On the Accessibility of Khoi’s Dilogarithm Identity Involving the Golden Ratio

Abstract

In this note, I make use of the Rogers five-term functional relation for the dilogarithm function with a suitable choice of arguments to show that a two-term dilogarithm identity involving the golden ratio $\phi = (1+\sqrt {5})/2$, as proposed by Khoi in a recent work, is accessible from the five-term relation.

1 Introduction

The classical dilogarithm function, defined as $\text {Li}_{2}{(z)} := {\sum }_{n=1}^{\infty } {z^{n}/n^{2}}$, which converges for all complex values of z with |z|≤1 [8],¹ was introduced by Leibniz in 1696 [7] and was thoroughly discussed by Euler in 1768 [3]. The analytic continuation of Li₂(z) is found by taking into account its integral definition, i.e.,

$$\text{Li}_{2}{(z)} = - {{\int}_{0}^{z}}{\frac{\ln{(1-u)}}{u} ~ du}, $$

which cannot be expressed in terms of elementary functions. In the complex plane, Li₂(z) has a branch point at z=1 and we are considering here the principal branch of this function, namely $z \in \mathbb {C} \backslash [1,\infty )$.² Since the times of Euler, the dilogarithm function was familiar only to a few enthusiasts, but in recent years it has become popular due to its appearance in hyperbolic geometry (e.g., in low-dimensional topology, when one computes the volume of hyperbolic 3-manifolds) and algebraic K-theory on one hand and in conformal field theory (e.g., in the computation of QED corrections to the electron giromagnetic ratio) and thermodynamics on the other hand [6, 13].

With respect to special values, the dilogarithm function has only eight values of $z\in \mathbb {C}\backslash (1,+\infty )$ for which both z and Li₂(z) can be written in closed form: {0,±1,1/2,±1/ϕ,1/ϕ ²,−ϕ}, where $\phi := (1+\sqrt {5})/2$ is the golden ratio [13].³ The function Li₂(z) can also be written in closed form for z=±i, 1±i, and (1±i)/2, but these values involve the Catalan’s constant $G := {\sum }_{n=0}^{\infty {(-1)^{n}/(2n+1)^{2}}}$, a real number whose irrationality remains unproven, so most purists prefer to omit them. In contrast to this paucity of special values, Li₂(z) satisfies many functional relations, which can be readily derived from its integral definition, as shown in [10](@var1@Section 3(a)@var1@). The main ones are the Euler reflection formula (1768)

$$ \text{Li}_{2}{(1-z)} = -\text{Li}_{2}{(z)} -\ln{(z)} \ln{(1-z)} +\frac{\pi^{2}}{6}, \quad z \not \in (-\infty,0] \cup [1,+\infty) $$

(1)

and the inversion formula

$$ \text{Li}_{2}\left( \frac{1}{z}\right) = -\text{Li}_{2}(z) -\frac{1}{2} \ln^{2}(-z) -\frac{\pi^{2}}{6}, \quad z \not \in [0,+\infty). $$

(2)

These functional equations can be used to derive the Landen’s formula (1780) [8](@var1@Eq. (1.12)@var1@)

$$ \text{Li}_{2}\left( \frac{z}{z-1}\right) = -\text{Li}_{2}(z) -\frac{1}{2}\ln^{2}{(1-z)}, \quad z \not \in [1,+\infty), $$

(3)

which is useful for transforming Li₂(z) with z<0 into Li₂(w) with w∈(0,1).

In an attempt to calculate the central charges of certain conformal field theories, many conjectures and problems have arisen since the late 1980’s involving some few-terms dilogarithm identities [2, 6, 9, 13]. A general theorem for functional equations involving dilogarithm values was then proved by Wojtkowiak in 1996 [12](@var1@Section 4@var1@), which establishes that any functional equation of the form ${\sum }_{n=1}^{N}{c_{n} \text {Li}_{2}{(f_{n}(z))}} = C$, with constants c ₁,…,c _N ,C and rational functions f ₁(z),…,f _N (z) is a consequence of the Hill five-term relation, our (6). It remains unknown whether this is true for algebraic functions.

