ALGEBRAIC SPACE-TIME CODES BASED ON DIVISION ALGEBRAS WITH A UNITARY INVOLUTION

In this paper, we focus on the design of unitary space-time codes achieving full diversity using division algebras, and on the systematic computation of their minimum determinant. We also give examples of such codes with high minimum determinant. Division algebras allow to obtain higher rates than known constructions based on finite groups.


Introduction
The problem addressed in the design of space-time codes in the coherent case (that is, in the case where the receiver knows the properties of the channel) can be summarized as follows: find a set C of complex n × n matrices such that the minimum determinant δ min (C) = inf X =X ∈C | det(X − X )| 2 is maximal. Of course, the first step is to ensure that δ min (C) is not zero. When this is the case, we will say that C is fully diverse. One natural way to achieve this is to use division algebras. Indeed, any division algebra D whose center k is a subfield of C may be identified to a subring of a matrix algebra M n (C). In particular, D × may be identified to a subgroup of GL n (C), and taking C to be a subset of D yields a fully diverse algebraic space-time code.
The use of division algebras for space-time coding started with the seminal work by B. A. Sethuraman and B. Sundar Rajan [13]. Number fields and cyclic algebras were discussed, which have been a favourite tool for space-time design. Some surveys are by now available [9,12], and we refer the interested reader to them for further details. Other algebras have also been explored, such as crossed product algebras [1] or non-associative algebras [11]. Recently, the optimality of algebraic codes obtained by Oggier et al. on cyclic algebras or crossed product algebras has been established ( [3]; see also [15]). Notice that the two surveys mentioned above focus on the coherent case.
In the non-coherent case, the problem has a different flavour: the minimum determinant still needs to be maximized, but the elements of the code C must be complex unitary matrices [4,5]. We will say that C is a unitary code.
The question of designing good unitary codes is far from being solved. The two main difficulties arising in the non-coherent case are the following: (a) fully diverse families of unitary matrices are hard to find; (b) contrary to the coherent case, no systematic way to compute or even estimate δ min (C) is known.
Question (a) has been addressed in [14] using unitary representations of finite fixedpoint free groups. Later on, Oggier proposed somme approach using cyclic algebras with a unitary involution ( [10,7]). We refer the reader to [8] for a survey of known results.
In this paper, we will give a method to construct unitary codes using division algebras carrying a unitary involution, generalizing in particular the work done by Oggier and Lequeu, and show how to compute the minimum determinant of such codes. The advantage of this approach compared to the group-theoretic approach is that division algebras allow to obtain higher rates (the rate corresponds, roughly speaking, to the cardinality of the code).
Notice that in the literature, the quantity which is asked to be maximal is not the minimum determinant, but the so-called diversity product ζ(C), defined by where n is the size of the matrices. In other words, we have The two optimization problems being obviously equivalent, we will focus essentially on the computation of the minimum determinant.
The structure of this paper is as follows. In Section 1, we recall some basic definitions on central simple algebras with unitary involutions, and provide some examples. In Section 2, we will provide a systematic way to construct unitary space-time codes using algebras with unitary involutions. Finally, in Section 3, we will explain how to compute the minimum determinant of these codes and provide examples.

Algebras with unitary involutions: definitions and examples
In this section, k is a field, and A is a central simple k-algebra. To simplify the exposition, we will assume that char(k) = 2. We will collect here some basic definitions and results on unitary involutions. We will assume that the reader is familiar with the theory of central simple algebras. We let the reader refer to [6] for the missing details and proofs concerning central simple k-algebras with involutions.
1. An involution on A is a ring anti-automorphism of A of order at most 2.
For example, the transposition is an involution on M n (k). Notice that Id A is never an involution unless A is commutative, which implies that A = k. Therefore, if A = k, an involution on A has order 2.
It is easy to check that for every λ ∈ k, we have σ(λ) ∈ k. Hence σ | k is an automorphism of order at most 2 of k.
We set We say that σ is an involution of the first kind if σ | k = Id k , that is if k = k 0 , and an involution of the second kind (or unitary) otherwise. In the latter case, k/k 0 is a quadratic field extension, and σ | k is the unique non-trivial k 0 -automorphism of k/k 0 . Conversely, if k/k 0 is a quadratic field extension, we will say that a unitary involution σ on a central simple k-algebra A is a k/k 0 -involution if σ | k is the unique non-trivial k 0 -automorphism of k.
We denote by Sym(A, σ) the set of symmetric elements of A, and by Skew(A, σ) the set of skew-symmetric elements of A. Both have a natural structure of a k 0 -vector space. We also set We say that two central simple k-algebras with involutions (A, σ) and (A , σ ) are isomorphic if there exists an isomorphism of k-algebras f : In this case, one may verify that σ and σ are involutions of the same kind. Moreover, f then induces isomorphisms of k 0 -vector spaces Sym(A, σ) Sym(A , σ ) and Skew(A, σ) Skew(A , σ ).
One may show that, if σ, σ are two k/k 0 -involutions of the second kind, there exists u ∈ Sym(A, σ) × , which is unique up to multiplication by an element of k × 0 , such that σ = Int(u) • σ, where Int(u) : Example 1.2. Let k/k 0 be a quadratic field extension, and let be its non-trivial k 0 -automorphism. If n ≥ 1, the map M n (k) −→ M n (k) We now would like to give a family of examples which will be useful in the sequel. First, we recall the notion of a crossed-product algebra. Definition 1.3. Let L/k be a finite Galois extension, with Galois group G. The group G acts by k-algebra automorphisms on L by Let us consider a 2-cocycle of G with values in L, that is a map satisfying ξ σ,Id = ξ Id,ρ = 1 for all σ, ρ ∈ G, and ξ σ,ρν ξ ρ,ν = ξ σρ,ν ξ ν σ,ρ for all σ, ρ, ν ∈ G.
Notice for later use that f Id is the unit element of (ξ, L/k, G), and that f σ is invertible for all σ ∈ G.
This is a central simple k-algebra of degree n (see [2, Lemma VI.1.7], for example).
Example 1.4. Let γ ∈ k × , let L/k be a cyclic extension of degree n, and let σ be a generator of its Galois group. Setting defines a 2-cocycle, and the corresponding crossed-product is simply the k-algebra (γ, L/k, σ) = n−1 i=0 e i L generated by one element e subject to the relations e n = γ, λe = eλ σ for all λ ∈ L.
Example 1.5. Let L/k be a biquadratic extension with Galois group G = σ, τ , let a, b, u ∈ L × satisfy and let ξ a,b,u : G × G −→ L × be defined by ξ a,b,u Id,Id = 1, ξ a,b,u Id,σ = 1, ξ a,b,u Id,τ = 1, ξ a,b,u Id,στ = 1, A lengthy case-by-case verification shows that ξ a,b,u is a 2-cocycle. It is easy to check that the corresponding crossed-product algebra is nothing but the k-algebra generated by two elements e and f satisfying subject to the relations The following result provides the family of examples we are aiming for, and generalizes the construction of a unitary involution of a cyclic algebra proposed in [10].
Lemma 1.6. Let k/k 0 be a quadratic field extension, and denote by its nontrivial k 0 -automorphism. Let L/k be a finite Galois extension with Galois group G.
Assume that there exists a ring automorphism α : L −→ L satisfying the following conditions: Let ξ : G×G −→ L × be a 2-cocycle satisfying (α•ξ)ξ = 1, and let B = (ξ, L/k, G) be the corresponding crossed-product algebra. Then there is a unique unitary involution τ on B satisfying for all σ ∈ G and τ | L = α.
Moreover, if M b is the matrix of left multiplication by b in the L-basis (f σ ) σ∈G , then we have where is the unitary involution on M n (L) defined by Proof. Assume that an involution τ satisfying the properties of the lemma exists. Using the fact that τ is an anti-automorphism, we get that This proves the uniqueness of τ . We now have to prove that the map τ defined by the formula above is indeed a unitary involution on B. Clearly, τ is additive, and for all x ∈ k, we have We now check that we have τ (xy) = τ (y)τ (x) for all x, y ∈ B.
