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Class Numbers of Orders in Quartic Fields

It is shown that the sum of class numbers of orders in totally complex quartic fields with no real quadratic subfield obeys an asymptotic law similar to the prime numbers, as the bound on the regulators tends to infinity. Here only orders which are maximal at a given set of primes containing an even number of elements are considered. The proof is accomplished by means of a Prime Geodesic Theorem for symmetric spaces formed as compact quotients of SL(4,R). This result extends work of Sarnak in the real quadratic case and of Deitmar and Hoffmann in the complex cubic case.


Introduction
Let D be the set of all natural numbers D ≡ 0, 1 mod 4 with D not a square. Then D is the set of all discriminants of orders in real quadratic fields. For D ∈ D the set is an order in the real quadratic field Q( √ D) with discriminant D. As D varies, O D runs through the set of all orders of real quadratic fields. For D ∈ D let h(O D ) denote the class number and R(O D ) the regulator of the order O D . It was conjectured by Gauss ([15]) and proved by Siegel ([34]) that, as x tends to infinity where ζ is the Riemann zeta function. For a long time it was believed to be impossible to separate the class number and the regulator in the summation. However, in [32], Theorem 3.1, Sarnak showed, using the Selberg trace formula, that as More sharply, The Prime Geodesic Theorem in this context gives an asymptotic formula for the number of closed geodesics on the surface SL 2 (Z)\H with length less than or equal to x > 0. This formula is analogous to the asymptotic formula for the number of primes less than x given in the Prime Number Theorem. The Selberg zeta function (see [33]) is used in the proof of the Prime Geodesic Theorem in a way analogous to the way the Riemann zeta function is used in the proof of the Prime Number Thoerem (see [5]). The required properties of the Selberg zeta function are deduced from the Selberg trace formula ( [33]).
It seems that following Sarnak's result no asymptotic results for class numbers in fields of degree greater than two were proven until in [9], Theorem 1.1, the first named author proved an asymptotic formula for class numbers of orders in complex cubic fields, that is, cubic fields with one real embedding and one pair of complex conjugate embeddings. This result can be stated as follows.
Let S be a finite set of prime numbers containing at least two elements and let C(S) be the set of all complex cubic fields F such that all primes p ∈ S are non-decomposed in F . Then as x → ∞ we have More sharply, This result was again proved by means of a Prime Geodesic Theorem, this time for symmetric spaces formed as a compact quotient of the group SL 3 (R). There exists a Selberg trace formula for such spaces ( [36]) by means of which the required properties of a generalised Selberg zeta function can be deduced ( [8]) in order to prove the Prime Geodesic Theorem. The class number formula is then deduced by means of a correspondence between primitive closed geodesics and orders in complex cubic number fields, under which the lengths of the geodesics correspond to the regulators of the number fields.
In this paper we follow the methods of [9]. We make use of the prime geodesic theorem in [13]. Let us quote this result as follows.
Let G = SL 4 (R) and let K be the maximal compact subgroup SO (4). Let Γ ⊂ G be discrete and cocompact. We then have a one to one correspondence between conjugacy classes in Γ and free homotopy classes of closed geodesics on the symmetric space X Γ = Γ\G/K. Let and B = SO(2) SO (2) .
Let E (Γ) be the set of primitive conjugacy classes [γ] in Γ such that γ is conjugate in G to an element a γ b γ of A − B. For γ ∈ Γ we write a γ also for the top left entry in the matrix a γ and define the length l γ of γ to be 8 log a γ . Let G γ , Γ γ be the centralisers of γ in G and Γ respectively and let K γ = K ∩ G γ . For x > 0 define the function π(x) = where χ 1 (Γ γ ) is the first higher Euler characteristic of the symmetric space X Γγ = Γ γ \G γ /K γ .
Theorem 0.1 (Prime Geodesic Theorem) For x → ∞ we have More sharply, log t dt is the integral logarithm.
We now state our result on class numbers. Let S be a finite, non-empty set of prime numbers containing an even number of elements. We define the sets C(S) and O(S) and the constants λ S (O) as above, replacing complex cubic fields with totally complex quartic fields in the definitions. A totally complex quartic field has at most one real quadratic subfield, as can be seen by comparing numbers of fundamental units. Let Then as x → ∞ we have Note the surprising factor 1/2! We give an explanation below.
Transfering the methods of [9] to the case of totally complex quartic fields does not proceed entirely smoothly, there are various technical difficulties to be overcome. For instance, the correspondence between primitive closed geodesics and orders is considerably more complicated than in either the real quadratic or the complex cubic case. As a result of this extra complexity our result is weaker than that obtained in the complex cubic case. We have had to impose the extra condition on the finite set S of prime numbers that it must contain an even number of elements and we have been unable to provide an error term in our final asymptotic.
Also we have had to restrict ourselves to counting the orders in fields without a real quadratic subfield, as the fields with a real quadratic subfield cannot be counted using our method. Indeed, any real quadratic field which occurs as a subfield of a totally complex quartic field may in fact occur as a subfield of infinitely many such fields. The fundamental units in the quadratic field are powers of fundamental units in the quartic fields, so the (possibly infinitely many) quartic fields all have regulator less than or equal to that of the quadratic field. Hence an asymptotic formula for a sum of class numbers of orders bounded by the regulators is not even possible in the case of totally complex quartic fields with a real subfield.
