, hence $C_k(L) =C_k(M)$. Working backwards ($kto k-1to...$) with the formula in the lemma gives $C_i(L) =C_i(M) subseteq M$ for all $i$, in particular $L=M$, contradiction to $M$ being a proper subalgebra.
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
add a comment |
For any Lie algebra $L$, the lower central series is defined by $C_0(L) := L$, $C_{n+1}(L):= [L, C_n(L)]$. It is well-known that $L$ is nilpotent if and only if $C_n(L) = 0$ for sufficiently high $n$.
Now let $M$ be a maximal proper subalgebra of a Lie algebra $L$. Then $M +C_1(L)$ is also a subalgebra, so by maximality we have either $M +C_1(L)=M$ or $M +C_1(L)=L$. In the first case, $M$ is an ideal. To see that the second case does not occur in our situation, first show via induction:
Lemma: $L=M +C_1(L) Rightarrow C_n(L)=C_n(M) +C_{n+1}(L)$ for all $n$.
Namely,
$C_{n+1}(L)\ =[L, C_n(L)]= [M+C_1(L), C_n(M) +C_{n+1}(L)] =\ underbrace{[M, C_n(M)]}_{C_{n+1}(M)} + underbrace{[M, C_{n+1}(L)] +[C_1(L), C_n(M)] + [C_1(L), C_{n+1}(L)]}_{C_{n+2}(L)}$
Now since $L$ is nilpotent, there is $k$ with $C_k(L) neq 0 = C_{k+1}(L)$, hence $C_k(L) =C_k(M)$. Working backwards ($kto k-1to...$) with the formula in the lemma gives $C_i(L) =C_i(M) subseteq M$ for all $i$, in particular $L=M$, contradiction to $M$ being a proper subalgebra.
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
add a comment |
For any Lie algebra $L$, the lower central series is defined by $C_0(L) := L$, $C_{n+1}(L):= [L, C_n(L)]$. It is well-known that $L$ is nilpotent if and only if $C_n(L) = 0$ for sufficiently high $n$.
Now let $M$ be a maximal proper subalgebra of a Lie algebra $L$. Then $M +C_1(L)$ is also a subalgebra, so by maximality we have either $M +C_1(L)=M$ or $M +C_1(L)=L$. In the first case, $M$ is an ideal. To see that the second case does not occur in our situation, first show via induction:
Lemma: $L=M +C_1(L) Rightarrow C_n(L)=C_n(M) +C_{n+1}(L)$ for all $n$.
Namely,
$C_{n+1}(L)\ =[L, C_n(L)]= [M+C_1(L), C_n(M) +C_{n+1}(L)] =\ underbrace{[M, C_n(M)]}_{C_{n+1}(M)} + underbrace{[M, C_{n+1}(L)] +[C_1(L), C_n(M)] + [C_1(L), C_{n+1}(L)]}_{C_{n+2}(L)}$
Now since $L$ is nilpotent, there is $k$ with $C_k(L) neq 0 = C_{k+1}(L)$, hence $C_k(L) =C_k(M)$. Working backwards ($kto k-1to...$) with the formula in the lemma gives $C_i(L) =C_i(M) subseteq M$ for all $i$, in particular $L=M$, contradiction to $M$ being a proper subalgebra.
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
For any Lie algebra $L$, the lower central series is defined by $C_0(L) := L$, $C_{n+1}(L):= [L, C_n(L)]$. It is well-known that $L$ is nilpotent if and only if $C_n(L) = 0$ for sufficiently high $n$.
Now let $M$ be a maximal proper subalgebra of a Lie algebra $L$. Then $M +C_1(L)$ is also a subalgebra, so by maximality we have either $M +C_1(L)=M$ or $M +C_1(L)=L$. In the first case, $M$ is an ideal. To see that the second case does not occur in our situation, first show via induction:
Lemma: $L=M +C_1(L) Rightarrow C_n(L)=C_n(M) +C_{n+1}(L)$ for all $n$.
Namely,
$C_{n+1}(L)\ =[L, C_n(L)]= [M+C_1(L), C_n(M) +C_{n+1}(L)] =\ underbrace{[M, C_n(M)]}_{C_{n+1}(M)} + underbrace{[M, C_{n+1}(L)] +[C_1(L), C_n(M)] + [C_1(L), C_{n+1}(L)]}_{C_{n+2}(L)}$
Now since $L$ is nilpotent, there is $k$ with $C_k(L) neq 0 = C_{k+1}(L)$, hence $C_k(L) =C_k(M)$. Working backwards ($kto k-1to...$) with the formula in the lemma gives $C_i(L) =C_i(M) subseteq M$ for all $i$, in particular $L=M$, contradiction to $M$ being a proper subalgebra.
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7012833
3,7012833
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
add a comment |
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
add a comment |
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Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central.
– Torsten Schoeneberg
Nov 25 at 19:44