The usual distributivity argument shows that it is enough to prove it for x = f σ λ, y = f ρ µ, σ, ρ ∈ G, λ, µ ∈ L. We have On the other hand, we have From the relation λf σ = f σ λ σ , we get Therefore, we get Since α commutes with the elements of G and α(ξ σ,ρ ) = ξ −1 σ,ρ by assumption, we get the desired equality. It remains to prove that τ 2 = Id B . Since τ is an antiautomorphism of rings, τ 2 is an automorphism of rings. Hence to prove that τ 2 is the identity map, it is enough to check that τ 2 (f σ ) = f σ for all σ ∈ G and that τ 2 | L = Id L , which is clear from the definition of τ . We finally prove the last assertion. We will index the entries of a matrix with coefficients in L with the elements of G.
Remark 1.7. The description of the involution τ in the lemma above may be made more explicit. As explained in the proof, we have Now we have f σ f σ −1 = ξ σ,σ −1 , and therefore Since α commutes with σ and α(ξ σ −1 ,σ )ξ σ −1 ,σ = 1 by assumption, we finally get that Example 1.8. Assume that k 0 is a number field. Let L/k be a finite Galois extension of k with Galois group G, and assume that complex conjugation induces a k 0 -automorphism α of L which commutes with elements of Gal(L/k). This automorphism satisfies the conditions of Lemma 1.6. In particular, if ξ : For example, if B = (γ, L/k, σ) is a cyclic k-algebra of degree n such that |γ| 2 = 1, then the unitary involution τ on B given by the previous lemma is defined by τ : as it may easily be seen by direct computations, or by using the remark above.
We recover in this way the involution obtained by Oggier and Lequeu [10].

Algebras with involutions and space-time coding
As briefly explained in the introduction, we would like to find a set C of n × n unitary matrices such that the minimum determinant As this has been done for the coherent case, we are going to use the theory of division algebras to construct unitary codes.
We now explain how we are going to proceed. First, we need a definition. Let k/k 0 be a quadratic field extension of number fields, whose non-trivial automorphism is given by the complex conjugation. Let (B, τ ) be a central simple k-algebra with a unitary k/k 0 -involution.
Notice that, if L is any subfield of C containing k and denotes the complex conjugation, the map τ ⊗ is a unitary involution on B ⊗ k L. Thus the following definition makes sense.
Definition 2.1. We say that (B, τ ) is positive definite if there exists a subfield L of C such that there exists an isomorphism of L-algebras with involutions In view of this definition, it does not seem to be very easy to check whether or not a given unitary involution is positive definite. In fact, one may show that τ is positive definite if and only if a certain hermitian form attached to (B, τ ) is positive definite. Since we will not need this criterion for our purpose, we postpone the statement and the proof of this criterion in the appendix.
Assume that τ is positive definite, and set U b = ϕ(b ⊗ 1) for all b ∈ B. Then the equality above may be rewritten as We may now prove an easy lemma.
is an injective morphism of k-algebras. Moreover, the induced group morphism Similarly, one shows that U b + U b = U b+b , and λU b = U λb for all λ ∈ k.
Moreover, U b = I n if and only if b = 1, since ϕ and the canonical map B −→ B ⊗ k L are injective. This concludes the proof.
Let us come back to the previous considerations. For all b ∈ B, we have . In particular, U b is unitary if and only if bτ (b) = 1. This motivates the following definition.
Definition 2.5. Let k/k 0 be any quadratic field extension, and let (B, τ ) be a central simple k-algebra with an arbitrary unitary k/k 0 -involution. We say that The set of unitary elements is easily seen to be a subgroup of B × , that we denote by U(B, τ ). Example 2.6. If k is a number field, B = M n (k) and τ is the conjugate transpose of matrices, a unitary element with respect to τ is nothing but a unitary matrix.
The previous results may then be summarized as follows.
Lemma 2.7. Let k/k 0 be a quadratic extension of number fields, whose non-trivial automorphism is the complex conjugation, and let (B, τ ) be a central simple kalgebra with a positive definite unitary k/k 0 -involution. The map induces an injective group morphism Let (B, τ ) be a central simple k-algebra with a positive definite unitary k/k 0involution. Keeping the previous notation, for any subgroup G of U(B, τ ), we get a unitary space-time code Hence, the main idea here is to take our unitary code C to be a finite subset of some In this case, if B is division, we will have δ min (C) > 0 (i.e. the code is fully diverse), and Of course, we still need to find a way to estimate δ min (C G ). This problem will be examined in the next section.
Example 2.8. Assume that B has a maximal subfield L ⊂ C, and that τ is positive definite. In this case, it is well-known that we have a unique isomorphism of L-algebras ϕ : where M b is the matrix of left multiplication by b with respect to a fixed L-basis of B ⊗ k L. In this case, for every b ∈ U(B, τ ), we will have U b = M b , and thus, for any subgroup G of U(B, τ ), we will get Example 2.9. Let k/k 0 be a quadratic extension of number fields, and L/k be a Galois extension of number fields with Galois group G, such that complex conjugation induces a k 0 -automorphism of L which commutes with the elements of G.
By Lemma 1.6, there exists a unique unitary involution τ on B such that Hence, for any subgroup G of U(B, τ ), we have It is about time to show how to find classes of unitary elements in a division algebra with a unitary involution (B, τ ) by looking at elements of norm 1 in some subfields of B. The following result has been proven is [10] in the case where B is a cyclic division k-algebra, but the result and its proof still hold in the general case (see also [2,Lemma IX.5.12]). Lemma 2.10. Let k be an arbitrary field, and let (B, τ ) be a division k-algebra with a k/k 0 -involution. Then for every x ∈ B, the following conditions are equivalent: 1.
x is unitary with respect to τ ; 2. there exists a subfield M of B containing x, such that τ restricts to a nontrivial k 0 -automorphism of M and N M/M τ (x) = 1; 3. there exist a subfield M of B containing x and u ∈ M × , such that τ restricts to a non-trivial k 0 -automorphism of M and x = uτ (u) −1 .
Consider the cyclic division algebra B = (j, L/Q(j), σ). Since |j| 2 = 1, by Example 1.8, there exists a positive definite unitary involution τ on B given by τ : Example 2.9 shows that the left multiplication matrix of any unitary element is a unitary matrix. Following the method explained above, we look for subfields M of B which are stable by τ . The first obvious subfield of B one can think of is L.
The restriction of τ on L is the complex conjugation. In this case, unitary elements contained in L are elements of the form zz −1 , z ∈ L × .
Let us consider now the subfield generated by e. Since 1, e, e 2 are linearly independent over L, they are also linearly independent over k. Therefore [k(e) : k] ≥ 3, and since e 3 = γ, we have [k(e) : k] ≤ 3. Thus k(e) is a subfield of B of degree 3 over k, and the minimal polynomial of e over k is X 3 − j. Thus we have an isomorphism where ζ 9 is a primitive 9 th -root of 1, this isomorphism mapping e onto ζ 9 . Since τ (e) = e −1 , the previous isomorphism maps τ (e) onto ζ −1 9 = ζ 9 . In other words, we have an isomorphism of k-algebras with involution (k(e), τ | k(e) ) ∼ =k (Q(ζ 9 ), ).
It follows that unitary elements in k(e) are mapped onto elements of the form uu −1 , u ∈ Q(ζ 9 ) × by this isomorphism.
Take for example the element u = 1+j +ζ 9 +ζ 2 9 j ∈ Q(ζ 9 ). This element corresponds to the element y = (1 + j) + e + e 2 j ∈ k(e), and the element u corresponds to the element τ (y). Set Y = M y . Then we have

Now we also have
Advances in Mathematics of Communications Volume 00, No. 0 (2013), which can be checked to be Y * . Then the element b = yτ (y) −1 is unitary, and its multiplication matrix U b = Y(Y * ) −1 is a unitary matrix, as we may check directly by computation.

