# Lectures on dilogarithm function

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## dilogarithm fuction

• Define

$\operatorname{Li}_ 2(z)= \sum_{n=1}^\infty {z^n \over n^2},\, |z|<1$

• extend domain

$\operatorname{Li}_ 2(z) = -\int_0^z{{\log (1-t)}\over t} dt,\, z\in \mathbb C\backslash [1,\infty)$

### functional equations

reflection properties

$\mbox{Li}_ 2(z)+\mbox{Li}_ 2(1/z)= -\frac{\pi^2}{6}-\frac{1}{2}\log^2(-z)$ $\mbox{Li}_ 2(z)+\mbox{Li}_ 2(1-z)= \frac{\pi^2}{6}-\log(z)\log(1-z)$

proof

$f(z): = \mbox{Li}_ 2(z)+\mbox{Li}_ 2(1/z)+\frac{1}{2}\log^2(-z)$ is constant as $$f'(z)$$ is $-\frac{\log (1-z)}{z}+\frac{\log \left(1-\frac{1}{z}\right)}{z}+\frac{\log (-z)}{z}=0$

When $$z=-1$$, $\mbox{Li}_ 2(z)+\mbox{Li}_ 2(1/z)+\frac{1}{2}\log^2(-z) = 2\mbox{Li}_ 2(-1)$

When $$z=1$$; $2\mbox{Li}_ 2(1)+\frac{1}{2}\log^2(-1) = 2\mbox{Li}_ 2(-1)$ $\frac{\pi^2}{3}-\frac{1}{2}\pi^2 = 2\mbox{Li}_ 2(-1)$

five-term relation

$\mbox{Li}_ 2(x)+\mbox{Li}_ 2(y)+\mbox{Li}_ 2 \left( \frac{1-x}{1-xy} \right)+\mbox{Li}_ 2(1-xy)+\mbox{Li}_ 2 \left( \frac{1-y}{1-xy} \right)=\text{something elementary}$

Let us state this in terms of the Rogers dilogarithm (no worry about the branches) $L(x): =\operatorname{Li}_ 2(x)+\frac{1}{2}\log x\log (1-x)=-\frac{1}{2}\int_{0}^{x}\left(\frac{\log(1-y)}{y}+\frac{\log(y)}{1-y}\right)dy,\, x\in (0,1)$

• $$0\leq x,y\leq 1$$

$L(x)+L(1-xy)+L(y)+L\left(\frac{1-y}{1-xy}\right)+L\left(\frac{1-x}{1-xy} \right)=\frac{\pi^2}{2}$

proof

Show that the partial derivatives of $$F(x,y):=L(x)+L(1-xy)+L(y)+L\left(\frac{1-y}{1-xy}\right)+L\left(\frac{1-x}{1-xy} \right)$$ are 0. Note $L(h(x))' = -\frac{h'(x) \log (1-h(x))}{2 h(x)}-\frac{h'(x) \log (h(x))}{2 (1-h(x))}.$

\begin{aligned} F_x = & \frac{1}{2} \left(\frac{\log (x)}{1-x}-\frac{\log (1-x)}{x}\right)+\frac{1}{2} \left(\frac{\log (1-x y)}{x}-\frac{y \log (x y)}{1-x y}\right)+0 \\ & +\frac{(1-y) \log \left(\frac{1-y}{1-x y}\right)+(1-x) y \log \left(\frac{(1-x) y}{1-x y}\right)}{2 (1-x) (1-x y)}-\frac{(1-x) \log \left(\frac{1-x}{1-x y}\right)+x (1-y) \log \left(\frac{x (1-y)}{1-x y}\right)}{2 (1-x) x (1-x y)} \\ & =\frac{1}{2} \log (x)\left(\frac{1}{1-x}+\frac{-y}{1-xy}+\frac{-(1-y)}{(1-x) (1-x y)} \right)+\dots \end{aligned} Do the same for $$F_y$$.

There is a more systematic way to control the cancellations.

Observe $\frac{d}{dx}L(h(x))=\frac{1}{2}[\log(h(x))\frac{d}{dx}\log (1-h(x))-\log(1-h(x))\frac{d}{dx}\log h(x)]$

For $$f,g\in \mathbb{Q}(x,y)$$, define (symbolically) $f\wedge g : = \log (f) d (\log (g))-\log (g) d (\log (f))$ where $$df = f_x dx + f_y dy$$.

For example, $$L'(x)dx =\frac{1}{2} x\wedge (1-x)$$

Then

• $$f\wedge g=-f \wedge g$$
• $$(f_1f_2)\wedge g=f_1\wedge g+f_2\wedge g$$

So $F_x dx+F_y dy = \frac{1}{2}\left(x\wedge (1-x)+(1-x y)\wedge (x y)+y\wedge (1-y)+\frac{1-y}{1-x y}\wedge \left(\frac{y(1-x)}{1-x y}\right)+\frac{1-x}{1-x y}\wedge \left(\frac{x(1-y)}{1-xy}\right) \right)=0$

remark
• recurrence relation

$1-x_{i}=x_{i-1}x_{i+1},\, x_0=x,\, x_2=y$

• 5-periodic solution

$x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1-x}{1-xy}$

remark
• we believe(?) all functional equations are coming from the five-term relation

remark

Zagier has $\frac{\pi^2}{6}-\log(x)\log(1-x)-\log(y)\log(1-y)+\log (\frac{1-x}{1-xy})\log (\frac{1-y}{1-xy})$ on the RHS, which is not correct

### special values

$$\mbox{Li}_{2}(0)=0$$

$$\mbox{Li}_{2}(1)=\frac{\pi^2}{6}$$

$$\mbox{Li}_{2}(-1)=-\frac{\pi^2}{12}$$

$$\mbox{Li}_{2}(\frac{1}{2})=\frac{\pi^2}{12}-\frac{1}{2}\log^2(2)$$

$$\mbox{Li}_{2}(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-\log^2(\frac{1+\sqrt{5}}{2})$$

$$\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})$$

$$\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})$$

$$\mbox{Li}_{2}(\frac{-1-\sqrt{5}}{2})=-\frac{\pi^2}{10}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})$$

## Bloch-Wigner dilogarithm

• $$\operatorname{Li}_2(z)$$ jumps by $$2\pi i \log|z|$$ as $$z$$ crosses the cut
• consider $$\operatorname{Li}_2(z)+i \log|z|\arg(1-z)$$, where $$-\pi <\arg z< \pi$$
• when it cross the line $$(1,\infty)$$, it becomes continuous
example

\begin{aligned} \operatorname{Li}_2(2+0.001i) & = 2.46583 + 2.17759 i \\ \operatorname{Li}_2(2-0.001i) & = 2.46583 - 2.17759 i \end{aligned} and \begin{aligned} \log |(2+0.001i)| \arg(1-(2+0.001i)) & = 0. - 2.17689 i \\ \log |(2-0.001i)| \arg(1-(2-0.001i)) & = 0. + 2.17689 i \end{aligned}

• define

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

• real analytic function on $$\mathbb{C}$$ except at 0 and 1, where it is continuous but not differentiable.
• $$D(\bar{z})=-D(z)$$, and vanishes on $$\mathbb{R}$$.
• It satisfies the following functional equations :

$$\label{functid1} D(x)+D(1-xy)+D(y)+D(\frac{1-y}{1-xy})+D(\frac{1-x}{1-xy})=0,$$ $$\label{functid2} D(x)+D(1-x) =D(x)+D(\frac{1}{x})=0.$$

## Bloch group

• Let $$\mathbb{F}$$ be a field
• $$\Lambda^2({\mathbb{F}^{\times}})$$ the set of all formal linear combinations $$x\wedge y$$, $$x,y\in\mathbb{F^{\times}}$$ subject to relations
• $$a\wedge b=-b \wedge a$$
• $$(x_1x_2)\wedge y=x_1\wedge y+x_2\wedge y$$
• $$\mathbb{Z}[\mathbb{F}^{\times}\backslash\{1\}]$$
• i.e. abelian group of formal sums $$[x_1]+[x_2]+\cdots+[x_n]$$, $$x_1,x_2,\cdots,x_n\in \mathbb{F}\backslash\{0,1\}$$
• $$\partial : \mathbb{Z}[\mathbb{F}^{\times}\backslash\{1\}] \to \Lambda^2({\mathbb{F}^{\times}})$$
• $$[x]\to x\wedge (1-x)$$
• Let $$\mathcal{A}(\mathbb{F})=\operatorname{ker}\partial$$ and $$\mathcal{C}(\mathbb{F})$$ subgroup of $$\mathcal{A}(\mathbb{F})=\operatorname{ker}\partial$$ generated by

$[x]+[1-xy]+[y]+[\frac{1-y}{1-xy}]+[\frac{1-x}{1-xy}]$

• The Bloch group is defined to be

$\mathcal{B}(\mathbb{F})=\mathcal{A}(\mathbb{F})/\mathcal{C}(\mathbb{F})$

• $$[x]+[1-x]$$ is in $$\mathcal{B}(\mathbb{F})$$
• Q. is $$[x]+[\frac{1}{x}]$$ in $$\mathcal{B}(\mathbb{F})$$?

example

$$F=\mathbb{Q}(\sqrt{-7})$$

$2[\frac{1+\sqrt{-7}}{2}]+[\frac{-1+\sqrt{-7}}{4}]\in \mathcal{B}(F)$

## values of Dedekind zeta at s=2

### Dedekind zeta

• Let $$F$$ be a number field with $$[F:\mathbb{Q}]=r_1+2r_2$$
• $$\zeta _F(s):= \zeta_F (s) = \sum_{I \subseteq \mathcal{O}_F} \frac{1}{(N_{F/\mathbf{Q}} (I))^{s}}$$
• functional equation

$\xi_{F}(s)=\left|d_F\right|{}^{s/2} 2^{r_2 (1-s)} \pi ^{\frac{1}{2} \left(-r_1-2 r_2\right) s}\Gamma \left(\frac{s}{2}\right)^{r_1} \Gamma (s)^{r_2}\zeta _F(s)$$\xi_{F}(s) = \xi_{F}(1 - s)$

• at $$s=-n, n=1,2\cdots$$, $$\zeta_F(s)$$ has zero of order $$r_2$$ or $$r_1+r_2$$ if $$n$$ is even or odd, respectively

$2^{(m+1) r_2} \pi^{-\frac{1}{2} m \left(-r_1-2 r_2\right)}\zeta_F(-m) \Gamma (-m)^{r_2}\Gamma(-\frac{m}{2} )^{r_1} \left| d_F\right| {}^{-\frac{m}{2}}\\ =2^{-m r_2} \pi ^{\frac{1}{2} (m+1) \left(-r_1-2 r_2\right)} \zeta _F(m+1)\Gamma(\frac{m+1}{2})^{r_1}\Gamma (m+1)^{r_2} \left| d_F\right| {}^{\frac{m+1}{2}}$

### Dirichlet class number formula

• residue at $$s=1$$

$\lim_{s\to 1} (s-1)\zeta_F(s)=\frac{2^{r_1}\cdot(2\pi)^{r_2}\cdot h_F\cdot R_F}{w_F \cdot \sqrt{|d_F|}}$

• equivalently, $$\zeta _F(s)$$ has zero of order $$r_1+r_2-1$$ at $$s=0$$

$\lim_{s\to 0}\frac{\zeta_F(s)}{s^{r_1+r_2-1}}=-\frac{h_F R_F}{w_F}$

### algebraic K-theory

• $$F$$ : number field
• $$K_0(F) = \mathbb{Z}$$
• $$K_1(F) = F^{\times}$$
• $$K_2(F) = F^{\times}\otimes F^{\times}/\langle x\otimes (1-x) \rangle$$
• $$K_0(\mathcal{O}_F) = \mathbb{Z}\oplus Cl_F$$
• $$K_1(\mathcal{O}_F) = (\mathcal{O}_F)^{\times}$$
• $$K_2(\mathcal{O}_F)$$ : finite group

### Borel's regulator

• Borel constructed a map

$K_{2i-1}(F) \to \mathbb{R}^{d_{i}},\, i\geq 2$ where $$d_i = r_2$$, or $$r_1+r_2$$ depending on the parity of $$i$$

• this can be used to show
• $$\operatorname{rank} K_3 =d_2 = r_2$$
• $$\operatorname{rank} K_5=d_3 = r_1+r_2$$
• $$\operatorname{rank} K_7=d_4 = r_2$$
• the covolume of the image of $$K_m,\, m=2i-1$$ under this regulator is a non-zero multiple of

$\frac{|d_{F}|^{1/2}}{\pi^{md_{i+1}}} \zeta_{F}(m)$

• for $$K_3$$, $$\frac{|d_{F}|^{1/2}}{\pi^{m(r_1+r_2)}} \zeta_{F}(2)$$
• this is a generalization of Dirichlet's class number formula
• the rational number given by the ratio is related to other $$K$$-groups (Lichtenbaum conjecture)

### Zagier, Bloch, Suslin

• The Bloch-Wigner dilogarithm $$D(z)$$ can be used to define a map from $$\mathcal{B}(\mathbb{C})$$ to $$\mathbb{R}$$.
• For $$\xi=\sum_{i} n_i[x_i] \in \mathcal{B}(\mathbb{C})$$, let $$D(\xi)=\sum_{i} n_i D(x_i)$$.
• by the 5-term relation satisfied by $$D$$, it is well-defined
• For an embedding $$\sigma : F\hookrightarrow \mathbb{C}$$ and $$\xi \in \mathcal{B}(F)$$, we may consider $$D\left(\sigma(\xi)\right)$$.
• If $$D\left(\sigma(\xi)\right)=0$$ for all such embeddings $$\sigma$$, then $$\xi \in \mathcal{B}(F)$$ is a torsion element in $$\mathcal{B}(F)$$.
• Bloch and Suslin connects $$K_3$$ with the Bloch group
• $$D$$ is compatible with Borel's regulator

$\frac{|d_{F}|^{1/2}}{\pi^{2(r_1 + r_2)}} \zeta_{F}(2) \sim_{\mathbb{Q^{\times}}} \det\left(D(\sigma_i(\xi_j))\right)_{1\leq i,j\leq r_2}$ where $$\xi_i,(i=1,\cdots, r_2)$$ is $$\mathbb{Q}$$-basis of $$\mathcal{B}(F)\otimes \mathbb{Q}$$ and $$a\sim_{\mathbb{Q^{\times}}} b$$ means $$a/b\in\mathbb{Q}$$

example

$$F=\mathbb{Q}(\sqrt{-7})$$

$\zeta_F(2) = \frac{4 \pi ^2}{21 \sqrt{7}} \left(D\left(\frac{-1+\sqrt{-7}}{4} \right)+2 D\left(\frac{1+ \sqrt{-7}}{2} \right)\right) =1.8948414489688\dots$

$$F=\mathbb{Q}(\sqrt{-23})$$

\begin{aligned} \zeta_F(2) & = \frac{4 \pi ^2}{69 \sqrt{23}}\left(21D\left(\frac{1}{2} \left(1+\sqrt{-23}\right)\right)+7D\left(2+\sqrt{-23}\right)+D\left(\frac{1}{2} \left(3+\sqrt{-23}\right)\right)+D\left(3+\sqrt{-23}\right)-3D\left(\frac{1}{2} \left(5+\sqrt{-23}\right)\right)\right) \\ & = 2.3081992895457\dots \end{aligned}

## hyperbolic 3-manifold

• $$\mathbb{H}^3$$ : upper-half space; $$SL_2(\mathbb{C})$$ acts as an isometry group
• hyperbolic 3-manifold : quotient of $$\mathbb{H}^3$$ by discrete isometry group
• ideal tetrahedron is a tetrahedron whose vertices are in $$\mathbb{C}\cup \{\infty\}$$
• let $$\tilde{D}(z_0,z_1,z_2,z_3) : = D\left(\frac{\left(z_0-z_2\right) \left(z_1-z_3\right)}{\left(z_1-z_2\right) \left(z_0-z_3\right)}\right)$$
fact

Let $$\Delta$$ an ideal tetrahedron with vertices $$z_0,z_1,z_2,z_3$$. Then its volume is given by $$\tilde{D}(z_0,z_1,z_2,z_3)$$.

fact
1. A complete oriented hyperbolic 3-manifold with finite volume can be triangulated into ideal tetrahedra $$\Delta_1,\dots, \Delta_{\nu}$$.
2. If we assume that the vertices of each tetrahedon are at $$\infty, 0,1$$ and $$z_{i}$$, then

$[z_1]+\dots + [z_{\nu}]\in \mathcal{B}_{\mathbb{C}}$ and the volume is given by $$\sum D(z_i)$$.

### five-term relation reinterpreted

• Pachner move : sum of two tetrahedra = sum of three tetrahedra
• it implies

$D\left(\frac{\left(z_0-z_2\right) \left(z_1-z_3\right)}{\left(z_1-z_2\right) \left(z_0-z_3\right)}\right)-D\left(\frac{\left(z_0-z_2\right) \left(z_1-z_4\right)}{\left(z_1-z_2\right) \left(z_0-z_4\right)}\right)+D\left(\frac{\left(z_0-z_3\right) \left(z_1-z_4\right)}{\left(z_1-z_3\right) \left(z_0-z_4\right)}\right)-D\left(\frac{\left(z_0-z_3\right) \left(z_2-z_4\right)}{\left(z_2-z_3\right) \left(z_0-z_4\right)}\right)+D\left(\frac{\left(z_1-z_3\right) \left(z_2-z_4\right)}{\left(z_2-z_3\right) \left(z_1-z_4\right)}\right)=0$

• if we set $$z_0=\infty,z_1= 0,z_2= 1,z_3= x,z_4 = x y$$, we get

$D(x)-D(x y)+D(y)+D\left(\frac{1-x y}{y(1-x)}\right)-D\left(\frac{1-x y}{1-x}\right) = 0$

example (volume of regular ideal tetrahedron)
• $$\Delta_4$$ : ideal tetrahedron with vertices $$\infty,0,1,e^{2\pi i/3}$$
• volume form on $$\mathbb{H}^3$$ $\frac{dx\,dy\,dz}{z^3}$. So the volume of $$\Delta_4$$ is

\begin{aligned} \int\int\int_{\Delta_4}\frac{dx\,dy\,dz}{z^3} & = \int\int_{\Delta_3}\int_{h(x,y)}^{\infty}\frac{1}{z^3}\,dz\,dx\,dy \\ & = \int\int_{\Delta_3}\frac{1}{2h(x,y)^2}\,dx\,dy \\ \end{aligned}

• this integral becomes

$\int\int_{\Delta_3'}\frac{1}{2(\frac{1}{3}-x^2-y^2)}\,dx\,dy = 1.01494\dots$ where $$\Delta_3'$$ is the triangle with vertices $$0,1,e^{2\pi i/3}$$ translated so that they lie on the circle with center $$(0,0)$$

## 메모

### Bloch-Suslin

• $$B(\mathbb{F})\simeq K_3^{\operatorname{ind}}(\mathbb{F})$$ ??
• $$0\to \tilde{\mu_{F}}\to K_3^{\operatorname{ind}}(\mathbb{F}) \to B(\mathbb{F})\to 0$$ where $$0\to \mathbb{Z}/\mathbb{Z}_2 \to \tilde{\mu_{F}} \to \mu_{F}\to 0$$ where \mu_{F} is the unit group of F
• $$K_3^{\operatorname{ind}}(\mathbb{F})$$ is a quotient of Milnor K3 by something else

• functional equation of $$\zeta_K$$ implies

$\pi^{-d_m} \lim_{s\to -m}\frac{\zeta_K(-m)}{(s+m)^{d_m}} \sim_{\mathbb{Q}^{\times}}\pi ^{-(m+1)(r_1+2 r_2)} \zeta _K(m+1)\left| d_K\right| {}^{\frac{1}{2}}$ where $$d_1 =d_3=\dots = r_2$$ or $$d_2=d_4=\dots =r_1+r_2$$