A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture

Question

The Rohrlich-Lang Conjecture for polynomial relations in Gamma values predicts that all polynomial relations between Gamma values over $\mathbb Q$ come from the functional equations satisfied by the Gamma function. An other statement, somewhat narrower: if a quotient of products of gamma values over $\mathbb Q$ is an algebraic number, then it can be evaluated by using only the well-known reflection and multiplication formulae.

Lang, S. – Relations de distributions et exemples classiques. Séminaire Delange-Pisot-Poitou, 19e année: 1977/78, Théorie des nombres, Fasc. 2, Exp. N° 40, 6 p. Collected Papers, vol. III, Springer (2000), 59–65.

I am wondering what that means for the following. Define $$A:=\prod_{k=0}^5{\Gamma(\frac{49^k \pmod {78}}{78})}= {\Gamma(\frac{1}{78})\Gamma(\frac{25}{78})\Gamma(\frac{43}{78})\Gamma(\frac{49}{78})\Gamma(\frac{55}{78})\Gamma(\frac{61}{78})} $$ and

$$B:=\prod_{k=0}^5 {\Gamma(\frac{7\cdot49^k\pmod {78}}{78})}= \Gamma(\frac{7}{78})\Gamma(\frac{19}{78})\Gamma(\frac{31}{78})\Gamma(\frac{37}{78})\Gamma(\frac{67}{78})\Gamma(\frac{73}{78}).$$ Then numerically, it appears that $$\frac AB=4.$$ Equivalently, as $AB=(2\pi)^6\;\sqrt[3\ \ \ ]{13}$ by the multiplication formula, we'd have to show $$A= 16\pi^3\;\sqrt[6\ \ \ ]{13}\ \ \text{ and / or }\ \ B=4\pi^3\;\sqrt[6\ \ \ ]{13}.$$ So the product of the $12$ Gamma terms $\Gamma(\frac{6j+1}{78})$ with $j=0,1,3,...,12$ (excluding $j=2$ where $\frac{13}{78}=\frac16$) splits into two halves, one $4$ times as big as the other.

How to "get rid of" the numerators, which are all $\equiv1\pmod6$? Applying the multiplication formula with step $\frac16$ to each term as suggested in this comment has the same effect (in terms of producing algebraic factors that can then be "ignored") as applying the reflection formula to each term: applying it for instance to $A$ will leave us with $\Gamma(\frac{77}{78})\Gamma(\frac{53}{78})\Gamma(\frac{35}{78})\Gamma(\frac{29}{78})\Gamma(\frac{23}{78})\Gamma(\frac{17}{78})$ in the denominator, no gain.

I am aware that there can be surprises like the fact that $\frac{ \Gamma\left(\frac{11}{42}\right)\Gamma\left(\frac{12}{42}\right)}{\Gamma\left(\frac{21}{42}\right)\Gamma\left(\frac2{42}\right)} $ and $\frac{ \Gamma\left(\frac{18}{42}\right)\Gamma\left(\frac{4}{42}\right)}{\Gamma\left(\frac{21}{42}\right)\Gamma\left(\frac1{42}\right)} $ are algebraic and I would not dare to claim that the identity $\frac AB=4$ provides a counterexample to the Rohrlich-Lang conjecture. It is just that the simplicity of the result, being an integer, is particularly intriguing.

Is there any other way to evaluate this, using only the functional equations of the Gamma function? Has anybody ever written a program for that (which is rather a combinatorial issue that should not be awfully hard)?

Answer 1

It turns out that triplication is not needed here: the recursion $\Gamma(z+1) = z \Gamma(z)$, the reflection formula $$ \Gamma(z) \Gamma(1-z) = \frac\pi{\sin \pi z}, $$ and the duplication formula $$ \Gamma(z) \Gamma\bigl(z+\frac12\bigr) = 2^{1-2z} \sqrt\pi \Gamma(2z) $$ suffice to reduce the gamma product to a trigonometric identity.

Apply duplication to $z = \{2^n/39\}$ with $0 \leq n < 12$ (where $\{\cdot\}$ is the fractional part) and multiply the resulting $12$ identities, using the recursion when $z>1/2$ to write $\Gamma(z+\frac12)$ and $\Gamma(2z)$ as rational multiples of $\Gamma(z-\frac12)$ and $\Gamma(2z-1)$. Then the factors $\Gamma(\{2^n/39\})$ cancel out. For the remaining factors $\Gamma(\{ \frac{2^n}{39} + \frac12 \})$, apply reflection when $n$ is odd, we get a formula for $A/B$.

Answer 2

A conceptual way to tackle this question is to look at universal distributions on $\mathbf{Q}/\mathbf{Z}$, studied by Kubert and Lang among others. Distributions arise naturally in number theory, see e.g. Lang's article Relations de distributions et exemples classiques.

There is a function $f$ on $\mathbf{Q}/\mathbf{Z}$ with values in an abelian group $A$ which satisfies the distribution relations $\sum_{a \in \mathbf{Z}/N\mathbf{Z}} f(x+a/N)=f(Nx)$ for every $x \in \mathbf{Q}/\mathbf{Z}$ and every $N \geq 1$, and which is universal for this property. The construction of $f$ and $A$ is obvious: take the free abelian group on $\mathbf{Q}/\mathbf{Z}$ and mod out by the distribution relations.

Similarly, you can look at functions on $\frac{1}{N}\mathbf{Z}/\mathbf{Z}$ satisfying the $M$-distribution relations for every $M$ dividing $N$. You get an abelian group $A_N$ and a canonical map $A_N \to A$.

Kubert has shown in the article The universal ordinary distribution that $A$ is a free abelian group, that the map $A_N \to A$ is injective, that $A_N$ is free of rank $\varphi(N)$, and has determined explicit free generators of $A_N$.

The $\Gamma$ function determines a distribution with values in $\mathbf{C}^\times/\overline{\mathbf{Q}}^\times$. This distribution is odd and the Rohrlich-Lang conjecture asserts that it is universal, in other words the induced map $\overline{\Gamma} : A_N^- \to \mathbf{C}^\times/\overline{\mathbf{Q}}^\times$ should be injective, where $A_N^-$ denotes the odd part of $A_N$.

Concretely, you can write your purported relation as a formal $\mathbf{Z}$-linear combination of the $a/N$ with $0 < a <N$ and check whether this divisor belongs to the subspace generated by the $M$-distribution relations for every $M|N$ and by the relations $[x]+[-x]=0$. This is just linear algebra in a $\mathbf{Q}$-vector space of dimension $N$ so should be doable for quite large $N$.

EDIT. In fact, you can simply use a known universal distribution and evaluate it on your divisor. For example, the Bernoulli distribution $$\tilde{B}_1(x) = \sum_{u \in (\mathbf{Z}/N\mathbf{Z})^\times} B_1(ux) [u]$$ taking values in $\mathbf{Q}[(\mathbf{Z}/N\mathbf{Z})^\times]$ is an odd universal distribution. Here $B_1(y) = \{y\}-\frac12$ for every $y \in \mathbf{R}/\mathbf{Z}$ is the Bernoulli polynomial. So you can simply check whether your divisor is in the kernel of $\tilde{B}_1$.