The minimum determinant of a unitary code
Let us summarize what we have done in the previous section. Let k/k 0 be a quadratic extension of number fields, whose non-trivial automorphism is given by complex conjugation. Let (B, τ ) be a central simple k-algebra of degree n with a positive definite unitary k/k 0 -involution, let L/k be a splitting field of B (L ⊂ C) and let be an isomorphism of L-algebras such that For any subgroup G of U (B, τ ), the set is a unitary algebraic code, which is fully diverse as soon as B is a division algebra.
As explained in the introduction, we would like to find a good estimation of the minimum determinant of our unitary code C G . The first step is, as in the coherent case, to find a more tractable expression of it. This is given by the next lemma. Proof. For all b, b ∈ U(B, τ ), b = b , using Lemma 2.7, we get Since the determinant of a unitary matrix is a complex number of modulus 1, we finally get that Thus, this equality may be rewritten as

Advances in Mathematics of Communications
Volume 00, No. 0 (2013), and therefore This concludes the proof.
Example 3.2. Let us keep the notation of Example 2.11. One may take the subgroup G = b of U(B, τ ) generated by b, and consider the unitary code C G . We then get an infinite unitary code. One way to see this is as follows: after computations, we get .
Hence, we have det(U b ) = e iθ , with cos(θ) = 11 38 . But one may show by induction that cos(2mθ) = 1 for all m ≥ 1. In particular, mθ is never a rational multiple of 2π. It follows that U m b = I 3 for all m ≥ 1, which is equivalent to saying that G is infinite. However, the minimum determinant of such a code is 0, as shown in the next proposition. Proof. Let b ∈ G be an element of infinite order. Since H = b ⊂ G, we have Hence, it is enough to prove that δ min (C H ) = 0. Notice that, by assumption on b, the corresponding matrix U b has infinite order, since the map is an injective group morphism by Lemma 2.7. Since U b is unitary, it can be diagonalized and all its eigenvalues have modulus 1.
Let e iθj , j = 1, . . . , n be the (not necessarily distinct) eigenvalues of U b . For all m ∈ Z, the matrix I n − U m b is similar to the diagonal matrix whose diagonal entries are 1 − e imθj = −2i sin( mθ j 2 )e i mθ j 2 , j = 1, . . . , n.
It follows easily that Now, since U b has infinite order, at least one θ j is not a rational multiple of 2π. For this θ j , the sequence (sin( mθj 2 )) m≥1 is dense in [−1, 1], so we may find an increasing sequence of integers (α m ) m≥1 such that lim m sin( α m θ j 2 ) = 0. This implies that δ min (C H ) = 0, and this concludes the proof.
We now prove a result which will allow us to compute the minimum determinant in terms of norms of cyclotomic extensions.
If n ≥ 1 is an integer, we denote by φ n the n th cyclotomic polynomial.

we have the equalities
Moreover, if D has prime degree and property (2) holds, then D × has an element of order m.
Proof. Let d ∈ D × be an element of order m, so we have d m = 1. Hence µ d,Q divides X m − 1, and therefore µ d,Q is a cyclotomic polynomial φ r , for some r | m. Since φ r | X r − 1, we have d r − 1 = 0, and therefore m | r. Hence r = m and µ d,Q = φ m . Now µ d,k | µ d,Q , so there exists ζ m ∈ C, a primitive m th -root of 1, such that µ d,k (ζ m ) = 0. Elementary Galois theory then shows that we have an isomorphism of k-algebras k(d) ∼ =k k(ζ m ), which maps d onto ζ m . This proves (1). Notice for later use that such an isomorphism preserves degrees and norms. Therefore, k(ζ m ) is isomorphic to a subfield of D. In particular, [k(ζ m ) : k] | n. If ζ m / ∈ k, k(ζ m )/k has degree at least 2, and it is well-known that D ⊗ k k(ζ m ) is not a division algebra (see [2, Proposition V.3.2] for example). Now assume that D has prime degree, and that [k(ζ m ) : k] | n. If ζ m ∈ k, then ζ m ∈ D × has order m. If D ⊗ k k(ζ m ) is not a division algebra, then k(ζ m )/k is an extension of degree at least 2 dividing n. Since D has prime degree, this implies that k(ζ m ) is isomorphic to a subfield of D. Such an isomorphism maps ζ m onto an element d ∈ D × of order m. This proves (2) and the last part of the proposition.

