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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 $y$,

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    $\begingroup$ Well...$[ax,x]=a[x,x]=0$ for all $x$ so the property is hard to satisfy. $\endgroup$ – lulu Oct 4 '18 at 12:39
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    $\begingroup$ If $x \neq 0$, then $[x, 0] = 0$, so any such Lie algebra is trivial. $\endgroup$ – Travis Oct 4 '18 at 12:45
  • $\begingroup$ Sorry, the question had a typo, please see the edit. $\endgroup$ – Fizikus Oct 4 '18 at 13:38
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    $\begingroup$ What is true is that if the dimension of a Lie algebra with this property is $\ge 2$, then its centre must be trivial, for otherwise take an element from the centre and another which is linearly independent from it. In particular, no nilpotent Lie algebra except the one-dimensional one has the property. This further implies (since subalgebras inherit the property) that a solvable Lie algebra which has this property must have $\dim D\mathfrak{g} =1$. $\endgroup$ – Torsten Schoeneberg Oct 8 '18 at 18:51
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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.


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?

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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 $[x,y]=x$.

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    $\begingroup$ It's immediate that in a Lie algebra with the property described in the OP, any abelian subalgebra must have dimension $\le 1$. Since CSAs of semisimple Lie algebras are abelian, this shortens the proof of $\dim(\mathfrak{h})=1$ considerably. $\endgroup$ – Torsten Schoeneberg Oct 4 '18 at 19:12
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    $\begingroup$ Up to isomorphism there is only one complex semisimple Lie algebra of rank 1, or in other words, $\mathfrak{sl}_2(\Bbb C) \simeq \mathfrak{so}_3(\mathbb{C}) \simeq \mathfrak{sp}_4(\mathbb{C})$. On the other hand, $\mathfrak{so}_2(\mathbb{C})$ is not simple, it is just the abelian Lie algebra of dimension 1 -- so it has the property, but for a different reason. The same is true for any alg. closed $F$ of char. 0 instead of $\Bbb C$. $\endgroup$ – Torsten Schoeneberg Oct 4 '18 at 19:29
  • $\begingroup$ @TorstenSchoeneberg Thank you. I forgot that $\mathfrak{sl}_2(F)$, $\mathfrak{so}_3(F)$, and $\mathfrak{sp}_4(F)$ are isomorphic (and that $\mathfrak{so}_2(F)$ is not simple). $\endgroup$ – user593746 Oct 7 '18 at 13:12
  • $\begingroup$ @TorstenSchoeneberg Do you have other nontrivial nonsemisimple examples? $\endgroup$ – user593746 Oct 7 '18 at 13:17
  • $\begingroup$ See my latest comment under the question, I have a feeling one can restrict the solvable case further, maybe all the way to the one example you gave. -- Actually, I am still very unclear but more interested in the semisimple case for non-algebraically closed base fields $k$ (even for $k = \Bbb R$). As in your proof, it's straightforward that only absolutely simple ones of $k$-rational rank $\le 1$ can have this property -- but do they all?? $\endgroup$ – Torsten Schoeneberg Oct 8 '18 at 18:53

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