# Nilpotence criterion for solvable Lie algebras

Let $$\mathfrak{g}$$ be solvable Lie algebra. Lie’s theorem states, that adjoint representation is a homomorphism $$\operatorname{ad}:\mathfrak{g}\to \mathfrak{t}$$, where $$\mathfrak{t}$$ is an algebra of upper-triangular matrices. Let $$\mathfrak{d}\subset\mathfrak{t}$$ be a subalgebra of diagonal matrices.

Is it true, that $$\mathfrak{g}$$ is nilpotent iff $$\operatorname{ad}^{-1}(\mathfrak{d})=Z(\mathfrak{g})$$? In one direction it is exactly the Engel’s theorem, but I cannot find counter examples or proof for the other direction.

UPD I want to rephrase my question in an equivalent way. Is it true, that all solvable, but not nilpotent subalgebras of upper-triangular algebra contain nonzero diagonal elements?

• As to your "UPD": The upper triangular matrices with all zeroes on the diagonal form a nilpotent Lie algebra, and all subalgebras of nilpotent algebras are nilpotent. By contraposition, any non-nilpotent subalgebra of some upper-triangular matrices must contain at least one element with at least one non-zero entry on the diagonal. – Torsten Schoeneberg May 25 '19 at 5:29
• But should it contain a diagonal matrix? – Boris Bilich May 25 '19 at 10:54
• No. $\lbrace \pmatrix{a&a&b\\0&0&0\\0&0&0}: a,b \in \Bbb C \rbrace$. However, that is not an adjoint representation; your update is not equivalent to the original question. – Torsten Schoeneberg May 25 '19 at 18:23

Consider the non commutative algebra of dimension $$2$$ generated by $$x,y$$ defined by $$[x,y]=x$$.
The matrix of $$ad_x$$ in the basis $$(x,y)$$ is $$\pmatrix{0&1\cr 0&0}$$ the matrix of $$ad_y$$ is $$\pmatrix{-1&0\cr 0&0}$$. $$ad_y$$ is diagonal, but this Lie algebra is solvable and not nilpotent and $$y$$ is not in the center.
• So you’ve just proved the statement: “If $\mathfrak{g}$ is nilpotent then it’s adjoint representation has no diagonal elements, except zero”. And I am interested if it is true in the opposite direction. – Boris Bilich May 24 '19 at 12:05
• So there is no contradiction with my statement. The algebra is not nilpotent and $\operatorname{ad}^{-1}(\mathfrak{d})$ is not center. – Boris Bilich May 24 '19 at 19:06