Thus, we have
Nrd It follows immediately that N k(ζm)/k (1 − ζ m ) = µ ζm,k (1), and this proves (4). This concludes the proof. Corollary 3.5. Let k be a number field, and let D be a central division k-algebra of degree n. Then any subgroup of D × is either finite or has an element of infinite order.
Proof. Let G be a subgroup of D × . Assume that every element of G has finite order. By the previous proposition, if g ∈ G has order m, then ϕ(m) | n[k : Q]. This implies that m may take only finitely many values. In particular, the least common multiple of the orders of the elements of G is finite, that is G has finite exponent. Now if L is a maximal subfield of D, the injective k-algebra morphism induces an injective group morphism D × → GL n (L). It follows that G is isomorphic to a subgroup of GL n (C) of finite exponent. By a celebrated theorem of Burnside, this implies that G is finite.
We now summarize our results on the minimum determinant of unitary codes in the following theorem.
Theorem 3.6. Let G be a subgroup of U(B, τ ), and assume that B is a division k-algebra of degree n. Then G is either finite or has an element of infinite order. Moreover, the following properties hold: 1. If G has an element of infinite order, then δ min (C G ) = 0; 2. If G is finite, we have where m b is the order of b.
Proof. This follows from Proposition 3.3, Proposition 3.4 and Corollary 3.5, since a subgroup of U(B, τ ) is a subgroup of B × .
Remark 3.7. If b ∈ G has finite order m b , Proposition 3.4 shows that that Nrd B (1− b) only depends on m b . In particular, δ min (C G ) only depends on the orders of the elements of G, and not on the group itself. Therefore, to compute the minimum determinant, one may proceed as follows: choose a subset S of G such that each element of S is obtained by a unique element of S; 3. the observation above shows that we have As a first application, we compute the exact value of the minimum determinant of a code presented in [7]. In fact, C is simply the unitary code C G , where G is the group of order 63, generated by e and ζ 21 . The possible values for the order of an element of G are 1, 3, 7, 9, 21, 63. Notice that G is not abelian, hence not cyclic, so G has no elements of order 63. We also look only at non-trivial elements of G, so we may also discard 1. One may also check that G has no element of order 9. By considering ζ 7 21 , ζ 3 21 and ζ 21 , we see that the other possible values are obtained.
The remark above shows that it is enough to compute | det(I 3 −D m )| 2 for m = 1, 3, 7.
Here, the minimum is obtained for m = 1, so Computing µ ζ21,K shows that the exact value is 5 − √ 21 2 .
Notice that we may extend this code by considering the group G = e, ζ 21 , −j = e, ζ 21 , −1 .
It is easy to check that G G × {±1}, so that Hence the orders of non-trivial elements of G are now 2, 3, 6, 7, 14, 21, 42, and −1, −ζ 7 21 , −ζ 3 21 and −ζ 21 are elements of order 2, 6, 14 and 42 respectively. One may compute that Remark 3.9. Let G be a finite subgroup of U(B, τ ). One way to get a group G whose cardinality is as large as possible is to ensure that G contains all the roots of unity lying in k. However, we will often get a small minimum determinant, as we proceed to show now.
Indeed, Theorem 3.6 shows in particular that, if ζ m ∈ k, then we have for any finite subgroup G of U(B, τ ) (where n is the degree of B over k), that is Hence, the diversity product satisfies Now, if m ≥ 7, this shows that The upper bound above also shows that ζ(C G ) will tend to be very small if the base field k contains roots of 1 of large order.
The next lemma, used together with the previous proposition, allows us to compute the minimum determinant of a unitary code C G when k/Q is a totally imaginary quadratic extension.
Lemma 3.10. Let k/Q be a totally imaginary quadratic extension, and let m ≥ 2.
Then we have otherwise .
