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 z z 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\} 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}) 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 z 0 = x z_0=x z 1 = y z_1=y
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}) ( 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.
Layer 1
0
0
1
1
z
z
∞
\infty
In hyperbolic space, an ideal tetrahedron, with vertices at 0 , 1 , ∞ , z 0,1,\infty,z D ( z ) D(z) z = z=\tfrac{}{} z z PSL 2 ( ℂ ) = Isom ( ℍ ) PSL_2(\mathbb{C})=\operatorname{Isom}(\mathbb{H}) ℙ 1 ( ℂ ) \mathbb{P}^1(\mathbb{C})
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))