Ramanujan’s Continued Fractions and the Platonic Solids
by Tito Piezas III
“ΑΓΕΩΜΕΤΡΗΤΟΣ ΜΗΔΕΙΣ ΕΙΣΙΤΩ” (“Let no one ignorant of geometry enter.” – sign at the entrance of Plato’s Academy)
Abstract: We give a short overview on how some jfunction formulas using three kinds of Ramanujan’s continued fractions are connected to the Platonic solids.
I. Introduction II. Platonic Solids, Dedekind eta function, and the jfunction III. Ramanujan’s octic continued fraction (p = 2) and the octahedral equation IV. Ramanujan’s cubic continued fraction (p = 3) and the tetrahedral equation V. RogersRamanujan continued fraction (p = 5) and the icosahedral equation
I. Introduction
It is wellknown that the continued fraction representation of quadratic irrationals is periodic. The simplest is for the golden ratio φ,
II. Platonic Solids, Dedekind eta function, and the jfunction
On one side of the coin are the Platonic solids, polyhedra with equivalent faces composed of regular polygons. There are only five such solids: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron, with 4, 6, 8, 12, 20 faces, respectively. However, for every polyhedron, there is a dual, that is, another polyhedron in which faces and vertices have complementary positions.
The dual of a Platonic solid is a Platonic solid. For 1) the tetrahedron, it is selfdual; for 2) the octahedron, it is the cube; and for 3) the icosahedron, it is the dodecahedron. Thus, for the symmetry groups of the Platonic solids, there are just three polyhedral groups: the tetrahedral group of order 12, the octahedral group of order 24, and the icosahedral group of order 60.
On the other side of the coin are Ramanujan’s three qcontinued fractions (and perhaps other similar ones) which can be expressed as special forms of the Dedekind eta function η(τ), where q = e^{2πiτ} = exp(2πiτ), in the form of the eta quotient, Not surprisingly, the number 24 appears again. Of course, there are only three p such that 24/(p^{2}1) is an integer, namely p = 2, 3, 5.
Fact: We have three polyhedral groups, and three eta quotients with the above form.
So here’s the interesting part: each of the three polyhedral groups can be associated with an eta quotient of order p, hence, to one (or other similar) form of qcontinued fractions. It was Ramanujan’s genius that enabled him to find an example for all three orders, though presumably he was not thinking about polyhedra. What can make this relationship clearer is the wellknown jfunction, j(τ). This is the modular function defined by,
j(τ)_{ }= 1/q + 744 + 196884q + 21493760q^{2} + 864299970q^{3} + … (A000521)
where (as in the rest of this paper), q = e^{2πiτ} = exp(2πiτ), and is responsible for remarkable approximations such as,
e^{π√163} ≈ 12^{3}(231^{2}1)^{3} + 743.9999999999992…
While there are formulas for j(τ)_{ }using the Jacobi theta functions and others, it turns out these qcontinued fractions can also be used in jfunction formulas where the numerator and denominator involve polynomial invariants of the Platonic solids, as will be seen in the next section.
