maximal subalgebras of nilpotent Lie algebra












1














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?










share|cite|improve this question
























  • 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
















1














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?










share|cite|improve this question
























  • 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














1












1








1


0





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?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 19:25

























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


















  • 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










1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • Thank you for your help.
    – Afsaneh
    Nov 26 at 19:47











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
});


}
});














draft saved

draft discarded


















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









1














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.






share|cite|improve this answer





















  • Thank you for your help.
    – Afsaneh
    Nov 26 at 19:47
















1














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.






share|cite|improve this answer





















  • Thank you for your help.
    – Afsaneh
    Nov 26 at 19:47














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 19:07









Torsten Schoeneberg

3,7012833




3,7012833












  • 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




Thank you for your help.
– Afsaneh
Nov 26 at 19:47


















draft saved

draft discarded




















































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.




draft saved


draft discarded














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





















































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







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei