THE ASYMMETRY OF AN ANTI-AUTOMORPHISM

ANNE CORTELLA AND JEAN-PIERRE TIGNOL

Abstract. The asymmetry of a nonsingular pairing on a vector space is an endomorphism of the space on which the classification of arbitrary pairings (not necessarily symmetric or skew-symmetric) is based. A general notion of asymmetry is defined for arbitrary anti-automorphisms on a central simple algebra, and conditions are given to characterize the elements which are the asymmetries of some anti-automorphism. The asymmetry is used to define the determinant of an anti-automorphism.

Introduction

The asymmetry of an arbitrary nonsingular pairing (not necessarily symmetric or skew-symmetric) on a finite-dimensional vector space V is an invertible endomorphism of V which is an important invariant of the pairing. It is 1 if and only if the pairing is symmetric and −1 if and only if it is skew-symmetric. This invariant was first considered by Williamson [9], and more recently by Riehm [6].

In the present paper, we determine under which conditions a linear map a ∈ GL(V) is the asymmetry of some nonsingular pairing on V: the map a must be conjugate to its inverse and satisfy some conditions on the generalized eigenspaces of eigenvalues +1 and −1, see Theorem 1. As pointed out by Ranicki, the property that a is an asymmetry could be rephrased by saying that a certain asymmetric Poincaré complex of dimension 1 is round simple null-cobordant. (See [5, Ch. 20] for background information on Poincaré complexes.)

In section 2, we define the asymmetry of an anti-automorphism σ on a central simple algebra A: it is an element a_{σ} ∈ A^{×} which is mapped, under scalar extension to a splitting field of A, to the asymmetry of any nonsingular pairing to which σ is adjoint. It is defined up to sign by the properties that σ^{2}(x) = a_{σ}xa_{σ}^{−1} for all x ∈ A and that σ(a_{σ}) = a_{σ}^{−1}. This element was incidentally used by Saltman [7, Lemma 3.3, Theorem 4.4] to show that if an Azumaya algebra A carries an anti-automorphism, then the ring of 2 × 2 matrices M_{2}(A) carries an involution, and that Azumaya algebras over connected semilocal rings which are isomorphic to their opposite have an involution. We show that in a central simple algebra of exponent 2, an invertible element is the asymmetry of some anti-automorphism if and only if it is conjugate to its inverse (Theorem 2). Albert’s theorem that every central simple algebra of exponent 2 has an involution is an immediate consequence, since involutions are the anti-automorphisms of asymmetry ±1. In the final section, the asymmetry is used to define the determinant of an anti-automorphism.

Date: July 1, 2000.

The authors warmly thank Andrew Ranicki for his comments on an earlier version of this paper.

They gratefully acknowledge support from the FNRS–CNRS cooperation agreement and from the TMR network “K-theory and linear algebraic groups” (contract ERB FMRX CT 97-0107). The second author is partially supported by the National Fund for Scientific Research (Belgium).

1. The asymmetry of a nonsingular pairing

Throughout this section, V denotes a finite-dimensional vector space over an arbitrary field F. We define the asymmetry and the adjoint anti-automorphism of a nonsingular pairing on V, and determine which linear transformations of V are asymmetries.

1.1. Definitions. Let V^{∗} = Hom_{F} (V, F) be the dual of V. Every pairing (or bilinear form) b: V × V → F induces a linear map b̂: V → V^{∗} which carries x ∈ V to b(x, •) ∈ V^{∗}. The transpose map b̂^{t}: V = V^{∗∗} → V^{∗} carries x ∈ V to b(•, x) ∈ V^{∗}.

Proposition 1. For a pairing b on V, the following conditions are equivalent:

(a) if x ∈ V is such that b(x, y) = 0 for all y ∈ V, then x = 0;

(b) if y ∈ V is such that b(x, y) = 0 for all x ∈ V, then y = 0;

(c) the map b̂ is bijective.

If these conditions hold, the pairing b is called nonsingular.

Proof. Condition (a) is equivalent to injectivity of b̂, and (b) to injectivity of b̂^{t}, hence also to surjectivity of b̂. Since dim V = dim V^{∗}, each of these conditions implies that b̂ is bijective.

All the pairings considered in the sequel are nonsingular. To every nonsingular pairing b on V we attach an anti-automorphism σ_{b} of End_{F} V and a linear transformation a_{b} ∈ GL(V) as follows:

Proposition 2. Let b be a nonsingular pairing on V. There is a unique map σ_{b}: End_{F} V → End_{F} V and a unique map a_{b}: V → V such that

(1)  b(f(x), y) = b(x, σ_{b}(f)(y))  for all x, y ∈ V, f ∈ End_{F} V

and

(2)  b(x, y) = b(y, a_{b}(x))  for all x, y ∈ V .

The map σ_{b} is an F -linear anti-automorphism of End_{F} V and the map a_{b} is linear and invertible. These maps satisfy the following properties:

(i) σ_{b}^{2}(f) = a_{b} ∘ f ∘ a_{b}^{−1} for all f ∈ End_{F} V; (ii) σ_{b}(a_{b}) = a_{b}^{−1}.

Proof. For f ∈ End_{F} V, let σ_{b}(f) = (b̂∘f ∘b̂^{−1})^{t}. Equality (1) is easily checked, and the fact that σ_{b} is an F -linear anti-automorphism of End_{F} V follows. Uniqueness of σ_{b} follows from the hypothesis that b is nonsingular.

On the other hand, let a_{b} = (b̂^{t})^{−1} ∘ b̂. This map is clearly linear and invertible, and it satisfies (2). Uniqueness of a_{b} is clear. To check the additional properties, observe that for f ∈ End_{F} V

σ_{b}^{2}(f) = (b̂ ∘ (b̂ ∘ f ∘ b̂^{−1})^{t} ∘ b̂^{−1})^{t} = ((b̂^{t})^{−1} ∘ b̂) ∘ f ∘ ((b̂^{t})^{−1} ∘ b̂)^{−1}

and

σ_{b}((b̂^{t})^{−1} ∘ b̂) = (b̂ ∘ ((b̂^{t})^{−1} ∘ b̂) ∘ b̂^{−1})^{t} = ((b̂^{t})^{−1} ∘ b̂)^{−1}.

