Lie algebra with additional property

Question

Let $[\cdot,\cdot]$ be the Lie bracket of some Lie algebra. Assume the following holds $$y \neq \lambda x \iff [x,y] \neq 0.$$ What is the name of Lie algebra with this property? This property holds, i.e., for the cross product of 3D vectors. Alternatively, this should be equivalent to the property $$[x,y]=0 \iff y = \lambda x$$ for nonzero $x$ and $y$,

Answer 1

First note two things: The property is equivalent to the statement that every abelian subalgebra has dimension $\le 1$; and if a Lie algebra has this property, then so does each of its subalgebras. With this we can show:

Let $k$ be a field of characteristic 0, and let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $k$ which has the described property. Then $\mathfrak{g}$ is isomorphic to either:

• the zero algebra $0$, or
• the one-dimensional abelian Lie algebra over $k$, or
• the two-dimensional Lie algebra $span_k\lbrace x, y\rbrace$ with $[x,y]=x$, or
• a three-dimensional simple Lie algebra over $k$ which after scalar extension to an algebraic closure $\bar k$ becomes isomorphic to $\mathfrak{sl}_2(\bar k)$; i.e., a $k$-form of $\mathfrak{sl}_2$;
• (or possibly, a non-central extension of $\mathfrak{sl}_2(k)$ by the two-dimensional Lie algebra described above; maybe this does not exist?)

Proof:

Case 1: $\mathfrak{g}$ is semisimple. Then it has a Cartan subalgebra (=maximal toral subalgebra) whose scalar extension is a split CSA in $\mathfrak{g}\otimes \bar k$ of the same dimension. This dimension is called the (absolute) rank of $\mathfrak{g}$. Since CSA's of semisimple Lie algebras are abelian, by the property the rank of $\mathfrak{g}$ is $1$, which implies that it is a $k$-form of $\mathfrak{sl}_2$. One can show that each such $k$-form has the described property.

(These $k$-forms can be parametrised by quaternion algebras over $k$. If $k =\bar k$, there is only the split form $\mathfrak{sl}_2(k)$ itself. For $k =\Bbb R$ or $p$-adic fields, there is up to isomorphism one more non-split one, the real one is usually called $\mathfrak{su}_2$ and is isomorphic to the cross product mentioned in the OP. Over $\Bbb Q$, there are infinitely many non-isomorphic ones, cf. https://math.stackexchange.com/a/2548338/96384.)

Case 2: $\mathfrak{g}$ is solvable. If the centre $Z$ of $\mathfrak{g}$ is non-trivial, then $\mathfrak{g}$ must be of dimension $1$ (and equal its centre), since otherwise $span\lbrace x,z\rbrace$ for $0 \neq z \in Z$ and $x$ linearly independent from $z$ is a two-dimensional abelian subalgebra.

Now $D\mathfrak{g} := [\mathfrak{g}, \mathfrak{g}]$ is known to be nilpotent, so it has a non-trivial centre, so by the above is either $0$ (in which case again $\dim_k \mathfrak{g} \le 1$) or one-dimensional.

In this last remaining case, say a basis vector of $D\mathfrak{g}$ is $x$. Now if there were two linearly independent vectors $y,z \in \mathfrak{g} \setminus D\mathfrak{g}$, then $[x,y] = ax$ and $[x,z] = bx$ with $a,b \neq 0$ (if one of them were $0$, it would violate the property). But then $[x, by-az] = 0$ violates the property. So $dim_k(\mathfrak{g}/D\mathfrak{g}) =1$ and up to isomorphism we are in the two-dimensional case of the list.

General case: By Levi-Maltsev, the only cases that could happen besides the ones on the list are semidirect products of a form of $\mathfrak{sl_2}$ with either the one- or the two-dimensional Lie algebras described (which would then be the radical of $\mathfrak{g}$). If the radical is one-dimensional however, the action of the simple part on it must be trivial, so it is central, contradiction as in case 2. If the radical is two-dimensional, the action on it is either trivial (contradiction again), or gives an isomorphism of the simple part to $\mathfrak{sl_2}$. This possibility I have not ruled out yet, but I don't know if it occurs either.

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Note: Regardless of whether that last case occurs or not, the list is quite short and the property does not deserve to have a name of its own so far. I would be interested if something more exciting happens in positive characteristic though. Experts?

Answer 2

I never heard if there is a particular name for such Lie algebras. However, I have a comment, which is too long to be put in the comment section, so I place it here.

First of all, if $\mathfrak{g}$ satisfies the condition $$\forall x\in \mathfrak{g},\text{ if }x\neq 0,\text{ then }\Big( \forall y\in\mathfrak{g}, [x,y]=0\text{ iff }\exists \lambda\in F, y=\lambda x\Big),\tag{1}$$ then $\mathfrak{g}$ is indecomposable (i.e., $\mathfrak{g}\neq \mathfrak{g}_1\oplus \mathfrak{g}_2$ for some nontrivial proper ideals $\mathfrak{g}_1$ and $\mathfrak{g}_2$ of $\mathfrak{g}$). If, in addition, $\mathfrak{g}$ is a finite-dimensional semisimple Lie algebra over an algebraically closed field $F$ of characteristic $0$, then $\mathfrak{g}$ must be simple as it is indecomposable. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$.

If $\mathfrak{h}$ is of dimension $n>1$, then there exist two linearly independent elements $h_1$ and $h_2$ of $\mathfrak{h}$. Pick an $\mathfrak{h}$-root $\alpha\in\mathfrak{h}^*$ of $\mathfrak{g}$ that separates $h_1$ and $h_2$. Let $x$ be a nonzero element of the $\mathfrak{h}$-root space $$\mathfrak{g}_\alpha(\mathfrak{h})=\big\{g\in\mathfrak{g}\big|\forall h\in\mathfrak{h}, [h,g]=\alpha(h)g\big\}.$$ Then, we have $$[h_1,x]=\alpha(h_1)x\text{ and }[h_2,x]=\alpha(h_2)x$$ so $$\big[\alpha(h_2)h_1-\alpha(h_1)h_2,x\big]=0.$$ As $x\notin\operatorname{span}\{h_1,h_2\}$, we must have $\alpha(h_2)h_1-\alpha(h_1)h_2=0$. But this means $$\alpha(h_1)=\alpha(h_2)=0,$$ as $h_1$ and $h_2$ are linearly independent. This is a contradiction (we chose $\alpha$ to separate $h_1$ and $h_2$, so they cannot simultaneously vanish), so $\dim \mathfrak{h}=1$ is the only possibility. (See comments below for a much easier proof that $\dim\mathfrak{h}=1$.)

In conclusion, if $\mathfrak{g}$ is a finite-dimensional semisimple Lie algebra over an algebraically closed field $F$ of characteristic $0$ such that the condition (1) holds, then $\mathfrak{g}$ is a simple Lie algebra of rank $1$. An example of such a Lie algebra is as given in your question. The cross-product structure on $\mathbb{R}^3$ can be complexified (i.e., extended to $\mathbb{C}^3$), which makes $\mathbb{C}^3$ with the cross product isomorphic to the Lie algebra $\mathfrak{so}_3(\mathbb{C})\cong\mathfrak{sl}_2(\mathbb{C})\cong\mathfrak{sp}_4(\mathbb{C})$.

There are however nonsemisimple Lie algebras with the required property. The one-dimensional Lie algebra is a trivial example. For another example, consider the $2$-dimensional nonabelian Lie algebra $\mathfrak{g}=\operatorname{span}\{x,y\}$ with $[x,y]=x$.