This version is completely unfinished but the first few pages are probably readable— I’d hoped to get more done tonight but I’m going to bed now and I’m going to spend the weekend in London so I won’t get anything done then; it’s also getting pretty big and a lot of what I’ve written already is not really that relevant for the computation. Still, I thought that at least you could have a look at the first few pages before it starts getting vague. I wouldn’t look at §? at all yet, I’ve realised there’s a better way to do it. Maybe I’ll finish it all on Monday night or something.

K

Computing modular forms on definite quaternion algebras.

Here are some notes on quaternion algebras, modular forms on definite quaternion algebras, and how they correspond to certain classical modular forms. The definition of modular forms on a definite quaternion algebra is essentially combinatorial, although somewhat long. The Jacquet-Langlands theorem gives a link between these new kinds of modular forms and certain classical modular forms, but in a rather abstract way. However it is possible to translate it down to a rather concrete statement, and that is the point of these notes.

Everything here is known to the experts, and the only thing these notes have going for them is their concreteness, something which is usually lacking in the more modern literature. Hopefully these notes will explain how one might attempt to actually do computations of spaces of modular forms in a way that as far as I know, noone has implemented. In fact, the initial motivation for these notes was to explain to William Stein (who has already written some programs for computing in spaces of modular forms) an algorithm for doing some more computations, especially computations which would work for any level k ≥ 2 and would perhaps be faster than the standard methods being used by people like Cremona. Whether they are in fact faster remains to be seen.

Of course, I learnt almost all of this from Richard Taylor. One reference for some of this stuff is his papers with Fred Diamond from about 1992. If anyone finds any inaccuracies in these notes or has any comments I’d love to hear about them, for example by email to

buzzard@dpmms.cam.ac.uk.

§1. Introduction.

This section is much more vague than the others, and is an attempt to explain what the point of this note is.

The Jacquet-Langlands theorem says that if you have two quaternion algebras, then certain automorphic forms for one of them are in canonical bijection with certain automorphic forms for the other. As it stands this statement is a bit general, as well as being a bit vague, and it also has the merit of being fairly incomprehensible to many number theorists. The point is that M_{2}(ℚ) is a quaternion algebra, and classical modular (eigen)forms can be viewed as automorphic forms for M_{2}(ℚ). So if one translates the statement of the JL theorem down a bit, one gets that certain classical modular forms should be related to certain “modular forms” on other quaternion algebras. In some cases, though, the definition of a modular form for a quaternion algebra is (surprisingly) much easier than the definition of a classical modular form.

In fact, if D is a definite quaternion algebra (all definitions are to come, of course) then the modular forms for D are rather concrete algebraic objects. Vaguely speaking, classical modular forms might be thought of as an “H^{1}” (via the Eichler-Shimura isomorphism), and automorphic forms for a definite quaternion algebra can be thought of as an “H^{0}” and we all know that H^{0}s are easier than H^{1}s. Another way of saying why definite quaternion algebras are easy is that classical modular forms can be thought of as H^{0} of a certain coherent sheaf on a (modular) curve, whereas modular forms for a definite quaternion algebra are H^{0} of a sheaf on a zero-dimensional thing (a finite set of points), and are hence no more than the sum of the fibres over each point.

My feeling is that computing with definite quaternion algebras must be much easier than with classical modular forms. Of course, one drawback is that one doesn’t see all classical forms in this way (for example one will never find Δ—remember that only some automorphic forms on one quaternion algebra will correspond to automorphic forms on another one) but one can see many, and at least it’s good for examples.

§2. Some definitions.

Now we become more precise. Let K be a field.

Definition. A quaternion algebra over K is a (necessarily non-commutative!) ring D equipped with an injective ring map K → D, satisfying the following 3 axioms:

1) (The image of) K is the centre of D

2) The dimension of D, considered as a K-vector space, is 4.

3) D has no non-trivial 2-sided ideals.

The canonical example of a quaternion algebra is D = M_{2}(K), with K being embedded as the scalar matrices.

Definition. A quaternion algebra over K which is isomorphic to M_{2}(K) is said to be split.

Of course, there are examples of non-split quaternion algebras. For example, ℍ = ℝ ⊕ ℝi ⊕ ℝj ⊕ ℝk with the usual relations holding on i, j and k, is a quaternion algebra over ℝ.

We sometimes abbreviate “Let D be a quaternion algebra over the field K” by “Let D/K be a quaternion algebra”.