We call σ_{b} the anti-automorphism adjoint to b. Using the Skolem-Noether theorem, it is easily seen that every F -linear anti-automorphism of End_{F} V is adjoint to some nonsingular pairing, see [4, p. 1]. The map a_{b} is called the asymmetry of b. From the definition, it is clear that the adjoint anti-automorphism and the asymmetry of any scalar multiple of b are the same as those of b. Moreover, the map a_{b} is determined up to sign by properties (i) and (ii).

We combine a_{b} and σ_{b} into a linear involution of End_{F} V as follows:

Proposition 3. Let b be a nonsingular pairing on V. There is a unique linear map γ_{b}: End_{F} V → End_{F} V such that

(3)  b(x, f(y)) = b(y, γ_{b}(f)(x))  for all x, y ∈ V, f ∈ End_{F} V.

This map satisfies the following additional properties:

(i) γ_{b}(f ∘ g ∘ h) = σ_{b}(h) ∘ γ_{b}(g) ∘ σ_{b}^{−1}(f) for f, g, h ∈ End_{F} V; (ii) γ_{b}^{2} = Id_{End V}; (iii) γ_{b}(Id_{V}) = a_{b}.

Proof. Set γ_{b}(f) = σ_{b}(f) ∘ a_{b} (= a_{b} ∘ σ_{b}^{−1}(f)) for f ∈ End_{F} V; then (iii) is clear and (3), (i), (ii) follow from the properties of σ_{b} and a_{b}.

We call γ_{b} the linear involution of End_{F} V associated to b. As for the adjoint anti-automorphism σ_{b} and the asymmetry a_{b}, it is clear that γ_{b} is also the linear involution associated to any scalar multiple of b.

Remark. There are corresponding notions for pairings on faithfully projective modules with values in invertible modules (over an arbitrary commutative ring R): see [3, Chap. III, (8.2)].

1.2. Characterization of asymmetries. The goal of this subsection is to answer the following question: Under which conditions on a map a ∈ GL(V) does there exist a nonsingular pairing b on V whose asymmetry is a, i.e., such that a_{b} = a? Identifying End_{F} V with a matrix algebra M_{n}(F) through the choice of a basis of V, this amounts to asking for which invertible matrices a ∈ GL_{n}(F) the equation a = (x^{t})^{−1}x has a solution x ∈ GL_{n}(F), in view of the definition of a in terms of b̂ in the proof of Proposition 2.

The conditions involve the following vector spaces: for an arbitrary integer m ≥ 1 and ε = ±1, we let

V_{m}^{ε} = \frac{ker(a − ε Id_{V})^{m}}{ker(a − ε Id_{V})^{m−1} + (a − ε Id_{V})(ker(a − ε Id_{V})^{m+1})}.

Theorem 1. Suppose char F ≠ 2. A map a ∈ GL(V) is the asymmetry of some nonsingular pairing on V if and only if the following conditions hold:

(1) a is conjugate to a^{−1} in GL(V);

(2) for every even integer m, dim V_{m}^{+1} is even;

(3) for every odd integer m, dim V_{m}^{−1} is even. If char F = 2, a map a ∈ GL(V) is the asymmetry of some nonsingular pairing on V if and only if conditions (1) and (2) hold.

Proof. We first show that the conditions are necessary. Suppose b is a nonsingular pairing on V such that a_{b} = a. Proposition 2 shows that σ_{b}(a) = a^{−1}. To see how this equality implies condition (1), we argue in terms of matrices. Using a basis of V, we identify End_{F} V with the matrix algebra M_{n}(F). Since the transpose map t is an anti-automorphism, σ_{b} ∘ t is a linear automorphism of M_{n}(F), hence the Skolem-Noether theorem yields an invertible matrix u such that σ_{b} ∘ t is the conjugation by u. Then σ_{b}(x) = ux^{t}u^{−1} for all x ∈ M_{n}(F). In particular, since σ_{b}(a) = a^{−1} it follows that a^{−1} is conjugate to a^{t}. But it is well-known that every matrix is conjugate to its transpose, hence condition (1) is proved.

To show that conditions (2) and (3) are necessary if char F ≠ 2, we show that the nonsingular pairing b induces a nonsingular skew-symmetric pairing on V_{m}^{+1} if m is even and on V_{m}^{−1} if m is odd. Conditions (2) and (3) follow because only even-dimensional vector spaces carry nonsingular skew-symmetric pairings if the characteristic of the base field is different from 2.

Fix some integer m and ε = ±1. For the convenience of notation, we let

U_{m}^{ε} = ker(a − ε Id_{V})^{m},

so V_{m}^{ε} = U_{m}^{ε}/(U_{m−1}^{ε} + (a − ε Id_{V})(U_{m+1}^{ε})). For x, y ∈ U_{m}^{ε}, define

b_{m}^{ε}(x, y) = b(x, (a − ε Id_{V})^{m−1}(y)).

Since y ∈ U_{m}^{ε}, we have

a ∘ (a − ε Id_{V})^{m−1}(y) = ε(a − ε Id_{V})^{m−1}(y), (4)

hence

b(y, (a − ε Id_{V})^{m−1}(x)) = εb(y, a ∘ (a − ε Id_{V})^{m−1}(x))

= εb((a − ε Id_{V})^{m−1}(x), y). (5)

On the other hand, equality (4) yields

(a − ε Id_{V})^{m−1}(y) = (εa^{−1})^{m−1}(a − ε Id_{V})^{m−1}(y) = (−1)^{m−1}σ_{b}(a − ε Id_{V})^{m−1}(y),

hence

b((a − ε Id_{V})^{m−1}(x), y) = (−1)^{m−1}b(x, (a − ε Id_{V})^{m−1}(y)). (6)

Comparing (5) and (6), we obtain

b_{m}^{ε}(y, x) = (−1)^{m−1}εb_{m}^{ε}(x, y).

Therefore, b_{m}^{ε} is a skew-symmetric bilinear form on U_{m}^{ε} if ε = +1 and m is even, and also if ε = −1 and m is odd.

To see that b_{m}^{ε} induces a nonsingular pairing on V_{m}^{ε}, we consider the radical of b_{m}^{ε}, which is

rad b_{m}^{ε} = {x ∈ U_{m}^{ε} | b(x, z) = 0 for all z ∈ (a − ε Id_{V})^{m−1}(U_{m}^{ε})}.