In comparison with the real quadratic and complex cubic cases one might have expected to get the asymptotic π S (x) ∼ e 4x 4x . In the result that we do in fact get there is an extra factor of one half. The fact that we have had to restrict the fields over which we are counting gives an explanation for this discrepancy.
We also prove the following asymptotic in which we introduce an extra factor into the summands.
where µ O is the number of roots of unity in O, the sum is over the 2µ O different fundamental units of O and the product is over the embeddings of O into C. We have: Then as x → ∞ we haveπ 2x .
An element γ in Γ is called regular if it centraliser in G is a torus and non-regular otherwise. The factor ν(O) was originally introduced in order to separate the contribution of non-regular elements from that of the regular elements in the Prime Geodesic Theorem. As it turned out, the complexity of the correspondence between geodesics and orders meant that we did not really gain anything from this approach. However, the factor ν(O) contains information about the arguments of the fundamental units in the order O which is interesting in its own right. In comparison with Theorem 0.2 we can see that "on average" the value of ν(O) as R(O) goes to infinity is 4. If ε is a fundamental unit in O with arguments θ, −θ, φ, −φ under the four embeddings of O into C then a simple calculation shows that Since this takes "on average" the value 4 we can see that there is a sense in which we can say that the arguments of the fundamental units in the orders O are evenly distributed about ±π/2 as R(O) goes to infinity.
In what follows we give a summary of the arguments and techniques used in the proof of our main results. In particular we shall point out the difficulties that arise in applying the methods of [9] to our situation.
In Section 1 we use a classification of the set of equivalence classes of division algebras over Q by means of a description of the Brauer group of Q ( [28], Theorem 18.5) to show the existence of a division algebra M (Q) whose maximal subfields include the fields in the set C c (S) and such that M (Q) ⊗ R ∼ = Mat 4 (R). In fact the maximal subfields of M (Q) are all quartic extensions of Q and the set of totally complex maximal subfields of M (Q) coincides with C(S). We further show that for an order O in O(S) the number of embeddings of O into M (Q), up to conjugation by the unit group of the maximal order M (Z), is λ S (O)h(O). The restriction on the set S of prime numbers that it has to contain an even number of elements is a consequence of the classification of division algebras over Q.
In Section 2 we prove a correspondence between the primitive, closed geodesics in the Prime Geodesic Theorem and the orders of totally complex quartic fields. Under this correspondence the lengths of the geodesics correspond to the regulators of the orders. It turns out that it is the regular, weakly neat elements of E (Γ) which correspond to orders in totally complex quartic fields with no real quadratic subfield. We use a "sandwiching" argument to isolate these elements in the Prime Geodesic Theorem.
We finish this introduction with a few remarks about the limitations of the method used here in terms of further applications and mention a couple of other recent results in the same direction. In order to be able to make use of the trace formula for compact spaces we have had to limit our sum over class numbers by means of the choice of a finite set of primes, as described above. In [12] Deitmar and Hoffmann have been able to use a different trace formula to prove that as where the sum is over all isomorphism classes of orders in complex cubic fields. In order to get the error term in the Prime Geodesic Theorem we have made use of the classification of the unitary dual of SL 4 (R). At present the unitary dual is not known for any higher dimensional groups so a Prime Geodesic Theorem with error term is not possible using our methods. Finally we mention that the correspondence between geodesics and orders actually works by identifying primitive geodesics with fundamental units in orders. By Dirichlet's unit theorem, an order in a number field F has a unique fundamental unit (up to inversion and multiplication by a root of unity) only if F is real quadratic, complex cubic or totally complex quartic. Hence an asymptotic of our form can be proven only in these three cases. In [10], the first named author has proved a Prime Geodesic Theorem for higher rank spaces, from which he deduces an asymptotic formula for class numbers of orders in totally real number fields of prime degree.
1 Division Algebras of Degree Four

Central simple algebras and orders
Let F be a field. Let C(F ) denote the set of isomorphism classes of finite dimensional central simple algebras over F and D(F ) the set of isomorphism classes of finite dimensional division algebras with centre F , then D(F ) ⊂ C(F ).
For convenience we collect together here a number of facts about central simple algebras and their orders, which will be used in the sequel. Let F be a field, K an extension of F and A a finite dimensional algebra over F .
Proof: All the statements of the proposition are proved in [28]. Statement (a) is Corollary 5.5b; (b) is Lemma 9.4; (c) is Proposition 12.4b(ii) and (d) is Theorem 12.7.
The next proposition gives some results about subfields of algebras.
Proof: All the statements of the proposition are proved in [31]. Statement (a) is Theorem 7.21; (b) is Theorem 8.6; (c) is Corollary 11.6 and (d) is Theorem 12.8.

Division algebras of degree four
Let A be a central simple algebra over a field F . By the Wedderburn Structure Theorem ( [28], Theorem 3.5) the algebra A is isomorphic to Mat n (D), where n ∈ N and D is a finite dimensional division algebra over F , which by [28], Proposition 12.5b is unique up to isomorphism. If B is another central simple algebra over F isomorphic to Mat m (E), where m ∈ N and E is a finite dimensional division algebra over F , then A and B are called Morita equivalent if and only if D ∼ = E. By [28], Proposition 12.5a, the tensor product over F induces a group structure on the set of equivalence classes. The group defined in this way is called the Brauer group.
