I want to know when certain expressions of the form

$ {\Gamma(r_1/m) \Gamma(r_2/m) \ldots \Gamma(r_j/m) \over \Gamma(s_1/m) \Gamma(s_2/m) \ldots \Gamma(s_j/m)} $

are algebraic numbers. These ratios of Γ functions occur in the asymptotic enumeration of certain classes of restricted partitions, but I don't think this is relevant. Also, In the partition problems I'm interested in, it's natural to have $r_1 + \ldots + r_j = s_1 + \ldots + s_j$ but this isn't necessary. This seems to happen with some frequency. For example,a note of Albert Nijenhuis (arXiv:0907.1689) shows that $\Gamma(1/14) \Gamma(9/14) \Gamma(11/14) = 4\pi^{3/2}$; the techniques of the same paper show that $\Gamma(3/14) \Gamma(5/14) \Gamma(13/14) = 2\pi^{3/2}$, so the quotient is in fact 2! Similarly, we can get the identity

$ {\Gamma(1/8) \Gamma(5/8) \Gamma(6/8) \over \Gamma(2/8) \Gamma(3/8) \Gamma(7/8)} = \sqrt{2}$

by applying the duplication formula

$ \Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z) $

to the first two factors in the numerator and the last two in the denominator. In trying to prove other identities of this type, the duplication formula, its generalization to the "multiplication formula"

$\Gamma(z) \Gamma(z+1/k) \cdots \Gamma(z+(k-1)/k) = (2\pi)^{(k-1)/2} k^{1/2-kz} \Gamma(kz)$

and the reflection formula

$ \Gamma(z) \Gamma(1-z) = \pi \csc(\pi z)$

are the most obvious tools. So this seems to be a problem in combinatorial number theory; given an expression of the form in the first displayed equation, when can we use the multiplication and reflection formulas to reduce it to a rational power of some integer times a product of trig functions of rational multiples of π?

  • $\begingroup$ I am pretty sure that it is even open whether $\Gamma(1/5)$ is irrational, much less algebraic, but I cannot find a reference. $\endgroup$ – Boris Bukh Dec 3 '09 at 1:00
  • $\begingroup$ That's a good point. The question I'm interested in, I suppose, is when we can actually write down a polynomial satisfied by a ratio of gamma functions. (In other words, assume Γ(z) is transcendental for all rational, non-integer z, which is morally true.) $\endgroup$ – Michael Lugo Dec 3 '09 at 1:10
  • $\begingroup$ -1 until the question is made precise. Is $\Gamma(1/173)\Gamma(4/61)$ morally irrational? Etc. $\endgroup$ – Boris Bukh Dec 3 '09 at 10:38

You should be interested when the solution for Monthly problem 11426 is published. A preview (credit to Albert Stadler):

$$ \frac{\Gamma(1/10)\Gamma(9/10)}{\Gamma(3/10)\Gamma(7/10)} = \frac{3+\sqrt{5}}{2}, $$

$$\frac{\Gamma(1/26)\Gamma(3/26)\Gamma(9/26)\Gamma(17/26) \Gamma(23/26)\Gamma(25/26)}{\Gamma(5/26)\Gamma(7/26)\Gamma(11/26) \Gamma(15/26)\Gamma(19/26)\Gamma(21/26)} = \frac{11+3\sqrt{13}}{2}, $$

$$\frac{\Gamma(1/34)\Gamma(9/34)\Gamma(13/34) \Gamma(15/34)\Gamma(19/34)\Gamma(21/34) \Gamma(25/34)\Gamma(33/34)}{\Gamma(3/34)\Gamma(5/34) \Gamma(7/34)\Gamma(11/34)\Gamma(23/34) \Gamma(27/34)\Gamma(29/34)\Gamma(31/34)} = 1 . $$

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  • $\begingroup$ Amer. Math. Monthly, November 2010, page 842 $\endgroup$ – Gerald Edgar Aug 20 '14 at 15:59

