16-DIMENSIONAL ALGEBRAS WITH INVOLUTION

Adrian R. Wadsworth

June 1990

These notes gives a simplified account of some of the very nice results in [KPS] — simplified because we consider algebras over a field instead of Azumaya algebras over an arbitrary commutative ring.

Let F be a field, and let E be a central simple F -algebra, with dim_{F} E = n^{2}. Until further notice, we assume char(F) ≠ 2. Set E^{∘} = group of units of E, and likewise F^{∘} = F \ {0}. Let ∗: E → E, a ↦ a^{∗} be an involution. All the involutions we work with are assumed to be of the first kind (i.e., α^{∗} = α for all α ∈ F). A good reference for the basic facts on central simple algebras with involution is Scharlau’s book [S]. Given the involution ∗ on E, we write S_{∗} = {s ∈ E | s^{∗} = s} and A_{∗} = {a ∈ E | a^{∗} = −a}. So, E = S_{∗} ⊕ A_{∗} as F -vector spaces (as char(F) ≠ 2). If ∗ is of orthogonal type, then dim_{F} S_{∗} = \frac{n(n+1)}{2} and dim_{F} A_{∗} = \frac{n(n−1)}{2}, while if ∗ is of symplectic type, then dim_{F} S_{∗} = \frac{n(n−1)}{2} and dim_{F} A_{∗} = \frac{n(n+1)}{2}.

Now let us fix two quaternion algebra Q_{i} = (\frac{α_{i},β_{i}}{F} ) over F, i = 1, 2. So, Q_{1} has its standard base 1, i_{1}, j_{1}, k_{1} satisfying

i_{1}^{2} = α_{1}, j_{1}^{2} = A_{1}, k_{1} = i_{1}j_{1} = −j_{1}i_{1}, with α_{1}, β_{1} ∈ F^{∘} .

Let −_{1} be the standard involution on Q_{1} (given by 1 ↦ 1, i_{1} ↦ −i_{1}, j_{1} ↦ −j_{1}), k_{1} ↦ −k_{1}) which is the unique involution of symplectic type on Q_{1}. Then A_{−_{1}} = F -span {i_{1}, j_{1}, k_{1}}, the “pure part” of Q_{1}. Likewise, by replacing the subscript 1 by 2, we have corresponding objects for Q_{2}.

Let E = Q_{1} ⊗_{F} Q_{2}. We will be looking at structures associated with orthogonal type involutions on E. The basic such involution, which we’ll work from, is chosen to be −_{1}⊗−_{2}, call this − (so \overline{a ⊗ b} = a^{−_{1}} ⊗ b^{−_{2}}). Set A = A_{−} and S = S_{−}. Clearly A = A_{−1} ⊕ A_{−_{2}} = F - span {i_{1}, j_{1}, k_{1}, i_{2}, j_{2}k_{2}} (we are identifying elements of the Q_{i} with their images in E) and S = F ·1⊕A_{−_{1}} ⊗A_{−_{2}} (dim S = 10). Define V_{1} = F -span {i_{1}, j_{1}, k_{1}} ⊆ E and V_{2} = F -span {i_{2}, j_{2}, k_{2}}, so that A = V_{1} ⊕ V_{2}.

Lemma 1. For any a = v_{1} + v_{2} ∈ A with v_{i} ∈ V_{i}, we have v_{i}^{2} ∈ F and

Nrd(a) = (v_{1}^{2} − v_{2}^{2})^{2} ∈ F^{2} .

Retypeset February 1995.

Proof. Since v_{1} ∈ A_{−_{1}} ⊆ Q_{1}, v_{1}^{2} = −Nrd_{Q_{1}}(v_{1}) ∈ F; likewise v_{2}^{2} ∈ F. Let K be the F -subalgebra of E generated by v_{1} and v_{2}. Assume, for simplicity, that v_{1} ≠ 0, v_{2} ≠ 0. Since v_{1}v_{2} = v_{2}v_{1} and v_{1}^{2}, v_{2}^{2} ∈ F, it is easy to see that K = F -span {1, v_{1}, v_{2}, v_{1}v_{2}}, dim_{F} K = 4, and K is a commutative Galois extension of F. The Galois group of K over F is evidently ℤ_{2} × ℤ_{2} with generators −_{1} ⊗ id and id ⊗ −_{2} restricted to K. The orbit of a under the Galois action is {v_{1} + v_{2}, v_{1} − v_{2}, −v_{1} + v_{2}, −v_{1} − v_{2}}. The reduced norm of E on K is just the norm from K to F, so

Nrd(a) = N_{K/F} (a) = (v_{1} + v_{2})(v_{1} − v_{2})(−v_{1} + v_{2})(−v_{1} − v_{2}) = (v_{1}^{2} − v_{2}^{2})^{2} ∈ F^{2} .

(This computation is valid even if K is not a field, since it is still a maximal separable subalgebra of E.) It’s easy to check that the formula for Nrd(a) is still valid if v_{1} = 0 or v_{2} = 0. □

Define an F -linear function p: A → A by p(v_{1} + v_{2}) = v_{1} − v_{2} for v_{i} ∈ V_{i}. Clearly for a = v_{1} + v_{2} ∈ A, ap(a) = p(a)a = v_{1}^{2} − v_{2}^{2} ∈ F (as v_{1} and v_{2} commute). So, setting q(a) = ap(a), we have a quadratic form q : A → F. Since we just showed Nrd(a) = q(a)^{2}, q(a) is called in[KPS] the Pfaffian of a, and p(a) is called the Pfaffian adjoint of a. Note that, relative to the orthogonal base {i_{1}, j_{1}, k_{1}, i_{2}, j_{2}, k_{2}}, q is the diagonal quadratic form ⟨α_{1}, β_{1}, −α_{1}β_{1}, −α_{2}, −β_{2}, α_{2}β_{2}⟩ which, up to sign, is the “Albert form” of Q_{1} ⊗_{F} Q_{2} (so called because Albert proved that E is a division algebra if and only if this form is anisotropic). It is important to know that the function p is unique up to scalar multiple:

Lemma 2. (cf. [KPS, Prop. 3.1]) If p′: A → A is an F -linear function such that ap′(a) ∈ F for all a ∈ A (or p′(a)a ∈ F, all a ∈ A), then p′ = λp for some λ ∈ F.

Proof. (For simplicity we’ll assume |F | > 3; the result is easily checked also when |F | = 3.) Take any v_{i} ∈ V_{i} with v_{i}^{2} ≠ 0, i = 1, 2. Then p′(v_{i}) ∈ F v_{i}^{−1} = F p(v_{i}); so p′(v_{i}) = λ_{i}p(v_{i}) for some λ_{i} ∈ F. Since |F | > 3 there exists μ ∈ F^{∘} with (μv_{2})^{2} ≠ v_{1}^{2}. Then v_{1}+μv_{2} ∈ E^{∘} since by Lemma 1, Nrd(v_{1} + μv_{2}) = (v_{1}^{2} − (μv_{2})^{2})^{2} ∈ F^{∘}. Hence, as with the v_{i}, p′(v_{1} + μv_{2}) = δp(v_{1} + μv_{2}) for some δ ∈ F. That is, λ_{1}p(v_{1}) + μλ_{2}p(v_{2}) = δp(v_{1}) + δμp(v_{2}). Since p(v_{1}) (∈ V_{1}) and p(v_{2}) (∈ V_{2}) are linearly independent over F, λ_{1} = δ and μλ_{2} = μδ, so λ_{2} = δ = λ_{1}, as μ ≠ 0. Since V_{1} and V_{2} each have a base consisting of units, this shows p′ of each base vector is the same scalar multiple (say λ) of p of that vector. So p′ = λp. □

Lemma 3. For every c ∈ E and a ∈ A, q(cac̄) = Nrd(c)q(a).

Proof. (Note that cac̄ ∈ A, as a ∈ A.) We have, by Lemma 1, q(cac̄)^{2} = Nrd(cac̄) = Nrd(c)^{2}Nrd(a) = (Nrd(c)q(a))^{2}. So q(cac̄) = ±Nrd(c)q(a). To see that the sign is always +, we use a homotopy argument. We may assume a ∈ E^{∘}, since otherwise q(a) = 0. Let d = cx + (1 − x) ∈ E(x) = E ⊗_{F} F (x) where x is an indeterminate. The function q: A → F induces by scalar extension q′: A ⊗_{F} F (x) → F (x), where A ⊗_{F} F (x) = A_{−′}, where −′ = − ⊗ id is the extension of − to E(x). The observations just made for q apply likewise to q′, showing q′(dad̄′)^{2} = (Nrd(d)q(a))^{2}. This is an equation in F [x]; so, for ε = ±1, q′(dad̄′) = εNrd(d)q(a). By specializing x → 0 we find d → 1, so q(a) = εNrd(1)q(a); hence, ε = 1 as q(a) ∈ F^{∘}. Now by specializing x → 1, as d → c we obtain q(cac̄) = εNrd(c)q(a) = Nrd(c)q(a). □

Now suppose ∗ is any involution of orthogonal type on E. By Skolem-Noether there is s ∈ E^{∘} with r^{∗} = sr̄s^{−1} for all r ∈ E. Fix this s, (s is uniquely determined up to a scalar multiple by ∗.) Since ∗ is an involution, s ∈ S ∪ A, and since ∗ has orthogonal type, s ∈ S. For any r ∈ E, we have (sr)^{∗} = s(r̄s̄)s^{−1} = sr̄; likewise (rs^{−1})^{∗} = r̄s^{−1}. Thus, for the skew-symmetric set with respect to ∗, A_{∗}, we have

A_{∗} = sA = As^{−1} .

We use the functions p and q on A to define analogous functions p_{∗}: A_{∗} → A_{∗} which is F -linear and q_{∗}: A_{∗} → F, which is a quadratic form, as follows: for b ∈ A_{∗} (so bs ∈ A) set

p_{∗}(b) = sp(bs) and q_{∗}(b) = bp_{∗}(b) = q(bs) .

From the properties of p we obtain corresponding properties of p_{∗}, i.e., p_{∗} is an F -linear isomorphism of A_{∗}; for all b ∈ A_{∗}, bp_{∗}(b) = p_{∗}(b)b = q_{∗}(b) ∈ F; and p_{∗} is uniquely determined up to a scalar multiple in the same way that p is in Lemma 2. (For, given a p′_{∗} on A_{∗}, we obtain p′ on A by p′(a) = s^{−1}p′_{∗}(as^{−1}).) Note that the map b → bs is an isometry between the quadratic form q_{∗} on A_{∗} and q on A. So, q_{∗} is also the Albert form for Q_{1} ⊗_{F} Q_{2}. Observe that for all b ∈ A_{∗},

Nrd(b) = Nrd(bs)Nrd(s^{−1}) = q(bs)^{2}Nrd(s^{−1})

= q_{∗}(b)^{2}Nrd(s^{−1}) . (1)

Thus, for every unit b ∈ A_{∗}, Nrd(b) maps to the same coset in F^{∘}/F^{∘2}, namely Nrd(s)F^{∘2}. Thus, Nrd(s)F^{∘2} is an invariant of ∗ (independent of the choice of s to describe ∗, and also independent of the initial choice of involution − on E). This square class Nrd(s)F^{∘2} is called the Pfaffian discriminant of the involution ∗, written pfd (∗).

Lemma 4. For all b ∈ A_{∗}, p_{∗}(p_{∗}(b)) = Nrd(s) · b.

Proof. We compute

p_{∗}(p_{∗}(b)) · p_{∗}(b) = q_{∗}(p_{∗}(b)) = q(p_{∗}(b)s) = q(sp(bs)s)

= Nrd(s)q(p(bs)) (by Lemma 3) = Nrd(s)bsp(bs) = Nrd(s)b · p_{∗}(b) .

Whenever b is a unit, so is p_{∗}(b), and we can cancel p_{∗}(b) in the formula above to obtain the desired equality. Because A_{∗} is spanned by units and p_{∗} ∘ p_{∗} is F -linear, the equality holds for all b ∈ A_{∗}. □

Lemma 5. Suppose ∗_{i} is an involution of orthogonal type on Q_{i} and ∗ = ∗_{1} ⊗ ∗_{2}. Then pfd (∗) = 1 · F^{∘2}.

Proof. Since ∗_{i} is of orthogonal type there is a unit t_{i} ∈ A_{−_{i}} (the pure part of Q_{i}) such that r^{∗_{i}} = t_{i}r^{−_{i}}t_{i}^{−_{i}} for all r ∈ Q_{i}. Then conjugation by s = t_{1} ⊗t_{2} induces ∗ = ∗_{1} ⊗∗_{2} from − = −_{1} ⊗ −_{2}. Let K = F (t_{1} ⊗ 1, 1 ⊗ t_{2}), a 4-dimensional commutative Galois extension of F in E. As in the proof of Lemma 1,

Nrd(s) = N_{K/F} (s) = (t_{1} ⊗ t_{2})(−t_{1} ⊗ t_{2})(t_{1} ⊗ −t_{2})(−t_{1} ⊗ −t_{2}) = t_{1}^{4} ⊗ t_{2}^{4} = (t_{1}^{2} · t_{2}^{2})^{2} ∈ F^{∘2} .

Hence, pfd (∗) = 1 · F^{∘2}. □

Theorem 1. (cf. [KPS, Cor. 5.5]) For the orthogonal type involution ∗ on E, the following are equivalent:

(i) There is an F -central quaternion algebra Q in E with Q^{∗} = Q.

(ii) There are F -central quaternion subalgebras Q′_{1} and Q′_{2} of E with Q_{i}^{′∗} = Q_{i},

i = 1, 2, and E = Q′_{1} ⊗_{F} Q′_{2}. (iii) pfd (∗) = 1 · F^{∘2} (i.e., Nrd(s) ∈ F^{∘2}).

Proof. (i) ⇒ (ii) This is immediate from the double centralizer theorem.

(ii) ⇒ (i) Clear.

(ii) ⇒ (iii) Let ∗_{i} be the restriction of ∗ to Q′_{i}, i = 1, 2. Since ∗ = ∗_{1} ⊗ ∗_{2} and ∗ has orthogonal type, ∗_{1} and ∗_{2} must be of the same type. If the ∗_{i} are of symplectic type, then ∗_{i} is the standard involution on Q′_{i}; then as ∗ = ∗_{1} ⊗ ∗_{2}, Lemma 1 applied to (E, ∗) instead of (E, −) shows Nrd(b) ∈ F^{2} for all b ∈ A_{∗}. Hence, pfd (∗) = 1 · F^{∘2}. If the ∗_{i} are of orthogonal type Lemma 5 applied with respect to the decomposition E = Q′_{1} ⊗_{F} Q′_{2} (instead of Q_{1}⊗_{F} Q_{2}) shows again Nrd(b) ∈ F^{2} for all b ∈ A_{∗}; hence again pfd (∗) = 1·F^{∘2}.

(iii) ⇒ (ii) Suppose pfd (∗) = 1 · F^{∘2}. Then Nrd(s) = λ^{2} for some λ ∈ F^{∘}. By Lemma 4, p_{∗} ∘ p_{∗} = λ^{2} · id. Hence, the only possible eigenvalues of p_{∗} are λ and −λ. Let Y_{ελ} = {b ∈ A_{∗} | p_{∗}(b) = ελb} for ε = ±1. Since p_{∗} is a diagonalizable linear transformation (as its minimal polynomial, dividing x^{2} − λ^{2}, has no repeated roots) A_{∗} = Y_{λ} ⊕ Y_{−λ}. Let Q′_{1} = F · 1 ⊕ Y_{λ} and Q′_{2} = F · 1 ⊕ Y_{−λ}. Clearly Q_{i}^{′∗} = Q′_{i}, i = 1, 2.