In a recent work, Khoi has found some new dilogarithm identities by calculating the Seifert volume of certain manifolds obtained by Dehn surgery on the figure-eight knot in two different ways [5]. In particular, his second identity (see the last Remark in [5]) reads

$$ L\left( \frac{1}{\phi(\phi+\sqrt{\phi})}\right) -L\left( \frac{\phi}{\phi+\sqrt{\phi}} \right) = \frac{\pi^{2}}{20}, $$

(4)

where L(z) is the Rogers dilogarithm function, defined as⁴

$$L(z) := \text{Li}_{2}(z) + \frac{1}{2} {\ln}(z){\ln}(1-z), \quad 0<z<1. $$

This function can be extended to the whole real line by setting L(0)=0, L(1) = π ²/6, and⁵

$$\begin{array}{@{}rcl@{}} L(x) = \left\{ \begin{array}{ll} - L\left( \frac{x}{x-1}\right),& \qquad x<0, \\ \pi^{2}/3 -L(1/x),& \qquad x>1, \end{array} \right. \end{array} $$

(5)

which yields a continuous function for all $x \in \mathbb {R}$ [10, 13]. In the next section, it will be shown that Khoi’s identity, above, is accessible from the five-term relation for the dilogarithm function, thus answering a question posed by Khoi himself at the end of [5].

2 The Five-Term Relation and the Corrected Version of Khoi’s Identity

There are several forms in which the two-variable five-term relation for the dilogarithm function can be written. For instance, Abel proved in 1826 (in a paper published posthumously, see [1]) that

$$\begin{array}{@{}rcl@{}} \text{Li}_{2}{\left( \frac{x}{1-x}~\frac{y}{1-y}\right)} &=& \text{Li}_{2}{\left( \frac{x}{1-y}\right)} +\text{Li}_{2}{\left( \frac{y}{1-x}\right)} -\text{Li}_{2}{(x)} -\text{Li}_{2}{(y)}\\ &&-\ln{(1-x)}\ln{(1-y)}, \end{array} $$

valid whenever x,y,x + y<1. On making the substitutions $x/(1-y) \rightarrow x$ and $y/(1-x) \rightarrow y$ and using the Landen’s formula in the last two dilogarithms, one promptly finds that

$$ \text{Li}_{2}(xy) = \text{Li}_{2}(x) +\text{Li}_{2}(y) +\text{Li}_{2}\left( \frac{xy-x}{1-x}\right) +\text{Li}_{2}\left( \frac{xy-y}{1-y}\right) +\frac12 \ln^{2}\left( \frac{1-x}{1-y}\right), $$

(6)

also valid for x,y,x y<1. This five-term relation is due to Hill (1828) [4]. Other functional relations involving five dilogarithm values are given in [8](@var1@pp. 7–11@var1@) and [6](@var1@pp. 88–89@var1@). As shown in Lewin’s book [8], any one of these five-term relations may be derived from any of the others by applying suitable transformations. A useful form due to Rogers (1907) is [11]

$$\begin{array}{@{}rcl@{}} \text{Li}_{2}(xy) &=& \text{Li}_{2}(x) +\text{Li}_{2}(y) - \text{Li}_{2}\left( \frac{x(1-y)}{1-xy}\right) - \text{Li}_{2}\left( \frac{y(1-x)}{1-xy}\right)\\ &&-\ln\left( \frac{1-x}{1-xy}\right) \ln{\left( \frac{1-y}{1-xy}\right)}, \qquad |x|<1 \text{ and } |y|<1, \text{ or } x y < 1. \end{array} $$

In the Rogers dilogarithm format, this relation reads

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

(7)

This is the only relation we shall use hereafter.

