p-adic modular forms on definite quaternion algebras

Kevin Buzzard UC Berkeley.

First draft. 27.3.1998

p-adic modular forms on definite quaternion algebras

Kevin Buzzard†

§0. Introduction.

Let N be a positive integer, let p be a prime not dividing N, and let k be a positive integer. The finite-dimensional complex vector space S_{k}(Γ_{1}(Np)) of cusp forms of weight k and level Np comes equipped with an action of a linear map U_{p}, the Hecke operator at p. The eigenvalues of this linear map are algebraic integers. Moreover we have a good understanding of their absolute values. For example, if λ is the eigenvalue of an eigenform which is old at p, then it is a theorem of Deligne that |λ|_{∞} = p^{(k−1)/2}. In the last few years, however, there has been much progress in understanding nature of the p-adic valuations of these eigenvalues. Firstly work of Hida, then conjectures of Gouvea and Mazur, and then theorems of Coleman and Wan, give us more insight into the behaviour, for fixed α, of the number of eigenvalues of U_{p} on S_{k}(Γ_{1}(Np)) with valuation α, as k grows.

Coleman uses the techniques of rigid analytic geometry to understand the spaces of classical modular forms. This idea goes back to Katz’ Antwerp paper on the subject. Coleman generalises results of Serre about p-adic Banach spaces, and deduces strong results about the behaviour of the valuations of the eigenvalues of U_{p}. Using Coleman’s results one can give many new examples of p-adic families of eigenforms, in a precise sense. Before Coleman’s results, the only examples of p-adic families of eigenforms known, to the author’s knowledge, fell into three types. There were the p-adic Eisenstein series constructed by Serre, the ordinary p-adic eigenforms constructed by Hida, and also some p-adic theta series. In particular, it was not known whether a classical eigenform was always a member of a p-adic family of eigenforms, before Coleman’s work.

Gouvea and Mazur actually made very precise conjectures, based on compelling numerical evidence calculated mostly by Mestre. One thing that has not been established yet is whether these precise conjectures are correct. In fact there still seems to be a rather large gap between the constants obtained by Coleman and Wan, and the constants suggested by computer calculation.

This paper began as an attempt to prove the analogues of the conjectures of some of these Gouvea-Mazur conjectures, for modular forms over definite quaternion algebras, where things were likely to be much simpler. Our hope was to be able to prove everything, even down to the analogues of the constants suggested by the computer calculations, in this setting. Our approach owes a lot more to Hida than Coleman and in fact is entirely elementary. To our surprise, it turned out to be a relatively easy task to prove results such as p-adic continuity of slope α eigenspaces of varying weights. Moreover, we could also construct, in a weak sense, p-adic families of eigenforms. But our simple-minded approach currently only gives us bounds of the order of magnitude of those of Coleman and Wan, although the author is still under the impression that this is because of his own incompetence at linear algebra. Having said that, the cohomological methods used in this

† The author would like to thank the Miller Institute and UC Berkeley for the financial support and hospitality they offered him whilst this article was being written. paper could possibly be adapted to prove weaker versions of the theorems of Coleman and Wan again using entirely low-level methods, and could also perhaps be used to prove boundedness results for high slope Siegel modular forms.

We remark finally that using the Jacquet-Langlands theorem, we can translate our results for definite quaternion algebras to results about classical modular forms. As a result of this, we can give a new construction of p-adic families of classical eigenforms.

§1. Definitions and notation.

Let D/ℚ be a definite quaternion algebra of discriminant d, and fix a maximal order O_{D} of D. For all primes l prime to d fix an isomorphism O_{D} ⊗_{ℤ} ℤ_{l} ≅ M_{2}(ℤ_{l}), and note that this induces an isomorphism D ⊗_{ℚ} ℚ_{l} ≅ M_{2}(ℚ_{l}). Fix once and for all a prime p not dividing d.

Let M denote the monoid consisting of matrices ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{2}(ℚ_{p}) such that

a, b, c, d ∈ ℤ_{p}, c ≡ 0 mod p and d ≡ 1 mod p. We consider M as a subset of (D ⊗_{ℚ} ℚ_{p})^{×} via our fixed isomorphism D ⊗_{ℚ} ℚ_{p} ≅ M_{2}(ℚ_{p}). Let R denote the monoid ring ℤ_{p}[M]. Then R is a non-commutative ring.

Write 𝔸_{f} for the finite adeles of ℚ, and D_{f} for the ring D ×_{ℚ} 𝔸_{f}. Write D_{f}^{×} for the group of units of this ring. Then D_{f}^{×} has a standard topology with respect to which it becomes a topological group. Let U be a compact open subgroup of D_{f}^{×} such that the projection from U to (D ⊗_{ℚ} ℚ_{p})^{×} lies in M. For u ∈ U let u_{p} ∈ M denote the image of u under this projection map.

Definition. If A is a left R-module, define

B(U, A) = {f : D^{×}\(D_{f})^{×} → A such that u_{p}f(gu) = f(g) ∀u ∈ U}.

Note that A is a ℤ_{p}-module and hence B(U, A) inherits the structure of a ℤ_{p}-module. We shall shortly define certain Hecke operators acting on B(U, A). But first a word on the actual nature of B(U, A).

The set D^{×}\(D_{f})^{×}/U is finite, of size σ(U) say, and hence we can write D_{f}^{×} =

∐_{i=1}^{σ(U)} D^{×}γ_{i}U for some γ_{i} ∈ D_{f}^{×}. The groups U ∩ γ_{i}^{−1}D^{×}γ_{i} are finite, and if U is small enough then they will all be trivial. Say U is sufficiently small if these groups are indeed all trivial. If U is sufficiently small, then the map B(U, A) → A^{σ(U)} defined by f ↦ (f(γ_{i}))_{1≤i≤σ(U)} is an isomorphism of ℤ_{p}-modules. This isomorphism is however noncanonical and it is hard to understand the Hecke action on B(U, A) this way. We now define this Hecke action.

For any g ∈ D_{f}^{×} such that g_{p} ∈ M we define a ℤ_{p}-linear map [UgU] : B(U, A) → B(U, A) as follows. Write UgU = ∐_{i} g_{i}U, and then for f in B(U, A) define [UgU]f by

([UgU]f)(h) = ∑_{i}(g_{i})_{p}f(hg_{i}).

This map is independent of choice of g_{i}, by definition of B(U, A).