The actual definition of a quaternion algebra looks a bit daunting, but we shall never actually use it, I just included it for completeness. What I shall do next is simply to state, for various fields, what all the quaternion algebras are up to isomorphism, and so if you are prepared to take things on trust then you never need even digest what the definition of a quaternion algebra is.

§3. First examples of quaternion algebras.

1) If K is algebraically closed, then there is only 1 quaternion algebra over K up to isomorphism, namely M_{2}(K), the split one.

2) If K is finite then there is also only one quaternion algebra over K and it’s of course the split one again.

3) If K is ℝ then there are two quaternion algebras (up to isomorphism), namely M_{2}(ℝ) and the quaternions ℍ as above.

4) If K is a finite extension of ℚ_{p} then there are again 2, namely M_{2}(K) and a very explicit second one which I shall write down just out of interest, although it’s not essential to know what it is. We know that finite unramified extensions of K correspond bijectively to finite extensions of the residue field of K. Now the residue field of K has a unique extension of degree 2, and hence there is a unique extension L/K of degree 2 which is unramified. Now let D be the subring of M_{2}(L) consisting of matrices

( \begin{matrix}a & b \\ πσ(b) & σ(a) \end{matrix} )

where a and b are arbitrary elements of L, π is a uniformiser of K (and hence of L too) and σ is the non-trivial element of Gal(L/K). This ring turns out to be a quaternion algebra over K which is not split, and as I say, it’s the only one, up to isomorphism.

Note that this quaternion algebra is usually not K ⊕ Ki ⊕ Kj ⊕ Kk with i, j and k satisfying the usual relations; even though this latter thing is a quaternion algebra, it is actually isomorphic to M_{2}(K) a lot of the time. For example, it’s a neat exercise to show that if K = ℚ_{p} for p odd, then this latter quaternion algebra is always M_{2}(K).

5) For more specific local fields, one can be a bit more concrete. For example, if K = ℚ_{p} with p odd, then here’s another description of the non-split quaternion algebra over K.

Choose \overline{u} ∈ (ℤ/pℤ)^{×} a non-square, and lift it to u ∈ ℤ_{p}. Then set D = ℚ_{p} ⊕ ℚ_{p}α ⊕ ℚ_{p}β ⊕ ℚ_{p}αβ, where we now have to explain how to multiply everything together to make this a ring. We ask that α^{2} = u, β^{2} = pu and αβ = −βα. It is now easily checked that we can extend this to a unique associative multiplication on D, and moreover it turns out that D is a non-split quaternion algebra over ℚ_{p}.

If K = ℚ_{2} then it turns out that the non-split quaternion algebra is isomorphic to ℚ_{2} ⊕ ℚ_{2}i ⊕ ℚ_{2}j ⊕ ℚ_{2}k with the usual relations.

§4 Base extension, traces and norms.

If D/K is a quaternion algebra over a field, and L is a field extension of K, then D ⊗_{K} L is a quaternion algebra over L.

Definition. Say that L splits D if D ⊗_{K} L is split, that is, isomorphic to M_{2}(L).

Note that D may well not be split, but could become split after a field extension. For example, one can take the non-split “usual” quaternions over ℝ and then tensor up to ℂ and they must become split, because there is only one quaternion algebra over ℂ. More generally, if D/K is a quaternion algebra and \overline{K} is the algebraic closure of K then \overline{K} must split D because there is only one quaternion algebra over \overline{K}.

Let K be an arbitrary field, and let D/K be a quaternion algebra. Choose an isomorphism D ⊗_{K} \overline{K} ≅ M_{2}(K), and note that this gives us a way of thinking of D as a subring of M_{2}(\overline{K}). In particular, elements of D now have traces and determinants.

Definition. If d ∈ D, define the trace of d to be the trace of d considered as an element of M_{2}(\overline{K}). Also define the norm of D to be the determinant of d considered as an element of M_{2}(\overline{K}).

Theorem. The norm and trace of an element of d are in K, and are independent of the choice of isomorphism D ⊗_{K} \overline{K} ≅ M_{2}(K).

§5 Quaternion algebras over number fields.

Let K be a number field and let D be a quaternion algebra over K. If v is a place of K then one can tensor up to the completion K_{v} of K at v, and ask whether or not this extension splits D. Let S(D) denote the set of places of K for which D does not split. Rather surprisingly, it turns out that knowing S(D) is all you need to know D, as explained by the following

Theorem.

(a) The set S(D) is a finite set of places of K, with an even number of elements, none of them complex.

(b) Two quaternion algebras D and E over K are isomorphic if and only if S(D) = S(E).

