Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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) : (Pages 10 - 11)

share|cite|improve this question
I guess you mean $z_i \in \mathbb{H}^3$. There is no need to create a $4$-dimensional space. Think about the simpler situation of a quadrilateral in the plane. This defines a $4$-simplex, and the alternated sum of areas of triangles is zero. –  François Brunault Oct 6 '12 at 18:09
The $z_i$ are in $CP1$, the boundary of $H3$. Thanks for your answer –  Trimok Oct 6 '12 at 18:14
You're right $z_i \in \partial \mathbb{H}^3$. Of course, the fact about volumes is true regardless whether the $z_i$'s are on the boundary or not. –  François Brunault Oct 6 '12 at 18:22

2 Answers 2

up vote 8 down vote accepted

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.

share|cite|improve this answer

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 $$\sum_{i=0}^4\left(-1\right)^i vol(\partial_i\sigma)=\sum_{i=0}^4\left(-1\right)^i \tilde{D}\left(z_0,\ldots,\hat{z}_i,\ldots,z_4\right).$$

share|cite|improve this answer
I agree, in fact here the non-trivial fact is that the five-term dilogarithm relation is linked to very specific 3-volumes in hyperbolic space (ABCD, ABCE, ABDE, ACDE, BCDE). So, yes, there is a 3-volume form which is closed. –  Trimok Nov 29 '12 at 8:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.