Now fix N a positive integer prime to dp. Let U_{1}(Np) ⊆ D_{f}^{×} be the subgroup of (O_{D}⊗_{ℤ}

ℤ̂)^{×} consisting of elements which are congruent to something of the form ( \begin{matrix}∗ & ∗ \\ 0 & 1 \end{matrix} ) modulo

Np (as usual via our fixed isomorphisms). Then U_{1}(Np) is a compact open subgroup of D_{f}^{×}. Moreover, if Np ≥ 4 then U_{1}(Np) is sufficiently small (in the precise sense above).

Let 𝕋 denote the (commutative) polynomial ring generated over ℤ_{p} by the following infinitely many symbols:

• T_{q} for all primes q not dividing Npd,

• S_{q} for all primes q not dividing Npd, and

• U_{q} for all primes q dividing Np.

For A a left R-module, define an action of 𝕋 on B(U_{1}(Np), A) thus. For a prime q not dividing d, define η_{q} to be the element of D_{f}^{×} which is the identity element at l for

primes l ≠ q, and equal to ( \begin{matrix}q & 0 \\ 0 & 1 \end{matrix} ) at q. Let T_{q} act as [Uη_{q}U] if q is prime to Npd, and

let U_{q} act as [Uη_{q}U] if q divides Np. If q is prime to Npd, let ϖ_{q} be the element of D_{f}^{×}

which is the identity element at l for primes l ≠ q, and equal to ( \begin{matrix}q & 0 \\ 0 & q \end{matrix} ) at q, and let

S_{q} act as [Uϖ_{q}U]. One can check that these operators all commute as endomorphisms of B(U_{1}(Np), A), and hence define an action of 𝕋 on B(U_{1}(Np), A).

Remark. The conscientious reader who compares our definition of Hecke operators

with that in [DT] may worry that we are using ( \begin{matrix}q & 0 \\ 0 & 1 \end{matrix} ) instead of ( \begin{matrix}1 & 0 \\ 0 & q \end{matrix} ) , which is

consistently used in [DT]. The reason for this that in [DT] only primes not dividing the

level are considered, and GL_{2}(ℤ_{l}) ( \begin{matrix}1 & 0 \\ 0 & l \end{matrix} ) GL_{2}(ℤ_{l}) = GL_{2}(ℤ_{l}) ( \begin{matrix}l & 0 \\ 0 & 1 \end{matrix} ) GL_{2}(ℤ_{l}), so it

does not matter which choice is made. For bad primes, however, it seems that the choice we have made is the correct one.

Note that if A and B are two left R-modules and α : A → B is a R-module homomorphism, then α induces a ℤ_{p}-linear map α̃ : B(U_{1}(Np), A) → B(U_{1}(Np), B) which is [U_{1}(Np)gU_{1}(Np)]-equivariant for all G ∈ D_{f}^{×} such that g_{p} ∈ M. Moreover, if Np ≥ 4 then by our explicit description of B(U_{1}(Np), A) as a direct sum of copies of A, we can see that the functor A ↦ B(U, A) from left R-modules to ℤ_{p}-modules is exact. In particular, if Np ≥ 4 then A ↦ B(U_{1}(Np), A) is an exact functor from left R-modules to 𝕋-modules.

Now we define the R-modules which are of most interest to us. If g ∈ ℤ_{≥0}, then define V_{g} to be the finite free ℤ_{p}-module of rank g + 1 consisting of polynomials over ℤ_{p} in two variables X and Y, which are homogeneous of total degree g. We give V_{g} a left M-action

as follows. For γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ M and f ∈ V_{g}, define

(γf)(X, Y) = f(aX + cY, bX + dY).

Of course V_{g} has a natural action of ℤ_{p}, so it inherits the structure of an R-module. We shall refer to the space B(U_{1}(Np), V_{g}) as the space of modular forms of level Np and weight g + 2 for D. One spectacular theorem about this space of modular forms is

Theorem (Jacquet-Langlands) Let g be a positive integer, and let k = g + 2. Fix an embedding ℤ_{p} ↪ ℂ. Then the 𝕋-module B(U_{1}(Np), V_{g}) ⊗_{ℤ_{p}} ℂ is isomorphic to the d-new subspace of the classical space S_{k}(Γ_{1}(Np) ∩ Γ_{0}(d)) of cusp forms over ℂ, equipped with its standard action of 𝕋.

Proof. This is just a concrete respelling of the theorem of Jacquet-Langlands. See Chapter 2 of [DT] for some more details of how to reduce the theorem to the usual statement of the Jacquet-Langlands theorem, a statement involving automorphic forms.

□

Remark. The statement of the theorem needs a slight modification if it is to work for the case k = 2, and since we are only interested in high weights, we will not go into the details.

Our goal in this paper is to achieve some understanding of the action of U_{p} on the space B(U_{1}(Np), V_{g}). We have a fairly explicit description of this space so it is perhaps not so surprising that this can be done. Let us end with one last definition.

Definition. For M prime to d, as usual, and k ≥ 2 an integer, let d(D, M, k, α) denote the number of eigenvalues of U_{p} acting on B(U_{1}(Np), V_{k−2}) with slope α.

More concretely, then, our goal is to prove p-adic continuity properties of d(D, M, k, α) considered as a function of k.

§2. The trick.

We assume from now on that Np ≥ 4. Define a certain left R-submodule K_{g,n} of V_{g} as follows. First note that ⨁_{g≥0} V_{g} = ℤ_{p}[X, Y]. Let J ⊆ ℤ_{p}[X, Y] denote the homogeneous ideal (p, X). Then for all n > 0, J^{n} is also homogeneous, so it can be written as J^{n} = ⨁_{g≥0} K_{g,n} with K_{g,n} ⊆ V_{g}. Explicitly, K_{g,n} is the ℤ_{p}-submodule of V_{g} generated by the n + 1 submodules p^{n−i}X^{i}V_{g−i} for 0 ≤ i ≤ n, where we write V_{g′} = 0 if g′ < 0. Note that J, and hence J^{n}, is a sub-left-R-module of ℤ_{p}[X, Y] and so K_{g,n} is a sub-left-R-module of V_{g}.

Set I_{g,n} = V_{g}/K_{g,n}. Then I_{g,n} also inherits a left R-module structure. Moreover, the structure of I_{g,n} as an abelian group is easily determined. For example, if g ≥ n − 1 then I_{g,n} = ⨁_{j=0}^{n−1}(ℤ/p^{j+1}ℤ)Y^{g−j}X^{j}.