By [28], Theorem 17.10, if F is a local, non-archimedean field, then there is a canonical isomorphism between the Brauer group of F and the group Q/Z. The Brauer invariant Inv A of A is defined to be the image in Q/Z of the Morita equivalence class of A under this isomorphism. The Schur index Ind A of A is defined to be the degree of D. By [28], Corollary 17.10a(iii), the order of the Brauer invariant of A in Q/Z equals Ind A.
The only (associative) division algebras over R are R itself, C and the quaternions H (see [28], Corollary 13.1c). Since C is itself a field, any algebra which is central simple over R cannot be isomorphic to a matrix algebra over C. The Brauer group of R is therefore isomorphic to Z/2Z. Let A be a central simple algebra over R. If A ∼ = Mat n (R) for some n ∈ N, then the Brauer invariant of A is defined to be zero and the Schur index of A is defined to be one. If A ∼ = Mat n (H) for some n ∈ N, then the Brauer invariant of A is one half and the Schur index of A is two.
Let M be a division algebra of degree 4 over Q. Let S be a finite, non-empty set of prime numbers with an even number of elements. We say that . For all primes p we define i p as follows. If p ∈ S define i p to be either 1 4 or 3 4 in such a way that p∈S i p ∈ Z. Note that we are able to do this since we have specified that S must contain an even number of elements. For all other p define i p = 0. Then [28], Theorem 18.5 tells us that we may choose M so that Inv M (Q p ) = i p for all primes p and Inv M (R) = 0. We can then see that the set of places at which M (Q) does not split coincides with the set S. More particularly, for p ∈ S we have that M (Q p ) is a division Q p -algebra, for p / ∈ S we have that M (Q p ) ∼ = Mat 4 (Q p ) and we also have M (R) ∼ = Mat 4 (R).
For a commutative ring R with unity, let Then G is a linear algebraic group, defined over Z. Let Γ = G (Z), then Γ forms a discrete subgroup of G = G (R) ∼ = SL 4 (R). Since M (Q) is a division algebra, it follows that G is anisotropic over Q and so Γ is cocompact in G (see [2], Theorem A). Let P be the parabolic subgroup .
be the subset of primitive conjugacy classes.
We say g ∈ G is regular if its centraliser is a torus and non-regular (or singular ) otherwise. Clearly, for γ ∈ Γ regularity is a property of the Γ-conjugacy class [γ], we denote by E p,reg P (Γ) the regular elements of E p P (Γ). By abuse of notation we sometimes write γ ∈ E p P (Γ) or γ ∈ E p,reg P (Γ) when we mean the conjugacy class of γ.
We will call a quartic field extension F/Q totally complex if it has two pairs of conjugate complex embeddings and no real embeddings into C.
The centraliser M (Q) γ of γ in M (Q) is a totally complex quartic field if and only if γ is regular. The centraliser M (Q) γ is a quaternion algebra over the real quadratic field Q[γ] if and only if γ is non-regular.
Depending on whether neither, one or both of the SO(2) components of b γ are ±I 2 , the dimension dim R M (R) aγ bγ is equal to 4, 6 or 8 respectively.

Field and order embeddings
In this section we prove a number of lemmas which will be needed later.
is quadratic or is totally complex quartic, or if u = ±1, then det (u) = 1.
The field norm N F |Q : F → Q can be written as a product over all embeddings of F into the complex field, N F |Q (x) = σ σx, and it is standard that u is a unit in O F if and only if N F |Q (u) = ±1. Proposition 16.2a of [28] tells us that where deg M (Q) = k[F : Q]. The first statement of the lemma follows. If u = ±1 (so that F = Q) or F is quadratic then by (1) we have det (u) = 1. Suppose F is a totally complex quartic extension. Let f (X) = X 4 + a 3 X 3 + a 2 X 2 + a 1 X + a 0 be the minimal polynomial of u over Q. Then det (u) = a 0 . If a 0 = det (u) = −1 then f must have a real root, which contradicts the supposition on F . Hence det (u) = 1.
The following lemma holds for any algebraic extension F of Q. Lemma 1.6 Let O be an order in the number field F/Q and p a prime number.
Then O p is the maximal order of F p for all but finitely many primes p.
Since F ⊂ pO M for only finitely many primes p, the lemma follows.
Let F be a number field. A prime number p is called non-decomposed in F if there is only one place in F lying above p.
Proof: The statement about the degree of F is Proposition 1.1 (c). The case F = Q is trivial.
Let F be a quartic field and suppose that F embeds into M (Q), then F is a strictly maximal subfield of M (Q). We say that a subfield K of M (Q) is a splitting field for M (Q) (or that K splits M (Q)) if M (Q) ⊗ K ∼ = Mat 4 (K). Proposition 1.2(d) tells us that a quartic field embeds into M (Q) if and only if it splits M (Q).
For p a rational prime, the field Q p is an extension of Q. Hence, by Proposition 1.1(c), the Q p -algebra M (Q p ) is central simple over Q p . Let p ∈ S. The Schur index, Ind M (Q p ), of M (Q p ) is, by [28], Corollary 17.10a(iii), equal to the order of [M (Q p )] in the Brauer group of Q p , where the square brackets denote the Morita equivalence class. By choice of M (Q) we have Ind M (Q p ) = 4. Let p i be the (finitely many) primes in F above p. Then Theorem 32.15 of [31] tells us that, since F splits M (Q), for all i. Corollary 8.4 of Chapter II of [27] says that which implies that there is only one prime p in F above p and that [F p : Q p ] = 4. Conversely, suppose that for each p ∈ S there is only one prime p in F above p. From (2) we get that For p / ∈ S, by choice of M (Q), the Schur index, ind M (Q p ), is equal to 1, so in this case (3) also holds for each p in F above p. Then, by [31], Theorem 32.15, the field F splits M (Q). By Proposition 1.2(d) we then deduce that F embeds into M (Q). Now let F be a quadratic field and suppose that F embeds into M (Q), then we consider F as a subfield of M (Q). By Proposition 1.2(c), the subfield F of M (Q) is not maximal so there exists a maximal subfield K of M (Q) properly containing F , which by the same proposition is quartic. It was shown above that every prime in S is non-decomposed in K and hence also in F .
Suppose p is non-decomposed in F for all p ∈ S. We choose a quaternion algebra A over F by specifying that the local Brauer invariant at p be 1 2 for all places p in F over some p ∈ S, and that the local Brauer invariant be 0 at all other places of F . Proposition 1.2(c) tells us that there exists a subfield K of A containing F with [K : F ] = 2. Then by the same argument as above, if p is a place of F over p for some p ∈ S, then p is non-decomposed in K. It follows that p is non-decomposed in K for all p ∈ S. It was shown above that the quartic extension K of Q, and thus also the subfield F of K, embeds into M (Q). The lemma is proven.
Let A fin denote the ring of finite adeles over Q, ie.
where p ranges over the rational primes and ′ denotes the restricted product, that is, almost all components of a given element are integral. LetẐ = p Z p ⊂ A fin . There is a natural embedding of Q into A fin , which maps q ∈ Q to the adele every one of whose components is q. From this embedding we derive an embedding of M (Q) in M (A fin ) and an embedding of G (Q) in G (A fin ).
Proof: We have the following commutative diagram with exact rows: for all x ∈ Z p . It follows that the map det : M (Ẑ) × →Ẑ × is surjective. From this we can then see that map det : there exist r ∈Ẑ × and q ∈ Q × such that w = rq. By the surjectivity of det , there existr ∈ M (Ẑ) × andq ∈ M (Q) × such that detr = r and detq = q. Letv =r −1 vq −1 . Then detv = 1, sov ∈ G (A fin ). By the Strong Approximation Theorem (see [25]), we have that G (Q) is dense in G (A fin ). Hence there existr ∈ G (Ẑ) andq ∈ G (Q) such that v =rq. Finally, we have v =rvq =r(rq)q = (rr)(qq), whererr ∈ M (Ẑ) × andqq ∈ M (Q) × . The lemma follows. Let σ : F → M (Q) be an embedding of the field F into M (Q). By Lemma 1.7 each prime p in the set S is non-split in F . Since for p ∈ S there is only one place in F above p, we can note that the p-adic completion F p = F ⊗ Q p is again a field (see [27], Chapter 1, Proposition 8.3). Firstly we need to show that for all p ∈ S, the completion O σ,p = O σ ⊗ Z p is the maximal order in F p . By Proposition 1.3(c), M (Z p ) is the maximal Z p -order in M (Q p ). Since M (Q p ) is a division algebra, M (Z p ) coincides with the integral closure of Z p in M (Q p ), by Proposition 1.3(d). We can extend σ to an embedding of F p in M (Q p ) and then O σ,p = σ −1 (σ(F p ) ∩ M (Z p )) is the integral closure of Z p in F p , which is the maximal order of F p .
For the converse, let O be an order of F such that the completion O p is maximal for each p ∈ S. Via σ we consider F to be a subfield of M (Q). For any u ∈ M (Q) The embedding u σ is then the one required by the lemma.
Let O 1 = F ∩ M (Z). Since O and O 1 are orders, by Lemma 1.6 they are both maximal at all but finitely many places. Let T be the set of primes p such that either O or O 1 is not maximal at p. Then T is finite and T ∩ S = ∅. Furthermore, if T F denotes the set of places of F lying over T , we have that for any place p of F with p / ∈ T F the completions O p and O 1,p are maximal in F p and thus equal, by uniqueness of maximal orders. Hence, for p / ∈ T we have Let p ∈ T , then p / ∈ S so M (Q p ) ∼ = Mat 4 (Q p ). Considering F p as a Q p -vector space we see that we can embed F p into Lin Qp (F p , F p ) by sending an element x ∈ F p to the linear map a → ax. We choose a Z p -basis of O p , which is then also a Q p -basis of F p . This gives us an isomorphism This isomorphism and the above embedding F p ֒→ Lin Qp (F p , F p ) then give us an embedding σ p : For p / ∈ T set u p = 1 and letũ = (u p ) ∈ A fin . By Lemma 1.8 so there exists an element u ∈ M (Q) × such that for all primes p we have for all primes p and we deduce that This completes the proof of the lemma. For a prime p ∈ Z let F p = F ⊗ Q p . Let p 1 , ...p n be the prime ideals of F above p. By [27], Chapter 1, Proposition 8.3 there exists a canonical Q p -algebra isomorphism defined by:

Counting order embeddings
where F pi denotes the completion of F with respect to the p i -adic absolute value and τ i the embedding Recall that S is a finite, non-empty set of prime numbers with an even number of elements. Let C(S) be the set of all totally complex quartic fields F such that all primes p in S are non-decomposed in F . Note in particular that by the isomorphism (5) this means that for each field F in C(S) and for each p ∈ S we have that F p = F ⊗ Q p is once again a field, namely the completion of F at the unique place of F above p. For F ∈ C(S) let O F (S) be the set of isomorphism classes of orders O in F which are maximal at all p ∈ S, ie. are such that the completion O p = O ⊗ Z p is the maximal order of the field F p for all p ∈ S. Let O(S) be the union of all O F (S), where F ranges over C(S).
