TITLE: The relationship between the dilogarithm and the golden ratio
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: Let $K$ be the field of real algebraic numbers. Denote by $B_2(K)_\mathbb Q$ the pre-Bloch group of raional numbers. It is quotient of free vector generated by $\{x\}_2, x\in K\backslash\{0,1\}$ by Abel five tern 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 some number of results(Borel theorem, Suslin theorem etc.). That this map is isomorphism. From the other side there is a conjecture that all the relations between values of dilogarithms comes from Abel 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 expessed 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 to see whether it zero or not.
Now in your case $x=Li_2\left(\dfrac{3-\sqrt{5}}{2}\right)$ 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.