Thus, rad b_{m}^{ε} is the intersection of U_{m}^{ε} with the orthogonal^{1} complement for the form b of

(a − ε Id_{V})^{m−1}(U_{m}^{ε}) = im(a − ε Id_{V})^{m−1} ∩ ker(a − ε Id_{V}),

which is ker σ_{b}(a − ε Id_{V})^{m−1} + im σ_{b}(a − ε Id_{V}). Since σ_{b}(a) = a^{−1}, we have ker σ_{b}(a − ε Id_{V})^{m−1} = ker(a − ε Id_{V})^{m−1} and im σ_{b}(a − ε Id_{V}) = im(a − ε Id_{V}), hence

1If b is not symmetric nor skew-symmetric, one has to distinguish orthogonality on the left and on the right; the orthogonal complements of a-invariant subspaces coincide, however.

rad b_{m}^{ε} = (U_{m−1}^{ε} + im(a − ε Id_{V})) ∩ U_{m}^{ε}
= U_{m−1}^{ε} + (im(a − ε Id_{V}) ∩ U_{m}^{ε})
= U_{m−1}^{ε} + (a − ε Id_{V})(U_{m+1}^{ε}).

Therefore, b_{m}^{ε} induces a nonsingular pairing on U_{m}^{ε}/(U_{m−1}^{ε} + (a − ε Id_{V})(U_{m+1}^{ε})) = V_{m}^{ε}.

Suppose now char F = 2. The arguments above still show that b_{m}^{ε} induces a nonsingular bilinear pairing on V_{m}^{ε}, but in characteristic 2 skew-symmetric pairings are symmetric, hence we cannot conclude that dim V_{m}^{ε} is even. To show that dim V_{m}^{+1} is even if m is even, we show that b_{m}^{+1} is in fact alternating if m is even. For x ∈ U_{m}^{+1} we have

(a − Id_{V})^{m−2}(x) ∈ ker(a − Id_{V})^{2} = ker(a^{2} − Id_{V}),

hence a^{2} ∘ (a − Id_{V})^{m−2}(x) = (a − Id_{V})^{m−2}(x). Since m is even, we obtain by induction

a^{m−2} ∘ (a − Id_{V})^{m−2}(x) = (a − Id_{V})^{m−2}(x),

hence

(a − Id_{V})^{m−2}(x) = a^{2−m} ∘ (a − Id_{V})^{m−2}(x) = σ(a − Id_{V})^{m−2}(x).

Therefore,

b(x, (a − Id_{V})^{m−2}(x)) = b((a − Id_{V})^{m−2}(x), x) = b(x, a ∘ (a − Id_{V})^{m−2}(x)).

It follows that b(x, (a − Id_{V})^{m−1}(x)) = 0, hence b_{m}^{+1} is alternating. This completes the proof that the conditions are necessary.

To prove that the conditions are sufficient, we shall make V into a module over the ring F [X, X^{−1}] of Laurent polynomials in one indeterminate X. As a preparation, we make some observations on the prime ideals of this principal ideal domain.

Let J be the automorphism of F [X, X^{−1}] which maps X to X^{−1}. We also denote by J the extension of this automorphism to the field of fractions F (X) and to the factor module E = F (X)/F [X, X^{−1}]. Every prime ideal P ⊂ F [X, X^{−1}] is generated by an irreducible polynomial of the form

π = a_{0} + a_{1}X + ⋯ + a_{d}X^{d} ∈ F [X]

such that a_{0}, a_{d} ≠ 0. If P^{J} = P, the Laurent polynomials π, π^{J} differ by a factor which is invertible in F [X, X^{−1}], hence π = αX^{d}π^{J} for some α ∈ F^{×}. Comparing coefficients, we have

a_{i} = αa_{d−i} for i = 0, …, d,

hence a_{d} = αa_{0} = α^{2}a_{d} and therefore α = ±1. If d is odd, then

π = ∑_{i=0}^{(d−1)/2} a_{i}(X^{i} + αX^{d−i}),

hence π is divisible by 1+αX. As π is irreducible, we may then choose π = X +1 if α = 1, and π = X −1 if α = −1. Suppose next d is even. If α = −1 and char F ≠ 2, then a_{d/2} = −a_{d/2} implies a_{d/2} = 0. In that case, we have

π = ∑_{i=0}^{d/2−1} a_{i}(X^{i} − X^{d−i}),

hence π is divisible by 1 − X. This is a contradiction, since π is assumed to be irreducible. Therefore, α = 1 and (X^{d/2}π^{−1})^{J} = X^{d/2}π^{−1}. We may then choose π of the form

π = 1 + a_{1}X + a_{2}X^{2} + ⋯ + a_{2}X^{d−2} + a_{1}X^{d−1} + X^{d}.

Let R_{1} be the set of irreducible polynomials of this form.

For each pair of prime ideals {P, P^{J}} with P^{J} ≠ P, we arbitrarily choose a generator π ∈ F [X] of one of P, P^{J} and denote by R_{2} the set of irreducible polynomials thus chosen. Thus, the set of prime ideals of F [X, X^{−1}] is {πF [X, X^{−1}]} where π runs over the set R_{1} ∪ R_{2} ∪ R_{2}^{J} ∪ {X − 1, X + 1}, and we have π^{J}F [X, X^{−1}] ≠ πF [X, X^{−1}] if and only if π ∈ R_{2} ∪ R_{2}^{J}.

Returning to the proof of Theorem 1, we define a structure of F [X, X^{−1}]-module on V by letting

X · v = a(v) for all v ∈ V.

Since F [X, X^{−1}] is a principal ideal domain, the F [X, X^{−1}]-module V decomposes as a (finite) direct sum of quotients of F [X, X^{−1}], as follows:

V ≃ ⨁_{π,m}(F [X, X^{−1}]/π^{m})^{μ(π,m)}

for some integers μ(π, m) which all vanish except a finite number, where π runs over R_{1} ∪ R_{2} ∪ R_{2}^{J} ∪ {X − 1, X + 1}, and m over the positive integers.

Condition (1) shows that the elementary divisors of a are the same as those of a^{−1}, hence

V ≃ ⨁_{π,m}(F [X, X^{−1}]/(π^{J})^{m})^{μ(π,m)}.