Let We define where f p (F ) is the inertia degree of p in F . Our aim is to prove the following proposition. We prepare for the proof of the propostition with the following discussion and then prove some lemmas, from which the proposition will follow.
Fix an embedding σ : Conversely, it is clear that for all u ∈ U we have u σ ∈ Σ(O), and that u σ = v σ if and only if u = vx for some x ∈ F × . Hence Then I u is a finitely generated O-module in F . We shall prove that the map is surjective and λ S (O) to one. This will be done in the following lemmas through localisation and strong approximation. Let p ∈ Z be a prime and let I(O p ) be the set of all finitely generated O p -submodules of F p and [I(O p )] the set of isomorphism classes. Let I p be a non-trivial, finitely generated O p -submodule of F p . Then, by [31], Theorem 10.6, the submodule I p is isomorphic as an O p -module to an ideal of O p so also has the property that I p ⊗ Q p = F p . If J p is another finitely generated O p -submodule of F p , such that J p ∼ = I p , then the isomorphism extends to an automorphism of F p considered as a module over itself, and As in the global case above, F × p acts on U p from the right and M (Z p ) × acts on U p from the left. For Then I up is a finitely generated O p -module in F p .

Lemma 1.11
Let p be a prime. Then |I(O p )/F × p | = 1.
Proof: Suppose p ∈ S. Then p is non-decomposed in F so there exists a unique prime ideal p of F above p, so by the isomorphism (5) we can see that F p ∼ = F p and hence F p is itself a field. By [27], Chapter II, Proposition 3.9, every ideal of O p is principal and hence all non-zero ideals of O p are isomorphic. The claim holds for p ∈ S. Suppose p / ∈ S. Let p 1 , ..., p n be the prime ideals of F over p. Under the isomorphism (5) an ideal in O p gives ideals in O p1 , ..., O pn , where O p1 denotes the ring of integers in the field F pi . As we saw above, these ideals are each unique up to isomorphism and hence the original ideal in O p is also unique up to isomorphism and the claim holds for p / ∈ S.
We now prove the surjectivity of Ψ.

Lemma 1.12
The map Ψ is surjective.
Proof: Let I ⊂ O be an ideal and for a prime p let I p = I ⊗ Z p . By Lemma 1.11, for all primes p the fin . By Lemma 1.8 there existz ∈ M (Ẑ) × and u ∈ M (Q) × such thatũ =zu. Then for all primes p we have that This is what was required to complete the proof of the lemma.
In the following series of lemmas we prove that Ψ is λ S (O) to one. Lemma 1.13 Let K be a non-archimedian local field and O K its ring of integers. For n ∈ N let ∆ n ⊂ Mat n (K) be the subspace of diagonal matrices and suppose that u ∈ GL n (K) is such that Then there exist z ∈ Mat n (O K ) × and x ∈ ∆ × n such that zu = x. If we suppose further that Proof: The first claim of the lemma will be proved by induction on n. For n = 1 we take z = 1 and x = u and the claim holds. Assume now that the claim holds for some n − 1 ∈ N. For any ring R let Upp n (R) denote the subalgebra of upper triangular matrices in Mat n (R). We can choose an element y ∈ Mat n (O k ) × such that u ′ = yu ∈ Upp n (K) is upper triangular. Then u ′ has the form where a ∈ K, b ∈ Mat 1×(n−1) (K) and c ∈ Upp n−1 (K). Then Since u satisfies ∆ n ∩Mat n (O K ) ⊂ ∆ n ∩u −1 Mat n (O K )u, it follows that u ′ satisfies ∆ n ∩Upp n ( where the matrix on the right is the n by n matrix with a one in the top left corner and all other entries zero. It then follows that α = 1, c −1 γc = 0 and b = −βc. Let Then so we have that It is then clear that c satisfies we get that zu = x and the proof of the first claim is complete.
If we also assume that u satisfies This can be the case only if x ∈ (∆ n ∩ Mat n (O K )) × so the lemma is proved.
Proof: Let p / ∈ S and let u p ∈ U p so that By Lemma 1.11, there exists λ p ∈ F × p such that F p ∩ M (Z p ) = F p ∩ M (Z p )u p λ p . Replacing u p with u p λ p we may assume that u p satisfies We shall show that there exist z p ∈ M (Z p ) × and x p ∈ F × p such that z p u p x p is the identity element in M (Q p ).
