1
$\begingroup$

Are there a non-abelian nilpotent Lie algebra $\mathfrak{n}$ over $\mathbb{R}$ and an automorphism $\alpha: \mathfrak{n} \to \mathfrak{n}$ such that:

  • $\alpha$ is periodic,
  • the fixed subspace of $\alpha$ is the origin, and
  • there is an $\alpha$-invariant lattice $L \subset \mathfrak{n}$ ?

REMARK: If $\mathfrak{n}$ is allowed to be abelian, then an example is $\alpha = -\mathrm{id}$.

$\endgroup$
6
  • $\begingroup$ What Heisenberg group do you mean in your second remark? $\endgroup$ Sep 6, 2011 at 4:39
  • $\begingroup$ I agree with Mariano: in Remark 2, there must be a confusion between the discrete and the continuous Heisenberg group. $\endgroup$ Sep 6, 2011 at 7:12
  • $\begingroup$ Remark 2 has been redacted. $\endgroup$
    – Qayum Khan
    Sep 7, 2011 at 3:32
  • $\begingroup$ Please, do not edit questions in such a way that you render parts of existing answers unconnected with them. Half of David's answer is concerned with what you deleted! $\endgroup$ Sep 7, 2011 at 3:33
  • $\begingroup$ Before edit of OP, there was a question about possibility of doing it with the real Heisenberg Lie algebra $Hei$. The answer is no. Indeed, if $\alpha$ is a periodic automorphism of $\Hei$, let $Ad(\alpha)\in GL_2(\mathbb{R})$ be the induced automorphism of $Hei/Z(Hei)$. If $\det(Ad(\alpha))=1$, then $\alpha$ fixes the center; if $\det(Ad(\alpha))=-1$, then $\alpha$ fixes a line in $Hei$. In both cases $\alpha$ has a non-trivial fixed subspace. David's answer shows that it is however possible with the COMPLEX Heisenberg Lie algebra (he could have taken $\omega =i$ and the Gaussian integers). $\endgroup$ Sep 7, 2011 at 9:00

1 Answer 1

5
$\begingroup$

There is no example with eigenvalues $-1$. More generally, suppose that $\mathfrak{g}$ is a Lie algebra, and $\alpha$ is an automorphism of order $2$ whose fixed subspace is trivial. Then I claim that $\mathfrak{g}$ is abelian.

Proof: Since $\alpha^2=\mathrm{Id}$ and $\alpha$ has no fixed points, we must have $\alpha = -\mathrm{Id}$. For any $u$ and $v$ in $\mathfrak{g}$, we have $[\alpha u, \alpha v] = \alpha [u, v]$, since $\alpha$ is an automorphism. But we just showed $\alpha = - \mathrm{Id}$, so this shows $[-u, -v] = - [u,v]$ and thus $[u,v]=0$. Since $u$ and $v$ were arbitrary, this shows that $\mathfrak{g}$ is abelian. QED

However, your bulleted conditions do not force $\alpha$ to have all eigenvalues $-1$. I will first construct an example over $\mathbb{C}$, then modify it to work over $\mathbb{R}$.

Let $\mathfrak{n}$ be the Lie algebra of strictly upper triangular $3 \times 3$ complex matrices. Let $\omega$ be a primitive cube root of unity and consider the automorphism $$\alpha : \begin{pmatrix} 0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix} \mapsto \begin{pmatrix} 0 & \omega x & \omega^2 y \\ 0 & 0 & \omega z \\ 0 & 0 & 0 \end{pmatrix}$$

This has order $3$, and has no nonzero fixed points. If you like, you can think of this as conjugation by the diagonal matrix $\mathrm{diag}(1, \omega, \omega^2)$.

To make an example over $\mathbb{R}$, just look at the same Lie algebra with the same automorphism and forget that it is a complex vector space, remembering only the real vector space structure. So it is now a $6$-dimensional real Lie algebra, with an automorphism of order $3$, and no fixed points.

Finally, I need to give an $\alpha$-stable lattice in $\mathfrak{n}$. Consider the lattice of matrices as above where each of $x$, $y$ and $z$ are in the ring $\mathbb{Z}[\omega]$ (sometimes called the Eisenstein integers).

$\endgroup$
3
  • $\begingroup$ Excellent, and thanks for pointing out my error about eigenvalues! $\endgroup$
    – Qayum Khan
    Sep 6, 2011 at 15:17
  • $\begingroup$ There is a nice paper of Higman dealing with such things. $\endgroup$ Sep 6, 2011 at 15:22
  • $\begingroup$ [Higman, Graham. Groups and rings having automorphisms without non-trivial fixed elements. J. London Math. Soc. 32 (1957), 321--334. MR0089204] $\endgroup$ Sep 6, 2011 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.