• Ramanujan’s first letter to Hardy:

\frac{e^{−2π/5}}{1 + \frac{e^{−2π}}{1 + \frac{e^{−4π}}{1 + ⋯}}}  = √\frac{5 + √5}{2} − \frac{√5 + 1}{2}

\frac{e^{−π/5}}{1 + \frac{e^{−π}}{1 + \frac{e^{−2π}}{1 + ⋯}}}  = √\frac{5 − √5}{2} − \frac{√5 − 1}{2}

• Hardy:

[These formulas ] defeated me completely. I had never seen anything in the least like this before. A single look at them is enough to show they could only be written down by a mathematician of the highest class. They must be true because no one would have the imagination to invent them.

• Where do these formulae come from, and why should they be true?

Let τ ∈ ℌ and q = e^{2πiτ}. We are lead to define

F (τ) =^{def} q^{1/5} (\frac{1}{1+} \frac{q}{1+} \frac{q^{2}}{1 + ⋯} ) .

What is known about F (τ)?

• Rogers-Ramanujan:

F = q^{1/5} ∞ ; ∏ ; n=1 \frac{(1 − q^{5n−4} )(1 − q^{5n−1} )}{(1 − q^{5n−2} )(1 − q^{5n−3} )} .

• Many “singular values.” For example, F (i), F (i/2), and

F (i/√10 ) = (√90(5 + 2√5) − 18 − 5√5) ^{1/5} .

• Modular equations: If x = F (τ) and y = F (3τ) then

(x − y^{3})(1 + xy^{3}) = 3x^{2}y^{2}.

• Arithmetic: If K is an imaginary quadratic field and τ ∈ ℌ ∩ K then F (τ) is a unit.

How are these things proved?

• Ramanujan’s theory of modular functions (mostly identities).

• Clever manipulation of q series.

• Modular forms and multiplier systems.

• Kronecker’s Limit Formula These methods are unsatisfactory as they do not provide any structure or framework in which to place the function F (τ). One cannot expect to prove more general results about F (τ) using these methods.

In this talk we will show:

• j_{5} =^{def} \frac{1}{F} is a modular function of full level 5, and hence an element of the function field of the modular curve X(5).

• The function field ℂ(X(5)) is rational, generated over ℂ by j_{5}.

This gives us the powerful interpretation of j_{5} (equivalently F) as coordinate on the genus 0 modular curve X(5). Using this viewpoint, we prove:

• x = j_{5}(τ) and y = j_{5}(nτ) satisfy a polynomial F_{n} ∈ ℤ[X, Y].

• When K is an imaginary quadratic field and τ ∈ K ∩ ℌ then j_{5}(τ) is a unit.

• The polynomial F_{n}(X, X) ∈ ℤ[X] satisfies simple congruences modulo primes p.

X(5) as Riemann Surface

• Any subgroup Γ of SL_{2}(ℤ) acts on ℌ^{∗} = ℌ∪ ℙ^{1}(ℚ) by fractional linear transformations.

• The quotient space X(Γ) =^{def} Γ\ℌ^{∗} admits the structure of a compact Riemann surface.

• We consider congruence subgroups, and in particular

Γ(N) =^{def} {α ∈ SL_{2}(ℤ) : α ≡ 1 mod N} and the associated Riemann surface

X(N) =^{def} X(Γ(N)).

• For any inclusion f : H ↪ G of congruence subgroups we get a field extension K(X(H))/K(X(G)) of degree [Ḡ : H̄].

• For 1 ≤ N ≤ 5 the genus of X(N) is 0, and hence ℂ(X(N)) ≃ ℂ(x).

Klein Forms

• L ⊂ ℂ a lattice with fixed ℤ basis ω_{1}, ω_{2}.

Put W = ( \begin{matrix}ω_{1} \\ ω_{2} \end{matrix} ) and for a ∈ ℚ^{2} set z = a·W.

• Weierstrass σ function:

σ(z, L) =^{def} z _{ω∈L−\{0\}}^{∏} (1 − \frac{z}{ω}) e^{z/ω+\frac{1}{2}(z/ω)^{2}} .

• Weierstrass η function: defined on L by

\frac{σ′}{σ}(z + ω, L) = \frac{σ′}{σ}(z, L) + η(ω, L).

• Define the Klein form

κ_{a}(W) =^{def} e^{−η(z,L)z/2} σ(z, L).

