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Geometry and String Theory
The Dilogarithm Function

Define

Li 2(z)≔∑ n≥1 z n n 2,|z|<1 \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,…m=1,2,\dots

Li m(z)≔∑ n≥1 z n n m,|z|<1 \operatorname{Li}_m(z) \coloneqq \sum_{n\geq 1} \frac{z^n}{n^m},\qquad |z|\lt 1

Note that

Li 1(z)=−log(1−z) \operatorname{Li}_1(z) = -\log(1-z)

and

d d z Li m(z)=Li m−1(z) \frac{d}{d z} \operatorname{Li}_m(z) = \operatorname{Li}_{m-1}(z)

So we get an analytic continuation

Li 2(z)=−∫ 0 z log(1−u)d u u \operatorname{Li}_2(z) = -\int_0^z \log(1-u) \frac{d u}{u}

where the path from 0 0 to z z is in ℂ∖[1,∞)\mathbb{C}\setminus [1,\infty)

Functional equations:

Li 1(1−x y)=Li 1(1−x)+Li 1(1−y) Li 2=5 terms (Spence 1809, Abel 1828, ...) \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 Li 2(x),log(x),1\operatorname{Li}_2(x),\log(x),1)

γ 0=(1 0 0 0 1 2 π i 0 0 1),γ 1=(1 −2 π i 0 0 1 0 0 0 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

(1 ℤ(1) ℤ(2) 0 1 ℤ(1) 0 0 1) \begin{pmatrix}1&\mathbb{Z}(1)&\mathbb{Z}(2)\\ 0&1&\mathbb{Z}(1)\\0&0&1\end{pmatrix}

## Bloch-Wigner Dilogarithm

D(z)≔Im Li 2(z)+arg(1−z)log|z| D(z) \coloneqq \operatorname{Im} \operatorname{Li}_2(z) + \arg(1-z)\log|z|

is real-analytic in ℂ∖{0,1}\mathbb{C}\setminus\{0,1\} and continuous in ℂ\mathbb{C}.

D(e i θ)=∑ n≥1 sin n θ n 2 D(z¯)=−D(z) \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 ℝ\mathbb{R}.

D(z) =D(1−z −1)=D((1−z) −1) −D(z −1)=−D(1−z)=−D(−z 1−z) \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 ℙ 1(ℂ)\mathbb{P}^1(\mathbb{C}) with a maximum at z=(1+−3)/2 z=(1+\sqrt{-3})/2: D(1+−3)/2)=1.0149…D(1+\sqrt{-3})/2)=1.0149\dots.

Define recursively

z n+1 z n−1=1−z n z_{n+1}z_{n-1} = 1-z_n

then z n+5=z n z_{n+5}=z_n. If we call z 0=x z_0=x, z 1=y z_1=y, then we find

x,y,1−y x,x+y−1 xy,1−x y x,y,\frac{1-y}{x},\frac{x+y-1}{xy},\frac{1-x}{y}

(Laurent phenomenon). (Cremona transformation of order 5 on ℙ 2(ℂ)\mathbb{P}^2(\mathbb{C}) is (x,y)↦(y,1−y x)(x,y)\mapsto\left(y,\tfrac{1-y}{x}\right).)

The 5-term recursion relation is

∑ j=0 4 D(z j)=0 \sum_{j=0}^4 D(z_j)=0

This can be explained geometrically.

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

0=∑ j=0 4(−1) j Vol((w 0,…,w^ j,…,w 4)) 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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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 π/2\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′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)(a_i, B_i, c_i, A_i, b_i), i=0,1,…,4 i=0,1,\dots,4, with (a 0,B 0,c 0,A 0,b 0)≡(a,B,c,A,b)(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)↦(a 1,B 1,c 1,A 1,b 1)=(A′,b′,a′,B,c′) (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)↦(A′,b,a,B′,c′) (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′ \sin a =\tan b \tan B' = \cos A' \cos c'

The surface

S:{1−z 1=z 2 z 0 1−z 2=z 3 z 1 ⋮ 1−z 0=z 1 z 4 Aut(S)≃S 5 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.

∑ j=0 4 z j=3−s,−s=∏ j=0 4 z j Schöne Gleichung (1−x)(1−y)(1−x−y)−s xy=0 \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)x_1(5).

Coxeter: 5-cycle

transmitted as mathematical gossip for a long time.

### Number Theory

Li 2(1)=ζ(2)=π 2 6(Euler 1768) \operatorname{Li}_2(1)=\zeta(2) = \frac{\pi^2}{6}\qquad\text{(Euler 1768)}

More generally,

L(χ,2)=∑ n≥1 χ(n)n 2 χ:(ℤ/N ℤ) ×→ℂ ×Dirichlet character \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}
Revised on October 04, 2010 00:46:06 by Jacques Distler