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Given a subring $A$ of $\mathbb{R}$, we can consider the set $\mathrm{PSL}_{2}(A)$ of elements in $\mathrm{PSL}_{2}(\mathbb{R})$ with entries in $A$ and the determinant of associated matrix is equal to one.

We define a hyperbolic elements from $\mathrm{PSL}_{2}(A)$ as elements which the trace of the associated matrix is greater than two, in absolute value.

Let us denote the set of all fixed points of all hyperbolic elements from $\mathrm{PSL}_{2}(A)$ by $\mathcal{P}_{A}$. Notice that $\mathcal{P}_{A}\subseteq \mathbb{R}\cup\{\infty\}$. If $A$ is a proper subring of $\mathbb{R}$, does it exist some result about $\mathcal{P}_{A}$ being a group?

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  • $\begingroup$ When you construct $\mathcal P_A$ as a set of fixed points, on what is $\mathrm{PSL}_2(A)$ acting? You say that $\mathcal P_A \subseteq \mathbb R$, but I don't know how to make $\mathrm{PSL}_2(A)$ act on $\mathbb R$. $\endgroup$
    – LSpice
    Commented Dec 13, 2018 at 2:42
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    $\begingroup$ @LSpice I think the action is defined by the isomorphism of $PSL_2(R)$ to fractional linear transformations acting on $R \cup \{\infty\}$. $\endgroup$
    – Neil Hoffman
    Commented Dec 13, 2018 at 5:44
  • $\begingroup$ If $A$ has infinite unit group (i.e. $A$ is not $\mathbb Z$) then $\infty$ is a fixed point of an hyperbolic element of $\mathrm{PSL}_2(A)$ so $\mathcal P_A$ is not contained in $\mathbb R$. On the other hand for $A =\mathbb Z$ zero is not in $\mathcal P_A$ (as the group is discrete and zero is a parabolic fixed point). $\endgroup$
    – Jean Raimbault
    Commented Dec 13, 2018 at 8:54
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    $\begingroup$ How could $\mathcal{P}_A$ be a group if it contains $\infty$? $\endgroup$
    – abx
    Commented Dec 13, 2018 at 10:26
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    $\begingroup$ As noted above this is the case only if $A = \mathbb Z$, and then $\mathcal P_A$ is not a subring (as noted in the deleted comment, $\sqrt 2, \sqrt 3 \in \mathcal P_A$ but not $\sqrt 2 + \sqrt 3$). So the answer to your question as asked is : never. It would help if you'd provide some context to explain what you are looking for exactly. $\endgroup$
    – Jean Raimbault
    Commented Dec 13, 2018 at 11:03

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