It remains to show that the Q_{i} are quaternion algebras over F which centralize each other. For this we use the quadratic form q_{∗}(b) = bp_{∗}(b) = p_{∗}(b)b. (For the last equality note that as q_{∗}(b) ∈ F, q_{∗}(b) = sq_{∗}(b)s^{−1} = s[q(bs)]s^{−1} = s[p(bs)bs]s^{−1} = sp(bs)b = p_{∗}(b)b. The associated bilinear form B_{∗} for q_{∗} is given by B_{∗}(b, d) = \frac{1}{2}[q_{∗}(b + d) − q_{∗}(b) − q_{∗}(d)], so

B_{∗}(b, d) = \frac{1}{2}[bp_{∗}(d) + dp_{∗}(b)] = \frac{1}{2}[p_{∗}(b)d + p_{∗}(d)b] .  (2)

If b ∈ Y_{λ} and d ∈ Y_{−λ}, this yields B_{∗}(b, d) = \frac{λ}{2}[−bd + db] = \frac{λ}{2}[bd − db]; since B_{∗}(b, d) = −B_{∗}(b, d), this yields B_{∗}(b, d) = 0 and bd = db. So, Y_{−λ} centralizes Y_{λ} and Y_{λ} ⊥ Y_{−λ} with respect to q_{∗}.

Because dim Y_{λ} + dim Y_{−λ} = 6, we may assume dim Y_{λ} ≥ 3. (Otherwise, interchange λ and −λ.) Since q_{∗} is a nondegenerate quadratic form on A_{∗} (as q is nondegenerate on A) and Y_{λ} ⊥ Y_{−λ}, q_{∗} must be nondegenerate on Y_{λ} and on Y_{−λ}. Therefore, we can choose two specific elements b, c ∈ Y_{λ} with q_{∗}(b) ≠ 0, q_{∗}(c) ≠ 0 (i.e., b and c are units) and B_{∗}(b, c) = 0 (i.e., from (2), bc = −cb). We have b^{2} = λ^{−1}bp_{∗}(b) = λ^{−1}q_{∗}(b) ∈ F^{∘} and likewise c^{2} ∈ F^{∘}. So, F (b, c) is a quaternion algebra over F. Now, (bc)^{∗} = c^{∗}b^{∗} = (−c)(−b) = cb = −bc; so bc ∈ A_{∗}. Further, as bcp_{∗}(bc) = q_{∗}(bc) ∈ F and bcbc = −b^{2}c^{2} ∈ F^{∘}, we see p_{∗}(bc) = μbc for μ ∈ F. That is, bc is an eigenvector for p_{∗}. But bc ∉ Y_{−λ} since bc anticommutes with b, while elements of Y_{−λ} commute with those of Y_{λ}. So, bc ∈ Y_{λ}. Hence, the 3-dimensional space F -span {b, c, bc} lies in Y_{λ}. Since q_{∗}|_{Y_{λ}} is nondegenerate, if dim Y_{λ} > 3 we could find y ∈ Y_{λ} with q_{∗}(y) ≠ 0 and 0 = B_{∗}(b, y) = B_{∗}(c, y) = B_{∗}(bc, y). By (2) this implies y anticommutes with b, c, and bc, forcing y = 0, which contradicts q_{∗}(y) ≠ 0. Thus, Y_{λ} = F - span {b, c, bc}, dim Y_{λ} = 3, and F (b, c) = Q′_{1}, showing Q′_{1} is indeed a quaternion algebra. Since dim Y_{−λ} = 6−dim Y_{λ} = 3, the same argument shows Q′_{2} is also a quaternion algebra. Since Y_{−λ} centralizes Y_{λ}, Q′_{2} centralizes Q′_{1}, completing the proof. □

Remark. There is another way to prove that the Q′_{i} in (iii) ⇒ (ii) are quaternion algebras centralizing each other. Suppose first that ∗ and − are isomorphic involutions on E, i.e., suppose there is an F -algebra isomorphism f : E → E such that f(μ^{∗}) = \overline{f(μ)}, for all r ∈ E. Then f(A_{∗}) = A; let p′ = f ∘ p_{∗} ∘ f^{−1} : A → A. Because bp_{∗}(b) ∈ F for all b ∈ A_{∗}, ap′(a) ∈ F for all a ∈ A. By Lemma 2, p′ = μ · p for some μ ∈ F. In fact, μ^{2} = λ^{2} since p′ ∘ p′ = λ^{2} · id (as p_{∗} ∘ p_{∗} = λ^{2}id) while p ∘ p = id. So, we may assume μ = λ. Then λ-eigenspace of p′ = 1-eigenspace of p = F -span {i_{1}, j_{1}, k_{1}}. So, f^{−1}(Q_{1}) = f^{−1}(F · 1 ⊕ λeigenspace of p′) = F · 1 + λ-eigenspace of p_{∗} = Q′_{1}. Because f is an algebra isomorphism, it follows that Q′_{1} is a quaternion algebra over F. Likewise, Q′_{2} = f^{−1}(Q_{2}) is a quaternion algebra, and Q′_{2} centralizes Q′_{1} since Q_{2} centralizes Q_{1}.

This completes the argument in case (E, ∗) ≅ (E, −). The condition for (E, ∗) ≅ (E, −) is easily checked to be that there exists y ∈ E^{∘} with ysȳ ∈ F^{∘} (cf. [S, Th. 7.4(iv), p. 302]). While this need not hold, it will hold after extending scalars to the separable closure of F.* Hence, there is a finite degree Galois extension K of F such that (E ⊗_{F} K, ∗⊗id) ≅ (E⊗_{F} K, −⊗id). The argument just given, applied to E⊗_{F} K, shows that Q′_{1}⊗_{F} K and Q′_{2} ⊗_{F} K are quaternion algebras over K centralizing each other. Since Q′_{1} is the set of fixed elements of Q′_{1} under the action of the Galois group G(K/F) (σ ∈ G(K/F) acts via id⊗σ: Q′_{1} ⊗_{F} K → Q′_{1} ⊗_{F} K, a ring automorphism), Q′_{1} is closed under multiplication. The ring Q′_{1} is a 4-dimensional central simple F -algebra since Q′_{1} ⊗_{F} K is a 4-dimensional central simple K-algebra. So Q′_{1} is a quaternion algebra over F, as likewise is Q′_{2}; finally, Q′_{2} centralizes Q′_{1} since Q′_{2} ⊗_{F} K centralizes Q′_{1} ⊗_{F} K.