There are other identities that are relevant, but are less systematically understood. For example,

$\Gamma \left(\frac{1}{7}\right) \Gamma \left(\frac{6}{7}\right)=\Gamma \left(\frac{3}{7}\right) \Gamma \left(\frac{4}{7}\right)+\Gamma \left(\frac{2}{7}\right) \Gamma \left(\frac{5}{7}\right).$

There's a known generalization of this with 7 replaced by $2^k-1$ (see my paper with Ron Graham) but it isn't known if this is all instances of cosecant sums being zero.

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All explained here:

Deligne, P. Valeurs de fonctions $L$ et périodes d'intégrales. (French) With an appendix by N. Koblitz and A. Ogus. Proc. Sympos. Pure Math., XXXIII, Automorphic forms, representations and $L$-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 2, pp. 313--346, Amer. Math. Soc., Providence, R.I., 1979. 12A70 (10D15 10D24 10H10)

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As a consequence of Euler's refectin and Gauss's multiplication formulas, all values $\Gamma(a)$ with $24a\in\mathbb{Z}$ or $60a\in\mathbb{Z}$ can be expressed algebraically in terms of these values: $$\Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac13\right), \quad \Gamma\!\left(\frac14\right), \quad \Gamma\!\left(\frac15\right), \quad \Gamma\!\left(\frac25\right), \quad \Gamma\!\left(\frac18\right),$$ $$\Gamma\!\left(\frac1{15}\right),\quad \Gamma\!\left(\frac1{20}\right), \quad \Gamma\!\left(\frac1{24}\right), \quad \Gamma\!\left(\frac1{60}\right), \quad \Gamma\!\left(\frac7{60}\right).$$ This is worked out in this paper, along with basic expressions for $a\in(0,1)$. For example, $$\Gamma\!\left(\frac7{10}\right)=\sqrt{\pi}\,2^{3/5}\,\Gamma\left(\frac15\right)^{\!-1}\,\Gamma\!\left(\frac2{5}\right),$$ $$\Gamma\!\left(\frac1{12}\right)=\frac{3^{3/8}\,\sqrt{\sqrt{3}+1}}{\sqrt{\pi}\,2^{1/4}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac14\right),$$ $$\Gamma\!\left(\frac{11}{15}\right)=2\pi\cdot 3^{3/10}\,\Gamma\!\left(\frac15\right)\,\Gamma\left(\frac25\right)^{\!-1}\,\Gamma\left(\frac1{15}\right)^{\!-1},$$ $$\Gamma\!\left(\frac{11}{20}\right)=2^{1/5}\,\sqrt{5+\sqrt5}\;\Gamma\!\left(\frac15\right)\,\Gamma\!\left(\frac25\right)\,\Gamma\left(\frac1{20}\right)^{\!-1},$$ $$\Gamma\!\left(\frac{49}{60}\right)=\frac{\sqrt{\pi}\,\sqrt3\,\sqrt{\sqrt3+1}\,\sqrt{5+\sqrt5}\,\sqrt{\sqrt5+\sqrt3}}{5^{1/24}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac1{60}\right)^{\!-1}.$$ Further algebraic independence of the listed 11 "basic" values (and more general answers) are yet to be settled.

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Taking cyclic subgroups seems promising for obtaining identities similar to the ones of Nijenhuis, and so I have looked at products of ${\Gamma\left(\dfrac{a^k }{N}\right)}/{\sqrt{2\pi}}$ with $k$ running over a certain range. The factor $\sqrt{2\pi}$ is motivated by the fact that

  • in Nijenhuis' formula as well as in the multiplication formula and even in the reflection formula, a factor $\sqrt{\pi}=\Gamma(\frac12)$ occurs with the same multiplicity as the gamma factors (so my products can be considered in fact as "well-poised" gamma quotients),
  • without including $\sqrt{2}$, those products, whenever they are integers, have a relatively big 2-valuation (see the factor $2^{b(A)}$ in Nijenhuis' formula $$\prod_{x\in A}{\Gamma\left(\frac{x }{2n}\right)}=2^{b(A)} \sqrt{\pi}^{\,\nu(n)},$$ where $A$ is the subgroup of the multiplicative group $\mathbb Z_{2n}^*$ generated by $n + 2$ or any one of its cosets, $\nu(n)$ its order and $b(A)$ the number of elements in $A$ that are larger than $n$).