Let ε be an integral element of F generating F over Q, and let L be an extension of Q containing all the zeros of the minimal polynomial of ε. Let L p = L⊗Q p , note that neither this nor F p are necessarily fields, rather a finite direct product of fields. The embedding F ֒→ M (Q) gives us an embedding F p ֒→ M (Q p ). Since p / ∈ S we have M (Q p ) ∼ = Mat 4 (Q p ) so we get an embedding Let O L be the ring of integers in L and let O Lp = O L ⊗ Z p . Let ∆ denote the subspace of diagonal matrices in Mat 4 (L p ). It follows from our choice of L that ε is diagonalisable in M (L p ) by an element A ∈ M (O Lp ) × . Since F p = Q p (ε), the whole of F p ⊗ L is simultaneously diagonalisable in M (L p ), that is, we have that A −1 (F p ⊗ L)A is a subspace of ∆, and by dimensional considerations we must have that A −1 (F p ⊗ L)A = ∆. Then, tensoring (6) and (7) with O L and conjugating by A we get and whereū p = u p A.
We are currently working in the matrix algebra M (L p ) ∼ = Mat 4 (L p ). Let p 1 , ..., p n be the prime ideals of L above p. By virtue of the isomorphism (5) we can consider separately each p i -adic component of the entries of the matrices. It then follows from Lemma 1.13 and equations (8) and (9) Consider now the trace map tr L/Q of the field extension L/Q. The image of O L under this map is an ideal in Z. Let ν ∈ Z be the greatest such that tr L/Q (O L ) ⊂ p ν Z. We define a linear map We denote by O L1 the ring of integers of the field L p1 . Let ξ ∈ (F ⊗ L p1 ) ∩ M (O L1 ) and let x 1 and z 1 be the p 1 components of x and z. Then Hence we can see that ξu −1 p ∈ M (O L1 ). We now consider the linear map defined by By Proposition 17.7 of [28] we have . Let e p (F ) be the ramification index of p in F . Then by [27], Chapter I, Proposition 8.2, and since p does not split in F , we have Let O be the valuation ring of F p and let Π = πO be the unique maximal ideal of O, where π ∈ O is a generator of the principal ideal Π. Let p = pZ p be the unique maximal ideal of Z p . Then p = Π ∩ Z p and p = π ep(F ) . Thus v M (π) = v M (p)/e p (F ), which implies that and hence from (10) we get Proof: Recall that O ∈ O F (S) for some F ∈ C(S). LetF = ′ p F p be the restricted product over all primes p, that is, all but finitely many components are integral, and letÛ be the set of allũ = . Let u ∈ U I for some I ∈ I(O). We denote byû the element of M (A fin ) each of whose components is u. Setting u p = u for all primes p, we see that for all p For each I ∈ I(O) we define the map We shall show that Φ I is a bijection. Since from Lemmas 1.14 and 1.15 we know that |M (Ẑ) × \Û /F × | = λ S (O), the lemma will then follow. First we show surjectivity. Letũ = (u p ) ∈Û , so that O p,up = O p for all primes p. For all primes p, by Lemma 1.11 we have I ⊗ Q p ∼ = F p ∩M (Z p )u p so there exists x p ∈ F × p such that I ⊗ Q p = F p ∩M (Z p )u p x p . We replace u p with u p x p . By Lemma 1.8 there existz = (z p ) ∈ M (Ẑ) × and u ∈ M (Q) × such thatũ =zû. It follows that O p,u = O p for all primes p and hence we get that O u = O, so that u ∈ U . Furthermore, since for all primes p, it follows that I = F ∩ M (Z)u, so u ∈ U I . We have proven that Φ I is surjective. For the proof of injectivity, suppose that u, v ∈ U I are such that Φ I (u) = Φ I (v). The assumption that u, v ∈ U I means that there is an isomorphism of O-modules F ∩ M (Z)u ∼ = F ∩ M (Z)v. As we saw above this implies that there exists The assumption that Φ I (u) = Φ I (v) means that there existz ∈ M (Ẑ) × andx ∈F × such thatû =zvx. Together with (11) this implies thatF and hencex ∈v −1 M (Ẑ) ×v . In other words, there existsỹ ∈ M (Ẑ) × such thatvx =ỹv. We deduce that Replacingz withzỹ ∈ M (Ẑ) × , we now have that Writing z =z ∈ M (Z) × we get that u = zv and injectivity is proven.
In this final section we complete the proof of the Main Theorem. We do this by using our results on orders in subfields of division algebras of degree four to interpret the Prime Geodesic Theorem algebraically. This will then yield information about the class numbers of those orders.

Regular geodesics
Let F be a number field with r real embeddings and s pairs of complex conjugate embeddings and let t = r + s − 1. Let O F denote the ring of integers of F and O × F the unit group therein. By Dirichlet's unit theorem ( [27], I Theorem 7.4), there exist units ε 1 , ..., ε t ∈ O × F such that every ε ∈ O × F can be written uniquely as a product ε = ζε ν1 1 · · · ε νt t , where ζ is a root of unity and ν i ∈ Z. The set {ε 1 , ..., ε t } is not uniquely determined. We shall refer to such a set of units as a system of fundamental units. If O ⊂ O F is an order of F and {ε 1 , ..., ε t } a system of fundamental units then there exist µ i ∈ N and unitsε i = ε µi i such that every ε ∈ O × can be written uniquely as a product ε = ζε ν1 1 · · ·ε νt t where ζ is a root of unity and ν i ∈ Z. Once again, the set {ε 1 , ...,ε t } is not uniquely determined. We shall call such a set of In the case that F is a totally complex quartic field we have t = 1, so we can choose a fundamental unit ε 1 such that for all ε ∈ O × F we have ε = ζε ν1 1 , where ζ is a root of unity and ν 1 ∈ Z. Then ε 1 is determined up to inversion and multiplication by a root of unity. In this case the regulator of an order O ⊂ O F of F with fundamental unitε 1 satisfies where τ is any embedding of F into C.