Therefore, we have μ(π, m) = μ(π^{J}, m) for all m if π ∈ R_{2}.

For all integers m and for ε = ±1 we have

dim V_{m}^{ε} = μ(X − ε, m).

Therefore, condition (2) says that μ(X − 1, m) is even for all m even, and condition (3) says that μ(X + 1, m) is even for all m odd. Assuming char F ≠ 2 and conditions (1), (2) and (3) hold, we may decompose V into a direct sum of six F [X, X^{−1}]-submodules

V = V_{1} ⊕ V_{2} ⊕ V_{3} ⊕ V_{4} ⊕ V_{5} ⊕ V_{6}

where

V_{1} ≃ ⨁_{π∈R_{1}} ⨁_{m}(F [X, X^{−1}]/π^{m})^{μ(π,m)},
V_{2} ≃ ⨁_{π∈R_{2}} ⨁_{m}(F [X, X^{−1}]/π^{m} ⊕ F [X, X^{−1}]/(π^{J})^{m})^{μ(π,m)},
V_{3} ≃ ⨁_{m odd}(F [X, X^{−1}]/(X − 1)^{m})^{μ(X−1,m)},
V_{4} ≃ ⨁_{m even}(F [X, X^{−1}]/(X − 1)^{m} ⊕ F [X, X^{−1}]/(X − 1)^{m})^{μ(X−1,m)/2},
V_{5} ≃ ⨁_{m even}(F [X, X^{−1}]/(X + 1)^{m})^{μ(X+1,m)},
V_{6} ≃ ⨁_{m odd}(F [X, X^{−1}]/(X + 1)^{m} ⊕ F [X, X^{−1}]/(X + 1)^{m})^{μ(X+1,m)/2}.

If char F = 2 and conditions (1), (2) hold, there is a similar decomposition

V = V_{1} ⊕ V_{2} ⊕ V_{3} ⊕ V_{4}

where V_{1}, …, V_{4} are as above. We shall show below (see Lemma 1) that there are nonsingular (−X)-hermitian forms with values in E (with respect to J) on

F [X, X^{−1}]/π^{m}  if π ∈ R_{1},
F [X, X^{−1}]/π^{m} ⊕ F [X, X^{−1}]/(π^{J})^{m}  if π ∈ R_{2},
F [X, X^{−1}]/(X − 1)^{m}  if m is odd,
(F [X, X^{−1}]/(X − 1)^{m})^{2}  if m is even,
F [X, X^{−1}]/(X + 1)^{m}  if m is even and char F ≠ 2,
(F [X, X^{−1}]/(X + 1)^{m})^{2}  if m is odd and char F ≠ 2.

(7)

The orthogonal sum of these forms yields a nonsingular (−X)-hermitian form

h: V × V → E

with respect to J. As Ischebeck-Scharlau [2] or Waterhouse [8], define an F -linear map T: E → F by observing that every element in E is represented by a unique rational fraction f which has a zero at ∞ and does not have a pole at 0, and letting

T (f + F [X, X^{−1}]) = f(0).

It is easily verified that T (r^{J}) = −T (r) for all r ∈ E. Moreover, for every nonzero r ∈ E there exists an integer k such that T (X^{−k}r) ≠ 0, hence T does not vanish on any nonzero F [X, X^{−1}]-submodule of E.

Let T_{∗}(h): V × V → F be the transfer bilinear map, defined by

T_{∗}(h)(x, y) = T (h(x, y)) for x, y ∈ V.

If x ∈ V is such that T_{∗}(h)(x, y) = 0 for all y ∈ V, then T vanishes on the F [X, X^{−1}]-submodule h(x, V), hence h(x, V) = {0} and therefore x = 0 since h is nonsingular. This shows that T_{∗}(h) is nonsingular.

Moreover, since h is (−X)-hermitian we have

T_{∗}(h)(y, x) = T ((−X)h(x, y)^{J}) = −T (Xh(x, y)^{J}) =

= T (X^{J}h(x, y)) = T (h(x, Xy)) = T_{∗}(h)(x, a(y))

for all x, y ∈ V. Therefore, a is the asymmetry of T_{∗}(h).

To complete the proof, we prove the existence of nonsingular (−X)-hermitian forms as asserted above.

Lemma 1. There are nonsingular (−X)-hermitian forms with values in E (with respect to J) on the modules listed in (7).

Proof. Suppose first π ∈ R_{1}, hence (X^{d/2}π^{−1})^{J} = X^{d/2}π^{−1}, where d is the degree of π. For u, v ∈ F [X, X^{−1}], let

h(u, v) = (X − 1)(X^{d/2}π^{−1})^{m}u^{J}v + F [X, X^{−1}] ∈ E.

This map induces a sesquilinear form on F [X, X^{−1}]/π^{m}. The induced form is (−X)-hermitian since (X − 1)^{J} = −X^{−1}(X − 1); it is nonsingular since h(1, v) = 0 implies π^{m} divides (X − 1)v in F [X, X^{−1}], hence v = 0 in F [X, X^{−1}]/π^{m} since π is prime to X − 1.

Next, suppose π ∈ R_{2}. For u_{1}, u_{2}, v_{1}, v_{2} ∈ F [X, X^{−1}], we let

h((u_{1}, u_{2}), (v_{1}, v_{2})) = π^{−m}u_{1}^{J}v_{2} − X(π^{J})^{−m}u_{2}^{J}v_{1} + F [X, X^{−1}] ∈ E.

Computation shows that this map induces a nonsingular (−X)-hermitian form on (F [X, X^{−1}]/π^{m}) × (F [X, X^{−1}]/(π^{J})^{m}).

Similarly, the following maps induce nonsingular (−X)-hermitian forms on the corresponding modules (where e is an arbitrary non-negative integer):

h(u, v) = X^{e−1}(X − 1)^{−2e−1}u^{J}v + F [X, X^{−1}] ∈ E on F [X, X^{−1}]/(X − 1)^{2e+1};

h((u_{1}, u_{2}), (v_{1}, v_{2})) = X^{e}(X − 1)^{−2e}(u_{1}^{J}v_{2} − Xu_{2}^{J}v_{1}) + F [X, X^{−1}] ∈ E

on (F [X, X^{−1}]/(X − 1)^{2e})^{2};

and if char F ≠ 2,

h(u, v) = (X − 1)X^{e}(X + 1)^{−2e}u^{J}v + F [X, X^{−1}] ∈ E on F [X, X^{−1}]/(X + 1)^{2e};

h((u_{1}, u_{2}), (v_{1}, v_{2})) = (X − 1)^{2e+1}(X + 1)^{−2e−1}(u_{1}^{J}v_{2} + Xu_{2}^{J}v_{1}) + F [X, X^{−1}] ∈ E

on (F [X, X^{−1}]/(X + 1)^{2e+1})^{2}.

