The relationship between the dilogarithm and the golden ratio

Question

Among the values for which the dilogarithm and its argument can both be given in closed form are the following four equations:

$Li_2( \frac{3 - \sqrt{5}}{2}) = \frac{\pi^2}{15} - log^2( \frac{1 +\sqrt{5}}{2} )$ (1)

$Li_2( \frac{-1 + \sqrt{5}}{2}) = \frac{\pi^2}{10} - log^2( \frac{1 +\sqrt{5}}{2} )$ (2)

$Li_2( \frac{1 - \sqrt{5}}{2}) = -\frac{\pi^2}{15} - log^2( \frac{1 +\sqrt{5}}{2} )$ (3)

$Li_2( \frac{-1 - \sqrt{5}}{2}) = -\frac{\pi^2}{10} - log^2( \frac{1 +\sqrt{5}}{2} )$ (4)

(from Zagier's The Remarkable Dilogarithm) where the argument of the logarithm on the right hand side is the golden ratio $\phi$. The above equations all have this (loosely speaking) kind of duality, and almost-symmetry that gets broken by the fact that $Li_2(\phi)$ fails to make an appearance. Can anyone explain what is the significance of the fact that $\phi$ appears on the right, but not on the left? Immediately one can see that the arguments on the lefthand side of (2)-(4) are related to $\phi$ as roots of a polynomial, but what other meaning does this structure have?

Answer 1

The identities $L_2(\frac{\sqrt{5}-1}{2}) = \frac{\pi^2}{10}$ and $L_2(\frac{3-\sqrt{5}}{2}) = \frac{\pi^2}{15}$ are due to J. Landen. The rest of the identities you wrote, I suppose, could be obtained by using some other ones, like $L_2(1) = \frac{\pi^2}{6}$, $L_2(-1) = -\frac{\pi^2}{12}$, $L_2(\frac{1}{2}) = \frac{\pi^2}{12}$, due to L. Euler, and the so-called five-term relation for the Wigner-Bloch dilogarithm function, which uses the analytic continuation of the dilogarithm initially defined by the known infinite series. This Wigner-Bloch's dilogarithm function combines the dilogarithm (the imaginary part of its analytic continuations, to be more precise) and the usual logarithm. I think that the paper by A. Kirillov could serve as a good reference. In short: I suggest that one of the relations comes from some computation, the others come from the five-term relation.

Also one may try the following way (in the case of the above identities): use the five-term relation for the usual dilogarithm $L_2$, and then apply A. Kirillov, 1.6. Exercises to Section 1. where identity (iii) involves both dilogarithm and logarithm squared.

I hope that the paper by Kirillov that I'm referring to, has a good background on the dilogarithm function and dilogarithm identities. Also, this paper has quite a rich reference list, worth seeing if one needs any additional bibliography on the subject.

Answer 2

Let $K$ be the field of real algebraic numbers. Denote by $B_2(K)_\mathbb Q$ the pre-Bloch group of rational numbers. It is the quotient of a free vector generated by $\{x\}_2, x\in K\backslash\{0,1\}$ by the Abel five-term relation. There is a map

$$\delta_2\colon B_2(K)_\mathbb Q\to \Lambda^2 K^\times\otimes\mathbb Q$$ which is given by the formula $\{x\}_2\mapsto x\wedge (1-x)$.

This map is well-defined. It follows from a number of results (including Borel's theorem and Suslin's theorem) that this map is an isomorphism.

On the other side, there is a conjecture that all the relations between values of dilogarithms come from Abel's five-term relations.

So practical application is as follows: if you have some element of the form $x=\sum\limits_{i=1}^nn_iLi_2(x_i)$ and you want to know whether this element can be expressed via elementary functions, you need to apply $\delta_2$ to the element $\widetilde x=\sum\limits_{i=1}^n n_i\{x_i\}_2$, and see whether $\delta_2(\widetilde{x})$ vanishes.

Now in your case $x=Li_2\left(\dfrac{3-\sqrt{5}}{2}\right)$, and then $\delta_2(\widetilde x)=\left(\dfrac{3-\sqrt{5}}{2}\right)\wedge \left(\dfrac{\sqrt{5}-1}{2}\right)$.

But this is equal to zero because $\left(\dfrac{\sqrt{5}-1}{2}\right)^2=\left(\dfrac{3-\sqrt{5}}{2}\right)$.

So we see that behind your identity there is some formula from algebraic number theory. I think the same computation can be done with other formulas.

Answer 3

According to Maple $$ {\rm polylog}\left(2, \dfrac{1+\sqrt{5}}{2} \right) = \dfrac{7 \pi^2}{30} - \dfrac{1}{2} \log^2 \left(\dfrac{1+\sqrt{5}}{2}\right) - \log \left(\dfrac{1+\sqrt{5}}{2}\right) \log \left(\dfrac{1-\sqrt{5}}{2}\right)$$

Of course $\log((1-\sqrt{5})/2) = \log(-1/\phi) = - \log (\phi) + i \pi$

Answer 4

Wikipedia says $$Li_2\left({1+\sqrt5\over2}\right)={\pi^2\over10}-\log^2{\sqrt5-1\over2}$$