Buzzard’s Group Presentation

March 14, 2001

Let p = 2. Consider the modular curve X_{0}(2) = (_{Γ_{0}(2)}\^{H})^{∗} which parametrizes pairs (E, C) where C is a subgroup of order 2. As an affine curve, X_{0}(2) is given by xy = 2^{12}. And we define

f := 1/x = \frac{Δ(q^{2})}{Δ(q)} = q ∏_{n=1}^{∞}(1 + q^{n})^{24}

Since Δ(q), Δ(q^{2}) are weight 12 level 2 modular forms, f is a modular function of weight 0 level 2. We have a cannonical map of degree 3 between X_{0}(2) and X_{0}(1), which maps the two cusps 0, ∞ of X_{0}(2) to the one cusp ∞ of X_{0}(1) and preserves q-expansions. (Picture was drawn here during the talk.) Since Δ(q) has a simple zero at ∞ and a double zero at 0 on X_{0}(2) and Δ(q^{2}) has a double zero at ∞ and a single zero at 0 on X_{0}(2), f has a simple zero at ∞ and a simple pole at 0. Therefore,

f : X_{0}(2) ⟶^{∼} ℙ^{1}

Now if we consider X_{0}(2) as a rigid analytic space, the collection {1, f, f^{2}, f^{3}, … } forms a basis of the space of funtions on {x ∈ X_{0}(2) : |f(x)| ≤ 1} such that ∑_{n=0}^{∞} a_{n}T^{n}, |a_{n}|_{2} → 0 as n → ∞, and where |T | < 1. The reduction of X_{0}(2)_{ℚ_{2}} to X_{0}(2)_{𝔽_{2}} takes an 2-adic annulus of supersingular points 2^{−12} > |x| > 1 to the singular point of xy = 0. (Picture could be drawn here.)

If we write f(q) = ∑_{n=1}^{∞} a_{n}q^{n}, then our goal is to compute what U_{2} looks like on \{1, f, f^{2}, f^{3}, … \}. If we let α = f(\sqrt{q}) and β = f(−\sqrt{q}), then we have

α + β = ∑ a_{n}q^{n/2} + ∑(−1)^{n}a_{n}q^{n/2}
= 2 ∑ a_{n}q^{n/2} + 2 ∑ a_{2n}q^{n}
= 2U_{2}f

Therefore U_{2}f = \frac{1}{2}(α + β), and U_{2}f : X_{0}(2) ⟶ ℙ_{\overline{ℚ}_{2}}^{1} is a map of degree 2.

If f(q) is a modular form for Γ, then f(δq) is a modular form for δ^{−1}Γδ. So f(\sqrt{q}) =

f(τ/2) = f(( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} )τ) is a modular form on ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} )^{−1}Γ_{0}(2)( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ) ⊇ Γ(2) = \{( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) mod 2\}

Note that we know the degree of α, β are 2, and looking at where our maps factor through we have

α, β : X(2)  \overline{2  ^{}}ℙ^{1}
[figure: 2; τ ↦ τ/2; 1; f]
X_{0}(2)

Thus deg(U_{2}f) ≤ 4 on X(2) and we have the following diagram

U_{2}f = 1/2(α + β) :X(2)  \overline{≤ 4  ^{}}ℙ^{1}
[figure: 2; ≤ 2]
X_{0}(2)

So U_{2}f will factor through X_{0}(2). Now the question becomes how far out do we need to look for f expansions?

By the above discussion we see that deg(U_{2}f) = 2 on X_{0}(2), so

U_{2}f = \frac{+ f + f^{2}}{+ f + f^{2}}

and we need to figure out what should go in the blanks. Since f : 0 → ∞ and f : ∞ → 0, U_{2}f = + f + f^{2}. By a computer calculation we see that U_{2}f = 24f + 2084f^{2}.

The matrix for 2U_{2} with respect to {1, f, f^{2}, f^{3}, … } will be denoted by (c_{ij}), where 2U_{2}f^{j} = ∑_{i} c_{ij}f^{i}. And this matrix actually begins

(\begin{matrix}2 & 0 & 0 & ⟶ \\ 0 & 48 & 2 \\ 0 & 2^{12} & 2304 & … \\ 0 & 0 & ⋮ \\ ↓ & ↓ & ⋮ \end{matrix})

So we are interested in a generating function. Let g = ∑ b_{n}q^{n}, then 2U_{2}g = 2 ∑ b_{2n}q^{n} = g(\sqrt{q}) + g(−\sqrt{q}). So if g = f^{j}, 2u_{2}f^{j} = α^{j} + β^{j} = ∑ c_{ij}f^{i}. Now

∑(αy)^{j} + (βy)^{j} = ∑ c_{ij}f^{i}y^{j}
= \frac{1}{1 − αy} + \frac{1}{1 − βy}
= \frac{2 − (α + β)y}{1 − (α + β)y + αβy^{2}}

In fact

αβ = \sqrt{q} ∏_{n} (1 + (\sqrt{q})^{n})^{24} · −\sqrt{q} ∏_{n} (1 + (−\sqrt{q})^{n})^{24}
= −q ∏_{n even} ((1 + q^{n})^{2})^{24} · ∏_{n odd}(1 − q^{n})^{24}

= −q ∏_{n} \begin{matrix}((1 + q^{n})^{2}) & 24 & (1 − q^{n}) & 24 \\ (1 − q^{2n}) & 24 \end{matrix}
= −q ∏_{n} (1 + q^{n})^{24}
= −f

So we have a generating function

\frac{2 − (48f + 2^{12}f^{2})y}{1 − (48f + 2^{12}f^{2})y − fy^{2}}

Can you figure out the c_{ij} from this generating function?

In order to make our argument rigorous, we need to show that 2U_{2} is a compact operator on the ρ-overconvergent modular forms with |ρ|_{2} < 1. We know that if M = (m_{ij}) is an infinite matrix giving an operator, then M is compact if and only if for γ_{i} = sup_{j} |m_{ij}|_{2}, we have that γ_{i} → 0 as i → ∞. The problem we face is that c_{i,2i} = 2, so we do not have a compact operator for the basis {1, f, f^{2}, f^{3}, … }.

Since we do not have a compact operator on the space of all 2-adic modular forms, we must restrict to ρ-overconvergent modular forms. Pick ω ∈ \overline{ℚ}_{2} such that |ω|_{2} < 1. We now adjust our basis: {1, (ωf), (ωf)^{2}, … } and look at 2U_{2} as a matrix with respect to this new basis

2U_{2}(ωf)^{i} = ω^{j} ∑_{i} c_{ij}f^{i}
= ∑_{i} c_{ij}ω^{j−i}(ωf)^{i}

So in terms of the new basis 2U_{2} = (d_{ij}) = (c_{ij}ω^{j−i}). Now letting ω = 2^{l} with l a positive rational and recalling the generating function for (c_{ij}), we have

∑ d_{ij}f^{i}y^{j} = ∑ c_{ij}ω^{j−i}(ωy)^{j}
= \frac{2 − (2^{4} · 3f + 2^{12}f^{2}ω^{−1})y}{1 − (2^{4} · 3f + 2^{12}f^{2}ω^{−1})y − ωfy^{2}}
= \frac{2 − (2^{4} · 3f + 2^{12−l}f^{2})y}{1 − (2^{4} · 3f + 2^{12−l}f^{2})y − 2^{l}fy^{2}}

which has a power series in O_{\overline{ℚ}_{2}}. Thus |d_{ij}|_{2} < |2^{il}|_{2} = 2^{−il} → 0 as i → 0. So 2U_{2} is compact on {1, (ωf), (ωf)^{2}, … } as desired.

We have been interested in computing (c_{ij}) and a result of Frank Calegari shows this matrix is computable for U_{2}.

Theorem. c_{ij} = \frac{2^{8i−4j}3j(i + j − 1)!}{(2i − j)!(2j − i)!} where c_{ij} = 0 whenever not well-defined.

Proof. There is a recurrence relation

F_{1} = 2U_{2}f = α + β = 48f + 4096f^{2}
F_{n} = 2U_{2}f^{n} = α^{n} + β^{n}
F_{n+1} = 2U_{2}f^{n+1} = α^{n+1} + β^{n+1} = (α + β)(α^{n} + β^{n}) − αβ(α^{n−1} + β^{n−1})

then

F_{n+1} = 2U_{2}f · 2U_{2}f^{n} + f(2U_{2})(f^{n−1})
= (48f + 4096f^{2})F_{n} + fF_{n−1}

so c_{i+1,j+1} = 48c_{i,j} + 4096c_{i−1,j} + c_{i,j−1} and the rest follows by an ugly calculation.

Ideally we would like to have a closed form, which we hope to derive from

\frac{2 − (α + β)y}{1 − (α + β)y + αβy^{2}}

but what we have been able to compute is that

c_{ij} = 2^{8i−4j}3^{2j−1} (( \begin{matrix}j \\ 2j − i \end{matrix} ) + j ∑_{k=1}^{[ \frac{2j−i}{3} ]} \frac{3^{−3k}}{k} ( \begin{matrix}j − (k + 1) \\ j − 2k \end{matrix} ) ( \begin{matrix}j − 2k \\ 2j − i − 3k \end{matrix} ))

but we have been unable to reduce this to Frank’s result.

There are many related open and fun problems. We can try the same calculations with E_{2}(q) = weight 2, level 2 Eisenstein series, then E_{2}(q), E_{2}(q^{2}) are overconvergent modular

forms of weight 2 and level 1. Let g = \frac{E_{2}(q)/E_{2}(q^{2}) − 1}{24} (which is overconvergent of weight 0 and level 2) try to compute the matrix in terms of the paramaters

U_{2}(g) = 0
U_{2}(g^{2}) = ∑ ((2i + 1)3^{i−2}2^{3(j−1)}(−1)^{i+1}) g^{i}
U_{2}(g^{3}) = 0
U_{2}(g^{4}) = ± (U_{2}(g^{2}))^{2}

There are many interesting problems for p = 3, 5, 7, 13, . . .. What are nice paramaters ... try to find nice formulae.