# Nilpotency and 2 dimensional abelian lie subalgebras

I am wondering is it really necessary to use the adjoint version of Engel’s theorem to prove that if every 2 dimensional lie sub-algebra of a given Lie algebra L is abelian then L is nilpotent? I wonder about that because you can write every 2 dimensional Lie algebra as the span of x and y in L and then by the assumption the bracket is zero for every two elements that I could choose. What am I missing here?

• It's false when the field is not algebraically closed, see my answer. – YCor Jul 31 '19 at 21:30

I think you are missing, that two arbitrary elements in $$L$$ will not span a subalgebra of $$L$$. For example, if $$L$$ is the Heisenberg Lie algebra with basis $$(x,y,z)$$ and Lie brackets determined by $$[x,y]=z$$, then the Lie bracket of each two elements spanning a subalgebra is zero, but not in general. $$x$$ and $$y$$ have a non-trivial Lie bracket and they do not span a subalgebra of $$L$$ (because the product is not "closed").

Proposition. Let $$\mathfrak{g}$$ be a finite-dimensional Lie algebra over an algebraically closed field $$K$$. Then $$\mathfrak{g}$$ is nilpotent iff every 2-dimensional subalgebra is abelian.

One direction is trivial (over an arbitrary field). For the nontrivial implication, suppose that $$\mathfrak{g}$$ is not nilpotent. By Engel's theorem (which holds over every field, see Jacobson), there exists $$x\in\mathfrak{g}$$ such that $$\mathrm{ad}(x)$$ is not nilpotent. Since the field is algebraically closed, this means that $$\mathfrak{x}$$ has a nonzero eigenvalue $$\lambda$$, say for en eigenvector $$y$$. Then $$(x,y)$$ span a non-abelian 2-dimensional subalgebra.

Remark. The above fails for every field $$K$$ that is not algebraically closed.

Indeed, choose an irreducible polynomial of degree $$d\ge 2$$. Let $$u$$ be the companion matrix of this polynomial. Then $$u$$ defines a semidirect product Lie algebra $$K^d\rtimes_uK$$ (of dimension $$d+1\ge 3$$). Then all its proper subalgebras are abelian (in particular, the 2-dimensional subalgebras are the 2-dimensional subspaces of the abelian ideal $$K^d$$).

Still there is a replacement, essentially following from the proof of the proposition. Say that an operator $$u$$ on $$K^d$$ is irreducible if $$d\ge 1$$ and $$u$$ preserves no subspace else than $$\{0\}$$ and $$K^d$$. So there exists such operators on $$K^d$$ if and only if there's an irreducible polynomial of degree $$d$$.

Proposition. Let $$\mathfrak{g}$$ be a finite-dimensional Lie algebra over a field field $$K$$. Then $$\mathfrak{g}$$ is nilpotent iff for any $$d\ge 1$$ it has no subalgebra of the form $$K^d\rtimes_u K$$ with $$u$$ nonzero irreducible operator.

For the real field, up to conjugation and nonzero scalar multiplication, the only possibilities for $$u$$, namely $$\begin{pmatrix}1\end{pmatrix}$$, $$\begin{pmatrix}t & -1\\1 & t\end{pmatrix}$$ for $$t\ge 0$$. So this yields

Proposition. Let $$\mathfrak{g}$$ be a finite-dimensional Lie algebra over $$\mathbf{R}$$. Then $$\mathfrak{g}$$ is nilpotent if and only if it has no subalgebra isomorphic to any of the following: the 2-dimensional non-abelian Lie algebra, and for any $$t\ge 0$$, the 3-dimensional Lie algebra with basis $$(h,x,y)$$ and brackets $$[h,x]=tx+y,\quad[h,y]=-x+ty,\quad [x,y]=0.$$

Finally, let me mention that the criterion is also false, over $$\mathbf{C}$$, in infinite dimension (finite dimension is too often implicit). Indeed the (finitely generated) Lie algebra with basis $$(x_n)_{n\ge 1}$$ and only nonzero brackets $$[x_1,x_i]=x_{i+1}$$ is residually nilpotent and hence its finite-dimensional subalgebras are nilpotent (actually they're even abelian, exercise). But it's clearly not nilpotent since $$\mathrm{ad}(x_1)$$ is not nilpotent. Non-nilpotent, locally nilpotent Lie algebras (e.g., a direct sum of nilpotent Lie algebras of unbounded nilpotency class) are also counterexamples.