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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 $\alpha = -\mathrm{id}$.

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What Heisenberg group do you mean in your second remark? –  Mariano Suárez-Alvarez Sep 6 '11 at 4:39
I agree with Mariano: in Remark 2, there must be a confusion between the discrete and the continuous Heisenberg group. –  Alain Valette Sep 6 '11 at 7:12
Remark 2 has been redacted. –  Qayum Khan Sep 7 '11 at 3:32
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! –  Mariano Suárez-Alvarez Sep 7 '11 at 3:33
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). –  Alain Valette Sep 7 '11 at 9:00
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1 Answer

up vote 4 down vote accepted

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).

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Excellent, and thanks for pointing out my error about eigenvalues! –  Qayum Khan Sep 6 '11 at 15:17
There is a nice paper of Higman dealing with such things. –  Mariano Suárez-Alvarez Sep 6 '11 at 15:22
[Higman, Graham. Groups and rings having automorphisms without non-trivial fixed elements. J. London Math. Soc. 32 (1957), 321--334. MR0089204] –  Mariano Suárez-Alvarez Sep 6 '11 at 16:20
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