We consider subfields of the field F . Apart from Q these are all quadratic. By Dirichlet's unit theorem, real quadratic fields contain a single fundamental unit (up to inversion and multiplication by ±1) and complex quadratic fields contain no fundamental unit. We can see then that F can have at most one real subfield. Let C r (S) and C c (S) be respectively the subsets of C(S) consisting of fields with and without a real quadratic subfield. Let O r (S) and O c (S) be respectively the subsets of O(S) consisting of orders in fields in C r (S) and C c (S). Then C(S) = C r (S) ∪ C c (S) and O(S) = O r (S) ∪ O c (S).
Let [γ] ∈ E p,reg P (Γ). Then by Lemma 1.4 the centraliser M (Q) γ is a totally complex quartic field which embeds into M (Q) and we denote this subfield of M (Q) by F γ . We say that an element γ ∈ Γ is weakly neat if the adjoint Ad(γ) has no non-trivial roots of unity as eigenvalues. Let E p,rw P (Γ) be the set of primitive, regular, weakly neat elements in E P (Γ).
Proof: Let O be the maximal order in F and let O × denote the group of units. By Dirichlet's unit theorem we can see that F γ contains a real quadratic subfield if and only if O × ∩ R = {±1}. Let γ 1 ∈ O × be a fundamental unit. The element γ is a unit F γ so there exists a root of unity ζ and n ∈ Z such that γ = ζγ n 1 . If there also exist a root of unity ξ and m ∈ Z {0} such that ξγ m 1 ∈ R then γ 1 is not weakly neat and so neither is γ.
Conversely, suppose γ is not weakly neat. We know that γ is conjugate in G to an element a γ b γ ∈ A − B. From the assumption that γ is not weakly neat it follows that there exists m ∈ N such that one of the components of b m γ is diagonal. Since the group AB is abelian we have that γ m is conjugate in G to a m γ b m γ . Since γ m ∈ E P (Γ) it follows from Lemma 1.4 that b m γ is diagonal and hence γ m generates a real quadratic subfield of F γ . The lemma follows.
Let γ, δ ∈ Γ. Then F δγδ −1 = δF γ δ −1 , and since by Lemma 1.5 the group Γ is a subgroup of the multiplicative group M (Z) × , we can conclude that Hence the map is well defined.

Proposition 2.2
The map θ is surjective.
Proof: Let O ∈ O c (S) be an order in the field F ∈ C c (S). Since deg F = 4, Proposition 1.2(b) implies that F is a maximal subfield of M (Q). Lemma 1.9 tells us that there exists an embedding σ : F → M (Q) such that Let ε be a fundamental unit in O × . Let γ be the image of ε under the embedding σ. The unit ε is integral in F , so γ ∈ M (Z).
The field F has two pairs of conjugate embeddings into C. Let σ 1 , σ 2 be two distinct, non-conjugate embeddings of F into C. The R-algebra F ⊗ Q R is isomorphic to the commutative R-algebra C ⊕ C via the map α ⊗ x → (xσ 1 (α), xσ 2 (α)) .
We note that O ⊗ Z R = F ⊗ Q R and we get the series of inclusions Thus we see that γ may be considered as an element, which we will denoteγ, of a commutative subalgebra of Mat 4 (C) isomorphic to C ⊕ C. The matrixγ has real entries and can be diagonalised in Mat 4 (C) to the matrix since the eigenvalues ofγ are the roots of its minimal polynomial, ie. the values of γ under its different embeddings into C. Let a, b ∈ R + ; θ, φ ∈ [−π, π] be such that σ 1 (γ) = ae iθ and σ 2 (γ) = be iφ . Since γ ∈ M (Z), we have σ σ(γ) = N F |Q (γ) = det γ = ±1, where N F |Q is the field norm and the product is over all embeddings σ of F into C ( [27], I Proposition 2.6(iii)). Hence a 2 b 2 = ±1. But a and b are both real so we must have a 2 b 2 = 1, and so b = a −1 and we have Without loss of generality we assume a < b so that a ∈ (0, 1). Let R be the 2 × 2 matrix Then X is conjugate in Mat 4 (C) via Y to the real matrix Soγ is conjugate in Mat 4 (C) to an element of A − B, hence by [26], XIV Corollary 2.3, the matrixγ is Then det Z = ±1 and Z −1γ Z = X ′ , so that γ is conjugate in SL ± 4 (R) to an element of A − B. Suppose that det Z = −1. Let W be the matrix Then ZW ∈ G and (ZW ) −1γ ZW = X ′′ , where Hence γ is conjugate in G to an element of A − B.