For brevity, define for coprime integers $a, N$ $$g(a,N):=\prod_{k=1}^{ind_a(N)} \frac{\Gamma\left(\dfrac{a^k \pmod N}{N}\right)}{\sqrt{2\pi}}.$$ Thus the product is taken over the subgroup $\langle a\rangle$ of $\mathbb Z_N^*$ generated by $a$, with all the arguments of the gamma factors between $0$ and $1$. For an integer $b$ coprime to $N$, define similarly the product over the corresponding coset $b\langle a\rangle=\langle a\rangle b$ as $$g_b(a,N):=\prod_{k=1}^{ind_a(N)} \frac{\Gamma\left(\dfrac{a^kb \pmod N}{N}\right)} {\sqrt{2\pi}}.$$

With this notation, the initial identity reads $g(9,14)=\sqrt{2}$, and we further have the somewhat complementary one $g_3(9,14)=1/\sqrt{2}$.

Once you know where to (ask a computer to) search, it is easy to find, in a few minutes, dozens of those products that appear numerically to be algebraic.

I have excluded the self-complementary groups, i.e. the groups for which $\langle a\rangle=\langle N-a\rangle $, because for those, we already know that $g(a,N)$ is algebraic by the reflection formula. Likewise I have excluded subgroups whose members form essentially an arithmetic sequence (more precisely, a set $\{\lambda s+t\}\cap \mathbb Z_N^*$ for given $s,t$), as those can be handled by the multiplication formula and yield again algebraic products. Call the remaining subgroups and the associated gamma products non-trivial.

A systematic search (with reasonably high numerical precision) shows that of the non-trivial algebraic gamma products, most are of the form $q^{u/v}$ with integers $u,v$. Generally, $u/v$ is positive and $q$ a prime dividing $N$. But there are exceptions like $g(24,203)=7^{-1/2}$ or $g(103,420)=(3^3\cdot5^3\cdot7)^{1/6}$.

For $n\le300$, there are $106$ non-trivial algebraic gamma products, $88$ of which can be written in form $q^{u/v}$. (With $v=1$ for 36 of them.) The $18$ remaining ones occur for $N= 60,105,120,140,156,180,220,231,255,285,300$ (note that all these $N$'s have at least three prime divisors) and have minimal polynomials of degrees $2,4,6$ or $8$. The latter holds for all $N\le 1000$.
Moreover so far all of them can be written with radicals, e.g. $g(103,105)=\sqrt[3]{3(1701+166\sqrt{105})}$ or $g(41,156)=2 \sqrt[4]{13}(2\sqrt{3}+\sqrt{13})$ or $g(83,120)=\sqrt{3(\sqrt{3}-1)(\sqrt{30}-5)}$.
A few of these identities, e.g. the one for $g(83,120)$, can be derived by using the standard formulas (thus in the way the OP asks next to the end). But I doubt this is possible where e.g. $\sqrt{13}$ occurs.

You can get more similar identities by taking the cosets of the same groups. I haven't looked systematically at them, but conjecturally, if $g(a,N)$ is algebraic, so is $g_b(a,N)$. Note that generally, $g_b(a,N)/g(a,N)$ does not seem to be algebraic.

The distribution of the non trivial subgroups seems to be at least as irregular as the distribution of the primes and not even very much correlated with the structure of $\mathbb Z_N^*$.

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