1

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?

Boris Bilich
  • 669
  • 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 05:29
  • But should it contain a diagonal matrix? – Boris Bilich May 25 '19 at 10:54
  • 1
    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

1 Answers1

-1

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.

Tsemo Aristide
  • 87,475
  • My question is incorrect, because zentrum is always a subset of the preimage. I’ll change it. – Boris Bilich May 24 '19 at 12:01
  • 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
  • @BorisBilich No, this is not true. The adjoint operators are all nilpotent, but still may have diagonal elements different from zero. – Dietrich Burde May 24 '19 at 18:43
  • 1
    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
  • Engel’s theorem states, that all operators of adjoint representation are nilpotent and the only nilpotent diagonal matrix is zero. – Boris Bilich May 24 '19 at 19:08