p-adic Modular Forms

K. Buzzard

3/13/01

Let E/R/R_{0} be an elliptic curve over an R_{0}-algebra R, where R_{0} = O_{K} with [K : ℚ_{p}] < ∞. Now consider E/K, then we have two cases:

v(E) ∈ \begin{cases}not defined & if E is very supersingular \\ [0,1) ∩ ℚ & otherwise \end{cases}  (1)

Theorem 1. (Katz-Lubin) If

v(E) < \begin{cases}\frac{p}{p+1} & if p ≥ 5 \\ \frac{p}{2(p+1)} & if p = 3 \\ \frac{p}{4(p+1)} & if p = 2 \end{cases}  (2)

then E has a ”canonical” subgroup of ord=p.

Remark 1. v(E) = 0 ⇔ E has ordinary reduction, and then the canonical subgroup is just the kernel of the reduction map on its p-torsions.

Assume v(ρ) < c_{p}, where c_{p} denotes the number on the right of (2) corresponding to different p’s. If (E/R, ω, Y) is a ρ-overconvergent test object, then v(E_{K}) ≤ v(ρ) < c_{p}. So E has a canonical subgroup H, and (E/R, ω, H) is a classical test object plus a subgroup of order p. A rule on these objects is a classical modular form of level p. Hence we get a map from classical modular forms of level p over K_{0} to ρ-overconvergent forms of level 1. So we also have a U_{p} operator acting on the ρ-overconvergent forms. If f is a ρ-overconvergent, then

Remark 2. Let E/K have v(E) < c_{p}, and H be the canonical subgroup, then

(1) If C is a subgroup of order n with (n, p) = 1 then v(E/C) = v(E),

(2) If C is not canonical then v(E/C) = \frac{1}{p}v(E),

(3) If v(E) < \frac{1}{p}c_{p} then v(E/C) = pv(E), so in fact U_{p} maps ρ-overconvergent forms to ρ^{P} -overconvergent forms.

Definition 1.

𝕄_{k}(K_{0}, ρ) = (ρ − overconvergent forms of weight k defined over R_{0}) ⊗ K_{0}.

Then 𝕄_{k}(K_{0}, ρ) is a p-adic Banach space over K_{0}.

As the remark indicates, we will have Hecke operators T_{l} for l ≠ p acting on 𝕄_{k}(K_{0}, ρ), and U_{p}: 𝕄_{k}(K_{0}, ρ) → 𝕄_{k}(K_{0}, ρ^{p}). While at the same time there is a natural inclusion

𝕄_{k}(K_{0}, ρ^{p}) ⟶ 𝕄_{k}(K_{0}, ρ)

where v(ρ) < \frac{1}{p}c_{p}. Hence we get a map

U_{p} : 𝕄_{k}(K_{0}, ρ) ⟶ 𝕄_{k}(K_{0}, ρ)

One can also get U_{p}(∑ a_{n}q^{n}) = ∑ a_{np}q^{n}.

Remark 3. T_{l}’s are continuous. U_{p} is even better than that! Let V be a big infinite dimensional p-adic Banach space, and assume e_{1}, e_{2}, … is a countable Banach basis of V. Then every v ∈ V can be written uniquely as

v = ∑ a_{i}e_{i}, with a_{m} → 0, a_{n} ∈ K_{0}

Let T : V → V be a continuous operator, and T (e_{i}) = ∑ c_{ji}e_{j}. So c_{ji} is the matrix of T with respect to the basis. Then the queation is: does this matrix have a trace? Of course one cannot expect an affirmative answer in general as the identity matrix has no trace. But the operator T : e_{i} → p^{i}e_{i} of V has a trace=∑ p^{i} = \frac{p}{1−p}.

Now denote L(V, V)=continuous linear maps:V → V. L(V, V) inherits a norm from V. Let F be the subspace consisting of the maps whose image is finite dimensional. We define compact operators to be the closure of these F ’s. Compact operators have traces, and even better, they have a spectral theory. Now say C is a compact linear operator, i.e. C = lim_{n→∞} C_{n}, where C_{n} : V → V have finite dimensional images. Put

P_{n}(X) = det(I − XC_{n}) = 1 − t_{n}X + ⋯ + (−1)^{n}det(C_{n})X^{n}

then P_{n}’s converge to a power series P ∈ K_{0}[[X]] called the characteristic power series of C.

Example:Let C_{n} = (\begin{matrix}a & 0 \\ 0 & 1 \end{matrix}), C = lim C_{n}. Then

P_{n}(X) = ∏_{i=1}^{n}(1 − p^{i}X)

therefore

P (X) = ∏_{i=1}^{∞}(1 − p^{i}X) ∈ K_{0}[[X]]

and P (x) converges ofor any x ∈ K_{0}.

