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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 $L \subset \mathfrak{n}$ ?

I think that all the eigenvalues of $\alpha$ must be $-1$ ?

REMARK1: If $\mathfrak{n}$ is allowed to be abelian, then an example is $\alpha = -\mathrm{id}$.

REMARK 2: The discrete 3-dimensional Heisenberg group $L=\mathrm{Hei}$ didn't work for me; it can be shown $$ \mathrm{Aut}(\mathrm{Hei}) = \mathbb{Z}^2 \rtimes GL_2(\mathbb{Z}), \text{ where } GL_2(\mathbb{Z}) = (C_4 *_{C_2} C_6) \rtimes C_2.$$ Generators of these four cyclic subgroups of $GL_2(\mathbb{Z})$ are: $$ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}. $$
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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 $L \subset \mathfrak{n}$ ?

Necessarily,

I think that all the eigenvalues of $\alpha$ must be $-1$.-1$ ?

REMARK 1: If $\mathfrak{n}$ is allowed to be abelian, then an example is $\alpha = -\mathrm{id}$.

REMARK 2: I couldn't get the The discrete 3-dimensional Heisenberg group $\mathrm{Hei}$ to L=\mathrm{Hei}$ didn't work for me; it can be shown that $$ \mathrm{Aut}(\mathrm{Hei}) = \mathbb{Z}^2 \rtimes GL_2(\mathbb{Z}), \text{ where } GL_2(\mathbb{Z}) = (C_4 *_{C_2} C_6) \rtimes C_2.$$ Generators of these four cyclic subgroups of $GL_2(\mathbb{Z})$ are: $$ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}. $$
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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 $L \subset \mathfrak{n}$ ?

Necessarily, all the eigenvalues of $\alpha$ must be $-1$.

REMARK 1: If $\mathfrak{n}$ is allowed to be abelian, then an example is $\alpha = -\mathrm{id}$.

REMARK 2: I couldn't get the 3-dimensional Heisenberg group $\mathrm{Hei}$ to work; it can be shown that $$ \mathrm{Aut}(\mathrm{Hei}) = \mathbb{Z}^2 \rtimes GL_2(\mathbb{Z}), \text{ where } GL_2(\mathbb{Z}) = (C_4 *_{C_2} C_6) \rtimes C_2.$$ Generators of these four cyclic subgroups of $GL_2(\mathbb{Z})$ are: $$ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}. $$
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