Consider $k$ a totally real finite extension of degree $n$ of $\mathbb{Q}$, i.e., all embeddings of $k$ in $\mathbb{C}$ have their image contained in the field of reals. Denote by $\mathcal{O}_k$ the ring of algebraic integers of $k$. Lets define $G = \text{PSL}_2(\mathcal{O}_k)$, this is what some people call the Hilbert modular group (although most people name $\text{SL}_2(\mathcal{O}_k)$ the Hilbert modular group).

Since $G$ acts on the hyperbolic plane $\mathbb{H}$ via Möbius transformations, we have three kinds of elements:

- Elliptic: Those who have exactly one fixed point in $\mathbb{H}$.
- Parabolic: Those who have exactly one fixed point in $\mathbb{R}\cup \{\infty\}$.
- Hyperbolic: Those who have exactly two fixed point in $\mathbb{R}\cup \{\infty\}$.

I am mainly interested in the normalizers of infinite cyclic subgroups of $G$, I mean $N_G(\langle g\rangle)$. Since elliptic elements are exactly those of finite order, then $g$ should be either parabolic or hyperbolic. Then we have two cases:

$g$ is parabolic. An easy calculation using matrices and the fact that the Hilbert modular group has a cusp (in the sense of Freitags book for instance), one shows that $N_G(\langle g\rangle)\cong \mathbb{Z}^n$. I think there is no problem here.

$g$ is hyperbolic. This is the case where my question arises. Using the fact that $N_G(\langle g\rangle)$ acts on the set of fixed points of $g$, one can prove that either $N_G(\langle g\rangle)\cong C_G(\langle g\rangle)\rtimes \mathbb{Z}/2$, where $C_G(\langle g\rangle)$ is the centralizer of $\langle g\rangle$, or $N_G(\langle g\rangle)\cong C_G(\langle g\rangle)$.

From now on I will suppose $g$ hyperbolic.

Example: If $g$ is represented by the matrix $ \left( \begin{array}{cc} a & 0 \\ 0 & a^{-1} \\ \end{array} \right) $ then it can be shown that $C_G(\langle g\rangle)\cong \mathbb{Z}^{n-1}$ using the Dirichlet unit theorem and a direct computation with matrices, The element in $G$ represented by $ \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right) $ lies in the normalizer $N_G(\langle g\rangle)$ and it conjugates $g$ to its inverse.

In general, I don't know how to deal with infinite cyclic subgroups generated by hyperbolic elements.

If $g$ fixes two points that belong to $k$ (i.e., points that are fixed by some parabolic elements), then $C_G(\langle g\rangle)\cong \mathbb{Z}^{n-1}$, and the only thing to prove is that there is an element of order two in the normalizer. This is equivalent to having a point in the unique geodesic fixed by $g$ with even isotropy group. Does anybody know if this is always the case?

It is even more mysterious to me what happens when the fixed points of $g$ are not fixed by parabolic elements. Does anybody know what happens in this case?

Thanks a lot

everyreal embedding. $\endgroup$ – YCor Feb 18 '17 at 0:591more comments