On the computation of trace forms of algebras with involution

Grégory Berhuy

Introduction: In this article, we determine up to isometry the trace form T_{σ} of central simple algebras with involution over a field k, which has at most one ordering and such that I^{3}(k) is torsion free. This quadratic form has been introduced by Weil in [W], and has been studied by Lewis [L] and Quéguiner [Q2].

Definitions and notation: Let A be a central simple algebra of degree n over a field k of characteristic different from 2. An involution on A is a ring antiautomorphism of order at most 2. An involution σ is of the first kind if σ|k = Id_{k}, and of the second kind if σ|k is a non trivial involution on k, denoted by¯ . In the last case, k is a quadratic extension of the subfield k_{0} fixed by¯ . So we have k = k_{0}(\sqrt{α}), α ∈ k_{0}^{∗}/k_{0}^{∗2}. The involution¯ is then defined by \overline{u + v\sqrt{α}} = u − v\sqrt{α}, with u, v ∈ k_{0}. If A is split, the involutions of the first kind are exactly the involutions adjoint to symmetric or skew symmetric bilinear forms. In the case where A is a central simple algebra with an involution of the first kind, and if L is a splitting field of A, A ⊗ L is a split algebra and σ ⊗ Id_{L} is an involution on A ⊗ L, then adjoint to a bilinear form b. We say that σ is of orthogonal type if b is symmetric , and of symplectic type if b is skew symmetric. This definition does not depend on the splitting field. Now consider the function (x, y) ∈ A × A ↦ Trd_{A}(σ(x)y). If the involution σ is of the first kind, this is a nondegenerate symmetric bilinear form of dimension n^{2} over k, and we denote by T_{σ} the corresponding quadratic form. If σ is of the second kind, this is a hermitian form over k, denoted by H_{σ}. Then we define T_{σ} by T_{σ} (x) = H_{σ} (x, x). This is a nondegenerate quadratic form over k_{0} of dimension 2n^{2}. If σ is of the first kind, we set Alt(σ)={ x − εσ(x), x ∈ A }, with ε = 1 if σ is orthogonal, and ε = −1 if σ is symplectic. Recall now the definition of the determinant of an involution of the first kind, given by Knus, Parimala and Sridharan:

Definition 1 (see [KPS]): Let A be an even dimensional central simple algebra with an involution σ of the first kind. The determinant of σ, denoted by d(σ) is the square class of Nrd_{A}(u), where u is any element of A^{∗} ⋂ Alt(σ), and where Nrd_{A} is the reduced norm.

We have d(σ) = 1 if σ is symplectic (take u = 1). If A is split, and σ is adjoint to b, then d(σ) = det(b).

In this paper, we denote by ℍ the two-dimensional hyperbolic plane, and by I^{3}(k) the cube of the fundamental ideal of the Witt ring of k. Finally, we will often use the following formula, which can be easily deduced from the definition of the Hasse-Witt invariant (see also [S], lemma 12.6):

w_{2}(q_{1} ⊥ q_{2}) = w_{2}(q_{1}) + (det(q_{1}), det(q_{2})) + w_{2}(q_{2}).

In the following, we determine the isomorphism class of T_{σ}, when k or k_{0} has at most one ordering and such that I^{3} is torsion free. For this, we use the well known property that quadratic forms over these fields are classified by dimension, determinant, Hasse-Witt invariant and the signature (see [EL]). Then we use the results of A. Quéguiner, who computed in [Q2] the the determinant and the Hasse invariant of T_{σ}. This last invariant has also been computed by Lewis in [L].

I. Case I^{3} = 0

In this part, we will prove the following result:

Theorem 1: Let A be a central simple algebra of degree n over k, with an involution σ.

1. If σ is of the second kind, set k = k_{0}(\sqrt{α}), and assume that I^{3}(k_{0}) = 0. Then T_{σ} ≃< (−α)^{n} >⊥ (2n^{2} − 1) < 1 >

2. If σ is of the first kind, assume that I^{3}(k) = 0. Then:

(a) If n is odd, A is a split algebra, σ is adjoint to a nondegenerate symmetric bilinear form, and we have T_{σ} ≃ n^{2} < 1 >.

(b) If n ≡ 0 [4], we have T_{σ} ≃ (n^{2} − 2) < 1 >⊥< d(σ), d(σ) >.

(c) If n ≡ 2 [4], then A is Brauer equivalent to a quaternion algebra Q. Moreover:

i. If σ is symplectic, set Q = (a, b). Then we get T_{σ} ≃< 1, −a, −b, ab >⊥ (n^{2} − 4) < 1 >.

ii. If n = 2, that is A = Q, and if σ is orthogonal, there exists b ∈ k^{∗} such Q = (a, b), with a = −d(σ), and such that the associated standard basis is orthogonal with respect to T_{σ}. Then we have T_{σ} ≃< 1, −a, b, −ab >.

iii. If n > 2 and if σ is orthogonal, set Q = (a, b). Then T_{σ} ≃< −d(σ), −d(σ), 1, −a, −b, ab >⊥ (n^{2} − 6) < 1 >.

We will divide the proof of this theorem into five lemmas.

A. Case of involutions of the second kind

Let σ be an involution on A of the second kind and k = k_{0}(\sqrt{α}). Then we have the following result (see [Q2]):

Proposition 1: We have det(T_{σ}) = (−α)^{n} and w_{2}(T_{σ}) = 0.

Assume now that I^{3}(k_{0}) = 0. The quadratic form T_{σ} is then uniquely determined by its determinant, dimension and Hasse invariant (see [EL]). Then we get:

Lemma 1: T_{σ} ≃< (−α)^{n} >⊥ (2n^{2} − 1) < 1 >.

B. Case of involutions of the first kind

Assume first that n is odd. Then it is well known that, if A is an algebra with an involution σ of the first kind, then A is split and σ is orthogonal (see [KMRT] Corollary 2.8). Thus σ is adjoint to a symmetric bilinear form b over k, which is uniquely determined up to similarity. Since n is odd, we have det(< λ > ⊗b) = λdetb, so we can assume that det(b) = 1, after multiplying b by a scalar. Moreover, we know that T_{σ} ≃ b ⊗ b (see [KMRT], Proposition 11.4). Since we have w_{2}(b ⊗ b) = (−1, det(b)) if b is symmetric (see [Q2]), we get w_{2}(T_{σ}) = (−1, 1) = 0. Finally, the two quadratic forms T_{σ} and n^{2} < 1 > have the same invariants, and we get:

Lemma 2: Let A be a central simple algebra over k of odd degree, with an involution σ of the first kind. Then T_{σ} ≃ n^{2} < 1 >, where n = degA.

Assume now that n = 2m. Then det(T_{σ}) = 1 for any involution (see [Q2]). Recall also that there exists an involution of the first kind on A if and only if A ⊗ A is split (see [J] Theorem 5.2.1). The class of A in the Brauer group of k, denoted by [A], is then an element of Br_{2}(k), which justifies the following statement (see [Q2]):

Proposition 2: We have w_{2}(T_{σ}) = \{ \begin{matrix}m(−1, −1) + m[A] if σ is symplectic \\ (−1, d(σ)) + m[A] if σ is orthogonal \end{matrix}

Assume that m is even. Since d(σ) = 1 if σ is symplectic, we get w_{2}(T_{σ}) = (−1, d(σ)) in all cases. Since the invariants of T_{σ} coincide with the invariants of the quadratic form (n^{2} − 2) < 1 >⊥< d(σ), d(σ) >, we have:

Lemma 3: Let A be a central simple algebra of degree n ≡ 0 [4], with an involution σ of the first kind. Then T_{σ} ≃ (n^{2} − 2) < 1 >⊥< d(σ), d(σ) >.

Assume now that m is odd. We know that the index of a central simple algebra of even degree with an involution of the first kind is a power of 2 (see [KMRT], Corollary 2.8). In our case, A is then Brauer equivalent to an algebra of degree 1 or 2. So A is equivalent to k or to a division quaternion algebra. Since k is equivalent to M_{2}(k) ≃ (1, 1), we have [A] = (a, b) in all cases. In the symplectic case, we get w_{2}(T_{σ}) = (−1, −1) + (a, b). An easy computation shows that w_{2}(< 1, −a, −b, ab >) = (−1, −1) + (a, b). Since w_{2}(< 1, −a, −b, ab >⊥ (n^{2} − 4) < 1 >) = w_{2}(< 1, −a, −b, ab >), and the determinant of this form is trivial, we get:

Lemma 4: Let A ≃ M_{m}(Q), where m is odd and Q is the quaternion algebra (a, b), with a symplectic involution σ. Then T_{σ} ≃< 1, −a, −b, ab >⊥ (n^{2} − 4) < 1 >.

Lemma 5: Let A ≃ M_{m}(Q), where m ≥ 1 is odd and Q is a quaternion algebra, with an orthogonal involution σ. If m = 1, there exists b ∈ k^{∗} such Q = (a, b), with a = −d(σ), and such that the associated standard basis is orthogonal with respect to T_{σ}. Then we have T_{σ} ≃< 1, −a, b, −ab >. If m > 1, set Q = (a, b). Then T_{σ} ≃< −d(σ), −d(σ), 1, −a, −b, ab >⊥ (n^{2} − 6) < 1 >.

Proof: If m > 1, set Q = (a, b). We have w_{2}(< −d(σ), −d(σ) >) = (d(σ), −1) + (−1, −1) and w_{2}(< 1, −a, −b, ab >) = (−1, −1) + (a, b). These two quadratic forms have a trivial determinant, so we easily get w_{2}(< −d(σ), −d(σ), 1, −a, −b, ab >⊥ (n^{2} − 6) < 1 >) = (−1, d(σ)) + (a, b). Since this last quadratic form is n^{2}-dimensional and has a trivial determinant, we have T_{σ} ≃< −d(σ), −d(σ), 1, −a, −b, ab >⊥ (n^{2} − 6) < 1 >. The case m = 1 can be found in [Q2]. Note that in [Q2], it is shown that T_{σ} ≃< 2, −2a, 2b, −2ab >, but it is easy to see that this last form is isometric to < 1, −a, b, ab > (it suffices to compare the invariants).

These five lemmas give the proof of theorem 1.

All these results are true in particular on local fields, function fields and non formally real number fields.

II. Case I^{3} torsion free and k uniquely ordered.

Recall first some facts about the signature of the form T_{σ}. If (k, P) is an ordered field, we denote by k_{P} the corresponding real closure, which is uniquely ordered by squares (see [S],Theorem 3.1.14). If q is a quadratic form over (k, P), then sign_{P} (q) = sign(q_{k_{P}}). Now let A be a central simple algebra over k. If σ is an involution of the first kind on A, then sign_{P} (T_{σ}) is a square. Moreover, if A ⊗ k_{P} ∼ 1 and σ is symplectic, or if A ⊗ k_{P} ≁ 1 and σ is orthogonal, then sign_{P} (T_{σ}) = 0. See [KMRT] for more details. If σ is of the second kind, assume that k_{0} is ordered by P, and k = k_{0}(\sqrt{α}). Then sign_{P} (H_{σ}) is a square and sign_{P} (T_{σ}) = 2sign_{P} (H_{σ}) (see [Q1]). Moreover, this signature is equal to 0 if α > 0. Now set sign_{P} (σ) = \sqrt{sign_{P} (T_{σ})} or

\sqrt{\frac{1}{2}sign_{P} (T_{σ})} respectively if σ is of the first or the second kind. In all cases, the signature and the degree of A have the same parity (see [KMRT] and [Q1]).

Before stating the results, recall that if the degree of A is odd, and if σ is an involution of the first kind on A, then A is split and σ = σ_{b} where b is symmetric. Moreover, when the determinant is defined, we have d(σ_{b}) = det(b), and d(σ) = 1 if σ is symplectic.

In the following, we will show:

Theorem 2: Let A be a central simple algebra of degree n, with an involution σ. We keep the notation of Theorem 1.

1. If σ is of the second kind, set k = k_{0}(\sqrt{α}) and sign(σ) = s (recall that s = 0 if α > 0). Assume that k_{0} is uniquely ordered and I^{3}(k_{0}) is torsion free. Then

T_{σ} ≃ \{ \begin{matrix}< (−α)^{n} >⊥ (n^{2} + s^{2} − 1) < 1 >⊥ (n^{2} − s^{2}) < −1 > if α < 0 \\ n^{2}ℍ if α > 0 and n is even \\ < 1, −α >⊥ (n^{2} − 1)ℍ if α > 0 and n is odd \end{matrix}

2. If σ is of the first kind, assume that k is uniquely ordered and that I^{3}(k) is torsion free, and set sign(σ) = s.

(a) If n is odd, then σ is adjoint to a symmetric bilinear form and we have T_{σ} ≃ \frac{n^{2}+s^{2}}{2} < 1 >⊥ \frac{n^{2}−s^{2}}{2} < −1 >.

(b) If n ≡ 0 [4], then we have:

i. If d(σ) > 0, we have s ≡ 0 [4] and T_{σ} ≃ (\frac{n^{2}+s^{2}}{2} − 2) < 1 >⊥ \frac{n^{2}−s^{2}}{2} < −1 >⊥< d(σ), d(σ) >

ii. If d(σ) < 0, we have s ≡ 2 [4] and T_{σ} ≃ \frac{n^{2}+s^{2}}{2} < 1 >⊥ (\frac{n^{2}−s^{2}}{2} − 2) < −1 >⊥< d(σ), d(σ) >

(c) Assume now that n ≡ 2 [4].

i. Assume that σ is symplectic.

A. If A ⊗ k_{P} ∼ 1, then we have s = 0 and T_{σ} ≃< 1, −a, −b, ab >⊥ \frac{n^{2}−4}{2}ℍ.

B. If A ⊗ k_{P} ≁ 1, then s ≡ 2 [4] and T_{σ} ≃< 1, −a, −b, ab >⊥

(\frac{n^{2}+s^{2}}{2} − 4) < 1 >⊥ \frac{n^{2}−s^{2}}{2} < −1 >. ii. If σ is orthogonal and n = 2, then T_{σ} ≃< 1, −a, b, −ab > and

s = 0 or 2 depending of the signs of a and b. iii. If σ is orthogonal and n > 2, then

A. If A ⊗ k_{P} ≁ 1, then s = 0, d(σ) > 0 and we have T_{σ} ≃ < −d(σ), −d(σ), 1, −a, −b, ab >⊥ (\frac{n^{2}}{2} − 4) < 1 > ⊥ (\frac{n^{2}}{2} − 2) < −1 >.

B. If A ⊗ k_{P} ∼ 1 and d(σ) > 0, then s ≡ 2 [4], and we have T_{σ} ≃< −d(σ), −d(σ), 1, −a, −b, ab >⊥ (\frac{n^{2}+s^{2}}{2} − 2) < 1 > ⊥ (\frac{n^{2}−s^{2}}{2} − 4) < −1 >.

C. If A ⊗ k_{P} ∼ 1 and d(σ) < 0, then s ≡ 0 [4], and we have T_{σ} ≃< −d(σ), −d(σ), 1, −a, −b, ab >⊥ (\frac{n^{2}+s^{2}}{2} − 4) < 1 > ⊥ (\frac{n^{2}−s^{2}}{2} − 2) < −1 >.

These results are true in particular over euclidean fields and over ℚ.

Proof: It suffices to show that the quadratic forms have the correct invariants. Note that there exist a unique division quaternion algebra on k_{P}, namely (−1, −1). Then (a, b) ⊗ k_{P} is a division algebra if and only if a < 0 and b < 0. Then sign_{P} (< 1, −a, −b, ab >) = 0 or 4 if A ⊗ k_{P} is a split algebra or a division algebra respectively. Then it is easy to verify that the signatures are the correct ones. Now we justify the conditions on s and d(σ): if q is a quadratic form over k, we have sign(q ⊗ k_{P})=sign_{P} (q), w_{2}(q ⊗ k_{P}) = w_{2}(q) ⊗ k_{P}. Now quadratic forms over k_{P} are the forms u < 1 >⊥ v < −1 >, and the unique quadratic form with dimension n^{2} and signature s^{2} is the form \frac{n^{2}+s^{2}}{2} < 1 >⊥ \frac{n^{2}−s^{2}}{2} < −1 >. If we compute the Hasse invariant of this form and if we compare it with the Hasse invariant of T_{σ} ⊗ k_{P}, we get the announced conditions.

Acknowledgments: I would thank the referee for his remarks, which have permitted to simplify the statements of the main results.

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