Define $$ \operatorname{Li}_2(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^2},\qquad |z|\lt 1 $$ More generally, the polylogarithm $m=1,2,\dots$ $$ \operatorname{Li}_m(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^m},\qquad |z|\lt 1 $$ Note that $$ \operatorname{Li}_1(z) = -\log(1-z) $$ and $$ \frac{d}{d z} \operatorname{Li}_m(z) = \operatorname{Li}_{m-1}(z) $$ So we get an analytic continuation $$ \operatorname{Li}_2(z) = -\int_0^z \log(1-u) \frac{d u}{u} $$ where the path from $0$ to $z$ is in $\mathbb{C}\setminus [1,\infty)$ Functional equations: $$ \begin{gathered} \operatorname{Li}_1(1-x y) = \operatorname{Li}_1(1-x) + \operatorname{Li}_1(1-y)\\ \operatorname{Li}_2 = \text{5 terms (Spence 1809, Abel 1828, ...)} \end{gathered} $$ Monodromy (on $\operatorname{Li}_2(x),\log(x),1$) $$ \gamma_0=\begin{pmatrix}1&0&0\\0&1&2\pi i\\0&0&1\end{pmatrix}, \gamma_1=\begin{pmatrix}1&-2\pi i&0\\0&1&0\\0&0&1\end{pmatrix} $$ generate a Heisenberg group $$ \begin{pmatrix}1&\mathbb{Z}(1)&\mathbb{Z}(2)\\ 0&1&\mathbb{Z}(1)\\0&0&1\end{pmatrix} $$ ##Bloch-Wigner Dilogarithm ## {:#Bloch-Wigner} $$ D(z) \coloneqq \operatorname{Im} \operatorname{Li}_2(z) + \arg(1-z)\log|z| $$ is real-analytic in $\mathbb{C}\setminus\{0,1\}$ and continuous in $\mathbb{C}$. $$ \begin{gathered} D\left(e^{i\theta}\right) = \sum_{n\geq 1} \frac{\sin n\theta}{n^2}\\ D(\overline{z}) = - D(z) \end{gathered} $$ hence vanishes on $\mathbb{R}$. $$ \begin{split} D(z)&= D\left(1-z^{-1}\right)= D\left({(1-z)}^{-1}\right)\\ & - D\left(z^{-1}\right) = - D(1-z) = -D\left(-\frac{z}{1-z}\right) \end{split} $$ So we have a continuous real-vaued function on $\mathbb{P}^1(\mathbb{C})$ with a maximum at $z=(1+\sqrt{-3})/2$: $D(1+\sqrt{-3})/2)=1.0149\dots$. Define recursively $$ z_{n+1}z_{n-1} = 1-z_n $$ then $z_{n+5}=z_n$. If we call $z_0=x$, $z_1=y$, then we find $$ x,y,\frac{1-y}{x},\frac{x+y-1}{xy},\frac{1-x}{y} $$ (Laurent phenomenon). (Cremona transformation of order 5 on $\mathbb{P}^2(\mathbb{C})$ is $(x,y)\mapsto\left(y,\tfrac{1-y}{x}\right)$.) The 5-term recursion relation is $$ \sum_{j=0}^4 D(z_j)=0 $$ This can be explained geometrically. $$ \begin{svg} <svg width="803" height="302" xmlns="http://www.w3.org/2000/svg" xmlns:se="http://svg-edit.googlecode.com" se:nonce="94701"> <g> <title>Layer 1</title> <path id="svg_94701_5" d="m208.5,93.625l-208,208l594,0l208.140259,-209.14032" stroke="#000000" fill="#ffeeee"/> <path id="svg_94701_3" d="m323.25,11l0,207c18,-51 84,-54 99,0c10,-54.333344 33,-75.666656 51,-52l0,-152.857872" stroke-width="2" stroke="#000000" fill="#eeffff"/> <line id="svg_94701_6" y2="217.021844" x2="422.25" y1="11" x1="422.25" stroke-width="2" stroke="#000000" fill="none"/> <foreignObject height="18" width="14" font-size="16" id="svg_94701_7" y="218" x="316.25"> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> <mrow> <mn>0</mn> </mrow> <annotation encoding="application/x-tex">0</annotation> </semantics> </math> </foreignObject> <foreignObject id="svg_94701_8" height="18" width="14" font-size="16" y="222" x="415.25"> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> <mrow> <mn>1</mn> </mrow> <annotation encoding="application/x-tex">1</annotation> </semantics> </math> </foreignObject> <foreignObject id="svg_94701_14" height="20" width="14" font-size="16" y="171" x="467.25"> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> <mrow> <mi>z</mi> </mrow> <annotation encoding="application/x-tex">z</annotation> </semantics> </math> </foreignObject> <foreignObject id="svg_94701_20" height="18" width="14" font-size="16" y="0" x="304.25"> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"> <semantics> <mrow> <mn>∞</mn> </mrow> <annotation encoding="application/x-tex">\infty</annotation> </semantics> </math> </foreignObject> </g> </svg> \end{svg}\includegraphics[width=602]{tetrahedron} $$ In hyperbolic space, an ideal tetrahedron, with vertices at $0,1,\infty,z$, has volume $D(z)$. 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Fix one to be $\pi/2$. Parametrize parts wiith cross ratios of 4 points taken out of 5. The sides of the pentagon are $A,B,b',c,a'$. Any equation among the parts remains valid after a cyclic permutation around the pentagon. Denote the five triangles by $(a_i, B_i, c_i, A_i, b_i)$, $i=0,1,\dots,4$, with $(a_0, B_0, c_0, A_0, b_0)\equiv(a, B, c, A, b)$. Under a cyclic permutation $$ (a,B,c,A,A,b)\mapsto (a_1,B_1,c_1,A_1,b_1) = (A',b',a',B,c') $$ and $$ (a,B',c',A',b)\mapsto (A',b,a,B',c') $$ Triangle can be solved if any two parts are known $$ \sin a =\tan b \tan B' = \cos A' \cos c' $$ The surface $$ S: \left\{\begin{gathered}1-z_1=z_2z_0\\ 1-z_2=z_3z_1\\ \vdots\\ 1-z_0 = z_1z_4 \end{gathered}\right. \qquad \operatorname{Aut}(S)\simeq S_5 $$ is a del Pezzo surface of degree 5. $$ \begin{gathered} \sum_{j=0}^4 z_j = 3-s,\quad -s=\prod_{j=0}^4 z_j\qquad\text{Schöne Gleichung}\\ (1-x)(1-y)(1-x-y) -s xy =0 \end{gathered} $$ universal elliptic curve with a 5-torsion point $x_1(5)$. Coxeter: 5-cycle transmitted as mathematical gossip for a long time. ###Number Theory### $$ \operatorname{Li}_2(1)=\zeta(2) = \frac{\pi^2}{6}\qquad\text{(Euler 1768)} $$ More generally, $$ \begin{gathered} L(\chi,2) = \sum_{n\geq 1} \frac{\chi(n)}{n^2}\\ \chi\colon {(\mathbb{Z}/N\mathbb{Z})}^\times \to \mathbb{C}^\times \qquad\text{Dirichlet character} \end{gathered} $$