• κ_{a}(W) depends on the choice of basis of L.

• When W = ( \begin{matrix}τ \\ 1 \end{matrix} ) for τ ∈ ℌ, we write κ_{a}(τ) for κ_{a}(W).

We have • κ_{a}(λW) = λκ_{a}(W) for any λ ∈ ℂ^{×}.

• For any b = (b_{1}, b_{2}) ∈ ℤ^{2} and a = (a_{1}, a_{2}) ∈

ℚ^{2},

κ_{a+b} (τ) = ϵe^{πi(a_{1}b_{2}−b_{1}a_{2})}κ_{a}(τ),

where

ϵ = \begin{cases}1 & if b · W ∈ 2L \\ −1 & otherwise \end{cases} .

• For α ∈ SL_{2}(ℤ), κ_{a}(αW) = κ_{aα}(W).

• Let q = e^{2πiτ} and q_{z} = e^{2πiz}. Then

κ_{a}(τ) = −\frac{q^{1/2(a_{1}^{2}−a_{1})}}{2πi} e^{πia_{2}(a_{1}−1)} (1 − q_{z})

× ∞ ; ∏ ; n=1 \frac{(1 − q^{n}q_{z})(1 − q^{n}/q_{z})}{(1 − q^{n})^{2}} .

Constructing Functions on X(N) for Odd N

• Use Klein forms to construct functions on ℌ invariant under Γ(N).

• Theorem 1[Kubert-Lang]: Let S ⊂ ℤ^{2} − \{0\} and A = \frac{1}{N} S. To each a = (a_{1}, a_{2}) ∈ A, associate an integer m(a) and let

f =^{def} _{a∈A}^{∏} κ_{a}^{m(a)} .

Then f is invariant under the action of Γ(N) if and only if

1. ∑_{a∈A} m(a) = 0

2. ∑_{a∈A} m(a)N^{2}a_{i}a_{j} ≡ 0 mod N for each pair (i, j) ∈ {(1, 1), (1, 2), (2, 2)}.

A Generator for ℂ(X(5)).

• Theorem 2: Let ζ = e^{2πi/5} and define

j_{5} =^{def} ζ^{−1} 4 ; ∏ ; k=0 \frac{κ_{(\frac{2}{5},\frac{k}{5})}}{κ_{(\frac{1}{5},\frac{k}{5})}} .

Then j_{5} is a modular function of level 5 and j_{5} = 1/F.

• Theorem 3: j_{5} as a function on X(5) has a single simple pole at ∞.

• Corollary 4: For any number field K, we have K(X(5)) = K(j_{5}).

• We have

j_{5}(τ + 1) = ζ^{−1}j_{5}(τ)
j_{5}(−1/τ) = \frac{1 + ^{\frac{1+√5}{2}} j_{5}(τ)}{j_{5}(τ) − ^{\frac{1+√5}{2}}} .

Proofs

• For Theorem 2, use Theorem 1 with N = 5,

A = \{(\frac{j}{5}, \frac{k}{5}) : j = 1, 2, 0 ≤ k ≤ 4\},

and m((\frac{j}{5}, ∗)) = (−1)^{j}.

• To prove Theorem 3, compute the order of j_{5} at each cusp of X(5). The product expansion makes it clear that j_{5} is holomorphic on ℌ.

Sample: The matrix α = ( \begin{matrix}−4 & 1 \\ −9 & 2 \end{matrix} ) ∈ SL_{2}(ℤ) takes the cusp represented by 2/9 to ∞. Using the properties of Klein forms:

j_{5} ∘ α^{−1} = −ζ^{2} \frac{κ_{(0,\frac{2}{5})} κ_{(\frac{2}{5},0)} κ_{(\frac{1}{5},\frac{1}{5})}κ_{(\frac{1}{5},\frac{2}{5})}κ_{(\frac{2}{5},\frac{1}{5})}}{κ_{(0,\frac{1}{5})} κ_{(\frac{1}{5},0)} κ_{(\frac{2}{5},\frac{2}{5})}κ_{(\frac{1}{5},\frac{3}{5})}κ_{(\frac{2}{5},\frac{4}{5})}} ,

and expanding this as a q series:

j_{5}∘α^{−1} = −(1+ζ^{4})+(1+3ζ +ζ^{2})q^{1/5} +…,

so the order at the cusp represented by 2/9 is 0.