We have a tautological exact sequence of R-modules

0 → K_{g,n} → V_{g} → I_{g,n} → 0

which induces an exact sequence

0 → B(U_{1}(Np), K_{g,n}) → B(U_{1}(Np), V_{g}) → B(U_{1}(Np), I_{g,n}) → 0

of 𝕋-modules. To get a hold on the middle term of this short exact sequence, we prove results (Lemma L and Lemma R) about the left and right hand terms.

Lemma L. If f ∈ B(U_{1}(Np), K_{g,n}) then U_{p}f ∈ p^{n}B(U_{1}(Np), V_{g}).

Proof. Note that if g ∈ U_{1}(Np)η_{p}U_{1}(Np) then g_{p} ∈ M and moreover g_{p} is of the form

( \begin{matrix}a & b \\ c & d \end{matrix} ) with both a and b multiples of p. Hence it suffices to check that if h ∈ K_{g,n} and

γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ M with p dividing a and b, then γh ∈ p^{n}V_{g}. But certainly γX is a multiple

of p. Hence γJ ⊆ pℤ_{p}[X, Y], and so γJ^{n} ⊆ p^{n}ℤ_{p}[X, Y] and this is what we want.

□

Before stating and proving Lemma R, we firstly establish

Useful Lemma. If m > 0, g ≥ 0 and f, f′ ∈ V_{g} with the property that f −f′ ∈ K_{g,m}, then for all s ≥ 0 we have f^{p^{s}} − f^{′p^{s}} ∈ K_{p^{s}g,s+m}.

Proof. We have f = f′+F for some polynomial F ∈ (p, X)^{m}, and raising both sides to their pth power gives us that f^{p} = f^{′p} +G for some polynomial G ∈ (pF, F^{p}) ⊆ (p, X)^{m+1}. The Useful Lemma follows by induction.

Lemma R. If g ≥ n − 1 then the isomorphism class of the R-module I_{g,n} depends only on g modulo p^{n−1}, and hence the isomorphism class of the 𝕋-module B(U_{1}(Np), I_{g,n}) depends only on g modulo p^{n−1}.

Proof. The second statement is immediate given the first, and so it suffices to check that for g ≥ n − 1 we have I_{g,n} ≅ I_{g+p^{n−1},n} as left R-modules. We shall show that multiplication by Y^{p^{n−1}} induces the required isomorphism. It is clear that multiplication by Y^{p^{n−1}} induces an isomorphism on the level of abelian groups for g ≥ n − 1, and so it suffices to check that the M-action is preserved.

Unravelling what this means, it suffices to check that if ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ M and f ∈ V_{g},

then f(aX + cY, bX + dY)(Y^{p^{n−1}} − (bX + dY)^{p^{n−1}}) ∈ (p, X)^{n}. It will do to show that in fact (Y^{p^{n−1}} − (bX + dY)^{p^{n−1}}) ∈ (p, X)^{n}. For n = 1 this is a consequence of the fact that d ≡ 1 mod p, and for n > 1 this follows from the n = 1 case using the Useful Lemma applied with m = 1 and s = n − 1.

□

Now we need a result which gives us information about the middle term of our short exact sequences, given the results we have about the left and right hand terms. This result is messy even to state in its exact form, so we leave a precise statement and the proof to the next section.

Proposition-to-be-made-precise Let L be a finite free ℤ_{p}-module, and let M ⊆ L be a submodule such that L/M is finite. Say n > 0 is an integer such that p^{n} annihilates L/M. Say ξ is a ℤ_{p}-linear endomorphism of L with the property that ξ(M) ⊆ p^{n}L (Note that p^{n}L ⊆ M by definition of n). Then there is a constant c depending only on n and the abelian group L/M, with the property that for all α < c, the number of eigenvalues of ξ on L with slope α depends only on the isomorphism class of the ℤ_{p}[ξ]-module L/M.

In its current state, this Proposition is essentially vacuous, because we have not given a precise definition of c. But later we shall make it more precise (and messier). We merely emphasize now that the value of c can be made explicit.

We apply these results to the case L = B(U_{1}(Np), V_{g}), M = B(U_{1}(Np), K_{g,n}), and ξ = U_{p}. Recall that d(D, Np, g + 2, α) is the number of eigenvalues of U_{p} on L with slope α. We will deduce (details again in the next section)

Corollary-to-be-made-precise. There is a constant c, depending only on D, N, n and p, such that for g ≥ n − 1, and for all α < c, d(D, Np, g + 2, α) depends only on g modulo p^{n−1}.

This is our first main result. We remark that later we shall show that c is O(\sqrt{n}), where all the constants will be made explicit and will depend only on N, D and p. We deduce that for n sufficiently large, we are getting p-adic continuity results for dimensions of slope α spaces of modular forms, for large α, and using only elementary methods.

§3 The gory details: the statement and proof of the Proposition.

Firstly we establish rather a lot of notation. Let L be a free ℤ_{p}-module of finite rank t, and let M ⊂ L be a ℤ_{p}-submodule of finite index greater than one. Write L/M ≅ ⨁_{i=1}^{r}(ℤ/p^{a_{i}}ℤ) where a_{1} ≥ a_{2} ≥ a_{3} ≥ … ≥ a_{r} > 0, and write a_{i} = 0 for i > r. Set b_{i} = n − a_{i} for all i ≥ 1. Then 0 ≤ b_{1} ≤ b_{2} ≤ … ≤ b_{r} < n = b_{r+1} = b_{r+2} = ….

Say n ≥ a_{1} is an integer, and let m be the smallest integer such that 2m ≥ n.

Define functions T and B : ℝ_{≥0} → ℝ_{≥0} as follows. For i ∈ ℤ_{≥0} set

B(i) = ∑_{j=1}^{i} b_{j}

and

T (i) = \{ \begin{matrix}m, & if i = 0, \\ m + B(i − 1), & if i ≥ 1. \end{matrix}

For convenience, we extend T and B piecewise linearly to functions ℝ_{≥0} → ℝ_{≥0}.

Definition. For L ⊆ M and n as above, define c(L/M, n) = inf_{i∈ℤ_{>0}}{T (i)/i}. Note that, as the notation suggests, c(L/M, n) only depends on the group L/M and the integer n.

