3 The ideal tetrahedron: an explicit volume integral

3.1 Lambda lengths and the ambient cross-ratio

Take the four labeled ideal vertices to be $0,1,\infty,z$, where $z=x+iy\in\mathbb C\setminus\{0,1\}$. For $y\ne0$ their convex hull is a genuine ideal tetrahedron. The two half-planes describe its two orientation types.

For a decorated ideal triangulation, set $\lambda_e=e^{\delta_e/2}$, where $\delta_e$ is the signed distance between the chosen horocycles along the edge. An overall fixed multiplier in the definition of $\lambda$ does not change its logarithmic differential. With cyclically ordered edge lengths $a,b,c$ on each oriented triangle, Penner’s formula in convention (1.1) is $$\omega_{\mathrm{WP}}=-2\sum_{\triangle} \bigl(\mathop{}\!\mathrm d\log a\wedge\mathop{}\!\mathrm d\log b+ \mathop{}\!\mathrm d\log b\wedge\mathop{}\!\mathrm d\log c+ \mathop{}\!\mathrm d\log c\wedge\mathop{}\!\mathrm d\log a\bigr).$$(3.1) This is a form on decorated Teichmüller space that descends after forgetting the horocycles [85]; its magnitude agrees with the cuspidal limit of Mondello’s arc-coordinate formula [79].

The four decorations can be chosen so that $$\lambda_{0\infty}=\lambda_{1\infty}=\lambda_{z\infty} =\lambda_{01}=1, \qquad \lambda_{0z}=r=|z|,\quad\lambda_{1z}=s=|1-z|.$$(3.2) For example, use equal Euclidean diameters for the three finite horospheres in the upper-half-space model and adjust their common scale together with the horosphere at infinity. The usual formula for the signed distance then gives (3.2). One may take diameter $1$ for each finite horosphere and height $1$ at infinity. The finite-edge formula is then $\lambda_{ij}=|z_i-z_j|$, and each edge to infinity has lambda length $1$. Overlapping horospheres cause no problem: the distances are signed, and smaller embedded decorations can first be chosen and then rescaled.

Every face containing $\infty$ has at most one varying lambda length, so its contribution in (3.1) vanishes in this section of the decoration bundle. The remaining face has lengths $1,r,s$. Consequently the positive WP measure is $$\boxed{ \mathop{}\!\mathrm dV_{\mathrm{WP}}=2\left|\mathop{}\!\mathrm d\log r\wedge\mathop{}\!\mathrm d\log s\right| =\frac{2|y|}{|z|^2|1-z|^2}\,\mathop{}\!\mathrm dx\,\mathop{}\!\mathrm dy.}$$(3.3) The Jacobian calculation is elementary: $$\mathop{}\!\mathrm d\log r=\frac{x\,\mathop{}\!\mathrm dx+y\,\mathop{}\!\mathrm dy}{r^2},\qquad \mathop{}\!\mathrm d\log s=\frac{(x-1)\,\mathop{}\!\mathrm dx+y\,\mathop{}\!\mathrm dy}{s^2}.$$ Their wedge is $y\,\mathop{}\!\mathrm dx\wedge\mathop{}\!\mathrm dy/(r^2s^2)$.

3.2 Integration in angle coordinates

For $y>0$, let $A,B,C$ be the interior angles of the Euclidean triangle with vertices $0,1,z$. They satisfy $$A>0,\quad B>0,\quad C>0,\quad A+B+C=\pi.$$ They are also the three interior dihedral angles of the ideal tetrahedron, each repeated on the opposite edge. Direct differentiation gives $$\left|\mathop{}\!\mathrm dA\wedge\mathop{}\!\mathrm dB\right| =\frac{y}{r^2s^2}\,\mathop{}\!\mathrm dx\,\mathop{}\!\mathrm dy, \qquad\text{hence}\qquad \mathop{}\!\mathrm dV_{\mathrm{WP}}=2\,\mathop{}\!\mathrm dA\,\mathop{}\!\mathrm dB.$$ One orientation sheet therefore contributes $$\int_{\substack{A>0,\ B>0\\A+B<\pi}}2\,\mathop{}\!\mathrm dA\,\mathop{}\!\mathrm dB=\pi^2.$$(3.4) Including the other sheet gives the promised direct calculation: $$\boxed{V_{0,4}=2\pi^2.}$$(3.5) The real configurations form the flat locus and contribute zero two-dimensional WP measure.

There is also a useful check on (2.5). In a tetrahedral angle cell, the four perimeter equations force opposite exterior angles to be equal, with three values $a,b,c$ satisfying $a+b+c=2\pi$. To see the disintegration factor explicitly, order the edges as $(01,02,03,12,13,23)$, keep $a=\theta_{01}$ and $b=\theta_{02}$, and eliminate the last four variables. The corresponding perimeter minor is $$\begin{pmatrix} 1&0&0&0\\0&1&1&0\\0&1&0&1\\1&0&1&1 \end{pmatrix},\qquad \det=-2.$$ It gives $(\theta_{03},\theta_{12},\theta_{13},\theta_{23}) =(2\pi-a-b,2\pi-a-b,b,a)$ and the measure $\frac12|\mathop{}\!\mathrm da\,\mathop{}\!\mathrm db|$. Up to permuting the three angle names, $a=\pi-A$ and $b=\pi-B$, so this is $\frac12\mathop{}\!\mathrm dA\,\mathop{}\!\mathrm dB$. There are two oriented cells, each with trivial label-preserving automorphism group. The factor $2^{4-2}=4$ in (2.5) converts their total map volume $\pi^2/2$ to $2\pi^2$.

3.3 What happens at the punctures of the parameter plane?

Finite volume requires integrability, not pointwise decay of a coordinate density. Near $z=0$, write $z=\rho e^{i\vartheta}$. Then $$\frac{2|y|}{|z|^2|1-z|^2} =\frac{2|\sin\vartheta|}{\rho}\,(1+O(\rho)), \qquad \mathop{}\!\mathrm dx\,\mathop{}\!\mathrm dy=\rho\,\mathop{}\!\mathrm d\rho\,\mathop{}\!\mathrm d\vartheta.$$ The singularity is integrable. The same is true at $1$. At infinity the density is $O(|z|^{-3})$, again integrable against Euclidean area. Thus the requested decay is best formulated as a condition on the measure, not on its density without its coordinate area element.

Nor do two ordinary closed angle triangles describe the full topology of the compactification without additional data. As $y\downarrow0$ with $0<x<1$, two of the triangle angles tend to zero and the third to $\pi$, independently of $x$. A whole family of flat quadrilaterals approaches one angle-simplex vertex. Retaining that modulus requires a refinement of the boundary, such as a real blow-up. This has no effect on the integral over the open cells.