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
1
active
oldest
votes
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
3,7412833
Thank you for your help.
– Afsaneh
Nov 26 at 19:47
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011965%2fmaximal-subalgebras-of-nilpotent-lie-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
3,7412833
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
3,7412833
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
3,7412833
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
3,7412833
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7412833
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7412833
answered Nov 26 at 19:07
Torsten Schoeneberg
3,7412833
3,7412833
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011965%2fmaximal-subalgebras-of-nilpotent-lie-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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