Remark. c(L/M, n) is the slope of the largest line through the origin that never lies above the graph of T. Moreover,we have T (i)/i → n as i → ∞, and hence c(L/M, n) > 0 because T (i) > 0 for all i > 0.

Before we proceed, let us work out better lower bounds on c(L/M, n) in two examples.

Example 1. Let L be a finite free ℤ_{p}-module and let M ⊆ L be a ℤ_{p}-submodule such that L/M ≅ (⨁_{i=1}^{n}(ℤ/p^{i}ℤ))^{σ} for some positive integer σ. Then we claim that c(L/M, n) ≥ \frac{2\sqrt{nσ+σ+1}−σ−2}{2σ}.

The proof of this is simply by an explicit calculation. In the notation above, we have

b_{i} = \{ \begin{matrix}j, & if jσ < i ≤ (j + 1)σ, for some j, 0 ≤ j < n; \\ n, & if i > nσ, \end{matrix}

so, for i ∈ ℤ_{≥0},

B(i) = \{ \begin{matrix}σj(j − 1)/2 + j(i − jσ), & if jσ ≤ i ≤ (j + 1)σ for some j, 0 ≤ j ≤ n; \\ σn(n − 1)/2 + n(i − nσ), & if i ≥ (n + 1)σ, \end{matrix}

and

T (i) = \{ \begin{matrix}m, & if i = 0; \\ m + σj(j − 1)/2 + j(i − 1 − jσ), & if jσ ≤ i − 1 ≤ (j + 1)σ for \\ & some j with 0 ≤ j ≤ n; \\ m + σn(n − 1)/2 + n(i − 1 − nσ), & if i ≥ (n + 1)σ + 1 \end{matrix}

(for convenience we have slightly overdetermined B and T).

We wish to minimise T (i)/i for integers i ≥ 1. We minimise separately over the two regions 1 ≤ i ≤ (n + 1)σ + 1 and (n + 1)σ + 1 ≤ i. For For i ≥ nσ + n + 1 an easy calculation shows that T (i)/i is minimised when i = nσ + n + 1 and hence it suffices to minimise T (i)/i in the region 1 ≤ i ≤ (n + 1)σ + 1. Next note that if P (x) denotes the polynomial m + \frac{(x−1)(x−1−σ)}{2σ} then P (i) ≤ T (i) for all integers i in this region, and so let us minimise P (i)/i instead. Recall the fact that if A > 0 is a real number, then x + A/x, x ∈ ℝ_{>0}, is minimised when x = \sqrt{A}. Using this fact we deduce easily that P (x)/x is always at least \frac{2\sqrt{2σm+σ+1}−σ−2}{2σ} for positive reals x, and hence c(L/M, n) ≥ \frac{2\sqrt{2σm+σ+1}−σ−2}{2σ} ≥ \frac{2\sqrt{nσ+σ+1}−σ−2}{2σ} as required.

Example 2. Say n′, σ ∈ ℤ_{>0} and β ∈ ℤ_{≥0}. Say
L/M ≅ (ℤ/p^{n′}ℤ)^{σ+βσ} ⊕ (⊕_{i=1}^{n′−1}(ℤ/p^{i}ℤ))^{σ}.

Then

c(L/M, n′) ≥ \frac{2\sqrt{σn′ + (1 + βσ)(1 + βσ + σ)} − 2 − 2βσ − σ}{2σ} .

The proof of this is essentially the same as for Example 1. Write m′ for the smallest integer at least n′/2. Then an elementary calculation gives

T (i) = \{ \begin{matrix}m′, & if 0 ≤ i ≤ σ + βσ + 1; \\ m′ + \frac{j(j−1)}{2}σ + j(i − 1 − βσ − jσ), & if βσ + jσ ≤ i − 1 ≤ βσ + (j + 1)σ, \\ & for some j with 0 ≤ j ≤ n′; \\ m′ + \frac{n′(n′−1)}{2}σ + n′(i − 1 − βσ − n′σ), & if i ≥ βσ + n′σ + 1, \end{matrix}

and for the first and third regions of the definition, the minimum of T (i)/i occurs on the boundary with the second region. Now we may proceed as before. If P (x) = \frac{2σm′+(x−1−βσ)(x−1−βσ−σ)}{2σ} then T (i) ≥ P (i) for σ + βσ ≤ i ≤ βσ + n′σ, and hence minimising P (x)/x in ℝ_{>0} we deduce

\begin{aligned}c(L/M, n′) & ≥ \frac{2\sqrt{2σm′ + (1 + βσ)(1 + βσ + σ)} − 2 − 2βσ − σ}{2σ} \\ & ≥ \frac{2\sqrt{σn′ + (1 + βσ)(1 + βσ + σ)} − 2 − 2βσ − σ}{2σ} . \end{aligned}

Now let ξ : L → L be a ℤ_{p}-linear endomorphism such that ξ(M) ⊆ p^{n}L (note that p^{n}L ⊆ M by our definition of n). Let \overline{L} denote L/M with its induced action of ξ. Let ∑_{s=0}^{t} d_{s}X^{t−s} be the characteristic polynomial of ξ acting on L, and write d_{s} = 0 for s > t. From now on we shall think of L as being a ℤ_{p}[ξ]-module, where ℤ_{p}[ξ] is thought of as an indeterminate over ℤ_{p}. Then M is a ℤ_{p}[ξ]-submodule of L.

Proposition.

(a) For i ∈ ℤ_{≥0}, (i) d_{i} ≡ 0 mod p^{B(i)} (ii) d_{i} modulo p^{T (i)} depends only on the isomorphism class of the ℤ_{p}[ξ]-module \overline{L}.

(b) Let c = c(L/M, n). Then for all α < c, the number of eigenvalues of ξ acting on L with slope α depends only on the isomorphism class of the ℤ_{p}[ξ]-module \overline{L}.