• Corollary 4 follows from the fact that j_{5} ∈ ℤ((q^{1/5} )).

• To prove the formulae for j_{5}(−1/τ), use the strategy in the proof of Theorem 3: we have

j_{5}(−1/τ) = −(ζ^{2}+ζ^{3})+(3+ζ+ζ^{4})q^{1/5} +…,

so that

\frac{3 + ζ + ζ^{4}}{j_{5}(−1/τ) + (ζ^{2} + ζ^{3})} − j_{5}(τ)

is a function on X(5) with no poles and is therefore constant. Inspection of the q series shows this constant to be ζ^{2} + ζ^{3}.

• In this way, we find value of j_{5} at each cusp. In particular, j_{5} has a simple zero at 2/5.

• We can now prove Ramanujan’s evaluation for F (i) = 1/j_{5}(i). Indeed, i is fixed by τ ↦ −1/τ so x = j_{5}(i) satisfies

x^{2} − (1 + √5)x − 1 = 0.

• Ramanujan’s continued fraction is coordinate on X(5).

• Gives analogy between j_{5} and j.

– For n > 1 and prime to 5, j_{5}(τ) and j_{5}(nτ) satisfy F_{n}(X, Y) ∈ ℤ[X, Y].

– If n is squarefree, H_{n}(X) = F_{n}(X, X) is monic.

• Because level is 5: more structure.

– There exists α ∈ SL_{2}(ℤ) such that j_{5} ∘ α = −1/j_{5}

– For K imaginary quadratic and τ ∈ K ∩ ℌ, j_{5}(τ) is a unit.

– For p ≡ ±1 mod 5

F_{p}(X, Y) ≡ (X^{p} − Y)(X − Y^{p}) mod p.

– For p ≡ ±2 mod 5

F_{p}(X, Y) ≡ (X^{p} − Y)(XY^{p} + 1) mod p.

An Involution on X(5)

.

• Let σ_{a} ∈ SL_{2}(ℤ) satisfy σ_{a} ≡ ( \begin{matrix}a^{−1} \\ & a \end{matrix} ) mod 5. Then

j_{5} ∘ σ_{a} = \begin{cases}j_{5} & a ≡ ±1 mod 5 \\ −\frac{1}{j_{5}} & a ≡ ±2 mod 5 \end{cases} .

• Observe that σ_{a} is in the normalizer of Γ(5) in SL_{2}(ℤ).

• We expect σ_{a} to be an involution since for any a, σ_{a}^{2} ∈ ±Γ(5).

• Proof is simple and follows from transformation properties of Klein forms.

– Reduce to case a ≡ 2 mod 5 since −1 acts trivially on ℌ.

Modular Equations for j_{5}

• Existence follows from algebraic geometry, but we proceed classically.

• (Double coset decomposition): Let

A = \{σ_{a} ( \begin{matrix}a & 5b \\ 0 & d \end{matrix} ) : (a, b, d) = 1, ad = n, 0 ≤ b ≤ d\} .

Then

Γ(5) ( \begin{matrix}1 & 0 \\ 0 & n \end{matrix} ) Γ(5) = _{α∈A}^{⋃} Γ(5)α.

• Let s_{k} be the k^{th} symmetric polynomial on {j_{5} ∘ α : α ∈ A}. Then s_{k} = P_{k}(j_{5})/j_{5}^{m_{k}} where P_{k} ∈ ℂ[X].

– s_{k} is invariant under Γ(5) action, so is a rational function of j_{5}.

– Key is that (n, 5) = 1 so j_{5} ∘ α has poles only at ∞, 2/5.

• Since j_{5}, 1/j_{5} ∈ ℤ((q^{1/5} )) we find j_{5} ∘ α ∈ ℤ[ζ]((q^{1/5} )).

– Galois action on coeffs. of q series permutes the set {j_{5} ∘ α : α ∈ A}.

– Hence P_{k}(X) ∈ ℤ[X] by Hasse Principle.

• Put

f_{n}(X) = j_{5}^{m_{0}} _{α∈A}^{∏}(X − j_{5} ∘ α).

– Have shown f_{n}(X) ∈ ℤ[X, j_{5}], so let F_{n}(X, Y) be such that F_{n}(X, j_{5}) = f_{n}(X). Hasse Principle shows F_{n}(X, Y) ∈ ℤ[X, Y].

