U.U.D.M. Project Report 2013:11

Examensarbete i matematik, 15 hpHandledare och examinator: Ernst DieterichJuni 2013

Department of MathematicsUppsala University

The (1, 2, 4, 8)-Theorem for CompositionAlgebras

Rasmus Précenth

The (1, 2, 4, 8)-Theorem for Composition

Algebras

Rasmus Précenth

June 2, 2013

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Contents

1 Introduction 31.1 Hurwitz Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 The (1,2,4,8)-Theorem for Real Division Algebras . . . . . . . . . 3

2 Preliminaries 42.1 Quadratic Forms and Bilinear Forms . . . . . . . . . . . . . . . . 42.2 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Composition Algebras 93.1 Definition and Basic Properties . . . . . . . . . . . . . . . . . . . 93.2 Purely Imaginary Elements and Conjugation . . . . . . . . . . . 123.3 Associativity Properties . . . . . . . . . . . . . . . . . . . . . . . 153.4 Doubling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4 The (1, 2, 4, 8)-Theorem 234.1 Proving Hurwitz Theorem . . . . . . . . . . . . . . . . . . . . . . 25

Appendices 28

Appendix A Hurwitz Problem 28

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1 Introduction

1.1 Hurwitz Problem

The topic of quadratic forms comes up in various parts of algebra. The so-calledHurwitz Problem, named after the german mathematician Adolf Hurwitz, asksfor what values of n the following identity holds(

n∑i=1

x2i

)(n∑

i=1

y2i

)=

n∑i=1

z2i (1.1)

where all xi, yi ∈ R and each zi is a linear combination of {xiyj | 1 < i, j ≤n}. The values of n where this is true is for n = 1, 2, 4, 8. This statement iscalled Hurwitz Theorem. This thesis will show that these are in fact the onlyvalid values of n no matter which field we’re considering. A theorem provedusing the framework of composition algebras by Nathan Jacobson for fields withcharacteristic not two (Jacobson 1958 [3]). Tonny Albert Springer completedthis theorem by proving it for characteristic two as well (Springer 1963 [4]).

The identities for 1 and 2 are relatively simple and are

x21y21 = (x1y1)

2 (1.2)

and(x21 + x

22)(y

21 + y

22) = (x1y1 − x2y2)2 + (x1y2 + x2y1)2. (1.3)

The ones for 4 and 8 are highly non-trivial. They are included in in AppendixA.

1.2 The (1,2,4,8)-Theorem for Real Division Algebras

A division algebra is a non-zero algebra1 A in which the linear operators La andRa are invertible for each a ∈ A \ {0}. If the base field is R then the algebra iscalled real. Well-known examples of real division algebras are;

• The algebra of real numbers R

• The algebra of complex numbers C

• The algebra of quaternions H

• The algebra of octinions O

These have dimensions 1,2,4 and 8 respectively.In 1877 the german mathematician Ferdinand Georg Frobenius proved the

following theorem for finite-dimensional real division algebras (Frobenius 1878[2]).

Theorem 1.1 (Theorem of Frobenius). There are, up to isomorphism, onlythree associative finite-dimensional real division algebras. They are R, C and Hwith H being the only non-commutative algebra among them.

1See section 2.2 for the definition of an algebra.

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Max Zorn, most famous for his lemma considering partial orders, later proveda theorem analogous to the theorem of Frobenius (Zorn 1931 [6]). In 1931 heproved the following.

Theorem 1.2. There are, up to isomorphism, only four alternative2 finite-dimensional real division algebras. They are R, C, H and O with O being theonly non-associative algebra among them.

This thesis will show that there is a similar theorem for composition algebras,where there is no restriction on the field.

2 Preliminaries

2.1 Quadratic Forms and Bilinear Forms

Definition Given a field k of any characteristic and a vector space V over saidfield. A quadratic form on V is a map N : V → k satisfying

N(λx) = λ2N(x) ∀λ ∈ k, x ∈ V (2.1)

and that the associated map 〈 , 〉N : V × V → k given by

〈x, y〉N = N(x+ y)−N(x)−N(y) (2.2)

is bilinear. It follows easily that N(0) = 0 and that 〈 , 〉N is symmetric. Therest of this thesis will only consider symmetric bilinear forms for this reason.That means that if nothing else is specified we assume that bilinear forms aresymmetric. For readability the subscript N will be omitted when it’s clear fromthe context what quadratic form the bilinear form refers to.

Example If V = Rn×1 then all quadratic forms on V are described by sym-metric matrices via the equation

N(x) = xTAx. (2.3)

Moreover, A is uniquely determined by N .

Proof. Given a symmetric matrix A ∈ Rn×n define the map N : V −→ k as in(2.3). The first property holds because

N(λx) = (λx)TA(λx) = λ2xTAx = λ2N(x)

and the second since

〈x, y〉 = N(x+ y)−N(x)−N(y)= (x+ y)TA(x+ y)− xTAx− yTAy= xTAx+ xTAy + yTAx+ yTAy − xTAx− yTAy= xTAy + yTAx

= xT (2A)y.

2See section 2.2 for the definition of alternative.

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The last equality holds since A is symmetric.

xTAy = (xTAy)T = yTAT (xT )T = yTAx

This shows that 〈 , 〉 is bilinear which proves that N is a quadratic form.The converse is shown using the fact that all symmetric bilinear forms 〈 , 〉 :

Rn×1 × Rn×1 → R are given by a symmetric matrix A in the following way

〈x, y〉 = xTAy.

The quadratic form is recovered from the bilinear form via the formula N(x) =12 〈x, x〉 since

〈x, x〉 = N(2x)− 2N(x) = 2N(x) (2.4)

and hence N(x) = 12xTAx = xTBx where B = 12A is symmetric.

