Is there an isometric embedding of the modular surface $X(1)=PSL(2,\mathbb{Z})\backslash \,\mathbb{H}$ into the Euclidean 3-space? For all I know this may be an open problem but I am also curious if anyone studied
it numerically or maybe even made a physical model of it. (Which would probably
look a little scary, with two horns (conical points) and a tail (cusp).)


P.S. To make it clear, I mean a $C^\infty$ embedding outside the conical points.



Isometric embedding of the modular surface
There is no isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of an argument:


First, let $\mathbb{H}\subset\mathbb{C}$ be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ on $\mathbb{H}$ is then defined by the inequalities $|z|\ge 1$ and $|x|\le \tfrac12$. Then one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z\_2 \equiv i$ (of order $2$) and $z\_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z\_1 \equiv +i\,\infty$.


Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z\_1,z\_2,z\_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ distinct from the three $z\_i$. There will be a hyperbolic disk $D\_r(z)$ of some radius $r>0$ about $z$ that does not contain any of the $z\_i$. Because the Gauss curvature of $X(1)$ is $K=-1$, the convex hull of the $f$-image of $D\_r(z)$ will contain an Euclidean ball of some positive radius $R>0$.


Meanwhile, let $\epsilon>0$ be a very small positive number and consider the subset $M\_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z\_2)\ge \epsilon$ and $d(z,z\_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M\_\epsilon$ is a compact smooth surface whose boundary $\partial M\_\epsilon$ consists of three disjoint circles:


1. $C\_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;
2. $C\_2$ (the points where $d(z,z\_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and
3. $C\_3$ (the points where $d(z,z\_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$.


In particular, when $\epsilon>0$ is taken to be sufficiently small, each of these curves has total length less than $4\epsilon$.


Thus, each $f(C\_i)$ must therefore lie in an Euclidean ball $B\_i$ of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M\_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$. (Just take a plane that passes through the centers of the three balls $B\_i$ of radius $4\epsilon$ and look at the $4\epsilon$-neighborhood of that plane.)


Now, because the Gauss curvature of $M\_\epsilon$ is strictly negative, the $f$-image of $M\_\epsilon$ must lie within the convex hull of the image of $\partial M\_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$.


However, if we take $\epsilon<R/4$ sufficiently small, the disk $D\_r(z)$ will lie entirely within $M\_\epsilon$ and hence the $f$-image of $D\_r(z)$, whose convex hull contains an Euclidean ball of radius $R$, must lie in an infinite slab of thickness at most $4\epsilon<R$, which is obviously impossible.


Thus, such an $f$ cannot exist.


