The question is in the title. Let's recall the definitions. A Fuchsian group $G$ is a discrete subgroup of $\operatorname{PSL}(2,\mathbb R)=\operatorname{SL}(2,\mathbb R)/\{\pm 1\}$. This acts on the upper half plane $\mathbb C^+=\{ z:\operatorname{Im}z>0\}$ by linear fractional transformations: $$ \begin{pmatrix}a & b\\ c& d\end{pmatrix} \cdot z = \frac{az+b}{cz+d} $$ The limit set $L$ can be defined as the collection of limits of the form $x=\lim A_n\cdot z\in\overline{\mathbb C^+}$, with distinct group elements $A_n\in G$ (and not necessarily distinct points $A_n\cdot z$, so for example a point with infinite stabilizer is in $L$).

Some basic observations about $L$ are: $L\subseteq\mathbb R^{\infty}$, $L$ is closed and invariant, and $L=\emptyset$, $L=\{ x\}$, $L=\{ x,y\}$, $L=\mathbb R^{\infty}$, or $L$ is a nowhere dense set with no isolated points.

My question is about the last case; in all other cases, it is easy to see that all possibilities occur. So, given such a $C\subseteq\mathbb R^{\infty}$, will there be a Fuchsian group $G$ with $L_G=C$?

I suspect this will be well known to experts; in this case, I would also appreciate pointers to the literature.

Edit: Moishe Kohan points out in the comments that the answer to the question as asked is no, due to "well known" (to those who know them well) extra restrictions on $L$, and suggests that there may be no good answer to the question of what sets $C$ exactly occur as limit sets.

Feel free of course to answer anyway (in the answer box) if you have something interesting to say.

Is every Cantor set $C\subseteq\mathbb R^{\infty}$ the limit set of a Fuchsian group?

Let's recall the definitions. A Fuchsian group $G$ is a discrete subgroup of $\operatorname{PSL}(2,\mathbb R)=\operatorname{SL}(2,\mathbb R)/\{\pm 1\}$. This acts on the upper half plane $\mathbb C^+=\{ z:\operatorname{Im}z>0\}$ by linear fractional transformations: $$ \begin{pmatrix}a & b\\ c& d\end{pmatrix} \cdot z = \frac{az+b}{cz+d} $$ The limit set $L$ can be defined as the collection of limits of the form $x=\lim A_n\cdot z\in\overline{\mathbb C^+}$, with distinct group elements $A_n\in G$ (and not necessarily distinct points $A_n\cdot z$, so for example a point with infinite stabilizer is in $L$).

Some basic observations about $L$ are: $L\subseteq\mathbb R^{\infty}$, $L$ is closed and invariant, and $L=\emptyset$, $L=\{ x\}$, $L=\{ x,y\}$, $L=\mathbb R^{\infty}$, or $L$ is a nowhere dense set with no isolated points.

My question is about the last case; in all other cases, it is easy to see that all possibilities occur. So, given such a $C\subseteq\mathbb R^{\infty}$, will there be a Fuchsian group $G$ with $L_G=C$?

I suspect this will be well known to experts; in this case, I would also appreciate pointers to the literature.

Edit: Moishe Kohan points out in the comments that the answer to the question as asked is no, due to "well known" (to those who know them well) extra restrictions on $L$, and suggests that there may be no good answer to the question of what sets $C$ exactly occur as limit sets.

Feel free of course to answer anyway (in the answer box) if you have something interesting to say.

The question is in the title. Let's recall the definitions. A Fuchsian group $G$ is a discrete subgroup of $\operatorname{PSL}(2,\mathbb R)=\operatorname{SL}(2,\mathbb R)/\{\pm 1\}$. This acts on the upper half plane $\mathbb C^+=\{ z:\operatorname{Im}z>0\}$ by linear fractional transformations: $$ \begin{pmatrix}a & b\\ c& d\end{pmatrix} \cdot z = \frac{az+b}{cz+d} $$ The limit set $L$ can be defined as the collection of limits of the form $x=\lim A_n\cdot z\in\overline{\mathbb C^+}$, with distinct group elements $A_n\in G$ (and not necessarily distinct points $A_n\cdot z$, so for example a point with infinite stabilizer is in $L$).

Some basic observations about $L$ are: $L\subseteq\mathbb R^{\infty}$, $L$ is closed and invariant, and $L=\emptyset$, $L=\{ x\}$, $L=\{ x,y\}$, $L=\mathbb R^{\infty}$, or $L$ is a nowhere dense set with no isolated points.

My question is about the last case; in all other cases, it is easy to see that all possibilities occur. So, given such a $C\subseteq\mathbb R^{\infty}$, will there be a Fuchsian group $G$ with $L_G=C$?

I suspect this will be well known to experts; in this case, I would also appreciate pointers to the literature.

The question is in the title. Let's recall the definitions. A Fuchsian group $G$ is a discrete subgroup of $\operatorname{PSL}(2,\mathbb R)=\operatorname{SL}(2,\mathbb R)/\{\pm 1\}$. This acts on the upper half plane $\mathbb C^+=\{ z:\operatorname{Im}z>0\}$ by linear fractional transformations: $$ \begin{pmatrix}a & b\\ c& d\end{pmatrix} \cdot z = \frac{az+b}{cz+d} $$ The limit set $L$ can be defined as the collection of limits of the form $x=\lim A_n\cdot z\in\overline{\mathbb C^+}$, with distinct group elements $A_n\in G$ (and not necessarily distinct points $A_n\cdot z$, so for example a point with infinite stabilizer is in $L$).

Some basic observations about $L$ are: $L\subseteq\mathbb R^{\infty}$, $L$ is closed and invariant, and $L=\emptyset$, $L=\{ x\}$, $L=\{ x,y\}$, $L=\mathbb R^{\infty}$, or $L$ is a nowhere dense set with no isolated points.

My question is about the last case; in all other cases, it is easy to see that all possibilities occur. So, given such a $C\subseteq\mathbb R^{\infty}$, will there be a Fuchsian group $G$ with $L_G=C$?

I suspect this will be well known to experts; in this case, I would also appreciate pointers to the literature.

Edit: Moishe Kohan points out in the comments that the answer to the question as asked is no, due to "well known" (to those who know them well) extra restrictions on $L$, and suggests that there may be no good answer to the question of what sets $C$ exactly occur as limit sets.

Feel free of course to answer anyway (in the answer box) if you have something interesting to say.