Lemma 1

(Abel’s formula) The Abel’s duplication formula

$$L(z) = L\left( \frac{1}{2-z}\right) +\frac{1}{2} L(2z - z^{2}) - \frac{\pi^{2}}{12} $$

holds for any z∈(0,1) and is accessible from the five-term relation only.

Proof

On taking x∈(0,1) and choosing x = z and y=2−z, so x y<1, in the Rogers five-term relation, (7), one finds

$$L(2z - z^{2}) = L(z) +L(2-z) -L\left( -\frac{z}{1-z}\right) -L\left( \frac{2-z}{1-z}\right). $$

The extended version of L(x), as stated in (5), yields $L\big (-\frac {z}{1-z}\big ) = L\big (\frac {z}{z -1} \big ) = -L(z)$ and, since 2−z>1, then $L(2-z) = \frac {\pi ^{2}}{3} -L\big (\frac {1}{2-z}\big )$. This reduces the above equation to

$$ L(2z - z^{2}) = 2L(z) +\frac{\pi^{2}}{3} -L\left( \frac{1}{2-z}\right) - L\left( \frac{2-z}{1-z}\right). $$

(8)

Again, since $\frac {2 -z}{1 -z} > 1$, the last dilogarithm value becomes $\frac {\pi ^{2}}{3} -L\big (\frac {1 -z}{2 -z} \big )$ and then (8), above, simplifies to

$$L(2z - z^{2}) = 2L(z) -L\left( \frac{1}{2-z}\right) + L\left( \frac{1-z}{2-z}\right). $$

Since $\frac {1-z}{2-z} = 1 -\frac {1}{2-z}$, then the Euler reflection formula, (1), which reads L(1−x)=−L(x) + π ²/6 in the Rogers format, applies to the last dilogarithm value, yielding the desired result. □

On inspecting Khoi’s identity, (4), numerically, one promptly finds evidence that it should be correct except by an opposite sign in the right-hand side. The same error of the sign is found in the other two identities stated by Khoi there in [5], which suggests it is not a typo. I have then noted that the derivative $L^{\prime }(x) = -\frac 12[\ln {(1-x)}/x +\ln {(x)}/(1-x)]$ is positive in all points of the open interval (0,1), which means that the function L(x) is monotone increasing in this interval, so L(a)−L(b)<0 whenever 0<a<b<1. Since 1<ϕ ², it is clear that 1/ϕ<ϕ, so the subtraction at the left-hand side of (4) has to be negative, which confirms that a minus sign is missing in its right-hand side. This is proved in the theorem below.

Theorem 1

(Main result) The two-term dilogarithm identity

$$L\left( \frac{1/\phi}{\phi+\sqrt{\phi}} \right) -L\left( \frac{\phi}{\phi+\sqrt{\phi}} \right) = -\frac{\pi^{2}}{20} $$

holds and is accessible from the five-term relation only.

Proof

Let $z = 1/[\phi (\phi +\sqrt {\phi })]$. Since z<1, being $1/(2-z) = \phi /(\phi +\sqrt {\phi })$ and 2z−z ²=1/ϕ ², then Abel’s duplication formula, our Lemma 1, yields

$$ L\left( \frac{1/\phi}{\phi+\sqrt{\phi}} \right) = L\left( \frac{\phi}{\phi +\sqrt{\phi}} \right) +\frac{1}{2} L\left( \frac{1}{\phi^{2}}\right) -\frac{\pi^{2}}{12}. $$

(9)

Now, note that $\text {Li}_{2}(1/\phi ^{2}) = \pi ^{2}/15 -\ln ^{2}{(1/\phi )}$ [13](@var1@Chap. 1@var1@) corresponds to $L(1/\phi ^{2}) = \pi ^{2}/15 -\ln ^{2}{(1/\phi )} +\frac {1}{2}\ln {(1/\phi ^{2})}\ln {(1 -1/\phi ^{2})} = \pi ^{2}/15$, in the Rogers dilogarithm format. Using this value in (9) and simplifying, gives the desired result. □