Department of Mathematics | - HistoricalBackgroundquadratic forms involved in these compositions are related to Pﬁster forms. In the 1960s Pﬁster found that for every m there do

Chapter 0

Historical Background

The theory of composition of quadratic forms over fields had its start in the 19thcentury with the search for n-square identities of the type

(x21 + x22 + · · · + x2n) · (y21 + y22 + · · · + y2n) = z21 + z22 + · · · + z2nwhere X = (x1, x2, . . . , xn) and Y = (y1, y2, . . . , yn) are systems of indeterminatesand each zk = zk(X, Y ) is a bilinear form inX and Y . For example when n = 2 thereis the ancient identity

(x21 + x22 ) · (y21 + y22 ) = (x1y1 + x2y2)2 + (x1y2 − x2y1)2.In this example z1 = x1y1 + x2y2 and z2 = x1y2 − x2y1 are bilinear forms in X, Ywith integer coefficients. This formula for n = 2 can be interpreted as the “law ofmoduli” for complex numbers: |α| · |β| = |αβ|where α = x1− ix2 and β = y1+ iy2.A similar 4-square identity was found by Euler (1748) in his attempt to prove

Fermat’s conjecture that every positive integer is a sum of four integer squares. Thisidentity is often attributed to Lagrange, who used it (1770) in his proof of that conjec-ture of Fermat. Here is Euler’s formula, in our notation:

(x21 + x22 + x23 + x24 ) · (y21 + y22 + y23 + y24 ) = z21 + z22 + z23 + z24where

z1 = x1y1 + x2y2 + x3y3 + x4y4z2 = x1y2 − x2y1 + x3y4 − x4y3z3 = x1y3 − x2y4 − x3y1 + x4y2z4 = x1y4 + x2y3 − x3y2 − x4y1.

After Hamilton’s discovery of the quaternions (1843) this 4-square formula wasinterpreted as the law of moduli for quaternions. Hamilton’s discovery came onlyafter he spent years searching for a way to multiply “triplets” (i.e. triples of numbers)so that the law of moduli holds. Such a product would yield a 3-square identity.Already in his Théorie des Nombres (1830), Legendre showed the impossibility ofsuch an identity. He noted that 3 and 21 can be expressed as sums of three squares ofrational numbers, but that 3× 21 = 63 cannot be represented in this way. It follows

2 0. Historical Background

that a 3-square identity is impossible (at least when the bilinear forms have rationalcoefficients). If Hamilton had known of this remark by Legendre he might have givenup the search to multiply triplets! Hamilton’s great insight was to move on to fourdimensions and to allow a non-commutative multiplication.Hamilton wrote to JohnGraves about the discovery of quaternions in October 1843

and within two months Graves wrote to Hamilton about his discovery of an algebraof “octaves” having 8 basis elements. The multiplication satisfies the law of moduli,but is neither commutative nor associative. Graves published his discovery in 1848,but Cayley independently discovered this algebra and published his results in 1845.Many authors refer to elements of this algebra as “Cayley numbers”. In this book weuse the term “octonions”.The multiplication of octonions provides an 8-square identity. Such an identity

had already been found in 1818 by Degen in Russia, but his work was not widelyread. After the 1840s a number of authors attempted to find 16-square identities withlittle success. It was soon realized that no 16-square identity with integral coefficientsis possible, but the arguments at the time were incomplete. These “proofs” werecombinatorial in nature, attempting to insert+ and− signs in the entries of a 16× 16Latin square to make the rows orthogonal.In 1898 Hurwitz published the definitive paper on these identities. He proved that

there exists an n-square identity with complex coefficients if and only if n = 1, 2, 4 or8. His proof involves elementary linear algebra, but these uses of matrices and linearindependence were not widely known in 1898. At the end of that paper Hurwitz posedthe general problem: Forwhich positive integers r , s, n does there exist a “compositionformula”:

(x21 + x22 + · · · + x2r ) · (y21 + y22 + · · · + y2s ) = z21 + z22 + · · · + z2nwhere X = (x1, x2, . . . , xr ) and Y = (y1, y2, . . . , ys) are systems of indeterminatesand each zk = zk(X, Y ) is a bilinear form in X and Y ?Here is an outline of Hurwitz’s ideas, givenwithout all the details. Suppose there is

a composition formula of size [r, s, n] as above. ViewX, Y and Z as column vectors.Then, for example, z21 + z22 + · · · + z2n = Z� ·Z, where the superscript� denotes thetranspose. The bilinearity condition becomes Z = AY where A is an n × s matrixwhose entries are linear forms in X. The given composition formula can then bewritten as

(x21 + x22 + · · · + x2r )Y� · Y = Z� · Z = Y�A�AY.

Since Y consists of indeterminates this equation is equivalent to

A� · A = (x21 + x22 + · · · + x2r )Is,where A is an n× s matrix whose entries are linear forms in X.

0. Historical Background 3

Of course Is here denotes the s × s identity matrix. Since the entries of A are linearforms we can express A = x1A1 + x2A2 + · · · + xrAr where each Ai is an n × s

matrix with constant entries. After substituting this expression into the equation andcanceling like terms, we find:

There are n× s matrices A1, A2, . . . , Ar over F satisfying

A�i · Ai = Is for 1 ≤ i ≤ r,

A�i · Aj + A�

j · Ai = 0 for 1 ≤ i, j ≤ r and i �= j.

This system is known as the “Hurwitz Matrix Equations”. Such matrices exist if andonly if there is a composition formula of size [r, s, n].Hurwitz considered these matrices to have complex entries, but his ideas work just

as well using any field of coefficients, provided that the characteristic is not 2. Thosematrices are square when s = n. In that special case the system of equations can begreatly simplified by defining the n × n matrices Bi = A−1

1 Ai for 1 ≤ i ≤ r . ThenB1, . . . , Br satisfy the Hurwitz Matrix Equations and B1 = In. It follows that:

There are n× n matrices B2, . . . , Br over F satisfying:

B�i = −Bi, for 2 ≤ i ≤ r;B2i = −In,

BiBj = −BjBi whenever i �= j .

Such a system of n × n matrices exists if and only if there is a composition formulaof size [r, n, n]. Hurwitz proved that the 2r−2 matrices Bi1Bi2 . . . Bik for 2 ≤ i1 ≤· · · ≤ ik ≤ r − 1 are linearly independent. This shows that 2r−2 ≤ n2 and in the caseof n-square identities (when r = n) quickly leads to the “1, 2, 4, 8 Theorem”.In 1922 Radon determined the exact conditions on r and n for such a system of

matrices to exist over the real field R. This condition had been found independentlyby Hurwitz for formulas over the complex field C and was published posthumouslyin 1923. They proved that:

A formula of size [r, n, n] exists if and only if r ≤ ρ(n),

where the “Hurwitz–Radon function” ρ(n) is defined as follows: if n = 24a+bn0where n0 is odd and 0 ≤ b ≤ 3, then ρ(n) = 8a+2b. There are several different ways


this function can be described. The following one is the most convenient for ourpurposes:

If n = 2mn0 where n0 is odd then ρ(n) =

2m+ 1 if m ≡ 0,2m if m ≡ 1,2m if m ≡ 2,2m+ 2 if m ≡ 3

(mod 4).

For example, ρ(n) = n if and only if n = 1, 2, 4 or 8, as expected from theearlier theorem of Hurwitz. Also ρ(16) = 9, ρ(32) = 10, ρ(64) = 12 and generallyρ(16n) = 8+ρ(n). New proofs of the Hurwitz–Radon Theorem for compositions ofsize [r, n, n] were found in the 1940s. Eckmann (1943b) applied the representationtheory of certain finite groups to prove the theorem over R, and Lee (1948) modi-fied Eckmann’s ideas to prove the result using representations of Clifford algebras.Independently, Albert (1942a) generalized the 1, 2, 4, 8 Theorem to quadratic formsover arbitrary fields, and Dubisch (1946) used Clifford algebras to prove the Hurwitz–Radon Theorem for quadratic forms over R (allowing indefinite forms). Motivatedby a problem in geometry, Wong (1961) analyzed the Hurwitz–Radon Theorem usingmatrix methods and classified the types of solutions over R. In the 1970s Shapiroproved the Hurwitz–Radon Theorem for arbitrary (regular) quadratic forms over anyfield where 2 �= 0, and investigated the quadratic forms which admit compositions.One goal of our presentation is to explain the curious periodicity property of theHurwitz–Radon function ρ(n):

Why does ρ(2m) depend only on m (mod 4)?

The explanation comes from the shifting properties of (s, t)-families as explained inChapter 2.Here are some of the questions which have motivated much of the work done in

Part I of this book. Suppose σ and q are regular quadratic forms over the field F ,where dim σ = s and dim q = n. Then σ and q “admit a composition” if there is aformula

σ(X)q(Y ) = q(Z),

where as usual X = (x1, x2, . . . , xs) and Y = (y1, y2, . . . , yn) are systems of inde-terminates and each zk is a bilinear form in X and Y , with coefficients in F . Thequadratic forms involved in these compositions are related to Pfister forms.In the 1960s Pfister found that for every m there do exist 2m-square identities,

provided some denominators are allowed. He generalized these identities to a widerclass: a quadratic form is a Pfister form if it expressible as a tensor product of binaryquadratic forms of the type 〈1, a〉. In particular its dimension is 2m for some m.Here we use the notation 〈a1, . . . , an〉 to stand for the n-dimensional quadratic forma1x

21 + · · · + anx2n .


Theorem (Pfister). If ϕ is a Pfister form andX, Y are systems of indeterminates, thenthere is a multiplication formula

ϕ(X)ϕ(Y ) = ϕ(Z),

where each component zk = zk(X, Y ) is a linear form in Y with coefficients in therational function field F(X). Conversely if ϕ is an anisotropic quadratic form over Fsatisfying such a multiplication formula, then ϕ must be a Pfister form.

The theory of Pfister forms is described in the textbooks by Lam (1973) andScharlau (1985). When dim ϕ = 1, 2, 4 or 8, such a multiplication formula existsusing no denominators, since the Pfister forms of those sizes are exactly the normforms of composition algebras. But if dim ϕ = 2m > 8, Hurwitz’s theorem impliesthat any such formula must involve denominators. Examples of such formulas can bewritten out explicitly (see Exercise 5).The quadratic forms appearing in the Hurwitz–Radon composition formulas have

a close relationship to Pfister forms. For any Pfister form ϕ of dimension 2m thereis an explicit construction showing that ϕ admits a composition with some form σ

having the maximal dimension ρ(2m). The converse is an interesting open question.

Pfister Factor Conjecture. Suppose q is a quadratic form of dimension 2m, and qadmits a composition with some form of the maximal dimension ρ(2m). Then q is ascalar multiple of a Pfister form.

This conjecture is one of the central themes driving the topics chosen for the firstpart of the book. In Chapter 9 it is proved true when m ≤ 5, and for larger values ofm over special classes of fields.The second part of this book focuses on the more general compositions of size

[r, s, n]. In 1898 Hurwitz already posed the question: Which sizes are possible?The cases where s = n were settled by Hurwitz and Radon in the 1920s. Furtherprogresswasmade around 1940when Stiefel andHopf applied techniques of algebraictopology to the problem, (for compositions over thefield of real numbers). In Part IIwediscuss these topological arguments and their generalizations, as well as consideringthe question for more general fields of coefficients. Further details are described inthe Introduction to Part II.

Exercises for Chapter 0

Note: For the exercises in this book, most of the declarative statements are to beproved. This avoids writing “prove that” in every problem.

1. In any (bilinear) 4-square identity, if z1 = x1y1 + x2y2 + x3y3 + x4y4 then z2, z3,z4 must be skew-symmetric. (Compare 4-square identity of Euler above.)


2. Doubling Lemma. From an [r, s, n]-formula over F construct an [r + 1, 2s, 2n]-formula.

(Hint. Given the Hurwitz Matrix Equations consider the 2n× 2s matrices

C1 =(A1 00 A1

), Cj =

(Aj 00 −Aj

)for 2 ≤ j ≤ r and Cr+1 =

(0 A1

−A1 0

).)

3. Let x1, x2, . . . , xn be indeterminates and consider the set S of all the vectors(±xα(1),±xα(2), . . . ,±xα(n)) where α is a permutation of the n subscripts and +/ −signs are arbitrary. What is the largest number of mutually orthogonal elements ofthis set of 2nn! vectors, using the usual dot product? Answer: The sharp bound is theHurwitz–Radon function ρ(n).

(Hint. If v1, . . . , vs is a set of mutually orthogonal vectors in S let A be the n × s

matrix with columns vj . Then A� · A = σIs where σ = x21 + x22 + · · · + x2n . Thisprovides a composition formula of size [n, s, n].)

4. Integer compositions. Suppose a composition formula of size [r, s, n] is givenwith integer coefficients. As above, let Z = AY where A is an n × s matrix whoseentries are linear forms in X, and A� · A = σIs where σ = x21 + x22 + · · · + x2r .(1) The columns of A are orthogonal vectors, and each column consists of the

entries ±x1, . . . ,±xr , placed (in some order) in r of the n positions, with zeroselsewhere. When r = n the columns of A are signed permutations of X.

(2) The matrix A for the 2-square identity is

(1 2−2 1

), where we write only the

subscript of each xi . Similarly express A for the 4-square identity. Try to constructthe 8-square identity directly in this way. (Explicit matrices A were listed by Hurwitz(1898).)(3) Write down the matrix A for a composition of size [3, 5, 7].

(Hint: (3) Express (x21 + x22 + x23 ) · (y21 + y22 + y23 + y24 ) as 4 squares.

5. Define DF (n) to be the set of those non-zero elements of the field F which areexpressible as sums of n squares of elements of F .

Theorem (Pfister). DF (n) is a group whenever n = 2m.

This result motivated Pfister’s theorem on multiplicative quadratic forms. Here isa proof.(1) If DF (n) is closed under multiplication then it is a group.(2) Lemma. Suppose c = c21 + c22 + · · · + c2n where n = 2m. Then there exists an

n×nmatrixC having first row (c1, c2, . . . , cn) and satisfyingC ·C� = C� ·C = cIn.(3) Proof of Theorem. Let c, d ∈ DF (n), with the corresponding matrices C, D

from the lemma. Then A = CD� satisfies AA� = cdIn, hence cd ∈ DF (n).


(4) Corollary. Let n = 2m and x1, , . . . , xn, y1, . . . , yn be indeterminates. Thenthere exist z1, . . . , zn ∈ F(X)[Y ] such that

(x21 + · · · + x2n) · (y21 + · · · + y2n) = z21 + · · · + z2n. (∗)In fact we can choose each zi to be a linear form in Y and z1 = x1y1 + · · · + xnyn.If n = 16, there exists such an identity where z1, , . . . , z8 are bilinear forms in

X, Y while z9, . . . , z16 are linear forms in Y . What denominators are involved in theterms zj ?

(Hint. (1) Note that F •2 ⊆ DF (n).(2) Express c = a + b where a, b are the sums of 2m−1 of the c2k . Let A, B be the

corresponding matrices which exist by induction. If a �= 0 define C =(A B

♦ A�),

where the entry ♦ is to befilled in. If b �= 0 useC =(A B

B� ♦). What if a = b = 0?

(4) When n = 16, start from a bilinear 8-square identity, then build up the ma-trix C.)

6. Hurwitz–Radon function.

(1) If n = 2m · (odd) then ρ(n) =

8a + 1 if m = 4a,8a + 2 if m = 4a + 1,8a + 4 if m = 4a + 2,8a + 8 if m = 4a + 3.

(2) Given r define v(r) to be the minimal n for which there exists an [r, n, n]formula. Then v(r) = 2δ(r) where δ(r) = min{m : r ≤ ρ(2m)}. Then ρ(2m) =max{r : δ(r) = m} and

δ(r) = #{k : 0 < k < r and k ≡ 0, 1, 2 or 4 (mod 8)}.Here are the first few values:

r 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16δ(r) 0 1 2 2 3 3 3 3 4 5 6 6 7 7 7 7

(3) r ≤ ρ(2m) if and only if δ(r) ≤ m. Equivalently: r ≤ ρ(n) iff 2δ(r)|n. Note:r ≤ 2δ(r) with equality if and only if r = 1, 2, 4, 8.

7. Vector fields on spheres. Let Sn−1 be the unit sphere in the euclidean space Rn.A (tangent) vector field on Sn−1 is a continuous map f : Sn−1 → Rn for which

f (x) is a vector tangent to Sn−1 at x, for every x ∈ Sn−1. Using 〈x, y〉 for the dotproduct on Rn, this says: 〈x, f (x)〉 = 0 for every x. Define a vector field to be linearif the map f is the restriction of a linear mapRn → Rn. A set of vector fields satisfiesa property if that property holds at every point x. For instance, a vector field f isnon-vanishing if f (x) �= 0 for every x ∈ Sn.(1) There is a non-vanishing linear vector field on Sn−1 if and only if n is even.


(2) There exist r independent, linear vector fields on Sn−1 if and only if r ≤ρ(n) − 1. Consequently, Sn−1 admits n − 1 independent linear vector fields if andonly if n = 1, 2, 4 or 8.

(Hint. (1) A linear vector field is given by an n × n matrix A with A� = −A. It isnon-vanishing iff A is nonsingular.(2) The vector fields, and Gram–Schmidt, provide r mutually orthogonal, unit

length, linear vector fields on Sn−1. These lead to the Hurwitz matrix equations.)

Remark. With considerable topological work, the hypothesis “linear” can beremoved here. For example, the fact that S2 has no non-vanishing vector field (the“Hairy Ball Theorem”) is a result often proved in beginning topology courses. Findingthe maximal number of linearly independent tangent vector fields on Sn−1 is a famoustopic in algebraic topology. Adams finally solved this problem in 1962, showing thatthe maximal number is just ρ(n) − 1. In particular, Sn−1 is “parallelizable” (i.e. ithas n− 1 linearly independent tangent vector fields) if and only if n = 1, 2, 4 or 8.

8. Division algebras. Let F be a field. An F -algebra is defined to be an F -vectorspace A together with an F -bilinear map m : A × A → A. (Note that there areno assumptions of associativity, commutativity or identity element here.) Writingm(x, y) as xy we see that the distributive laws hold and the scalars can be movedaround freely. For a ∈ A define the linear maps La,Ra : A→ A by La(x) = ax andRa(x) = xa. Define A to be a division algebra if La and Ra are bijective for everya �= 0.(1) Suppose A is a finite dimensional F -algebra. Then A is a division algebra if

and only if it has no zero-divisors.(2) Suppose D is a division algebra over F , choose non-zero elements u, v ∈ D,

and define a new multiplication ♥ on D by: x ♥ y = (R−1u (x))(L−1

v (y)). Then(xu) ♥ (vy) = xy and (D, ♥) is a division algebra over F with identity element vu.(3) F -algebras (A, ·) and (B, ∗) are defined to be isotopic (writtenA ∼ B) if there

exist bijective linear maps α, β, γ : A → B such that γ (x · y) = α(x) ∗ β(y) forevery x, y ∈ A. Isotopy is an equivalence relation on the category of F -algebras, andA ∼= B implies A ∼ B. If A is a division algebra and A ∼ B then B is a divisionalgebra. Every F -division algebra is isotopic to an F -division algebra with identity.

Remark. Since we are used to the associative law, great care must be exercisedwhen dealing with such general division algebras. Suppose A is a division algebrawith 1 and 0 �= a ∈ A. There exist b, c ∈ A with ba = 1 and ac = 1. These left andright “inverses” are unique but they can be unequal. Even if b = c it does not followthat b · ax = x for every x. See Exercise 8.7.

9. Division algebras and vector fields. Suppose there is an n-dimensional realdivision algebra (with no associativity assumptions).(1) There is such a division algebraD with identity element e. (Use Exercise 8(2).)


(2) If d ∈ D let fd(x) be the projection of d ∗ x to (x)⊥. Then fd is a vector fieldon Sn−1 which is non-vanishing if d /∈ R · e. A basis of D induces n − 1 linearlyindependent vector fields on Sn−1.(3)Deduce the “1, 2, 4, 8Theorem” for real division algebras fromAdams’ theorem

on vector fields (mentioned in Exercise 7).

10. Hopf maps. Suppose there is a real composition formula of size [r, s, n]. Thenthere is a bilinear map f : Rr × Rs → Rn satisfying |f (x, y)| = |x| · |y| for everyx ∈ Rr and y ∈ Rs . Construct the associated Hopf map h : Rr × Rs → R × Rn byh(x, y) = (|x|2 − |y|2, 2f (x, y)).(1) |h(x, y)| = |(x, y)|2, using the usual Euclidean norms.(2) h restricts to a map on the unit spheres h0 : Sr+s−1 → Sn.(3) If x ∈ Sr−1 and y ∈ Ss−1 then (cos(θ) · x, sin(θ) · y) ∈ Sr+s−1 ⊆ Rr × Rs .

This provides a covering S1 × Sr−1 × Ss−1 → Sr+s−1. The Hopf map h0 carries

(cos(θ) · x, sin(θ) · y) �→ (cos(2θ), sin(2θ) · f (x, y)).(4) When r = s = n = 1 the map h0 : S1 → S1 wraps the circle around itself

twice. When r = s = n = 2 the map h : C × C → R × C induces h0 : S3 → S2.This map is surjective and each fiber is a circle. Further properties of Hopf maps aredescribed in Chapter 15.

Notes on Chapter 0

Further details on the history of n-square identities appear in Dickson (1919), van derBlij (1961), Curtis (1963), Halberstam and Ingham (1967), Taussky (1970), (1981)and van der Waerden (1976), (1985). The interesting history of Hamilton and hisquaternions is described in Crowe (1967). Veldkamp (1991) also has some historicalremarks on octonions.The terminology for the 8-dimensional composition algebra has changed over

the years. The earliest name was “octaves” used by Graves and others in the 19thcentury. This term remained in use in German articles (e.g. Freudenthal in the 1950s),and some authors followed that tradition in English. Many papers in English used“Cayley numbers”, the “Cayley algebra” or the “Cayley–Dickson algebra”. The term“octonions” came into use in the late 1960s, and may have first appeared in Jacobson’sbook (1968). Several of Jacobson’s students were using “octonions” shortly after thatand this terminology has become fairly standard. Jacobson himself says that thisterm was motivated directly from “the master of terminology” J. J. Sylvester, whointroduced quinions, nonions and sedenions. See Sylvester (1884).In proving his 1, 2, 4, 8 Theorem, Hurwitz deduced the linear independence of

the 2r−2 matrices formed by products of subsets of {B2, . . . , Br−1}. His argumentused the skew-symmetry of the Bj . The more general independence result, proved in


(1.11), first appeared in Robert’s thesis (1912), written under the direction of Hurwitz.It also appears in the posthumous paper of Hurwitz (1923). Compare Exercise 1.12.It is interesting to note that the law of multiplication of quaternions had been

discovered, but not published, by Gauss as early as 1820. See the reference in van derWaerden (1985), p. 183.The theory of Pfister forms appears in a number of books, including Lam (1973)

and Scharlau (1985) and in Knebusch and Scharlau (1980). The basic multiplicativeproperties of Pfister forms are derived in Chapter 5.Basic topological results (Hairy Ball Theorem, vector fields on spheres, etc.) ap-

pear inmany topology texts, including Spanier (1966) andHusemoller (1975). Furtherinformation on these topics is given in Chapter 12.

Exercise 5. This property of DF (2m) was foreshadowed by the 8 and 16 squareidentities of Taussky (1966) and Zassenhaus and Eichhorn (1966). The multiplicativeproperties of DF (n) were discovered by Pfister (1965a) and generalized in (1965b).The simple matrix proof here is due to Pfister and appears in Lam (1973), pp. 296–298and inKnebusch and Scharlau (1980), pp. 12–14. Pfister’s analysis of themore generalproducts DF (r) ·DF (s) is presented in Chapter 14.Exercise 6. The function δ(r) arises later in (2.15) and Exercise 2.3, and comes

up in K-theory as mentioned in (12.17) and (12.19).

Exercise 7. Further discussion of Adams’ work in K-theory is mentioned in Chap-ter 12.

Exercise 8. (2) This trick to get an identity element goes back to Albert (1942b)and was also used by Kaplansky (1953).(3) Isotopies were probably first introduced by Albert (1942b). He mentions that

this equivalence relation on algebras was motivated by some topological results ofSteenrod.

Exercise 9. The statement of the 1, 2, 4, 8 Theorem on real division algebras ispurely algebraic, but the only known proofs involve non-trivial topology or analysis.The original proofs (due to Milnor and Bott, and Kervaire in 1958) were simplifiedusing themachinery of K-theory. Gilkey (1987) found an analytic proof. Real divisionalgebras are also mentioned in Chapter 8 and Chapter 12. Hirzebruch (1991) providesa well-written outline of the geometric ideas behind these applications of algebraictopology.

Exercise 10. The Hopf map S3 → S2 is an important example in topology. Hopf(1931) proved that this map is not homotopic to a constant, providing early impetus forthe study of homotopy theory and fiber bundles. Further information on Hopf mapsappears in Gluck, Warner and Ziller (1986), and Yiu (1986). Also see Chapter 15.

Part I

Classical Compositions and Quadratic Forms

Chapter 1

Spaces of Similarities

In order to understand the Hurwitz matrix equations we formulate them in the moregeneral context of bilinear forms and quadratic spaces over a field F . The first step isto use linear transformations in place of matrices, and adjoint involutions in place oftransposes. Then we will see that a composition formula becomes a linear subspaceof similarities. This more abstract approach leads to more general results and simplerproofs, but of course the price is that readers must be familiar with more notations andterminology.Rather than beginning with the Hurwitz problem here, we introduce some standard

notations from quadratic form theory, discuss spaces of similarities, and then remarkin (1.9) that these subspaces correspond with composition formulas. The chapter endswith a quick proof of the Hurwitz 1, 2, 4, 8 Theorem.We follow most of the standard notations in the subject as given in the books by

T. Y. Lam (1973) and W. Scharlau (1985). Throughout this book we work with afield F having characteristic not 2. (For if 2 = 0 then every sum of squares is itself asquare, and the original question about composition formulas becomes trivial.)Suppose (V , q) is a quadratic space over F . This means that V is a (finite dimen-

sional) F -vector space q : V → F is a regular quadratic form. To explain this, wedefine

B = Bq : V × V → F by 2B(x, y) = q(x + y)− q(x)− q(y).Then q is a quadratic form if this Bq is a bilinear map and if q(ax) = a2q(x) forevery a ∈ F and x ∈ V . The form q (or the bilinear form B) is regular if V ⊥ = 0,that is:

if x ∈ V and B(x, y) = 0 for every y ∈ V, then x = 0.

Ifq is not regular it is called singular. (Regular forms are sometimes called nonsingularor nondegenerate.) Since 2 is invertible in F , the bilinear form B can be recoveredfrom the associated quadratic form q by:

q(x) = B(x, x) for all x ∈ V.Depending on the context we use several notations to refer to such a quadratic space.It could be called (V , q) or (V , B), or sometimes just V or just q.

14 1. Spaces of Similarities

To get the matrix interpretation of a quadratic space (V , q), choose a basis{e1, . . . , en} ofV . The form q can then be regarded as a homogeneous degree 2 polyno-mial by setting q(X) = q(x1, . . . , xn) = q(x1e1+· · ·+xnen). TheGram matrix of qisM = (B(ei, ej )), an n×n symmetric nonsingular matrix. Then q(X) = X� ·M ·X,where X is viewed as a column vector and X� denotes the transpose. Another basisfor V furnishes a matrix M ′ congruent to M , that is: M ′ = P� ·M · P where P isthe basis-change matrix.Here is a list of some of the terminology used throughout this book. Further

explanations appear in the texts by Lam and Scharlau.

1V denotes the identity map on a vector space V . Then 1V ∈ End(V ).F • = F − {0}, the multiplicative group of F .F •2 = {a2 : a ∈ F •}, the group of squares.〈c〉 = cF •2 is the coset of c in the group of square classes F •/F •2. We also use〈c〉 to denote the 1-dimensional quadratic form with a basis element of length c.〈a1, . . . , an〉 = 〈a1〉 ⊥ · · · ⊥ 〈an〉 is the quadratic space (V , q) where V hasa basis whose corresponding Gram matrix is diag(a1, . . . , an). Interpreted as apolynomial this form is q(X) = a1x

21 + · · · + anx2n .

〈〈a1, a2, . . . , an〉〉 = 〈1, a1〉 ⊗ 〈1, a2〉 ⊗ · · · ⊗ 〈1, an〉 is the n-fold Pfister form.It is a quadratic form of dimension 2n. For example 〈〈a, b〉〉 = 〈1, a, b, ab〉.det(q) = 〈d〉 if the determinant of a Gram matrix Mq equals d. Choosing adifferent basis alters det(Mq) by a square, so det(q) is a well-defined elementin F •/F •2.dq = (−1)n(n−1)/2 det(q) when dim q = n.

q1 ⊥ q2 and q1 ⊗ q2 are the orthogonal direct sum and tensor product of thequadratic forms q1 and q2.

q1 � q2 means that q1 and q2 are isometric.

q1 ⊂ q2 means that q1 is isometric to a subform of q2.

nq = q ⊥ · · · ⊥ q (n terms). In particular, n〈1〉 = 〈1, 1, . . . , 1〉.〈a〉q = 〈a〉 ⊗ q is a form similar to q.

DF (q) = {a ∈ F • : q represents a} is the value set.GF (q) = {a ∈ F • : 〈a〉q � q} is the group of “norms”, or “similarity factors”of q.

1.1 Definition. Let (V , q) be a quadratic space over F with associated bilinear formB. If c ∈ F , a linear map f : V → V is a c-similarity if

B(f (x), f (y)) = cB(x, y)

1. Spaces of Similarities 15

for every x, y ∈ V . The scalar c = µ(f ) is the norm (also called the multiplier,similarity factor or ratio) of f . A map is a similarity if it is a c-similarity for somescalar c. An isometry is a 1-similarity.

Let Sim(V , q) be the set of all such similarities. Sometimes it is denoted bySim(V ), Sim(V , B) or Sim(q). It is easy to check that the composition of similaritiesis again a similarity and the norms multiply: µ(fg) = µ(f )µ(g). If f is a similarityand b is a scalar: µ(bf ) = b2µ(f ).Suppose f ∈ Sim(V ) and µ(f ) = c �= 0. Then f must be bijective (using the

hypothesis that V is regular to conclude that f is injective), so that f induces anisometry from (V , 〈c〉q) to (V , q). Then 〈c〉q � q and the norm c = µ(f ) lies in thegroup GF (q).Define Sim•(V , q) = {f ∈ Sim(V , q) : µ(f ) �= 0}, the set of invertible elements

in Sim(V , q). Then Sim•(V , q) is a group containing the non-zero scalar maps andthe orthogonal group O(V , q) as subgroups.1 The norm map µ : Sim•(V , q) → F •is a group homomorphism yielding the exact sequence:

1 −→ O(V , q) −→ Sim•(V , q) −→ GF (q) −→ 1.

The subgroupF • O(V , q) consists of all elementsf ∈ Sim•(V , q)whereµ(f ) ∈ F •2.For instance if F is algebraically closed, everything in Sim•(V , q) is a scalar multipleof an isometry. The notation “Sim(V , q)” is an analog of “End(V )”, emphasizing theadditive structure of the similarities, so it is important to include 0-similarities.The adjoint involution Iq is essential to our analysis of similarities. This map Iq

is a generalization of the transpose of matrices.2

1.2 Definition. Let (V , q) be a quadratic space with associated bilinear form B. Theadjoint map Iq : End(V )→ End(V ) is characterized by the formula:

B(v, Iq(f )(w)) = B(f (v), w)

for f ∈ End(V ) and v,w ∈ V . When convenient we write f instead of Iq(f ).

It follows from the regularity of the form that Iq is a well-defined map which is anF -linear anti-automorphism of End(V ) whose square is the identity. If a basis of Vis chosen we obtain the Gram matrix M of the form q, and the matrix A of the mapf ∈ End(V ). Then the matrix of f = Iq(f ) is just M−1A�M , a conjugate of thetranspose matrix A�. If q � n〈1〉 = 〈1, . . . 1〉 and an orthonormal basis is chosen,then Iq coincides with the transpose of matrices.

1.3 Lemma. Let (V , q) be a quadratic space.

(1) If f ∈ End(V ) then f is a c-similarity if and only if f f = c1V .

1 There seems to be no standard notation for this group. Some authors call it GO(V , q).2 The notation Iq should not cause confusion with the n×n identity matrix In (we hope).


(2) If f, g ∈ Sim(V ) the following statements are equivalent:

(i) f + g ∈ Sim(V ).(ii) af + bg ∈ Sim(V ) for every a, b ∈ F .

(iii) f g + gf = c1V for some c ∈ F .

Proof. Part (1) follows easily from the definitions and (2) is a consequence of (1). ��

Similarities f, g ∈ Sim(V ) are called comparable if f +g ∈ Sim(V ) as well. Thelemma implies that if f1, f2, . . . , fr are pairwise comparable similarities in Sim(V )then the whole vector space S = span{f1, f2, . . . , fr} is inside Sim(V ). For if g ∈ Sthen g is a linear combination of the maps fi , and gg is a linear combination of theterms fifi and fifj + fj fi . Since these terms are scalars by hypothesis, gg is also ascalar and g ∈ Sim(V ).Such a subspace S ⊆ Sim(V ) is more than just a linear space. Themapµ restricted

to S induces a quadratic form σ on S. (To avoid notatonal confusion we have not usedthe letter µ again for this form on S.) Generally if f, g ∈ Sim(V ) are comparable,define Bµ(f, g) ∈ F by:

f g + gf = 2Bµ(f, g)1V .

Then Bµ is bilinear whenever it is defined, and µ(f ) = Bµ(f, f ). This map µ :Sim(V ) → F has all the properties of a quadratic form except that its domain is notnecessarily closed under addition.The induced quadratic form σ on a subspace S ⊆ Sim(V ) could be singular. Of

course, this can occur only when there are non-trivial 0-similarities of V . A mapf ∈ End(V ) is a 0-similarity if and only if the image f (V ) is a totally isotropicsubspace of V (i.e. the quadratic form vanishes identically on f (V )). Then if (V , q)is anisotropic, the only 0-similarity is the zero map.