We omit the straightforward verifications.

Remark. The theory of hermitian forms over principal ideal domains can also be used to show that the conditions in Theorem 1 are necessary.

2. The asymmetry of an anti-automorphism

2.1. Definition. Let A be a (finite-dimensional) central simple algebra over an arbitrary field F, and let σ: A → A be an F -linear anti-automorphism of A. Our goal is to attach to σ a unit a_{σ} ∈ A^{×} which plays the same rôle as the asymmetry a_{b} of a nonsingular pairing b with respect to the adjoint anti-automorphism σ_{b}. The key to the definition is an analogue of the linear involution γ_{b}, which we now define.

Proposition 4. There is a unique linear map γ_{σ}: A → A which satisfies the following property: for any splitting field K of A, any isomorphism

θ: A_{K} = A ⊗_{F} K → End_{K} V

and any nonsingular pairing b on V such that σ_{b} = θ ∘ (σ ⊗ Id_{K}) ∘ θ^{−1},

θ ∘ (γ_{σ} ⊗ Id_{K}) ∘ θ^{−1} = γ_{b}.

This map satisfies the following additional properties:

(i) γ_{σ}(xyz) = σ(z)γ_{σ}(y)σ^{−1}(x) for x, y, z ∈ A; (ii) γ_{σ}^{2} = Id_{A}.

Proof. It suffices to prove the existence of γ_{σ}. Uniqueness is then clear, and the additional properties follow from those of γ_{b} in Proposition 3.

Let T_{σ}: A × A → F be the nonsingular pairing defined by

T_{σ}(x, y) = Trd_{A}(σ(x)y) for x, y ∈ A,

where Trd_{A} is the reduced trace. Let (e_{i})_{i∈I} be a basis of A and let (e_{i}^{♯})_{i∈I} be the dual basis with respect to the pairing T_{σ}, so that

T_{σ}(e_{i}^{♯}, e_{j}) = δ_{ij} for i, j ∈ I.

We let

γ_{σ}(x) = ∑_{i∈I} e_{i}xe_{i}^{♯} for x ∈ A.

In other words, γ_{σ} is the image of ∑_{i∈I} e_{i} ⊗e_{i}^{♯} ∈ A⊗_{F} A under the “sandwich” map Sand: A ⊗_{F} A → End_{F} A defined by Sand(x ⊗ y)(z) = xzy. Observe that γ_{σ} does not depend on the choice of the basis (e_{i})_{i∈I} since ∑_{i∈I} e_{i} ⊗e_{i}^{♯} is the element which corresponds to Id_{A} under the bijection Id_{A} ⊗T̂_{σ}: A ⊗_{F} A → A ⊗_{F} A^{∗} = End_{F} A.

As a consequence, for every field extension K/F, the map γ_{σ⊗Id_{K}}: A ⊗ K → A ⊗ K satisfies

γ_{σ⊗Id_{K}} = γ_{σ} ⊗ Id_{K}

since for x ∈ A ⊗ K,

γ_{σ⊗Id_{K}} (x) = ∑_{i∈I}(e_{i} ⊗ 1)x(e_{i}^{♯} ⊗ 1) = (γ_{σ} ⊗ Id_{K})(x).

To show that γ_{σ} is as required, assume that A is split: let A = End_{F} V and let b be a nonsingular pairing on V such that σ = σ_{b}. We have to show that γ_{σ} = γ_{b}. To prove this equality, we use the identification V ⊗_{F} V = End_{F} V defined by the linear isomorphism Id_{V} ⊗b̂: V ⊗_{F} V → V ⊗_{F} V^{∗} = End_{F} V. Then (v ⊗ w)(x) = vb(w, x) for v, w, x ∈ V and moreover

f ∘ (v ⊗ w) = f(v) ⊗ w, σ(v ⊗ w) = a_{b}(w) ⊗ v and Trd(v ⊗ w) = b(w, v)

for v, w ∈ V and f ∈ End_{F} V. Let (v_{i})_{1≤i≤n} be a basis of V and let (v′_{i})_{1≤i≤n} be the dual basis for the pairing b, so that

b(v′_{i}, v_{j}) = δ_{ij} for i, j = 1, …, n. (8)

Then (v_{i} ⊗ v_{j})_{1≤i,j≤n} is a basis of End_{F} V, and the dual basis with respect to T_{σ} is given by

(v_{i} ⊗ v_{j})^{♯} = v′_{i} ⊗ v′_{j}.

Therefore, we have for f ∈ End_{F} V

γ_{σ}(f) = ∑_{i,j=1}^{n}(v_{i} ⊗ v_{j}) ∘ f ∘ (v′_{i} ⊗ v′_{j})

= ∑_{i,j=1}^{n} v_{i} ⊗ v′_{j} b(v_{j}, f(v′_{i}))

= ∑_{i,j=1}^{n} v_{i} ⊗ v′_{j} b(v′_{i}, γ_{b}(f)(v_{j})).

For all x ∈ V we have x = ∑_{i=1}^{n} v_{i}b(v′_{i}, x), hence ∑_{i=1}^{n} v_{i} b(v′_{i}, γ_{b}(f)(v_{j})) = γ_{b}(f)(v_{j}) for all j, and the last equality above simplifies to

γ_{σ}(f) = ∑_{j=1}^{n} γ_{b}(f)(v_{j}) ⊗ v′_{j} = γ_{b}(f) ∘ (∑_{j=1}^{n} v_{j} ⊗ v′_{j}).

Since ∑_{j=1}^{n} v_{j} ⊗ v′_{j} = Id_{V}, it follows that γ_{σ}(f) = γ_{b}(f).

