Lie Algebras : Showing $L$ is nilpotent if every maximal Lie subalgebra of $L$ is an ideal.
Given a finite dimensional Lie algebra $L$, suppose that each maximal lie subalgebra of $L$ is an ideal. Suppose the adjoint map, $operatorname{ad}_y$ is not nilpotent.
Then pick a maximal subalgebra $M$ containing $L_{0,y} := {x in L |(operatorname{ad}_y)^n(x) = 0 space text{for some} space nin mathbb{N}}$
Then $M$ is an ideal by assumption. And $dim(L/M) = 1$.
But $y in L_{0,y} leq M$ and so $[y,x] in M quad forall xin L$.
So $operatorname{ad}_y$ acts like $0$ on $L/M$.
$bf{Question}$ : How can I then deduce that $operatorname{ad}_y$ has eigenvalue $0$ on $L/L_{0,y}$?
lie-algebras ideals
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
add a comment |
Given a finite dimensional Lie algebra $L$, suppose that each maximal lie subalgebra of $L$ is an ideal. Suppose the adjoint map, $operatorname{ad}_y$ is not nilpotent.
Then pick a maximal subalgebra $M$ containing $L_{0,y} := {x in L |(operatorname{ad}_y)^n(x) = 0 space text{for some} space nin mathbb{N}}$
Then $M$ is an ideal by assumption. And $dim(L/M) = 1$.
But $y in L_{0,y} leq M$ and so $[y,x] in M quad forall xin L$.
So $operatorname{ad}_y$ acts like $0$ on $L/M$.
$bf{Question}$ : How can I then deduce that $operatorname{ad}_y$ has eigenvalue $0$ on $L/L_{0,y}$?
lie-algebras ideals
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x in L$.
– Samuel Reid
May 26 '13 at 5:25
What is $y$? It appears all of a sudden without having been introduced.
– Mariano Suárez-Álvarez
May 27 '13 at 3:48
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument.
– user58514
May 27 '13 at 5:08
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help?
– Vishal Gupta
May 28 '13 at 3:45
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this?
– user58514
May 28 '13 at 4:03
add a comment |
Given a finite dimensional Lie algebra $L$, suppose that each maximal lie subalgebra of $L$ is an ideal. Suppose the adjoint map, $operatorname{ad}_y$ is not nilpotent.
Then pick a maximal subalgebra $M$ containing $L_{0,y} := {x in L |(operatorname{ad}_y)^n(x) = 0 space text{for some} space nin mathbb{N}}$
Then $M$ is an ideal by assumption. And $dim(L/M) = 1$.
But $y in L_{0,y} leq M$ and so $[y,x] in M quad forall xin L$.
So $operatorname{ad}_y$ acts like $0$ on $L/M$.
$bf{Question}$ : How can I then deduce that $operatorname{ad}_y$ has eigenvalue $0$ on $L/L_{0,y}$?
lie-algebras ideals
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
Given a finite dimensional Lie algebra $L$, suppose that each maximal lie subalgebra of $L$ is an ideal. Suppose the adjoint map, $operatorname{ad}_y$ is not nilpotent.
Then pick a maximal subalgebra $M$ containing $L_{0,y} := {x in L |(operatorname{ad}_y)^n(x) = 0 space text{for some} space nin mathbb{N}}$
Then $M$ is an ideal by assumption. And $dim(L/M) = 1$.
But $y in L_{0,y} leq M$ and so $[y,x] in M quad forall xin L$.
So $operatorname{ad}_y$ acts like $0$ on $L/M$.
$bf{Question}$ : How can I then deduce that $operatorname{ad}_y$ has eigenvalue $0$ on $L/L_{0,y}$?
lie-algebras ideals
lie-algebras ideals
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
edited Nov 27 at 13:25
Davide Giraudo
125k16150259
125k16150259
asked May 26 '13 at 5:17
user58514
Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x in L$.
– Samuel Reid
May 26 '13 at 5:25
What is $y$? It appears all of a sudden without having been introduced.
– Mariano Suárez-Álvarez
May 27 '13 at 3:48
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument.
– user58514
May 27 '13 at 5:08
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help?
– Vishal Gupta
May 28 '13 at 3:45
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this?
– user58514
May 28 '13 at 4:03
add a comment |
Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x in L$.
– Samuel Reid
May 26 '13 at 5:25
What is $y$? It appears all of a sudden without having been introduced.
– Mariano Suárez-Álvarez
May 27 '13 at 3:48
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument.
– user58514
May 27 '13 at 5:08
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help?
– Vishal Gupta
May 28 '13 at 3:45
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this?
– user58514
May 28 '13 at 4:03
Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x in L$.
– Samuel Reid
May 26 '13 at 5:25
Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x in L$.
– Samuel Reid
May 26 '13 at 5:25
What is $y$? It appears all of a sudden without having been introduced.
– Mariano Suárez-Álvarez
May 27 '13 at 3:48
What is $y$? It appears all of a sudden without having been introduced.
– Mariano Suárez-Álvarez
May 27 '13 at 3:48
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument.