(c) If S is any finite set of places of K containing no complex places, and S has an even number of elements, then there is exactly one quaternion algebra D/K such that S(D) = S.

Put another way, there is a bijection between the set of isomorphism classes of quaternion algebras over K and the set of finite sets of non-complex places of K with even size.

For example, if D = M_{2}(K) then D splits at every completion of K and hence S(D) is the empty set. As another example, if K = ℚ and D = ℚ ⊕ ℚi ⊕ ℚj ⊕ ℚk, then D is not split, and it turns out that D splits over ℚ_{p} if and only if p is odd, and so S(D) = {2, ∞}.

To simplify things, let’s now set K = ℚ. So let D/ℚ be a quaternion algebra.

Definition. The discriminant d = disc(D) of D is the product of the primes p in S(D).

Of course, by a prime, I mean a finite prime! By the theorem above, we see that disc(D) uniquely determines D, and moreover that every square-free positive integer d is the discriminant of a unique quaternion algebra over ℚ.

Given a square-free positive integer d, it is easy to work out whether or not D ⊗_{ℚ} ℝ is split or not, because one knows that the total number of places in S(D) is even, and so one can count the number of primes dividing d, and this number is even iff ∞ ∉ S(D).

Definition. A quaternion algebra D/ℚ is definite if ∞ ∈ S(D), that is, if D ⊗_{ℚ} ℝ ≢ M_{2}(ℝ).

Karsten once pointed out to me that a way to remember this is that definite quaternion algebras are definitely not M_{2}(ℚ).

Once I wondered how one could “write down” a quaternion algebra with discriminant any given squarefree number d. I managed to do it—it’s probably well-known, but I don’t know a reference other than my notes. My first construction was to find an explicit subring of M_{2}(\overline{ℚ}) that worked. More recently I noticed that one could look instead for quaternion algebras of the form D = ℚ ⊕ ℚα ⊕ ℚβ ⊕ ℚαβ with α^{2} = a, β^{2} = b, αβ = −βα, and a and b cunningly chosen. For example, setting a = −1 and b = −p where p is a prime congruent to 3 mod 4 yields the quaternion algebra of discriminant p, and as I’ve already mentioned, setting a = b = −1 gives the quaternion algebra of discriminant 2.

§6. Orders.

Our aim in the next couple of sections is to define modular forms over definite quaternion algebras. As I remarked above, this is actually not too hard, although it will take a while. First I need to talk about orders. I shall be lazier now and just talk about quaternion algebras over ℚ and ℚ_{p} from now on, because this is all that we shall need.

Let p be a prime and let D/ℚ_{p} be a quaternion algebra.

Definition. An order in D is a subring O ⊂ D such that

(a) O ≅ ℤ_{p}^{4} as an abelian group, and

(b) O ⊗_{ℤ_{p}} ℚ_{p} = D.

An order is said to be maximal if it is not properly contained in any larger order.

It is a theorem that maximal orders exist (and this is not a Zorn argument!). For example, M_{2}(ℤ_{p}) is a maximal order in M_{2}(ℚ_{p}), and so are all its conjugates. In fact, any maximal order of M_{2}(ℚ_{p}) is conjugate to M_{2}(ℤ_{p}). If D is the non-split quaternion algebra over K then D has a unique maximal order, consisting of the elements of D with integral norm.

Now let’s consider the case K = ℚ.

Definition. An order in a quaternion algebra D/ℚ is a subring O of D such that (a) O ≅ ℤ^{4} as an abelian group, and

(b) O ⊗_{ℤ} ℚ = D.

It is a theorem that every quaternion algebra over ℚ has maximal orders.

As an example, if D = M_{2}(ℚ) then O = M_{2}(ℤ) is a maximal order, as are its conjugates. In fact, if D is any indefinite quaternion algebra over ℚ then all the maximal orders of D are conjugate.

As another example, if D = ℚ ⊕ ℚi ⊕ ℚj ⊕ ℚk with the usual relations on i, j and k, then O = {a + bi + ck + dk | a, b, c, d ∈ ℤ} is an order in D, and one might expect that it’s a maximal order, just as one might expect that ℤ[\sqrt{5}] might be the integers in ℚ(\sqrt{5}) when one first starts doing algebraic number theory. However O is not maximal: there is a bigger order, namely \{a + bi + cj + dk | either a, b, c, d ∈ ℤ or a, b, c, d ∈ ℤ + \frac{1}{2}\}, and in fact this latter order is maximal.

If D/ℚ is definite, then there are only a finite number of conjugacy classes of maximal orders, but there could be more than one.

