Pfister Involutions

E. Bayer–Fluckiger and R. Parimala

Introduction

Pfister forms play a prominent role in the algebraic theory of quadratic forms. On the other hand, involutions on central simple algebras share many properties with quadratic forms. Hence it is natural to look for an analog of the notion of Pfister form in the framework of algebras with involution.

The aim of the present paper is to propose such a notion. An n–fold Pfister involution (or Pfister involution, for short) will be by definition a central simple algebra with an orthogonal involution which is a tensor product of n quaternion algebras with involution. We show that if n = 4, then after passing to any splitting field of the algebra, the involution is induced by a Pfister form. For n ≤ 3, this was already proved by D. Tao. We also compute cohomological invariants of 2-fold Pfister involutions, and raise some open questions.

1. Definitions and notation

Let F be a field of characteristic ≠ 2. Let D be a central simple algebra over F. We say that an involution σ : D → D is of the first kind if its restriction to F is the identity. After extension to a splitting field of D, any involution of the first kind is induced by a symmetric or by a skew-symmetric form. We say that the involution is of the orthogonal type in the first case, and of the symplectic type in the second case.

Definition. Let (D, σ) be a central simple algebra endowed with an involution of the first kind. We say that (D, σ) is an n–fold Pfister involution if σ is of the orthogonal type, and if there exist quaternion algebras H_{1},…,H_{n}, and involutions σ_{i} : H_{i} → H_{i} such that

(D, σ) ≃ (H_{1}, σ_{1}) ⊗ ... ⊗ (H_{n}, σ_{n}).

Let F_{s} be a separable closure of F, and set H^{n}(F) = H^{n}(Gal(F_{s}/F), Z/2Z).

2. Statement of results and open questions

Let (D, σ) be a central simple algebra with an orthogonal involution of the first kind. Let K be a splitting field of D. Then after tensoring with K, the involution is induced by a quadratic form q defined over K. It is natural to ask whether (D, σ) is a Pfister involution if and only if q is similar to a Pfister form. This is proved by Tao [5] for algebras of degree ≤ 8 (cf. [4], [2], p.150).

In the present paper, we show that for a 4–fold Pfister involution the associated quadratic form is a 4–fold Pfister form (see §5, th. 2). It would be interesting to know whether the converse also holds : this is still an open question.

It is also natural to try to define n–dimensional cohomological invariants of n–fold Pfister involutions. More precisely, suppose that (D, σ) is an n–fold Pfister involution. Let X_{D} be the Brauer–Severi variety of D and q_{F (X_{D})} the associated quadratic form. For n ≤ 4, we know that q_{F (X_{D})} is an n–fold Pfister form. Let E_{n}(D) = Ker(H^{n}(F) → H^{n}(F (X_{D})). Does there exist an invariant e_{n}(D, σ) ∈ H^{n}(F)/E_{n}(D) with the property that e_{n}(q_{F (X_{D})}) is equal to the image of e_{n}(D, σ) in H^{n}(F (X_{D})) ? In §4, we define such an invariant for 2–fold Pfister involutions.

3. A lemma

The following lemma will be used in §4 and §5 :

Lemma 1. Let τ be an orthogonal involution on D. Let ϕ_{τ} : D ⊗_{F} D → End_{F} (D) be the isomorphism induced by ϕ_{τ} (v, w)(x) = vxτ(w). Let u ∈ D^{∗} be such that τ(u) = u. Then the involution τ ⊗ Int(u)oτ transports under ϕ_{τ} to σ_{q}, where σ_{q} is the adjoint involution on End_{F} (D) induced by the quadratic form q : D → F given by q(x) = Trd(xu^{−1}τ(x)).

Proof. See [2], II.1, page 133.

4. Invariants of 2-fold Pfister involutions

Let (H_{1}, τ_{1}) and (H_{2}, τ_{2}) be two quaternion algebras with involution, and set (D, τ) = (H_{1}, τ_{1}) ⊗ (H_{2}, τ_{2}). Suppose that τ is an orthogonal involution. By [1] we know that disc(τ) = 1 and that (D, τ) = (H_{3}, τ_{3}) ⊗ (H_{4}, τ_{4}), where H_{3}, H_{4} are quaternion algebras, and τ_{3}, τ_{4} are their canonical involutions. We have C(D, τ) = H_{3} × H_{4} and the factors are unique up to switch.