– user58514
May 27 '13 at 5:08
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument.
– user58514
May 27 '13 at 5:08
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help?
– Vishal Gupta
May 28 '13 at 3:45
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help?
– Vishal Gupta
May 28 '13 at 3:45
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this?
– user58514
May 28 '13 at 4:03
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this?
– user58514
May 28 '13 at 4:03
add a comment |
1 Answer
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$DeclareMathOperator{ad}{ad}$ To prove the statement Vishal Gupta mentioned assume that $K subset L$ is an ideal, $L/K$ is nilpotent and $ad_K x := (ad_L x)_{|K}$ is nilpotent, for all $x in L$. Let $pi: L to L/K$ the projection. We know (use induction) that $(L/K)^n = pi(L^n)$, hence there is some $N in mathbb N$ for which $L^N subset K$. Let $x in L$ arbitrary. Let $M in mathbb N$ such that $(ad_K x)^M = 0$. We now have
$$
(ad_L x)^{M+N}(L) subset (ad_L x)^M(K) = (ad_K x)^M(K) = 0.
$$ Note that we used that $(ad_L x)^N(L) subset L^N subset K$. Now use Engel's theorem to conclude that $L$ is nilpotent.
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
add a comment |
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$DeclareMathOperator{ad}{ad}$ To prove the statement Vishal Gupta mentioned assume that $K subset L$ is an ideal, $L/K$ is nilpotent and $ad_K x := (ad_L x)_{|K}$ is nilpotent, for all $x in L$. Let $pi: L to L/K$ the projection. We know (use induction) that $(L/K)^n = pi(L^n)$, hence there is some $N in mathbb N$ for which $L^N subset K$. Let $x in L$ arbitrary. Let $M in mathbb N$ such that $(ad_K x)^M = 0$. We now have
$$
(ad_L x)^{M+N}(L) subset (ad_L x)^M(K) = (ad_K x)^M(K) = 0.
$$ Note that we used that $(ad_L x)^N(L) subset L^N subset K$. Now use Engel's theorem to conclude that $L$ is nilpotent.
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
add a comment |
$DeclareMathOperator{ad}{ad}$ To prove the statement Vishal Gupta mentioned assume that $K subset L$ is an ideal, $L/K$ is nilpotent and $ad_K x := (ad_L x)_{|K}$ is nilpotent, for all $x in L$. Let $pi: L to L/K$ the projection. We know (use induction) that $(L/K)^n = pi(L^n)$, hence there is some $N in mathbb N$ for which $L^N subset K$. Let $x in L$ arbitrary. Let $M in mathbb N$ such that $(ad_K x)^M = 0$. We now have
$$
(ad_L x)^{M+N}(L) subset (ad_L x)^M(K) = (ad_K x)^M(K) = 0.
$$ Note that we used that $(ad_L x)^N(L) subset L^N subset K$. Now use Engel's theorem to conclude that $L$ is nilpotent.
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
add a comment |
$DeclareMathOperator{ad}{ad}$ To prove the statement Vishal Gupta mentioned assume that $K subset L$ is an ideal, $L/K$ is nilpotent and $ad_K x := (ad_L x)_{|K}$ is nilpotent, for all $x in L$. Let $pi: L to L/K$ the projection. We know (use induction) that $(L/K)^n = pi(L^n)$, hence there is some $N in mathbb N$ for which $L^N subset K$. Let $x in L$ arbitrary. Let $M in mathbb N$ such that $(ad_K x)^M = 0$. We now have
$$
(ad_L x)^{M+N}(L) subset (ad_L x)^M(K) = (ad_K x)^M(K) = 0.
$$ Note that we used that $(ad_L x)^N(L) subset L^N subset K$. Now use Engel's theorem to conclude that $L$ is nilpotent.
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
$DeclareMathOperator{ad}{ad}$ To prove the statement Vishal Gupta mentioned assume that $K subset L$ is an ideal, $L/K$ is nilpotent and $ad_K x := (ad_L x)_{|K}$ is nilpotent, for all $x in L$. Let $pi: L to L/K$ the projection. We know (use induction) that $(L/K)^n = pi(L^n)$, hence there is some $N in mathbb N$ for which $L^N subset K$. Let $x in L$ arbitrary. Let $M in mathbb N$ such that $(ad_K x)^M = 0$. We now have
$$
(ad_L x)^{M+N}(L) subset (ad_L x)^M(K) = (ad_K x)^M(K) = 0.
$$ Note that we used that $(ad_L x)^N(L) subset L^N subset K$. Now use Engel's theorem to conclude that $L$ is nilpotent.
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
answered Jun 1 '15 at 16:10
Epsilon
3,47832238
3,47832238
add a comment |
add a comment |
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Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x in L$.
– Samuel Reid
May 26 '13 at 5:25
What is $y$? It appears all of a sudden without having been introduced.
– Mariano Suárez-Álvarez
May 27 '13 at 3:48
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument.
– user58514
May 27 '13 at 5:08
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help?
– Vishal Gupta
May 28 '13 at 3:45
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this?
– user58514
May 28 '13 at 4:03