By an old theorem of Albert, every 16-dimensional central simple F -algebra E with involution is a tensor product of quaternion algebras. The first example of such an E with an involution ∗ which does not admit a ∗-compatible decomposition into quaternion algebras (in the sense of (ii) of Theorem 1) was given in [ART]. Theorem 1 shows that “almost every” E has an involution which does not admit a compatible quaternionic decomposition. It suffices to find a symmetric element s with respect to any orthogonal type involution on E such that Nrd(s) ∉ F^{2}. In fact, the unusual case is to have an E such that every orthogonal type involution admits a compatible quaternionic decomposition. Of course this occurs if F is quadratically closed (F = F^{2}); but then, E ≅ M_{4}(F). Here is an example of such an E which is a division algebra.

Example. Let F be a field with Henselian valuation v, with residue field F̄, char(F̄) ≠ 2, and let E be a 16-dimensional F -central division algebra with involution. Suppose E is totally ramified over F with respect to the unique extension of v to a valuation of E. Then Nrd(r) ∈ F^{2} for each r ∈ E. Hence, every orthogonal type involution on E admits a compatible quaternionic decomposition. (For an explicit example of this, let F_{0} be any field (charF_{0} ≠ 2), let F be the iterated Laurent power series field F = F_{0}((x_{1}))((x_{2}))((x_{3}))((x_{4})), and let E = (\frac{x_{1},x_{2}}{F} ) ⊗_{F} (\frac{x_{3},x_{4}}{F} ).)

Proof. See [TW] for more on totally ramified valuations. In particular, if Γ_{E} (resp. Γ_{F}) is the value group of the valuation on E (resp. F), then the order of E in the Brauer group of F equals the exponent of the factor group Γ_{E}/Γ_{F}. Hence, for r ∈ E, v(r^{2}) = 2v(a) ∈ Γ_{F}. So, r^{2} = aλ with λ ∈ F and u ∈ E with v(u) = 0. E has the same residue field as F since E is totally ramified over F; so, we may modify λ and u so that the residue of u is 1. Write u = 1 + m with v(m) > 0. Because the valuation on the field F (r) extending v on F is Henselian, 1 + m ∈ F (r)^{2}, say 1 + m = (1 + n)^{2} with n ∈ F (r) and v(n) > 0. So, r = cd where c = r(1+u)^{−1} and α = 1+n ∈ F (r). Since c^{2} = λ ∈ F, Nrd(c) = c^{4} ∈ F^{2}. Because the Henselian valuation v extends uniquely to F (r), N_{N(r)/F} (d) = 1 + μ with v(u) > 0; but 1 + u ∈ F^{2} is v is Henselian. Since Nrd(d) is a power of N_{N(r)/F} (d) Nrd(d) ∈ F^{2}. Thus Nrd(r) = Nrd(c)Nrd(d) ∈ F^{2}. The rest follows by Theorem 1. □

*Suppose L is a separably closed field and ∗ is any orthogonal type involution on the matrix algebra M_{4}(L). Since ∗ is of the same type as the transposed involution t there is a z ∈ M_{4}(L)^{∘} with z^{t} = z and r^{∗} = zr^{t}z^{−1} for all μ ∈ M_{4}(L). The condition for (M_{4}(L), ∗) ≅ (M_{4}(L), t) is that for some unity y, y = y^{t} = some δ ∈ L. That is, the symmetric bilinear form associated to z is isometric to the diagonal form ⟨δ, δ, …, δ⟩. By Gram-Schmidt, the form is diagonalizable; as L is separably closed, it can be diagonalized to ⟨1, 1, …, 1⟩. So, (M_{4}(L), ∗) ≅ (M_{4}(L), t).

So far, we have considered only orthogonal type involutions on E = Q_{1} ⊗_{F} Q_{2}. But the approach given above can be adapted with few modifications for symplectic type involutions, as we now describe. The main point is that Theorem 1 still holds, but further, condition (iii) of the Theorem (hence also (i) and (ii)) is true for every symplectic type involution on E.

With − still denoting our specified (orthogonal type) involution −_{1} ⊗ −_{2} on E = Q_{1} ⊗_{F} Q_{2}, let ⋆ be any symplectic type involution on E. Fix a unit z ∈ A = A_{−} with μ^{⋆} = zμ̄z^{−1} for all r ∈ E. The appropriate 6-dimensional subspace to work with here is the symmetric set S_{⋆}. We have S_{⋆} = zA = Az^{−1}. So we obtain p_{i}: S_{⋆} → S_{⋆} and q_{⋆}: S_{⋆} → F from p and q on A as before: for b ∈ S_{⋆}, set p_{⋆}(b) = zp(bz) and q_{⋆}(b) = bp_{⋆}(b) = q(bz). As before, p_{⋆} has the same uniqueness property as p in Lemma 2; as in (1) on p. 3, Nrd(b) = q_{⋆}(b)^{2}Nrd(z)^{−1} for all b ∈ S_{⋆}, and the analogue to Lemma 4 holds, with the same proof. So, we again get a well-defined Pfaffian discriminant pfd (⋆) = Nrd(z)F^{∘2} ∈ F^{∘}/F^{∘2}. But Lemma 1 shows Nrd(z) ∈ F^{∘2}. So, pfd (⋆) = 1·F^{∘2} for every symplectic type involution on E.

In proving Theorem 1 for the symplectic ⋆, (ii) ⇒ (iii) is trivial since we have just seen that (iii) alway holds. For (iii) ⇒ (ii) we must change the way of getting Q′_{1} and Q′_{2} because of the different eigenspace structure of p_{⋆} (see below). We still have p_{⋆}^{2} = Nrd(z)·id = λ^{2}·id, so S_{⋆} has the eigenspace decomposition S_{⋆} = Y_{λ} ⊕ Y_{−λ}. The analogue to (2) on p. 4 for B_{⋆} shows that Y_{λ} ⊥ Y_{−λ} with respect to q_{⋆}, so q_{⋆} |_{Y_{±λ}} is nondegenerate. With λ chosen (from ±λ) so that dim Y_{λ} ≥ 3, we pick b, c ∈ Y_{λ} such that q_{⋆}(b), q_{⋆}(c) ≠ 0 (so b^{2}, c^{2} ∈ F^{∘}) and B_{⋆}(b, c) = 0 (so by (2) bc = −cb). Let Q′_{1} = F (b, c), which is clearly a quaternion algebra over F; furthermore, Q′_{1}^{⋆} = Q′_{1} such that b^{⋆} = b and c^{⋆} = c. Then let Q′_{2} be the centralizer of Q′_{1} in E; then Q′_{2} is a ⋆-invariant quaternion algebra with E = Q′_{1} ⊗_{F} Q′_{2}, as desired.

In the orthogonal case there was a canonical choice of the Q′_{i} for (ii) of Theorem 1 coming from the eigenspaces of p_{∗}. But in the symplectic case there is no such canonical choice. To see why, let us make a specific choice z = i_{1}, so r^{⋆} = i_{1}r̄i_{1}^{−1}. Then, easy calculations show: S_{⋆} = F -span {1, j_{1}, k_{1}, i_{1}i_{2}, i_{1}j_{2}, i_{1}k_{2}}, p_{⋆} has eigenvalues ±α_{1} with eigenspaces Y_{α_{1}} = F -span {1} and Y_{−α_{1}} = F -span {j_{1}, k_{1}, i_{1}i_{2}, i_{1}j_{2}, i_{1}k_{2}}. The eigenspaces have dimension 1 and 5, not 3 and 3! By a scalar extension argument like the one in the remark on p. 5, we can see that for any symplectic type involution ⋆ on E, p_{⋆} will have eigenspaces of dimension 1 and 5. In fact, the Pfaffian arises again for any unit z of A; if ⋆ is the symplectic type involution r ↦ zr̄z^{−1}, then p_{i}^{2} = Nrd(z) · id = q(z)^{2} · id, so the eigenvalues of p_{i} are ±q(z). The 1-dimensional eigenspace is that of q(z), and the 5-dimensional eigenspace is that of −q(z).

There is also a nice proof in [KPS] of Jacobson’s theorem [J, Th. 3.12] relating 16 dimensional algebras with involution and their Albert forms, which we turn to now.

Lemma 6. (cf. [KPS, Prop. 3.3]) With E = Q_{1} ⊗_{F} Q_{2} and − = −_{1} ⊗ −_{2} as above, let q(a) = ap(a): A → F be the Albert form (= Pfaffian form) on A = A_{−}, as above. For any λ ∈ F^{∘}, let C(λq) be the Clifford algebra of λq. Then, C(λq) ≅ M_{2}(E).

Proof. Define h: A → M_{2}(E) by a ↦ ( \begin{matrix}0 & a \\ λp(a) & 0 \end{matrix} ). Then h(a)^{2} = λap(a) · 1 = λq(a).

From the universal mapping property for Clifford algebras, h induces an F -algebra homomorphism f: C(λq) → M_{2}(E). This f is injective, as C(λq) is simple, so an isomorphism by dimension count. □

Theorem 2. (Jacobson) (cf. [KPS, Th. 3.4]). Let E = Q_{1} ⊗_{F} Q_{2} and Ẽ = Q̃_{1} ⊗_{F} Q̃_{2}, where Q_{1}, Q_{2}, Q̃_{1}, Q̃_{2} are quaternion algebras over F. Then E ≅ Ẽ if and only if the Albert forms q and q̃ are similar (i.e., q̃ is isometric to λq for some λ ∈ F^{∘}).

Proof. Suppose q̃ is isometric to λq. Then, by Lemma 6, M_{w}(Ẽ) ≅ C(q̃) ≅ C(λq) ≅ M_{2}(E). Hence, Ẽ ≅ E. Conversely, suppose there is an isomorphism f : Ẽ → E. The standard involution ∼ on Ẽ (relative to the decomposition Ẽ = Q̃_{1} ⊗_{F} Q̃_{2}) induces an involution ∗ of orthogonal type on E, given by r^{∗} = f\binom{}{f^{−1}(r)}. Then f(A_{∼}) = A_{∗}, and

the Pfaffian adjoint p̃ on A_{∼} induces a function p′ on A_{∗} (p′ = f ∘ p̃ ∘ f^{−1}) such that bp′(b) = bf(q̃(f^{−1}(b))) ∈ F, for all b ∈ A_{∗}. Hence, by the uniqueness Lemma 2, p′ = λp_{∗}. Then for s ∈ S_{−} such that r^{∗} = sr̄s^{−1} all r ∈ E for any b ∈ A_{∗}, f(q̃(f^{−1}(b))) = bp′(b) = λbp_{∗}(b) = λq(bs). So, a ↦ f(a)s gives an isometry between q̃ and λq. □

We turn finally to characteristic 2. In [KPS] the Pfaffian adjoint is defined for any Azumaya algebra of rank 16 with involution over any commutative ring. We show how it plays out when the ring is a field F and char(F) = 2. Because char(F) = 2, every involution ∗ on E is of orthogonal type in that dim S_{∗} = 10 where S_{∗} = {s ∈ E | s^{∗} = s}. Since ∗∘∗ = id the linear map ∗: E → E satisfies the equation 0 = x^{2} −1 = (x−1)^{2}. So, 1 is the only possible eigenvalue of ∗, but ∗ is not diagonalizable since its minimal polynomial had a repeated root. So, E doesn’t decompose as an orthogonal sum of eigenspaces of ∗. By dimension count (since the 1-eigenspace of ∗ has dimension 10), the Jordan canonical

form of ∗ must consist of 6 2 × 2 blocks ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) together with 4 |x| blocks (1). To get

a 6-dimensional subspace, we define A_{∗} = {b − b^{∗} | b ∈ E}. Since rank (id − ∗) = 6 from the Jordan form, dim A_{∗} = 6. So, A_{∗} ⊆ S_{∗}, instead of A_{∗} and S_{∗} being complementary subspaces.

Suppose Q = [\frac{α,β}{F} ) is a quaternion algebra over our char 2 field F, i.e., Q is the F -algebra with F -base {1, i, j, k} with relations

i^{2} + i = α ∈ F, j^{2} = β ∈ F^{∘}, k = ij = ji + j .

Note the further relations

ik = ki + k and jk = kj + β and k^{2} = αβ .

The standard involution − on Q is given by ¯1 = 1, ī = i + 1, j̄ = j, k̄ = k. So the symmetric set S_{−} = F -span {1, j, k} while the A-set A_{−} = F -span {1}. For x = μ + νi + σj + τk, we have

Nrd(x) = xx̄ = x̄x = μ^{2} + μν + αν^{2} + βσ^{2} + βσ^{2} + βστ + βατ^{2} .

Thus, Nrd(x) = x^{2} whenever x = x̄, i.e., x ∈ S_{−} = F -span {1, j, k}. Note also that the reduced norm quadratic form is [1, α] ⊥ β[1, α], where [1, α] denotes the nondegenerate 2-dimensional quadratic form 1 · μ^{2} + 1 · μν + α · ν^{2}.

Now suppose E = Q_{1}⊗_{F} Q_{2} where Q_{1} = [\frac{α_{1},β_{1}}{F} ) (resp. Q_{2} = [\frac{α_{2},β_{2}}{F} )), and 1,i_{1},j_{1},k_{1} (resp. 1,i_{2},j_{2},k_{2}) is the standard base of Q_{1} (resp. Q_{2}) and −_{1} (resp. −_{2}) is its standard involution. Write 1 for 1⊗1 and i_{1}, …, k_{2} for i_{1} ⊗1, …, 1⊗k_{2} in E. Let A = A_{−}. Clearly S_{−_{1}} ⊗ 1 and 1 ⊗ S_{−_{2}} lie in A_{−}, but they don’t span it. An added generator of A_{−} is given by i_{1} + i_{2} since i_{1}(i_{2} + 1) − \overline{i_{1}(i_{2} + 1)} = i_{1}(i_{2} + 1) − (i_{2})(i_{1} + 1) = i_{1} + i_{2}. Thus, an F -base of A_{−} is {1, j_{1}, k_{1}, j_{2}, k_{2}, i_{1} + i_{2}}. The analogue of Lemma 1 above is:

Lemma 7. For x = μ · 1 + r + s + ν · (i_{1} + i_{2}) ∈ A, where r ∈ F -span {j_{1}, k_{1}}, s ∈ F -span {j_{2}, k_{2}}, and μ, ν ∈ F, we have

x^{2} + νx ∈ F and Nrd(x) = (x^{2} + νx)^{2} ∈ F^{2} .

Proof. Let z = i_{1} + i_{2}. Easy calculations show z^{2} = z + (α_{1} + α_{2}), zr + rz = r, and zs + sz = s. Thus, as rs = sr and r^{2}, s^{2} ∈ F,

\begin{aligned}x^{2} + νx & = [μ^{2} + μ^{2} + s^{2} + ν^{2}z^{2} + ν[rz + zr] + ν[sz + zs]] + ν[μ + r + s + νz] \\ & = μ^{2} + r^{2} + s^{2} + μν + ν^{2}(α_{1} + α_{2}) ∈ F . \end{aligned}

The equations z^{2} + z = α_{1} + α_{2}, (r + s)z = z(r + s) + (r + s), and (r + s)^{2} (= r^{2} + s^{2}) ∈ F show that z and (r + s) generate a quaternion algebra Q′ isomorphic to [\frac{α_{1}+α_{2},r^{2}+s^{2}}{F} ) inside E.* Thus, Nrd_{E}(x) (reduced norm relative to E) is the square of Nrd_{Q′}(x). Since x + v is the image of x under the standard involution on Q′, Nrd_{Q′}(x) = x(x + v), and Nrd_{E}(x) = Nrd_{Q′}(x)^{2} = [x(x + v)]^{2} ∈ F^{2}. □

From this lemma it is clear we can take the Pfaffian adjoint p: A → A to be p(x) = x + v · 1 for x = μ · 1 + r + s + v · z as above. As in Lemma 2, the function p is unique to a scalar multiple, and the Pfaffian quadratic form q: A → F is q(x) = xp(x) = p(x)x = μ^{2} + μν + ν^{2}(α_{1} + α_{2}) + r^{2} + s^{2}. Here, r^{2} gives the quadratic form β_{1}[1, α_{1}] and s^{2} gives β_{2}[1, α_{2}]. So, the form q is isometric to [1, α_{1} + α_{2}] ⊥ β_{1}[1, α_{1}] ⊥ β_{2}[1, α_{2}]. This is the form called in [MS] the Albert form associated to E = Q_{1} ⊗_{F} Q_{2}. Mammone and Shapiro proved the analogue to Jacobson’s theorem, Theorem 2 above, for char(F) = 2 using this Albert form. Notice that p∘p = id, but p is not diagonalizable on A, as the only eigenvalue of p is 1, and the 1-eigenspace is F -span {1, j_{1}, k_{1}, j_{2}, k_{2}}, of dimension 5. Observe that the analogue to Lemma 3 holds here, and the proof is easier since there is no sign ambiguity about the square root.

*There is a technical difficulty with this if r^{2} = s^{2}. For then Q′ is not a quaternion algebra (not even semisimple). But the lemma is still valid, as can be seen by first working over the rational function field K = F (X_{1}, X_{2}) with the algebra [ \frac{α_{1},x_{1}}{K} ) ⊗_{K} [ \frac{α_{2},x_{2}}{K} ), where the argument is valid (except if r − s = 0, which is easily treated separately) then specializing x_{1} ↦ β_{1}, x_{2} ↦ β_{2}.

Now, take any unit s ∈ S_{−}, and form a new involution ∗ on E by r^{∗} = sr̄s^{−1}. From the equations (sr)^{∗} = sr̄ and sr − (sr)^{∗} = s(r − r̄), we see S_{∗} = sS_{−} = S_{−}s^{−1} and A_{∗} = sA = As^{−1}. Thus, we may again define a Pfaffian adjoint p_{∗} : A_{∗} → A_{∗} by p_{∗}(b) = sp(bs), and define q_{∗} : A_{∗} → F by q_{∗}(b) = bp_{∗}(b) = q(bs). So q_{∗} is a quadratic form on A_{∗} isometric to q on A. Then p_{∗} has the same type of uniqueness as p in Lemma 2. Since Lemma 6 clearly still holds, it follows that the proof of Theorem 2 works also in characteristic 2, yielding the Mammone-Shapiro version of Jacobson’s theorem. (Of course, this is still just another case of [KPS, Th. 3.4], which is valid for rank 16 Azumaya algebras over any commutative ring.)

Formula (1) above holds also when char(F) = 2: for all b ∈ A_{∗},

Nrd(b) = q_{∗}(b)^{2}Nrd(s)^{−1} .

Hence, again there is a Pfaffian discriminant for ∗, pfd (∗) = Nrd(s)F^{∘2} ∈ F^{∘}/F^{∘2}. One can ask about Theorem 1 for char(F) = 2, and the answer seems not completely settled. One can see, as in Lemma 5 that for any involution ∗ on E arising as ∗_{1} ⊗ ∗_{2} where ∗_{i} is an involution on Q_{i}, pfd (∗) = 1 · F^{∘2}. Thus, the argument for (ii) ⇒ (iii) of Theorem 1 works when char(F) = 2. This allows us to construct examples of E with an involution ∗ admitting no ∗-invariant quaternionic decomposition. Now, Lemma 4 also holds in characteristic 2 (with the same proof), so for any involution ∗ on E, if pfd (∗) = 1 · F^{∘2}, then p_{∗} has a (unique) eigenvalue in F^{∘}. If the eigenspace Y for p_{∗} is 5-dimensional, then one can always find a 2-dimensional subspace of Y on which q_{∗} is nondegenerate, and this can be used to build a ∗-invariant quaternion subalgebra of E; so then we have (iii) ⇒ (ii) of Theorem 1. But, I think pfd (∗) = 1 and dim(Y) = 5 if and only if s ∈ A. However, it can also occur that dim(Y) = 3 and the associated bilinear form B_{∗} of q_{∗} (given by (b, c) ↦ q_{∗}(b + c) − q_{∗}(b) − q_{∗}(c)) vanishes identically on Y × Y_{∘}. (I think this occurs whenever s ∈ S_{−} but s ∉ A (and Nrd(s) ∈ F^{∘2}).) But it is not clear in this case whether one can always construct a ∗-invariant quaternion subalgebra of E.

References

[ART] S. A. Amitsur, L. H. Rowen, and J.-P. Tignol, Division algebras of degree 4 and 8 with involution, Israel J. Math. 33 (1979), 133–148.

[J] N. Jacobson, Some applications of Jordan norms to involutorial simple associative algebras, Adv. in Math. 48 (1983), 149–165.

[KPS] M.-A. Knus, R. Parimala, and R. Sridharan, Pfaffians, central simple algebras, and similitudes, Math. Z. 206 (1991), 589–604.

[MS] P. Mammone and D. B. Shapiro, The Albert quadratic form for an algera of degree 4, Proc. Amer. Math. Soc. 105 (1989), 525–530.

[S] W. Scharlau, Quadratic and Hermitian forms, Springer, 1985.

[TW] J.-P. Tignol and A. R. Wadsworth, Totally ramified valuations on finite-dimensional division algebras, Trans. Amer. Math. Soc. 302 (1987), 223–250.

9500 Gilman Dr., Dept. of Mathematics, La Jolla, CA 92093-0112 E-mail address: arwadsworth@ucsd.edu