III. Ramanujan’s octic continued fraction (p = 2) and the octahedral equation
The general form is given by, (The example given in the Introduction, as well as the two others, was just the case τ = √1.) For convenience, let u(τ) = u. For p = 2, since 24/(p^{2}1) = 24/3 = 8, then a formula for u in terms of eta quotients is, or equivalently, or just, This qcontinued fraction can be used in the jfunction formula, Or let u = √v, and we get,
Its connection to the 8faced octahedron, can be made apparent when one looks at the octahedral equation, a set of related equations derived from its projective geometry. Assume an octahedron centered (0,0,0) on the intersection of three planes (x,y,z). Using a stereographic projection, by projecting the vertices with unit inradius, circumradius, and midradius, respectively, and expressing these vertex locations as roots of an algebraic equation, one gets the following,
Inradius: z^{8} + 14z^{4} + 1 = 0 Circumradius: z^{5} – z = 0 Midradius: z^{12} – 33z^{8} – 33z^{4} + 1 = 0 For a clearer presentation, please refer to the link given above. Do the first two polynomials look familiar? These are the numerator and denominator of the jfunction formula using Ramanujan’s octic continued fraction in the variable v. And the third polynomial can be derived using the formula, –Numerator + 12^{3}Denominator, hence,
the same formula which will appear in the tetrahedral and icosahedral equations as well. Also, considering the fact that the tetrahedron is composed of 8 triangles, perhaps it is not surprising that the polynomial x^{8}+14x^{4}y^{4}+y^{8} can be expressed in terms of the Pythagorean triple {x^{2}y^{2}, 2xy, x^{2}+y^{2}} as the sum of three 8th powers,
Lastly, the jfunction formula in v has 24 roots, and the octahedral group is of order 24.
IV. Ramanujan’s cubic continued fraction (p = 3) and the tetrahedral equation
The general form is, As usual, let c(τ) = c for convenience. For p = 3, since 24/(p^{2}1) = 24/8 = 3, then,
This qcontinued fraction can also used in a jfunction formula. Define, then, Its connection to the tetrahedron, can be seen in the tetrahedral equation (which is just analogous to the one in the previous section),
Circumradius: z^{4} – 2√2z = 0 Inradius: 2√2z^{3} + 1 = 0 Midradius: z^{6} + 5√2z^{3} – 1 = 0 which again involves the numerator and denominator of the jfunction formula and where the polynomial for the midradius can be derived as –Numerator + 12^{3}Denominator, hence, The associated jfunction formula has 12 roots, and the tetrahedral group is of order 12.
V. RogersRamanujan continued fraction (p = 5) and the icosahedral equation
The general form is, Let r(τ) = r. For p = 5, since 24/(p^{2}1) = 24/24 = 1, then,
with jfunction formula, As expected, its connection to the icosahedron, is given by the icosahedral equation,
Inradius: z^{20} – 228z^{15} + 494z^{10} + 228z^{5} + 1 = 0 Circumradius: z(z^{10} + 11z^{5} – 1) = 0 Midradius: z^{30} + 522z^{25} – 10005z^{20} – 10005z^{10} – 522z^{5} + 1 = 0 where –Numerator + 12^{3}Denominator yields, The jfunction formula has 60 roots, and the icosahedral group is of order 60.
Thus, the relationship between the jfunction j(τ) and the continued fractions u(τ), c(τ), r(τ) imply that, for arguments τ = √d or τ = (1+√d)/2 for some positive integer d, since it is well known that j(τ) is an algebraic number, then so are u(τ), c(τ), r(τ). Also, since the three continued fractions involve the expressions q^{1/8}, q^{1/3}, q^{1/5}, respectively, then these in fact are multivalued functions, having n = {8, 3, 5} values, some of which are complex numbers. This is particularly relevant in the case τ = (1+√d)/2 since the eta quotient is a complex number.
For a more technical discussion of some of the points discussed in this paper, pls refer to Continued Fractions and Modular Equations by W. Duke (Bulletin of the AMS, Vol. 42, 2005)
 END 
P.S. One of my favorite episodes of the Twilight Zone. In Her Pilgrim Soul, a scientist creates an advanced holographic projector. To his surprise, the machine generates an intelligent being. But there is more to this entity than what is first apparent...
The closing narration quotes the last paragraph of the poem, When You Are Old, given below. After you watch the episode (parts 1,2,3 are in YouTube) and then read the poem, you will understand why it was entitled Her Pilgrim Soul. They don't make TV shows like these anymore. :(
When You Are Old by William Butler Yeats (18651939) WHEN you are old and gray and full of sleep, How many loved your moments of glad grace, And bending down beside the glowing bars, © Oct 2010, Tito Piezas III
You can email author at tpiezas@gmail.com