We will restrict attention to those S ⊆ Sim(V )

whose induced quadratic form is regular.

If S is a subspace of similarities with induced (regular) quadratic form σ , we write(S, σ ) ⊆ Sim(V , q). If σ is a quadratic form over F we write

σ < Sim(V , q)

if there exists a subspace S ⊆ Sim(V , q) whose induced quadratic form is isometricto σ .

Given q what is the largest possible σ?

Given σ what is the smallest possible q?

Various aspects of these questions comprise Part I of this book.


1.4 Proposition. Suppose (S, σ ) ⊆ Sim(q) is a regular subspace of similarities.

(1) If a ∈ DF (q) then 〈a〉σ ⊂ q. In particular, dim σ ≤ dim q.

(2) If σ is isotropic then q is hyperbolic.

Proof. (1) For any v ∈ V , the evaluation map S → V sending f �→ f (v) is aq(v)-similarity. If q(v) �= 0 then this map must be injective, since (S, σ ) is regular.(2) If dim V = n then any totally isotropic subspace ofV has dimension≤ n/2, and

(V , q) is hyperbolic if and only if there exist totally isotropic subspaces of dimensionequal to n/2. We are given a 0-similarity f ∈ S with f �= 0. Since f f = 0,image f = f (V ) is totally isotropic. Also ker f is totally isotropic, for if v ∈ V hasq(v) �= 0 then the proof of part (1) shows that f (v) �= 0. Then dim image f anddim ker f are both ≤ n/2, but their sum equals n. Therefore these dimensions equaln/2 and (V , q) is hyperbolic. ��

We can also generalize the Hurwitz Matrix Equations mentioned in Chapter 0.

1.5 Lemma. Let (V , q) be a quadratic space and (S, σ ) ⊆ Sim(V ). If the form σ onS has Gram matrix Mσ = (cij ), then there exist fi ∈ S satisfying

fifj + fj fi = 2cij1V , for all 1 ≤ i, j ≤ s.

Conversely if maps fi ∈ End(V ) satisfy these equations then they span a subspaceS ⊆ Sim(V ) where the induced form has Gram matrix M = (cij ).

Proof. By definition of the Gram matrix, there is a basis {f1, . . . , fs} of S such thatcij = Bµ(fi, fj ). The required equations are immediate from the definition of Bµ.Conversely, given fi satisfying those equations, the remarks after Lemma 1.3 implythat S = span{f1, . . . , fn} is a subspace of Sim(V , q). ��

Suppose that we choose an orthogonal basis {f1, . . . , fn} of S. Then σ �〈a1, . . . , as〉 and

fifi = ai1V

fifj + fj fi = 0

for 1 ≤ i ≤ s,

for 1 ≤ i, j ≤ s.

When σ = s〈1〉 = 〈1, 1, . . . , 1〉 and q = n〈1〉 = 〈1, 1, . . . , 1〉 and we use matrices,these equations become the Hurwitz Matrix Equations mentioned in Chapter 0.Here are some ways to manipulate subspaces of Sim(V ).

1.6 Lemma. Let (S, σ ) ⊆ Sim(V , q) over the field F .

(1) If (W, δ) is another quadratic space then σ < Sim(q ⊗ δ).

(2) If K is an extension field of F then σ ⊗K < Sim(q ⊗K).

(3) If f, g ∈ Sim•(V , q) then f Sg ⊆ Sim(V , q).


Proof. (1) Use S ⊗ 1w acting on V ⊗W .(2) If fi ∈ S and ci ∈ K then Lemma 1.3 implies that

∑fici lies in Sim(V ⊗K).

(3) This is clear since Sim(V , q) is closed under composition. ��

The ideas developed so far can be generalized to subspaces of Sim(V ,W) for twoquadratic spaces (V , q) and (W, q ′). (See Exercise 2.) In this case the matrices arerectangular and the involution is replaced by the map J = JVW : Hom(V ,W) →Hom(W, V ). Since our main concern is the case V = W , we will not pursue thisgenerality here. The main advantage of this restriction is that we can arrange theidentity map 1V to be in S.

1.7 Proposition. Let σ � 〈1, a2, . . . , as〉 be a quadratic form representing 1.Then σ < Sim(V , q) if and only if there exist maps f2, . . . , fs in End(V ) satisfy-

ing:

fi = −fifor 2 ≤ i ≤ s,

f 2i = −ai1Vfifj = −fjfi whenever i �= j .

Proof. Given S ⊆ Sim(V ) with induced form σ we can choose maps fi as above.Replacing S by the isometric space f−1

1 S we may assume that f1 = 1V . Then theequations above reduce to those given here. The converse follows similarly. ��

The conditions above correspond to the second form of the Hurwitz Matrix Equa-tions. With this formulation an experienced reader will notice that the algebra gener-ated by the fi is related to the Clifford algebra C(〈−a2, . . . ,−as〉). This connectionis explored in Chapter 4.

1.8 Example. Let q = 〈1, a〉 with corresponding basis {e1, e2} of V . The adjointinvolution Iq on End(V ) is translated into matrices as follows:

if f =(x y

z w

)then f =

(x az

a−1y w

).

Let f2 =(0 −a1 0

), g1 =

(1 00 −1

)and g2 =

(0 a

1 0

). One can quickly check

that f2=−f2 and f 22 =−a1V . Then S = span{1V , f2} ={(

x −ayy x

): x, y ∈ F

}is a subspace of Sim(〈1, a〉) and the induced form is (S, σ ) � 〈1, a〉. Similarly g1and g2 are comparable (in fact, orthogonal) similarities and T = span{g1, g2} is alsoa subspace of Sim(〈1, a〉) with induced form 〈1, a〉.


This 2-dimensional example contains the germs of several ideas exploited below.One way to understand this example is to interpret 〈1, a〉 as the usual norm form onthe quadratic extension K = F(θ) where θ2 = −a. (When −a /∈ F •2, K is a field.)Let L : K → End(K) be the regular representation: L(x)(y) = xy. With {1, θ} asthe F -basis of K , the matrix ofL(θ) is f2, and the subspace S above is justL(K) ⊆Sim(K). Similarly to get T define the twisted representation L′ : K → End(K) byL′(x)(y) = xy. Then T = L′(K) ⊆ Sim(K).More generally suppose there is anF -algebraA furnished with a “norm” quadratic

form q which is multiplicative: q(xy) = q(x)q(y) for all x, y ∈ A. Using the leftregular representation in a similarway,wegetq < Sim(q). For instance the quaternion

algebra A =(−a,−bF

)with basis {1, i, j, k} has the norm form q � 〈1, a, b, ab〉.

Conversely if q < Sim(q) then q does arise as the norm form of some “compositionalgebra” A. Any subspace σ < Sim(q) can be viewed as coming from a “partialmultiplication” S × V → V .

1.9 Proposition. Suppose (V , q) and (S, σ ) are quadratic spaces over F . The fol-lowing conditions are equivalent:

(1) σ < Sim(q).

(2) There is a bilinear pairing ∗ : S × V → V satisfying q(f ∗ v) = σ(f )q(v) forevery f ∈ S and v ∈ V .

(3) There is a formula σ(X)q(Y ) = q(Z) where each zk is a bilinear form in thesystems of indeterminates X, Y with coefficients in F .

Proof. (1) ⇐⇒ (2). Given S ⊆ Sim(V , q)where the induced form on S is isometricto σ . For f ∈ S and v ∈ V , define f ∗ v = f (v). Since f is a σ(f )-similarity weget required equation. Conversely, the map ∗ induces a linear map λ : S → End(V )by λ(f )(v) = f ∗ v. For each f ∈ S, λ(f ) is a σ(f )-similarity and therefore λ is anisometry from (S, σ ) to the subspace λ(S) ⊆ Sim(V , q). Since (S, σ ) is regular, thisλ is injective and σ < Sim(q).(1) �⇒ (3). Given (S, σ ) ⊆ Sim(V , q) as before, choose bases {f1, . . . , fs} of S

and {v1, . . . , vn} of V . LetM = Mq be the n× n Gram matrix of q,Mσ = (cij ) thes × s Gram matrix of σ , and Ai the n × n matrix of the map fi . Then the matrix offi isM−1A�

i M and the equations given in (1.5) become:

A�i MAj + A�

j MAi = 2cijM, for 1 = i, j = s.

Let A = x1A1 + · · · + xsAs . Since the xi are indeterminates the system of equationsabove is equivalent to the single equation:

A�MA = σ(X)M.

Since q(Y ) = Y�MY we see thatZ = AY satisfies the condition: q(Z) = σ(X)q(Y ).(3) �⇒ (1). Reverse the steps above. ��


If σ < Sim(q) and dim σ ≥ 2, then dim q must be even. One way to see this isto scale σ to represent 1 and get a map f = f2 as in (1.7). Since f is nonsingularand skew-symmetric, the dimension must be even. (In matrix notation in the proof of(1.9) we see thatMA is a skew-symmetric matrix.) For a different proof, let K be anextension field of F where σ becomes isotropic. Since σ ⊗K < Sim(q ⊗K), (1.4)implies q ⊗K hyperbolic and hence of even dimension. The next proposition showsthat more can be said about the structure of q.

1.10 Proposition. (1) If 〈1, a〉 < Sim(q) then q � 〈〈a〉〉 ⊗ ϕ for some form ϕ.(2) If 〈1, a, b〉 < Sim(q) then q � 〈〈a, b〉〉 ⊗ ψ for some form ψ .

Proof. (1) By hypothesis there is f ∈ Sim(V ) with µ(f ) = a and f = −f . Thisskew-symmetry implies B(v, f (v)) = 0 for every v ∈ V . Choose v with q(v) �= 0.Then the lineU = Fv is a regular subspace such thatU and f (U) are orthogonal. LetU0 bemaximal among such regular subspaces. ThenU0 ⊥ f (U0) is a regular subspaceof V . If this subspace is proper, choosew ∈ (U0 ⊥ f (U0))

⊥ with q(w) �= 0 and notethat U0+Fw contradicts the maximality. Therefore V = U0 ⊥ f (U0) � 〈1, a〉⊗ϕ,where ϕ is the quadratic form on U0.(2) Given a basis 1V , f , g corresponding to 〈1, a, b〉, chooseW0 maximal among

the regular subspacesW for whichW , f (W), g(W) and fg(W) are mutually orthog-onal. The argument above generalizes to show that V = W0 ⊥ f (W0) ⊥ g(W0) ⊥fg(W0) � 〈1, a, b, ab〉 ⊗ ψ , where ψ is the quadratic form onW0. ��

This elementary proof in part (1) can perhaps be better understood by consideringthe given f as inducing an action of K = F(

√−a) on V . The formation of U0corresponds to choosing a K-basis of V . Similarly part (2) corresponds to the action

of the quaternion algebra(−a,−bF

)on V . These ideas are explored in Chapter 4 when

we view V as a module over a certain Clifford algebra.We conclude this chapter with an independence argument, due to Hurwitz, which

suffices to prove the “1, 2, 4, 8 Theorem”. Let us first set up a multi-index notation.Let F2 = {0, 1} be the field of 2 elements. If δ ∈ F2 define

aδ ={1 if δ = 0a if δ = 1.

Let � = (δ1, . . . , δn) be a vector in Fn2. If f1, f2, . . . , fn are elements of some ring,define

f� = fδ11 . . . f δnn .

By convention f 0 = 1 here. Define |�| to be the number of indices i such that δi = 1.Then f� is a product of |�| elements fi .

1.11 Proposition. Suppose A is an associative F -algebra with 1 and {f1, . . . , fn}is a set of pairwise anti-commuting invertible elements of A. If n is even then{f� : � ∈ Fn2} is a linearly independent set of 2n elements of A.


Proof. Suppose there exists a non-trivial dependence relation and let∑c�f

� = 0be such a relation having the fewest non-zero coefficients c� ∈ F . We may assumec0 �= 0 by multiplying the relation by (f �)−1 for some � where c� �= 0. For fixedi, conjugate the given relation by fi and subtract, noting that fif

�f−1i = ±f�.

The result is a shorter relation among the f� (since the f 0 terms cancel). By theminimality of the given relation, this shorter one must be trivial. Therefore if c� �= 0then f� must commute with fi . Since this holds for every index i it follows thateither � = 0 or else � = (1, 1, . . . , 1) and n is odd. Since n is even the dependencerelation must have just one term: c0f 0 = 0, which is absurd. ��

1.12 Corollary. If σ < Sim(V , q) where dim σ = s and dim q = n, then 2s−2 ≤ n2.Consequently, s = n is possible only when n = 1, 2, 4 or 8.

Proof. By (1.7) there are s − 1 anti-commuting invertible elements in End(V ). Sincedim End(V ) = n2, Proposition 1.11 provides the inequality. If s = n the inequality2n−2 ≤ n2 implies that n ≤ 8. The restrictions given in (1.10) show that nmust equal1, 2, 4 or 8. ��

Appendix to Chapter 1. Composition algebras

This appendix contains another proof of the 1, 2, 4, 8 Theorem. This approach usesthe classical “doubling process” to construct the composition algebras, so it providesinformation on the underlying algebras and not just their dimensions. This appendix isself-contained, using somewhat different notation for the quadratic and bilinear forms.The theorem here is well- known, first proved by Albert (1942a), who also handled thecase when the characteristic is 2. The organization of the ideas here follows a lectureby J. H. Conway (1980).SupposeF is afieldwith characteristic �= 2 and letA be anF -algebra. This algebra

is not assumed to be associative or finite dimensional: A is simply an F -vector spaceand the multiplication is an F -bilinear map A × A → A. We do assume that A hasan identity element 1.

A.1 Definition. An F -algebra A with 1 is called a composition algebra if there is aregular quadratic form A→ F , denoted a �→ [a], such that

[a] · [b] = [ab] for every a, b ∈ A. (∗)

Let [a, b] denote the associated symmetric bilinear form: 2[a, b] = [a+b]−[a]−[b]. This differs from our previous notation: [a] = q(a) and [a, b] = B(a, b). Thisnotation will not be used elsewhere in this book because the square brackets stand forso many other things (like quaternion symbols and cohomology classes).


We will determine all the possible composition algebras over F . The classicalexamples over the field R include the real numbers (dim = 1), the complex numbers(dim = 2), the quaternions (dim = 4) and the octonions (dim = 8). We assume noknowledge of these examples and consider an arbitrary composition algebra A overF . This appendix is organized as a sequence of numbered statements, with hints fortheir proofs.

A.2. [ac, ad] = [a] · [c, d]. (Set b = c + d in (∗)).Symmetrically, [ac, bc] = [a, b] · [c].

A.3. The “Flip Law”: [ac, bd] = 2[a, b] · [c, d]− [ad, bc].(Replace a by a + b in (A.2)).

A.4. Define “bar” by: c = 2[c, 1]− c.Then [ac, b] = [a, bc]. (Apply (A.3) with d = 1.)Symmetrically, [ca, b] = [a, cb].

Repeating property (A.4) yields a “braiding sequence” of six equal quantities:

· · · = [a, bc] = [ac, b] = [c, ab] = [cb, a] = [b, ca] = [ba, c] = [a, bc] = · · ·

Basic Principle: to prove X = Y show that [X, t] = [Y, t] for every t ∈ A.

A.5. Properties of “bar”:

a + b = a + b.c is scalar if and only if c = c.

[a, b] = [a, b]. (Apply the braiding sequence when c = 1.)¯c = c. (Use the Basic Principle.)

bc = cb. (Use [a, bc] = [cb, a] from braiding.)

b · ac = 2[a, b]c − a · bc. (Use (A.3), isolate d, apply the Basic Principle:[b · ac, d] = [2[a, b]c, d]− [a · bc, d].)

a · ab = ba · a = [a]b. (Set a = b in previous line.)

aa = aa = [a]. (Set b = 1 in previous line.)

Since a = 2[a, 1] − a we have a · ab = a2b and ba · a = ba2. These are the“Alternative Laws”, a weak version of associativity.Now suppose that H ⊆ A is a composition subalgebra (that is, H is an F -

subalgebra on which the quadratic form is regular, i.e. H ∩ H⊥ = 0.) SupposeH �= A. Then we may choose i ∈ H⊥ with [i] = α �= 0.


A.6. H and Hi are orthogonal, H ′ = H + Hi is a subspace on which the form isregular and dimH ′ = 2 · dimH . Moreover, if a, b, c, d ∈ H then:

(a + bi) · (c + di) = (ac − αdb)+ (da + bc)i.Consequently H ′ is also a composition subalgebra of A.

Proof. H is invariant under “bar” since 1 ∈ H . If a, b ∈ H then [a, bi] = [ba, i] = 0.Then H and Hi are orthogonal and H ∩Hi = {0} since the form is regular.To verify the formula for products it suffices to analyze three cases.

[bi · c, t] = [bi, t c] = −[bc, ti] using the Flip Law (A.3),(1)

= [bc · i, t].Hence bi · c = bc · i.(2) If x ∈ H then x · iy = i · xy, since [iy, xt] = −[it, xy] by the Flip Law. In

particular, xi = ix. Hence if a, d ∈ H then a · di = a · id = i · ad = i · da = da · i.

[bi · di, t] = [di, bi · t] = [d · i,−bi · t] since [bi, 1] = 0,(3)

= [d · t, bi · i] by the Flip Law,

= −[dt · i, bi] = −α[dt, b] = [t,−αdb].Hence bi · di = −αdb. ��

This observation imposes severe restrictions on the structure of a compositionalgebra A. The smallest composition subalgebra is A0 = F . If A �= F there mustbe a 2-dimensional subalgebra A1 ⊆ A built as A1 = F + Fi for some i ∈ F⊥. IfA �= A1 there must be a 4-dimensional subalgebra A2 ⊆ A built as A2 = A1 + A1j

for some j ∈ A⊥2 . If A �= A2 there must be an 8-dimensional subalgebra A3 ⊆ A,

etc. This doubling process cannot continue very long.

A.7. Suppose H is a composition subalgebra of A, and H ′ is formed as in (A.6).Then:

H is associative.

H ′ is associative if and only if H is a commutative and associative.

H ′ is commutative and associative if and only if H = F .

Proof. We know that H ′ = H ⊕ Hi is a composition algebra. Then [(a + bi) ·(c + di)] = [a + bi] · [c + di], for every a, b, c, d ∈ H . The left side equals[(ac − αdb) + (da + bc)i] = [ac − αdb] + α[da + bc] and the right side equals([a]+ α[b]) · ([c]+ α[d]). Expanding the two sides and canceling like terms yields:[ac, db] = [da, bc]. Then [d ·ac, b] = [da ·c, b] and we conclude that d ·ac = da ·cand H is associative. The converse follows since these steps are reversible.


The proofs of the other statements are similar direct calculations. ��

A.8 Theorem. If A is a composition algebra over F then A is obtained from thealgebra F by “doubling” 0, 1, 2 or 3 times. In particular, dimA = 1, 2, 4 or 8.

Proof. As remarked before (A.7), if dimA /∈ {1, 2, 4, 8} then there is a chain ofsubalgebras F = A0 ⊂ A1 ⊂ A2 ⊂ A3 ⊂ A4 ⊆ A, where dimAk = 2k . Applyingthe statements in (A.7) repeatedly we deduce that A1 is commutative and associativebut not equal to F ; A2 is associative but not commutative; A3 is not associative andthereforeA4 cannot exist inside the composition algebraA. This contradiction provesthe assertion. ��

This argument does more than compute the dimensions. It characterizes all thecomposition algebras. These are

(0) F ,

(1) F 〈α〉 � F [x]/(x2 − α), a quadratic extension of F ,(2)

(α,β

F

), a quaternion algebra over F ,

(3)(α,β,γ

F

), an octonion algebra over F .

These algebras are defined recursively by the formulas in (A.6). Further propertiesof quaternion algebras are mentioned in Chapter 3. Further properties of octonionsalgebras appear in Chapters 8 and 12.This relationship between H ′ and H in (A.6) is clarified by formalizing the idea

of doubling an F -algebra. A map a �→ a on an F -algebra is called an involution ifit is F -linear, ¯a = a and ab = ba. Suppose H is an F -algebra with 1, a �→ [a] is aregular quadratic form on H , and a �→ a is an involution on H . Let α ∈ F •.

Definition. The α-double, Dα(H), is the F -algebra with underlying vector spaceH × H , and with multiplication given by the formula in (A.6), viewing 1 = (1, 0)and i = (0, 1). For (a, b) = a + bi ∈ Dα(H), define a + bi = a − bi, and[a + bi] = [a]+ α[b].

A.9. Dα(H) is anF -algebra containingH as a subalgebra; dimDα(H) = 2·dimH ;[a] is a regular quadratic form on Dα(H); and a �→ a is an involution on Dα(H). Ifthe elements of H satisfy the following properties then so do the elements of Dα(H):

a = a if and only if a ∈ F ;

[a, b] = [a, b];

[ax, b] = [a, bx] and [xa, b] = [a, xb];

aa = [a] = aa;

ab + ba = 2[a, b] = ab + ba;


T (x) = x + x satisfies T (ab) = T (ba) and T (a · bc) = T (ab · c).

Proof. These are straightforward calculations. ��

The algebras built from F by repeated application of this doubling process arecalled Cayley–Dickson algebras. Given a sequence of scalars α1, . . . , αn ∈ F • define

A(α1, . . . , αn) = Dαn . . .Dα1(F ).

When n ≤ 3 we obtain the composition algebras mentioned above. For example,

A(α, β) �(−α,−β

F

)is the quaternion algebrawith norm form 〈〈α, β〉〉 = 〈1, α, β, αβ〉.

Every Cayley–Dickson algebra A(α1, . . . , αn) satisfies the properties listed in(A.9), even though it is not a composition algebra when n > 3. Further proper-ties are given in Exercise 25. Here are a few properties of composition algebras, notall valid for larger Cayley–Dickson algebras.

A.10. If A is a composition algebra and a, b, x, y, z ∈ A then:

a · ba = ab · a. (The “Flexible Law”. Hence, it is unambiguous to write aba.)aba · x = a · (b · ax), (These are the Moufang identities.)a(xy)a = ax · ya.

Proof. From (A.5) b ·ac = 2[a, b]c− a ·bc. Substitute a for b, b for a and ax for c, todeduce: a ·(b ·ax) = (2[a, b]a−[a]b)·x. When x = 1 this is a ·ba = 2[a, b]a−[a]b,so that: a · (b · ax) = (a · ba) · x. Apply “bar” to the formula for a · ba and replacea, b by a, b to find: ab · a = 2[a, b]a − [a]b = a · ba.The second Moufang identity also follows directly from the Flip Law.

[ax · ya, t] = [a · x, t · ya]= 2[a, t] · [xy, a]− [a] · [y, tx] by the Flip Law and (A.5),

= 2[a, t] · [xy, a]− [a] · [xy, t].Then for fixed a the value ax ·ya depends only on the quantity xy. The stated formulais clear from this independence. ��

These identities hold in any alternative ring. See Exercise 26.


1. Composition algebras. Suppose S ⊆ Sim(V , q) with dim S = dim V . Explainhow this yields a composition algebra, as defined in (A.1) of the appendix.

(Hint. Scale q to assume it represents 1 and choose e ∈ V with q(e) = 1. IdentifyS with V by f �→ f (e). (1.9) provides a pairing. Apply Exercise 0.8 to obtain analgebra with 1.)


2. Let (V , q) and (W, q ′) be two quadratic spaces. Define the “adjoint” map J =JVW : Hom(V ,W)→ Hom(W, V ).(1) The inverse of JVW is JWV . How does J behave on matrices?(2) Define subspaces of Sim(V ,W) and generalize (1.3), (1.5), (1.6) and (1.9).(3) σ < Sim(q, q ′) if and only if q< Sim(σ, q ′).

3. Let (V , q) be a quadratic space and f, g ∈ End(V ).(1) f is a c-similarity if and only if q(f (v)) = cq(v) for every v ∈ V .(2) f g + gf = 0 if and only if f (v) and g(v) are orthogonal for every v ∈ V .(3) If f , g are comparable similarities, then B(f (v), g(v)) = Bµ(f, g)q(v). In

euclidean space over R, the notion of the angle between two vectors is meaningful.Similarities f and g are comparable iff the angle between f (v) and g(v) is constant,independent of v.

4. (1) Suppose q � 〈1, a〉. If f ∈ Sim•(q) then either f =(x −ayy x

)or f =(

x ay

y −x)for some x, y ∈ F with µ(f ) = x2 + ay2. Consequently, every (regular)

subspace of Sim(q) is contained in one of the spaces S, T in Example 1.8. If q �〈1,−1〉, list all the 0-similarities.(2) Let D =

(−1,−1F

)be the quaternion algebra with norm form q = 〈1, 1, 1, 1〉.

Let L0 = L(D) ⊆ Sim(D, q) arising from the left regular representation of D.

Then L0 is the set of all

a b c d

−b a d −c−c −d a b

−d c −b a

. Similarly R0 = R(D) is the set

of all

a b c d

−b a −d c

−c d a −b−d −c b a

. Then L0 and R0 are subalgebras of End(D) whichcommute with each other and they are 4-dimensional subspaces of Sim(q). Also,R0 · J = J · L0 where J = “bar”. These subspaces of similarities are unique in astrong sense:

Lemma. IfU ⊆ Sim(D, q) is a (regular) subspace and f ∈ U•, eitherU ⊆ f ·L0or U ⊆ f · R0.(3) What are the possibilities for the intersections f · L0 ∩ L0 and f · L0 ∩ R0?

(Hint. (2) Reduce to the case 1V ∈ U . IfA2 ∈ F • for a 4× 4 skew-symmetric matrixA show that A ∈ L0 or A ∈ R0. Deduce that U ⊆ L0 or U ⊆ R0.

5. Characterizing similarities. Let (V , q) be a quadratic space and f ∈ End(V ).(1) If f preserves orthogonality (i.e. B(v,w) = 0 implies B(f (v), f (w)) = 0),

then f is a similarity.


(2) If q is isotropic and f preserves isotropy (i.e. q(v) = 0 implies q(f (v)) = 0),then f is a similarity. (This also follows from Exercise 14, at least if F is infinite.)(3) If q represents 1 and f preserves the unit sphere (i.e. q(v) = 1 implies

q(f (v)) = 1), then f is an isometry.

6. Multiplication formulas. (1) Suppose q = 2m〈1〉. If q represents c ∈ F • then〈c〉q � q. Equivalently, DF (q) = GF (q).(2) Follow the notations of Proposition 1.9. There is an equivalence:(i) There is a formula σ(X)q(Y ) = q(Z) such that each zk ∈ F(X)[Y ] is a linear

form in Y .(ii) σ(X) ∈ GF(X)(q ⊗ F(X)).

(Hint. (1) The matrix in Exercise 0.5 provides a c-similarity.)

7. Let (V , B) be an alternating space, that is, B is a nonsingular skew-symmetricbilinear form on the F -vector space V .(1) The adjoint involution IB is well defined. The set Sim(V , B) is defined as usual

and if S ⊆ Sim(V , B) is a linear subspace then the norm map induces a quadraticform σ on S.(2)When dim V = 2 then Sim(V , B) = End(V ) and the mapµ : End(V )→ F is

the determinant. That is, the 2-dimensional alternating space admits a 4-dimensionalsubspace of similarities.(3) dim V = 2m is even and (V , B) � mH � H ⊥ · · · ⊥ H , where H is the

2-dimensional alternating space where the Gram matrix of the form is

(0 −11 0

). If

B is allowed to be singular then (V , B) � r〈0〉 ⊥ mH .

(4) Let Jr,m be the n × n matrix

( 0r 0 00 0m −Im0 Im 0m

)where n = r + 2m. Any

skew-symmetric n × n matrix is of the type P� · Jr,m · P for some r , m and someP ∈ GLn(F ). In particular rankM = 2m is even and detM is a square.

8. Jordan forms. Let (V , q) be a quadratic space over an algebraically closed fieldF . Let f ∈ O(V , q), that is, f f = 1V . For each a ∈ F define the general eigenspaceV ((a)) = {x ∈ V : (f − a1V )k(x) = 0 for some k}.(1) If ab �= 1 then V ((a)) and V ((b)) are orthogonal. In particular if a �= ±1 then

V ((a)) is totally isotropic.(2) Therefore V = V ((1)) ⊥ V ((−1)) ⊥ ⊥(V ((a))⊕ V ((a−1))), summed over

some scalars a �= ±1. Then det f = (−1)m where m = dim V ((−1)).(3) If f ∈ End(V ), what condition on the Jordan form of f ensures the existence

of a quadratic form q on V having f ∈ O(V , q)? Certainly f must be similar to f−1.Proposition. If f ∈ GL(V ) then f ∈ O(V , q) for some regular quadratic form q

if and only if f ∼ f−1 and every elementary divisor (x ± 1)m form even occurs witheven multiplicity.


(Hint. (3) (⇐�) Iff hasmatrix(B 00 B−�

)in block form,wecanuseM =

(0 11 0

)as the Gram matrix of q. By canonical form theory we need only findM in the casef has the single elementary divisor (x − 1)m for odd m.(�⇒) Trickier to prove. See Gantmacher (1959), §11.5, Milnor (1969), §3, or

Shapiro (1992).)

9. More Jordan forms. Let f ∈ End(V ). What conditions on the Jordan formof f are needed to ensure the existence of a quadratic form q on V for which f issymmetric: Iq(f ) = f ? (Answer. Such q always exists. In matrix terms this saysthat for any square matrix A there is a nonsingular symmetric matrix S such thatSAS−1 = A�. This result was proved by Frobenius (1910) and has appeared in theliterature in various forms. Similar questions arise: When does there exist a quadraticform q such that Iq(f ) = −f ? When does there exist an alternating form B on Vsuch that IB(f ) = ±f ? These are questions are addressed in Chapter 10.)

10. Determinants of skew matrices. If (V , q) is a quadratic space then all theinvertible skew-symmetric matrices have the same determinant, modulo squares. Infact:

if f ∈ GL(V ) and Iq(f ) = −f then 〈det f 〉 = det q in F •/F •2.

(Hint. The determinant of a skew-symmetric matrix is always a square.)

11. Multi-indices. In the notation of Proposition 1.11:(1) For any �, � note that f� · f � = ±f � · f�. Determine this sign explicitly.(2) Suppose further that f 2i = ai ∈ F •. Define a� = a

δ11 . . . a

δnn . Then (f �)2 =

±a�. Determine exactly when that sign is “+”. Further investigation of such signsappears in Exercise 3.18 below.(3) The “multi-index” � can be viewed as the subset of {1, 2, . . . , n} consisting

of all indices i such that δi = 1. Listing this subset as � = {i1, . . . , ik} such thati1 < i2 < · · · < ik , we have f� = fi1fi2 . . . fik . Then |�| is the cardinality of �,�� = � ∩ �, and �+ � = (� ∪ �)− (� ∩ �), the symmetric difference.(Hint. (1) The answer involves only |�|, |�| and |��|.)

12. Anticommuting matrices. (1) Proposition. Suppose n = 2mn0 where n0 isodd. There exist k elements of GLn(F ) which anticommute pairwise, if and only ifk ≤ 2m+ 1.(2) Suppose f1, . . . , f2m+1 is a maximal anticommuting system in GLn(F ) as

above. If n = 2m then f 2i must be a scalar. For other values of n this claim may fail.

(Hint. (1) (⇐�) Inductively construct suchf1, . . . , f2m+1 inGL(n)withf 2i = ±1. Toget the system in GL2n(F ) use block matrices:

(1 00 −1

),

(0 11 0

),

(0 fi

−fi 0

).


(�⇒) We may assume F is algebraically closed and k = 2k0 is even. To show:2k0 |n. Use notations ofExercise 8withV = Fn. ThenV is a direct sumof theV ((a))’sand fj : V ((a)) → V ((−a)) is a bijection, for every j > 1. Let na = dim V ((a)),so that n = ∑

2na , summed over certain eigenvalues a. The maps f2fj inducek − 2 anticommuting elements of GL

(V ((a))

), and hence 2k0−1|na , by induction

hypothesis.(2) By (1.11) the matrices f� span all ofMn(F ) and f 2i commutes with every fj .)

13. Trace forms. The map µ : Sim(V ) → F induces a quadratic form on everylinear subspace.(1) In factµ extends to a quadratic form onEnd(V ), at least if dim V is not divisible

by the characteristic of F .(2) If (V , q) is a quadratic space define τ : End(V ) → F by τ(f ) = trace(f f ).

Then (End(V ), τ ) � q ⊗ q.(3) Let A = {f ∈ End(V ) : f = −f }, the space of skew-symmetric maps.

Compute the restriction τA of the trace form τ to A× A.(4) If (V , B) is a alternating space what can be said about the trace form τ?(5) Suppose (V , α) and (W, β) are quadratic spaces and use the isomorphism

θα : V → V to find an isomorphism V ⊗ W → Hom(V ,W). Does the quadraticform α ⊗ β get carried over to the trace form τ on Hom(V ,W) defined by τ(f ) =trace(J VW (f ) � f )? (See Exercise 2 for JVW .)

(Hint. (1) Consider trace(f f ).(2) The bilinear form b = Bq induces an isomorphism θ : V → V where V is the

dual vector space. Then ϕ : V ⊗ V → V ⊗ V → End(V ) is an isomorphism. Verifythat(i) ϕ(v1 ⊗ w1) � ϕ(v2 ⊗ w2) = b(w1, v2) · ϕ(v1 ⊗ w2).(ii) trace(ϕ(v ⊗ w)) = b(v,w).(iii) Iq(ϕ(v ⊗ w)) = ϕ(w ⊗ v). Show that ϕ carries q ⊗ q to τ .(3) If q � 〈a1, . . . , an〉 define P2(q) = 〈a1a2, a1a3, . . . , an−1an〉. Then

τA � 〈2〉P2(q). To see this let {v1, . . . , vn} be the given orthogonal basis and notethat ϕ−1(A) is spanned by vi ⊗ vj − vj ⊗ vi for i < j .) Compare Exercise 3.13.)

14. Geometry lemma. Let X = (x1, . . . , xn) be indeterminates. If f ∈ F [X]define the zero-set Z(f ) = {a ∈ Fn : f (a) = 0}. If f |p then p vanishes onZ(f ), or equivalently, Z(f ) ⊆ Z(p). The converse is false in general even if fis irreducible. (Hilbert’s Nullstellensatz applies when F is algebraically closed.)However, the converse does hold in some cases:(1) Lemma. Let F be an infinite field and let x, Y = (y1, . . . , yn) be indetermi-

nates. Suppose f (x, Y ) = x ·g(Y )+h(Y ) ∈ F [x, Y ] where g(Y ), h(Y ) are non-zeroand relatively prime in F [Y ]. Then f is irreducible and if p ∈ F [x, Y ] vanishes onZ(f ) then f |p.


Proof outline. Express p = c0(Y )xd + · · ·+ cd(Y ). Since xg ≡ −h (mod f ), we

have gdp ≡ Q (mod f ) where Q = c0(Y )(−h(Y ))d + · · · + cd(Y )g(Y )d ∈ F [Y ].

Since p vanishes onZ(f ) it follows thatQ(B) = 0 for every B ∈ Fn with g(B) �= 0.Therefore g ·Q vanishes identically on Fn. HenceQ = 0 as a polynomial and f |gdpand we conclude that f |p.(2) Suppose degp = d and deg f = m above. If F is finite and |F | ≥ (m + 1)·

(d + 1), the conclusion still holds.(3) Suppose q is an isotropic quadratic form over an infinite fieldF , and dim q > 2.

If p ∈ F [X] vanishes on Z(q), then q|p. In particular if q, q ′ are quadratic formswith dimension > 2 and if Z(q) ⊆ Z(q ′) then q ′ = c · q for some c ∈ F . For whatfinite fields F do these statements hold?

(Hint. (2) If k(Y ) ∈ F [Y ] vanishes on Fn and if |F | > deg k then k = 0.(3) Change variables to assume q(X) = xy + h(Z) where h is a quadratic form

of dim ≥ 1. Analyze (1) to show that degQ ≤ 2d. The argument works if |F | ≥2 · degp + 2.)

15. Transversality. Suppose F is a field with more than 5 elements.(1) If a ∈ F • there exist non-zero r, s ∈ F such that r2 + as2 = 1.(2) Suppose q � 〈a1, . . . , an〉 represents c. That is, there exists 0 �= v ∈ Fn such

that q(v) = ∑aiv

2i = c. Then there exists w ∈ Fn such that q(w) = c and wi �= 0

for every i.(3) Transversality Lemma. Suppose (V , α) and (W, β) are quadratic spaces

over F and α ⊥ β represents c. Then there exist v ∈ V and w ∈ W such thatc = α(v)+ β(w) and α(v) �= 0 and β(w) �= 0.(4) Generalize (2) to non-diagonal forms. The answer reduces to the case c = 0:

Proposition. Let (V , q) be an isotropic quadratic space with dim q = n ≥ 3and let H1, . . . , Hn be linearly independent hyperplanes in V . Then there exists anisotropic vector v ∈ V such that v /∈ H1 ∪ · · · ∪Hn.For example, relative to a given basis the coordinate hyperplanes are linearly

independent. If F is infinite we can avoid any finite number of hyperplanes in V .(Use Exercise 14 with p a product of linear forms.) If F is finite the result followsfrom Exercise 14 provided that |F | ≥ 2n + 2. The stated result (valid if |F | > 5) isdue to Leep (unpublished) and seems quite hard to prove.

(Hint. (1) Use the formulas r = t2−at2+a and s = 2t

t2+a .(2) Suppose n = 2. Re-index to assume v1 �= 0 and scale to assume a1 = 1. If

v2 = 0 scale again to assume c = 1. Apply (1). What if n > 2?(3) If α, β are anisotropic diagonalize them and apply (2). If α is isotropic choose

w so that β(w) �= 0, c and note that α represents c − β(w).)

16. Conjugate subspaces. Lemma. Suppose S, S′ ⊆ Sim(V , q) are subspacescontaining 1V and such that S � S′ as quadratic spaces (relative to the inducedquadratic form). If dim S ≤ 3 then S′ = gSg−1 for some g ∈ O(V , q).


(Hint. Compare (1.10). Induct on dim V . Restate the hypotheses in the case dim S =2: For i = 1, 2we have isometric spaces (Vi, qi) and fi ∈ End(Vi) such that fi = −fiand f 2i = −a · 1V . Choose vi ∈ Vi such that q1(v1) = q2(v2) �= 0 and let Wi =span{vi, fi(vi)}. Define g : W1 → W2 with g(v1) = v2 and g(f (v1)) = f (v2). Theng is an isometry and f2 = gf1g

−1 onW2. Apply induction toW⊥i .)

17. Proper similarities. If f ∈ Sim•(V , q) where dim V = n then (det f )2 =µ(f )n.(1) If n is odd then µ(f ) ∈ F •2 and f ∈ F • O(V ).(2) Suppose n = 2m. Define f to be proper if det f = µ(f )m. The proper

similarities form a subgroup Sim+(V ) of index 2 in Sim•(V ). This is the analog ofthe special orthogonal group O+(n) = SO(n).(3) Suppose f = −f . If g = a1V + bf for a, b ∈ F then g is proper.(4) Wonenburger’s Theorem. Suppose f, g ∈ Sim•(V ) and f = −f . If g

commutes with f , then g is proper. If g anticommutes with f and 4 ||n then g is proper.(5) Let L0, R0 ⊆ Sim(〈1, 1, 1, 1〉) be the subspaces described in Exercise 4(2).

LetG be the group generated byL•0 andR

•0. ThenG is the set of all maps g(x) = axb

for a, b ∈ D•.Lemma. G = Sim+(q).

(Hint. (1) Show µ(f ) ∈ F •2.(4) AssumeF algebraically closed and f 2 = 1V . The eigenspacesU+ andU− are

totally isotropic of dimension m. Examine the matrix of g relative to V = U+ ⊕U−

using the Gram matrix

(0 11 0

).

(5) G ⊆ Sim+(V ) by Wonenburger. Conversely it suffices to show that SO(q) ⊆G. The maps τa generate O(q), where τa is the reflection fixing the hyperplane (a)⊥.Therefore the maps τaτ1 generate SO(q). Writing [a] for q(a) as in the appendix, wehave τa(x) = x − 2[x,a]

[a] · a = x − [a]−1(xa + ax)a = −[a]axa. Then τaτ1(x) =[a]−1axa so that τaτ1 lies in G.)

18. Zero-similarities. Let h ∈ End(V ) where (V , q) is a quadratic space.(1) (image h)⊥ = ker h and (ker h)⊥ = image h.(2) If h ∈ Sim(V ), it is possible that h /∈ Sim(V ). However if h is comparable to

some f ∈ Sim•(V ) then h ∈ Sim(V ).(3) Suppose h ∈ S ⊆ Sim(V ) where µ(h) = 0 and S is a regular subspace. Then

hh = hh = 0 and dim ker h = dim image h, as in the proof of (1.4). Conversely ifh ∈ End(V ) satisfies these conditions then h is in some regular S ⊆ Sim(V ).

(Hint: (3) U = ker h = image h and U ′ = image h = ker h are totally isotropic ofdimension n/2. Replace h by gh for a suitable g ∈ O(V , q), to assume U = U ′.Choose a totally isotropic complementW and use matrices relative to V = U ⊕W toconstruct a 0-similarity k with hk + kh = 1V .)


19. Singular subspaces. (1) Find an example of a regular quadratic space (V , q)admitting a (singular) subspace S ⊆ Sim(V , q) where dim S > dim V .(2) Find such an example where 1V ∈ S.

(Hint. Let (V , q) = mH be hyperbolic and choose a basis so that the matrix of q is

M =(0 11 0

)in m×m blocks. If f =

(a b

c d

)then f =

(d� b�c� a�

). Consider

the cases f =(a b

0 0

).)

20. Extension Theorem. (1) Lemma. Suppose (V , q) is a quadratic space,W ⊆ V

is a regular subspace and f : W → V is a similarity. If µ(f ) ∈ GF (q). Then thereexists f ∈ Sim(V , q) extending f .(2) Corollary. If D is an n × r matrix over F satisfying D� · D = aIr and

a ∈ GF (n〈1〉) then D can be enlarged to an an n × n matrix D = (D|D′) whichsatisfies D� · D = aIn.

(Hint. (1) There exists g ∈ Sim(V , q) with µ(g) = µ(f ). Then g−1 � f : W → V

is an isometry and Witt’s Extension Theorem for isometries applies. See Scharlau(1985) Theorem 1.5.3.)

21. Bilinear terms in Pfister Form Formulas. According to Pfister’s Theory, ifn = 2m and X, Y are systems of n indeterminates over F , there exists a formula

(x21 + x22 + · · · + x2n)(y21 + y22 + · · · + y2n) = z21 + z22 + · · · + z2n, (∗)where each zk is a linear form in Y with coefficients in the rational function fieldF(X). In Exercise 0.5 we found formulas where several of the zk’s were also linearin X. Question. How many of the terms zk can be taken to be bilinear in X, Y ?

Proposition. Suppose F is a formally real field, n = 2m and X, Y are systems ofn indeterminates. There is an n-square identity (∗) as above with each zk linear in Yand with z1, . . . , zr also linear in X if and only if r ≤ ρ(n).

Proof outline. Suppose z1, . . . , zr are also linear inX. Then Z = AY for an n×nmatrixA over F(X) such that the entries in the first r rows ofA are linear forms inX.Then (∗)becomes: A�·A = (∑

x2i

)·In. ConsequentlyA·A� = (∑x2i

)·In. ExpressA in block form asA =

(B

C

)whereB is an r×nmatrixwhose entries are linear forms

inX andC is an (n−r)×nmatrix overF(X). ThenB ·B� = (∑x2i

) ·Ir , so thatB�satisfies the Hurwitz conditions for a formula of size [n, r, n]. The Hurwitz–RadonTheorem implies r ≤ ρ(n).Conversely if r ≤ ρ(n) there exists an r × nmatrix B, linear inX, with B ·B� =(∑x2i

)·Ir . Note that (∑ x2i

)is inGF(X)(n〈1〉) by Pfister’s theorem. (Use Exercise 6

or invoke (5.2) (1) below.) Apply Exercise 20(2) toB� overF(X) to find thematrixA.


22. Isoclinic planes. If U ⊆ Rn is a subspace and � is a line in Rn let � (�, U) bethe angle between them. If � �⊆ U⊥ then � (�, U) is the angle between the lines � andπU(�), where πU : Rn → U is the orthogonal projection. Subspaces U,W ⊆ Rn areisoclinic if � (�, U) is the same for every line � ⊆ W .(1) U ,W are isoclinic if and only if πU is a similarity when restricted toW .(2) If dimU = dimW this relation “isoclinic” is symmetric.(3) Suppose R2n = U ⊥ U ′ where dimU = dimU ′ = n. If f ∈ Hom(U,U ′)

define its graph to be U [f ] = {u + f (u) : u ∈ U} ⊆ R2n. Let W ⊆ R2n be asubspace of dimension n.

Lemma. U ,W are isoclinic if and only if eitherW = U ′ orW = U [f ] for somesimilarity f .

Proposition. Suppose f, g ∈ Sim(U,U ′). Then U [f ], U [g] are isoclinic if andonly if f , g are comparable similarities.

(4) If T ⊆ Sim(U,U ′) is a subspace define S(T ) = {U [f ] : f ∈ T }∪{U ′}. ThenS(T ) is a set of mutually isoclinic n-planes in R2n. It is called an isoclinic spheresince it is a sphere when viewed as a subset of the Grassmann manifold of n-planesin 2n-space.(5) If n ∈ {1, 2, 4, 8} there exists such T with dim T = n. In these cases S(T ) is

“space filling”: whenever 0 �= x ∈ R2n there is a unique n-plane in S(T ) containing x.

23. Normal sets of planes. Define two n-planes U , V in R2n to be normally relatedif U ∩V = {0} = U ∩V ⊥. A set of n-planes is normal if every two distinct elementsare either normally related or orthogonal.(1) Suppose S is a maximal normal set of n-planes inR2n. IfU ∈ S thenU⊥ ∈ S.

A linear map f : U → U⊥ has a corresponding graph U [f ] ⊆ U ⊥ U⊥ = R2n,as in Exercise 22. If W ∈ S then either W = U⊥ or W = U [f ] for some bijectivef . Note that U [f ]⊥ = U [−f−�], and that U [f ] and U [g] are normally related ifff − g and f + g−� are bijective.(2) Let O = Rn × {0} be the basic n-plane. If T ⊆ Mn(R) define

S(T ) = {O[A] : A ∈ T } ∪ {O⊥}.Anymaximal normal set of n-planes inR2n containingO equals S(T ) for some subsetT such that: T is nonsingular (i.e. every non-zero element is in GLn(Rn)), T is anadditive subgroup and T is closed under the operation A �→ A−�.(3) Consider the case T ⊆ Mn(R) is a linear subspace. (If S(T ) is maximal normal

must T be a subspace?)

Proposition. If T ⊆ Mn(R) is a linear subspace such that S(T ) is a maximalnormal set of n-planes, then T ⊆ Sim(Rn) and S(T ) is a maximal isoclinic sphere.

(Hint. (3) If 0 �= A ∈ T express A = PDQ� where P,Q ∈ O(n) and D is diagonalwith positive entries. (This is a singular value decomposition.) If a, b ∈ R thenaA + bA−� ∈ T . Then if aD + bD−� is non-zero it must be nonsingular. DeduceD = scalar so that A ∈ Sim(Rn).


24. Automorphisms. Suppose C is a composition algebra and a ∈ C.(1) If a is invertible, must x �→ axa−1 be an automorphism of C?Define La(x) = ax, Ra(x) = xa and Ba(x) = axa.(2)Ba(xy) = La(x) ·Ra(y). ConsequentlyBaBb(xy) = LaLb(x) ·RaRb(y), etc.

These provide examples of maps α, β, γ : C → C satisfying: γ (xy) = α(x) · β(y).(3) For such α, β, γ if α(1) = β(1) = 1 then α = β = γ is an automorphism of

C. If C is associative this automorphism is the identity. Suppose C is octonion andγ = Ba1Ba2 . . . Ban , etc. There exist non-trivial automorphisms built this way (butthey require n ≥ 4).(4) Suppose C1 and C2 are composition algebras. If ϕ : C1 → C2 is an isomor-

phism then ϕ commutes with the involutions: ϕ(x) = ϕ(x), and ϕ is an isometry ofthe norm forms [x]. Conversely if the norm forms of C1 and C2 are isometric then thealgebras must be isomorphic. (However not every isometry is an isomorphism.)

(Hint. (2) Moufang identities.(3) α(1) = 1 �⇒ γ = β, and β(1) = 1 �⇒ γ = α. Case n = 2: ab = 1 implies

a · bx = x since alternative. Case n = 3: Given a · bc = 1 = cb · a, show a, b, clie in a commutative subalgebra. Then a, b, c, x lie in an associative subalgebra andLaLbLc(x) = x.(4) a ∈ C is pure (i.e. a = −a) iff a /∈ F and a2 ∈ F . Any isomorphism preserves

“purity”. Given an isometry η : C1 → C2 assume inductively that Ci is the double ofa subalgebraBi , that η(B1) = B2 and that the restriction of η toB1 is an isomorphism.By Witt Cancellation the complements B⊥

i are isometric.)

25. Suppose A = A(α1, . . . , αn) is a Cayley–Dickson algebra. Then elements of Asatisfy the identities in (A.9).(1) The elements of A also satisfy:

yx · x = x · xya · ba = ab · a (the flexible law)

[ab] = [ab] = [ba]

an · am = an+m (A is power-associative).

(2) If ab = 1 in A does it follow that ba = 1? If a, b, c ∈ A does it follow that[ab · c] = [a · bc]?(3) dimA = 2n and A is a composition algebra if and only if n ≤ 3. Define

A◦ = ker(T ), the subspace of “pure” elements, so that A = F ⊥ A◦. If a, b areanisotropic, orthogonal elements of A◦ (using the norm form) then a2, b2 ∈ F • andab = −ba. These elements might fail to generate a quaternion subalgebra.(4) There exists a basis e0, e1, . . . , e2n−1 of A such that:e0 = 1

e1, . . . , e2n−1 ∈ A◦ pairwise anti-commute and e2i ∈ F •.For each i, j : eiej = λek for some index k = k(i, j) and some λ ∈ F •.


The basis elements are “alternative”: eiei · x = ei · eix and xei · ei = x · eiei .(5) The norm form x �→ [x] on A is the Pfister form 〈〈−α1, . . . ,−αn〉〉.(6) Let An = A(α1, a2, . . . αn). We proved that the identity a · bc = ab · c holds

in every A2 and fails in every A3. Similarly the identity a · ab = aa · b holds in A3and fails in A4.

Open question. Is there some other simple identity which holds in A4 and failsin A5?

(Hints. (1) If a = −a we know [x, ax] = 0 = [x, xa]. The flexible law for Dα(H)

follows from the properties of H , after some calculation. It suffices to prove power-associativity for pure elements, that is, when a = −a. But then a2 ∈ F .(2) The ab = 1 property holds, at least when the form [x] is anisotropic: Express

a = α + e and b = β + γ e+ f where 1, e, f are orthogonal and [e], [f ] �= 0. Then{1, e, f, ef } is linearly independent.(4) Construct the basis inductively.)

26. Alternative algebras. Suppose A is a ring. If x, y, z ∈ A define the associator(x, y, z) = xy · z− x · yz. Suppose A is alternative: (a, a, b) = (a, b, b) = 0.(1) (x, y, z) is an alternating function of the three variables. In particular, a · ba =

ab · a, which is the “Flexible Law”.(2) xax · y = x(a · xy)

y · xax = (yx · a)x (the Moufang identities)bx · yb = b · xy · b

(3) (y, xa, z)+ (y, za, x) = −(y, x, a)z− (y, z, a)x.(4) Proposition. Any two elements of A generate an associative subalgebra.

(Hint. (2) xax ·y−x(a ·xy) = (xa, x, y)+ (x, a, xy) = −(x, xa, y)− (x, xy, a) =−x2a · y − x2y · a + x(xa · y + xy · a) = −(x2, a, y)− (x2, y, a)+ x · [(x, a, y)+(x, y, a)] = 0. For the second, use the opposite algebra. Finally, bx ·yb−b ·xy ·b =(b, x, yb) − b(x, y, b) = −(b, yb, x) − b(x, y, b) = b · [(y, b, x) − (x, y, b)] = 0,using the first identity.(3) (y, xa, x) = −(y, x, a)x by (2). Replace x by x + z.(4) If u, v ∈ A, examine “words” formed from products of u’s and v’s. It suffices

to show that (p, q, r) = 0 for any such words p, q, r . Induct on the sum of the lengthsof p, q, r . Rename things to assume that words q and r begin with u. Apply (3) whenx = u.)

27. The nucleus. The nucleus N (A) of an algebra A is the set of elements g ∈ Awhich associate with every pair of elements in A. That is, xy · z = x · yz wheneverone of the factors x, y, z is equal to g. Then N (A) is an associative subalgebra of A.(1) If A is alternative it is enough to require g · xy = gx · y for every x, y.(2) If A is an octonion algebra over F then N (A) = F .(3) Does this hold true for all the Cayley–Dickson algebras An when n > 3?


28. Suppose A is an octonion algebra and a, b ∈ A. Then: (a, b, x) = 0 for everyx ∈ A iff 1, a, b are linearly dependent.

(Hint. (�⇒) a, b ∈ H for some quaternion subalgebra H . Use (A.6) to deduce thatab = ba.)

Notes on Chapter 1

The independence result in (1.11) was proved by Hurwitz (1898) for skew symmetricmatrices. The general result (for matrices) is given in Robert’s thesis (1912) and inDickson (1919). Similar results are mentioned in the Notes for Exercise 12.The notation 〈〈a1, . . . , an〉〉 for Pfister forms was introduced by T. Y. Lam. Other

authors reverse the signs of the generators, writing 〈〈a〉〉 for 〈1,−a〉. In particular thisis done in the monumental work on involutions by Knus, Merkurjev, Rost and Tignol(1998). We continue to follow Lam’s notation, hoping that readers will not be undulyconfused.The topics in the appendix have been described in several articles and textbooks.

The idea of the “doubling process”, building the octonions from pairs of quaternions,is implicit in Cayley’s works, but it was first formally introduced in Dickson’s 1914monograph. Dickson was also the first to note that the real octonions form a divisionring. E. Artin conjectured that the octonions satisfy the alternative law. This was firstproved by Artin’s student M. Zorn (1930).Perhaps the first study of the Cayley–Dickson algebras of dimension> 8was given

by Albert (1942a). He analyzed the algebras An of dimension 2n over an arbitraryfield F (allowing F to have characteristic 2), and proved a general version of TheoremA.8. Properties ofAn appear in Schafer (1954), Khalil (1993), Khalil and Yiu (1997),andMoreno (1998). Further information on composition algebras appears in Jacobson(1958), Kaplansky (1953), Curtis (1963).Are there infinite dimensional composition algebras? Kaplansky (1953) proved

that every composition algebra with (2-sided) identity element must be finite dimen-sional. However there do exist infinite dimensional composition algebras having onlya left identity element. Further information is given in Elduque and Pérez (1997).Define an algebraA over the real fieldR to be an absolute-valued algebra ifA is a

normed space (in the sense of real analysis) and |xy| = |x| · |y|. Urbanik and Wright(1960) proved that an absolute-valued algebra with identity must be isomorphic to oneof the classical composition algebras. Further information and references appear inPalacios (1992).

Exercise 3. (3) Let V • = {v ∈ V : q(v) �= 0} be the set of anisotropic vectorsin (V , q). If x, y ∈ V • define the angle-measure � (x, y) = B(x,y)

q(x)q(y). If f ∈ End(V )

preserves all such angle-measures, mustf be a similarity? Alpers and Schröder (1991)investigate this question (without assuming f is linear).


Exercise 5. (1) In fact if A,B : V × W → F are bilinear forms such thatA(x, y) = 0 implies B(x, y) = 0, then B is a scalar multiple of A. This is proved inRothaus (1978). A version of this result forp-linearmaps appears in Shaw andYeadon(1989). In a different direction, Alpers and Schröder (1991) study maps f : V → V

(not assumed linear) which preserve the orthogonality of the vectors in V • (the set ofanisotropic vectors).(2) See Samuel (1968), de Géry (1970), Lester (1977).

Exercise 6. This is part of the theory of Pfister forms. See Chapter 5.

Exercise 9. Compare Exercise 10.13. Also see Gantmacher (1959), §11.4,Taussky and Zassenhaus (1959), Kaplansky (1969), Theorem 66, Shapiro (1992).

Exercise 12. This generalizes (1.11). Variations include systems where the ma-trices are skew-symmetric, or have squares which are scalars. Results of these typeswere obtained by Eddington (1932), Newman (1932), Littlewood (1934), Dieudonné(1953), Kestelman (1961), Gerstenhaber (1964), and Putter (1967). The skew-sym-metric case appears in Exercise 2.13. A system of anticommuting matrices whosesquares are scalars becomes a representation of a Clifford algebra, and the boundsare determined by the dimension of an irreducible module. Another variation on theproblem is considered by Eichhorn (1969), (1970).

Exercises 14. This Geometry Lemma was pointed out to me by A. Wadsworthwith further comments by D. Leep.

Exercise 15 (1), (2) appears in Witt (1937). There is a related Transversality The-orem for quadratic forms over semi-local rings, due to Baeza and Knebusch.

Exercise 16. Compare (7.16).

Exercise 17. (4) Wonenburger (1962b).(5) This lemma appears in Coxeter (1946). It is also valid in more general quater-

nion algebras.

Exercise 21. Remark. There is very little control on the denominators that arisein the process of extending the similarity B to the full matrix A. Even writing outan explicit 16-square identity having 9 bilinear terms seems difficult. There are somany choices for extending the given 9× 16 matrix to a 16× 16 matrix that nothinginteresting seems to arise.One can generalize these results to formulas ϕ(X)ϕ(Y ) = ϕ(Z) for any Pfister

form ϕ. There are similar results on multiplication formulas for hyperbolic forms, butdifficulties arise in cases of singular spaces of similarities.

Exercise 22. Details and further results about isoclinic planes are given in: Wong(1961), Wolf (1963), Tyrrell and Semple (1971), Shapiro (1978b), andWong andMok(1990). Isoclinic spaces are briefly mentioned after (15.23) below.

Exercise 23. These ideas follow Wong and Mok (1990), and Yiu (1993). Yiuproves:


Theorem. Every maximal subspace in Sim(Rn) is maximal as a subspace ofnonsingular matrices.

The proof uses homotopy theory. Related results appear in Adams, Lax and Philips(1965).

Exercise 24. Information on automorphisms appears in Jacobson (1958). Jacobsondefines the inner automorphisms of an octonion algebra to be the ones defined in(3), and he proves that every automorphism is inner. That construction of innerautomorphisms works for any alternative algebra. Also see Exercise 8.16 below.Automorphisms may also be considered geometrically. For the real octonion divisionalgebraK the group Aut(K) is a compact Lie group of dimension 14, usually denotedG2. The map σa(x) = axa−1 is not often an automorphism of an octonion algebra.In fact, H. Brandt proved that if [a] = 1 then σa is an automorphism⇐⇒ a6 = 1.Proofs appear in Zorn (1935) and Khalil (1993).

Exercise 25. See Schafer (1954), Adem (1978b) andMoreno (1998). The Cayley–Dickson algebras are also mentioned in Chapter 13. The 2n basis elements ei ofExercise 25 (4) can instead be indexed by V = Fn2, using notation as in (1.11). Ife2i = −1 then: e� · e� = (−1)β(�,�)e�+� for some map β : V × V → F2. Gordanet al. (1993) discuss the following question: If n = 3, which maps β yield an octonionalgebra? These results can also be cast in terms of intercalate matrices defined inChapter 13. A related situation for Clifford algebras is mentioned in Exercise 3.19.

Exercise 26. The proposition (due to Artin) follows Schafer (1966), pp. 27–30.See also Zhevlakov et al. (1982), p. 36.

Chapter 2

Amicable Similarities

Analysis of compositions of quadratic forms leads us quickly to the Hurwitz–Radonfunction ρ(n) as defined in Chapter 0. This function enjoys a property sometimescalled “periodicity 8”. That is: ρ(16n) = 8 + ρ(n). This and similar propertiesof the function ρ(n) can be better understood in a more general context. Instead ofconsidering s − 1 skew symmetric maps as in the original Hurwitz Matrix Equations(1.7), we allow some of the maps to be symmetric and some skew symmetric. Thisformulation exposes some of the symmetries of the situation which were not evidentat the start. For example we can use these ideas to produce explicit 8-square identitieswithout requiring previous knowledge of octonion algebras.

2.1 Definition. Two (regular) subspaces S, T ⊆ Sim(V , q) are amicable if

f g = gf for every f ∈ S and g ∈ T .In this case we write (S, T ) ⊆ Sim(V , q). If σ and τ are quadratic forms, the notation

(σ, τ ) < Sim(V , q)

means that there is a pair (S, T ) ⊆ Sim(V , q) where the induced quadratic forms onS and T are isometric to σ and τ , respectively.

It follows from the definition that if (S, T ) ⊆ Sim(V ) and h, k ∈ Sim•(V ), then(hSk, hT k) ⊆ Sim(V ). On the level of quadratic forms this says:

If (σ, τ ) < Sim(q) and d ∈ GF (q) then (〈d〉σ, 〈d〉τ) < Sim(q).If S �= 0 we may translate by some f to assume 1V ∈ S. Since this normalization isso useful we give it a special name.

2.2 Definition. An (s, t)-family on (V , q) is a pair (S, T ) ⊆ Sim(V , q) wheredim S = s, dim T = t and 1V ∈ S. If (σ, τ ) < Sim(V , q) and σ represents 1,we abuse the notation and say that (σ, τ ) is an (s, t)-family on (V , q).

2.3 Lemma. Suppose σ � 〈1, a2, . . . , as〉 and τ � 〈b1, . . . , bt 〉. Then (σ, τ ) <Sim(V , q) if and only if there exist f2, . . . , fs; g1, . . . , gt in End(V ) satisfying the

40 2. Amicable Similarities

following conditions:

fi = −fi,gj = gj ,

f 2i = −ai1Vg2j = bj1V

for 2 ≤ i ≤ s.

for 1 ≤ j ≤ t .

The s + t − 1 maps f2, . . . , fs; g1, . . . , gt pairwise anti-commute.

Proof. If (σ, τ ) < Sim(V , q) let (S, T ) ⊆ Sim(V , q) be the corresponding amicablepair. Since σ represents 1, we may translate by an isometry to assume 1V ∈ S.Let {1V , f2, . . . , fs} and {g1, . . . , gt } be orthogonal bases of S and T , respectively,corresponding to the given diagonalizations. The conditions listed above quicklyfollow. The converse is also clear. ��

An (s, t)-family corresponds to a system of s + t − 1 anti-commuting matriceswhere s−1 of them are skew-symmetric and t of them are symmetric. Stated that wayour notation seems unbalanced. The advantage of the terminology of (s, t)-familiesis that s and t behave symmetrically:

(σ, τ ) < Sim(V , q) if and only if (τ, σ ) < Sim(V , q).

Example 1.8 provides an explicit (2, 2)-family on a 2-dimensional space:(〈1, d〉, 〈1, d〉) < Sim(〈1, d〉). More trivially, (〈1〉, 〈1〉) < Sim(V ) for every spaceV .

2.4 Basic sign calculation. Continuing the notation of (2.3), let z = f2 . . . fsg1 . . . gt .Let det(σ ) det(τ ) = 〈d〉. Then z = ±z and µ(z) = zz = d, up to a square factor. Infact, z = z if and only if s ≡ t or t + 1 (mod 4).

Proof. The formula for zz is clear since ˜ reverses the order of products. If e1, e2, . . . , enis a set of n anti-commuting elements, then en . . . e2e1 = (−1)ke1e2 . . . en wherek = (n− 1)+ (n− 2)+ · · · + 2+ 1 = n(n− 1)/2. Since z is a product of s + t − 1anti-commuting elements and the tilde ˜ produces another minus sign for s−1 of thoseelements, we find that z = (−1)s−1(−1)kz where k = (s + t − 1)(s + t − 2)/2. Thestated calculation of the sign is now a routine matter. ��

2.5 Expansion Lemma. Suppose (σ, τ ) < Sim(V ) with dim σ = s and dim τ = t .Let det(σ ) det(τ ) = 〈d〉.

If s ≡ t − 1 (mod 4), then (σ ⊥ 〈d〉, τ ) < Sim(V ).If s ≡ t + 1 (mod 4), then (σ, τ ⊥ 〈d〉) < Sim(V ).

If (σ ′, τ ′) is the amicable pair obtained from (σ, τ ) then dim σ ′ ≡ dim τ ′ (mod 4)and dσ ′ = dτ ′.

Proof. First suppose σ represents 1, arrange 1V ∈ S and choose bases as in (2.3). Letz = f2 . . . fsg1 . . . gt as before. Then z ∈ Sim•(V ) and µ(z) = d, up to a square

2. Amicable Similarities 41

factor. If s + t is odd, then z anti-commutes with each fi and gj . If z = −z thenz can be adjoined to S while if z = z then it can be adjoined to T . The congruenceconditions follow from (2.4) and the properties of (σ ′, τ ′) follow easily. ��

2.6 Shift Lemma. Let σ , τ , ϕ, ψ be quadratic forms and suppose dim ϕ ≡dimψ (mod 4). Let 〈d〉 = (det ϕ)(detψ). Then:

(σ ⊥ ϕ, τ ⊥ ψ) < Sim(V , q) if and only if (σ ⊥ 〈d〉ψ, τ ⊥ 〈d〉ϕ) < Sim(V , q).

This remains valid when ϕ or ψ is zero. That is, if α is a quadratic form withdim α ≡ 0 (mod 4) and 〈d〉 = det α then:

(σ ⊥ α, τ) < Sim(q) if and only if (σ, τ ⊥ 〈d〉α) < Sim(q).This shifting result exhibits someof theflexibility of these families: an (s+4, t)-familyis equivalent to an (s, t + 4)-family.

Proof of 2.6. Suppose (S ⊥ H, T ⊥ K) ⊆ Sim(V , q) and a ≡ b (mod 4), wherea = dimH and b = dimK . We may assume S �= 0. (For if T �= 0 interchange theeigenspaces. If S = T = 0 the lemma is clear.) Scale by suitable f to assume 1V ∈ S.Choose orthogonal bases {1V , f2, . . . , fs, h1, . . . , ha} and {g1, . . . , gt , k1, . . . , kb}and define y = h1h2 . . . hak1 . . . kb. Then y = y and y commutes with elements of Sand T , and anticommutes with elements ofH andK . Therefore (S+yK, T +yH) ⊆Sim(V , q). The converse follows since the same operation applied again leads backto the original subspaces. ��

2.7 Construction Lemma. Suppose (σ, τ ) < Sim(q)where σ represents 1. If a ∈ F •then

(σ ⊥ 〈a〉, τ ⊥ 〈a〉) < Sim(q ⊗ 〈〈a〉〉).

Proof. Recall that 〈〈a〉〉 = 〈1, a〉 is a binary form. Let (S, T ) be an (s, t)-familyon (V , q) corresponding to (σ, τ ), and express S = F1V ⊥ S1. Recall the (2, 2)-family constructed in Example 1.8 given by certain 2 × 2 matrices f2, g1 and g2in Sim(〈〈a〉〉). We may verify that S′ = F(1V ⊗ 1) + S1 ⊗ g1 + F(1V ⊗ f2) andT ′ = T ⊗ g1+F(1V ⊗ g2) does form an (s+ 1, t + 1)-family on q⊗〈〈a〉〉. CompareExercise 1. ��

2.8 Corollary. If ϕ � 〈〈a1, . . . , am〉〉 is a Pfister form of dimension 2m, then there isan (m+ 1,m+ 1)-family (σ, σ ) < Sim(ϕ), where σ � 〈1, a1, . . . , am〉.

Proof. Starting from the (1, 1)-family on 〈1〉, repeat the Construction Lemmam times. ��


The construction here is explicit. The 2m + 2 basis elements of this family canbe written out as 2m × 2m matrices. Each entry of one of these matrices is either 0 or±ai1 . . . air for some 1 ≤ i1 < · · · < ir ≤ m. In particular if every ai = 1 then thematrix entries lie in {0, 1,−1}.These lemmas can be used to construct large spaces of similarities. Suppose ϕ is an

m-fold Pfister form as above. Then there is an (m+1,m+1)-family (σ, σ ) < Sim(ϕ).Now use the Shift Lemma to shift as much as possible to the left. The resulting sizesare:

(2m+ 1, 1) if m ≡ 0

(2m, 2) if m ≡ 1(mod 4)

(2m− 1, 3) if m ≡ 2

(2m+ 2, 0) if m ≡ 3

Ignoring the “t” parts of these families we get some large subspaces of Sim(ϕ). Inthe case m ≡ 2 we have 2m− 1 ≡ 3 (mod 4), and the Expansion Lemma furnishes asubspace of dimension 2m. We have found subspaces of dimension ρ(2m).

2.9 Proposition. Suppose n = 2mn0 where n0 is odd. Suppose q is a quadraticform of dimension n expressible as q � ϕ ⊗ γ where ϕ is an m-fold Pfister form anddim γ = n0. Then there exists σ < Sim(q) with dim σ = ρ(n).

Proof. From the definition of ρ(n) given in Chapter 0 and the remarks above we seethat there exists σ < Sim(ϕ) with dim σ = ρ(2m) = ρ(n). Then also σ < Sim(q)as mentioned in (1.5). ��

Using a little linear algebra we prove the following converse to the ConstructionLemma.

2.10 Eigenspace Lemma. Suppose (σ ⊥ 〈a〉, τ ⊥ 〈a〉) < Sim(q) is an (s+1, t+1)-family, where s ≥ 1. Then q � ϕ ⊗ 〈〈a〉〉 for some quadratic form ϕ such that(σ, τ ) < Sim(ϕ).

Proof. Translating the given family if necessarywemay assume it is given by (S, T ) ⊆Sim(V , q) where {1V , f2, . . . , fs, f } and {g1, . . . , gt , g} are orthogonal bases andµ(f ) = µ(g) = a. Then f = −f and f 2 = −a1V , g = g and g2 = a1V . Then h =f−1g = −a−1fg satisfies h = hand h2 = 1V . Let U and U ′ be the ±1-eigenspacesfor h. Since h = h these spaces are orthogonal and V = U ⊥ U ′. Let ϕ, ϕ′ be thequadratic formsonU andU ′ induced byq, so thatq � ϕ ⊥ ϕ′. Sincef anti-commuteswith h we have f (U) = U ′ and f (U ′) = U . Consequently dimU = dimU ′ =12 dim V , ϕ � 〈a〉ϕ and q � ϕ ⊗ 〈〈a〉〉. Furthermore 1V , f2, . . . , fs, g1, . . . , gtall commute with h so they preserve U . Their restrictions to U provide the family(σ, τ ) < Sim(U, ϕ). ��


We now have enough information to find all the possible sizes of (s, t)-families onquadratic spaces of dimension n.

2.11 Theorem. Suppose n = 2mn0 where n0 is odd. There exists an (s, t)- family onsome quadratic space of dimension n if and only if s ≥ 1 and one of the followingholds:

(1) s + t ≤ 2m,

(2) s + t = 2m+ 1 and s ≡ t − 1 or t + 1 (mod 8),(3) s + t = 2m+ 2 and s ≡ t (mod 8).

Proof. (“if” ) Suppose that there exist numbers s′, t ′ such that s ≤ s′, t ≤ t ′, s′ + t ′ =2m+2 and s′ ≡ t ′ (mod 8). Then there is an (s′, t ′)-family and hence an (s, t)-familyon some quadratic space of dimension n. To see this first use the Construction Lemmato get an (m + 1,m + 1)-family on a space of dimension 2m and tensor it up (as in(1.6)) to get such a family on an n-dimensional space (V , q). A suitable applicationof the Shift Lemma then provides an (s′, t ′)-family in Sim(q).If s, t satisfy one of the given conditions then such s′, t ′ do exist, except in the

case s + t = 2m and s ≡ t + 4 (mod 8). In this case, suppose s ≥ 2. Thens−1 ≡ t+3 (mod 8) and the work above implies that there is an (s−1, t+3)-familyin Sim(q) for some n-dimensional form q. Restrict to an (s − 1, t)-family and applythe Expansion Lemma 2.5 to obtain an (s, t)-family. Similarly if t ≥ 1 there is an(s + 3, t − 1)-family which restricts and expands to an (s, t)-family.(“only if” ) Suppose there is an (s, t)-family on some n-dimensional space and

proceed by induction on m. If m = 0 Proposition 1.10 implies s, t ≤ 1 and therefores + t ≤ 2. If s + t = 2 then certainly s = t . Similarly, if m = 1 Proposition 1.10implies s, t ≤ 2 so that s + t ≤ 4. If s + t = 4 then s = t .Suppose m ≥ 2. If s + t ≤ 4 the conditions are satisfied. Suppose s + t > 4

and apply the Shift Lemma and the symmetry of (s, t) and (t, s) to arrange s ≥ 2 andt ≥ 1. If (σ, τ ) < Sim(q) is the given (s, t)-family where dim q = n, pass to anextension field of F to assume σ and τ represent a common value. The EigenspaceLemma then implies that there is an (s−1, t −1)-family on some space of dimensionn/2. The induction hypothesis implies the required conditions. ��

2.12 Corollary. If σ < Sim(q) where dim q = n then dim σ ≤ ρ(n).

Proof. Suppose s = dim σ > ρ(n) where n = 2mn0 and n0 is odd. Ifm ≡ 0 (mod 4)then s > ρ(n) = 2m+ 1 so there is a (2m+ 2, 0)-family in Sim(q). But 2m+ 2 ≡2 (mod 8) contrary to the requirement in the Theorem 2.11. The other cases followsimilarly. ��


This corollary is the generalization of the Hurwitz–Radon Theorem to quadraticforms over any field (of characteristic not 2). A refinement of Theorem 2.11 appearsin (7.8) below.Theorem 2.11 contains all the information about possible sizes of families. This

information can be presented in a number of ways. For example, given n and t wecan determine the largest s for which an (s, t)-family exists on some n-dimensionalquadratic space.

2.13 Definition. Given n and t , define ρt (n) to be the maximum of 0 and the valueindicated in the following table. Here n = 2mn0 where n0 is odd.

m (mod 4) ρt (n)

m ≡ t 2m+ 1− tm ≡ t + 1 2m− tm ≡ t + 2 2m− tm ≡ t + 3 2m+ 2− t

2.14 Corollary. Suppose (V , q) is an n-dimensional quadratic space.

(1) If there is an (s, t)-family in Sim(V , q) then s ≤ ρt (n).

(2) Suppose s = ρt (n). Then there is some (s, t)-family in Sim(V , q), provided thatq can be expressed as a product ϕ ⊗ γ where ϕ is a Pfister form and dim γ isodd.

The proof is left as an exercise for the reader.Here is another way to codify this information. Given s, t we determine the

minimal dimension of a quadratic space admitting an (s, t)-family.

2.15 Corollary. Let s ≥ 1 and t ≥ 0 be given. The smallest n such that there is an(s, t)-family on some n-dimensional quadratic space is n = 2δ(s,t) where the valuem = δ(s, t) is defined as follows.Case s + t is even:

if s ≡ t (mod 8) then s + t = 2m+ 2 and δ(s, t) = s+t−22 ,

if s �≡ t (mod 8) then s + t = 2m and δ(s, t) = s+t2 .

Case s + t is odd:

if s ≡ t ± 1 (mod 8) then s + t = 2m+ 1 and δ(s, t) = s+t−12 ,

if s ≡ t ± 3 (mod 8) then s + t = 2m− 1 and δ(s, t) = s+t+12 .

Proof. This is a restatement of Theorem 2.11. The value δ(r) = δ(r, 0)was calculatedin Exercise 0.6. Further properties of δ(s, t) appear in Exercise 3. ��


If q = ϕ ⊗ γ where ϕ is a large Pfister form, then (2.14) implies that Sim(V , q)admits a large (s, t)-family. We investigate the converse. By Proposition 1.10 sub-spaces of Sim(q) provide certain Pfister forms which are tensor factors of q. Thefollowing consequence of the Eigenspace Lemma is a similar sort of result: certain(s, t)- families in Sim(q) provide Pfister factors of q.

2.16 Corollary. Suppose (σ ⊥ α, τ ⊥ α) < Sim(q) where dim σ ≥ 1 and α �〈a1, . . . , ak〉. Then q � 〈〈a1, . . . , ak〉〉 ⊗ γ for some quadratic form γ such that(σ, τ ) < Sim(γ ).

Proof. Repeated application of (2.10). ��

Suppose dim q = 2mn0 where n0 is odd. If there is an (m + 1,m + 1)- family(σ, τ ) < Sim(q)with σ � τ � 〈1, a1, . . . , am〉 then q � 〈〈a1, . . . , am〉〉⊗γ for someform γ of odd dimension n0. The Pfister Factor Conjecture is a stronger version ofthis idea.

2.17 Pfister Factor Conjecture. Suppose that q is a quadratic form of dimensionn = 2mn0 where n0 is odd. If there exists an (m + 1,m + 1)-family in Sim(q) thenq � 〈〈a1, . . . , am〉〉 ⊗ γ for some quadratic form γ of dimension n0.

This conjecture is true form ≤ 2 by Proposition 1.10. We cannot get much furtherwithout more tools. We take up the analysis of this conjecture again in Chapter 9.The Decomposition Theorem and properties of unsplittable modules (Chapters 4–7)reduce the Conjecture to the case dim q = 2m. Properties of discriminants and Wittinvariants of quadratic forms (Chapter 3) can then be used to prove the conjecturewhen m ≤ 5. The answer is not known when m > 5 over arbitrary fields, but overcertain nice fields (e.g. global fields) the conjecture can be proved for all values ofm.After learning some properties of the invariants of quadratic forms (stated in (3.21)),the reader is encouraged to skip directly to Chapter 9 to see how they relate to thePfister Factor Conjecture.The Conjecture can be stated in terms of compositions of quadratic forms in the

earlier sense: If dim q = n = 2mn0 as usual and if there is a subspace σ < Sim(q)where dim σ = ρ(n), then there is an (m + 1,m + 1)-family in Sim(q). In factin Proposition 7.6 we prove that dim σ ≥ 2m − 1 suffices to imply that there is an(m + 1,m + 1)-family in Sim(q). This “expansion” result seems to require someknowledge of algebras and involutions.



1. Construction Lemma revisited. (1) Write out the elements of the (s + 1, t + 1)-family in (2.7) using block matrices: if S has basis {1, h2, . . . , hs} then for examplehi ⊗ g1 =

(hi 00 −hi

)while 1⊗ f2 =

(0 −a1 0

).

(2) Let (S, T ) and (G,H) be commuting pairs of amicable subspaces of Sim(V ).That is, every element of S ∪ T commutes with every element of G ∪ H . Chooseanisotropic f ∈ S, g ∈ G and define S1 = (f )⊥ and G1 = (g)⊥. Then(f S1 + gH, gG1 + f T ) is an amicable pair in Sim(V ).(3) An (s, t)-family on V and an (a, b)-family on W where s, a ≥ 1 yield an

(s + b − 1, t + a − 1)-family on V ⊗W .

2. When σ does not represent 1. Define the norm group GF (q) = {a ∈ F • :〈a〉q � q}, the group of norms of all elements of Sim•(q). For any form q,GF (q)·DF (q) ⊆ DF (q). If σ < Sim(q) then DF (σ) ⊆ GF (q). Let σ , τ bequadratic forms over F .(1) Suppose c ∈ DF (σ) and let σ0 = 〈c〉σ and τ0 = 〈c〉τ . Then (σ, τ ) < Sim(q)

if and only if c ∈ GF (q) and (σ0, τ0) < Sim(q).(2) The Expansion Lemma 2.5 remains true without assuming σ represents 1.(3) If (S ⊥ A, T ) ⊆ Sim(V , q) where dimA = 4, there is a shifted amicable

family (S, T ⊥ A′) ⊆ Sim(V , q). If {h1, . . . , h4} is an orthogonal basis of A, whatis an explicit basis of A′?(4) The Construction Lemma remains true without assuming σ represents 1.

(Hint. (4) If c ∈ GF (q) then q ⊗ 〈〈ca〉〉 � q ⊗ 〈〈a〉〉.)

3. (1) Deduce the following formulas directly from the early lemmas about (s, t)-families.

δ(s, t) = δ(t, s) δ(s + 1, t + 1) = 1+ δ(s, t) δ(s + 4, t) = δ(s, t + 4).(2) Recall the definition of δ(r) from Exercise 0.6: r ≤ ρ(n) iff 2δ(r) ||n. Note that

δ(r + 8) = δ(r)+ 4 and use this to extend the definition to δ(−r).Lemma. δ(s, t) = t + δ(s − t).

(3) δ(s) =

s−22 if s ≡ 0s−12 if s ≡ ±1s2 if s ≡ ±2, 4s+12 if s ≡ ±3

(mod 8).

4. More Shift Lemmas. (1) If dim ϕ ≡ 2+ dimψ (mod 4) in the Shift Lemma then(σ ⊥ ϕ, τ ⊥ ψ) is equivalent to (σ ⊥ 〈d〉ϕ, τ ⊥ 〈d〉ψ).


(2) Let (σ, τ ) = (〈1, a, b, c〉, 〈x, y〉) and d = abcxy. Then we can shift it to(σ ′, τ ′) = (〈1, a, b, c, dx, dy〉, 0). Similarly if (σ, τ ) = (〈1, a, b〉, 〈x, y, z〉), andd = abxyz we can shift it to (σ ′, τ ′) = (〈1, a, bd〉, 〈x, y, zd〉).(3) Generalize this idea to other (s, t)-pairs where s + t is even. If s ≡ t (mod 4)

explain why δ(s, t + 2) = δ(s + 2, t).

5. (1) If (〈1, a, b〉, 〈x〉) < Sim(q) then 〈1, abx〉 < Sim(q).(2)Generalize this observation to (σ, τ ) < Sim(q)where s ≡ t+2 or t+3 (mod 4).

(Hint. (1) Examine the element z as in (2.4).)

6. Alternating spaces. (1) Lemma. Let σ , τ and α be quadratic forms and V avector space. Suppose dim α ≡ 2 (mod 4). Then (σ ⊥ α, τ) < Sim(V , q) for somequadratic form q on V if and only if (σ, τ ⊥ α) < Sim(V , B) for some alternatingform B on V .(2) Theorem. Suppose n = 2mn0 where n0 is odd. There exists an (s, t)- family

on some alternating space of dimension n if and only if one of the following holds:(i) s + t ≤ 2m,(ii) s + t = 2m+ 1 and s ≡ t − 3 or t + 3 (mod 8),(iii) s + t = 2m+ 2 and s ≡ t + 4 (mod 8).(3) Let δ′(s, t) be the corresponding function for alternating spaces. Then

δ′(s + 2, t) = δ(s, t + 2). Note that δ(s, t) = δ′(s, t) iff s ≡ t ± 2 (mod 8). Howdoes Exercise 4 help “explain” this? Does δ′(s, t) = t + δ′(s − t)?(4) Let ρ′(n) be the Hurwitz–Radon function for alternating spaces. Note that

ρ′(1) = 0 and ρ′(2) = 4. (See Exercise 1.7.) The formula for ρ′(n) in terms ofm (mod 4) is a “cycled” version of the formula for ρ(n). In other words, ρ′(4n) =4+ ρ(n) whenever n ≥ 1.

(Hint:. (1) Let (S ⊥ A, T ) ⊆ Sim(V , q) and z be given as in the Shift Lemma2.6. Define the form B ′ on V by: B ′(u, v) = B(u, z(v)). Then (S, T ⊥ A) ⊆Sim(V , B ′).)

7. Hurwitz–Radon functions. (1) Write out the formulas for ρ′t (n), the alternatingversion of the functions ρt (n) given in (2.13) and prove the analog of (2.14).(2) We write ρλ(n) to denote ρ(n) if λ = 1 and ρ′(n) if λ = −1. The fol-

lowing properties of the Hurwitz–Radon functions are consequences of the formulas,assuming in each case that the function values are large enough.

ρλt+1(2n) = 1+ ρλt (n),ρλt (n) = 2+ ρ−λt+2(n),

ρλt (n) = 4+ ρλt+4(n), ρ−λt (4n) = 4+ ρλt (n),max{ρ(n), ρ′(n)} =

{ 2m+ 1 if m is even,2m+ 2 if m is odd.

(3) Explain each of these formulas more theoretically.


8. That element “z”. The element z = z(S) · z(T ) was used in (2.4) and (2.5). Whatif a different orthogonal basis is chosen for S? Is there a suitable definition for z whenS does not contain 1V ? We use the following result originally due to Witt:

Chain-Equivalence Theorem. Let (V , q) be a quadratic space with two or-thogonal bases X and X′. Then there exists a chain of orthogonal bases X =X0,X1, . . . ,Xm = X′ such that Xi and Xi−1 differ in at most 2 vectors.

Proofs appear in Scharlau (1985), pp. 64–65 andO’Meara (1963), p. 150. CompareSatz 7 of Witt (1937).(1) Let S ⊆ Sim(V , q) be a subspace of dimension s. IfB = {f1, f2, . . . , fs} is an

orthogonal basis of S, define z(B) = f1 · f2 ·f3 · f4 . . . andw(B) = f1 ·f2 · f3 ·f4 . . . .Lemma. If B ′ is another orthogonal basis, then z(B ′) = λ · z(B) and w(B ′) =

λ · w(B) for some λ ∈ F •.Define z(S) = z(B) and w(S) = w(B). These values are uniquely determined

by the subspace S, up to non-zero scalar multiple. Note that z(S) · z(S)∼ = w(S) ·w(S)∼ = det(σ ) · 1V .Let z = z(B) and w = w(B).(2) If s is odd: w = (−1)(s−1)/2 · z. For every f ∈ S•, z · f = f · z.If s is even: z = (−1)s/2 · z and w = (−1)s/2 · w. For every f ∈ S•, zf = fw.Consequently if s is even then z2 = w2 = dσ · 1V .(3) Suppose ϕ : S → S is a similarity. Then ϕ(B) is another orthogonal basis and

z(ϕ(B)) = (det ϕ) · z(B).(4) Let g, h ∈ Sim•(V , q) with α = µ(g)µ(h).If s is odd: z(gBh) = α(s−1)/2 · g · z(B) · h.If s is even: z(gBh) = αs/2 · g · z(B) · g−1.(5) Analyze the Expansion and Shift Lemmas using these ideas. (Compare Exer-

cise 2.)

9. Symmetric similarities. (1) Lemma (Dieudonné). Suppose f ∈ Sim•(V , q)where dim V = 2m is even. Then there exists g ∈ O(V , q) and a decompositionV = V1 ⊥ · · · ⊥ Vm such that gf (Vi) = Vi , dim Vi = 2 and gf = gf . Furthermore,given anisotropic v ∈ V , there is such a decomposition with v ∈ V1.(2) 〈a〉 < Sim(q) if and only if (〈1〉, 〈a〉) < Sim(q).(3) If dim q is even then GF (q) ⊆ DF (〈〈−dq〉〉).

(Hint. (1) Assume µ(f ) /∈ F •2 and let V1 = span{v, f (v)}. Then V1 is a regular2-plane and there exists g1 ∈ O(V ) with g1f (v) = f (v) and g1f 2(v) = µ(f )v.Then g1f preserves V1. Apply induction to construct g. Note that (gf )2 = µ(f )1V .(3) If a ∈ GF (q) with a /∈ F •2 then q � 〈x1〉〈〈−d1〉〉 ⊥ · · · ⊥ 〈xm〉〈〈−dm〉〉 where

〈〈−dj 〉〉 represents a. Then 〈〈−a〉〉 represents 〈d1d2 . . . dm〉 = dq.)

10. An orthogonal design of order n and type (s1, . . . , sk) is an n× n matrix A withentries from {0,±x1, . . . ,±xk} such that the rows of A are orthogonal and each row


of A has si entries of the type ±xi . Here the xi are commuting indeterminates and siis a positive integer. Then A · A� = σ · In where σ =∑k

i=1 six2i .Proposition. (1) If there is such a design then 〈s1, . . . , sk〉 < Sim(n〈1〉). In

particular k ≤ ρ(n).(2) If the si are positive integers and s1+· · ·+sk ≤ ρ(n) then there is an orthogonal

design of order n and type (s1, . . . , sk).

(Hint. The Construction, Shift and Expansion Lemmas provide Ai ∈ Mn(Z), for1 ≤ i ≤ ρ(n), which satisfy the Hurwitz Matrix Equations. This yields an integercomposition formula as in Exercise 0.4, hence an orthogonal design of order n andtype (1, 1, . . . , 1). Set some of the variables equal.)

11. Constructing composition algebras. (1) From the Construction and ExpansionLemmas there is a 4-dimensional subspace 〈1, a, b, ab〉 < Sim(〈〈a, b〉〉). This inducesa 4-dimensional composition algebra (see Exercise 1.1). This turns out to be the usualquaternion algebra.(2) The Construction and Shift Lemmas provide an explicit σ < Sim(〈〈a, b, c〉〉)

with dim σ = 8, and we get an induced 8-dimensional composition algebra. Thisturns out to be the standard octonion algebra.

12. Amicable spaces of rectangular matrices. Let V , W be two regular quadraticspaces and consider subspaces S, T ⊆ Sim(V ,W) as in Exercise 1.2. Suppose S andT are “amicable” in the sense generalizing (2.1). If dim S = s, dim T = t , dim V = r

and dimW = n we could call this an (s, t)-family of n× r matrices.(1) If there is an (s, t)-family of n×r matrices then there is an (s+1, t+1)-family

of 2n× 2r matrices. This is the analog of the Construction Lemma.(2) Why do the analogs of the Expansion and Shift Lemmas fail in this context?

13. Systems of skew-symmetric matrices. Definition. αF (n) = max{t : there existA1, . . . , At ∈ GLn(F ) such that A�

i Aj + A�j Ai = 0 whenever i �= j}. Certainly

ρ(n) ≤ αF (n).

Open question. Is this always an equality?

(1) αF (n) − 1 = max{k : there exist k skew-symmetric elements of GLn(F )which pairwise anticommute}.(2) If n = 2mn0 where n0 is odd, then αF (n) ≤ 2m+ 2.(3) If n0 is odd then αF (n0) = 1 and αF (2n0) = 2.(4) Proposition. αF (2m) = ρ(2m).(5) Let {I, f2, . . . , fs} be a system overF as above. Let V = Fn with the standard

inner product, and view fi ∈ End(V ). Define W ⊆ V to be invariant if fi(W) ⊆ W

for every i. The system is “decomposable” if V is an orthogonal sum of non-zeroinvariant subspaces.

Lemma. If {fi} is an indecomposable system over R, then f 2i = scalar.

(6) Proposition. If F is formally real then αF (n) = ρ(n).


(Hint. (2) See Exercise 1.12.(3) For 2n0 apply the “Pfaffians” defined in Chapter 10. If f , g are nonsingular,

skew-symmetric, anti-commuting, then pf (fg) = pf (gf ) = pf (−fg).(4) If αF (2m) > ρ(2m) there exist 2m skew-symmetric, anti-commuting elements

fi ∈ GL2m(F ). Then f 2i = scalar as in Exercise 1.12(2), and Hurwitz–Radon applies.(5) The Spectral Theorem implies V is the orthogonal sum of the eigenspaces of

the symmetric matrix f 2j . Since every fi commutes with f2j these eigenspaces are

invariant.(6) Generalize (5) to real closed fields and note that F embeds in a real closed

field. Apply Hurwitz–Radon. Over R we could apply Adams’ theorem (see Exercise0.7) to the orthogonal vector fields f2(v), . . . , fs(v) on Sn−1.)

14. AHadamard design of order n on k letters {x1, . . . , xk} is an n×nmatrixH suchthat each entry is some ±xj and the inner product of any two distinct rows is zero.If there is such a design then there exist n × n matrices H1, . . . , Hk with entries in{0, 1,−1} such that Hj ·H�

i +Hi ·H�j = 0 if i �= j and Hi ·H�

i = diagonal.

Proposition. If there exists a Hadamard design of order n on k letters, each ofwhich occurs at least once in every column of H , then k ≤ ρ(n).

(Hint. Note that each Hi is nonsingular and apply Exercise 13.)

15. Hermitian compositions. A hermitian (r, s, n)-formula is:

(|x1|2 + · · · + |xr |2) · (|y1|2 + · · · + |ys |2) = |z1|2 + · · · + |zn|2

whereX = (x1, . . . , xr ) and Y = (y1, . . . , ys) are systems of complex indeterminantsover C, and each zk is bilinear in X, Y . Such a formula can be viewed as a bilinearmap f : Cr × Cs → Cn satisfying |f (x, y)| = |x| · |y|. We consider three versionsof bilinearity:Type 1: each zk is bilinear in (X, X) and (Y, Y ).Type 2: each zk is bilinear in (X, X) and Y .Type 3: each zk is bilinear in X and Y .Note that if X = X1 + iX2 then z is linear (over C) in (X, X) if and only if z is

C-linear in the system of real variables (X1, X2). For example z1 = x1y1 + x2y2 andz2 = x1y2 − x2y1 provides a (2, 2, 2)-formula of types 1 and 2.

Proposition. (1) A hermitian (r, s, n)-formula of type 1 exists if and only if thereexists an ordinary (2r, 2s, 2n)-formula over R.(2) A hermitian (r, s, n)-formula of type 2 exists if and only if there exist two ami-

cable subspaces of dimension r in Simherm(Cs ,Cn), the set of hermitian similarities.(3) A hermitian (r, s, n)-formula of type 3 exists if and only if n ≥ rs.

(Hints. (2) The formula exists iff there is an n × s matrix A whose entries are lin-ear forms in (X, X) and satisfying A∗ · A = (∑r

1 xj xj)Is , where ∗ denotes the

conjugate-transpose. Express xj = uj +vj√−1 where uj , vj are real variables. Then


A = ∑r1 ujBj + vj

√−1Cj , where Bj and Cj are n × s matrices over C. ThenS = span{B1, . . . , Br} and T = span{C1, . . . , Cr} are the desired subspaces.(3) Here we get the same equation for A, where A = x1A1+ · · · + xrAr and each

Aj is a complex n × s matrix. Then A∗j Aj = Is and A∗

j Ak = 0 if j �= k. Choose

A1 =(Is0

).)

16. Consider the analogous composition formulas of the type

(x21 + · · · + x2r ) · (|y1|2 + · · · + |ys |2) = |z1|2 + · · · + |zn|2

whereX is a system overR, Y is a system overC and each zk is bilinear inX, Y . Is theexistence of such a formula equivalent to the existence ofA2, . . . , Ar ∈ GLn(C)whichare anti-hermitian (A∗

j = −Aj), unitary (A∗j · Aj = 1) and pairwise anticommute?

Notes on Chapter 2

The notion of amicable similarities was pointed out to me by W. Wolfe in 1974 (seeWolfe (1976)). He introduced the term “amicable” in analogy with a related idea incombinatorics.The idea of allowing some symmetric elements in the Hurwitz Matrix Equations

has occurred independently in Ławrynowicz and Rembielinski (1990). They do thisto obtain some further symmetries in their approach to the theory.

Exercise 9 follows Dieudonné (1954). Compare the appendix of Elman and Lam(1974). The decomposition of V is closely related to the “β-decomposition” in Corol-lary 2.3 of Elman and Lam (1973b). See Exercise 5.7 below.

Exercise 10. Orthogonal designs are investigated extensively in Geramita andSeberry (1979).

Exercise 13. The lemma in (5) follows Putter (1967). Further information onanticommuting matrices is given in the Notes on Chapter 1.

Exercise 14 generalizes ideas of Storer (1971).

Exercise 15–16. The observation on C-bilinear hermitian compositions in 15 (3)is due to Alarcon and Yiu (1993). Hermitian compositions are discussed furtherin Exercise 4.7. Compositions as in Exercise 16 are also considered in Furuoya etal. (1994). They work with a slightly more general situation, allowing forms ofarbitrary signature over R. Compare Exercise 4.7.

Chapter 3

Clifford Algebras

This is essentially a reference chapter, containing the definitions and basic properties ofClifford algebras andWitt invariants, alongwith some related technical results thatwillbe used later. The reader should have some acquaintance with central simple algebras,the Brauer group Br(F ) and the Witt ring W(F ). This background is presented in anumber of texts, including Lam (1973), Scharlau (1985).Clifford algebras have importance in algebra, geometry and analysis. We need the

basic algebraic properties of Clifford algebras over an arbitrary field F . We includethe proofs of some of the basic results, assuming familiarity with the classical theoryof central simple algebras. The exposition is simplified since we assume that thecharacteristic of F is not 2.Every F -algebra considered here is a finite dimensional, associative F -algebra

with an identity element denoted by 1. The field F is viewed as a subset of thealgebra. An unadorned tensor product ⊗ always denotes ⊗F , the tensor productover F .The first non-commutative algebra was the real quaternion algebra discovered by

Hamilton in 1843. Thatmotivates the general definition of a quaternion algebra overF .

Definition. If a, b ∈ F •, the quaternion algebra A =(a,bF

)is the associative

F -algebra with generators i, j satisfying the multiplication rules:

i2 = a, j2 = b and ij = −ji.

The associativity implies that A is spanned by {1, i, j, ij} and it follows thatdimA = 4. An element of A is called “pure” if its scalar part is 0. Then the setA◦ = A0 of pure quaternions is the span of {i, j, ij}. Direct calculation shows that

A0 = {u ∈ A : u /∈ F • and u2 ∈ F }.Consequently A0 is independent of the choice of generators i, j . Define the “bar”map on A to be the linear map which acts as 1 on F and as −1 on A0. Then ¯a = a

and: a = a if and only if a ∈ F . Another calculation shows: uv = v · u. Then “bar”is the unique anti-automorphism of A with i = −i and j = −j .

3. Clifford Algebras 53

The norm N : A → F , defined by N(a) = a · a, is a quadratic form on A andcalculation on the given basis shows that (A,N) � 〈〈−a,−b〉〉. Moreover, N(ab) =N(a) ·N(b) so that A is a composition algebra as described in Chapter 1, Appendix.

3.1 Lemma. (1)(a,bF

)is split if and only if its norm form 〈〈−a,−b〉〉 is hyperbolic.

(2) The isomorphism class of(a,bF

)is determined by the isometry class of

〈〈−a,−b〉〉.

Proof. An explicit isomorphism(1,−1F

)→ M2(F ) is provided by sending i �→(

0 11 0

)and j �→

(0 1−1 0

). It suffices to prove (2). An isomorphism of quaternion

algebras preserves the pure parts, so it commutes with “bar” and is an isometry forthe norms. If A is a quaternion algebra and (A,N) � 〈〈−a,−b〉〉, Witt Cancellationimplies (A0, N) � 〈−a,−b, ab〉, so there exist orthogonal elements i, j in A0 withN(i) = −a and N(j) = −b. Therefore i2 = a, j2 = b and ij + ji = 0 and these

generators provide an isomorphism A ∼=(a,bF

). ��

These results on quaternion algebras, together with the systems of equationsin (1.6) and (2.3), help to motivate the investigation of algebras having generators{e1, e2, . . . , en} which anticommute and satisfy e2i ∈ F •. An efficient method fordefining these algebras is to use their universal property.Suppose (V , q) is a quadratic space over F and A is an F -algebra. A linear map

ι : V → A is compatible with q if it satisfies:

ι(v)2 = q(v) for every v ∈ V .

For such a map ι, the quadratic structure of (V , q) is related to the algebra structureof A. For example, if v,w ∈ V are orthogonal then ι(v) and ι(w) anticommute inA. (For 2Bq(v,w) = q(v + w) − q(v) − q(w) = ι(v + w)2 − ι(v)2 − ι(w)2 =ι(v)ι(w)+ ι(w)ι(v).)The Clifford algebra C(V, q) is the F -algebra universal with respect to being

compatible with q. More formally, define a Clifford algebra for (V , q) to be anF -algebra C together with an F -linear map ι : V → C compatible with q and suchthat for any F -algebra A and any F -linear map ϕ : V → A which is compatible withq, there exists a unique F -algebra homomorphism ϕ : C → A such that ϕ = ϕ � ι.

C

ϕ

��

V

ι

��ϕ

�� A

54 3. Clifford Algebras

3.2 Lemma. For any quadratic space (V , q) over F there is a Clifford algebra(C(V, q), ι), which is unique up to a unique isomorphism.

Proof. The uniqueness of the Clifford algebra follows by the standard argument foruniversal objects. (See Exercise 1.) To prove existence we use the tensor algebra

T (V ) = ⊕kTk(V ) where T k(V ) is the k-fold tensor product of V .

Then T (V ) = F ⊕ V ⊕ (V ⊗ V )⊕ (V ⊗ V ⊗ V )⊕ · · · . Let C = T (V )/I where Iis the two-sided ideal of T (V ) generated by all elements v ⊗ v − q(v) · 1 for v ∈ V .Let ι : V → C be the canonical map V → T (V )→ T (V )/I = C.

Claim. (C, ι) is a Clifford algebra for (V , q). For if ϕ : V → A is anF -linear mapcompatible with q then the universal property of tensor algebras implies that ϕ extendsto a unique F -algebra homomorphism ϕ : T (V )→ A. Since ϕ is compatible with qwe find that ϕ(I) = 0 and therefore ϕ induces a unique homomorphism ϕ : C → A

such that ϕ = ϕ � ι. ��

Since theClifford algebra is uniquewe are often sloppy about the notations, writingC(q) rather thanC(V, q). A major advantage of using the universal property to defineClifford algebras is that the “functorial” properties follow immediately:

3.3 Lemma. (1) An isometry f : (V , q)→ (V ′, q ′) induces a unique algebra homo-morphism C(f ) : C(V, q) → C(V ′, q ′). Consequently if q � q ′ thenC(q) ∼= C(q ′).(2) If K is an extension field of F then there is a canonical isomorphism

C(K ⊗F (V, q)) ∼= K ⊗F C(V, q).

Proof. Exercise 1. ��

The isomorphism class ofC(q) depends only on the isometry class of the quadraticform q. It is natural to ask the converse question: If C(q) ∼= C(q ′) does it follow thatthe quadratic forms q and q ′ are isometric? The answer is “no” in general, but thestudy of those quadratic forms which have isomorphic Clifford algebras is one of themajor themes of this theory.With the universal definition of Clifford algebras given above it is not immedi-

ately clear what the dimensions are. We spend some time presenting a proof that ifdim q = n then dimC(q) = 2n.

3.4 Lemma. (1) C(V, q) is an F -algebra generated by ι(V ).(2) If dim q = n then dimC(q) ≤ 2n.

Proof. (1) Exercise 1.


(2) If {v1, . . . , vn} is an orthogonal basis of V let ei = ι(vi). As mentioned earlier,e1, . . . , en anticommute and e2j = q(vj ) ∈ F . By part (1), C(q) is spanned by the

products eδ11 . . . eδnn where each δi = 0 or 1. There are 2n of these products. ��

3.5 Lemma. Suppose q is a quadratic form on V , A is an F -algebra and ϕ : V → A

is a linear map compatible with q such that A is generated by ϕ(V ). If dim q = n

and dimA = 2n then A ∼= C(q). If such an algebra A exists then dimC(q) = 2n andι : V → C(q) is injective.

Proof. There is an algebra homomorphism ϕ : C(q)→ A such that ϕ = ϕ � ι. Sinceι(V ) generates C(q), ϕ(V ) = ϕ(ι(V )) generates ϕ(C(q)) ⊆ A. By hypothesis ϕ(V )generates A and we see that ϕ is surjective. Then dimC(q) = 2n + dim(ker ϕ) and(3.4) implies that dimC(q) = 2n and ker ϕ = {0}. Therefore ϕ is an isomorphism.If ι is not injective then dim ι(V ) < n and the argument in (3.4) would imply thatdimC(q) < 2n. ��

This criterion can be used to provide some explicit examples of Clifford algebras.If a ∈ F • define the quadratic extension to be F 〈√a〉 = F [x]/(x2 − a). If 〈a〉 �� 〈1〉(that is, a is not a square in F ) this algebra is just the quadratic field extension F(

√a).

However if 〈a〉 � 〈1〉 then F 〈√a〉 ∼= F × F , the direct product of two copies of F .

3.6 Examples. (1) C(〈a〉) ∼= F 〈√a〉, the quadratic extension.(2) C(〈a, b〉) ∼=

(a,bF

), the quaternion algebra.

(3) If q is the zero form on V then C(V, q) = �(V ) is the exterior algebra.

Proof. (1) The space 〈a〉 is given by V = Fe with q(xe) = ax2. If A = F 〈√a〉define ϕ : Fe→ A by ϕ(xe) = x

√a. Then ϕ is compatible with q and (3.5) applies.

(2) The space 〈a, b〉 is given by V = Fe1+Fe2 with q(xe1+ ye2) = ax2+ by2.Define ϕ : V →

(a,bF

)by ϕ(xe1+ ye2) = xi+ yj . Then ϕ is compatible with q and

(3.5) applies.(3) If q = 0 the definition of the Clifford algebra coincides with the definition of

the exterior algebra. ��

Perhaps the best way to prove the general result that dimC(q) = 2n is to developthe theory of graded tensor products, prove that C(α ⊥ β) ∼= C(α) ⊗ C(β) and useinduction. That approach has the advantage that it gives a unified treatment of thetheory, combining the “even” and “odd” cases. Furthermore it is valid for quadraticforms over a commutative ring, provided that the forms can be diagonalized. Gradedtensor products are discussed in the books by Lam and Scharlau, but see the bookletsby Chevalley (1955) and Knus (1988) for further generality.Rather than repeating that treatment, we provide an elementary, direct argument

using the independence result (1.11). This method works only over fields of charac-


teristic not 2, but that is the case of interest here anyway. The quadratic form q maybe singular here.

3.7 Proposition. If q is a quadratic form on V with dim q = n then dimC(q) = 2n

and the map ι : V → C(q) is injective.

Proof. First suppose q is a regular form of even dimension n. Choose an orthogonalbasis {v1, . . . , vn} yielding the diagonalization q � 〈a1, . . . , an〉. ThenC(q) containselements ei = ι(vi) which anticommute and with e2i = ai ∈ F •. Then (1.11) impliesthat the 2n elements e� are linearly independent. Therefore 2n ≤ dimC(q) and (3.4)and (3.5) complete the argument.If n is odd let q ′ = q ⊥ 〈1〉 and choose an isometry ψ : (V , q) → (V ′, q ′).

Then (V ′, q ′) has an orthogonal basis {w1, . . . , wn,wn+1} where wi = ψ(vi) fori = 1, . . . , n. Let ei = ι(wi) in C(q ′). As before (1.11) implies that the elementse� are linearly independent. Let A be the subalgebra generated by e1, . . . , en, so thatdimA = 2n. Then ι � ψ : V → A is compatible with q and (3.5) completes theargument.If (V , q) is singular the same idea works. Choose an isometry ϕ : (V , q) →

(V ′, q ′) where q ′ is regular. (Why does such ϕ exist? See Exercise 20.) First step: if{w1, . . . , wm} is any basis of V ′ (not necessarily orthogonal) and fi = ι(wi) then the2n elements f� are linearly independent. Second step: choose {w1, . . . , wm} so thatwi = ϕ(vi) for 1 ≤ i ≤ n and let A be the subalgebra generated by f1, . . . , fn. ThendimA = 2n and ι � ψ : V → A is compatible with q. ��

Since ι : V → C(V, q) is always injective we simplify the notation by consideringV as a subset of C(V, q). If U ⊆ V is a subspace and q induces the quadratic formϕ on U , then C(U, ϕ) is viewed as a subalgebra of C(V, q). If {e1, . . . , en} is a basisof V then {e� : � ∈ Fn2} is sometimes called the derived basis of C(V, q).If f : (V , q) → (V ′, q ′) is an isometry the universal property implies that there

is a unique algebra homomorphism C(f ) : C(V, q)→ C(V ′, q ′) extending f . Con-sequently, if g ∈ O(V , q) there is an automorphism C(g) of C(V, q) extending g.When g = −1V we get an automorphism α = C(−1V ) of particular importance.

3.8 Definition. The canonical automorphism α of C(V, q) is the automorphism withα(x) = −x for every x ∈ V . Define C0(V , q) to be the 1-eigenspace of α andC1(V , q) to be the (−1)-eigenspace of α. This subalgebra C0(V , q) is called the evenClifford algebra (or the second Clifford algebra) of q.

This notation α for an automorphism should not cause confusion even though wesometimes use α to denote a quadratic form. The meaning of the ‘α’ ought to be clearfrom the context.Note that α2 = α � α is the identity map on C(q) and therefore C(q) = C0(q)⊕

C1(q). Note that α(v1v2 . . . vm) = (−1)mv1v2 . . . vm for any v1, v2, . . . , vm ∈ V .


Therefore C0(V , q) is the span of all such products where m is even. Suppose{e1, e2, . . . , en} is an orthogonal basis of V corresponding to q � 〈a1, a2, . . . , an〉.Then

α(e�) = (−1)|�|e�.

Therefore {e� : |�| is even} is a basis of C0(q) while {e� : |�| is odd} is a basis ofC1(q). In particular, dimC0(q) = dimC1(q) = 2n−1.Furthermore,C0 ·C1 ⊆ C1 andC1 ·C1 ⊆ C0. If u ∈ C1(q) is an invertible element

then C1(q) = C0(q) · u. The next lemma shows that this subalgebra C0(q) can itselfbe viewed as a Clifford algebra, at least if q is regular.

3.9 Lemma. If q � 〈a〉 ⊥ β and a �= 0 then C0(q) � C(〈−a〉β).

Proof. Let {e1, e2, . . . , en} be an orthogonal basis of V where e1 corresponds to〈a〉. Then the elements e1ei ∈ C0(q) anticommute and (e1ei)2 = −a · q(ei) fori = 2, . . . , n. Then the inclusion map from W = span{e1e2, . . . , e1en} to C0(q)is compatible with the form 〈−a〉β on W and the universal property provides analgebra homomorphism ϕ : C(〈−a〉β) → C0(q). Since a �= 0 the elements e1eigenerate C0(q) so that ϕ is surjective. Counting dimensions we conclude that ϕ is anisomorphism. ��

Let us now restrict attention again to regular quadratic forms. The next goal is todefine the “Witt invariant” c(q) of a regular quadratic form and to derive some of itsproperties. The first step is to prove that if q has even dimension thenC(q) is a centralsimple algebra. Then c(q) will be defined to be the class of C(q) in the Brauer groupBr(F ).To begin this sequence of ideas we determine the centralizer of C0(q) in C(q).

The argument here is reminiscent of the proof of (1.11).

3.10 Definition. If {e1, . . . , en} is an orthogonal basis of (V , q) define the elementz(V, q) = e1e2 . . . en ∈ C(V, q).Define the subalgebra Z(V, q) = span{1, z(V , q)} ⊆ C(V, q).

We sometimes write z(q) or z(V ) for the element z(V, q). If e2i = ai ∈ F • thenz(q)2 = (e1 . . . en) · (e1 . . . en) = (−1)n(n−1)/2a1 . . . an. Abusing notation slightlywe have

z(q)2 = dq and Z(q) ∼= F 〈√dq〉.

Then if dq �� 〈1〉 the subalgebra Z(q) ∼= F(√dq) is a field. If dq � 〈1〉 then

Z(q) ∼= F × F . From the next proposition it follows that Z(q) is independent ofthe choice of basis. Furthermore, the element z(q) is unique up to a non-zero scalarmultiple.

3.11 Proposition. Suppose q is a regular form on V and dim q = n.


(1) Z(V, q) is the centralizer of C0(V , q) in C(V, q).

(2) The center of C(V, q) is

{F if n is even,Z(V, q) if n is odd.

(3) The center of C0(V , q) is

{Z(V, q) if n is even,F if n is odd.

Proof. (1) Suppose c is an element of that centralizer. Let {e1, e2, . . . , en} be theorthogonal basis used in (3.10) and express c = ∑

c�e� for coefficients c� ∈ F .

Since c commutes with every eiej and (eiej )−1e�(eiej ) = ±e� it follows that ifc� �= 0 then e� commutes with every eiej . If� = (δ1, . . . , δn) and δr �= δs for somer , s then eres anticommutes with e�. Therefore either � = (0, . . . , 0) and e� = 1or � = (1, . . . , 1) and e� = z(q). Hence c is a combination of 1 and z(q) so thatc ∈ Z(q).(2) If c is in the center then c ∈ Z(q) by part (1). If n is even every ei commutes

with z(q) and the center is Z(q). If n is odd every ei anticommutes with z(q) and thecenter is F .(3) z(q) is in C0(q) if and only if n is even. ��

3.12 Lemma. If n is odd and q is regular then C(V, q) ∼= C0(V , q)⊗ Z(V, q).

Proof. Since C0(q) and Z(q) are subalgebras of C(q) which centralize each otherthere is an induced algebra homomorphism ψ : C0(q) ⊗ Z(q) → C(q). Since n isodd these two subalgebras generateC(q) so thatψ is surjective. Counting dimensionswe see that ψ is an isomorphism. ��

3.13 Structure Theorem. Let C = C(V, q) be the Clifford algebra of a regularquadratic space over the field F . Let C0 = C0(V , q) and Z = Z(V, q).

(1) If dim V is even then Cis a central simple algebra over F , and C0 has center Z.

(2) If dim V is odd then C0 is a central simple algebra over F and C ∼= C0 ⊗ Z.

If dq �� 〈1〉 then C is a central simple algebra over Z. If dq = 〈1〉 thenC ∼= C0 × C0.

Proof. Suppose n = dim V so that dimC = 2n. The centers of these algebras aregiven in (3.11).(1) Suppose n is even and I is a proper ideal ofC so that C = C/I is an F -algebra

with dim C = 2n−dim I. If {e1, . . . , en} is an orthogonal basis of (V , q), the imagese1, . . . , en are anticommuting invertible elements of C. Since n is even (1.11) impliesthat dim C ≥ 2n and therefore I = 0. Hence C is simple.(2) Apply (3.9), part (1) and (3.12). The final assertions follow since Z is a field

if dq �� 〈1〉 and Z ∼= F × F if dq � 〈1〉. ��


If q is a regular quadratic form of even dimension then C(q) is a central simplealgebra. In fact it is isomorphic to a tensor product of quaternion algebras. Beforebeginning the analysis of the Witt invariant we describe an explicit decomposition ofC(q) which will be useful later on. This decomposition provides another proof thatC(q) is central simple, since the class of central simple F -algebras is closed undertensor products.

3.14 Proposition. If q � 〈a1, . . . , a2m〉 define uk = e1e2 . . . e2k−1 and vk = e2k−1e2kfor k = 1, 2, . . . , m. The subalgebra Qk generated by uk and vk is a quaternionalgebra and C(q) ∼= Q1 ⊗ · · · ⊗Qm.

Proof. Check thatuk anticommuteswith vk but commuteswith everyui andwith everyvj for j �= k. Since u2k = (−1)k−1a1a2 . . . a2k−1 and v2k = −a2k−1a2k are scalars,each Qk is a quaternion subalgebra. The induced map Q1 ⊗ · · · ⊗ Qm → C(q)

is injective since the domain is simple. By counting dimensions we see it is anisomorphism. ��

IfA is anF -algebra, the “opposite algebra”Aop is the algebra defined as the vectorspace A with the multiplication ∗ given by: a ∗ b = ba. An algebra isomorphismϕ : A→ Aop can be interpreted as an anti-automorphism ϕ : A→ A. That is, ϕ is avector space isomorphism and ϕ(ab) = ϕ(b)ϕ(a) for every a, b ∈ A. If (V , q) is aquadratic space and g ∈ O(V , q) then the map ι � g : V → C(V, q)op is compatiblewith q. The universal property provides a homomorphism ϕ : C(V, q)→ C(V, q)op.This map is surjective and therefore it is an isomorphism. As above we may interpretthis map as an anti-automorphism

C′(g) : C(V, q)→ C(V, q).

This is the unique anti-automorphism of C(V, q) which extends g : V → V .The involutions of C(V, q) will be particularly important for our work. By defini-

tion, an involution of an F -algebra is an F -linear anti-automorphism whose square isthe identity. (These are sometimes called “involutions of the first kind”.) For example,the transpose map on the matrix algebra Mn(F ) and the usual “bar” on a quaternionalgebra are involutions. If g ∈ O(V , q) satisfies g2 = 1V then C′(g) is an involutionon C(V, q).

3.15 Definition. If V = R ⊥ T , define JR,T to be the involution on C(V, q) whichis −1R on R and 1T on T . The canonical involution (denoted J0 or � ) is J0,V andthe bar involution (denoted J1 or ¯ ) is JV,0.

This JR,T is the anti-automorphism of C = C(V, q) extending the reflection(−1R)⊥ (1T ) on V = R ⊥ T . These involutions will be important when we analyze(s, t)- families using Clifford algebras.


The canonical involution acts as the identity on V . If v1, v2, . . . , vm ∈ V then(v1v2 . . . vm)

� = vm . . . v2v1, reversing the order of the product. A simple signcalculation shows that (e�)� = (−1)k(k−1)/2e� where k = |�|. The bar involutionis the composition of α and �, and similar remarks hold. For the quaternion algebras,the bar involution is the usual “bar”.For the discussion of the Witt invariant below we assume some familiarity with

the theory of central simple algebras. If A is a central simple algebra over F , thenWedderburn’s theory implies that A is a matrix ring over a division algebra. Thatis, A ∼= Mk(D) where D is a central division algebra over F , which is uniquelydetermined (up to isomorphism) byA. Two central simple algebrasA,B are equivalent(A ∼ B) if their corresponding division algebras are isomorphic. That is, A ∼ B ifand only if Mm(A) ∼= Mn(B) for some m, n. Let [A] denote the equivalence classof a central simple algebra A. Each such class contains a unique division algebra.The Brauer group Br(F ) is the set of these equivalence classes, with a multiplicationinduced by ⊗.The inverse of [A] in Br(F ) is [Aop], the class of the opposite algebra. (For there

is a natural isomorphism A ⊗ Aop ∼= EndF (A).) Since quaternion algebras possessinvolutions, [A]2 = 1 whenever A is a tensor product of quaternion algebras. LetBr2(F ) be the subgroup of Br(F ) consisting of all [A] with [A]2 = 1.

3.16 Definition. Let (V , q) be a (regular) quadratic space over F . TheWitt invariantc(q) is the element of Br2(F ) defined as:

c(q) ={[C(V, q)] if dim V is even,[C0(V , q)] if dim V is odd.

By the Structure Theorem the indicated algebras are central simple, so c(q) is well-defined. To derive the basic properties of c(q) we must recall some of the propertiesof quaternion algebras. The class of the quaternion algebra is written [a, b].

3.17 Lemma. (1) [a, b] = [a′, b′] if and only if 〈〈−a,−b〉〉 � 〈〈−a′,−b′〉〉. Inparticular, [a, b] = 1 if and only if 〈〈−a,−b〉〉 is hyperbolic if and only if 〈a, b〉represents 1.(2) [a, b] = [b, a], [a, a] = [−1, a], [a, b]2 = 1 and [a, b] · [a, c] = [a, bc].

Proof. (1) This is a standard result about quaternion algebras: the isomorphism class

of the algebra(a,bF

)is determined by the isometry class of its norm form 〈〈−a,−b〉〉.

In particular the symbol [a, b] depends only on the square classes 〈a〉 and 〈b〉.(2) The first and second statements follow from (1), and the third one is clear. The

last statement is equivalent to the isomorphism:(a, bF

)⊗(a, cF

) ∼=(a, bcF

)⊗ M2(F ).


To prove this let i, j be the generators of the algebra B =(a,bF

)so that i2 = a

and j2 = b. Similarly let i′, j ′ be the generators of the algebra C =(a,cF

)so that

i′2 = a and j ′2 = c. Then the subalgebra of B ⊗ C generated by i ⊗ 1 and j ⊗ j ′ isisomorphic to

(a,bcF

)and the subalgebra generated by i⊗ i′ and 1⊗ j ′ is isomorphic

to(a2,cF

) ∼= M2(F ). Since those subalgebras centralize each other and together span

the algebra B ⊗ C the claim follows. ��

The same sorts of arguments are used to prove the various isomorphisms of Cliffordalgebras stated below.

3.18 Lemma. (1) If dim α and dim β are even then c(α ⊥ β) = c(α) · c((dα) · β).(2) If dim α is odd and dim β is even then c(α ⊥ β) = c(α) · c((−dα) · β).

Proof. (1) We must prove that C(α ⊥ β) ∼= C(α) ⊗ C((dα) · β). In fact thisholds whenever dim α is even. Let v1, . . . , vr , w1, . . . , ws be an orthogonal basiscorresponding to α ⊥ β. The subalgebra of C(α ⊥ β) generated by {v1, . . . , vr} isisomorphic to C(α). Since r is even the element u = z(α) = v1 . . . vr anticommuteswith each vi , commutes with each wj and 〈u2〉 = dα. Then the subalgebra generatedby {uw1, . . . , uws} is isomorphic to C((dα)β) and centralizes C(α). Since thesesubalgebras together generate the whole algebra C(α ⊥ β) the isomorphism follows.(2) We must show that C0(α ⊥ β) ∼= C0(α) ⊗ C((dα) · β). In fact this holds

whenever dim α is odd. Either use an argument similar to (1) above or apply (1) andLemma 3.9. We omit the details. ��

By applying (3.18) (1) successively to q � 〈a1, . . . , a2m〉 we find that C(q) is iso-morphic to a tensor product of quaternion subalgebras. These are the same subalgebrasfound in (3.14) above.

3.19 Proposition. Let α, β be (regular) quadratic forms over F and let x, y, z ∈ F •.

(1) c(α ⊥ β) ={c(α) · c(β) · [dα, dβ] if dim α and dim β are both even

or both odd,c(α) · c(β) · [−dα, dβ] if dim α is odd and dim β is even.

(2) c(〈x〉α) ={c(α)[x, dα] if dim α is even,c(α) if dim α is odd.

(3) c(α ⊥ H) = c(α) where H = 〈1,−1〉 is the hyperbolic plane.

(4) c(〈〈x〉〉 ⊗ α) = [−x, dα]. Hence c(〈〈x, y〉〉) = [−x,−y] and c(〈〈x, y, z〉〉) = 1.

Proof. (3) follows immediately from (3.18) when β = H. To prove (2) first note thatC0(〈x〉q) ∼= C0(q) for any form q. This settles the case dim α is odd. For the evencase of (2) we apply (3.18) in two ways: c(α ⊥ 〈−1, x〉) = c(α) · c((dα) · 〈−1, x〉) =


c(〈−1, x〉) · c(〈x〉 · α). Therefore c(〈x〉 · α) = c(α)[−dα, (dα) · x] · [−1, x] =c(α) · [dα, x], using the properties in (3.17).(1) If dim α and dim β are even then c(α ⊥ β) = c(α) · c((dα) ·β) = c(α) · c(β) ·

[dα, dβ] by (3.18) and part (2). A similar argument works when dim α is odd anddim β is even. Suppose dim α and dim β are odd. One way to proceed is to expressα = 〈a〉 ⊥ α′ and β = 〈b〉 ⊥ β ′ so that c(α ⊥ β) = c(〈a, b〉 ⊥ α′ ⊥ β ′) andc(α) = c(〈−a〉α′), c(β) = c(〈−b〉β ′) by (3.9). The desired equality follows afterexpanding both sides using the properties for forms of even dimension. Alternatively

we can prove the isomorphism C(α ⊥ β) ∼= C0(α) ⊗ C0(β) ⊗(dα,dβF

)by directly

examining basis elements. Further details are left to the reader.The formula for c(〈〈x〉〉 ⊗ α) = c(α ⊥ 〈x〉α) in (4) follows from (1) and (2). ��

Recall that if a (regular) quadratic form q is isotropic, then q � H ⊥ α for someform α. A form ϕ is hyperbolic if ϕ � mH � H ⊥ H ⊥ · · · ⊥ H. Then everyform q can be expressed as q � q0 ⊥ H where q0 is anisotropic andH is hyperbolic.(This is the “Witt decomposition” of q.) Witt’s Cancellation Theorem implies thatthis form q0 is uniquely determined by q (up to isometry). Two forms α and β areWitt equivalent (written α ∼ β) if α ⊥ −β is hyperbolic. If dim α > dim β andα ∼ β then α � β ⊥ H for some hyperbolic spaceH . Consequently every Witt classcontains a unique anisotropic form. These classes form the elements of the Witt ringW(F ), where the addition is induced by ⊥ and the multiplication by ⊗.We often abuse the notations, using symbols like q, ϕ, α to stand for regular

quadratic forms, and writing q ∈ W(F) rather than stating that the Witt class of q liesinW(F).Define IF to be the ideal in the Witt ring W(F ) consisting of all quadratic forms

of even dimension. (This is well-defined since dimH is even.) Then IF is additivelygenerated by all the forms 〈〈a〉〉 = 〈1, a〉 for a ∈ F •. The square I 2F is additivelygenerated by the 2-fold Pfister forms 〈〈a, b〉〉, and similarly for higher powers.The determinant det α ∈ F •/F •2 does not generally induce a map on W(F ).

The “correct” invariant is the signed discriminant dα = (−1)n(n−1)/2 det α, becaused(α ⊥ H) = dα. This discriminant induces a map d : W(F )→ F •/F •2. The idealI 2F is characterized by this discriminant: α ∈ I 2F if and only if dim α is even anddα = 〈1〉.The Witt invariant c(q) induces a map c : W(F) → Br(F ). The formulas in

(3.19) imply that the restriction c : I 2F → Br(F ) is a homomorphism. One naturalquestion is: Which quadratic forms q have all three invariants trivial? That is:

dim q = even, dq = 〈1〉, c(q) = 1.

It is easy to check that any 3-fold Pfister form 〈〈a, b, c〉〉 has trivial invariants. Con-sequently so does everything in the ideal I 3F . For convenience we let J3(F ) be theideal of elements inW(F ) which have trivial invariants. That is:

J3(F ) = ker(c : I 2F → Br(F )).


Does J3(F ) = I 3F for every fieldF ? This was amajor open question untilMerkurjevproved it true in 1981using techniques fromK-theory (which arewell beyond the scopeof this book). Before 1981 the only result in this direction valid over arbitrary fieldswas the one of Pfister (1966): if q ∈ J3(F ) and dim q ≤ 12 then q ∈ I 3F . Animportant tool used in the proof is the following easy lemma about the behavior offorms under quadratic extensions.

3.20 Lemma. Let q be an anisotropic quadratic form over F .

(1) q ⊗ F(√d) is isotropic iff q � 〈x〉〈1,−d〉 ⊥ α for some x ∈ F • and some formα over F .

(2) q ⊗ F(√d) is hyperbolic iff q � 〈〈−d〉〉 ⊗ β for some form β over F .

Proof. Exercise 8. ��

With this lemma, and some clever arguments with quaternion algebras, Pfister(1966) characterized the small forms in I 3F up to isometry.

3.21 Pfister’s Theorem. Let ϕ be a regular quadratic form over F with dim ϕ even,dϕ = 〈1〉 and c(ϕ) = 1.

(1) If dim ϕ < 8 then ϕ ∼ 0.

(2) If dim ϕ = 8 then ϕ � 〈a〉〈〈x, y, z〉〉 for some a, x, y, z ∈ F •.(3) If dim ϕ = 10 then ϕ is isotropic.

(4) If dim ϕ = 12 then ϕ � 〈〈x〉〉 ⊗ δ for some x ∈ F • and some quadratic form δ

where dim δ = 6 and dδ = 〈1〉. Furthermore if 〈a〉〈〈−b〉〉 ⊂ ϕ then ϕ � ϕ1 ⊥ϕ2 ⊥ ϕ3 where dim ϕi = 4, dϕi = 〈1〉 and 〈a〉〈〈−b〉〉 ⊂ ϕ1.

The proof is given in Exercises 9 and 10.The three basic invariants described above induce group homomorphisms

dim : W(F)/IF → Z/2Z, d : IF/I 2F → F •/F •2, c : I 2F/I 3F → Br2(F ).

The first two maps above are easily seen to be isomorphisms. Milnor (1970) conjec-tured that these maps are the first three of a sequence of well defined isomorphismsen : InF/In+1F → Hn(F ), where Hn is a suitable Galois cohomology group.Many mathematicians have worked on various aspects of these conjectures. Merkur-jev (1981) proved that e2 = c is always an isomorphism. Recently Voevodsky provedthat Milnor’s conjecture is always true. For an outline of these ideas and furtherreferences see Morel (1998).



1. Universal property. (1) If f : (V , q) → (V ′, q ′) is an isometry then there is aunique algebra homomorphism ψ : C(V, q) → C(V ′, q ′) such that ψ � ι = ι′ � f .If f is bijective then ψ is an isomorphism. The uniqueness of the Clifford algebrafollows.(2) 1 ⊗ ι : K ⊗ V → K ⊗ C(V, q) is compatible with K ⊗ q and satisfies the

universal property. This proves (3.3) (2).(3) C(V, q) is generated as an F -algebra by ι(V ). This proves (3.4) (1):

(Hint. (3) Apply the proof of (3.2). Or directly from the definitions: Let A be thesubalgebra generated by ι(V ) so that ι induces a map ι′ : V → Awhich is compatiblewith q. Apply the definition of C(q) to get an induced algebra homomorphism ψ :C(q)→ C(q) with ψ � ι = ι. The uniqueness implies ψ = 1C.)

2. Homogeneous components. (1) Let {e1, e2, . . . , en} be an orthogonal basis of(V , q), and define the subspace V(k) = span{e� : |�| = k}. Then dim V(k) =

(n

k

).

For instance, V(0) = F and V(1) = V . Each V(k) is independent of the choice oforthogonal basis.(2) SinceV(n) = F ·z(q), the element z(q) inDefinition 3.10 is uniquely determined

up to a non-zero scalar multiple. For any subspace U ⊆ V , the line spanned by z(U)is uniquely determined.

(Hint. (1) Use the Chain-Equivalence Theorem stated in Exercise 2.8.)

3. Prove (3.19) (1), (2) by exhibiting explicit algebra isomorphisms, similar to theproof of (3.18). For instance if dim α and dim β are odd,

C(α ⊥ β ⊥ H) ∼= C(α)⊗ C(β)⊗(dα, dβ

F

).

4. Discriminants. Suppose dim α = m and dim β = n.(1) d(α ⊥ β) = (−1)mndα · dβ. In particular d(α ⊥ H) = dα.(2) d(〈c〉α) = 〈c〉m · dα.(3) d(α ⊗ β) = (dα)n · (dβ)m.(4) If q � p〈1〉 ⊥ r〈−1〉 define the signature sgn(q) = p − r . Then

dq ={ 〈1〉 if sgn(q) ≡ 0, 1〈−1〉 if sgn(q) ≡ 2, 3

(mod 4).

5. Witt invariant calculations. (1) c(〈a, b〉)= [a, b], and c(〈a, b, c〉)= [−ab,−ac].(2) If dα = dβ, c(α) = c(β) and dim α = dim β ≤ 3 then α � β.(3) Let β = 〈1〉 ⊥ β1. Then d(β) = d(−β1) and c(β) = c(−β1).


(4) c(α ⊗ β) ={ [dα, dβ] if dim α and dim β are even,c(α)c(β) if dim α and dim β are odd,c(β)[dα, dβ] if dim α is odd and dim β is even.

(5) If q � p〈1〉 ⊥ r〈−1〉 define the signature sgn(q) = p − r . Then

c(q) ={1 if sgn(q) ≡ −1, 0, 1, 2[−1,−1] if sgn(q) ≡ 3, 4, 5, 6

(mod 8).

6. Hasse invariant. If q � 〈a1, . . . , an〉 define the Hasse invarianth(q) =

∏i≤j[ai, aj ].

(1) If dim q = n then h(q) = c(q ⊥ n〈−1〉). Consequently h(q) depends only onthe isometry class of q.(2) This independence of h(q) can be proved without Clifford algebras. Use the

Chain-Equivalence Theorem stated in Exercise 2.8.(3) h(α ⊥ β) = h(α) · h(β) · [det α, det β].(4) Another version of the Hasse invariant is given by s(q) =∏

i<j [ai, aj ]. Howare h(q) and s(q) related?

(Hint. (1) Apply (3.14) to q ⊥ n〈−1〉 � 〈a1,−1, a2,−1, . . . , an,−1〉.)

7. (1) Prove from the definition of I 2F that:If dim q is even then q ≡ 〈1,−dq〉 (mod I 2F).If dim q is odd then q ≡ 〈dq〉 (mod I 2F).(2) q ∈ I 2F if and only if dim q is even and dq = 〈1〉.(3) If dim q is even and 〈a〉q � q then a ∈ DF (〈〈−dq〉〉).(4) If α is a form of even dimension then 〈〈dα〉〉 ⊗ α ∈ I 3F .(5) Suppose q ≡ 〈〈x〉〉 ⊗ α (mod I 3F), where dq = 〈1〉 and c(q) = 1. Then

q ∈ I 3F .(Hint. (3) Use Witt invariants. Compare Exercise 2.9(3).)

8. Quadratic extensions. (1) Prove Lemma 3.20.(2) If A is a central simple F -algebra and K = F(

√d), then:

A⊗K ∼ 1⇐⇒ A ∼(d, xF

)for some x ∈ F •.

(3) c(q) = quaternion ⇐⇒ c(qK) = 1 in Br(K) for some quadratic extensionK/F .

(Hint. (1) If q(v) = 0 for v ∈ V ⊗ K , express v = v0 + v1√d and conclude:

q(v0) + dq(v1) = 0 and Bq(v0, v1) = 0. The second statement follows by repeatedapplication of the first.


(2) The theory of division algebras implies: If D is a central F -division algebraof degree n and K is a splitting field, then [K : F ] ≥ n. A central simple algebras ofdegree 2 must be quaternion.)

9. Trivial invariants. Suppose ϕ is a quadratic form over F .(1) If dim ϕ = 2, 4 or 6, dϕ = 〈1〉 and c(ϕ) = 1, then ϕ ∼ 0.(2) Suppose dim ϕ = 6, dϕ = 〈1〉 and c(ϕ) = quaternion, then ϕ is isotropic.(3) Prove the cases where dim ϕ ≤ 10 in (3.21).

(Hint. (1) If dim ϕ = 6, express ϕ � 〈a〉〈1,−b〉 ⊥ ψ . If K = F(√b) then

ϕK ∼ 0 because ψK has trivial invariants over K . If ϕ is anisotropic, (3.20) impliesϕ � 〈〈−b〉〉 ⊗ β. Discriminants show that 〈b〉 = 〈1〉 so that ϕ ∼ 0.(2) c(ϕ) is split over some K = F(

√b). Finish as in part (1).

(3) Let dim ϕ = 8, anisotropicwith trivial invariants. Expressϕ � 〈a〉〈〈−b〉〉 ⊥ ψ ,so that dψ = 〈b〉 and c(ψ) = [−a, b]. Part (1) over F(√b) and (3.20) imply thatψ � 〈〈−b〉〉⊗〈u, v,w〉. Witt invariants show that [b, auvw] = 1 and 〈〈−b〉〉 representsauvw. Therefore ψ � 〈〈−b〉〉 ⊗ 〈u, v, auv〉 and ϕ � 〈a〉〈〈−b, au, av〉〉.Let dim ϕ = 10, anisotropic with trivial invariants. Express ϕ � 〈w〉〈1, a, b, c〉 ⊥

ψ . Then dimψ = 6, dψ = 〈abc〉. Let K = F(√abc) so that c(ψK) = quaternion.

Part (2) implies ψK is isotropic, so that ψ � 〈u〉〈〈−abc〉〉 ⊥ δ where dim δ = 4and dδ = 〈1〉. Then ϕ ∼= ω ⊥ δ where ω = 〈w〉〈1, a, b, c〉 ⊥ 〈u〉〈〈−abc〉〉. Sincedω = 〈1〉 and c(ω) = c(−δ) is quaternion, (2) leads to a contradiction.)

10. Linked Pfister forms. If a Pfister form ϕ � 〈1〉 ⊥ ϕ′ then ϕ′ is called thepure part of ϕ. For instance, if ϕ = 〈〈a, b〉〉 then ϕ′ = 〈a, b, ab〉. In this case, if ϕ′represents d then we may use d as one of the “slots”: ϕ � 〈〈d, x〉〉 for some x. The2-fold Pfister forms ϕ and ψ are linked if there is a common slot: ϕ � 〈〈a, x〉〉 andψ � 〈〈a, y〉〉 for some a, x, y ∈ F •. This occurs if and only if ϕ′ and ψ ′ represent acommon value, if and only if ϕ′ ⊥ −ψ ′ is isotropic.(1) ϕ and ψ are linked⇐⇒ c(ϕ)c(ψ) = quaternion.(2) Suppose ψ1, ψ2, ψ3 are 2-fold Pfister forms and c(ψ1)c(ψ2)c(ψ3) = 1. Then

ψ1 � 〈〈a, x〉〉, ψ2 � 〈〈a, y〉〉 and ψ3 � 〈〈a,−xy〉〉 for some a, x, y ∈ F •.(3) Suppose β is anisotropic, dim β = 8, dβ = 〈1〉 and c(β) = quaternion. Then

β � 〈〈a〉〉 ⊗ γ for some a ∈ F • and some form γ .(4) Complete the proof of (3.21).(5) Let Q1 and Q2 be quaternion algebras with norm forms ϕ1 and ϕ2. Let

α = ϕ′1 ⊥ −ϕ′2. ThenQ1 ⊗Q2 is a division algebra if and only if α is anisotropic.

(Hint. (1) q = ϕ′ ⊥ −ψ ′ has dim q = 6, dq = 〈1〉 and c(q) = c(ϕ)c(ψ). ApplyExercise 9(2).(2) c(ψ1)c(ψ2) = c(ψ3) and (1) imply ψ1 � 〈〈a, x〉〉, ψ2 � 〈〈a, y〉〉 for some a, x,

y. Witt invariants then imply ψ3 � 〈〈a,−xy〉〉.(3) Let β = 〈x〉〈〈−b〉〉 ⊥ δ. Exercise 9(2) over K = F(

√b) and (3.20) imply

〈y〉〈〈−b〉〉 ⊂ δ for some y. Then β � ϕ1 ⊥ ϕ2 where dim ϕi = 4 and dϕi = 〈1〉.


(Here ϕ1 � 〈〈−b〉〉⊗〈x, y〉.) Express ϕi � 〈xi〉ψi for some xi and some 2-fold Pfisterforms ψi . Apply part (1).(4) Suppose dim ϕ = 12, and ϕ is anisotropic with trivial invariants. Express

ϕ = 〈a〉〈〈−b〉〉 ⊥ α. By (3.21) (3), α is isotropic overK = F(√b). Then ϕ � ϕ1 ⊥ β

where ϕ1 = 〈〈−b〉〉 ⊗ 〈a, u〉, dim β = 8, dβ = 〈1〉 and c(β) = [b,−au]. Part (3)implies β = ϕ2 ⊥ ϕ3 where dim ϕi = 4 and dϕi = 〈1〉. Express ϕi = 〈xi〉ψi andapply part (2).)

11. Graded algebras. Let A be an associative F -algebra with 1. Then A is graded(or more precisely, “Z/2Z-graded”) ifA = A0⊕A1, a direct sum as F -vector spaces,such thatAi ·Aj ⊆ Ai+j , where the subscripts are taken mod 2. It follows that 1 ∈ A0and A0 is a subalgebra of A. Every Clifford algebra C(q) is a graded algebra usingC(q) = C0(q)⊕C1(q). An element of A is homogeneous if it lies in A0 ∪A1. If a ishomogeneous define the degree ∂(a) = i if a ∈ Ai . The graded F -algebras A and Bare graded-isomorphic, if there exists f : A→ B which is anF -algebra isomorphismsatisfying f (Ai) = Bi for i = 0, 1.Define the graded tensor product A⊗B by taking the vector spaceA⊗B with the

new multiplication induced by: (a ⊗ b) · (a′ ⊗ b′) = (−1)∂(a)∂(b)aa′ ⊗ bb′. ThenA ⊗ B is a graded algebra with (A ⊗ B)0 = A0 ⊗ B0 + A1 ⊗ B1 and (A ⊗ B)1 =A1 ⊗ B0 + A0 ⊗ B1. Then A ⊗ 1 and 1 ⊗ B are graded subalgebras of A ⊗ B.In the category of graded F -algebras the graded tensor product is commutative andassociative, and it distributes through direct sums.(1)Lemma. If α, β are quadratic forms thenC(α ⊥ β) ∼= C(α)⊗C(β) as graded

algebras.(2) Lemma. The Clifford algebras C(α) and C(β) are graded-isomorphic if and

only if

dim α = dim β, dα = dβ and c(α) = c(β).

(3) LetA be a graded F -algebra. AnA-module V is a gradedA-module if V has adecompositionV = V0⊕V1 such thatAi ·Vj ⊆ Vi+j whenever i, j ∈ Z/2Z. LetC bethe Clifford algebra of some regular quadratic form. If V is a graded C-module thenV ∼= C ⊗C0 V0. Thus there is a one-to-one correspondence: {graded C-modules} ↔{C0-modules}.12. 4-dimensional forms. Suppose dim α = 4, dα = 〈d〉, and L = F(

√d).

(1) c(α ⊗ L) = 1 iff α is isotropic.(2) c(α) = quaternion iff α ⊥ 〈−1, d〉 is isotropic.(3) Suppose dim α = dim β = 4, dα = dβ = 〈d〉, and L = F(

√d).

Then α and β are similar over F if and only if α⊗L and β⊗L are similar over L.(4) If dim α = dim β = 4, dα = dβ and c(α) = c(β) then α and β are similar.(5) If α and β are 4-dimensional forms then α, β are similar if and only if

C0(α) ∼= C0(β).


(Hint. (3) Scale α, β to assume α � 〈1〉 ⊥ α′ and β � 〈1〉 ⊥ β ′. Since αL andβL are similar Pfister forms they are isometric and Exercises 8(3) and 9(2) imply thatα′ ⊥ −β ′ is isotropic (see Exercise 10). If α � 〈〈u〉〉 ⊥ 〈a〉〈〈ud〉〉 and β � 〈〈u〉〉 ⊥〈b〉〈〈ud〉〉 then by (3.20), 〈1,−ab, u,−abud〉 is isotropic. Choose x ∈ DF (〈〈u〉〉) ∩DF (〈ab〉〈〈ud〉〉), to obtain 〈x〉α � β.(4) If 〈d〉 �= 〈1〉 apply (3).)

13. (1) Suppose dim ϕ = 5. Then ϕ ⊂ 〈a〉ρ for some 3-fold Pfister form ρ if andonly if ϕ represents dϕ, if and only if c(ϕ) = quaternion.(2) Suppose dim ϕ = 6. Then ϕ ⊂ 〈a〉ρ for some 3-fold Pfister form ρ if and

only if ϕ � 〈〈x〉〉 ⊗ δ for some x ∈ F • and some form δ if and only if c(ϕ) is split byF(

√dϕ).

14. Trace forms. Let C = C(V, q) be a Clifford algebra of dimension 2m. Definethe “trace” map � : C → F to be the scalar multiple of the regular trace having�(1) = 1. Then for a derived basis {e�} we have �(e�) = 0 whenever� �= 0. Recallthat “bar” = J0 is the involution extending −1V . Define the associated trace formB0 : C × C → F by B0(x, y) = �(xy).(1) B0 is a regular symmetric bilinear form on C. If q � 〈a1, . . . , am〉 then

(C, B0) � 〈〈−a1, . . . ,−am〉〉. In particular the isometry class of this Pfister form isindependent of the basis chosen for q.(2) Suppose β � 〈b1, . . . , bm〉 and define P(β) = 〈〈b1, . . . , bm〉〉, the associated

Pfister form.

Lemma. If β � γ then P(β) � P(γ ).

This follows from part (1). Also prove it usingWitt’s Chain-Equivalence Theorem(Exercise 2.8), without mentioning Clifford algebras.(3) Define Pi(β) to be the “degree i” part of the Pfister form P(β), so that

dim Pi(β) =(mi

). For example, P0(β) = 〈1〉, P1(β) = β and P2(β) = 〈b1b2, b1b3,

. . . , bm−1bm〉. The lemma generalizes: If β � γ then Pi(β) � Pi(γ ) for each i.(4) If C = C(−α ⊥ β) and J = JA,B is the involution extending (−1) ⊥ (1) on

−α ⊥ β, define the trace form B as before. Then (C, B) � P(α ⊥ β).

15. More trace forms. LetC = C(V, q)with the associated trace formB0 : C×C →F defined in Exercise 14.(1) Let L and R : C → EndF (C) be the left and right regular representa-

tions. If c ∈ C then L(c) is a similarity of (C, B0) if and only if cc ∈ F . Con-sequently L(F + V ) ⊆ Sim(C, B0) is a subspace of dimension m+ 1. Similarly forR(F+V ). These two subspaces can be combined to provide an (m+1,m+1)-family(L(F + V ),R(F +V ) � α) where α is the canonical automorphism of C. How doesthis compare to the Construction Lemma 2.7?(2) Clifford and Cayley. Let C = C(〈a1, a2, a3〉) be an 8-dimensional Clifford

algebra. Let U = span{1, e1, e2, e3} so that C = U + Uz. Shift the (4, 4)-familyconstructed in (1) to an (8, 0)-family and identify it with C as in (1.9). This provides


a new multiplication " on C given by: (x + y) " c = xc+ α(c)y, for x ∈ U , y ∈ Uzand c ∈ C. Using N0(c) = B0(c, c) we have N0(ab) = N0(a)N0(b). This " definesan octonion algebra. Compare the multiplication tables of the two algebras to see thatthey differ only by a few ± signs.(3) LetW ⊆ C be a linear subspace spanned by elements of degree≡ 2, 3 (mod 4).

If ww = ww ∈ F for every w ∈ W then L(W) + R(W) � α ⊆ Sim(C, B0). Forexample we find 〈〈a1〉〉 ⊗ 〈1, a2, . . . , am〉 < Sim(〈〈a1, . . . , am〉〉).

16. Clifford division algebras. Let α = 〈1〉 ⊥ α1 where dim α = m + 1, and letC = C(−α1). Then dimC = 2m. Note that dα = d(−α1).(1) Here are necessary and sufficient conditions for C to be a division algebra:m = 1: α is anisotropic.m = 2: α is anisotropic.m = 3: α and 〈1,−dα〉 are anisotropic.m = 4: α ⊥ 〈−dα〉 is anisotropic.m = 5: 〈1,−dα〉 is anisotropic and α ⊗ F(

√da) is anisotropic.

No similar result is known for m = 6.(2) Let q be a form over F and t an indeterminate. Then C(q ⊥ 〈t〉) is a division

algebra over F(t) if and only if C(q) is a division algebra over F .

(Hint. (1) Form = 4 use Exercise 10(5). (2) IfA = C(q) overF , letA(t) = A⊗F(t).ThenC = C(q ⊥ 〈t〉) = A(t)⊕A(t)e where e2 = t and e−1xe = α(x) for x ∈ A(t).SupposeA is a division algebra andC is not, and choose c = x(t)+y(t)ewith c2 = 0.Relative to a fixed basis of A assume that x(t) and y(t) have polynomial coefficientsof minimal degree. From x2 + yα(y)t = 0 argue that t divides everything, contraryto the minimality.)

17. Albert forms. (1) Suppose q is a form with dim q = 6 and dq = 〈1〉. Thenq ⊥ H � ϕ1 ⊥ −ϕ2, where ϕ1 and ϕ2 are 2-fold Pfister forms, and c(q) is a tensorproduct of two quaternion algebras.(2) Lemma. Suppose α and β are forms with dim α = dim β = 6, dα = dβ and

c(α) = c(β). Then α and β are similar.(3) Suppose A = Q1 ⊗Q2 is a tensor product of two quaternion algebras whose

norm forms are ϕ1 and ϕ2. Define the Albert form α = αA = ϕ′1 ⊥ −ϕ′2. Thendim α = 6, dα = 〈1〉, c(α) = [A]. Also: A is a division algebra if and only if αA isanisotropic (by Exercise 10(5)).

Proposition. The similarity class of αA depends only on the algebraA and not onQ1 and Q2.

(4) If A = C(q) where dim q = 4 then αA = q ⊥ 〈−1, dq〉.(Hint. (2) Given α ≡ β (mod J3(F )), we may assume α � 〈1〉 ⊥ α1 and β � 〈1〉 ⊥β1. By (3.12), α1 ⊥ −β1 is isotropic, so there exists d such that α1 � 〈d〉 ⊥ α2 andβ1 � 〈d〉 ⊥ β2, where α2 and β2 are 4-dimensional. Apply Exercise 12(4).)


18. Generalizing Albert forms. (1) If dim q = 2m then C(q) is isomorphic to atensor product of m quaternion algebras. Conversely, if A is a tensor product of mquaternion algebras thenA ∼= C(q) for some form q with dim q = 2m. PossiblyC(q)is similar to a product of fewer than m quaternion algebras (e.g. if q is isotropic).(2) An F -algebra A is similar to a tensor product of m− 1 quaternion algebras if

and only if [A] = c(ϕ) for some form ϕ with dim ϕ = 2m and dϕ = 〈1〉. Such a formϕ is called an Albert form for A. Compare Exercise 17.(3) SupposeA is a tensor product of quaternion algebras. IfA is a division algebra

then every Albert form for A is anisotropic.(4) There exists some A having two Albert forms which are not similar. There

exists someAwhich is not a division algebra but every Albert form ofA is anisotropic.

(Hint. (2) (⇐�) Express ϕ � 〈a〉 ⊥ α, note that dα = 〈−a〉 and compute c(ϕ) =c(α) = [C0(α)]. (�⇒) Express A ∼= C(q) where dim q = 2m − 2. Let ϕ = q ⊥〈−1, dq〉.(4) For the second statement let D be the example of Amitsur, Rowen and Tignol

(1979) mentioned in Theorem 6.15(2) below. Then A = M2(D) is a product of 4quaternion algebras but D contains no quaternion subalgebras. If ϕ is an isotropicAlbert form for A then [D] = [A] = c(ϕ) is a product of 3 quaternion algebras inBr(F ).)

19. The signs in C(V, q). Let {e1, . . . , en} be a basis of (V , q) corresponding toq � 〈a1, . . . , an〉, and with {e�} the derived basis of C = C(V, q). Then e�e� =±a��e�+� where a� = a

δ11 . . . a

δnn ∈ F •, as in Exercise 1.11.

(1) This± sign is (−1)β(�,�) where β : Fn2 ×Fn2 → F2 is a bilinear form. In fact,if {ε1, . . . , εn} is the standard basis for Fn2 then β(εi, εj ) = 1 if i > j and 0 otherwise.(2) Conversely, given a bilinear form β on Fn2 define an F -algebraA(β) of dimen-

sion 2n with basis {e�} using the formula above. Then A(β) is an associative algebrawith 1. (For the β specified above, this observation leads to a proof of the existenceof the Clifford algebra.)(3) Let Qβ be the quadratic form on F2n defined by β, that is, Qβ(x) = β(x, x).

If β ′ is another form andQβ = Qβ ′ then A(β) ∼= A(β ′). Our algebra A(β) could becalled A(Q).(4) If Q is a regular form (i.e. the bilinear form Q(x + y) − Q(x) − Q(y) is

regular) then the algebra A(Q) is central simple over F . (For the simplicity, supposeI is an ideal and choose 0 �= c ∈ I with c = ∑

c�e� of minimal length. Compare

Proposition 1.11.)(5) If Q � Q1 ⊥ Q2 then A(Q) ∼= A(Q1) ⊗ A(Q2). Therefore if Q is regular,

A(Q) is a tensor product of quaternion algebras.

20. Singular forms. (1) Suppose q is a quadratic form on V and define the radical ofV to be V ⊥ = {v ∈ V : B(v, x) = 0 for every x ∈ V }. Then q induces a quadraticform q on the quotient space V = V/V ⊥ and q is regular. If W is any complement


of V ⊥ in V (i.e. V = V ⊥ ⊕W as vector spaces) the restriction (W, q|W) is isometricto (V , q). Then q � r〈0〉 ⊥ q1 for a regular form q1, unique up to isometry.(2) For q as above there exists an isometry (V , q)→ (V ′, q ′)where q ′ is a regular

form. In fact we can use q ′ = rH ⊥ q1. (Note: isometries are injective by definition.)What is the minimal value for dim V ′ here? Is that minimal form q ′ unique?(3) Complete the proof of (3.7).(4) If q = r〈0〉 ⊥ q1 for a regular form q1, what is the center ofC(q)? In particular

what is the center of the exterior algebra �(V )?

Notes on Chapter 3

In 1878 W. K. Clifford defined his algebras using generators and relations. He ex-pressed such an algebra as a tensor product of quaternion algebras and the center (butusing different terminology). Clifford’s algebras were used by R. Lipschitz in hisstudy of orthogonal groups in 1884. These algebras were rediscovered in the casen = 4 by the physicist P. A. M. Dirac, who used them in his theory of electron spin.The algebraic theory of Clifford algebras was presented in a more general form byBrauer and Weyl (1935), by E. Cartan (1938) and by C. Chevalley (1954). Also com-pare Lee (1945) and Kawada and Iwahori (1950). Most of the algebraic informationabout Clifford algebras that we need appears in various texts, including: Bourbaki(1959), Lam (1973), Jacobson (1980), Scharlau (1985) and Knus (1988).Terminologies and notations often vary with the author, and sometimes the no-

tations are inconsistent within one book. For example we used A◦ for the “pure”part of the Cayley–Dickson algebra in Exercise 1.25 (3) and we used C0 for the evenClifford algebra in (3.8). (Some authors useA+ andC+ for these objects.) Moreover,a quaternion algebra A is both a Cayley–Dickson algebra and a Clifford algebra, butwe sometimes write A0 for the set of pure quaternions.Here are some further examples of confusing notations in this subject. As men-

tioned earlier, the Pfister form 〈〈a1, . . . , an〉〉 in this book (and in Lam (1973) andScharlau (1985)) is written as 〈〈−a1, . . . ,−an〉〉 in Knus et al. (1998). For a quadraticform q, Lam uses dq for our determinant det(q) and d±q for the discriminant. InClifford algebra theory, the names for the canonical automorphism α, and for theinvolutions x� = J0(x) and x = J1(x) are even less standard. Their names andnotations vary widely in the literature.The quaternion algebra decompositions in (3.14) follow Dubisch (1940). Similar

explicit formulas were given by Clifford (1878).Proofs of (3.18) also appear in Lam (1973), p. 121 or implicitly in Scharlau (1985),

p. 81, Theorem 12.9.The ideal J3(F ) coincides with the ideal defined by Arason and Knebusch using

the “degree” of a quadratic form. See Scharlau (1985), p. 164.


Theproof ofTheorem3.21first appeared inSatz 14ofPfister (1966). It is also givenin Scharlau (1985), pp. 90–91. Some work has been done recently on 14-dimensionalforms with trivial invariants. See the remarks before (9.12) below.

Exercise 2. There is a more abstract treatment of the subspaces V(k). It usesthe canonical bijection �(V ) → C(V, q), and the exterior power �k(V ) corre-sponds to V(k). In the usual product on C(V, q), if x ∈ V(r) and y ∈ V(s) thenxy ∈ V(r+s) + V(r+s−2) + · · · . Using that bijection to transfer the exterior product“∧” to C(V, q), it turns out that x ∧ y is exactly the V(r+s)-component of xy. SeeBourbaki (1959), Wonenburger (1962a) or Marcus (1975).

Exercise 6. This first appeared in Satz 9 of Witt (1937).

Exercise 8. These are old results going back at least to Albert (1939). The simpleproof of part (1) appears in Lam (1973), p. 200 and Scharlau (1985), p. 45.

Exercise 10. (5) Albert (1931) first discovered when a tensor product of twoquaternion algebras is a division algebra. IfQ1 ⊗Q2 is not a division algebra then αis isotropic and Q1 and Q2 have a common maximal subfield (as in 10(1)). A moredirect proof of this was given by Albert (1972). A different approach appears in Knuset al. (1998) in Corollary 16.29.

Exercise 11. Further information on such graded algebras appears in Lam (1973),Chapter 4 and in Knus (1988), Chapter 4.

Exercise 12 followsWadsworth (1975). See Exercise 5.23 below for another proof.A different method and related results appear in Knus (1988), pp. 76–78. Tignol hasfound a proof involving the corestriction of algebras.

Exercise 14. Some trace forms in the split case are considered in Exercise 1.13.See also Exercise 7.14.

Exercise 16. (1) Compare Edwards (1978).(2) A more general result is stated in Mammone and Tignol (1986).

Exercise 17. Albert forms were introduced by Albert (1931) in order to charac-terize when a tensor product of two quaternion algebras is a division algebra. Theproposition about the similarity class of αA was first proved by Jacobson (1983) usingJordan structures. The quadratic form proof here was extended to characteristic 2 inMammone and Shapiro (1989). The Albert form arises more naturally as the formon the alternating elements of A induced by a “Pfaffian” as mentioned in Chapter 9below. This application of Pfaffians originated with Knus, Parimala and Sridharan(1989). This whole theory is also presented in Knus et al. (1998), §16.

Exercise 18 is part of the preliminary results for Merkurjev’s construction of anon-formally real field F having u(F ) = 2n. See Lam (1989) and Merkurjev (1991).

Exercise 19 follows ideas told to me by F. Rodriguez-Villegas.

Chapter 4

C-Modules and the Decomposition Theorem

An (s, t)-family on (V , q) provides V with the structure of a C-module, for a certainClifford algebraC. Moreover, the adjoint involution Iq on End(V ) is compatible withan involution J on C. We examine a more general sort of (C, J )-module, discusshyperbolic modules and derive the basic Decomposition Theorem for (s, t)-families.Before pursuing these general ideas let us state the main result. If (σ, τ ) is a pair

of quadratic forms over F we say that a quadratic space (V , q) is a (σ, τ )-module if(σ, τ ) < Sim(V , q). A (σ, τ )-module (V , q) is unsplittable if there is no decompo-sition (V , q) � (V1, q1) ⊥ (V2, q2) where each (Vi, qi) is a non-zero (σ, τ )-module.Clearly every (σ, τ )-module is isometric to an orthogonal sum of unsplittables.

4.1 Decomposition Theorem. Let (σ, τ ) be a pair of quadratic forms where σ rep-resents 1. All unsplittable (σ, τ )-modules have the same dimension 2k , for some k.

Without the condition that σ represents 1 the result fails. Examples appear inExercise 5.9.The proof of this theorem involves the development of the theory of quadratic

(C, J )-modules, where C is a Clifford algebra with an involution J . In order topinpoint the properties of the Clifford algebras used in the proof, we will develop thetheory of quadratic modules over a semisimple F -algebra with involution. But beforeintroducing those ideas we point out some simple consequences of (4.1).First of all, this theorem explains why the Hurwitz–Radon function ρ(n) depends

only on the 2-power part: if n = 2mn0 where n0 is odd, then ρ(n) = ρ(2m). In fact,suppose (σ, τ ) < Sim(q) where dim q = n. By the theorem, an unsplittable (σ, τ )-module (W, ϕ)must have dimension 2k dividing n. Then k ≤ m and (σ, τ ) < Sim(q ′)for some form q ′ of dimension 2m. Here q ′ can be taken to be 2m−kϕ.The theorem also provides a more conceptual proof of Proposition 1.10. Suppose

(W, α) is unsplittable for 〈1, a〉. (I.e., it is an unsplittable (〈1, a〉, 0)-module.) By(1.9) we know that 〈x〉〈〈a〉〉 ⊂ α for some x ∈ F •, and in particular dim α ≥ 2. Byexplicit construction, 〈1, a〉 < Sim(〈〈a〉〉), so 〈〈a〉〉 is an unsplittable module for 〈1, a〉.The Decomposition Theorem then implies that dim α = 2, so that α � 〈x〉〈〈a〉〉. Itquickly follows that if 〈1, a〉 < Sim(q) then q is a sum of unsplittables, or equivalentlyq � 〈〈a〉〉 ⊗ ϕ for some form ϕ.

74 4. C-Modules and the Decomposition Theorem

Several small (s, t)-families can be analyzed in this way. These arguments arepresented in Chapter 5, and the interested reader can skip there directly. For the restof this Chapter we discuss quadratic modules.Suppose (S, T ) ⊆ Sim(V , q) is an (s, t)-family and σ and τ are the quadratic

forms induced on S and T . Then 1V ∈ S and we define S1 = (1V ) ⊥. If f ∈ S1 andg ∈ T then:

Iq(f + g) = −f + g(f + g)2 = f 2 + g2 = (−σ(f )+ τ(g))1V .

Therefore the inclusion map S1 ⊥ T → End(V ) is compatible with the quadraticform on −σ1 ⊥ τ on S1 ⊥ T . By the definition of Clifford algebras, there is aninduced F -algebra homomorphism

π : C → End(V ),

whereC = C(−σ1 ⊥ τ) is the Clifford algebra of dimension 2s+t−1. This π providesan action of C on V , making V into a C-module.To be a little more careful let us define the space S1 ⊥ T to be the generating

space of C , so that S1 = π(S1) and T = π(T ). Setting S = F · 1 + S1 ⊆ C, wehave S = π(S). For example, suppose σ � 〈1, a2, . . . , as〉 and τ � 〈b1, . . . , bt 〉 ,and let f2, . . . , fs, g1, . . . , gt be given as in (2.1). The algebra C = C(−σ1 ⊥ τ)

has generators e2, . . . , es, d1, . . . , dt corresponding to the given diagonalization of−σ1 ⊥ τ . Then S = span{1, e2, . . . , es}, T = span{d1, . . . , dt } and π(ei) = fi ,π(dj ) = gj . We often identify S with S and T with T to simplify notations. Wedo this even though the identification of S and T with subspaces of C is sometimesmisleading, since this embedding depends on the representation π .

4.2 Proposition. Let (V , q) be a quadratic space of dimension n = 2mn0, where n0is odd. Then any (s, t)-family on (V , q) must have s + t ≤ 2m+ 2.

Proof. Suppose (σ, τ ) < Sim(V , q) and let C = C(−σ1 ⊥ τ) be the associatedClifford algebra with representation π : C → End(V ). By the Structure Theorem 3.6eitherπ(C) orπ(C0) is a central simple subalgebra of End(V ). The double centralizertheorem implies that the dimension of this subalgebramust divide dim(End(V )) = n2.Then 2s+t−2 divides n2 = 22mn20 so that s + t − 2 ≤ 2m. ��

The original 1, 2, 4, 8 Theorem is an immediate consequence of this inequality.(Compare Exercise 1.1.) The proof of this proposition uses only the information thatV is aC-module. To get sharper information we will take the involutions into accountand analyze the “unsplittable components” of (V , q).

4.3 Definition. Define JS = JS1,T

to be the involution on C as in Definition 3.15.

Then JS acts as −1 on S1 and as 1 on T .

4. C-Modules and the Decomposition Theorem 75

This involution is chosen to match the behavior of the involution Iq on End(V ).That is, the action of Iq on the maps fi , gj in End(V )matches the action of JS on thegenerators ei , dj in C. Therefore, for every c ∈ C, π(JS(c)) = Iq(π(c)). Hence, πis a homomorphism of algebras-with-involution:

π : (C, JS)→ (End(V ), Iq).

Such a map π is sometimes called a similarity representation or a spin representation.The map π is usually not written explicitly. We write the action of an element

c ∈ C as a multiplication: cv = π(c)(v), for v ∈ V . Then the compatibility of theinvolutions says exactly that:

Bq(cv,w) = Bq(v, JS(c)w), (∗)for every c ∈ C and v,w ∈ V . Conversely, ifV is aC-module andq is a quadratic formon V satisfying the compatibility condition (∗), then (σ, τ ) < Sim(V , q). Therefore(V , q) is a (σ, τ )-module if and only if V is a C-module and the form q satisfies thecompatibility condition (∗) above.Different C-module structures on V can arise from the same family (S, T ) in

Sim(V , q). To see this let δ be an automorphism of C which preserves the subspacesS1 and T . If π is a similarity representation coming from (S, T ), define π ′ = π � δ.Then π ′ is another similarity representation associated to (S, T ). This ambiguityshould cause little trouble, since we usually fix one representation.Let us now set up the theory of quadratic modules. To see where the special

properties of Clifford algebras are used, we describe the theory for a wider class ofalgebras.

Notation. LetC be a finite dimensional semisimpleF -algebra with involution J . Theinvolution is often written simply as “bar”. Unless explicitly stated otherwise, everymodule is a left C-module which is finite dimensional over F .

It is useful to consider alternating spaces in parallel with quadratic spaces. Tohandle these cases together, let λ = ±1 and define a λ-form B on a vector space V tobe a bilinear formB : V ×V → F which is λ-symmetric, that is: B(y, x) = λB(x, y)

for every x, y ∈ V . The λ-form B is regular if V ⊥ = 0, (or equivalently if the inducedmap θB from V to its dual is a bijection). Then a λ-space (V , B) is a vector space Vwith a regular λ-form B.Since 2 �= 0 in F , a quadratic space is the same as a 1-space. That is, a quadratic

form q is determined by its associated symmetric bilinear form Bq . An alternatingspace is another name for a (−1)-space. For any λ-space (V , B) there is an associatedadjoint involution IB as in (1.2).It is well known that alternating spaces over the field F must have even dimension,

and any two alternating spaces of the same dimension are isometric. However in thecategory of alternating spaces admitting C, such an easy characterization no longerapplies.


4.4 Definition. Let C be an algebra with involution as above. Suppose B is a regularλ-form on a C-module V . Then B admits C if

B(cu, v) = B(u, cv)

for every u, v ∈ V and c ∈ C. In this case (V , B) is called a λ-space admitting C,or a λ-symmetric (C, J )-module. If V , V ′ are λ-spaces admitting C, then they areC-isometric (written V ≈ V ′) if there is a C-module isomorphism V → V ′ which isalso an isometry.

We will extend the standard definitions and techniques of the theory of quadraticforms over F to this wider context. Much of this theory can be done more generally,allowing F to be a commutative ring having 2 ∈ F • and considering λ-hermitianmodules. For example see Fröhlich and McEvett (1969), Shapiro (1976), or formore generality, Quebbemann et al. (1979). The category of λ-spaces admitting C isequivalent to the category of λ-hermitian C-modules. This equivalence is proved inthe appendix to this chapter.If U ⊆ V then U⊥ is the “orthogonal complement” in the usual sense:

U⊥ = {x ∈ V : B(x,U) = 0}. Then a subspace U ⊆ V is regular if and onlyif U ∩ U⊥ = 0, and is totally isotropic iff U ⊆ U⊥.

4.5 Lemma. Let (V , B) be a λ-space admitting C, and let T ⊆ V be a C-submodule.

(1) Then T ⊥ is also a C-submodule.

(2) If T is a regular subspace of V then V ≈ T ⊥ T ⊥.

(3) If T is an irreducible submodule then either T is regular or T is totally isotropic.

Proof. The same argument as in the classical cases. For (3) consider T ∩ T ⊥. ��

Wecan consider these bilinear forms in terms of dual spaces. Let V =HomF (V, F )be the dual vector space. The dual pairing is written

〈 | 〉 : V × V → F.

The space V has a natural structure as a right C-module, defined by: 〈v|vc〉 = 〈cv|v〉.We will use the involution to change hands and make V into a left C-module.

4.6 Definition. Let V be a left C-module. For v ∈ V and c ∈ C define cv by:〈v|cv〉 = 〈cv|v〉 for all v ∈ V.

A λ-formB on V induces a linear map θB : V → V by defining θB(v) = B(−, v).That is, 〈u|θB(v)〉 = B(u, v). By definition, B is regular if and only if θB is bijective.Furthermore the definitions imply that: B admits C if and only if θB is a C-moduleisomorphism.


Consequently if (V , B) is a λ-space admitting C then V ∼= V as C-modules.Conversely, if V ∼= V must there exist such a λ-form? Here is one special case whena form always exists.

4.7 Definition. Let T be a C-module and λ = ±1. DefineHλ(T ) to be the C-moduleT ⊕ T together with the λ-form BH defined by: BH(s + s, t + t ) = 〈s|t 〉 + λ〈t |s 〉,for s, t ∈ T and s, t ∈ T . A λ-space admitting C is C-hyperbolic if it is C-isometricto some Hλ(T ).

One easily checks that BH is regular, λ-symmetric and admits C. If T ∼= T ′ asC-modules then Hλ(T ) ≈ Hλ(T

′). Also Hλ(S ⊕ T ) ≈ Hλ(S) ⊥ Hλ(T ).

4.8 Lemma. Let (V , B) be a λ-space admitting C.

(1) V is C-hyperbolic iff V = T1 + T2 where T1 and T2 are totally isotropic sub-modules.

(2) V ⊥ 〈−1〉V ≈ Hλ(V ) is C-hyperbolic.

Proof. (1) If V = Hλ(T ) let T1 = T ⊕ 0 and T2 = 0 ⊕ T . Conversely, supposeV = T1 + T2. Then Ti ⊆ T ⊥

i , and since V⊥ = 0 we have T ⊥

1 ∩ T ⊥2 = 0. The map

ψ : T2 → T1, defined by ψ(x) = B(−, x), is a C-homomorphism and kerψ = 0.The surjectivity ofψ follows from the surjectivity of θB (or by comparing dimensions).Then ψ is a C-isomorphism and 1⊕ ψ is a C-isometry V → Hλ(T1).(2) We are given a space U = V ⊕ V ′ and a C-isomorphism f : V → V ′ which

is also a (−1)-similarity. Let T+ = {x + f (x) : x ∈ V } ⊆ U be the graph of f , andsimilarly let T− be the graph of −f . It is easy to check that T+ and T− are totallyisotropic submodules and U = T+ + T− so part (1) applies. ��

4.9 Proposition. Let (V, B) be a λ-space admitting C and T ⊆ V a totally isotropicsubmodule. Then there is another totally isotropic submodule T ′ ⊆ V with T + T ′ ≈Hλ(T ), a regular submodule of V .

Proof. Since C is semisimple there exists a submoduleW of V complementary to thesubmodule T ⊥. Then the induced pairingB0 : W×T → F induces aC-isomorphismW ∼= T . If W is totally isotropic then T +W ≈ Hλ(T ) by Lemma 4.8. Therefore theproposition is proved once we find a totally isotropic complement to T ⊥.Given W , all other complements to T ⊥ appear as graphs of C-homomorphisms

f : W → T ⊥. Define the homomorphism f : W → T ⊆ T ⊥ by the equation:

B0(w1, f (w2)) = − 12B(w1, w2),

for w1, w2 ∈ W . The nonsingularity of the induced pairing B0 above shows that f iswell-defined and C-linear. The graph of this f provides the complement we need. ��


This proposition gives information about the structure of unsplittables. We definea λ-space V admitting C to be C-unsplittable if there is no expression V ≈ V1 ⊥ V2where V1 and V2 are non-zero submodules. Equivalently, V is C-unsplittable if andonly if V has no regular proper submodules. The following result contains most ofthe information needed to prove the Decomposition Theorem 4.1.

4.10 Theorem. Let (V , B) be an unsplittable λ-space admitting C. Then eitherV is irreducible or V ≈ Hλ(T ) for an irreducible module T . Moreover, Hλ(T ) isunsplittable if and only if T is irreducible and possesses no regular λ-form admittingC.

Proof. IfV is reducible letT ⊆ V be an irreducible submodule. SinceV is unsplittableT must be singular, and hence totally isotropic by (4.5) (3). Then by (4.9) we haveT ⊆ H ⊆ V , where H ≈ Hλ(T ). Since V is unsplittable, V = H .Now suppose T is an irreducible C-module. If T possesses a regular λ-form

admitting C, then by (4.8) (2) we have Hλ(T ) ≈ T ⊥ 〈−1〉T is splittable. ��

Let (V , B) be any λ-space admitting C. Then there is a decomposition

V = Hu ⊥ Hs ⊥ Va,

where Hu is an orthogonal sum of unsplittable C-hyperbolic subspaces, and Hs is anorthogonal sum of splittable hyperbolic subspaces, and Va is C-anisotropic. Here wedefine a λ-space to be C-anisotropic if it has no totally isotropic irreducible submod-ules. From (4.10) we conclude that no irreducible submodule ofHu can be isomorphicto a submodule of Hs ⊥ Va . Therefore Hu and Hs ⊥ Va are uniquely determinedsubmodules of V .Moreover, as in the classical case, the submodules Hs and Va are unique up to

isometry because there is a Cancellation Theorem:

If U,V andW are λ-spaces admitting C and U ⊥ V ≈ U ⊥ W then V ≈ W.

We omit the proof of this theorem because it does not seem to have a direct applicationto the study of (s, t)-families. Proofs ofmore general results appear inMcEvett (1969),Shapiro (1976), and Quebbemann et al. (1979). Knowing the Cancellation Theoremit is natural to investigate the Witt ring Wλ(C, J ) of λ-spaces admitting (C, J ). Wewill not pursue this investigation here.When does an irreducible C-moduleW possess a regular λ-form admitting C? If

such a form exists then certainly W ∼= W as C-modules.

4.11 Lemma. If W is an irreducible C-module with W ∼= W , then W has a regularλ-form admitting C, for some sign λ.

Proof. If g : U → V is a C-module homomorphism, the transpose g� : V → U

is defined by the equation 〈u|g�(v)〉 = 〈g(u)|v〉. Any C-isomorphism θ : W → W


induces a regular bilinear form B on W by setting B(x, y) = 〈x|θ(y)〉. After identi-fyingW with its double dual, we see thatB is a regular λ-form if and only if θ� = λθ .Now from the given C-isomorphism θ , define θλ = 1

2 (θ + λθ�). These two maps areC-homomorphisms and θ�λ = λ · θλ. They are not both zero since θ = θ+ + θ−, so atleast one of them must be an isomorphism, by the irreducibility. The correspondingform B is then a regular λ-form admitting C. ��

We now return to the original situation of Clifford algebras. Let (σ, τ ) be a pair offorms as usual andC = C(−σ1 ⊥ τ)with the involution JS . For a λ-space (V , B)wesee that (σ, τ ) < Sim(V , B) if and only if V can be expressed as a C-module whereB admits (C, JS). If C is simple then up to isomorphism there is only one irreduciblemodule, and the ideas above are easy to apply. The non-simple case requires anadditional remark.Suppose C is not simple. Then s+ t is even, d(−σ1 ⊥ τ) = 〈1〉 and C0 is simple.

We can choose z = z(S1 ⊥ T ) with z2 = 1. Then Z = F + Fz is the center of Cand the non-trivial central idempotents are eε = 1

2 (1 + εz) for ε = ±1. Then C =Ce+×Ce− ∼= C0×C0 and there are two natural projection maps p+, p− : C → C0.Let V be an irreducible C0-module with associated representation π : C0 → V .Define Vε to be the C-module associated to the representation πε = π �pε. It followsthat every irreducible C-module is isomorphic to either V+ or V−. These two modulestructures differ only by an automorphism of C.

4.12 Lemma. Let C, JS be as above. Let λ = ±1 be a fixed sign.

(1) Suppose C is not simple. If s ≡ t (mod 4) then V+ ∼= V+ and V− ∼= V−. Ifs ≡ t + 2 (mod 4) then V+ ∼= V−.

(2) If one irreducible C-module possesses a regular λ-form admitting C, then theyall do.

Proof. (1) By the definition of W as a left C-module, z acts on W as JS(z) acts onW . Therefore if JS(z) = z then Vε ∼= Vε. If JS(z) = −z then Vε ∼= Vε. A signcalculation completes the proof.(2) If C is simple the claim is trivial. Suppose C is not simple and Vε possesses

a regular λ-form B admitting C. The module V−ε can be viewed as the same vectorspace V with a twisted representation: π−ε(c) = πε(α(c)) where α is the canonicalautomorphism of C. It follows that the form B admits this twisted representation(since α and JS commute), so V−ε also has a regular λ-form admitting C. ��

4.13 Corollary. Let (C, JS) be the Clifford algebra with involution as above. Letλ = ±1 be a fixed sign. Then all theC-unsplittable λ-spaces have the same dimension2k , for some k.


Proof. By the remarks above, all irreducible C-modules have the same dimension 2m,for some m. It is a power of 2 since C is a direct sum of irreducibles and dimC =2r+s−1. If one irreducible module possesses a regular λ-form admitting C, then theyall do and by (4.10) every unsplittable is irreducible of dimension 2m. Otherwise(4.10) implies that every unsplittable is isometric toHλ(T ) for some irreducible T , sothat the unsplittables all have dimension 2m+1. ��

Finally we can complete the proof of the original Decomposition Theorem 4.1.The only remaining step is to show that the two notions of “unsplittable” coincide.The problem is that the definition of unsplittable (σ, τ )-module involves isometryof quadratic spaces over F (written �) while the definition of unsplittable quadraticspace admitting C involves C-isometry (written ≈). If a space is (σ, τ )-unsplittableit certainly must be C-unsplittable. Conversely suppose (V , q) is C-unsplittable, butV � V1 ⊥ V2 for some non-zero (σ, τ )-modules Vi . Then Vi is a quadratic spaceadmitting C, so it is C-isometric to a sum of C-unsplittables. Comparing dimensionswe get a contradiction to (4.13). This completes the proof. ��

4.14 Definition. An (s, t)-pair (σ, τ ) is of regular type if an unsplittable quadratic(σ, τ )-module is irreducible. Otherwise it is of hyperbolic type. In working withalternating forms we say that (σ, τ ) is of (−1)-hyperbolic type or of (−1)-regulartype.

By (4.12) this condition does not depend on the choice of unsplittable module. If(σ, τ ) is of hyperbolic type then (4.10) implies that every unsplittable (σ, τ )-moduleisHλ(T ) for some irreducible C-module T . Consequently, every (σ, τ )-module is C-hyperbolic and (σ, τ )-modules are easy to classify. For example, if s ≡ t + 2(mod 4)then (4.12) implies that (σ, τ ) is of hyperbolic type (and of (−1)-hyperbolic type).The other cases are not as easy to classify. Further information about these types isobtained in Chapter 7.WhenC is a division algebra with involution the irreducible module is justC itself.

In this case it is sometimes useful to analyze all the λ-forms on C which admit C.This can be done by comparing everything to a given trace form.

4.15 Definition. LetC be an F -algebra with involution. An F -linear map � : C → F

is an involution trace if

(1) �(c) = �(c) for every c ∈ C;(2) �(c1c2) = �(c2c1) for every c1, c2 ∈ C;(3) The F -bilinear form L : C × C → F , defined L(c1, c2) = �(c1c2), is regular.

For example if C = End(V ) and J is any involution on C, then the ordinary traceis an involution trace. Suppose C = C(−σ1 ⊥ τ) is a Clifford algebra as above and


J = JS . Then the usual trace is an involution trace. Generally every semisimpleF -algebra with involution does possess an involution trace. (See Exercise 14 below.)

4.16 Proposition. Let C be an F -algebra with involution J and with an involutiontrace �. If B is a regular λ-form on C such that (C, B) admits C then there existsd ∈ C• such that J (d) = λd and B(x, y) = �(xdJ (y)) for all x, y ∈ C.

Proof. Let B1(x, y) = �(xJ (y)). Since � is an involution trace, B1 is a regular 1-form on C. It is easy to check that (C, B1) admits C. Since B1 is regular, the givenbilinear form B can be described in terms of B1: there exists f ∈ EndF (C) suchthat B(x, y) = B1(f (x), y) for all x, y ∈ C. Since B is regular this f is invertible.The condition that B admits C becomes B1(f (cx), y) = B(cx, y) = B(x, J (c)y) =B1(f (x), J (c)y) = B1(cf (x), y). Therefore f (cx) = cf (x), and f is determinedby its value d = f (1). Therefore B(x, y) = B1(xd, y) = �(xdJ (y)). Since B is aλ-form we find J (d) = λd. ��

For example, one such trace form was considered in Exercise 3.14. In the case that

C =(−a,−bF

)is a quaternion division algebra with the usual involution J = “bar”

there is a unique unsplittable quadratic (C, J )-module. This follows since the onlyJ -symmetric elements of C are the scalars. The “uniqueness” here is up to a C-similarity. In terms of the quadratic forms over F this says that if 〈1, a, b〉 < Sim(q)is unsplittable then q is similar to 〈〈a, b〉〉. This re-proves Proposition 1.10.

Appendix to Chapter 4. λ-Hermitian forms over C

In this appendix we return to the more general set-up where C is a semisimple F -algebra with involution J and describe the one-to-one correspondence between theλ-spaces admittingC and theλ-hermitian forms overC. This equivalence of categoriesprovides a different viewpoint for this whole theory. Since these ideas are not heavilyused later we just sketch them in this appendix.We could allow the involutions here to be non-trivial on the base ring F . In that

case there can be λ-hermitian forms for any λ ∈ F with λλ = 1. The details wereworked out by Fröhlich andMcEvett (1969), and the reader is referred there for furtherinformation.

A.1 Definition. Let V be a C-module and λ = ±1. A λ-hermitian form on V (overC) is a mapping h : V × V → C satisfying

(1) h is additive in each slot,

(2) h(cx, y) = ch(x, y) for every x, y ∈ V and c ∈ C,(3) h(y, x) = λh(x, y) for every x, y ∈ V .


It follows that h(x, cy) = h(x, y)c. For a simple example let V = C and fora ∈ C define ha : C×C → C by ha(x, y) = xay. If a = λa then ha is a λ-hermitianform.Define the C-dual module V = HomC(V,C). For fixed x ∈ V the map θh(v) =

h(−, v) is in V . This map θh : V → V is F -linear. The form h is said to be regular ifθh is bijective. Define a λ-hermitian space over C to be a C-module V equipped witha regular λ-hermitian form h. One can now define isometries, similarities, orthogonalsums and tensor products in the category of λ-hermitian spaces.In analogy with our treatment of the F -dual, we write the dual pairing as

[ | ] : V × V → C.

Then [ | ] is F -bilinear, and by definition [cx|x] = c[x|x]. With this notation wehave: [u|θh(v)] = h(u, v).As before we use the involution onC to change hands andmake V a leftC-module:

for x ∈ V and c ∈ C define cx by[x|cx] = [x|x]c

for all x ∈ V . Therefore if h is a λ-hermitian form then θh : V → V is a homomor-phism of left C-modules.The hyperbolic functor Hλ can be introduced in this new context, in analogy with

the discussion for λ-spaces admitting C. All the results proved above for λ-spacesadmitting C can be done for λ-hermitian modules. In fact, when there is an involutiontrace map on C, these two contexts are equivalent. This equivalence arises becausethe two notions of “dual” module, V and V , actually coincide.

A.2 Proposition. Let C be a semisimple F-algebra with involution J and possessingan involution trace �.

(1) Let (V , h) be a λ-hermitian space over C and define B = � � h. Then (V , B) isa λ-space admitting C, denoted by �∗(V , h).

(2) Let (V, B) be a λ-space admitting C. Then there is a unique regular λ-hermitianform h on V having �∗(V , h) = (V , B).

(3) This correspondence �∗ preserves isometries, orthogonal sums and the Hλ con-struction.

Proof sketch. (This is Theorem 7.11 of Fröhlich and McEvett.) (1) It is easy to seethat B is F -bilinear, λ-symmetric and admits C. Composition with � induces a mapon the dual spaces �0 : V → V , that is: 〈x|�0(x)〉 = �([x|x]). The properties of �imply that this �0 is an isomorphism of left C-modules. Furthermore �0 � θh = θB ,and we conclude that B is regular.(2) Given B we can construct θh as �

−10 � θB . It then follows that h is λ-hermitian

and � � h = B.(3) Checking these properties is routine. ��


By using hermitian forms over C we sometimes obtain a better insight into aproblem. For instance the simplest hermitian spaces over C are the “1-dimensional”forms obtained from the left C-module C itself. If a ∈ C• satisfies a = λa then theform

ha : C × C → C defined by ha(x, y) = xay,

is a regular λ-hermitian form on C. Let 〈a〉C denote this λ-hermitian space (C, ha).In this notation, the trace form considered in Proposition 4.16 is just �∗(〈d〉C). Thistransfer result (A.2) quickly yields another proof of (4.16).

A.3 Lemma. Let C be an F -algebra with involution and a, b ∈ C• with a = λa andb = λb. Then 〈a〉C � 〈b〉C if and only if b = cac for some c ∈ C•.

Proof. Suppose ϕ : (C, hb) → (C, ha) is an isometry. Then ϕ is an isomorphism ofleft C-modules, so that ϕ(x) = xc where c = ϕ(1) ∈ C•. Since ha(ϕ(x), ϕ(y)) =hb(x, y) the claim follows. The converse is similar. ��

A.4 Corollary. Suppose D is an F -division algebra with involution.

(1) Every hermitian space over D has an orthogonal basis.

(2) If the involution is non-trivial then every (−1)-hermitian space over D has anorthogonal basis.

Proof. (1) The irreducible left D-module D admits a regular hermitian form (e.g.〈1〉D). Then by (A.2) and (4.10) we conclude that every hermitian space overD is anorthogonal sum of unsplittable submodules, each of which is 1-dimensional over D.These provide an orthogonal basis.(2) There exists a ∈ D• such that a = −a. Then 〈a〉D is a regular (−1)-hermitian

form on D. The conclusion follows as before. ��

Of course this corollary can be proved more directly, without transferring to thetheory of λ-spaces admitting D.


1. Group rings. Suppose G be a finite group with order n and the characteristic ofF does not divide n. Then the group algebra C = F [G] is semisimple (Maschke’stheorem). Define the involution J on C by sending g �→ g−1 for every g ∈ G. Thereis a one-to-one correspondence between orthogonal representations G → O(V , q)and quadratic (C, J )-modules. These algebras provide examples where unsplittable(C, J )-modules may have different dimensions.


2. Proposition. Suppose V , W are C-anisotropic λ-spaces admitting C. If V ⊥ W

is C-hyperbolic then V ≈ 〈−1〉W . This is a converse of (4.8) (2).

(Hint. Let V ⊥ W = H and suppose T ⊆ H is a totally isotropic submodule with2 · dim T = dimH . Examine the projections to V andW to see that T is the graph ofsome f : V → W . This f must be a (−1)-similarity.)

3. Averaging Process. (1) Let F be an ordered field and suppose σ , τ are positivedefinite forms over F with σ � 〈1〉 ⊥ σ1 as usual. Let C = C(−σ1 ⊥ τ) and let Vbe a C-module. Then there exists a positive definite quadratic form q on V making(σ, τ ) < Sim(q).(2) Let R be the real field, and C = C((r − 1)〈−1〉). Then r〈1〉 < Sim(n〈1〉)

over R if and only if there is a C-module of dimension n. This explains why theHurwitz–Radon Theorem can be done over R without considering the involutions.

(Hint. (1) Let ϕ be a positive definite form on V and define q by averaging. Forinstance if σ � 〈1, a2, . . . , an〉 and τ = 0, define q(x) =∑

a−1� ϕ(e�x).)

4. Commuting similarities. What are the possible dimensions of two subspaces ofSim(V , q) which commute elementwise?

Definition. κr(n) = max{s : there exist commuting subspaces of dimensions rand s in Sim(V , q) for some n-dimensional quadratic space (V , q)}.Let R, S ⊆ Sim(V , q) be commuting subspaces which we may assume contain

1V . Let D be the Clifford algebra corresponding to R so that V is a left D-module.Define SimD(V, q) and note that S ⊆ SimD(V, q). If C is the Clifford algebra forS then this occurs if and only if there is a homomorphism C ⊗D → End(V ) whichpreserves the involutions.(1) If n = 2m· (odd) then κr(n) = κr(2m).(2) Define (s, t)-families in SimD(V, q) and prove the Expansion, Shift and Con-

struction Lemmas.(3) Define κ ′r (n) analogously for alternating forms (V , B).

(i) κ ′r (4n) ≥ 4+ κr(n) and κr(4n) ≥ 4+ κ ′r (n).(ii) If r ≡ 2 (mod 4) then κ ′r (n) = κr(n).

(4) κ2(2m) = 2m.(5) Proposition. Suppose κr(n) > 0 where n = 2m· (odd). Then

κr(n) ={ρ(n)+ 1− r if r ≡ 1 (mod 4)2m+ 2− r if r ≡ 2 (mod 4)1+ ρr(n) if r ≡ 3 (mod 4).

(Hint. (3) (i) If σ < SimD(q) then (σ ⊥ 〈1, 1〉, 〈1, 1〉) < Sim(〈〈1, 1〉〉 ⊗ q). Shiftby 2 as in Exercise 2.6(1). (ii) Let R, S ⊆ Sim(V , q). Then z = z(R1) commuteswith R and S, z = −z and R, S ⊆ Sim(V , B ′) where B ′(u, v) = B(u, zv). In fact an(s, t)-family in this case is equivalent to an (s + t, 0)-family.


(4) κ2(2m) ≤ 2m for otherwise the representation C → EndF (V ) is surjectiveand there is no room for D. Check κ2(1) = 0 and κ2(2) = 2, then apply (3), (4) andinduction.(5) When r is odd, consider commuting R, S ⊆ Sim(V ) and examine z(R1) � S.

For the even case use (3) to see κr(n) ≤ min{κr−1(n), κ ′r−1(n)} = 2m + 2 − r .Equality follows as in (4).)

5. More commuting similarities. Let D = C(〈−1〉α1) where α = 〈1〉 ⊥ α1 is agiven form with dim α = r . We examine subspaces S ⊆ SimD(V, q).

Definition. κ(α; n) = max{s : there exists a subspace of dimension s inSimD(V, q) for some n-dimensional quadratic space for which α < Sim(V , q)}.Then certainly κ(α; n) ≤ κr(n) with equality if F is algebraically closed.(1) If n = 2m· (odd) then κ(α; n) = κ(α; 2m). Also κ−λ(α; 4n) ≥ 4+ κλ(α; n).(2) If α � 〈1, a〉 then κ(α; n) = κ2(n). If α � 〈1, a, b〉 or α � 〈1, a, b, ab〉 then

κ(α; n) = κ3(n).(3) Suppose K is a field with a non-trivial involution and F is the fixed field of

that involution, so that K = F(√−a). If (V , h) is a hermitian space over K define

SimK(V, h) and note that comparable similarities span F -quadratic spaces. Definethe corresponding Hurwitz–Radon function ρherm(n), where n = dimK(V ).Let B = � � h be the underlying F -bilinear form and let g be the action of√−a on V . Then {1V , g} spans a subspace R ⊆ Sim(V , B) with R � 〈1, a〉.

Then S ⊆ SimK(V, h) if and only if S ⊆ SimF (V, B) and S commutes with R.Consequently, if n = 2m· (odd) then

ρherm(n) = κ(〈1, a〉; 2n) = κ2(2n) = 2m+ 2.(4)Howdoes this analysis generalize to hermitian forms over a quaternion algebra?

(Hint. (2) The equalities for small values follow by considering the quadratic andquaternion algebras with prescribed norm forms.)

6. More Hurwitz–Radon functions. (1) Define ρ+(n) = max{k : C((k−1)〈1〉) hasan n-dimensional module over R}. Let n = 24a+bn0 where n0 is odd and 0 ≤ b ≤ 3.According to Lam (1973), p. 132, Theorem4.8′, ρ+(n) = 8a+b+[b/3]+2. (Here [x]denotes the greatest integer function.) In our notation, ρ+(n) = 1+ρ1(n). “Explain”this result as in Exercise 3.(2) Let D be the quaternion division algebra over R. Let ρD(n) = max{r :

C((r − 1)〈−1〉) has an n-dimensional module over D}. Then ρD(n) = 8a + 2b +12 (b + 2)(3 − b), according to Wolf (1963a), p. 437. Modify Exercise 3 to see thatρD(n) = max{r : r〈1〉 < SimD(n〈1〉)}. Use Exercises 3, 4, 5 to show ρD(n) =κ(3〈1〉; n) = κ3(4n) = 1+ ρ3(4n), which coincides with Wolf’s formula.

7. Hermitian compositions again. Suppose n = 2m· (odd).(1) Recall the compositions (over C) considered in Exercise 2.15. The formula

has type 2 if each zk if bilinear in (X, X) and Y .


Proposition. A type 2 composition of size (r, n, n) exists if and only if r ≤ m+ 1.(2) s ≤ ρherm(n) = 2m+ 2 if and only if there is a formula

(x21 + x22 + · · · + x2s ) · (|y1|2 + · · · + |yn|2) = |z1|2 + · · · + |zn|2

where X = (x1, . . . , xs) is a system of real variables, Y = (y1, . . . , yn) is a systemof complex variables and each zk is C-bilinear in X, Y . Write out some examples ofsuch formulas.

(Hint. (1) From Exercise 2.15 such a composition exists iff there is an (r, r)-family inSimherm(V , h) where V = Cn and h is the standard hermitian form. Clifford algebrarepresentations imply 2r − 2 ≤ 2m. Conversely constructions over R provide an(m+ 1,m+ 1)-family in Simherm(V , h).)

8. Matrix factorizations. Suppose σ(X) = σ(x1, . . . , xr ) is a quadratic form. Recallthat: σ < Sim(n〈1〉) if and only if there exists an n × n matrix A whose entries arelinear forms inX satisfying: A�·A = σ(X)·In. (Compare (1.9).) Define a somewhatweaker property: σ admits amatrix factorization in order n if there exist n×nmatricesA, B whose entries are linear forms inX satisfying: A ·B = σ(X) · In. If σ has sucha factorization over F then so does every quadratic form similar to σ .(1) Proposition. Let σ be a quadratic form over F . Then σ admits a matrix

factorization in order n if and only if there is a C0(σ )-module of dimension n.In fact, any regular quadratic form σ possesses “essentially” just one matrix fac-

torization.(2) If σ = 〈1〉 ⊥ σ1 represents 1 then C0(σ ) ∼= C(−σ1) as ungraded F -algebras.

Suppose F is an ordered field and σ is a positive definite form over F . Then σadmits a matrix factorization in order n over F if and only if σ < Sim(q) for somen-dimensional positive definite form q over F .

(Hint. (1) (�⇒): Let (S, σ ) be the given space. ViewA, B as linear maps α, β : S →End(V ), where dim V = n. Define λ : S → End(V ⊕ V ) by: λ =

(0 α

β 0

). Then

λ(f )2 = σ(f ) · 1V⊕V for every f ∈ S so that V ⊕ V becomes C(S, σ )-module. Itis a graded module as in Exercise 3.10(3), determined by the C0(σ )-module V .(2) See Exercise 3.)

9. Conjugate families. Let C and JS be as usual. Suppose (V , q) and (V ′, q ′) arequadratic spaces admitting (C, JS), with associated (s, t)-families (S, T ) ⊆ Sim(V , q)and (S′, T ′) ⊆ Sim(V ′, q ′). Then V and V ′ are C-similar if and only if (S′, T ′) =(f Sf−1, f Tf−1) for some invertible f ∈ Sim(V , V ′).

10. Quaternion algebras. Let A be a quaternion algebra over F , with the usualbar-involution. Recall that the norm and trace on A are defined by: N(a) = a · a andT (a) = a + a. Let ϕ be the norm form of A, so that DF (ϕ) = {N(d) : d ∈ A•} isthe group of all norms. Let A0 be the subspace of “pure” quaternions.


(1) If a, b ∈ F • then 〈a〉A � 〈b〉A if and only if the classes of a, b coincide inF •/DF (ϕ).(2) Two λ-hermitian spaces (Vi, hi) are similar if (V2, h2) � (V1, r ·h1) for some

r ∈ F •.

Lemma. Let a, b ∈ A•0. The following statements are equivalent.

(i) 〈a〉A and 〈b〉A are similar as skew-hermitian spaces.

(ii) b = t · dad for some t ∈ F • and some d ∈ A•.(iii) 〈N(a)〉 = 〈N(b)〉 in F •/F •2.

(3) It is harder to characterize isometry. The lemma above reduces the question todetermining the “similarity factors” of the space 〈a〉A. Suppose a ∈ A•

0 is given andlet x = N(a), so that the norm form is ϕ � 〈〈x, y〉〉 for some y ∈ F •.

If t ∈ F • then: 〈t · a〉A � 〈a〉A if and only if t ∈ DF (〈〈x〉〉) ∪ −y ·DF (〈〈x〉〉).In particular: 〈t · a〉A � 〈a〉A for every t ∈ F • if and only if DF (〈〈x〉〉) is a subgroupof index 1 or 2 in F •.(4) Let D be the quaternion division algebra over R with the “bar” involution.

Isometry of hermitian spaces overD is determined by the dimension and the signature.Isometry of skew-hermitian spaces over D is determined by the dimension.

(Hint. (2) (iii) �⇒ (ii). Given N(b) = s2 · N(a) for s ∈ F •, alter a to assumeN(b) = N(a).

Claim. There exists u ∈ A• such that b = uau−1. (If A is split this is standardlinear algebra. Suppose A is a division algebra. If b = −a choose u ∈ F(a)⊥,otherwise let u = a + b.)(3) Choose b ∈ A0 with ab = −ba and N(b) = y. From the isometry we have

t · a = dad for some d ∈ A•. Then t = λN(d) where λ = ±1, so that λad = da. Ifλ = 1 then d ∈ F(a) while if λ = −1 then d ∈ b · F(a).)

11. Associated Hermitian Forms. Supposeσ � 〈1, a2, . . . , as〉 and letC = C(−σ1)and J = JS as usual. Let � be the trace map with �(1) = 1 (as in Exercise 3.14).Then J (e�)e� = a� where {e�} is the derived basis. If (V , h) is a λ-hermitian spaceover C, the associated bilinear form B = � � h is defined on V as in Proposition A.2above. Given B we can reconstruct the hermitian form h explicitly as: h(x, y) =∑B((e�)−1x, y)e� =∑

B(x, e�y)J (e�)−1.

12. Let (V , B) be a regular λ-symmetric bilinear space over F . Then the ring E =End(V ) acts on V as well, and the form B admits (End(V ), IB). Use Proposition A.2(with the usual trace map on End(V )) to lift the form B : V × V → F to a uniqueλ-hermitian form h : V × V → End(V ). Exactly what is this form h?(Answer: h(u, v)(x) = B(x, v)u for all x, u, v ∈ V . Compare Exercise 1.13.)


13. Let C be semisimple F -algebra with involution. Then C ∼= A1× · · · ×Ak wherethe Aj are the simple ideals of C. Let ej be the identity element of Aj and let Vj bean irreducible Aj -module, viewed as a C-module by setting AiVj = 0 if i �= j .(1) Every C-module is isomorphic to a direct sum of some of these Vj .(2) Any involution J on C permutes {e1, . . . , ek}. If J (ei) = ej then Vi ∼= Vj .(3) Under what conditions do all the unsplittable λ-spaces admitting (C, J ) have

the same dimension?

14. Existence of an involution trace. Generalize Definition 4.15, allowingK to be afieldwith involution (“bar”), requiring the involution on the algebraC to be compatible(rc = r · c), and replacing condition (4.15) (1) by: �(c) = �(c)

Proposition. Every semisimple K-algebra C with involution has an involutiontrace C → K .

The proof is done in several steps:(1) Every central simpleK-algebraC with involution has an involution traceC →

K .(2) IfE/K is a finite extension of fields with involution, there is an involution trace

E → K . Consequently, if C is an E-algebra with involution having an involutiontrace C → E then there is an involution trace C → K .(3) The proposition follows by considering the simple components.

(Hint. (1) The reduced trace Trd always works. EveryK-linear map � : C → K with�(xy) = �(yx) must be a scalar multiple of Trd. (For [C,C] = span{xy − yx} hascodimension 1.)

15. Homometric elements. Let A be a ring with involution J = “bar”. Elementsa, b ∈ A are called homometric if aa = bb. If u ∈ A is a “spectral unit”, that isif uu = 1, then a and ua are homometric. We say that (A, J ) has the homometricproperty if the converse holds: if aa = bb then a and b differ by a spectral unit.(1) If A is a division ring then (A, J ) has the homometric property.(2) Suppose (A, J ) has the homometric property and a ∈ A. If aa is nilpotent,

then a = 0. Consequently A has no non-trivial nil ideals.(3) Let A = End(V ) and J = Ih, where (V , h) is an anisotropic hermitian space

over a field F with involution. (For example, A ∼= Mn(C) with J = conjugate-transpose.) Then (A, J ) has the homometric property.(4) What other semisimple rings with involution have the homometric property?

(Hint. (3)Givenf, g ∈ End(V ). Then ker(f f ) = ker(f ) and hence f f (V ) = f (V ).If f f = gg then f (V ) = g(V ). Construct an isometry σ : f (V )→ g(V ). By WittCancellation, σ extends to an isometry ϕ : V → V . Then ϕϕ = 1 and ϕf = g.)


Notes on Chapter 4

The proof of Proposition 4.9 follows Knebusch (1970), especially (3.2.1) and (3.3.2).

Exercises 4 and 5 are derived from §6 of Shapiro (1977a).

Exercise 8. Matrix factorizations of quadratic forms are investigated in Buchweitz,Eisenbud and Herzog (1987), where they are related to graded Cohen–Macauley mod-ules over certain rings. A similar situation was studied by Eichhorn (1969), (1970).Matrix factorizations of forms of higher degrees are analyzed using generalized Clif-ford algebras by Backelin, Herzog and Sanders (1988).

Exercise 14. This follows Shapiro (1976), Proposition 4.4.

Exercise 15. See Rosenblatt and Shapiro (1989).

Chapter 5

Small (s, t)-Families

As a break from the general theory of algebraswith involutionwe present some explicitexamples of (σ, τ )-modules for small (s, t)-pairs. Since we are concerned with these(σ, τ )-modules up to F -similarity, we will work only with quadratic forms over F .Good information is obtained for (σ, τ )-modules when the unsplittables have di-

mension at most 4. In these cases we can classify the (σ, τ )-modules in terms ofcertain Pfister factors. The smallest case where non-Pfister behavior can occur is for(2, 2)-families. The unsplittable (〈1, a〉, 〈x, y〉)-modules are analyzed using a new“trace” method. We obtain concrete examples where the unsplittable module is notsimilar to a Pfister form.As a convenience to the reader we provide the proofs of the basic properties of

Pfister forms, even though this theory appears in a number of texts. If q is a quadraticform recall that the value set and the norm group are:

DF (q) = {c ∈ F • : q represents c} and GF (q) = {c ∈ F • : 〈c〉q � q}.One easily checks that GF (q) · DF (q) ⊆ DF (q). A (regular) quadratic form ϕ isdefined to be round if GF (ϕ) = DF (ϕ). In particular this implies that the value setDF (n) is a multiplicative group. We will prove below that every Pfister form is round.We need the notion of “divisibility” of quadratic forms:

α || β means that β � α ⊗ δ, for some quadratic form δ.

For anisotropic forms, we have seen in (3.20) that divisibility by a binary form 〈1, b〉 =〈〈b〉〉 is determined by behavior under a quadratic extension. We restate that result heresince it is so important for motivating some of the later work.

5.1 Lemma. Let q be an anisotropic quadratic form over F . If q ⊗ F(√−b) is

isotropic then q � 〈x〉〈1, b〉 ⊥ q1 for some x ∈ F • and some form q1 over F .Consequently, q ⊗ F(

√−b) is hyperbolic iff 〈〈b〉〉 || q.

Proof. See Lemma 3.20. ��

5.2 Proposition. Let ψ be a round form over F , and ϕ = 〈〈a〉〉 ⊗ψ for some a ∈ F •.(1) Then ϕ is also round. Consequently, every Pfister form is round.

5. Small (s, t)-Families 91

(2) If ϕ is isotropic then it is hyperbolic.

(3) Suppose ϕ is a Pfister form and define the pure subform ϕ′ by ϕ � 〈1〉 ⊥ ϕ′. Ifb ∈ DF (ϕ′) then 〈〈b〉〉 || ϕ.

