Geometry and String Theory
The Dilogarithm Function (Rev #4, changes)

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Define

Li 2(z) n1z nn 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) n1z nn 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(1z) \operatorname{Li}_1(z) = -\log(1-z)

and

ddzLi m(z)=Li m1(z) \frac{d}{d z} \operatorname{Li}_m(z) = \operatorname{Li}_{m-1}(z)

So we get an analytic continuation

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

where the path from 00 to zz is in [1,)\mathbb{C}\setminus [1,\infty)

Functional equations:

Li 1(1xy)=Li 1(1x)+Li 1(1y) 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 2πi 0 0 1),γ 1=(1 2πi 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 (1) (2) 0 1 (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)ImLi 2(z)+arg(1z)log|z| D(z) \coloneqq \operatorname{Im} \operatorname{Li}_2(z) + \arg(1-z)\log|z|

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

D(e iθ)= n1sinnθn 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 \mathbb{R}.

D(z) =D(1z 1)=D((1z) 1) D(z 1)=D(1z)=D(z1z) \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 1()\mathbb{P}^1(\mathbb{C}) with a maximum at z=(1+3)/2z=(1+\sqrt{-3})/2: D(1+3)/2)=1.0149D(1+\sqrt{-3})/2)=1.0149\dots.

Define recursively

z n+1z n1=1z n z_{n+1}z_{n-1} = 1-z_n

then z n+5=z nz_{n+5}=z_n. If we call z 0=xz_0=x, z 1=yz_1=y, then we find

x,y,1yx,x+y1xy,1xy x,y,\frac{1-y}{x},\frac{x+y-1}{xy},\frac{1-x}{y}

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

The 5-term recursion relation is

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

This can be explained geometrically.

Layer 1 0 0 1 1 z z \infty

In hyperbolic space, an ideal tetrahedron, with vertices at 0,1,,z0,1,\infty,z, has volume D(z)D(z). (z=z=\tfrac{}{} is the regular tetrahedron, more generally, zz is the cross ratio of the 4 vertices , which is invariant under PSL 2()=Isom()PSL_2(\mathbb{C})=\operatorname{Isom}(\mathbb{H})) The 5-term recursion relation comes from taking 5 points in 1()\mathbb{P}^1(\mathbb{C}) and constructing five tetrahedra by taking the points 4 at a time

0= j=0 4(1) jVol((w 0,,w^ j,,w 4)) 0 = \sum_{j=0}^4 {(-1)}^{j}\operatorname{Vol}((w_0,\dots,\hat{w}_j,\dots,w_4))