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:

  1. $g$ 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.

  2. $g$ 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

  • 1
    $\begingroup$ "elliptic elements are those of finite order": this is not true, or you're unclear at some point. You seem to consider "the action" on the hyperbolic plane, which suggests you've been fixing a real embedding. Then as soon as $k\neq\mathbf{Q}$, your embedding of the group into $PSL_2(\mathbf{R})$ has a dense image and hence has dense image in restriction to any torsion-free subgroup of finite index, and thus you get plenty of elliptic elements, since this is a condition of nonempty interior (in the real topology). $\endgroup$
    – YCor
    Feb 18, 2017 at 0:58
  • 1
    $\begingroup$ ... and actually (since $k$ it totally real), finite-order elements are those that are elliptic for every real embedding. $\endgroup$
    – YCor
    Feb 18, 2017 at 0:59
  • $\begingroup$ Yes, you are right. Then there exist more cases than those I were considering... $\endgroup$
    – Luis
    Feb 18, 2017 at 1:11
  • $\begingroup$ For parabolic elements the normaliser is actually $U \rtimes \mathbb Z^n$ where $U$ is the group of units of $\mathcal O_k$ (which is inifinite as soon as $k \not \mathbb Q$). $\endgroup$ Feb 19, 2017 at 8:31
  • $\begingroup$ It would be nice if you could rephrase your question using the terminology of Isaac Y. Efrat's paper, "The Selberg trace formula for $\text{PSL}_2(\mathbb{R})^n$" (1987). Efrat considers all real embeddings of $k$, i.e., $G \subset \text{PSL}_2(\mathbb{R})^n$ acts on $\mathbb{H}^n$. Thus you would have elliptic/parabolic/hyperbolic components, which addresses ycor's comment. As indicated by Jean Raimbault, it seems that Theorem 5.7 page 26 should answer your question $\endgroup$
    – Luc Guyot
    Feb 20, 2017 at 22:28

1 Answer 1


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)

  • $\begingroup$ I didn't find the Section 4.3 in your preprint. Looking at Isaac Y. Efrat's paper, "The Selberg trace formula for $\text{PSL}_2(\mathbb{R})^n$" (1987), it seems that Theorem 5.7 page 26 answers OP's question. $\endgroup$
    – Luc Guyot
    Feb 20, 2017 at 22:23
  • $\begingroup$ Yes, in Theorem 5.7 of Efrat's paper there is a complete description of the centralizer of a mixed element. The only task left would be to decide whether there is an order two element in the normalizer or not... I think. $\endgroup$
    – Luis
    Feb 21, 2017 at 22:35

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