# Surface of a Ideal Tetrahedron in Hyperbolic Space H3

The hyperbolic space $\mathbb{H}^3$, has a boundary $\mathbb{CP}^1$.

A ideal tetrahedron in $\mathbb{H}^3$, is a tetrahedron, where the four vertices are on the boundary $\mathbb{CP}^1$.

The four vertices of the tetrahedron may be parametrized by four complex numbers $z_1, z_2, z_3, z_4$.

What is the surface of this ideal tetrahedron, as function of $z_1, z_2, z_3, z_4$?

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Do you mean "What is the surface area of the boundary of the tetrahedron?" Or do you mean "What is the conformal class of the induced metric on the boundary?" The first is answered by Igor and the second by Ian. –  Sam Nead Oct 7 '12 at 11:54
This is the first one : What is the surface area of the boundary of the tetrahedron? And Ok, the answer was given by Igor –  Trimok Oct 8 '12 at 9:25

The answer is $4\pi.$
It is the Lobachevsky function of the dihedral angle, the Bloch dilogarithm if you use the cross ratio, but the surface of a tetrahedron is the union of four ideal triangles, all of which have area $\pi.$ –  Igor Rivin Oct 6 '12 at 16:59
The surface of the tetrahedron is a 4-punctured sphere, made of 4 ideal triangles. In fact, this is a complete hyperbolic metric, as may be seen by sending $z_1$ to $\infty$ (in fact, we may assume $z_1=\infty, z_2=0, z_3=1, z_4=z$, for $z$ the cross-ratio of the 4 points).
The hyperbolic metrics on the 4-punctured sphere may be parameterized by shearing coordinates for the triangulation. For the edge connecting $0$ and $\infty$, the shearing coordinate will be $\ln(|z|)$, since one triangle is scaled by a factor of $|z|$. Opposite edges will have the same shear coordinate because of the symmetries, and the other cross-ratios are $\frac{1}{1-z}$ and $1-1/z$, from which one obtains the other shearing coordinates. The fact that the product of the cross-ratios $|z \cdot \frac{1}{1-z} \cdot (1-1/z)|=1$ corresponds to the fact that the cusps will be complete.