Ideal of ideal is ideal of Lie algebra

Question

Q: Let L be a Lie algebra, J be its ideal and I is an ideal of J, then I is an ideal of L.

My attempt: We know that for any ideal J of Lie algebra L: [L, J] and [J, L] are in J. So i in I, j in J and l in L: [i, [j, l]] =-[j, [l, i]] - [l, [i, j]] The RHS is in I, thus LHS as well. My question is can we deduce that [l, i] is in I (so Q is proven)?

In this older question it is also stated that the result is false in general, but true in a special case. — Jyrki Lahtonen, Mar 11 '22 at 15:20
You could (possibly should?) have left the question in its earlier form. The way it is currently written leaves it in a form that is not quite up to the standards outlined in our guide for new askers. Nobody here will think less of you for your original thinking leading to a dead end. On the other hand, if the question looks like an attempt to outsource an assignment, that may attract negative attention. It may not feel natural to a new user, but that's the way site policies have developed to work. Don't worry! — Jyrki Lahtonen, Mar 12 '22 at 10:10

score 5 · Accepted Answer · answered Mar 11 '22 at 15:02

5

The claim is false as stated.

Consider the strictly upper triangular $3\times 3$ matrices. They are spanned by the elementary matrices $e_{12}$, $e_{13}$ and $e_{23}$, where the subscripts indicate the position of the $1$ (the rest of the entries are all zeros). They have the commutator relations $$[e_{12},e_{13}]=0,\quad [e_{12},e_{23}]=e_{13},\quad [e_{23},e_{13}]=0.$$

Let $L=\langle e_{12},e_{13},e_{23}\rangle$, $J=\langle e_{12},e_{13}\rangle$, $I=\langle e_{12}+e_{13}\rangle$. Then $[L,L]\subset J$, so $J$ is an ideal. The ideal $J$ is an abelian Lie algebra, so $I\subset J$ is an ideal. But $I$ is not an ideal of $L$ as $$[e_{12}+e_{13},e_{23}]=e_{13}\notin I.$$

answered Mar 11 '22 at 15:02

Jyrki Lahtonen

133,153

It is easier to construct examples, when you have a similar sequence of groups at hand. I just recalled a fitting example I cooked up for my group theory course last Spring. $G_3\unlhd G_2\unlhd G_1$ such that $G_3$ is not a normal subgroup of $G_1$. These are the matching Lie algebras. – Jyrki Lahtonen Mar 11 '22 at 15:06
I am surprised if this has not been covered already. I could not find it. Alas, I no longer frequent the Lie algebra tag, so I'm rusty with useful buzzwords for the search engine :-( – Jyrki Lahtonen Mar 11 '22 at 15:10
1

Jyrki, I just found this post saying that it only works if the ideal is characteristic. But this has no concrete counterexample (except for a comment by Yves). So I have upvoted your post. – Dietrich Burde Mar 11 '22 at 15:17
Thanks @Dietrich. I feel a bit better about this if you could not find a matching old thread either. – Jyrki Lahtonen Mar 11 '22 at 15:18

Ideal of ideal is ideal of Lie algebra

1 Answers1