Lawren Smithline’s NTS talk, March 22, 2000

William A. Stein^{∗}

March 23, 2000

Newton polygon of U_{p}.

Wan’s result. For p ≥ 5, for all weights k ∈ Z, there is a parabola below the Newton polygon for U_{p}.

More precise statement about Wan’s result.

det(1 − tU_{p}) = ∑_{i=0}^{∞} a_{i}t^{i}.

New Theorem. In level N, p ∤ N, the newton polygon of U_{p} is bounded below by a parabola with quadratic term depending on

X(Γ_{1}(N) ∩ Γ_{0}(p)) = X(Γ).

The dimensions dim S_{k}(Γ) grow essentially linearly in terms of k, say roughly s_{Γ} · k.

Fix p ≥ 5. What is our space of modular forms? Overconvergent p-adic modular forms of weight k, level N, p ∤ N, coefficients in R, growth condition r:

M_{k}(N, R)

Here’s how to build the above spaces from the classical spaces M_{k}(N, R). My R is O_{K}, where [K : Q_{p}] < ∞. Katz observed that

M_{k+p−1} = E_{p−1} · M_{k} + W_{k+p−1}(N, B).

Then

M_{k} = \{b_{0} + ∑_{j=1}^{∞} r^{j} \frac{b_{j}}{E_{p−1}^{j}} \}

where b_{0} ∈ M_{k}(N, R) and b_{j} ∈ W_{k+(p−1)j}. When v_{p}(−r) > 0, these are overconvergent modular forms.

Action of U_{p}. U_{p} acts on q-expansions by

U(∑ b_{n}q^{n}) = ∑ b_{np}q^{n}.

∗University of California, Berkeley, U.S.A., was@mail.math.berkeley.edu

It turns out that

U ∈ End(M_{k}(N, R, p^{\frac{1}{p+1}} ) ⊗ K).

Let

P_{k}(t) = det(1 − tU) = ∑_{i=0}^{∞} a_{i}t^{i} = ∑ tr Λ^{i}U_{p} · t^{i}.

Theorem (Wan). The Newton polygon of U_{p} (i.e., P_{k}(t)) is bounded below by a parabola, independent of k, dependent on N and p. Coleman’s Twist:

U_{((k)+(p−1)j)} ∼ U_{(k)} (\frac{E_{p−1}(q)}{E_{p−1}(q^{p})})^{j}

Note

G = \frac{E_{p−1}(q)}{E_{p−1}(q^{p})} ∈ M_{0}(N, R, p^{\frac{1}{p+1}} ),

so multiplying by G doesn’t change the weight. Here we abuse notation: the p is “understood.” “Similiar” is in the sense that there is an isomorphism of K-vector spaces so that the appropriate diagram commutes. “Abstractly, the spaces of various weights are all isomorphic.” Using this we make k + (p − 1)j, in range 0 … p − 1.

Let b_{i,s} be a basis for W_{k+j(p−1})(N, R). Let

e_{i,s} = \frac{r^{i}}{E_{p−1}} b_{i,s}.

Then

U · G^{j}(e_{i,s}) = \frac{1}{p} ∑(\frac{r^{p}}{E_{p−1}} )^{u}b_{u}(i, s, j)

for b_{u} ∈ W_{u}(N, R) from classical weight k +(p−1)u. Here little “u” is an index; it should be i, j, …, but he ran out.

U · G^{j}(e_{i,s})\frac{1}{p} ∑ r^{p−1}(\frac{r^{p}}{E_{p−1}} )^{a}b_{u}(i, s, j)

U · G^{j}(e_{i,s})\frac{1}{p} = ∑ A_{i,s}^{a,v}e_{u,v}

v_{p}(A_{i,s}^{u,v}) ≥ u(p − 1)v_{p}(r) − 1

v_{p}(A_{i,s}^{u,v}) ≥ \frac{u(p − 1)}{p + 1} − 1

Now let d_{ℓ} = dim M_{a(p−1)ℓ}(N, B). Then

v_{p}(a_{n}(k + j(p − 1))) ≥ \frac{p − 1}{p + 1} ∑_{0}^{ℓ} um_{u} + (ℓ? − 1)(n − d_{ℓ}) − n

for d_{ℓ} ≤ n ≤ d_{ℓ+1}, where m_{ℓ} = d_{ℓ} − d_{ℓ−1}.

Ogus attacks!!: “Draw the diagram which describes the Coleman Twist!” Lawren responds:

U · f(q^{p}) = f(q) “Frobenius linear”

U_{p} ∼ U_{p} \frac{E_{p−1}}{V (E_{p−1})}

(E_{p} − 1)^{−1}U_{p}E_{p−1}

p = 2, 3 Katz expansion in powers of E_{4}. “The message of this talk is that p = 2 and p = 3 are not so special as they

think they are.”

E.g., N = 1, p ≡ 1 (mod 12). In this case we have

dim M_{k}(1, R) = \begin{cases}floor of k/12 \\ floor of (k/12) + 1 & if k ≡ 2 (mod 12) \end{cases}

m_{j} = \frac{p−1}{12} so “parabola with quadratic term 6/(p + 1).

Ogus attacks again!! The important part must be to go from overconvergent to classical forms! How do you do that? Lawren responds: “I’m not going to tell you.” What is the conjecture!?

The conjecture of GM says: “close weights imply close slopes”