• When n is squarefree, lead term in q series of j_{5}−j_{5}∘α is a unit. Follows that H_{n}(X) = F_{n}(X, X) is monic.

• H_{n}(j_{5}(τ)) = 0 iff j_{5}(ατ) = j_{5}(τ) for some α ∈ A.

– For some β ∈ A and γ ∈ Γ(5), we have σ_{2}α = γβσ_{2}.

– Follows that j_{5}(σ_{2}τ) = j_{5}(βσ_{2}τ).

– Hence H_{n}(−1/j_{5}(τ)) = 0.

• Hence for any root z of H_{n}(X), −1/z is also a root.

• By classical methods, if K is imaginary quadratic and for τ ∈ K ∩ ℌ then j_{5}(τ) is a root of H_{n} for some squarefree n. It follows that j_{5}(τ) is a unit.

F_{p}(X, Y) mod p.

• For a prime p ≠ 5, let α_{b} = ( \begin{matrix}1 & 5b \\ 0 & p \end{matrix} ) and

α_{p} = σ_{p} ( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} ). Then A = \{α_{b} : 0 ≤ b ≤ p\}.

• Expanding as q series, find for 0 ≤ b < p

j_{5} ∘ α_{0} − j_{5} ∘ α_{b} ≡ 0 mod (1 − ζ_{p}).

• Similarly j_{5}∘α_{p} ≡ j_{5}^{p} mod p if p ≡ ±1 mod 5 and j_{5} ∘ α_{p} ≡ −_{\frac{1}{j_{5}^{p}}} mod p if p ≡ ±2 mod 5.

• Finally, (j_{5} ∘ α_{0})^{p} ≡ j_{5} mod p.

• Since (1 − ζ_{p}) is the unique prime of ℤ[ζ_{p}] over p, we can piece these congruences together.

Arithmetic of Singular Values

• Well known result [Shimura] that for K imaginary quadratic and τ ∈ K ∩ ℌ the field K(ζ, f(τ)) is a certain class field of K, where we adjoin all values f(τ) for modular functions f of level 5 with coefficients in ℚ.

• By Corollary 4, we know that K(ζ, j_{5}(τ)) is a class field.

• In fact, if τ = t_{1}/t_{2} with (t_{1}, t_{2}) ⊂ O_{K} then K(ζ, j_{5}(τ)) is the ray class field of K of conductor 5.

Computations

• Can compute F_{p}(x, y) for small p using linear algebra. We find:

F_{2}(x, y) = yx^{3} − y^{3}x^{2} + x + y^{2}

F_{3}(x, y) = yx^{4} − y^{4}x^{3} + 3y^{2}x^{2} + x − y^{3}

F_{5}(x, y) = y^{5}(x^{4} − 2x^{3} + 4x^{2} − 3x + 1)

− (x^{5} + 3x^{4} + 4x^{3} + 2x^{2} + x)

F_{7}(x, y) = yx^{8} + (−y^{8} + 7y^{3})x^{7} + 7y^{5}x^{6} + (−7y^{7} + 7y^{2})x^{5} + 35y^{4}x^{4} + (−7y^{6} − 7y)x^{3}

− 7y^{3}x^{2} + (7y^{5} + 1)x − y^{7}

F_{11} (x, y) = x^{12} +(−y^{11} +11y^{6}−11y)x^{11} +66y^{2}x^{10}

− 220y^{3}x^{9} + 495y^{4}x^{8} − 792y^{5}x^{7}+ (11y^{11} + 803y^{6} − 11y)x^{6} − 792y^{7}x^{5} + 495y^{8}x^{4} −220y^{9}x^{3}+66y^{10} x^{2}+(−11y^{11} −11y^{6}−y)x+y^{12}

Questions

• Why are the coefficients of F_{n} so small?

• Let K be imaginary quadratic, τ ∈ K ∩ ℌ and L = K(ζ, j_{5}(τ)). Let O_{L}^{×} be the full group of units in O_{L} and let U be the subgroup of O_{L}^{×} generated by all values j_{5}(τ) ∈ L with τ ∈ K ∩ H. What is the index [O_{L}^{×} : U]?

Preprint and References

• www.math.lsa.umich.edu/∼bcais/papers.html