About normalizers of infinite cyclic subgroups of Hilbert modular group 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
 A: This is not a complete answer but it's too long to fit in a comment, so here it goes. In case $g \in \mathrm{SL}_2(\mathcal O_k)$ is semisimple there are two possibilities : 


*

*As you observed, if $g$ fixes a "cusp" then the centraliser is the unit group of $\mathcal O_k$ modulo $\pm 1$, which is isomorphic to $\mathbb Z^{n-1}$. 

*In other cases, the centraliser can be described as a subgroup of units in a quadratic extension of $k$, and its rank is $n - m$ where $m$ is the number of real embeddings where $|\mathrm{tr}(g)| < 2$. 
This describes the normaliser up to finite index, you'd have to do a bit more to get the full description. A reference for the second case is https://arxiv.org/abs/1311.5375, section 3.3 (see also I. Efrat's paper quoted there, The Selberg trace formula for $\mathrm{PSL}_2(\mathbb R^n)$, Memoirs AMS. 65, 1987, http://www.ams.org/mathscinet-getitem?mr=874084). 
(edit : corrected ref. to preprint following Luc Guyot's comment, added reference and MR link for Efrat's paper)
