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The Bloch-Wigner function $D(z)$ gives the volume of an ideal tetrahedron in the hyperbolic space $\mathbb{H}^3$. Here $z$ is the cross-ratio $(z_1,z_2,z_3,z_4)$ parametrizing the tetrahedron in $\mathbb{C}P^1$.
Put $\tilde D(z_1,z_2,z_3,z_4) = D(z)$.

The five-term relation for the dilogarithm could be interpreted as the fact that the signed sum of some volumes of tetrahedra vanishes:

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

Here the $z_i$ are 5 points in $\mathbb{C}P^1$, 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 what space this simplex exists (hyperbolic space ?) ?

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

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    $\begingroup$ 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. $\endgroup$
    – François Brunault
    Commented Oct 6, 2012 at 18:09
  • $\begingroup$ The $z_i$ are in $CP1$, the boundary of $H3$. Thanks for your answer $\endgroup$
    – Trimok
    Commented Oct 6, 2012 at 18:14
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    $\begingroup$ 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. $\endgroup$
    – François Brunault
    Commented Oct 6, 2012 at 18:22

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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.

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

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    $\begingroup$ 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. $\endgroup$
    – Trimok
    Commented Nov 29, 2012 at 8:45

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