An octonion algebra originating in combinatorics

C.H. Yang discovered a polynomial version of the classical Lagrange identity expressing the product of two sums of four squares as another sum of four squares. He used it to give short proofs of some important theorems on composition of delta-codes (now known as T-sequences). We investigate the possible new versions of his polynomial Lagrange identity. Our main result shows that all such identities are equivalent to each other.


Introduction
C.H. Yang [8,9,10] discovered a polynomial version (see below) of the classical Lagrange identity on the product of two sums of four squares. He used his identity to give short and elegant proofs of some important theorems [10] in combinatorics of binary and ternary sequences. These results provide new methods for the construction of several classes of combinatorial objects such as T -sequences, orthogonal designs, and Hadamard matrices [6,10]. Our motivation and the main goal was to investigate the possible new versions of the polynomial Lagrange identity. However, our main result shows that all such identities are equivalent to each other (for the precise statement see Theorem 4.1).
Let A = Z[z, z −1 ] be the Laurent polynomial ring over the integers Z. For any f = f (z) = k a k z k ∈ A, a k ∈ Z, we define its conjugate as f * = f (z −1 ). We also say that a 0 is the constant term of f and write CT(f ) = a 0 . Note that CT(f f * ) = k a 2 k . Let A 0 be the fixed subring of A under the conjugation, i.e., A 0 = {f ∈ A | f = f * }. It is easy to see that A 0 = Z[z + z −1 ]. We embed the polynomial ring Z[t] into A by sending t → z + z −1 and view A as a Z[t]-algebra. It is easy to check that A = A 0 ⊕ A 0 z. Thus A is a free Z[t]-module of rank 2.

D.Ž. D -OKOVIĆ AND K. ZHAO
We are mainly interested in the free A-module E = A 4 . When viewed as a Z[t]-module it is again free but now its rank is 8. For any x = (x 0 , x 1 , x 2 , x 3 ) ∈ E we define its norm as N(x) = k x k x * k . Thus N : E → A 0 is a quadratic form on the A 0 -module E.
Now we state the Lagrange identity for Laurent polynomials (modified Theorem 1 in [10]). Let x = (x k ), y = (y k ) ∈ E and define (p, q, r, s) ∈ E by Yang's formulae: We can use these formulae to define an A 0 -bilinear multiplication "•" on E by x • y = (p, q, r, s). It is straightforward to verify that the Lagrange identity N(x • y) = N(x)N(y) is indeed valid.
We shall see in the next section that (E, •) is in fact an octonion algebra over Z[t]. In Section 3 we give an explicit description of the orthogonal group O(N) of the pair (E, N), see Theorem 3.4. In Section 4, we show that all A 0 -bilinear multiplications on E satisfying the Lagrange identity are equivalent, in the sense defined there, to the above multiplication "•".
Our main result can probably be generalized by replacing Z with a more general commutative ring. We decided not to pursue this here in order to preserve the essentially combinatorial flavour of the original problem.
We are grateful to the referee for correcting a couple of errors in the original proof of Lemma 4.3, for giving the stronger result in Theorem 4.3, and for his other detailed comments and suggestions.

Yang formulae define an octonion algebra
We shall often identify Z[t] with its image A 0 under this homomorphism. According to the definition in [1, Chapter III, §2], A is a quadratic Z[t]-algebra. By using the basis {1, z}, we see that its type is (−1, t). Indeed we have z 2 = −1 + tz. Moreover, (A, * ) is an example of a Cayley algebra (see loc. cit.).
Let H = A × A, a free A-module of rank 2. We shall also view it as a free Z[t]-module of rank 4 with basis (1, 0), (z, 0), (0, 1), (0, z). We make H into an associative noncommutative algebra by using the Cayley-Dickson process, i.e., we define the multiplication in H by The involution " * " on A extends to an involutory anti-automorphism of H by setting Thus H is an example of a quaternion algebra over Z[t], see loc. cit.
Let us write the above octonion multiplication in terms of the Abasis {e 0 , e 1 , e 2 , e 3 }. For x = (x 0 , x 1 , x 2 , x 3 ) and y = (y 0 , y 1 , y 2 , y 3 ), we find that is an isomorphism of this octonion algebra with the algebra (E, •) defined by the Yang formulae.

Orthogonal group
It is obvious that U 1 = {x ∈ A | xx * = 1} is the group of invertible elements of A. It consists of the elements ±z k , k ∈ Z. In our proofs below we shall often use the following obvious fact: The subset U 1 generates A as an additive group.

D.Ž. D -OKOVIĆ AND K. ZHAO
The ring homomorphisms ϕ : A → C are parametrized by nonzero complex numbers w: By definition, the homomorphism ϕ w corresponding to w sends the indeterminate z to w. The homomorphism ϕ w is compatible with the involutions (conjugation on A and complex conjugation on C), i.e, ϕ w (x * ) = ϕ w (x) for all x ∈ A, if and only if |w| = 1.
The following lemma will be used in the next section. Proof. Assume that f f * = m for some f ∈ A. By applying the homo- We also remark that, for x ∈ A, CT(xx * ) = 0 implies x = 0, and CT(xx * ) = 1 implies xx * = 1.
Denote the general linear group of the A 0 -module E by GL(A 0 , E). We introduce the orthogonal groups Note that Q is a symmetric A 0 -bilinear O(N)-invariant form, and that Q is nondegenerate, i.e., its kernel is 0. Let us begin with a useful remark.
2. An A 0 -linear map ϕ : E → E preserving N is automatically bijective, and so ϕ ∈ O(N). Indeed, as Q is nondegenerate, the injectivity is obvious. If S is the matrix of Q with respect to some A 0 -basis of E, then det(ϕ) 2 det(S) = det(S) = 0, forcing det(ϕ) = ±1, and ϕ is bijective.
The following result is crucial to this paper.
ij we see that, for each i, exactly one of a ij is ±1 and the other ones vanish. The rest follows easily.
(b) By applying the homomorphisms ϕ ±1 to xx * = −(z − z −1 ) 2 , we conclude that x is divisible by z − 1 and z + 1, and so we have x k x * k = 1. By comparing the constant terms, we conclude that exactly one of the x k belongs to U 1 while the other vanish. ( , which by the above remark implies that exactly one of the x k s is nonzero. This nonzero component belongs to ( Note that O 4 (Z) = E 16 Σ 4 is the semidirect product of the elementary abelian group E 16 of order 16 which acts on Z 4 by multiplying the coordinates with ±1 and the symmetric group Σ 4 of degree 4 which permutes the coordinates.
For  Proof. Let ϕ ∈ O(N). Since ϕ(U 4 ) = U 4 , for any index i there exist u i ∈ U 1 and an index i ′ such that ϕ(e i ) = u i e i ′ . Since the form Q is ϕ-invariant, the map i → i ′ must be a permutation of {0, 1, 2, 3}.
For any a ∈ A we have a + a * ∈ A 0 and so Now let a ∈ U 1 , a = ±1. Then each of the terms ϕ(ae i ), ϕ(a * e i ), au i e i ′ and a * u i e i ′ belongs to U 4 , and the sum of the first two terms is not 0. It follows that ϕ(ae i ) is equal to au i e i ′ or a * u i e i ′ .
Assume that there exist a, b ∈ U 1 different from ±1 such that ϕ(ae i ) = au i e i ′ while ϕ(be i ) = b * u i e i ′ . Then ϕ((a + b)e i ) = (a + b * )u i e i ′ , and by taking the norms we obtain the contradiction (a − a * )(b − b * ) = 0.

D.Ž. D -OKOVIĆ AND K. ZHAO
Consequently, one of the following two identities: ϕ(ae i ) = au i e i ′ or ϕ(ae i ) = a * u i e i ′ must hold for all a ∈ A. Equivalently, there exists The composite β = σ −1 u ϕτ is A-linear and we have σ −1 u ϕτ (e i ) = e i ′ for all i. Hence, ϕ = σ u βτ with σ u ∈ T , β ∈ Σ 4 and τ ∈ Γ.

Composition algebra structures on E
In this section we determine all A 0 -bilinear multiplications "·" on E which satisfy the (polynomial) Lagrange identity (4.1) N(x · y) = N(x)N(y).
Since E is a free A 0 -module of rank 8 and the form Q is nondegenerate, such algebra (E, ·) will be a composition algebra in the sense of the definition in [4, p. 305] provided that it has an identity element. We say that two A 0 -bilinear multiplications ⋆ and ⋄ are equivalent if there exist σ 1 , σ 2 , τ ∈ O(N) such that x ⋄ y = τ (σ 1 (x) ⋆ σ 2 (y)) for all x, y ∈ E.
The following theorem is our main result.
Theorem 4.1. Any A 0 -bilinear multiplication "·" on E satisfying the Lagrange identity (4.1) is equivalent to the multiplication "•" defined by Yang's formulae.
Proof. We shall reduce the proof to the special case where e 0 is the identity element of the A 0 -algebra (E, ·). Then the assertion of the theorem follows from the next theorem. Define A 0 -linear maps L, R : E → E by L(u) = e 0 · u and R(u) = u · e 0 . The Lagrange identity implies that L and R preserve N. By the remark 3.2, L, R ∈ O(N).
Next we use a well known argument due to, at least, Kaplansky [3]. We have N( Thus e 0 · e 0 is the identity element of the algebra (E, ⋆). As e 0 · e 0 ∈ U 4 , we have e 0 · e 0 = ae i for some a ∈ U 1 and some index i. Hence, there exists σ ∈ O(N) such that σ(e 0 ) = e 0 · e 0 . If x ⋄ y = σ −1 (σ(x) ⋆ σ(y)), then e 0 is the identity element of the algebra (A, ⋄).
We assume now that e 0 is the identity element of (E, ·). Consequently, this is a composition algebra and the alternative laws (4.2) x · (x · y) = (x · x) · y, (x · y) · y = x · (y · y) are valid [4, p. 306]. The reader should consult this book for additional properties of composition algebras. We state only a few properties that we need. Define the A 0 -linear map trace T : E → A 0 by T (x) = Q(e 0 , x). Then for any x ∈ E we have (4.3) x · x − T (x)x + N(x)e 0 = 0.
Linearizing gives Clearly Ae 0 ⊥ A 3 with respect to Q, where A 3 = Ae 1 + Ae 2 + Ae 3 . Recall that we have extended the conjugation * from A to E in Section 2. Moreover, note that It is easy to verify that The fact that in the next theorem one can assert that the two algebras are actually isomorphic is due to the referee. We shall break the proof into several lemmas. 1 Lemma 4.3. By replacing "·" with an isomorphic multiplication, we may also assume that (4.6) (ae 0 ) · y = ay for all a ∈ A and y ∈ E.
Proof. Let us fix an index i and b ∈ U 1 . Then the right multiplication by be i belongs to O(N), and Theorem 3.4 yields an index j as well as u j ∈ U 1 such that (ae 0 ) · (be i ) = u j ae j for all a ∈ A or (ae 0 ) · (be i ) = u j a * e j for all a ∈ A. Specializing a to 1 implies that j = i and u j = b, hence (ae 0 ) · (be i ) = abe i for all a ∈ A or (ae 0 ) · (be i ) = a * be i for all a ∈ A.

