Skip to main content
1 of 3
Pengfei
  • 2.2k
  • 17
  • 31

Lie algebra admitting some hyperoblic automorphism is nilpotent

Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{R}$ and $\phi:\mathfrak{g}\to\mathfrak{g}$ be a Lie algebra automorphism.

Viewing $\mathfrak{g}$ as a linear space and $\phi$ a linear automorphism, we can say $\phi$ is hyperbolic if the eigenvalues of $\phi$ are disjoint from $\lbrace z\in\mathbb{C}:|z|=1\rbrace$.

Then Proposition 3.6 in Smale's paper (here) says that:

  • Suppose that $\phi:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism which is hyperbolic as a linear map. Then $\mathfrak{g}$ must be nilpotent.

He also mentioned the following result in (Exercise in Bourbaki with hints: Algebras de Lie, Ex. 21b, p. 124.):

  • Let $\mathfrak{g}$ be a finite dimensional Lie algebra having an automorphism $\phi$, no eigenvalue of which is a root of unity, then $\mathfrak{g}$ is nilpotent.

Do you have ideas how to prove these results?

Thanks!

Pengfei
  • 2.2k
  • 17
  • 31