Remark. We feel it is worth trying to explain what this Proposition really says. One diagram would do it but we’ve never learnt how to put diagrams into TeX files, so it’ll have to be 1000 words instead. If one wants to work out the slopes of the linear map ξ on the space L, then one way to do it is as follows. For 0 ≤ s ≤ t plot the point (s, v_{p}(d_{s})) (where v_{p} denotes p-adic valuation, and if d_{s} = 0 then consider v_{p}(d_{s}) as being +∞), and then consider the convex closure of these points. It is a standard fact that the “lower faces” of this Newton polygon correspond to the slopes of the eigenvalues of ξ, in the sense that if ξ has a eigenvalues with slope α < ∞ then there will be a side of this Newton Polygon with slope α and whose projection onto the x-axis has length a. We would like to say something about the Newton Polygon of ξ in our case. The Proposition above says two things about this polygon. Firstly it says that it lies entirely above the function i ↦ B(i). Secondly it says that the line segments of the polygon lying completely below the function i ↦ T (i) only depend on \overline{L}. This latter fact is the real key (in fact we will make no use of (a)(i) and have just included it for completeness). Hence the functions B and T form the bottom and the top boundaries of a region where “the Newton Polygon of ξ depends only on \overline{L}”, and the fact that we shall utilise later is that if ξ has some eigenvalues of small slope, then the side of the Newton Polygon of ξ corresponding to this slope will have to lie in this region, and hence will only depend on \overline{L}.

Proof.

(a) Choose a ℤ_{p}-basis e_{i}, 1 ≤ i ≤ t of L such that p^{a_{i}}e_{i}, 1 ≤ i ≤ t is a ℤ_{p}-basis of M. With respect to the basis e_{i} we consider the matrix representation (u_{j,k})_{j,k} of ξ, so ξ(e_{k}) = ∑_{j=1}^{t} u_{j,k}e_{j}. The fact that ξ(M) ⊆ p^{n}L translates to the fact that p^{b_{k}} divides u_{j,k}. Moreover the knowledge of the action of ξ on \overline{L} translates to the knowledge of u_{j,k} mod p^{a_{j}}.

Unravelling the definition of ∑_{s=0}^{t} d_{s}X^{t−s} = det(X.Id−(u_{j,k})) we find that for s ≥ 0,

d_{s} = (−1)^{s} ∑_{{J⊆\{1,2,...t\} \atop of size s}} ∑_{σ∈Symm(J)} sgn(σ) ∏_{j∈J} u_{j,σj}.

Recall that p^{b_{σj}} divides u_{j,σj} and hence

d_{s} = (−1)^{s} ∑_{{J⊆\{1,2,...t\} \atop of size s}} p∑_{j∈J} b_{j} ∑_{σ∈Symm(J)} sgn(σ) ∏_{j∈J}(u_{j,σj}/p^{b_{σj}}).

This proves (i), as for all J of size s, ∑_{j∈J} b_{j} ≥ b_{1} + b_{2} + … b_{s}.

To prove (ii), we firstly establish some notation. Let L′ be another free ℤ_{p}-module, of rank t′, equipped with a ℤ_{p}-linear action of ξ, and let M′ be a submodule. Say ξ(M′) ⊆ p^{n}L′ ⊆ M′ and moreover that the ℤ_{p}[ξ]-modules L/M and L′/M′ are isomorphic. Let ∑_{s=0}^{t′} d′_{s}X^{t′−s} be the characteristic polynomial of ξ acting on L′, and write d′_{s} = 0 for s > t. We have to show that for all i ≥ 0, d_{i} and d′_{i} are congruent modulo p^{T (i)} (note that the definition of T (i) only depends on L/M).

Let us fix an isomorphism of ℤ_{p}[ξ]-modules \overline{L} ≅ \overline{L′}. We have already chosen a basis e_{i} of L. Define, just for 1 ≤ i ≤ r, f_{i} ∈ \overline{L} to be the reduction of the element e_{i} ∈ L. Say that via our isomorphism \overline{L} ≅ \overline{L′}, f_{i} becomes identified with the element f′_{i} ∈ \overline{L′}. Fix a lifting e′_{i} ∈ L′ of f′_{i}, 1 ≤ i ≤ r. It is not difficult to check that the elements p^{a_{i}}e′_{i}, 1 ≤ i ≤ r, are not only linearly independent in M′, but can be extended to a ℤ_{p}-basis (p^{a_{i}}e′_{i})_{1≤i≤t′} of M′ (recall that a_{i} = 0 if i > r). Moreover, one can check that (e′_{i})_{1≤i≤t′} are a ℤ_{p}-basis of L′. Say ξ(e′_{k}) = ∑_{j=1}^{t′} u′_{j,k}e′_{j}. Then p^{b_{k}} divides u′_{j,k} and moreover we have u_{j,k} ≡ u′_{j,k} mod p^{a_{j}} because \overline{L} and \overline{L′} are isomorphic. Set u_{j,k} = 0 if max\{j, k\} > t, and set u′_{j,k} = 0 if max{j, k} > t′. It will suffice to show that for all s ≥ 0 and for all J ⊆ ℤ_{>0} of size s we have ∏_{j∈J} u_{j,σj} ≡ ∏_{j∈J} u′_{j,σj} modulo p^{T (s)}. All that we have done here is written down what we are trying to prove. The proof is in fact not so hard.

Writing c_{j,k} = max{a_{j} − b_{k}, 0}, we see that u_{j,k}/p^{b_{k}} ≡ u′_{j,k}/p^{b_{k}} mod p^{c_{j,k}}. Hence for J ⊆ ℤ_{>0} of size s we see ∏_{j∈J} u_{j,σj} ≡ ∏_{j∈J} u′_{j,σj} modulo p^{N_{J}} where N_{J} = ∑_{j∈J} b_{j} + min_{j∈J}{c_{j,σj}} ≥ ∑_{j∈J} b_{j} + c_{j_{0},j_{0}} where j_{0} = max_{j∈J} j. We deduce N_{J} ≥ ∑_{j∈J,j≠j_{0}} b_{j} + max{a_{j_{0}}, b_{j_{0}}} ≥ b_{1} + b_{2} + … + b_{s−1} + m = T (s). Hence, summing over all J, d_{s} ≡ d′_{s} mod p^{T (s)}.

(b) The function T is convex, and c ≤ T (i)/i for all integers i ≥ 1 and hence for all reals i > 0, as T (0) = T (1) = m > 0 and T (i) is piecewise linear.

It is immediate that if f is the Newton Polygon of ξ and f and has a side of slope α < c then the side with that slope must lie entirely below the graph of T. Hence the endpoints of this side are (s_{0}, v_{p}(d_{s_{0}})) and (s_{1}, v_{p}(d_{s_{1}})), and we have v_{p}(d_{s_{i}}) < T (s_{i}) for i = 0, 1. Hence by (a)(ii) and a quick sketch, the existence of the side of the Newton Polygon of ξ with this slope and length is assured given only \overline{L}.

