# Geometry and String Theory The Dilogarithm Function

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

$D(z) \coloneqq \operatorname{Im} \operatorname{Li}_2(z) + \arg(1-z)\log|z|$

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.

$0$

In hyperbolic space, an ideal tetrahedron, with vertices at $0,1,\infty,z$, has volume $D(z)$. ($z=\tfrac{1+\sqrt{-3}}{2}$ is the regular tetrahedron; more generally, $z$ is the cross ratio of the 4 vertices , which is invariant under $PSL_2(\mathbb{C})=\operatorname{Isom}(\mathbb{H})$) The 5-term recursion relation comes from taking 5 points in $\mathbb{P}^1(\mathbb{C})$ 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.

$C$

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 $\pi/2$.

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,

$\begin{gathered} L(\chi,2) = \sum_{n\geq 1} \frac{\chi(n)}{n^2}\\ \chi\colon {(\mathbb{Z}/N\mathbb{Z})}^\times \to \mathbb{C}^\times \qquad\text{Dirichlet character} \end{gathered}$