We also saw above that det γ = 1 and so γ ∈ Γ and hence γ ∈ E P (Γ). It then follows from Lemma 1.4 that γ, and hence also ε, generates either a real quadratic or a totally complex quartic field over Q. The field generated by ε over Q is a subfield of F . However F ∈ C c (S) so has no real quadratic subfields. Hence Q(γ) ∼ = Q(ε) is a totally complex quartic field and Lemma 1.4 tells us that γ is regular. It then follows from Lemma 2.1 that γ is weakly neat.
It remains to show that γ is primitive in Γ. Note that Q(ε) is a quartic field contained in and hence equal to the field F . Recall that we write F γ = M (Q) γ for the centraliser of γ in M (Q). Then we have However we also have γ 0 ∈ F γ and so γ 0 ∈ F γ ∩ M (Z) × , which is a contradiction. Hence γ is primitive in Γ and we have γ ∈ E p,reg P (Γ). This concludes the proof of the surjectivity of θ. Proof: Let ε be a fundamental unit in O with minimal polynomial f (x) over Q. Suppose there exists another fundamental unit δ ∈ O × with δ = ε which is also a root of f (x). Then there exist embeddings α and β of F into C such that α(ε) = β(δ).
Let η ∈ O × also be a fundamental unit in O with minimal polynomial g(x) over Q. Then η = ζε ±1 for some root of unity ζ ∈ O × . We have Let n be the order of the root of unity ζ. The order O is a ring so contains all roots of unity of order n and hence there exists an n th root of unity ξ ∈ O × such that β(ξ) = α(ζ). Let θ be the fundamental unit ξδ ±1 ∈ O × . Then α(η) = β(θ) and so θ is also a root of g(x). The lemma follows. (Γ) such that F γ ∩M (Z) ∼ = O. We have seen that the choice of fundamental unit is determined up to inversion and multiplication by a root of unity so there are 2µ O choices for ε. We saw in the previous proposition that for a given embedding σ of F into M (Q) and choice of a fundamental unit ε ∈ O there is a conjugacy class [γ σ,ε ] ∈ E p,reg P (Γ) with σ(ε) = γ σ,ε . However, we saw above that if there exists an automorphism α of F such that δ = α(ε), then setting τ = σ • α −1 we have [γ σ,ε ] = [γ τ,δ ]. Such an automorphism exists if and only if δ is a root of the minimal polynomial of ε over Q. The claim of the proposition folows.
Proof: The element γ is a fundamental unit in O γ . We shall write f (x) for its minimal polynomial. The roots of f (x) are the eigenvalues of γ, which are α 1 = ae iθ , α 2 = ae −iθ , α 3 = a −1 e iφ , α 4 = a −1 e −iφ . Suppose κ(O γ ) > 1. Then there exist i, j ∈ {1, 2, 3, 4}, i = j and a root of unity ζ such that α i = ζα ±1 j . If α i = ζα j then α i α −1 j = ζ and either {i, j} = {1, 2} or {i, j} = {3, 4}. Without loss of generality take i = 1 and j = 2, then ζ = α 1 α −1 2 = e i2θ . It follows that if n ∈ N is such that ζ n = 1 then α 2n 1 = a n ∈ (0, 1) and so F γ has a real quadratic subfield. However, by Lemma 2.1 this contradicts the assumption that γ ∈ E p,rw P (Γ), so we must have α i = ζα −1 j . By considering the possible values of i and j which would satisfy α i = ζα −1 j we see that either θ + φ or θ + φ is equal to qπ, where q ∈ Q is such that ζ = e iqπ . It remains to show what the possible values for q are, that is, which roots of unity may occur in O γ . We saw in the proof of Lemma 2.3 that the only roots of unity which may occur in a quartic field are ±1 and roots of order 3, 4, 5, 6, 8, 10 or 12. If ζ is a root of order 5, 8, 10 or 12 then ζ generates a totally complex quartic field, however, Q(ζ) has the real quadratic subfield Q(ζ + ζ −1 ), which possibility is excluded in our case. So we are left with the possibility that ζ = ±1 or ζ is a root of order 3, 4 or 6, which implies the claim of the lemma.
Proof: By Proposition 1.10 of [13], where Γ ′ ⊂ Γ is a torsion free subgroup of finite index. In the case under consideration we have that Γ γ is isomorphic as a group to the group of norm one elements in the order O γ = F γ ∩ M (Z). By Lemma 1.5 this is equal to the group of units O × γ . By Dirichlet's unit theorem and Lemma 2.3 it follows that for some generator ε and where µ(γ) is the finite cyclic group of roots of unity in O × γ . It then follows that Γ ′ γ ∼ = ε kZ , for some k ∈ N. Since every torsion element in Γ γ is sent to the identity under the projection onto Γ γ,A we also have the isomorphisms Γ γ,A ∼ = ε Z and Γ ′ γ,A ∼ = ε kZ .
The lemma follows.

Class numbers of orders in totally complex quartic fields
We are now in a position to prove our main theorem. Then, as x → ∞ we have π S (x) ∼ e 4x 8x .
We can also prove the following: It follows from Lemma 2.10 of [13] that 0 ≤ trσ(b γ ) ≤ 16 for all γ ∈ E P (Γ), and hence we have We can also deduce from (12) and (13) The theorem then follows from Proposition 2.8 using the same arguments as in the proof of the previous theorem.
The methods we have used to deduce the asymptotic result for class numbers from the Prime Geodesic Theorem were not sharp enough to preserve the error term which was proven there. However, we make the following conjecture: