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for 2 ¤ k ¤ n = deg A,
—k k
rdim(A sk ) =
0 for k > n = deg A.
If A is split: A = EndF (V ), A—k = EndF (V —k ), then there is a natural isomor-
phism of A—k -modules:
k
A—k sk = HomF ( V , V —k ).
116 II. INVARIANTS OF INVOLUTIONS


Proof : Since the reduced dimension does not change under scalar extension, it
su¬ces to prove the second part. Under the correspondence between left ideals in
A—k = EndF (V —k ) and subspaces of V —k (see §??), we have
A—k sk = HomF (V —k / ker sk , V —k ).
From the description of gk (π) in the split case, it follows that ker sk contains the
subspace of V —k spanned by the products v1 — · · · — vk where vi = vj for some
k
V ’ V —k / ker sk .
indices i = j. Therefore, there is a natural epimorphism
To prove that this epimorphism is injective, pick a basis (e1 , . . . , en ) of V . For
the various choices of indices i1 , . . . , ik such that 1 ¤ i1 < i2 < · · · < ik ¤ n, the
images sk (ei1 — · · · — eik ) are linearly independent, since they involve di¬erent basis
vectors in V —k . Therefore,
n
dim(V —k / ker sk ) = dim im sk ≥ ,
k

and the epimorphism above is an isomorphism.

(10.4) De¬nition. Let A be a central simple algebra of degree n over a ¬eld F .
For every integer k = 2, . . . , n we de¬ne the k-th »-power of A as
»k A = EndA—k (A—k sk ).
We extend this de¬nition by setting »1 A = A. Note that for k = 2 we recover the
de¬nition of »2 A given in §?? (see (??)).
The following properties follow from the de¬nition, in view of (??), (??) and (??):
(a) »k A is a central simple F -algebra Brauer-equivalent to A—k , of degree
n
deg »k A = .
k

(b) There is a natural isomorphism:
k
»k EndF (V ) = EndF ( V ).
(10.5) Corollary. If k divides the index ind A, then ind A—k divides (ind A)/k.
Proof : By replacing A by a Brauer-equivalent algebra, we may assume that A is
a division algebra. Let n = deg A = ind A. Arguing by induction on the number
of prime factors of k, it su¬ces to prove the corollary when k = p is a prime
number. If K is a splitting ¬eld of A of degree n, then K also splits »k A, hence
ind »p A divides n. On the other hand, ind »p A divides deg »p A = n , and the
p
n p
greatest common divisor of n and p is n/p, hence ind » A divides n/p. Since
»p A is Brauer-equivalent to A—p , we have ind »p A = ind A—p , and the proof is
complete.

10.B. The canonical involution. Let V be a vector space of even dimen-
n
sion n = 2m over a ¬eld F of arbitrary characteristic. Since dim V = 1, the
composition of the exterior product
m m n
§: V— V’ V
n m

with a vector-space isomorphism V ’ F is a bilinear form on
’ V , which is
uniquely determined up to a scalar factor.
§10. THE DISCRIMINANT ALGEBRA 117


(10.6) Lemma. The bilinear map § is nonsingular. It is symmetric if m is even
and skew-symmetric if m is odd. If char F = 2, the map § is alternating for
all m. Moreover, the discriminant of every symmetric bilinear form induced from
n
§ through any isomorphism V F is trivial.
Proof : Let (e1 , . . . , en ) be a basis of V . For every subset of m indices
S = {i1 , . . . , im } ‚ {1, . . . , n} with i1 < · · · < im
m
we set eS = ei1 § · · · § eim ∈ V . As S runs over all the subsets of m indices,
m
the elements eS form a basis of V.
Since eS § eT = 0 when S and T are not disjoint, we have for x = xS e S
x § eT = ±xT e1 § · · · § en ,
where T is the complementary subset of T in {1, . . . , n}. Therefore, if x § eT = 0
for all T , then x = 0. This shows that the map § is nonsingular. Moreover, for all
subsets S, T of m indices we have
2
eS § eT = (’1)m eT § eS ,
hence § is symmetric if m is even and skew-symmetric if m is odd. Since eS §eS = 0
for all S, the form § is alternating if char F = 2.
n
Suppose m is even and ¬x an isomorphism V F to obtain from § a sym-
m m
metric bilinear form b on V . The space V decomposes into an orthogonal
direct sum:

m
V= ES
S∈R

where ES is the subspace spanned by the basis vectors eS , eS where S is the
complement of S and R is a set of representatives of the equivalence classes of
subsets of m indices under the relation S ≡ T if and only if S = T or S = T . (For
instance, one can take R = { S | 1 ∈ S }.)
On basis elements eS , eS the matrix of b has the form ± ± for some ± ∈ F — .
0
0
m d/2 —2
Therefore, if d = dim V we have det b = (’1) · F , hence
disc b = 1.


m
Since § is nonsingular, there is an adjoint involution γ on EndF ( V ) de¬ned
by
γ(f )(x) § y = x § f (y)
m m
for all f ∈ EndF ( V ), x, y ∈ V . The involution γ is of the ¬rst kind and its
discriminant is trivial; it is of orthogonal type if m is even and char F = 2, and it
is of symplectic type if m is odd or char F = 2. We call γ the canonical involution
m
on EndF ( V ).
Until the end of this subsection, A is a central simple F -algebra of degree
n = 2m. Our purpose is to de¬ne a canonical involution on »m A in such a way as
to recover the de¬nition above in the split case.

We ¬rst prove a technical result concerning the elements sk ∈ A—k de¬ned
in the preceding section:
118 II. INVARIANTS OF INVOLUTIONS


(10.7) Lemma. Let A be a central simple F -algebra of degree n = 2m. Since
sm ∈ A—m , we may consider sm — sm ∈ A—n . Then
sn ∈ A—n (sm — sm ).
Proof : In the symmetric group Sn , consider the subgroup Sm,m Sm — Sm con-
sisting of the permutations which preserve {1, . . . , m} (and therefore also the set
{m + 1, . . . , n}). The split case shows that gn (π1 , π2 ) = gm (π1 ) — gm (π2 ) for π1 ,
π2 ∈ Sm . Therefore,
sm — s m = sgn(π)gn (π).
π∈Sm,m

Let R be a set of representatives of the left cosets of Sm,m in Sn , so that each
π ∈ Sn can be written in a unique way as a product π = ρ —¦ π for some ρ ∈ R and
some π ∈ Sm,m . Since gn is a homomorphism, it follows that
sgn(π)gn (π) = sgn(ρ)gn (ρ) · sgn(π )gn (π ),
hence, summing over π ∈ Sn :
sn = sgn(ρ)gn (ρ) sm — sm .
ρ∈R



Recall from (??) that »m A = EndA—m (A—m sm ). There is therefore a natural
isomorphism:
»m A —F »m A = EndA—n A—n (sm — sm ) .
Since (??) shows that sn ∈ A—n (sm — sm ), we may consider
I = { f ∈ EndA—n A—n (sm — sm ) | sf = {0} }.
(10.8) n

This is a right ideal in EndA—n A—n (sm — sm ) = »m A —F »m A.
(10.9) Lemma. If A = EndF (V ), then under the natural isomorphisms
m m m m
»m A —F »m A = EndF ( V ) —F EndF ( V ) = EndF ( V— V)
the ideal I de¬ned above is
m m
I = { • ∈ EndF ( V— V ) | § —¦• = 0}
m
where § is the canonical bilinear form on V , viewed as a linear map
m m n
§: V— V’ V.
m
Proof : As observed in (??), we have A—m sm = HomF ( V, V —m ), hence
m m
A—n (sm — sm ) = HomF ( V , V —n ).
V—
n
Moreover, we may view sn as a map sn : V —n ’ V —n which factors through V:
there is a commutative diagram:
s
V —n ’ ’ ’ V —n
’n’
¦
¦ ¦
¦
sn

§
m m n
V— V ’’’
’’ V
m m
The image of sn ∈ A—n (sm — sm ) in HomF ( V, V —n ) under the
V—
identi¬cation above is then the induced map sn .
By (??) every endomorphism f of the A—n -module A—n (sm — sm ) has the form
xf = x —¦ •
§10. THE DISCRIMINANT ALGEBRA 119

m m
for some uniquely determined • ∈ EndF ( V— V ). The correspondence
f ” • yields the natural isomorphism
m m
»m A — »m A = EndA—n A—n (sm — sm ) = EndF ( V— V ).
Under this correspondence, the elements f ∈ EndA—n A—n (sm — sm ) which vanish
m m
on sn correspond to endomorphisms • ∈ EndF ( V— V ) such that sn —¦ • =
0. It is clear from the diagram above that ker sn = ker §, hence the conditions
sn —¦ • = 0 and § —¦ • = 0 are equivalent.
(10.10) Corollary. The right ideal I ‚ »m A —F »m A de¬ned in (??) above sat-
is¬es the following conditions:
(1) »m A —F »m A = I • (1 — »m A).
0
(2) I contains the annihilator (»m A — »m A) · (1 ’ g) , where g is the Goldman
element of »m A —F »m A, if m is odd or char F = 2; it contains 1 ’ g but not
0
(»m A — »m A) · (1 ’ g) if m is even and char F = 2.
Proof : It su¬ces to check these properties after scalar extension to a splitting ¬eld
of A. We may thus assume A = EndF (V ). The description of I in (??) then shows
that I is the right ideal corresponding to the canonical involution γ on EndF ( m V )
under the correspondence of (??).
(10.11) De¬nition. Let A be a central simple F -algebra of degree n = 2m. The
canonical involution γ on »m A is the involution of the ¬rst kind corresponding to
the ideal I de¬ned in (??) under the correspondence of (??).
The following properties follow from the de¬nition by (??) and (??), and by
scalar extension to a splitting ¬eld in which F is algebraically closed:
m
(a) If A = EndF (V ), the canonical involution γ on »m A = EndF ( V ) is the
m
adjoint involution with respect to the canonical bilinear map § : V—
m n
V’ V.
(b) γ is of symplectic type if m is odd or char F = 2; it is of orthogonal type if
m is even and char F = 2; in this last case we have disc(γ) = 1.
In particular, if A has degree 2 (i.e., A is a quaternion algebra), then the
canonical involution on A = »1 A has symplectic type, hence it is the quaternion
conjugation.
10.C. The canonical quadratic pair. Let A be a central simple F -algebra
of even degree n = 2m. As observed in (??), the canonical involution γ on »m A
is symplectic for all m if char F = 2. We show in this section that the canonical
involution is actually part of a canonical pair (γ, f ) on »m A for all m ≥ 2 if
char F = 2. (If char F = 2, a quadratic pair is uniquely determined by its involution;
thus »m A carries a canonical quadratic pair if and only if γ is orthogonal, i.e., if
and only if m is even).
We ¬rst examine the split case.
(10.12) Proposition. Assume char F = 2 and let V be an F -vector space of di-
mension n = 2m ≥ 4. There is a unique quadratic map
m n
q: V’ V
which satis¬es the following conditions:
(1) q(v1 § · · · § vm ) = 0 for all v1 , . . . , vm ∈ V ;
120 II. INVARIANTS OF INVOLUTIONS

m m n
(2) the polar form bq : V— V’ V is the canonical pairing §. In
particular, the quadratic map q is nonsingular.
Moreover, the discriminant of q is trivial.
Proof : Uniqueness of q is clear, since decomposable elements v1 § · · · § vm span
m
V . To prove the existence of q, we use the same notation as in the proof
m
of (??): we pick a basis (e1 , . . . , en ) of V and get a basis eS of V , where S runs
over the subsets of m indices in {1, . . . , n}. Fix a partition of these subsets into two
classes C, C such that the complement S of every S ∈ C lies in C and conversely.
(For instance, one can take C = { S | 1 ∈ S }, C = { S | 1 ∈ S }.) We may then
/
m
de¬ne a quadratic form q on V by
q xS e S = xS xS e 1 § · · · § e n .
S S∈C
The polar form bq satis¬es
bq xS e S , yT eT = xS y S + x S y S e 1 § · · · § e n
S T S∈C
= xS y S e 1 § · · · § e n .
S

Since the right side is also equal to ( S xS eS ) § ( T yT eT ), the second condition
is satis¬ed.
It remains to prove that q vanishes on decomposable elements. We show that q
m’1
actually vanishes on all the elements of the type v§·, where v ∈ V and · ∈ V.
n
Let v = i=1 vi ei and · = I ·I eI , where I runs over the subsets of m ’ 1
indices in {1, . . . , n}, so that
v§· = vi ·I e{i}∪I = vi ·S eS .
{i}
i,I,i∈I S i∈S
We thus get
q(v § ·) = vi ·S vj ·S .
{i} {j}
S∈C i∈S j∈S
The right side is a sum of terms of the form vi vj ·I ·J where I, J are subsets of
m ’ 1 indices such that {i, j} ∪ I ∪ J = {1, . . . , n}. Each of these terms appears
twice: vi vj ·I ·J appears in the term corresponding to S = {i} ∪ I or S = {j} ∪ J
(depending on which one of these two sets lies in C) and in the term corresponding
to S = {i} ∪ J or {j} ∪ I. Therefore, q(v § ·) = 0.
To complete the proof, we compute the discriminant of q. From the de¬nition,
it is clear that q decomposes into an orthogonal sum of 2-dimensional subspaces:
q = ⊥S∈C qS ,
where qS (xS eS + xS eS ) = xS xS e1 § · · · § en . It is therefore easily calculated that
disc q = 0.
(10.13) Remark. The quadratic map q may be de¬ned alternately by representing
m
V as the quotient space F V m /W , where F V m is the vector space of formal
linear combinations of m-tuples of vectors in V , and W is the subspace generated
by all the elements of the form
(v1 , . . . , vm )
where v1 , . . . , vm ∈ V are not all distinct, and
(v1 , . . . , vi ± + vi ± , . . . , vm ) ’ (v1 , . . . , vi , . . . , vm )± ’ (v1 , . . . , vi , . . . , vm )±
where i = 1, . . . , m and v1 , . . . , vi , vi , . . . , vm ∈ V , ±, ± ∈ F . Since m-
tuples of vectors in V form a basis of F V m , there is a unique quadratic map
§10. THE DISCRIMINANT ALGEBRA 121

n
q: F V m ’ V whose polar form bq factors through the canonical pairing §
and such that q(v) = 0 for all v ∈ V m . It is easy to show that this map q factors
through the quadratic map q.
By composing q with a vector-space isomorphism n V ’ F , we obtain a


m
quadratic form on V which is uniquely determined up to a scalar factor. There-
m
fore, the corresponding quadratic pair (σq , fq ) on EndF ( V ) is unique. In this
pair, the involution σq is the canonical involution γ, since the polar form bq is the
canonical pairing.
Given an arbitrary central simple F -algebra A of degree n = 2m, we will
construct on »m A a quadratic pair (γ, f ) which coincides with the pair (σq , fq ) in
the case where A = EndF (V ). The ¬rst step is to distinguish the right ideals in
»m A which correspond in the split case to the subspaces spanned by decomposable
elements.
The following construction applies to any central simple algebra A over an
arbitrary ¬eld F : if I ‚ A is a right ideal of reduced dimension k, we de¬ne
ψk (I) = { f ∈ »k A = EndA—k (A—k sk ) | sf ∈ I —k · sk } ‚ »k A.
k

This set clearly is a right ideal in »k A.
(10.14) Lemma. If A = EndF (V ) and I = Hom(V, U ) for some k-dimensional
k k
subspace U ‚ V , then we may identify ψk (I) = Hom( V, U ). In particular,
rdim ψk (I) = 1.
Proof : If I = Hom(V, U ), then I —k = Hom(V —k , U —k ) and
k
I —k · sk = Hom( V, U —k ).
k
V ), we have sf ∈ I —k · sk if and only if the
Therefore, for f ∈ »k A = EndF ( k
image of the composite map
f s
k k
V ’’ V —k
k
V ’’
is contained in U —k . Since s’1 (U —k ) = k U , this condition is ful¬lled if and only
k
k
if im f ‚ U . Therefore, we may identify
k k
ψk Hom(V, U ) = Hom( V, U ).
k
Since dim U = 1 if dim U = k, we have rdim ψk (I) = 1 for all right ideals I of
reduced dimension k.
In view of the lemma, we have
ψk : SBk (A) ’ SB(»k A);
if A = EndF (V ), this map is the Pl¨cker embedding
u
k
ψk : Grk (V ) ’ P( A)
which maps every k-dimensional subspace U ‚ V to the 1-dimensional subspace
k k
U‚ V (see §??).
Suppose now that σ is an involution of the ¬rst kind on the central simple
F -algebra A. To every right ideal I ‚ A, we may associate the set I · σ(I) ‚ A.
(10.15) Lemma. Suppose σ is orthogonal or char F = 2. If rdim I = 1, then
I · σ(I) is a 1-dimensional subspace in Sym(A, σ).
122 II. INVARIANTS OF INVOLUTIONS


Proof : It su¬ces to prove the lemma in the split case. Suppose therefore that
A = EndF (V ) and σ is the adjoint involution with respect to some nonsingular
bilinear form b on V . Under the standard identi¬cation •b , we have A = V — V
and I = v — V for some nonzero vector v ∈ V , and σ corresponds to the switch
map. Therefore, I · σ(I) = v — v · F , proving the lemma.

Under the hypothesis of the lemma, we thus get a map
• : SB(A) ’ P Sym(A, σ)
which carries every right ideal I ‚ A of reduced dimension 1 to I · σ(I). If A =
EndF (V ), we may identify SB(A) = P(V ) and P Sym(A, σ) = P(W ), where W ‚
V — V is the subspace of symmetric tensors; the proof above shows that • : P(V ) ’
P(W ) maps v · F to v — v · F .
The relevance of this construction to quadratic pairs appears through the fol-
lowing lemma:
(10.16) Lemma. Suppose char F = 2 and σ is symplectic. The map (σ, f ) ’
ker f de¬nes a one-to-one correspondence between quadratic pairs (σ, f ) on A and
hyperplanes in Sym(A, σ) whose intersection with Symd(A, σ) is ker Trpσ .
Proof : For every quadratic pair (σ, f ), the map f extends Trpσ (this is just condi-
tion (??) of the de¬nition of a quadratic pair, see (??)); therefore
ker f © Symd(A, σ) = ker Trpσ .
If U ‚ Sym(A, σ) is a hyperplane whose intersection with Symd(A, σ) is ker Trpσ ,
then Sym(A, σ) = U +Symd(A, σ), hence there is only one linear form on Sym(A, σ)
with kernel U which extends Trpσ .

Suppose now that A is a central simple algebra of degree n = 2m over a ¬eld
F of characteristic 2. We consider the composite map
ψm •
SBm (A) ’’ SB(»m A) ’’ P Sym(»m A, γ) ,
where γ is the canonical involution on »m A.
(10.17) Proposition. If m ≥ 2, there is a unique hyperplane in Sym(»m A, γ)
which contains the image of • —¦ ψm and whose intersection with Symd(»m A, γ) is
ker Trpσ .
Proof : The proposition can be restated as follows: the subspace of Sym(»m A, γ)
spanned by the image of • —¦ ψm and ker Trpσ is a hyperplane which does not

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