Dilogarithm, tetrahedrons, and hyperbolic space - MathOverflow

Dilogarithm, tetrahedrons, and hyperbolic space

2012-10-06T17:45:11Z

The Bloch-Wigner function $D(z)$, gives the volume of a ideal tetrahedron in hyperbolic space $\mathbb H3$, where z is the cross-ratio (z1,z2,z3,z4) parametrizing the tetrahedron in $\mathbb CP1$.
$D(z)= \tilde D(z1,z2,z3,z4)$

Relating to this, the five-term relation for the dilogarithm, could be interpreted, as the fact that the signed sum of some volume of tetrahedrons, is null :

$$\sum^4_{i=0} (-1)^i \tilde D(z_0, ...., \hat z_i, ... z_4) = 0$$

Here the $z_i$ are 5 points in $\mathbb CP1$, and the notation $\hat z_i$ means that we don't take the vertex $z_i$ in account. The above equation looks like some function of a boundary of some 5-simplex.

But what is this 5-simplex (which, I think, corresponds to a 4-volume), and in which space this simplex exists (hyperbolic space ?) ?

Reference (Zagier) : http://maths.dur.ac.uk/~dma0hg/dilog.pdf (Pages 10 - 11)

Answer by Igor Rivin for Dilogarithm, tetrahedrons, and hyperbolic space

2012-10-06T18:20:56Z

The five term relation comes from the fact that the sum of the volumes of tetrahedra $ABCD$ and $ABCE$ equals the sum of the volumes of the three tetrahedra $ABDE, ACDE, BCDE.$ One can think of $ABCDE$ as a degenerate four-dimensional simplex.

Answer by unknown (google) for Dilogarithm, tetrahedrons, and hyperbolic space

2012-11-28T06:12:30Z

In fact this follows from Stokes' theorem. Consider the 4-simplex $\sigma$ with vertices ABCDE. Since the volume form $\omega$ is closed we have $$\int_{\partial\sigma}\omega=\int_{\sigma} d\omega=0.$$ But the integral of the volume form over $\partial\sigma$ is exactly the alternating sum $$\sum_{i=0}^4\left(-1\right)^i vol(\partial_i\sigma)=\sum_{i=0}^4\left(-1\right)^i \tilde{D}\left(z_0,\ldots,\hat{z}_i,\ldots,z_4\right).$$