Dilogarithm, tetrahedrons, and hyperbolic space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:47:58Z http://mathoverflow.net/feeds/question/109015 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109015/dilogarithm-tetrahedrons-and-hyperbolic-space Dilogarithm, tetrahedrons, and hyperbolic space Trimok 2012-10-06T17:45:11Z 2012-11-28T07:24:18Z <p>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 $\mathbb CP1$.<br> $D(z)= \tilde D(z1,z2,z3,z4)$</p> <p>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 :</p> <p>$$\sum^4_{i=0} (-1)^i \tilde D(z_0, ...., \hat z_i, ... z_4) = 0$$</p> <p>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.</p> <p>But what is this 5-simplex (which, I think, corresponds to a 4-volume), and in which space this simplex exists (hyperbolic space ?) ?</p> <p>Reference (Zagier) : <a href="http://maths.dur.ac.uk/~dma0hg/dilog.pdf" rel="nofollow">http://maths.dur.ac.uk/~dma0hg/dilog.pdf</a> (Pages 10 - 11)</p> http://mathoverflow.net/questions/109015/dilogarithm-tetrahedrons-and-hyperbolic-space/109016#109016 Answer by Igor Rivin for Dilogarithm, tetrahedrons, and hyperbolic space Igor Rivin 2012-10-06T18:20:56Z 2012-10-06T18:20:56Z <p>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.</p> http://mathoverflow.net/questions/109015/dilogarithm-tetrahedrons-and-hyperbolic-space/114734#114734 Answer by unknown (google) for Dilogarithm, tetrahedrons, and hyperbolic space unknown (google) 2012-11-28T06:12:30Z 2012-11-28T07:24:18Z <p>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 $$\sum_{i=0}^4\left(-1\right)^i vol(\partial_i\sigma)=\sum_{i=0}^4\left(-1\right)^i \tilde{D}\left(z_0,\ldots,\hat{z}_i,\ldots,z_4\right).$$</p>