D.Ž. D -OKOVIĆ AND K. ZHAO
Assume that (ae 0 ) · (b 1 e i ) = ab 1 e i holds for all a ∈ U 1 and (ae 0 ) · (b 2 e i ) = a * b 2 e i holds for all a ∈ U 1 , for some units b 1 , b 2 different from ±1. Then (ae 0 ) · ((b 1 + b 2 )e i ) = (ab 1 + a * b 2 )e i holds for all a ∈ U 1 . By taking norms on both sides, we obtain that . This is clearly a contradiction.
Since U 1 generates A as an abelian group, we conclude that either (ae 0 ) · (be i ) = abe i for all a, b ∈ A or (ae 0 ) · (be i ) = a * be i for all a, b ∈ A. In particular, (ae 0 ) · (be i ) is A-linear in b.
If (ae 0 ) · (be 0 ) = a * be 0 for all a, b ∈ A, then setting b = 1 gives the contradiction: ae 0 = a * e 0 for all a ∈ A. Thus we have (ae 0 ) · (be 0 ) = abe 0 for all a, b ∈ A. For i = 0, there is an ε i ∈ {0, 1} such that If τ is the product of the τ ε i i with i = 0, then τ 2 = 1 and the multiplication "⋆" defined by x ⋆ y = τ (τ (x) · τ (y)) satisfies all the requirements.
We assume from now on that the identity (4.6) holds. This property will be shared by the modified multiplication which will be introduced in Lemma 4.6.
Proof. Recall that Ae 1 + Ae 2 + Ae 3 is the kernel of the trace T and is perpendicular to Ae 0 relative to Q. By using (4.4), we obtain that Lemma 4.6. By replacing "·" with an isomorphic multiplication, we may assume that (4.7) e i · e j = e k = −e j · e i for any cyclic permutation (i, j, k) of (1, 2, 3).
Proof. Let (i, j, k) be a cyclic permutation of (1, 2, 3). Since (e 0 + e i ) · (e 0 + e j ) = e 0 + e i + e j + e i · e j and N(e 0 + e j ) = N(e 0 + e j ) = 2, we have N(e 0 + e i + e j + e i · e j ) = 4. As e i · e j ∈ U 4 , Lemma 3.1 implies that e i · e j ∈ U 1 e k . A similar argument shows that e j · e i ∈ U 1 e k . In particular we have e 1 · e 2 = ue 3 for some u ∈ U 1 . The multiplication "⋆" defined by x ⋆ y = σ −1 a (σ a (x) · σ a (y)), where a = (1, 1, 1, u), has the previously stated property and sends (e 1 , e 2 ) → e 3 . By replacing "·" with this new multiplication, we may assume that e 1 · e 2 = e 3 .
Since (e i + e j ) · (e i + e j ) = e i · e j + e j · e i − 2e 0 , we have N(e i · e j + e j · e i − 2e 0 ) = 4. Since e i · e j + e j · e i ∈ Ae k , it follows that this sum is 0, i.e., e j · e i = −e i · e j . In particular, we have e 2 · e 1 = −e 3 . Thus the assertion of the lemma is valid if (i, j, k) = (1, 2, 3).
To finish the proof, we point out that the map σ a : (E, ⋆) → (E, ·) is an isomorphism of unital A 0 -algebras.
In view of the last lemma, we may assume now (and we do) that the identities (4.7) are valid for any cyclic permutation (i, j, k) of (1, 2, 3). Lemma 4.7. For any cyclic permutation (i, j, k) of (1, 2, 3) we have for all a, b ∈ A. Consequently, the multiplications "·" and "•" coincide.
Proof. In view of (4.4), it suffices to prove the first equality. We first treat the case a = 1. By applying the linearized left alternative law, we find that e i · (be j ) = e i · ((be 0 ) · e j ) = (e i · (be 0 ) + (be 0 ) · e i ) · e j − (be 0 ) · (e i · e j ) = (b + b * )e k − be k = b * e k .
The referee supplied an alternative proof of Theorem 4.2. His proof proceeds first by changing scalars from the base ring Z[t] to its quotient field Q(t), and then applying a theorem of Thakur [7] (see also [5, Chapter VIII, Exercise 6]) which establishes a connection between octonion algebras and ternary hermitian forms. In the case of Yang algebra, let us write E = Ae 0 ⊕ A 3 , where A 3 is the column A-space. Extending the conjugation of A componentwise to A 3 , we have a hermitian form h : A 3 × A 3 → A given by h(v, w) = v t w * . With this notation, the Yang multiplication is given by where a, b ∈ A and v, w ∈ A 3 . The sign × stands for the ordinary cross product in 3-space.

Addendum
Alberto Elduque [2] has found a simpler proof of Theorem 4.2. Here we present his simplification. We point out that the involution " * " coincides with the the involution " − " in [4, p. 305] and so it is an involution of the algebra (E, ·).
The first paragraph of the proof of Lemma 4.3 implies that Ae 0 · e i = Ae i for each i. By applying the involution " * " we obtain that also e i · Ae 0 = Ae i for each i. By (4.3) we have (ze 0 ) · (ze 0 ) = −e 0 + tze 0 = z 2 e 0 . It follows that (xe 0 ) · (ye 0 ) = xye 0 for all x, y ∈ A. By (4.3), we also have e 1 · e 1 = −e 0 . By applying [4, Lemma (7.1.7)], with B = Ae 0 and v = e 1 , we conclude that P = Ae 0 + Ae 1 is a subalgebra of (E, ·). From the proof of that lemma it also follows that the multiplications "·" and "•" coincide on P .
Note that this proof does not use any of the lemmas 4.4-7, and uses only the first paragraph of the proof of Lemma 4.3.