Proof. Since k/Q is a totally quadratic imaginary extension, we have Therefore, we have Notice now that µ 1−ζm,Q (X) = (−1) ϕ(m) φ m (1 − X). It follows that we have If p is a prime number, we have the well-known relations It follows easily that φ m (1) = p if m = p r , r ≥ 1 and φ m (1) = 1 otherwise. This concludes the proof.
Remarks 3.11. Assume that k/Q is a totally imaginary quadratic extension. Let G be a subgroup of U(B, τ ), and assume that B is a division k-algebra of degree n.
1. It follows from Theorem 3.6 that if G contains an element of order m we have If G is finite and contains an element whose order is not a prime power, then we have Indeed, this is an immediate consequence of Theorem 3.6 and Lemma 3.10.
If we want to find subgroups G of U(B, τ ) such that δ min (C G ) > 0, Theorem 3.6 says that all elements of G need to have finite (multiplicative) order. Such elements may be found as follows: choose a subfield M of B which is stable by τ , and look for unitary elements among 'roots of 1 in M ', that is elements b ∈ M such that µ b,Q = φ m for some m ≥ 1. Moreover, a list of possible values for m may be found using points (2) and (3) of Proposition 3.4.
However, the product of elements of finite order is not necessarily an element of finite order. Hence, once we found several unitary elements of finite order, we are still not ensured that the group they generate only has elements of finite order. The next lemma shows how to avoid this problem.
Lemma 3.12. Assume that k/Q is a totally quadratic imaginary extension. Let Λ be a subring of B which is also a free O k -module of finite rank, where O k is the ring of integers of k. Then U(B, τ ) ∩ Λ × is finite.
Proof. Let n = deg(B). By Lemma 2.7, the map ψ : is an injective group morphism. Therefore, ψ(Λ) is also a free O k -module of finite rank. Since k/Q is quadratic imaginary, O k is a full Z-lattice of the R-vector is finite, and therefore so is U(B, τ )∩Λ × . This concludes the proof.
Remark 3.13. Such a subring Λ always exists. One may even assume that Λ contains a k-basis of B. For example, let e 1 = 1 B , e 2 , . . . , e n 2 be a k-basis of B containing 1 B . For all 1 ≤ i, j ≤ n 2 , there exist m ij ∈ Z such that Let m be the least common multiple of the m ij s. Then we have By construction, Λ is a subring of B, which contains a k-basis of B, and which a free O k -module of finite rank.
Example 3.14. We will use here the notation introduced in Example 1.5. Let k = Q(i), and consider the central simple k-algebra where σ and ρ are defined in a unique way by As shown in [1], this is a division k-algebra. By Example 1.5, the values of the cocycle corresponding to the algebra (a, b, u, L/k, σ, ρ) will have modulus 1 if and only if a, b and u have modulus 1. All these conditions are fulfilled here, so by Lemma 1.6, there is an involution τ on B such that τ | L is the complex conjugation, τ (e) = e −1 , and τ (f ) = f −1 , where e, f are the generators of B.
The elements e and f are unitary and e has finite order. However, f has infinite order. Since √ 5 and f commute, M = k(f, √ 5) is a subfield of B which is stable by τ . Let α ∈ C such that α 2 = 1 + 2i √ 5 . Notice that (αα) 2 = 1, and thus αα = 1.
We then have an isomorphism of k-algebras which maps f onto α and √ 5 onto √ 5.
Since τ (f ) = f −1 is mapped onto α −1 = α, it easily follows that we have an isomorphism of k-algebras with involution . One may check that the element satisfies ζ 5 = i, that is ζ is a primitive 20 th -root of 1. In particular, ζζ = 1. Using the isomorphism above, this yields an element which is unitary and which has order 20.
Let us give another example.
Example 3.15. Let k = Q(j), and let L = k(ζ 7 ). Then L/k is a cyclic extension of degree 6, a generator σ of Gal(L/k) being given by σ : Consider the cyclic k-algebra B = (−j, k(ζ 7 )/k, σ). One may show that B is a division k-algebra (see [2,Example IX.5.28] for a proof). Since B fulfills all the assumptions of Lemma 1.6, we may consider the unitary involution τ described in this lemma.
