Lectures on dilogarithm function
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 $z=1$; \[2\mbox{Li}_ 2(1)+\frac{1}{2}\log^2(-1) = 2\mbox{Li}_ 2(-1)\] \[\frac{\pi^2}{3}-\frac{1}{2}\pi^2 = 2\mbox{Li}_ 2(-1)\]
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- 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
\[x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1-x}{1-xy}\]
3.png)
- 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 :
\begin{equation}\label{functid1} D(x)+D(1-xy)+D(y)+D(\frac{1-y}{1-xy})+D(\frac{1-x}{1-xy})=0, \end{equation} \begin{equation}\label{functid2} D(x)+D(1-x) =D(x)+D(\frac{1}{x})=0. \end{equation}
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
$F=\mathbb{Q}(\sqrt{-7})$
$$ \zeta_F(2) = \frac{4 \pi ^2}{21 \sqrt{7}} \left(D\left(\frac{-1+\sqrt{-7}}{4} \right)+2 D\left(\frac{1+ \sqrt{-7}}{2} \right)\right) =1.8948414489688\dots $$
$F=\mathbb{Q}(\sqrt{-7})$
$$ \begin{aligned} \zeta_F(2) & = \frac{4 \pi ^2}{69 \sqrt{23}}\left(21D\left(\frac{1}{2} \left(1+\sqrt{-23}\right)\right)+7D\left(2+\sqrt{-23}\right)+D\left(\frac{1}{2} \left(3+\sqrt{-23}\right)\right)+D\left(3+\sqrt{-23}\right)-3D\left(\frac{1}{2} \left(5+\sqrt{-23}\right)\right)\right) \\ & = 2.3081992895457\dots \end{aligned} $$
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
- A complete oriented hyperbolic 3-manifold with finite volume can be triangulated into ideal tetrahedra $\Delta_1,\dots, \Delta_{\nu}$.
- 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 $(0,0)$