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Remark. This is the whole point of everything. These last two sides of linear algebra have just been to establish this fact. The value of c only depends on the a_{i} and n, and frequently lower bounds for c can be easily obtained using only linear algebra, as the examples above show. Moreover, we are actually losing out here, for the sake of getting a general result. It seems clear to us that in any given example, if one wants the strongest results possible, then rather than applying the Proposition, one should actually draw the “diagram” of the Proposition, that is, work out the chunk of the Newton Polygon below T, and draw conclusions from this. This Proposition merely says that it is always possible to conclude something, but in any given case one could probably conclude more.

Now we apply this to the situation we are interested in and see what we get.

Corollary. Let σ denote the size of the finite set D^{×}\D_{f}^{×}/U_{1}(Np). If 0 ≤ α <

\frac{2\sqrt{nσ+σ+1}−σ−2}{2σ} and g ≥ n − 1 then the dimension of the slope α eigenspace of U_{p} on B(U_{1}(Np), V_{g}) only depends on g modulo p^{n−1}.

Proof. Set L = B(U_{1}(Np), V_{g}) and M = B(U_{1}(Np), K_{g,n}). We know that L/M ≅ B(U_{1}(Np), I_{g,n}) as a 𝕋-module and that B(U_{1}(Np), I_{g,n}) depends only on g mod p^{n−1} for g ≥ n − 1. By the Proposition, it suffices to prove that

\frac{2\sqrt{nσ + σ + 1} − σ − 2}{2σ} ≤ c(L/M, n).

Now as an abelian group we have that I_{g,n} ≅ V_{g}/K_{g,n} ≅ ⨁_{i=1}^{n}(ℤ/p^{i}ℤ), and hence B(U_{1}(N), I_{g,n}) ≅ I_{g,n}^{σ} ≅ ⨁_{i=1}^{n}(ℤ/p^{i}ℤ)^{σ}. Hence we are done by Example 1 above.

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We can rearrange the variables and rephrase the Corollary. If r ∈ ℝ, write ⌊r⌋ for the largest integer which is at most r.

Corollary (rephrased). Say α ∈ ℚ_{≥0}, and set n(α) = ⌊σα^{2} + (σ + 2)α + \frac{1}{4}σ⌋ + 1. Then if k, k′ ≥ n(α) + 1 and k ≡ k′ mod p^{n(α)−1} we have d(D, Np, k, α) = d(D, Np, k′, α).

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We remark that if we firstly fix Np ≥ 4 and then pick n large then we are in fact getting information about slope α eigenspaces for arbitrarily large α, using entirely lowlevel methods. To end this section, let us translate these results, via the Jacquet-Langlands theorem, to statements about classical modular forms.

As mentioned in §1, for g ≥ 1 and Np ≥ 4, the space B(U_{1}(Np), V_{g})⊗_{ℤ_{p}}ℂ is isomorphic as a 𝕋-module to the subspace of the space of classical cusp forms S_{g+2}(Γ_{1}(Np) ∩ Γ_{0}(d)) consisting of those forms which are new at l for all primes l dividing d. Hence for α < \frac{2\sqrt{nσ+σ+1}−σ−2}{2σ}, the slope α subspace of S_{g+2}(Γ_{1}(Np)∩Γ_{0}(d))^{d−new} depends only on g mod p^{n−1} assuming g is sufficiently large. We should say that results of this form can certainly be deduced from work of Coleman and Wan (one can deal with the d-new condition by applying their results to Γ_{1}(Np) ∩ Γ_{0}(d′) for all divisors d′ of d). See [C] and the main theorem of [W]. Our method for getting a quadratic term in the rephrased corollary seems to bear some similarity to that of Wan.

§4. Some p-adic eigenforms.

What we shall show in this section is that, again using completely elementary techniques, we can in certain cases construct certain non-ordinary p-adic families of eigenforms. Again, all these results can be deduced from work of Coleman and Wan: the existence of p-adic families is due to Coleman and the constants that we have are slightly weaker than those obtained by Wan in the classical case. However, we feel that perhaps this elementary approach deserves a mention as it does seem to be perhaps more amenable to generalisation to, for example, Siegel modular forms.

The Corollary of the previous section proves that certain vector spaces (the slope α parts of certain spaces of modular forms) all have the same dimension. We shall further investigate this phenomenon under the additional assumption that this dimension is one. More precisely, say that α ∈ ℚ_{≥0} and n_{0} = ⌊σα^{2} + (σ + 2)α + \frac{1}{4}σ⌋ + 1. Then by the Corollary of §3, there is an integer d_{0} such that for every g ≥ n_{0} − 1 such that g ≡ g_{0} mod p^{n_{0}−1} we have d(D, Np, g + 2, α) = d_{0}. From now on we are interested in the case when d_{0} = 1. Then the unique (up to scalar) eigenvector f_{g} ∈ B(U_{1}(Np), V_{g}) for U_{p} with eigenvalue of slope α must be an eigenvector for all t ∈ 𝕋. In this section we show that, in a sense to be made precise, the f_{g} form a family of eigenforms. In fact this result again just is a corollary of a general result from linear algebra, the gory details of which are about to appear. As in §3 we do the algebra in a general setting and then deduce the concrete results about modular forms easily.

Before we state the linear algebra proposition, we make the necessary definitions. Let L be a ℤ_{p}[ξ, ψ]-module (ξ and ψ indeterminates) which as a ℤ_{p}-module is finite and free. In applications, L will be a space of modular forms over a definite quaternion algebra as before, ξ will be the Hecke operator U_{p}, and ψ will be an arbitrary Hecke operator.

Let M be a submodule of L of finite index. Let n be an integer such that p^{n} annihilates L/M. Assume that ξ(M) ⊆ p^{n}L. Let L′ be another ℤ_{p}[ξ, ψ]-module, again finite and free as a ℤ_{p}-module, and let M′ be a submodule of finite index such that ξ(M) ⊆ p^{n}L. Say that L/M ≅ L′/M′ as a ℤ_{p}[ξ, ψ]-module. Let c = c(L/M, n) be as in §1. Say 0 ≤ α < c and the characteristic polynomial P_{L} of ξ : L → L has a unique root of slope α. Then by the Proposition of the previous section, the characteristic polynomial P_{L′} of ξ on L′ also has a unique root of slope α. Because Galois-conjugate elements of \overline{ℚ}_{p} have the same valuation, and the characteristic polynomials that we are considering are defined over ℚ_{p}, we see that the roots of P_{L} and P_{L′} of slope α are in fact in ℤ_{p}, and in particular α ∈ ℤ_{≥0}. To go further we must make sure that α is sufficiently small.