If e is the canonical generator of B, then e is a unitary element of order 36. Moreover, z = ζ 7 is a unitary element of order 7. It follows from the equality ze = ez σ = ez 3 that the subgroup G of U(B, τ ) generated by e and z is a finite group of order 36·7 = 252. Theorem 3.6 and Lemma 3.10 then show that ζ(C G ) = 1 2 .
In other words, the unitary code Notice that the method using fixed-point free groups in [14] does not provide an example of group constellations for n = 6, and provides a non-group constellation C of 72 matrices with diversity product equal to 1 2 .
In this section, we mainly focused on the computation of infinite unitary codes built using a subgroup G of U(B, τ ). Theorem 3.6 thus tells us that we have to exclude elements of infinite order to ensure that δ min (C G ) > 0. However, in practice, we only need finite subsets C of C G . Therefore, we may use these elements to extend further the codes obtained using the techniques developed above, and we are not restricted to consider only finite groups.
Example 3.16. Let us keep the notation of Example 3.14, and consider the code Notice that the method using fixed-point free groups in [14] yields a non-group constellation C of 289 matrices with diversity product approximatively equal to 0.31.
As promised, we give in this appendix an explicit criterion to decide whether or not a given unitary involution is positive definite. First, we need a lemma. All the fields here have characteristic different from 2.
Lemma 3.17. Let k/k 0 be a quadratic extension of arbitrary fields, let be its nontrivial k 0 -automorphism, and let (B, τ ) be a central simple k-algebra with a unitary k/k 0 involution. Then we have where Trd B is the reduced trace.
Proof. Let L/k be a splitting field of B, so we have an isomorphism of L-algebras It is easy to check that τ is a unitary involution of M n (L). By Example 1.2, there exists an invertible hermitian matrix H ∈ M n (L) such that τ = Int(H)• * . In other words, we have Thus, for all b ∈ B, we get By definition of the reduced trace, we have Trd B (τ (b)) = tr(ϕ(τ (b)⊗1)). Therefore, we get This concludes the proof.
Notice now that for all b ∈ B, τ (b)b is τ -symmetric. In view of this lemma, the map is a hermitian form on B with respect to (k, ).
Let L/k be a field extension, and let α : L −→ L be a ring automorphism of L extending . In particular, α = Id L . If L 0 denotes the subfield of L fixed by α, then L 0 contains k 0 and L/L 0 is a quadratic extension.
If h : V × V −→ k is a hermitian form on a finite dimensional k-vector space V with respect to (k, ) , we denote by h (L,α) the unique hermitian form on V ⊗ k L with respect to (L, α) satisfying We then have the following lemma. (2) the map τ ⊗ α is a unitary L/L 0 -involution on B ⊗ k L, and we have (3) Let B = M n (k) and let τ = Int(H)• * , for some invertible hermitian matrix H ∈ M n (k). Finally, let h H be the hermitian form on k n defined by h H : If h H ∼ =k λ 1 , . . . , λ n , λ i ∈ k × 0 , then T (B,τ ) ∼ =k 1, λ 1 λ −1 2 , . . . , λ j λ −1 i , . . . . Proof.
Since for b ∈ B the elements b ⊗ 1 span B ⊗ k L as an L-vector space, this yields the desired result.
(3) Let (X 1 , . . . , X n ) be an h H -orthogonal basis of k n , and let P ∈ GL n (k) be the matrix whose columns are X 1 , . . . , X n . The matrix D = P * HP is a diagonal invertible matrix (with diagonal entries lying in k × 0 ). For all M ∈ M n (k), easy computations show that we have We are now ready to state and prove the desired criterion.
Theorem 3.19. Assume that k/k 0 is a quadratic extension of number fields, whose non-trivial k 0 -automorphism is the complex conjugation. In particular, k 0 ⊂ R. It follows that for all non-zero x ∈ B ⊗ k L, we have (T (B,τ ) ) (L, ) (x, x) > 0.
In particular, for all non-zero b ∈ B, we get Conversely, assume that Trd B (τ (b)b) > 0 for all b ∈ B \ {0} and take L = C.