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Jul 10, 2020 at 12:26 comment added Robert Bryant @BS.: Actually, I just realized (silly me) that an isometric immersion of $Y(3)$ is ruled out by Hilbert's Theorem. The point is that the simply-connected cover of $Y(3)$ is the entire Poincaré disk, and Hilbert's Theorem (1901) is that there is no isometric immersion of the entire Poincaré disk into $\mathbb{E}^3$. (Hilbert's Theorem didn't apply to $X(1)$ because of the two conical points. In particular, the composition $\mathbb{H}\to X(1)\to \mathbb{E}^3$ is not an immersion.)
Jul 9, 2020 at 21:20 comment added BS. That would be a great example !
Jul 9, 2020 at 18:31 comment added BS. What about $Y(3)$ ? Conformally, it is a regular tetrahedron without its vertices, quotient of the upper half plane by the torsion-free group $\Gamma(3)$, kernel of the surjective morphism $PSL_2(\mathbb{Z})\to PSL_2(\mathbb{Z}/3)\simeq A_4$.
Jul 5, 2020 at 0:43 history edited Robert Bryant CC BY-SA 4.0
Fixed some typos and run-on sentences. Hopefully improved the readability of the proof.
Jul 4, 2020 at 9:42 vote accept Alex Gavrilov
Jul 3, 2020 at 14:22 comment added Robert Bryant @AlexGavrilov: Are you asking about "For an isometric immersion of a compact surface with negative Gauss curvature into $\mathbb{E}^3$, the image lies in the convex hull of the image of the boundary"? This is a common exercise: If $f:M\to\mathbb{E}^3$ is an isometric immersion and $h:\mathbb{E}^3\to\mathbb{R}$ is a linear function such that $h\circ f$ is not positive on $\partial M$, then $h\circ f$ is not positive on $M$ since, otherwise, if $h\circ f$ has a positive maximum at $p\in M$, then $f(M)$ will lie on one side of the image tangent plane at $f(p)$, which is impossible since $K<0$.
Jul 3, 2020 at 12:25 comment added Alex Gavrilov About the convex hull, could you give a reference?
Jul 2, 2020 at 21:45 history edited Robert Bryant CC BY-SA 4.0
Fixed some typos
Jul 2, 2020 at 21:26 history answered Robert Bryant CC BY-SA 4.0