Lie algebras with a single ideal and with all ideals

Question

Consider Lie algebra $\mathfrak{L}_2$ over field $F$ of characteristics zero, given by basis $\{x,y\}$ and Lie bracket being defined with $[x,y] = x$ (or, to be formal, with $[ax+by,cx+dy] = (ad-bc)x$, $a,b,c,d \in F$). This Lie algebra has ideals $0,\mathfrak{L}_2$ and ideal $I$ generated with $x$. That are all ideals in $\mathfrak{L}_2$. This algebra is a nice example of:

Lie algebra having only one proper (different then $0$ and itself) ideal

Lie algebra having an ideal of dimension $k$, for every $k \leq n$, and not being Abelian.

In the first case, classifying all such Lie algebras is certainly impossible: Take for example Lie algebra generated with $\{x_1,...,x_n\}$, take $[x_1,x_i] = x_1$ for all $2 \leq i \leq n$, now if $L_1$ is any Lie algebra generated by $\{x_2,...,x_n\}$ then whatever ideal might be in $L_1$ it is not ideal in $L$, so $L$ is given by ANY Lie algebra of dimension $n-1$+ something additional. So, softcore, this problem depends on a problem of finding all Lie algebras of given dimension, a task that is currently unknown (up to my poor knowledge). Did anyone do anything with respect to these Lie algebras? I am not certain about any progression in classification of the other case. I myself do not have a clue about this. Thankieees!

Thanks to a comment by @DietrichBurde, first case can see some progress. Suppose $L$ is a Lie algebra with only one proper ideal $I$, then $Z(L)$ is an ideal in $L$ and we have three options:

(1) $Z(L) = 0$

(2) $Z(L) = I$, but then if $I$ has basis $\{x_1,...,x_k\}$ and $\{x_1,...,x_n\}$ is basis of $L$, then Lie subalgebra $L_1$ of $L$ with basis $\{x_{k+1},...,x_n\}$ cannot have any nontrivial ideals, since if $J$ is an ideal in $L_1$, it is indeed an ideal in $L$, so $L$ has at least two proper ideals: $I$ and $J$. So, $L_1$ has to be simple. $Z(L)$ is already Abelian. Any ideal in $Z(L)$ is automatically an ideal in $L$, so $Z(L)$ can not have non-trivial ideals of its own. On the other hand, $Z(L)$ is Abelian Lie algebra, and Abelian Lie algebra has one proper ideal if and only if it is $\mathrm{Ab}_2$, Abelian algebra of dimension 2 (a bit of thinking makes it clear that all Abelian Lie algebras of dimension $n$ are mutually isomorphic). So, between $Z(L)$ being $\mathrm{Ab}_2$, and $L_1$ being simple, us working over field of characteristics zero, this case is solved.

(3) $Z(L) = L$ are Abelian Lie algebras anyway and all such that have one proper ideal are $\mathrm{Ab}_2$ as discussed in previous paragraph.

We can hence refine our question:

Can we know something more about Lie algebras with one proper ideal and with $Z(L) = 0$?

Answer 1

We are trying to classify those Lie algebras $L$ (if needed: of finite dimension over a characteristic $0$ field $k$) which have exactly one proper ideal $0\subsetneq I \subsetneq L$. Note that this implies that the quotient $L/I$ has no proper ideals, i.e. it is either simple or one-dimensional abelian.

1. $Z(L)=0$. Because if not, either $Z(L)=L$ i.e. $L$ is abelian, but a one-dimensional abelian Lie algebra has no proper ideals, while abelian Lie algebras of dimension $\ge 2$ have $\ge 3$ proper ideals (over any field). Or $Z(L)=I$; then $L/I$ cannot be one-dimensional without $L$ being abelian, contradiction, so $L/I$ is simple. But that makes $L$ a central extension of a simple Lie algebra, which are known to be trivial i.e. $L \simeq L/I \oplus Z(L)$ as Lie algebras and we have more than one proper ideal.

2. If $L$ is solvable, it is of the following form: $I=V$ is a nonzero abelian Lie algebra a.k.a. $k$-vector space, and there is an automorphism $T \in GL(V)$ whose characteristic polynomial is irreducible (i.e. such that $V$ has no nonotrivial $T$-invariant subspaces, cf. 1, 2, 3); $L$ is the vector space $kx_1 \oplus V$ where $ad(x_1)$ operates on $V$ via $T$. This includes the two-dimensional non-abelian Lie algebra mentioned as example $\mathfrak{L}_2$ by you, and if $k$ is algebraically closed, that's the only option; over the real numbers, $I$ can have dimension $2$ as well but nothing more; over other fields, any dimension might be possible, e.g. see 4. --- To prove this: Since $0\neq [L,L] \neq L$, we have $[L,L] = I$, but by solvability, $[I,I]=0$ i.e. $I$ is abelian; also by solvability, $L/I$ must be one-dimensional, spanned by $x_1$, say; then $T:= ad(x_1)$ must operate as automorphism on $I$ (its kernel is contained in $Z(L)$), and any nontrivial ideals inside $I$ would have to be nontrivial $T$-invariant subspaces.

3. Remains the case $Rad(L) = I$, because if $Rad(L)=0$, $L$ would either be simple i.e. have no proper ideals, or semisimple but not simple, i.e. have at least two proper ideals. Necessarily $L$ is perfect i.e. $[L,L]=L$ since otherwise $[L,L]$ would be contained in $I$ which makes $L$ solvable; also, since $I$ is solvable, $[I,I]$ must be $0$ i.e. $I=V$ is an abelian Lie algebra again. Further, $L/I$ must be simple, because if it were abelian, $L$ would be solvable. Now using Levi-Maltsev, $L$ contains a subalgebra $S \simeq L/I$; the adjoint action of $S$ on $V$ makes $V$ an $S$-module, which must be irreducible because otherwise its components would be further nontrivial ideals inside $I=V$. Hence, in this case, $L$ is isomorphic to a semidirect product $S \ltimes V$ of a simple Lie algebra operating on one of its irreducible representations. This is the case alluded to by Dietrich Burde in his comment, and more generally, in his answer to perfect Lie algebra but not semisimple (just as, as said, we need $S$ to be simple not just semisimple here).

So basically, your own example and Dietrich's example, with the outlined generalizations they allow, are all possibilities. In fact, note that both of them can be phrased as: "A Lie algebra without proper ideals operating on one of its indecomposable representations".