Proof. (1) Suppose ϕ represents c, say c = x + ay where x, y ∈ DF (ψ) ∪ {0}.Suppose x, y �= 0. (The other cases are easier and are left to the reader.) Comparingdeterminants we see that 〈x, ay〉 � 〈c, caxy〉. Since ψ is round DF (ψ) = GF (ψ)

is a group so that x, y and xy lie in GF (ψ). Then ϕ � 〈〈a〉〉 ⊗ ψ � 〈x, ay〉 ⊗ψ � 〈c〉〈1, axy〉 ⊗ ψ � 〈c〉〈〈a〉〉 ⊗ ψ � 〈c〉ϕ and consequently ϕ is round. Aninduction proof now shows that a Pfister form ϕ is round, for if dim ϕ = 2m > 1 thenϕ ∼= 〈〈a〉〉 ⊗ ψ where ψ is another Pfister form.(2) Since ϕ is isotropic there exist x, y ∈ DF (ψ) such that x + ay = 0. Then

−a = xy−1 ∈ GF (ψ) so that ϕ � 〈〈−xy−1〉〉 ⊗ ψ � 〈〈−1〉〉 ⊗ ψ is hyperbolic.(3) We are given ϕ � 〈〈a〉〉 ⊗ ψ for a Pfister form ψ . Note that ψ remains round

under any field extension. By hypothesis, ϕ � 〈1, b, . . .〉. If ϕ is isotropic then by (2)it is hyperbolic and the conclusion is clear. Otherwiseϕ is anisotropic butϕ⊗F(√−b)is isotropic. But then ϕ ⊗ F(√−b) is hyperbolic by (2) applied over this larger field,and (5.1) implies 〈〈b〉〉 || ϕ. ��

The fundamental fact here is that Pfister forms are round. That is, a Pfister form ϕhas multiplicative behavior:

DF (ϕ) is a subgroup of F •.

Applying this to the form 2m〈1〉 we see that the set DF (2m) of all non-zero sums of2m squares in F is closed under multiplication. (See Exercise 0.5 for another proof ofthis fact.)Ifϕ is anym-foldPfister formoverF , the elementϕ(X) inF(X) = F(x1, . . . , x2m)

is represented by the form ϕ ⊗ F(X), and the proposition implies that ϕ(X) lies inGF(X)(ϕ ⊗ F(X)). Writing V for the underlying space of ϕ ⊗ F(X), this says thatthere exists a linear mapping f : V → V with ϕ(f (v)) = ϕ(X)ϕ(v) for every v ∈ V .Writing this out in terms of matrices as done in Chapter 0, we obtain a multiplicationformula

ϕ(X) · ϕ(Y ) = ϕ(Z),

where each entry zk is linear in Y with coefficients in F(X). Further informationappears in the texts by Lam and Scharlau.When the unsplittable (σ, τ )-modules have dimension ≤ 4 we characterize the

unsplittables in terms of certain Pfister forms. In the discussion below we use thequadratic forms over F rather than working directly with modules over the Cliffordalgebras. The module approach provides more information but the proofs tend to belonger.

92 5. Small (s, t)-Families

5.3 Proposition. In the following table, every unsplittable (σ, τ )-module is F -similarto one of the forms q listed.

(σ, τ ) q

(〈1〉, 〈x〉) where 〈x〉 �� 〈1〉 q � 〈〈w〉〉 where x ∈ GF (q)(〈1, a〉, 0) q � 〈〈a〉〉(〈1, a, b〉, 0) q � 〈〈a, b〉〉(〈1, a〉, 〈x〉) where

〈1, a,−x〉 is anisotropic q � 〈〈a,w〉〉 where x ∈ GF (q)(〈1, a〉, 〈x, y〉) where 〈axy〉 �� 〈1〉

and 〈1, a,−x,−y〉 is isotropic q � 〈〈a,w〉〉 where 〈〈xy〉〉 || q(〈1, a, b, c〉, 0) where 〈abc〉 �� 〈1〉 q � 〈〈a, b,w〉〉 where abc ∈ GF (〈〈w〉〉)(〈1, a, b〉, 〈x〉) where〈1, a, b,−x〉 is anisotropic q � 〈〈a, b,w〉〉 where 〈〈abx〉〉 || q

Proof. We will do a few of these cases in detail, leaving the rest to the reader. Firstsuppose (σ, τ ) � (〈1, a, b〉, 0), and σ < Sim(q) is unsplittable where q represents1. Then (1.9) implies σ ⊂ q so that dim q ≥ 3. Since σ < Sim(〈〈a, b〉〉), theDecomposition Theorem shows that dim q = 4. Therefore q � 〈1, a, b, d〉 for somed ∈ F •. Since 〈1, a〉 < Sim(q) we know from an earlier case (or from (1.10)) that〈〈a〉〉 || q, so that det q = 〈1〉. Then 〈d〉 = 〈ab〉 and q � 〈〈a, b〉〉.Suppose (σ, τ ) = (〈1, a〉, 〈x, y〉) where 〈axy〉 �� 〈1〉 and 〈1, a,−x,−y〉 is

isotropic. Then 〈1, a〉 and 〈x, y〉 represent some common value e. Scaling by ewe get an equivalent pair of forms (〈e〉〈1, a〉, 〈e〉〈x, y〉) � (〈1, a〉, 〈1, xy〉). Thereexist 4-dimensional (σ, τ )-modules, for example 〈〈a, xy〉〉. Since 〈axy〉 �� 〈1〉 theunsplittables cannot have dimension 2, and the Decomposition Theorem implies thatevery unsplittable q has dim q = 4. Since 〈1, a〉 < Sim(q) we know q � 〈〈a,w〉〉for some w. Also since 〈1, xy〉 < Sim(q) we have 〈〈xy〉〉 || q. Conversely supposeq � 〈〈a,w〉〉 and 〈〈xy〉〉 || q. Then the form 〈a,w, aw〉 represents xy, so we can ex-press xy = ar2 + wd, for some r ∈ F and d ∈ DF (〈〈a〉〉) ∪ {0}. Then d �= 0,since 〈axy〉 �� 〈1〉, so that q � 〈〈a,wd〉〉 and 〈a,wd〉 represents xy. Therefore(〈1, a〉, 〈1, xy〉) ⊂ (〈1, a, wd〉, 〈1, a, wd〉) < Sim(q), and the result follows.Suppose (σ, τ ) � (〈1, a, b, c〉, 0) where 〈abc〉 �� 〈1〉. There exist (σ, τ )-modules

of dimension 8, for instance 〈〈a, b, c〉〉. If q is an unsplittable module which represents1, then 〈1, a, b, c〉 ⊂ q and since 〈1, a, b〉 < Sim(q) we also have 〈〈a, b〉〉 || q. Ifdim q = 4 we contradict the hypothesis 〈abc〉 �� 〈1〉. Therefore dim q = 8 andq � 〈〈a, b, u〉〉 for some u. Since 〈1, a, b, c〉 ⊂ q we find 〈ab〉 ⊥ 〈u〉〈〈a, b〉〉 representsc and therefore 〈1〉 ⊥ 〈u〉〈〈a, b〉〉 represents abc. Express abc = r2 + ue wherer ∈ F and e ∈ DF (〈〈a, b〉〉) ∪ {0}. Since 〈abc〉 �� 〈1〉 we find e �= 0 and thereforeq � 〈〈a, b, ue〉〉 and 〈1, ue〉 represents abc. The desired shape for q follows when weset w = ue. The converse follows as before. ��


In the small cases analyzed above we can go on to characterize arbitrary (σ, τ )-modules. For instance it immediately follows from (5.2) that 〈1, a〉 < Sim(q) if andonly if 〈〈a〉〉||q, and that 〈1, a, b〉 < Sim(q) if and only if 〈〈a, b〉〉||q. For the other caseswe need a decomposition theorem for Pfister factors analogous to the DecompositionTheorem 4.1.

5.4 Definition. Let M be a set of (isometry classes of) quadratic forms over F . Aquadratic form q ∈ M isM-indecomposable if there is no non-trivial decompositionq � q1 ⊥ q2 where qi ∈ M.

Certainly any form q in M can be expressed as q � q1 ⊥ · · · ⊥ qk where eachqj isM-indecomposable. We will get some results about theM-indecomposables forsome special classesM. Generally if the ϕi are round forms and bj ∈ F • we considerthe classes of the type

M = M(ϕ1, . . . , ϕk, b1, . . . , bn) = {q : ϕi || q and bj ∈ GF (q) for every i, j}.TheM-indecomposables are easily determined in a few small cases. For instance,

for a single round form ϕ we see that q is M(ϕ)-indecomposable if and only if q issimilar to ϕ. For a single scalar b ∈ F • where 〈b〉 �� 〈1〉, everyM(b)-indecomposablehas dimension 2. (This is Dieudonné’s Lemma of Exercise 2.9; also see Exercise 7.)The Proposition 5.6 below generalizes these two cases.We first prove a lemma about “division” by round forms which is of some interest

in its own right. Recall thatH = 〈1,−1〉 is the hyperbolic plane and that any quadraticform q has a unique “Witt decomposition” q = qa ⊥ qh where qa is anisotropic andqh � mH is hyperbolic.

5.5 Lemma. Suppose ϕ is a round form.

(1) If ϕ || q and a ∈ DF (q) then q � ϕ ⊗ α for some form α which represents a.If ϕ || q and q is isotropic with dim q > dim ϕ then q � ϕ⊗ α for some isotropicform α.

(2) If ϕ || α ⊥ β and ϕ || α then ϕ || β.

(3) Suppose ϕ is anisotropic. Then: ϕ ||mH if and only if dim ϕ ||m. If ϕ || q thenϕ || qa and ϕ || qh, where q = qa ⊥ qh is the Witt decomposition.

Proof. (1) If q � ϕ ⊗ 〈b1, . . . , bn〉 represents a then a = b1x1 + · · · + bnxn forsome xj ∈ DF (ϕ) ∪ {0}. Define yj = xj if xj �= 0 and yj = 1 if xj = 0 andset α = 〈b1y1, . . . , bnyn〉. Then α represents a and since ϕ ⊗ 〈yj 〉 � ϕ we haveq � ϕ ⊗ α. If q is isotropic we use a non-trivial representation of a = 0. If the termsxi above are not all 0 the previous argument works. Otherwise the non-triviality of therepresentation implies that ϕ must be isotropic and hence universal. Since ϕ is roundthis implies that 〈c〉ϕ � ϕ for every c ∈ F •. In particular, ϕ⊗〈b1, b2〉 � ϕ⊗〈1,−1〉and the result follows.


(2) Apply induction on dim α. If a ∈ DF (α) then part (1) implies that α ⊥ β �〈a〉ϕ ⊥ δ and α � 〈a〉ϕ ⊥ α0 for some forms δ and α0 such that ϕ || δ and ϕ || α0.Cancelling we find that δ � α0 ⊥ β and the induction hypothesis applies.(3) Let k = dim ϕ. The “if” part is clear since ϕ ⊗ H � kH. For the “only if”

part we use induction onm, assuming ϕ ||mH. Since ϕ is anisotropic we know k ≤ m.Part (1) implies that mH � ϕ ⊗ α where α is isotropic. Expressing α � H ⊥ α′ wehave mH � kH ⊥ (ϕ ⊗ α′). If k = m we are done. Otherwise, ϕ || (m− k)H and theinduction hypothesis applies.For the last statement let qh � mH and use induction on m. We may assume

m > 0. Then q is isotropic and dim q > dim ϕ (since ϕ is anisotropic). Part (1)implies that q � ϕ ⊗ α for some isotropic α. Expressing α � α′ ⊥ H we haveqa ⊥ mH � q � (ϕ ⊗ α′) ⊥ kH where k = dim ϕ. Therefore k ≤ m andcancellation implies ϕ || (qa ⊥ (m − k)H). The result follows using the inductionhypothesis. ��

5.6 Proposition. Suppose ϕ is a Pfister form and b ∈ F •. Then all M(ϕ, b)-indecomposables have the same dimension and all M(ϕ, 〈〈b〉〉)-indecomposables havethe same dimension.

Proof. We will consider the case M = M(ϕ, 〈〈b〉〉) here, leaving the other to thereader. Suppose q is anM-indecomposable which represents 1. If 〈〈b〉〉 || ϕ it is clearthat q � ϕ.Suppose 〈〈b〉〉 � ϕ. By (5.5), q � ϕ ⊥ q1 and 〈〈b〉〉 ⊂ q. Then b ∈ DF (ϕ′ ⊥ q1)

so that b = x + y where x ∈ DF (ϕ′) ∪ {0} and y ∈ DF (q1) ∪ {0}. If y = 0 thenb = x ∈ DF (ϕ

′) and (5.2) (3) implies that 〈〈b〉〉 || ϕ, contrary to hypothesis. Theny �= 0 and by (5.5) again we have q1 � 〈y〉ϕ ⊥ q2 where ϕ || q2, and thereforeq � ϕ ⊗ 〈〈y〉〉 ⊥ q2. Since α = ϕ ⊗ 〈〈y〉〉 is a Pfister form and α′ = ϕ′ ⊥ 〈y〉ϕrepresents b we know that 〈〈b〉〉 || α so that α ∈ M. By (5.5) (2) we also have q2 ∈ M.Since q isM-indecomposable, q2 must be 0 and dim q = 2 · dim ϕ. ��

Since ϕ is a Pfister form here we see that everyM-indecomposable is also a Pfisterform.

5.7 Proposition. (1) (〈1〉, 〈x〉) < Sim(q) iff x ∈ G(q).(2) 〈1, a〉 < Sim(q) iff 〈〈a〉〉 || q.

(3) 〈1, a, b〉 < Sim(q) iff 〈〈a, b〉〉 || q.

(4) (〈1, a〉, 〈x〉) < Sim(q) iff 〈〈a〉〉 || q and x ∈ G(q).(5) If 〈1, a,−x,−y〉 is isotropic, then (〈1, a〉, 〈x, y〉) < Sim(q) iff 〈〈a〉〉 || q and

〈〈xy〉〉 || q.

(6) 〈1, a, b, c〉 < Sim(q) iff q � 〈〈a, b〉〉 ⊗ γ where abc ∈ GF (γ ).(7) (〈1, a, b〉, 〈x〉) < Sim(q) iff 〈〈a, b〉〉 || q and 〈〈abx〉〉 || q.


Proof. Wewill prove the last two, omitting the others. Suppose 〈1, a, b, c〉 < Sim(q).If 〈abc〉 � 〈1〉 the result is easy, so suppose 〈abc〉 �� 〈1〉. If 〈1, a, b, c〉 < Sim(q)then q is a sum of unsplittables of the type listed in (5.3), and it follows that q �〈〈a, b〉〉 ⊗ γ where abc ∈ GF (γ ). Conversely suppose q is given in this way. Since〈abc〉 �� 〈1〉, Proposition 5.6 implies that the M(abc)-indecomposables all havedimension 2. Therefore γ � γ1 ⊥ · · · ⊥ γr where dim γj = 2 and γj ∈ M(abc).Then γj � 〈uj 〉〈〈wj 〉〉 where abc ∈ GF (〈〈wj 〉〉) and we get q � q1 ⊥ · · · ⊥ qrwhere qj � 〈uj 〉〈〈a, b,wj 〉〉. Again by Proposition 5.3 we conclude that 〈1, a, b, c〉 <Sim(q).Now consider the case (〈1, a, b〉, 〈x〉). If 〈1, a, b〉 represents x then 〈〈abx〉〉||〈〈a, b〉〉

and 〈1, a, b〉 < Sim(q) if and only if (〈1, a, b〉, 〈x〉) < Sim(q). Therefore we mayassume 〈1, a, b,−x〉 is anisotropic. If (〈1, a, b〉〈x〉) < Sim(q) then q � q1 ⊥· · · ⊥ qr where each (〈1, a, b〉, 〈x〉) < Sim(qj ) is unsplittable. By Proposition 5.3we have 〈〈a, b〉〉 || qj and 〈〈abx〉〉 || qj , and the claim follows. Conversely supposethat q ∈ M = M(〈〈a, b〉〉, 〈〈abx〉〉). Then q � q1 ⊥ · · · ⊥ qr where each qj isM-indecomposable. Since 〈1, a, b,−x〉 is anisotropic we see that 〈〈abx〉〉 � 〈〈a, b〉〉and Proposition 5.6 implies that dim qj = 8. Therefore qj � 〈uj 〉〈〈a, b,wj 〉〉 where〈〈abx〉〉 || qj . Apply (5.3) again to conclude that (〈1, a, b〉, 〈x〉) < Sim(qj ) for each jand therefore (〈1, a, b〉, 〈x〉) < Sim(q). ��

The rest of this chapter is concerned with the more difficult case of (2, 2)-families.Let (σ, τ ) = (〈1, a〉, 〈x, y〉). The case when 〈1, a,−x,−y〉 is isotropic is includedin Proposition 5.7. If 〈axy〉 � 〈1〉 then (〈1, a〉, 〈x, y〉) < Sim(q) iff (〈1, a〉, 〈x〉) <Sim(q), by the Expansion Lemma. This case is also included in (5.7). Therefore letus assume that

〈1, a,−x,−y〉 is anisotropic and 〈axy〉 �� 〈1〉.Let C = C(〈−a, x, y〉) be the associated Clifford algebra. Then the center of Cis isomorphic to the field E = F(

√axy) and C is a quaternion algebra over E. It

follows thatC is a division algebra (this is part of Exercise 3.16) and every unsplittable(σ, τ )-module has dimension 8. Let JS denote the usual involution on C.If (〈1, a〉, 〈x, y〉) < Sim(V , q) then we have 〈〈a〉〉 || q, 〈〈xy〉〉 || q and x ∈ GF (q).

It is not so clear whether the converse holds: do those “divisibility” conditions on qalways imply the existence of the (2, 2)-family? Those conditions do provide somemotivation for the following approach.To say that (〈1, a〉, 〈x, y〉) < Sim(V , q) is equivalent to saying that (V , q) is a

quadratic (C, JS)-module. In this case we have f2, g1, g2 ∈ End(V )which satisfy thefamiliar rules listed in Lemma 2.3. Then f = f2 satisfies f = −f and f 2 = −a ·1 soit corresponds to the subspace 〈1, a〉 < Sim(q). Similarly g = g1g2 satisfies g = −gand g2 = −xy ·1 so that g corresponds to 〈1, xy〉 < Sim(q). Since f and g commute,they induce an action of the field K = F(

√−a,√−xy) on the vector space V . LetJ be the involution of K sending

√−a and √−xy to their negatives. Then (V , q)becomes a quadratic (K, J )-module. Naturally we may view K as a subfield of C


where J is the restriction of JS to K . Note that E = F(√axy) is the subfield of K

which is fixed by J . We often write “bar” for J when there is no ambiguity.Conversely, if (V , q) is a quadratic (K, J )-module, what further information is

needed to make it a (C, JS)-module?

5.8 Lemma. Suppose (K, J ) is the field with involution described above and (V , q)is a quadratic (K, J )-module. This structure extends to make (V , q) into a (C, JS)-module if and only if there exists k ∈ EndF (V ) such that k is (K, J )-semilinear, k = k,and k2 = x · 1.

Proof. The (K, J )-semilinearity of k means that k(αv) = αk(v) for every α ∈ K andv ∈ V . This is equivalent to saying that k anticommutes with f and g. If the (C, JS)-module structure is given, just define k = g1. Conversely, given the (K, J )-modulestructure and given k, define f2 = f , g1 = k, g2 = k−1g and verify that they providethe desired (2, 2)-family. ��

Suppose (V , q) is a (K, J )-module. Then the symmetric bilinear formbq : V × V → F is the “transfer” of a hermitian form over K . This is a special caseof Proposition A.2 of Chapter 4, applied to the algebra C = K . In fact if s : K → F

is an involution trace then there exists a unique hermitian form h : V × V → K suchthat bq = s � h. (See Exercise 16.) This hermitian space (V , h) is an orthogonal sumof some 1-dimensional spaces over K:

(V , h) � 〈θ1, θ2, . . . , θm〉K for some θi ∈ E•.

These diagonal entries θi lie in E since they must be symmetric: θi = θi .To do calculations we must choose an involution trace. First note that K =

E(√−a) and define tr : K → E by setting tr(1) = 1 and tr(

√−a) = 0. (This is theunique involution trace from K to E, up to scalar multiple.) Since the involution onE = F(

√axy) is trivial there are many involution traces from E to F . We will use

the standard one employed in quadratic form theory, namely � : E → F defined bysetting �(1) = 0 and �(

√axy) = 1. If θ ∈ E• then the 1-dimensional hermitian space

〈θ〉K over K transfers down to a 4-dimensional quadratic space � � tr(〈θ〉K) over F .

5.9 Lemma. Suppose θ ∈ E•. Then � � tr(〈θ〉K) � 〈s〉〈〈a,−Nθ〉〉 over F .

Proof. Here N is the field norm NE/F . If θ = r + s√axy then N(θ) = r2 − axys2.The 1-dimensional hermitian space 〈θ〉K can be viewed as the K-vector space Kwith the form h : K × K → K given by h(x, y) = θxy. If b = tr �h thenb(u, v) = θ · tr(uv). If u = u1 + u2

√−a and v = v1 + v2√−a we find that

tr(uv) = u1v1 + au2v2, so that tr(〈θ〉K) � 〈θ, aθ〉E as quadratic forms over E.To transfer the quadratic form 〈θ〉E from E = F(

√axy) down to F , we compute

the inner products (relative to this form) of the basis elements {1,√axy} to find the


Grammatrix

(s r

r axys

). Since it represents s and has determinant−N(θ), we have

�(〈θ〉E) � 〈s〉〈〈−Nθ〉〉, provided s �= 0. If s = 0 that form is isotropic, hence is H.The result now follows since � � tr(〈θ〉K) = �(〈θ〉E) ⊥ 〈a〉�(〈θ〉E). ��

Remark. If θ, θ ′ ∈ E• then 〈θ〉K � 〈θ ′〉K as hermitian spaces over K if and only ifθ ′ = ααθ for some α ∈ K•. (For a K-linear map K → K must be multiplication bysome α ∈ K .) If such α exists then the transferred quadratic forms over F must alsobe isometric.

Our goal is to construct unsplittable modules (V , q) for the (2, 2)-pair(〈1, a〉, 〈x, y〉). Then dimF V = 8 and dimK V = 2 using the induced (K, J )-action, and we view (V , q) as the transfer of a hermitian space (V , h) = 〈θ1, θ2〉K .Given such a hermitian space over K , we will find conditions on θ1 and θ2 whichimply that this (K, J )-action can be extended to an action of (C, JS).

5.10 Lemma. Suppose (V , q) is the transfer of the hermitian space (V , h) = 〈θ1, θ2〉K .The following statements are equivalent, where θ = θ1θ2.

(1) (V , q) is a (C, JS)-module in a way compatible with the given (K, J )-action.

(2) 〈1, θ〉K represents x, that is, αα + θββ = x for some α, β ∈ K•.(3) 〈1, a, θ, aθ〉 represents x over E.

(4) −θ ∈ DE(〈〈a,−x〉〉).

Proof. We use a matrix formulation of (1) to show its equivalence with (2). By (5.8),condition (1) holds if and only if there exists k ∈ EndF (V )which is (K, J )-semilinear,k = k and k2 = x · 1. We are given a K-basis {v1, v2} of V such that h(vi, vi) = θiand h(v1, v2) = 0. Representing a vector v = x1v1 + x2v2 in V as a column vector

X =(x1x2

), the (K, J )-semilinear map k is represented by a matrix A =

(α γ

β δ

)where α, β, γ, δ ∈ K . This is done so that k(v) = x′1v1 + x′2v2 is represented by thecolumn vector X′ = AX.The adjoint map k is also (K, J )-semilinear and has matrix A = M−�A�M

where M =(θ1 00 θ2

)is the matrix of the hermitian form. (See Exercise 15 for

more details.) The symmetry condition k = k is equivalent to the symmetry of the

matrixMA =(θ1α θ1γ

θ2β θ2δ

). This condition holds if and only if θ2β = θ1γ . Define

θ = θ2/θ1 (which has the same square class in E• as θ1θ2). Then the symmetrycondition becomes: γ = θβ.The condition k2 = x ·1 becomes the matrix equationAA = xI (see Exercise 15).

On multiplying this out when A =(α θβ

β δ

)we find it to be equivalent to the


following equations:

αβ = −βδ, αα = δδ, αα + θββ = x.

If β = 0 then αα = x so that x ∈ DE(〈1, a〉) and 〈1, a,−x〉E is isotropic overE. Weknow that C =

(−a,xE

)is a quaternion division algebra, so its norm form 〈〈a,−x〉〉E

is anisotropic over E. This is a contradiction.Therefore β �= 0 and δ = −αββ−1. With this formula for δ the first two equations

above are automatic. Therefore statement (1) holds if and only if there exist α, β ∈ Ksuch that αα + θββ = x. This is statement (2).The equivalence of statements (2) and (3) is clear since {αα : α ∈ K} =

DE(〈1, a〉). Finally note that (3) holds if and only if 〈〈a,−x, θ〉〉 is hyperbolic overE,if and only if 〈〈a,−x〉〉 represents −θ over E. Therefore (3) and (4) are equivalent.��

In the statement of Lemma 5.10 the symmetry between x and y is not apparent.However, since 〈axy〉E � 〈1〉E we may note that 〈1, a,−x,−y〉E � 〈〈a,−x〉〉E �〈〈a,−y〉〉E .The payoff of these calculations can now be summarized.

5.11 Proposition. Suppose a, x, y ∈ F • such that 〈axy〉 �� 〈1〉 and 〈1, a,−x,−y〉 isanisotropic. Let E = F(

√axy) and let N = NE/F be the norm. If q is a quadratic

form over F with dim q = 8, then the following statements are equivalent:

(1) (〈1, a〉, 〈x, y〉) < Sim(q).(2) There exist θi = ri + si√axy in E• such that

q � 〈s1〉〈〈a,−Nθ1〉〉 ⊥ 〈s2〉〈〈a,−Nθ2〉〉 and such that −θ1θ2 ∈ DE(〈〈a,−x〉〉).

Proof. Here, as before, if si = 0 the corresponding term in the expression for q isinterpreted as 2H. This equivalence is obtained by combining (5.9) and (5.10). ��

As one immediate consequence we see that (〈1, a〉, 〈x, y〉) < Sim(4H) for everya, x, y. To prove the next corollary we use the following “Norm Principle”.

5.12 Lemma. Let E = F(√d) be a quadratic extension, let ϕ be a Pfister form over

F and let θ ∈ E•. Then θ ∈ F • ·DE(ϕE) if and only if Nθ ∈ DF (ϕ).

Proof. See Elman and Lam (1976). The proof is outlined in Exercise 17. ��

5.13 Corollary. Suppose a, x, y ∈ F • as above and suppose ϕ is a 2-fold Pfisterform. The following are equivalent.

(1) (〈1, a〉, 〈x, y〉) < Sim(ϕ ⊥ 2H).

(2) ϕ ∼= 〈〈a,−c〉〉 for some c ∈ DF (〈〈−axy〉〉) ∩DF (〈〈a,−x〉〉).


Proof. (1) �⇒ (2). By (5.5) 〈〈a〉〉 || ϕ so that ϕ � 〈〈a,w〉〉 for some w. Furthermoreϕ ⊥ 2H � 〈s1〉〈〈a,−Nθ1〉〉 ⊥ 〈s2〉〈〈a,−Nθ2〉〉 for θi as in (5.11). Computing Wittinvariants we find that ϕ � 〈〈a,−c〉〉 where c = N(θ1θ2) ∈ DF (〈〈−axy〉〉). Since−θ1θ2 ∈ DE(〈〈a,−x〉〉) the lemma implies that c ∈ DF (〈〈a,−x〉〉).(2) �⇒ (1). We may express c = Nθ ∈ DF (〈〈a,−x〉〉). If 〈c〉 � 〈1〉 the claim

is vacuous, so we may assume that θ �∈ F . By the lemma we find that θ = t · θ1where t ∈ F • and −θ1 ∈ DE(〈〈a,−x〉〉). Let θ2 = 1 and apply (5.11) to concludethat (〈1, a〉, 〈x, y〉) < Sim(q) where q � 〈s1〉〈〈a,−Nθ1〉〉 ⊥ 2H. Then 〈s1〉q �〈〈a,−c〉〉 ⊥ 2H and the result follows. ��

Example 1. Let a = 1, x = −1 and y = −2 over the rational field Q. Then〈1, a,−x,−y〉 � 〈1, 1, 1, 2〉 is anisotropic and 〈axy〉 � 〈2〉 �� 〈1〉. If ϕ is a 2-foldPfister form then (5.13) says: (〈1, 1〉, 〈−1,−2〉) < Sim(ϕ ⊥ 2H) if and only ifϕ � 〈〈1,−c〉〉 for some c ∈ DQ(〈〈−2〉〉) ∩DQ(〈〈1, 1〉〉). For example

(〈1, 1〉, 〈−1,−2〉) < Sim(q) where q = 〈〈1,−7〉〉 ⊥ 2H.

To get anisotropic examples we use the criterion in (5.11). To deduce that a form〈1, a,−x,−ax, θ〉E is isotropic over an algebraic number field we need only check(by the Hasse–Minkowski Theorem) that it is indefinite relative to every orderingof E.

Example 2. Let a = 1, x = 7 and y = 14 over Q. Then 〈axy〉 � 〈2〉 �� 〈1〉 and〈1, a,−x,−y〉 � 〈1, 1,−7,−14〉 is anisotropic. (In fact it is anisotropic over thefield Q7.) Furthermore E = Q(

√2), and K = Q(

√−1,√−2). For any θ ∈ E• theform 〈〈a,−x〉〉 ⊥ 〈−θ〉 = 〈1, 1,−7,−7, θ〉 is isotropic over E = Q(

√2) since it is

indefinite at both orderings. Using θ1 = 1+√2 and θ2 = 1+ 2√2 we find:

(〈1, 1〉, 〈7, 14〉) < Sim(q) where q = 〈〈1, 1〉〉 ⊥ 〈〈1, 7〉〉 � 〈〈1〉〉 ⊗ 〈1, 1, 1, 7〉.This form q is anisotropic but is not similar to a Pfister form.

Many further examples can be constructed along these lines. Non-Pfister unsplit-tables for larger (s, t)-families can be found using the Construction and Shift Lemmas.The smaller families considered earlier in this chapter were all characterized by certain“division” properties of quadratic forms. If (〈1, a〉, 〈x, y〉) < Sim(q) then certainly〈〈a〉〉 || q, 〈〈xy〉〉 || q and x ∈ GF (q). Is it possible that these three independent con-ditions suffice? Certainly this converse fails for simple reasons of dimensions: the4-dimensional form 2H always satisfies these divisibility conditions, but the dimen-sions of the unsplittables may be 8. We modify the conjecture as follows:

5.14 Question. If dim q = 8 and 〈〈a〉〉 ||q, 〈〈xy〉〉 ||q and x ∈ GF (q) does it follow that(〈1, a〉, 〈x, y〉) < Sim(q)?


We may assume that 〈1, a,−x,−y〉 is anisotropic and 〈axy〉 �� 〈1〉 since the othercases are settled by Proposition 5.7. The answer is unknown in general. In Chapter 10we succeed in proving the answer to be “yes” when F is a global field. As the ideasused to prove (5.7) indicate, the following question is relevant.

5.15 Question. If M = M(〈〈a〉〉, 〈〈xy〉〉, x) what are the dimensions of the M-indecomposables?

We will see in Chapter 10 that over a global field the indecomposables must havedimension 2 or 4. It is unknown what dimensions are possible over arbitrary fields.The following observation is interesting (and perhaps surprising) in light of the

F ⊆ E ⊆ K set-up used above. To simplify notations we use b in place of xy here.

5.16 Proposition. Suppose 〈a〉 �� 〈b〉 over F . Let K = F(√−a,√−b) with involu-

tion J as above. The following statements are equivalent for a quadratic space (V , q)over F .

(1) 〈〈a〉〉 || q and 〈〈b〉〉 || q.

(2) (V, q) can be made into a (K, J )-module.

Proof. (2) �⇒ (1). Given the (K, J )-module (V , q) let f = L(√−a) be the mul-

tiplication map on V . Then f ∈ End(V ) and f 2 = −a · 1. Since q admits (K, J )and J (

√−a) = −√−a we know that f = −f . Therefore {1V , f } span a space ofsimilarities 〈1, a〉 < Sim(V , q) and we conclude that 〈〈a〉〉 || q by (1.10). Similarlyusing g = L(

√−b) we find 〈〈b〉〉 || q.(1) �⇒ (2). It suffices to settle the case dim q = 4. This follows from (5.6) since

theM(〈〈a〉〉, 〈〈b〉〉)-indecomposables are 4-dimensional. Given dim q = 4 and 〈〈a〉〉 ||qwe know there exists f ∈ End(V ) with f = −f and f 2 = −a · 1. Similarly since〈〈b〉〉 || q there exists g ∈ End(V ) with g = −g and g2 = −b · 1. The difficulty is tofind such f , g which commute.We may assume q represents 1. By hypothesis, q � 〈〈a, c〉〉 � 〈〈b, d〉〉 for some

c, d ∈ F •. In fact we may assume c = d, (see Exercise 19) so that

q � 〈〈c, a〉〉 � 〈〈c, b〉〉 and 〈a, ac〉 represents b.Let {v1, w1, v2, w2} be the orthogonal basis corresponding to q � 〈1, c, a, ac〉

and define f by setting f (v1) = v2, f (v2) = −av1 and similarly for the wi’s. Thenf = −f and f 2 = −a · 1. The matrix of f can be expressed as a Kronecker product:f =

(0 −a1 0

)⊗(1 00 1

).

Let {v1, w1, v2, w2} be a basis corresponding to q � 〈1, c, b, bc〉 and define ganalogously using this new basis, so that g = −g and g2 = −b · 1. To compute gexplicitly, express b = ax2 + acy2 for some x, y ∈ F , and use v2 = xv2 + yw2

Documents

Department of Mathematics | - HistoricalBackgroundquadratic forms involved in these compositions are related to Pﬁster forms. In the 1960s Pﬁster found that for every m there do