In view of property (i), we have

γ_{σ}(x) = σ(x)γ_{σ}(1) = γ_{σ}(1)σ^{−1}(x) for all x ∈ A. (9)

Therefore, γ_{σ} is completely determined by the element γ_{σ}(1) ∈ A^{×}.

Definition. The asymmetry of the anti-automorphism σ is the element a_{σ} = γ_{σ}(1) ∈ A^{×}, where γ_{σ} is the linear involution defined in Proposition 4.

If A = End_{F} V and σ = σ_{b} is the anti-automorphism adjoint to some nonsingular pairing b on V, it follows from Proposition 4 and property (iii) of Proposition 3 that a_{σ} is the asymmetry of the nonsingular form b, i.e.,

a_{σ} = a_{b}.

In the general case, equation (9) shows that

σ^{2}(x) = a_{σ}xa_{σ}^{−1} for all x ∈ A. (10)

Moreover, since γ_{σ}^{2} = Id_{A} we have

1 = γ_{σ}(a_{σ}) = σ(a_{σ})a_{σ}. (11)

The element a_{σ} is uniquely determined up to sign by (10) and (11).

Recall that an anti-automorphism σ is called an involution if σ^{2} = Id_{A}.

Proposition 5. A linear anti-automorphism is an involution if and only if its asymmetry is +1 or −1.

Proof. If a_{σ} = ±1, equation (10) shows that σ^{2} = Id_{A}. Conversely, if σ is an involution, (10) shows that a_{σ} ∈ F^{×}. It then follows from (11) that a_{σ}^{2} = 1, hence a_{σ} = ±1.

If char F ≠ 2, a linear involution σ is called orthogonal (resp. symplectic) if after scalar extension to a splitting field it is adjoint to a symmetric (resp. skewsymmetric) bilinear pairing. Therefore, orthogonal involutions are exactly the linear anti-automorphisms with asymmetry +1, and symplectic involutions are those with asymmetry −1. Therefore, equations (10) and (11) are not sufficient to determine the type of the involution. This observation suggests that the sign of a_{σ} is meaningful for arbitrary anti-automorphisms.

The following proposition yields an alternative definition of the asymmetry a_{σ}, without reference to the linear involution γ_{σ} and without scalar extension to a splitting field.

Let σ_{∗}: A ⊗_{F} A → End_{F} A be the F -algebra homomorphism defined by

σ_{∗}(a ⊗ b)(x) = axσ(b) for a, b, x ∈ A,

and recall (from [4, (3.5)], for instance) the Goldman element of A: this is the element g ∈ A ⊗_{F} A such that Sand(g)(x) = Trd_{A}(x) for all x ∈ A. Thus, there is a well-defined linear endomorphism σ_{∗}(g): A → A.

Proposition 6. The asymmetry of σ is the unique element a_{σ} ∈ A^{×} such that

σ(σ_{∗}(g)(f)) = a_{σ}f

for all f ∈ A.

Proof. It suffices to prove that a_{σ} satisfies the property above, since uniqueness is clear. To do this, we may extend scalars to a splitting field. Therefore, we may assume A = End_{F} V for some F -vector space V, and σ = σ_{b} is the antiautomorphism adjoint to some nonsingular pairing b on V.

For all f ∈ A and all x, y ∈ V we have

b(f(x), y) = b(y, a_{σ} ∘ f(x)),

by definition of the asymmetry (see (2)), hence we have to show

b(f(x), y) = b(y, σ(σ_{∗}(g)(f))(x))

or, equivalently (by definition of σ = σ_{b}),

b(f(x), y) = b(σ_{∗}(g)(f)(y), x) (12)

for all f ∈ A and all x, y ∈ V.

In order to compute the right-hand side, we identify A = End_{F} V to V ⊗_{F} V via the linear isomorphism Id_{V} ⊗b̂: V ⊗_{F} V → V ⊗_{F} V^{∗} = End_{F} V, as in the proof of Proposition 4. If (v_{i})_{1≤i≤n} is a basis of V and (v′_{i})_{1≤i≤n} is the dual basis for the pairing b (see (8)), then the Goldman element is

g = ∑_{i,j}(v_{i} ⊗ v′_{j}) ⊗ (v_{j} ⊗ v′_{i})

since it is easily computed that for all u, w ∈ V

Sand(g)(u ⊗ w) = ∑_{i,j}(v_{i} ⊗ v′_{j}) ∘ (u ⊗ w) ∘ (v_{j} ⊗ v′_{i}) =

= (∑_{i} v_{i} ⊗ v′_{i})(∑_{j} b(v′_{j}, u)b(w, v_{j})) = b(w, u) ∑_{i} v_{i} ⊗ v′_{i} = Trd(u ⊗ w) Id_{V} .

Now, for u, w ∈ V,

σ_{∗}(g)(u ⊗ w) = ∑_{i,j}(v_{i} ⊗ v′_{j}) ∘ (u ⊗ w) ∘ σ(v_{j} ⊗ v′_{i}).

Since (u ⊗ w) ∘ σ(f) = u ⊗ f(w) for f ∈ End_{F} V, the right-hand side of the last equality simplifies to

∑_{i,j}((v_{i} ⊗ v′_{j}(u)) ⊗ ((v_{j} ⊗ v′_{i})(w)) = ∑_{i,j} v_{i} ⊗ v_{j}b(v′_{j}, u)b(v′_{i}, w),

hence

σ_{∗}(g)(u ⊗ w) = w ⊗ u.

Therefore, for u, w, x, y ∈ V,

b(σ_{∗}(g)(u ⊗ w)(y), x) = b((w ⊗ u)(y), x) = b(w, x)b(u, y).

Since we also have b((u ⊗ w)(x), y) = b(u, y)b(w, x), equation (12) holds for f = u ⊗ w. Since End_{F} V = V ⊗_{F} V, it follows that (12) holds for all f ∈ A, and the proof is complete.

Remark. Asymmetries can be defined on the same model for anti-automorphisms of Azumaya algebras; one may avoid the use of a basis of A in Proposition 4 by defining γ_{σ} = Sand(ξ_{σ}) where ξ_{σ} ∈ A⊗A is the element mapped to Id_{A} by Id_{A} ⊗T̂_{σ}. Alternatively, we may set ξ_{σ} = (Id_{A} ⊗σ^{−1})(g) where g ∈ A ⊗ A is the Goldman element. This is the approach taken by Saltman in [7] (see also [3, Chap. III, §8]).

