Isometric embedding of the modular surface 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.
 A: 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:

*

*$C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;


*$C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and


*$C_3$ (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.
A: Correction: The following idea doesn't work as stated, because (as Robert points out) the cusp and cones allow points where mean curvature is not defined.
If you could isometrically and smoothly embed the modular surface, you would smoothly and locally isometrically immerse the hyperbolic plane, which is impossible by a theorem of Hilbert.
