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?

  • $\begingroup$ 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. $\endgroup$ – Torsten Schoeneberg May 25 '19 at 5:29
  • $\begingroup$ But should it contain a diagonal matrix? $\endgroup$ – Boris Bilich May 25 '19 at 10:54
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    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ My question is incorrect, because zentrum is always a subset of the preimage. I’ll change it. $\endgroup$ – Boris Bilich May 24 '19 at 12:01
  • $\begingroup$ 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. $\endgroup$ – Boris Bilich May 24 '19 at 12:05
  • $\begingroup$ @BorisBilich No, this is not true. The adjoint operators are all nilpotent, but still may have diagonal elements different from zero. $\endgroup$ – Dietrich Burde May 24 '19 at 18:43
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    $\begingroup$ So there is no contradiction with my statement. The algebra is not nilpotent and $\operatorname{ad}^{-1}(\mathfrak{d})$ is not center. $\endgroup$ – Boris Bilich May 24 '19 at 19:06
  • $\begingroup$ Engel’s theorem states, that all operators of adjoint representation are nilpotent and the only nilpotent diagonal matrix is zero. $\endgroup$ – Boris Bilich May 24 '19 at 19:08

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