So from now on, assume that α has the property that not only 0 ≤ α < c(L/M, n), but also that 2α < n, and that for all integers n′ with n − 2α ≤ n′ ≤ n we have α < c(L/(M + p^{n′}L), n′).

Now let κ be any positive integer such that κ ≤ n − 2α and furthermore such that α < c(L/(M + p^{n′}L), n′) for all integers n′ in the range n − 2α − κ < n′ ≤ n.

Now we continue with our analysis of L and L′. Say that p^{α}u is the root of P_{L} with slope α, and p^{α}u′ is the root of P_{L′} with slope α. Then there exists F ∈ L such that ξF = p^{α}uF and H′ ∈ L′ such that ξH′ = p^{α}u′H′. By rescaling if necessary, we may assume that F ∉ pL and H ∉ pL′. Moreover, note that F and H′ must be eigenvectors for ψ as well, because ψ and ξ commute.

Let us fix an isomorphism L/M ≅ L′/M′ of ℤ_{p}[ξ]-modules. Choose once and for all H in L such that H + M becomes identified with H′ + M′ under this isomorphism.

What we shall do in this section is compare the lines in L generated by F and ξH. The guts of the proof consist of unravelling the following thought. If these lines are sufficiently congruent modulo a high power of p, then the eigenvalues of ψ on F and H′ must also be highly congruent. If the lines are however not sufficiently congruent, then we have two lines in L, fairly far apart in some sense, where ξ is acting approximately by a scalar of slope α. We then will define a new ℤ_{p}-linear map ξ̃: L → L such that ξ̃ really does have at least two eigenvalues of slope α and furthermore ξ̃ and ξ are highly congruent. We will however then use the Proposition of the previous section to show that the number of eigenvalues of slope α should be the same for ξ and ξ̃. This contradiction will establish that in fact the lines generated by F and ξH are close and hence the eigenvalues of ψ on F and H are also close, p-adically.

Let a be the eigenvalue of ψ acting on F, and let a′ be the eigenvalue of ψ on H′. Then we shall in fact prove

Proposition. a and a′ are congruent modulo p^{κ}.

Proof.

Firstly observe the following fact. If p^{n−α−1}F ∈ M then applying ξ we deduce that p^{n−1}F ∈ p^{n}L, contrary to our assumption that F is not divisible by p in L. Hence p^{n−α−1}F ∉ M, and in particular, we cannot write F = p^{α+1}l + m for ∈ L and m ∈ M. Similarly, p^{n−α−1}H ∉ M.

Next we see that p^{n−2α−1}ξH ∉ M. For if this were false then p^{n−2α−1}ξH′ ∈ M′ and hence p^{n−α−1}H′ ∈ M′ and hence p^{n−α−1}H ∈ M, contradicting the previous paragraph. In particular, we deduce that we cannot write ξH = p^{2α+1}l + m for l ∈ L and m ∈ M.

We have that ψH′ = a′H′ and hence ψH − a′H ∈ M. Applying ξ we deduce ψ(ξH) − a′(ξH) ∈ p^{n}L. We also have that ψF = aF.

Let us now consider the sublattice Λ ⊆ L defined by Λ = (ℚ_{p}F + ℚ_{p}ξH) ∩ L, then intersection taking place in L ⊗_{ℤ_{p}} ℚ_{p}. This is a lattice of rank 1 or 2. We can easily deduce the Proposition in the case that the lattice has rank one. For in this case, ξH is a multiple of F, and hence ψξH = aξH. Hence (a − a′)ξH ∈ p^{n}L ⊆ M and hence p^{n−2α} divides (a − a′) by a calculation above, and we are done because n − 2α ≥ κ.

We may hence assume that the lattice Λ has rank two. Because we have chosen F such that F ∉ pL, we can extend F to a basis {F, G} of Λ. Write ξH = λF + μG for λ, μ ∈ ℤ_{p}. We have already dealt with the case μ = 0 so we may write μ = p^{β}u′′ with u′′ a unit. Our strategy is to show that β is sufficiently big, i.e. that ξH is close to the line ℤ_{p}F. Let us firstly show why this suffices, and make it more precise.

Say for now that β ≥ 2α + κ. We know that we cannot write ξH = p^{2α+1}l + m for l ∈ L and m ∈ M, and so we deduce that p^{2α+1} ∤ λ. Now we have from above that ψ(ξH)−a′(ξH) ∈ p^{n}L and ψF = aF. Applying ψ to the formula ξH = λF +μG, we deduce that a′ξH ≡ λaF + μψG mod p^{n}L, and substituting back for ξH gives us λa′F + μa′G ≡ λaF + μψG mod p^{n}L. Reducing mod p^{2α+κ} we deduce that (a′ − a)λF ∈ p^{2α+κ}L and hence that p^{2α+κ} divides (a′ − a)λ. But p^{2α+1} ∤ λ and hence p^{κ} | (a′ − a) as required.

The case that we have to eliminate is the case β < 2α + κ. As mentioned above, we show that this in fact cannot happen as it would contradict the Proposition of §3, and this will finish the proof. We shall firstly show how, if β < 2α+κ, to construct a map ξ̃: L → L with the property that ξ = ξ̃ mod p^{n−β} and such that ξ has at least two eigenvalues with slope α.

We have ξH = λF + μG. Applying ξ gives us that p^{α}u′ξH ≡ λp^{α}uF + μξG mod p^{n}, and substituting back for ξH gives us p^{α}(u′ − u)λF + p^{α}u′μG ≡ μξG mod p^{n}. The right hand side of this equation is zero mod μ and hence so is the left hand side, from which we deduce that μ divides p^{α}(u′ − u)λ. Write ν = p^{α}(u′ − u)λ/μ. Then we have νF + p^{α}u′G ≡ ξG mod p^{n−β}. By our definition of Λ, we have that L/Λ is torsion-free, and hence we can extend {F, G} to a basis {F, G, l_{1}, l_{2}, …, l_{r}} of L. Now we define a ℤ_{p}-linear map ξ̃: L → L by ξ̃F = p^{α}uF = ξF ξ̃G = νF + p^{α}u′G and ξ̃l_{i} = ξl_{i} for all i, 1 ≤ i ≤ r.