2.2. Characterization of asymmetries. In this subsection, we show that in a central simple algebra of exponent 2, every unit which is conjugate to its inverse is the asymmetry of some anti-automorphism.

We first compare the asymmetries of two anti-automorphisms σ, τ on a central simple algebra A. The Skolem-Noether theorem shows that the automorphism τ ∘ σ^{−1} is the conjugation by some unit u ∈ A^{×}, i.e.,

τ(x) = uσ(x)u^{−1} for all x ∈ A. (13)

Proposition 7. Let σ, τ be anti-automorphisms of a central simple algebra A, and let u ∈ A^{×} be such that (13) holds. The asymmetries a_{σ}, a_{τ} of σ and τ are related by

a_{τ} = uσ(u)^{−1}a_{σ}.

Proof. We use the definition of asymmetry provided by Proposition 6. For a, b, x ∈ A, we have

τ_{∗}(a ⊗ b)(x) = axτ(b) = axuσ(b)u^{−1}

hence

τ_{∗}(a ⊗ b)(x) = σ_{∗}(a ⊗ b)(xu)u^{−1}.

Therefore, denoting by r_{u}: A → A the linear map of multiplication on the right by u, we have

τ_{∗}(a ⊗ b) = (r_{u})^{−1} ∘ σ_{∗}(a ⊗ b) ∘ r_{u}

for all a, b ∈ A, hence also

τ_{∗}(g) = (r_{u})^{−1} ∘ σ_{∗}(g) ∘ r_{u}

for g the Goldman element of A. It follows that for all f ∈ A,

τ_{∗}(g)(f) = σ_{∗}(fu)u^{−1}. (14)

By Proposition 6, the asymmetry a_{τ} satisfies

a_{τ} f = τ(τ_{∗}(g)(f)) for all f ∈ A.

Using (14), we obtain

a_{τ} f = τ(σ_{∗}(g)(fu)u^{−1}) = uσ(σ_{∗}(g)(fu)u^{−1})u^{−1} = uσ(u)^{−1}σ(σ_{∗}(g)(fu))u^{−1}.

Proposition 6 also yields σ(σ_{∗}(g)(fu)) = a_{σ}fu, hence

a_{τ} f = uσ(u)^{−1}a_{σ}f for all f ∈ A.

The proposition follows.

Theorem 2. Let A be a central simple algebra of exponent 2 over an arbitrary field F. A unit is the asymmetry of some anti-automorphism of A if and only if it is conjugate to its inverse.

Proof. Suppose a ∈ A^{×} is the asymmetry of some anti-automorphism σ. We have to show that the F -vector space

U = {x ∈ A | xa = a^{−1}x}

contains an invertible element. This amounts to proving that the restriction of the reduced norm polynomial Nrd_{A} does not vanish on U. Theorem 1 shows that this polynomial does not vanish on U ⊗ K, for any splitting field K of A, since a is the asymmetry of σ ⊗ Id_{K}. Therefore, the reduced norm does not vanish on U, since F is an infinite field. (Note that every central simple algebra over a finite field is split, hence of exponent 1.)

For the converse, suppose a ∈ A^{×} is conjugate to a^{−1}. Let K be a splitting field of A; identify A ⊗ K = End_{K} V for some K-vector space V. We first show, by using Theorem 1, that a (= a ⊗ 1) is the asymmetry of some anti-automorphism of End_{K} V. With the same notation as in Theorem 1, we have to prove that dim_{K} V_{m}^{+1} is even if m is even, and moreover that dim_{K} V_{m}^{−1} is even if m is odd and char F ≠ 2. For every integer m ≥ 1 and ε = ±1, we have an exact sequence of K-vector spaces

0 → \frac{ker(a − ε Id_{V})^{m+1}}{ker(a − ε Id_{V})^{m}} ⟶^{a−ε Id_{V}} \frac{ker(a − ε Id_{V})^{m}}{ker(a − ε Id_{V})^{m−1}} → V_{m}^{ε} → 0,

hence

(15) dim V_{m}^{ε} = rk(a − ε Id_{V})^{m−1} − 2 rk(a − ε Id_{V})^{m} + rk(a − ε Id_{V})^{m+1},

where rk denotes the rank.

For all b ∈ A we have

rk b = \frac{dim_{K} b(A ⊗ K)}{deg(A ⊗ K)} = \frac{dim_{F} bA}{deg A},

hence rk b is divisible by the Schur index ind A (see [4, (1.9)]). Since A has exponent 2, ind A is even, by [1, Theorem 5.17]. Therefore, rk b is even for all b ∈ A, and equation (15) shows that dim V_{m}^{ε} is even for every integer m and for ε = ±1. By Theorem 1, it follows that a is the asymmetry of some anti-automorphism θ of A ⊗ K.

Now, fix some anti-automorphism σ of A. Let a_{σ} be its asymmetry and consider the F -vector space

W = {x ∈ A | xa = σ(x)a_{σ}}.

If u ∈ (A ⊗ K)^{×} is such that θ(x) = u(σ ⊗ Id_{K})(x)u^{−1} for all x ∈ A ⊗ K, then u^{−1} ∈ W ⊗ K, by Proposition 7. Therefore, the same arguments as in the first part of the proof show that W contains an invertible element w. Using Proposition 7 again, we see that a is the asymmetry of the anti-automorphism x ↦ w^{−1}σ(x)w.

Corollary 1 (Albert). Every central simple algebra of exponent 2 carries an involution. Moreover, if the characteristic of the base field is different from 2, every central simple algebra of exponent 2 carries involutions of both orthogonal and symplectic types.

Proof. It readily follows from Theorem 2 that +1 and −1 are asymmetries of some anti-automorphisms. These anti-automorphisms are involutions, by Proposition 5.

2.3. The determinant of an anti-automorphism. Let σ be a linear antiautomorphism of a central simple algebra A over an arbitrary field F. Let a_{σ} ∈ A^{×} be the asymmetry of A and γ_{σ} the linear involution of Proposition 4. Consider the vector spaces

Alt(A, σ) = {x − σ(x)a_{σ} | x ∈ A} = {x − γ_{σ}(x) | x ∈ A}

and

Sk(A, σ) = {x ∈ A | σ(x) + xa_{σ}^{−1} = 0} = {x ∈ A | γ_{σ}(x) = −x}.