Let X_{D} be the Severi–Brauer variety of D, and set

E_{2}(D) = Ker(H^{2}(F) → H^{2}(F (X_{D})).

Then E_{2}(D) is the subgroup of H^{2}(F) generated by the class of D. Let us denote by [H_{3}] the class of H_{3} in H^{2}(F)/E_{2}(D). Then [H_{3}] is an invariant of (D, τ).

If D is split, then H_{3} ≃ H_{4}, and (D, τ) = (H_{3}, τ_{3}) ⊗ (H_{4}, τ_{4}) ≃ (End(H_{3}), τ_{q}), where q : H_{3} → F is the quadratic form q(x) = Trd(xτ_{3}(x)) = 2Nrd(x), cf. Lemma 1. Thus q ≃ 2N_{H_{3}}, where N_{H_{3}} is the norm form of the quaternion algebra H_{3}. This implies that e_{2}(q) ∈ H^{2}(F) is the class of H_{3}. Set e_{2}(D, τ) = [H_{3}] ∈ H^{2}(F)/E(D). This is an invariant of the algebra with involution (D, τ), and it coincides with e_{2}(q) if D ≃ M_{4}(F) and τ = τ_{q}.

Theorem 1. Let (D, τ) = (H_{1}, Int(u)oτ_{1}) ⊗ (H_{2}, Int(v)oτ_{2}) where the H_{i} are quaternion algebras, τ_{i} the canonical involution of H_{i} and u ∈ H_{1}, v ∈ H_{2} satisfy τ_{1}(u) = −u, τ_{2}(v) = −v. Then e_{2}(D, τ) = [H_{1}] + (−Nrd(u)) ∪ (−Nrd(v)) in H^{2}(F)/E(D).

Proof. Suppose that H_{1} ⊗ H_{2} is split. We may assume that H_{1} = H_{2}. Then we have

(D, τ) = (H_{1}, Int(u)oτ_{1}) ⊗ (H_{1}, Int(θ)oInt(u)oτ_{1}),

where θ = vu^{−1}. Then τ = τ_{q}, where q : H_{1} → F is the quadratic form

q(x) = Trd(xθ^{−1}Int(u)oτ_{1}(x)) = Trd(xuv^{−1}uτ_{1}(x)u^{−1}),

cf. lemma 1. Let us define b_{q} : H_{1} × H_{1} → F by b_{q}(x, y) = Trd(xuv^{−1}uτ_{1}(y)u^{−1}). Set λ = Trd(uv^{−1}).

If λ = 0, then q(1) = Trd(uv^{−1}) = 0. Thus q is isotropic and has discriminant 1, hence it is hyperbolic. As λ = 0, the elements 1, u, v, uv form a quaternionic basis for H_{1}, so that H_{1} = (−N(u)) ∪ (−N(v)) and [H_{1}] + (−N(u)) ∪ (−N(v)) = 0. This concludes the proof in case λ = 0.

Let us suppose that λ ≠ 0. Then 1, u and vu^{−1} − \frac{2}{λ} are mutually orthogonal.

Indeed, b_{q}(1, u) = −Trd(uv^{−1}u^{2}u^{−1}) = −\frac{Trd(uv^{−1}u^{−1})}{u^{2}} = 0, as Trd(v) = 0. We have b_{q}(1, vu^{−1} − \frac{2}{λ}) = Trd(uv^{−1}u(u^{−1}v − \frac{2}{λ})u^{−1}) = Trd(1− \frac{2}{λ}uv^{−1}) = 0. Further, b_{q}(u, vu^{−1} − \frac{2}{λ}) = Trd(u^{2}v^{−1}u(u^{−1}v − \frac{2}{λ})u^{−1}) = Trd(−N(u)u^{−1} + \frac{2}{λ}N(u)v^{−1}) = 0.

We have q(1) = λ, q(u) = Trd(−u^{2}v^{−1}u) = N(u)λ, and

q(vu^{−1} − \frac{2}{λ}) = Trd[(vu^{−1} − \frac{2}{λ})uv^{−1}u(u^{−1}v − \frac{2}{λ})u^{−1})]

= Trd[(u − \frac{2}{λ}uv^{−1}u)(u^{−1}vu^{−1} − \frac{2}{λ}u^{−1})].

Hence q(vu^{−1} − \frac{2}{λ}) = Trd(vu^{−1} − \frac{4}{λ} + \frac{4}{λ^{2}} uv^{−1}) = Trd(vu^{−1}) − \frac{4}{λ} = \frac{λ}{N(uv^{−1})} − \frac{4}{λ}.

Since disc(q) = 1, we have

q ≃ λ < 1, N(u), \frac{1}{N(uv^{−1})} − \frac{4}{λ^{2}} , N(u)(\frac{1}{N(uv^{−1})} − \frac{4}{λ^{2}} ) > .

We have to check that

(−N(u)) ∪ (−N(v)) + [H_{1}] = (−N(u)) ∪ −(\frac{1}{N(uv^{−1})} − \frac{4}{λ^{2}} ).

Since u, v are trace zero elements, there exists a trace zero element w such that 1, u, w, uw is a quaternionic basis for H_{1}, and v = au + bw, [H_{1}] = (−N(u)) ∪ (−N(w)).

Hence we have :

(−N(u)) ∪ (−N(v)) + [H_{1}] = (−N(u) ∪ (N(v)N(w)).

It suffices to show that N(v)N(w)(\frac{4}{λ^{2}} − \frac{1}{N(uv^{−1})}) is a value of < 1, N(u) >.

We have N(v)N(w)(\frac{4}{λ^{2}} − \frac{1}{N(uv^{−1})}) = (a^{2}N(u) + b^{2}N(w))N(w)(\frac{4}{Trd(uv^{−1})^{2}} − \frac{1}{N(uv^{−1}} )).

Note that

Trd(uv^{−1}) = \frac{Trd(vu^{−1})}{N(vu^{−1})} = \frac{2a}{a^{2} + b^{2}N(wu^{−1})},

and

N(uv^{−1}) = \frac{1}{N(vu^{−1})} = \frac{1}{a^{2} + b^{2}N(wu^{−1})}.

We have

N(v)N(w)(\frac{4}{λ^{2}} − \frac{1}{N(uv^{−1})}) = (a^{2}N(u)+b^{2}N(w))N(w)(a^{2}+b^{2}N(wu^{−1}))[\frac{a^{2} + b^{2}N(wu^{−1})}{a^{2}} −1]

= \frac{[a^{2}N(u) + b^{2}N(w)]^{2}}{N(u)^{2}a^{2}} N(w)^{2}b^{2},

which is a square, hence a norm. Therefore we have e_{2}(q) = (−N(u)) ∪ (−N(v)) + [H_{1}], as claimed.

5. 4–fold Pfister involutions

The aim of this section is to prove the following result :

Theorem 2. Let (D, σ) be a 4–fold Pfister involution. Let K be a splitting field of D, and let q = q_{K} be the quadratic form induced by σ over K. Then q is a Pfister form.

The proof will make use of the following lemmas :

Lemma 2. Let H_{1} and H_{2} be two quaternion algebras, and set D = H_{1} ⊗ H_{2}. Let τ be an orthogonal involution with trivial discriminant on D, and let u ∈ D^{∗} with τ(u) = u. Then the quadratic form q : D → F defined by q(x) = Trd(xuτ(x)) is in I^{3}(F).

Proof. By lemma 1, the involution τ ⊗(Intuoτ) on D ⊗D transports into σ_{q} on End_{F} (D). Thus disc(q) = 1 and C_{0}(q) ≃ C(D ⊗ D, τ ⊗ (Intuoτ)) is split (see for instance [2],p. 150). This implies that C(q) is split and that q is in I^{3}(F).

Lemma 3. Let H_{1} and H_{2} be two quaternion algebras, and set D = H_{1} ⊗H_{2}. Let τ_{i} be the canonical involution on H_{i}, and set τ = τ_{1} ⊗ τ_{2}. For any u ∈ D^{∗} such that Nrd(u) ∈ F^{∗2}, we define the quadratic form q_{u} : D → F by q_{u}(x) = Trd(xuτ(x)). Then q_{u} is a Pfister form.

Proof. It suffices to show that if q_{u} is isotropic then it is split. Let us suppose q_{u} isotropic. By a general position argument, there exists y ∈ D^{∗} such that q_{u}(y) = 0. The quadratic forms q_{u} and q_{yuτ(y)} are isometric under the map D → D, x ↦ xy^{−1}. Hence replacing u by yuτ(y), we may assume that u has the additional properties Trd(u) = 0, τ(u) = u.

For x ∈ H_{1}, we have τ(x) = x^{−1}Nrd(x) so that q_{u}(x) = Trd(xux^{−1}Nrd(x)) = Nrd(x)Trd(xux^{−1}) = 0. Thus H_{1} is totally isotropic for q_{u} : indeed, H_{1} is totally isotropic for q_{z} for any z ∈ D^{∗} with Trd(z) = 0, τ(z) = z.

The involution Int(u)oτ on D has trivial discriminant, hence D = H′_{1} ⊗ H′_{2} and Int(u)oτ = τ′_{1} ⊗ τ′_{2}, where τ′_{i} denotes the canonical involution on H′_{i}. Let j ∈ H′_{2} with Trd(j) = 0 and let (H′_{1})^{0} be the set of trace zero elements in H′_{1}. Then W = (H′_{1})^{0}j is a 3–dimensional subspace of D elementwise fixed by Int(u)oτ such that for every y ∈ W, we have y^{2} ∈ F. Set T = {y ∈ D, Trd(uτy) = 0}. Then dim(T) = 15 and dim(T ∩ W) ≥ 2. Let V_{j} ⊂ T ∩ W be a 2–dimensional subspace. Then for y ∈ V_{j}, we have Trd(uτ(y) = 0, y^{2} ∈ F and Int(u)oτ(y) = y. This implies that uτ(y) = yu, hence τ(uτ(y) = yu = uτ(y), i.e. uτ(y) is symmetric under τ. Since for y ∈ V_{j} we have y^{2} ∈ F, we have Trd(yuτ(y) = Trd(y^{2}u) = y^{2}Trd(u) = 0. Thus V_{j} is totally isotropic for q_{u}.

Further for y ∈ V_{j}, the element uτ(y) being symmetric for τ with trace zero, H_{1} is totally isotropic for q_{uτ(y)}. Hence for x, z ∈ H_{1}, we have Trd(xuτ(y)τ(z)) = 0. In particular, setting z = 1, we get, for x ∈ H_{1}, Trd(xuτ(y)) = 0. Hence H_{1} is orthogonal to V_{j} with respect to q_{u}.

Let i, j, k be a basis of trace zero elements in H′_{2} with i = jk, jk = −kj. Suppose that V_{j} ⊂ H_{1}. Then we have V_{j} ⊂ H_{1}^{0}. If V_{k} ⊂ H_{1}, then we have V_{j} + V_{k} ⊂ H_{1}^{0}, and V_{j} ∩ V_{k} = 0. Hence we get dim(V_{j} + V_{k}) = 4, contradicting dim(H_{1}^{0}) = 3.

Thus there exists y ∈ V_{j} or V_{k} such that y ∉ H_{1}, y totally isotropic for q_{u} and Trd(xuτ(y)) = 0 for all x ∈ H_{1}. Thus H_{1} ⊕ F y is a 5–dimensional totally isotropic subspace for q_{u}. Therefore the anisotropic rank of q_{u} is at most 6 and by Lemma 2, we have q_{u} ∈ I^{3}(F). Hence q_{u} is hyperbolic.

Proof of theorem 2. Since (D, σ) is Pfister, we may write (D, σ) as

(D_{1}, σ_{1}) ⊗ (D_{2}, σ_{2}) ⊗ (D_{3}, σ_{3}) ⊗ (D_{4}, σ_{4}),

where the D_{i}’s are quaternion algebras and σ_{i}’s are orthogonal involutions. Using [1], we may write (D_{1}, σ_{1})⊗(D_{2}, σ_{2}) as (H_{1}, τ_{1})⊗(H_{2}, τ_{2}), where the H_{i}’s are quaternion algebras and the τ_{i}’s their canonical involutions. To prove the theorem, we may assume that D is split.

Using [1] again, we may write (D_{3}, σ_{3})⊗(D_{4}, σ_{4}) = (H_{1} ⊗H_{2}, Int(u^{−1})o(τ_{1} ⊗τ_{2}), with u ∈ H_{1} ⊗ H_{2} satisfying Nrd(u) ∈ F^{∗2}. By lemma 1, the form q associated to σ is given by q(x) = Trd(xuτ_{1} ⊗ τ_{2}(x)). By lemma 3, the form q is indeed a 4–fold Pfister form.

Bibliography

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E. Bayer Fluckiger Département de Mathématiques Ecole Polytechnique Fédérale de Lausanne 1015 Lausanne Switzerland eva.bayer@epfl.ch

R. Parimala Tata Institute of Fundamental Research Homi Bhambha Road Mumbai 400005 India parimala@math.tifr.res.in