Definition. If D/ℚ is a quaternion algebra, and p is a prime, then write D_{p} for the quaternion algebra D ⊗_{ℚ} ℚ_{p} over ℚ_{p}. If O ⊂ D is an order, write O_{p} for the order O ⊗_{ℤ} ℤ_{p} of D_{p} (it’s not hard to check that O_{p} is in fact an order in D_{p}).

It’s a neat exercise to prove that if D/ℚ is a quaternion algebra and O ⊂ D is an order, then O is maximal iff O_{p} is maximal in D_{p} for all p.

§7 Enter the adeles.

Let D be a quaternion algebra over ℚ and fix a maximal order O of D. Later we shall assume that D is definite but at the minute we don’t need to. Let 𝔸^{f} denote the finite adeles, that is, the restricted product of ℚ_{p} for all primes p. Even more frightening, let D^{f} denote D ⊗_{ℚ} 𝔸^{f}. We shall now give a more down-to-earth description of D^{f}. Choose an order O ⊂ D. Then an element of D^{f} can be thought of as the choice of an element d_{p} ∈ D_{p} for every prime p, subject to the restriction that d_{p} ∈ O_{p} for but finitely many p. Note that this does not depend on the choice of O because any two orders will differ only at a finite number of primes.

I should say something about topologies, but I fear it will be unhelpful. The adeles have a topology on them. If I choose a ℚ-basis for D then D^{f} becomes isomorphic to (𝔸^{f})^{4} and we can give it the product topology induced from the adeles. This makes D^{f} into a topological ring. Now one follows the usual procedure to make (D^{f})^{×} into a topological group—one embeds it into (D^{f})^{2} via x ↦ (x, x^{−1}) and then takes the subspace topology of the product topology.

On a more practical level, one can work prime by prime. Choose an order O ⊂ D. Then O_{p} ≅ ℤ_{p}^{4}, and so one can put a p-adic topology on O_{p} to make this an isomorphism of topological groups. Give (O_{p})^{×} the subspace topology, and set K = ∏_{p}(O_{p})^{×} with the product topology. Note that K ⊂ D^{f} as a set, and in fact the topology that we just put on K is the subspace topology. In particular, K is compact, it being the product of compact spaces.

Moreover, it turns out that K is a compact open subgroup of D^{f}. Knowing this actually tells us exactly what the topology on D^{f} is, because D^{f} is just the disjoint union of translates of K.

To be even more concrete, we could take D = M_{2}(ℚ) and O = M_{2}(ℤ). Then K = GL_{2}(ℤ̂) is a compact open subgroup.

In fact, there are a wealth of compact open subgroups of D^{f}. For example, choose any finite set of primes and for each such prime choose an open compact subgroup K_{p} of (D_{p})^{×}. So for example if p was split then (D_{p})^{×} ≡ GL_{2}(ℚ_{p}) and K_{p} could be the matrices in GL_{2}(ℤ_{p}) which are congruent to the identity modulo p^{n} for some n. For the primes not in this chosen finite set, set K_{p} = (O_{p})^{×}. Then again the product K = ∏_{p} K_{p} will be an open compact subgroup of (D^{f})^{×}.

In practice, again, one doesn’t really need to understand exactly what the topology is, one just needs to know some examples. By the way, not all the open compact subgroups will be expressible in this “product” form, but essentially all of the open compact subgroups that we are interested in will be.

Finally, let D/ℚ be a definite quaternion algebra, and let K ⊂ (D^{f})^{×} be an open compact subgroup. The important result is that

Theorem. The double coset space D^{×}\(D^{f})^{×}/K is finite.

This finite set is the analogue of a modular curve in the definite quaternion algebra set, although I’m actually too lazy to explain why at the minute. It turns out that the space of modular forms of weight k and “level K” can just be thought of as the complex vector space of functions from this finite set to a certain finite-dimensional complex vector space. Moreover, the space of “mod p modular forms for D of weight k and level K” can be thought of as the space of functions from this finite set to a certain finite-dimensional vector space over 𝔽_{p}.

§7. Examples.

