Geometry and String Theory
The Dilogarithm Function (Rev #5, changes)

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Define

Li 2(z)≔∑ n≥1z 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)≔∑ n≥1z 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(1−z) \operatorname{Li}_1(z) = -\log(1-z)

and

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

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

Functional equations:

Li 1(1−xy)=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 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 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)≔ImLi 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 iθ)=∑ n≥1sinnθn 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(−z1−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)/2z=(1+\sqrt{-3})/2: D(1+−3)/2)=1.0149…D(1+\sqrt{-3})/2)=1.0149\dots.

Define recursively

z n+1z n−1=1−z 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,1−yx,x+y−1xy,1−xy 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−yx)(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(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) 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))

The cancellation is the 3-2 Pachner move.