Define
\operatorname{Li}_2(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^2},\qquad |z|\lt 1More generally, the polylogarithm
\operatorname{Li}_m(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^m},\qquad |z|\lt 1Note 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 to is in
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 )
\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}D(z) \coloneqq \operatorname{Im} \operatorname{Li}_2(z) + \arg(1-z)\log|z|is real-analytic in and continuous in .
\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 .
\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 with a maximum at : .
Define recursively
z_{n+1}z_{n-1} = 1-z_nthen . If we call , , 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 is .)
The 5-term recursion relation is
\sum_{j=0}^4 D(z_j)=0This can be explained geometrically.
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\end{svg}\includegraphics[width=602]{tetrahedron}In hyperbolic space, an ideal tetrahedron, with vertices at , has volume . ( is the regular tetrahedron; more generally, is the cross ratio of the 4 vertices , which is invariant under ) The 5-term recursion relation comes from taking 5 points in and constructing five tetrahedra by taking the points 4 at a time
0 = \sum_{j=0}^4 {(-1)}^{j}\operatorname{Vol}((w_0,\dots,\hat{w}_j,\dots,w_4))The cancellation is the 3-2 Pachner move.
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\end{svg}\includegraphics[width=180]{pentagon}Napier: Mirifici logorithorum canonis descriptio
Rule of circular parts
Spherical trigonometry (navigation mathematics): 6 quantities (3 angles + 3 lengths or, equivalently, 6 angles). Fix one to be .
Parametrize parts wiith cross ratios of 4 points taken out of 5. The sides of the pentagon are .
Any equation among the parts remains valid after a cyclic permutation around the pentagon. Denote the five triangles by , , with . 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_5is 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 .
Coxeter: 5-cycle
transmitted as mathematical gossip for a long time.
\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}