Now it is clear that ξ̃l ≡ ξl mod p^{n−β} for all l ∈ L, as ξ̃G ≡ ξG mod p^{n−β} by the above calculation. Write ñ = n − β and M̃ = M + pñL. We have ξ̃(M̃) ⊆ pñL ⊆ M̃ and ξ(M̃) ⊆ pñL, and the actions of ξ and ξ̃ on L/M̃ agree. Hence we are in a position to apply the Proposition of §3, to L and M̃ with their structure of ℤ_{p}[ξ] and ℤ_{p}[ξ̃]-modules. We deduce that if α < c(L/M̃, ñ) then the number of eigenvalues of ξ and ξ̃ with slope α must be equal. But on the subspace Λ of L we see that ξ̃ already has 2 eigenvalues of slope α, whereas by assumption ξ only has one eigenvalue of slope α on L. We deduce that there is a contradiction if α < c(L/M̃, ñ), and so it suffices to show that this inequality holds. But by the very definition of κ this inequality does hold, as n − 2α − κ < ñ ≤ n.

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We apply this Proposition to our setting. We choose α ∈ ℚ_{≥0} and g_{0} ∈ ℤ. Let n_{0} = ⌊σα^{2} + (σ + 2)α + \frac{1}{4}σ⌋ + 1. Then we have seen that there exists d_{0} such that for all g ≥ n_{0} − 1 such that g ≡ g_{0} mod p^{n_{0}−1} we have d(D, Np, g + 2, α) = d_{0}. Assume that d_{0} = 1. Say n is any integer greater than 5σα^{2} + (σ + 4)α + \frac{1}{4}σ. Then n ≥ n_{0}. Set

κ = ⌊\frac{n−5σα^{2}−(σ+4)α− \frac{1}{4} σ}{2σα+1} ⌋ . Let g and g′ be integers congruent to g_{0} modulo p^{n_{0}−1} such that

g, g′ ≥ n − 1 and g ≡ g′ mod p^{n−1}. Then d(D, Np, g + 2, α) = d(D, Np, g′ + 2, α) = 1, and so up to scalar there are unique eigenforms f in B(U_{1}(Np), V_{g}) and f′ in B(U_{1}(Np), V_{g′}) with slope α. These eigenforms for U_{p} must be eigenforms for all t ∈ 𝕋. Now choose t ∈ 𝕋. Let a be the eigenvalue of t on f, and let a′ be the eigenvalue of f′. Then a, a′ ∈ ℤ_{p}.

Corollary. a ≡ a′ mod p^{κ}.

Proof. Let L = B(U_{1}(Np), V_{g}), M = B(U_{1}(Np), K_{n,g}), L′ = B(U_{1}(Np), V_{g′}), M′ = B(U_{1}(Np), K_{n,g′}). We give L and L′ the structre of ℤ_{p}[ξ, ψ]-modules by letting ξ act as U_{p} and ψ act as t. All we have to do is check that we are in a position to apply the Proposition of this section. If κ = 0 then there is nothing to prove, so we shall assume κ > 0. All the assumptions about L, M, L′ and M′ were verified in §2, and so it suffices to prove that 2α + κ ≤ n and that for all integers n′ with n − 2α − κ < n′ ≤ n we have α < c(L/(M + p^{n′}L), n′). Again, this is very easy given what we know, but we shall just indicate some of the details.

By definition, we have (2σα+1)κ < n and hence 2α+κ < n. Moreover, if n−2α−κ < n′ ≤ n then writing β = n−n′ we have L/(M +p^{n′}L) ≅ (ℤ/p^{n′}ℤ)^{σ+βσ} ⊕(⊕_{i=1}^{n′−1}(ℤ/p^{i}ℤ))^{σ}, and hence by Example 2 of §3 we have that

c(L/(M + p^{n′}L)) ≥ \frac{2\sqrt{σn′ + (1 + βσ)(1 + βσ + σ)} − 2 − 2βσ − σ}{2σ} .

It suffices to show that this latter quantity is always greater than α, when n − 2α − κ < n′ ≤ n. Equivalently, we must verify that if 0 ≤ β < 2α + κ, then

α < \frac{2\sqrt{σ(n − β) + (1 + βσ)(1 + βσ + σ)} − 2 − 2βσ − σ}{2σ} .

The untrusting reader might like to check this, it is elementary but long.

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Again we can rephrase the Corollary (without the square roots).

Corollary (rephrased). Say α ∈ ℚ_{≥0} and k_{0} ∈ ℤ. Let n_{0} = ⌊σα^{2}+(σ+2)α+ \frac{1}{4}σ⌋+1. Then there exists d_{0} such that for all k sufficiently large with k ≡ k_{0} mod p^{n_{0}−1}, we have d(D, Np, k, α) = d_{0}. Say d_{0} = 1 and let F_{k} denote the unique up to scalar form of slope α. Then F_{k} is an eigenform for all t ∈ 𝕋. Say a_{k}(t) is the eigenvalue of t on F_{k}.

Then for all integers k ≥ 0, if n ≥ (2σα + 1)κ + 5σα^{2} + (σ + 4)α^{2} + \frac{1}{4}σ, and k, k′ are two sufficiently large integers congruent to k_{0} mod p^{n_{0}−1} such that k ≡ k′ mod p^{n−1}, we have for all t ∈ 𝕋, a_{k}(t) ≡ a_{k′}(t) mod p^{κ}.

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Remark. It does not seem inconcievable to us that the methods of this section could be adapted to show the existence of families in the sense described above, in the case when the slope α subspace has dimension greater than one.

References.

[AS] A. Ash and G. Stevens, Modular forms in characteristic l and special values of their L-functions, Duke Math. J. 53 (1986), 849-868.

[C] R. Coleman, p-adic Banach Spaces and Families of Modular Forms, preprint.

[DT] F. Diamond and R. Taylor, Non-optimal levels for mod l modular representations of Gal(\overline{ℚ}/ℚ), Invent. Math. 115 (1994), 435-462.

[W] D. Wan, Dimension variation of Classical and p-adic Modular Forms, preprint.