Rather than muck about with the “easiest” case of M_{2}(ℚ), I shall just start with an explicit definite quaternion algebra and a compact open subgroup, and try to work out the finite set mentioned at the end of the last section. Motivated by the fact that one of the reasons I am typing all this up is a detailed study of the quaternion algebra over ℚ of discriminant 11, I shall use this one as our example. So let D be the quaternion algebra ℚ ⊕ ℚi ⊕ ℚβ ⊕ ℚiβ where i^{2} = −1, β^{2} = −11 and βi = −iβ. Ones first guess at a maximal order is ℤ ⊕ ℤi ⊕ ℤβ ⊕ ℤiβ, but again we are slightly wrong. The localisation of this order is maximal at all odd primes, but not at p = 2. It turns out that we can adjoin γ := (1 + β)/2 and iγ = (i + iβ)/2, and consider O = ℤ ⊕ ℤi ⊕ ℤγ ⊕ ℤiγ. Note that γ^{2} = γ − 3 and γi = i − iγ so O is a ring.

The group D^{×} is just D\{0}, because D is not split (maybe I should mention this earlier?). To make things easier, let’s make the following

Definition. If K ⊂ (D^{f})^{×} is an open compact subgroup, let X(K) denote the finite set D^{×}\(D^{f})^{×}/K.

We have chosen D to be the quaternion algebra with discriminant 11. Now let’s choose a compact open K. Firstly we can try the “easiest” K, namely K = ∏_{p}(O_{p})^{×}. We ask ourselves the following question: what is X(K) in this instance?

The object of this section is to show that it is actually possible in some cases to explicitly work out the set D^{×}\(D^{f})×/K for K an explicitly-given compact open subgroup.

STUFF STILL TO BE WRITTEN UP HERE

§?. Modular forms for definite quaternion algebras.

STUFF STILL TO BE WRITTEN HERE. DO THE COMPLEX CASE FIRST. THEN DO EVERYTHING MOD P MAYBE INSTEAD OF OVER ZP?

Fix D/ℚ and O and now let’s assume that D is definite.

Choose once and for all a prime p for which D is unramified (that is, such that ℚ_{p} splits D). I’ll define now a module which is free of finite rank over ℤ_{p} which can be thought of as the space of modular forms over D. This space is analogous to the space of modular forms for a congruence subgroup of

Choose a maximal order O in D. Choose an isomorphism D ⊗ ℚ_{p} ≡ M_{2}(ℚ_{p}) which identifies O ⊗_{ℤ} ℤ_{p} with M_{2}(ℤ_{p}) (this is always possible, because first of all you choose an isomorphism D ⊗ ℚ_{p} ≡ M_{2}(ℚ_{p}) and then you can conjugate it until it sends the order where you want it to go).

Note that (the multiplicative monoid!) M_{2}(ℤ_{p}) acts (on the left) on (ℤ_{p})^{2} and hence on the symmetric powers of (ℤ_{p})^{2}. This action can be seen in a more down-to-earth way as follows. Let g ≥ 0 be an integer, and let V_{g} denote the space of homogenous polynomials over ℤ_{p} in 2 variables X and Y with total degree g. Then V_{g} is isomorphic to the gth

symmetric power of (ℤ_{p})^{2}. One can think of ( \begin{matrix}a & b \\ c & d \end{matrix} ) as acting on V_{g} by sending X to

aX + cY and Y to bX + dY, and then extending in the obvious way, sending X^{n} to (aX + cY)^{n} etc. So let U a compact open subgroup of (D ⊗_{ℚ} 𝔸^{f})^{×}, and let k ≥ 2 be an integer. Define L_{k}(U; ℤ_{p}) to be the set of functions f : (D ⊗_{ℚ} 𝔸^{f})^{×} → Symm^{k−2}((ℤ_{p})^{2}) such that

1) f(tg) = f(g) for all t ∈ D^{×}, and

2) f(gu) = (u_{p})^{−1}f(g) for all u ∈ U.

Here (u_{p}) denotes the component of u at p, and we’re identifying D ⊗_{ℚ} ℚ_{p} with M_{2}(ℚ_{p}). I guess I’d better assume that the p-part of U is a subgroup of GL_{2}(ℤ_{p}) or I don’t have an action, so assume this.

Well, what is this mess? It’s certainly a ℤ_{p}-module. Moreover, it’s actually finite and free. This is because D^{×}\(D ⊗_{ℚ} 𝔸^{f})^{×}/U is a finite set. This set is some kind of artificial realisation of the H^{0} of a certain l-adic sheaf on the 0-dimensional non-connected complex manifold corresponding to this finite set.

Well, I can define Hecke operators on this space, and also prove a comparison theorem between this space and another finite-dimensional space, namely that of modular forms of level “U” in some sense which are special or supercuspidal at all primes dividing d. I’ve drifted off again into vagueland so here’s some explanation of what’s going on here.

MORE STUFF TO BE WRITTEN HERE

Kevin Buzzard, May 1998.