Now we have a very nice result

Theorem 2. If v(ρ) ∈ (0, \frac{1}{p}c_{p}), then U_{p} : 𝕄_{k}(K_{0}, ρ) ⟶ 𝕄_{k}(K_{0}, ρ) is compact.

Re-interpretation of G-M: Fix ρ such that 0 < v(ρ) < \frac{1}{p}c_{p}. Recall that M_{k}(Γ_{0}(p), K_{0}) denotes the classical modular forms with weight k of level p over K_{0}. Then we have a U_{p}-covariant linear injection

M_{k}(Γ_{0}(p), K_{0}) ⟶ 𝕄_{k}(K_{0}, ρ)

M_{k}(Γ_{0}(p), K_{0}) = (old part) ⊕ (new part). U_{p} acts differently on these two parts:

(1) if f ∈(old part), then U_{p}(f) = a_{p}f and U_{p} has eigenvalues as roots of X^{2} − a_{p}X + p^{k−1}, both of which have valuation≤ k − 1,

(2) if f ∈(new part), then U_{p} has eigenvalues ±p^{\frac{p−2}{2}} . Therefore if λ is a U_{p}-eigenvalue on the classical forms, then v(λ) ≤ k − 1. The converse is almost true!

Theorem 3 (Coleman). Assume f ∈ 𝕄_{k}(K_{0}, ρ) is an eigenform for U_{p}, T_{l}, and the U_{p}-eigenvalue is λ. If v(λ) < k − 1 then f ∈ the image of M_{k}(Γ_{0}(p), K_{0}).

Definition. v(λ) is called the slope of the overconvergent form f. Hence one can retrieve classical forms as being ”overconvergent forms of small slope”. Gouvea-Mazur Conjecture. Let k ∈ 2ℤ, α ∈ ℚ, 𝕄_{k}(K_{0}, ρ), and d(k, α) = ♯{eigenvalues of U_{p} with valuation α}. Then k_{1} ≡ k_{2} (mod (p − 1)p^{m}), for m ≥ α, will imply that d(k_{1}, α) = d(k_{2}, α).

Theorem 4 (Coleman). If P_{k}(X)=char power series of U_{p} acting on 𝕄_{k}(K_{0}, ρ), then P_{k} varies analytically with k.

This theorem implies that d(k, α) is a ”locally constant” function of k.

Proposition 2. If k_{1} ≡ k_{2} (mod (p − 1)p^{m}), and α < O(\sqrt{m}), then d(k_{1}, α) = d(k_{2}, α).

Example of the Spectrum of U_{p}. Let’s seek the structure of U_{2} on 𝕄_{0}(K_{0}, ρ) (i.e. k = 0, N = 1). Let the char power series of U_{2} be

∑_{n≥0} a_{n}X^{n} = ∏_{i≥0}(1 − λ_{i}X).

The question is: what are the valuations of λ_{i}? Inspired by a method of Kilford, we find that:

Theorem 5. (Buzzard, Calegari) The valuations are 3,7,13,15,17,…, where the ith term is given by

1 + 2v_{2} (\frac{(3i)!}{i!}) .

Proof. Let’s write down a basis for 𝕄_{0}(K_{0}, ρ) (the basis depends on ρ although the characteristic p.s. of ρ does not), say,

1, αf, α^{2}f^{2}, α^{3}f^{3}, ⋯

where

f = \frac{Δ(q^{2})}{Δ(q)} = q + 24q^{2} + ⋯

and α = α(ρ), α ∈ ℚ̄_{2}, |α| < 1. The matrix of U_{2} is:

U_{2}(f^{m}) = ∑_{n=⌈ \frac{m}{2} ⌉}^{2m} s_{m,n}f^{n}

where

s_{m,n} = 2^{8n−4m−1} · 3m(m + n − 1)!/(2n − m)!(2m − n)!

Write U_{2} = A · B, where A is lower triangular, B is upper triangular, with 1’s on both diagonals. Actually we can compute the entries A_{ij} and B_{ij}. Now let A = C · D with D diagonal, then

D_{ii} = 2^{1+2v((3i)!/i!)}

Once we take α = 2^{6}: it changes C_{ij} and B_{ij} by 2^{6(j−i)}. Then the following lemma concludes the proof.

Lemma 3. After making the change if C ≡ B ≡ Id mod 2, then the slopes of the characteristic power series of U_{2} and D are the same.