From equations (10) and (11), it follows that Alt(A, σ) ⊂ Sk(A, σ). Moreover, we have x − γ_{σ}(x) = 2x for all x ∈ Sk(A, σ), hence Alt(A, σ) = Sk(A, σ) if char F ≠ 2.

Lemma 2. Suppose σ, τ are anti-automorphisms of A, and let u ∈ A^{×} be such that

τ(x) = uσ(x)u^{−1} for all x ∈ A.

Then

Alt(A, τ) = u Alt(A, σ) and Sk(A, τ) = u Sk(A, σ).

Proof. Proposition 7 yields a_{τ} = uσ(u)^{−1}a_{σ} and a_{σ} = u^{−1}τ(u)a_{τ}. Therefore, for all x ∈ A we have

x − τ(x)a_{τ} = u(u^{−1}x − σ(u^{−1}x)a_{σ}) and u(x − σ(x)a_{σ}) = ux − τ(ux)a_{τ} ,

proving that Alt(A, τ) = u Alt(A, σ). The proof that Sk(A, τ) = u Sk(A, σ) is along the same lines.

Lemma 3. If deg A is even, Alt(A, σ) contains invertible elements. Moreover, the square class Nrd_{A}(x) · F^{×2} ∈ F^{×}/F^{×2} does not depend on the choice of x ∈ A^{×} ∩ Alt(A, σ).

Proof. Let τ be an anti-automorphism of A with asymmetry +1 and let u ∈ A^{×} be such that

τ(x) = uσ(x)u^{−1} for all x ∈ A.

By Lemma 2, we have

Alt(A, σ) = u^{−1} Alt(A, τ). (16)

Since τ is an involution, Corollary (2.8) of [4] shows that Alt(A, τ) contains invertible elements if deg A is even, hence Alt(A, σ) also contains invertible elements. Moreover, from [4, (7.1)], it follows that all the invertible elements have the same reduced norm up to a square of F; therefore, if v ∈ A^{×} ∩Alt(A, τ) it follows from (16) that Nrd_{A}(x) ∈ Nrd_{A}(u^{−1}v) · F^{×2} for all x ∈ A^{×} ∩ Alt(A, σ).

This last lemma allows us to define the determinant of an anti-automorphism σ of a central simple algebra A of even degree, as follows:

det σ = Nrd_{A}(x) · F^{×2} ∈ F^{×}/F^{×2}

for any x ∈ A^{×} ∩ Alt(A, σ).

This definition is consistent with [4, (7.2)], where the determinant of an orthogonal involution is defined.

Example 1. Since clearly 1 − a_{σ} ∈ Alt(A, σ), we have

det σ = Nrd_{A}(1 − a_{σ}) · F^{×2}

if 1 − a_{σ} is invertible. Therefore, the determinant of σ is entirely determined by its asymmetry in this particular case.

Example 2. The transpose involution on a matrix algebra M_{n}(F) (with n even) has trivial determinant. Indeed, the matrix

(\begin{matrix}m_{1} & & & & 0 \\ & . & . & . \\ 0 & & & & m_{n/2} \end{matrix})  where m_{1} = ⋯ = m_{n/2} = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

is in Alt(M_{n}(F), t) and has determinant 1.

Proposition 8. Let σ, τ be anti-automorphisms of a central simple algebra A of even degree, and let u ∈ A^{×} be such that

τ(x) = uσ(x)u^{−1} for all x ∈ A.

Then

det τ = Nrd_{A}(u) det σ.

Proof. This readily follows from Lemma 2.

Proposition 9. Let V be an even-dimensional vector space over an arbitrary field F and let b be a nonsingular pairing on V. For every basis (v_{i})_{1≤i≤n} of V,

det σ_{b} = det(b(v_{i}, v_{j}))_{1≤i,j≤n} · F^{×2}.

Proof. Identify End_{F} V with the matrix algebra M_{n}(F) by means of the basis (v_{i})_{1≤i≤n}. The anti-automorphism σ_{b} is then given by

σ_{b}(m) = u^{−1}m^{t}u for all m ∈ M_{n}(F),

where u = (b(v_{i}, v_{j}))_{1≤i,j≤n} ∈ M_{n}(F). Therefore, Proposition 8 yields

det σ_{b} = det u^{−1} det t.

Since it was observed in Example 2 above that det t is trivial, the proposition follows.

References

[1] A.A. Albert, Structure of Algebras, Coll. Pub. 24, Amer. Math. Soc., Providence, RI, 1939.

[2] F. Ischebeck, W. Scharlau, Hermitesche und orthogonale Operatoren über kommutativen Ringen, Math. Ann. 200 (1973), 327–334.

[3] M.-A. Knus, Quadratic and Hermitian Forms over Rings, Grundlehren der Mathematischen Wissenschaften, vol. 294, Springer-Verlag, Berlin, 1991.

[4] M.-A. Knus, A.S. Merkurjev, M. Rost and J.-P. Tignol, The Book of Involutions, Coll. Pub. 44, Amer. Math. Soc., Providence, RI, 1998.

[5] A. Ranicki, High-dimensional Knot Theory, Springer Monographs in Mathematics, Springer-Verlag, Berlin-Heidelberg-New York, 1998.

[6] C. Riehm, The equivalence of bilinear forms, J. Algebra 31 (1974), 45–66.

[7] D.J. Saltman, Azumaya algebras with involution, J. Algebra 52 (1978), 526–539.

[8] W.C. Waterhouse, A nonsymmetric Hasse-Minkowski theorem, Amer. J. Math. 99 (1977), 755–759.

[9] J. Williamson, On the algebraic problem concerning the normal form of linear dynamical systems, Amer. J. Math. 58 (1936), 141–163.

Anne Cortella

e-mail: cortella@math.univ-fcomte.fr

UMR CNRS 6623 Laboratoire de Mathématiques de Besançon Université de Franche-Comté 16, route de Gray F-25030 Besançon Cedex France

Jean-Pierre Tignol

e-mail: tignol@agel.ucl.ac.be
web: http://www.math.ucl.ac.be/tignol/

Départment de Mathématiques Université catholique de Louvain B-1348 Louvain-la-Neuve Belgium