Finally assume that for some quadratic form N on V we have

N(x) = xTAx = xTBx

for A 6= B. It follows that 0 = xTAx−xTBx = xT (A−B)x which implies thatA−B = 0 and hence A = B.

The equation in (2.4) is true for all quadratic forms on all vector spacesand gives a way to recover the quadratic form from the associated bilinear formprovided that char k 6= 2.

Proposition 2.1. If char k 6= 2 all quadratic forms N on V can be recoveredfrom the bilinear form 〈 , 〉N via the formula

N(x) =1

2〈x, x〉N . (2.5)

However, if char k = 2 we have 〈x, x〉N = 0 for all x ∈ V .

Proof.〈x, x〉N = N(2x)− 2N(x) = 4N(x)− 2N(x) = 2N(x) (2.6)

If char k 6= 2 we get N(x) = 12 〈x, x〉N and if char k = 2 we have 〈x, x〉N =2N(x) = 0.

We now move on to properties of symmetric bilinear forms.

Definition Two vectors x, y ∈ V are said to be orthogonal with respect to abilinear form 〈 , 〉 on V if 〈x, y〉 = 0. We denote this with x ⊥ y (or y ⊥ x since〈x, y〉 = 〈y, x〉). Two subsets U1, U2 ⊆ V are said to be orthogonal if all theirvectors are orthogonal, i.e

U1 ⊥ U2 ⇔ x ⊥ y ∀x ∈ U1, y ∈ U2.

The notation x ⊥ U is short for {x} ⊥ U . The orthogonal complement U⊥ of asubset U ⊆ V is the set consisting of all vectors orthogonal to all vectors in U ,i.e

U⊥ = {x ∈ V | x ⊥ U}

or equivalently, the largest subset U⊥ of V satisfying U⊥ ⊥ U .

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Proposition 2.2. Given a bilinear form 〈 , 〉 : V ×V −→ k and a subset U ⊆ V ,then U⊥ is a subspace of V and U ⊆ (U⊥)⊥.

Proof. (SS1) 〈0, x〉 = 〈0 + 0, x〉 = 〈0, x〉+ 〈0, x〉. This shows that 〈0, x〉 = 0 forall x ∈ U so 0 must belong to U⊥.

(SS2) Assume that x and y lie in U⊥. Then 〈x, u〉 = 〈y, u〉 = 0 for all u ∈ U . Itfollows that 〈x+ y, u〉 = 〈x, u〉+ 〈y, u〉 = 0 + 0 = 0 for all u ∈ U . Hencex+ y ∈ U⊥.

(SS3) Assume that x ∈ U⊥. Then, since 〈x, u〉 = 0 for all u ∈ U , we have〈λx, u〉 = λ〈x, u〉 = λ0 = 0 for all u ∈ U and this shows the final subspaceaxiom, namely λx ∈ U⊥.

This proves that U⊥ is a subspace of V . Now assume that x ∈ U , thenx ⊥ U⊥ by definition. But that is also the definition of an element in (U⊥)⊥,hence x ∈ (U⊥)⊥.

The converse (U⊥)⊥ ⊆ U is not true for all bilinear forms and subspaces U . Fora subspace to behave that way we need a property on both the subspace andthe form to make sure it behaves nicely.

Definition A bilinear form 〈 , 〉 is called non-degenerate if the only vector inV orthogonal to all other vectors is 0, i.e V ⊥ = {0} or equivalently

〈x, y〉 = 0, ∀y ∈ V =⇒ x = 0. (2.7)

A quadratic form N on V is said to be non-degenerate if its associated bilinearform 〈 , 〉N is non-degenerate.

If the restriction of 〈 , 〉 to a subspace U of V is non-degenerate we callthat subspace non-singular . The notations 〈 , 〉|U×U and ⊥U will denote therestricted form and the restricted orthogonality relation, respectively.

Proposition 2.3. If the form 〈 , 〉 is non-degenerate and 〈a, y〉 = 〈b, y〉 for ally ∈ V then a = b.

Proof. Using (2.7) we get

〈a, y〉 = 〈b, y〉 ∀y ∈ V⇐⇒ 〈a− b, y〉 = 0 ∀y ∈ V=⇒ a− b = 0⇐⇒ a = b

The last proposition will be very useful later when proving identities in compo-sition algebras. By proving that 〈x, z〉 = 〈y, z〉 holds for all z one can show thatx = y.

One thing useful for proving things for bilinear forms is the following equiv-alence.

Proposition 2.4. If 〈 , 〉 is a bilinear form on a vectorspace V and U is asubspace of V , then the following are equivalent.

(i) U is non-singular.

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(ii) U ∩ U⊥ = {0}.

Proof. (i)⇒ (ii) If U is non-singular, then 〈 , 〉|U×U is non-degenerate. Whichis the same as U⊥U = {0}. On the other hand U⊥U = {x ∈ U | x ⊥ U} =U ∩ U⊥. So U ∩ U⊥ = {0}.

(ii)⇒ (i) Assume U ∩ U⊥ = {0}. Let x ∈ U and assume

〈x, y〉 = 0, ∀y ∈ U.

If we can show x = 0, then we are done. The above property is the sameas x ⊥ U , so x ∈ U⊥. That means that x ∈ U ∩U⊥ = {0} so x must be 0.

Lemma 2.5. Let V be a finite-dimensional vector space and 〈 , 〉 a non-degenerate bilinear form on V . Then the linear map λ : V −→ V ∗, v 7→ λv = 〈v, ·〉is an isomorphism.

Proof. Let V be a vector space over k and dimV = n for some n ∈ N. Also let〈 , 〉 : V × V −→ k be a non-degenerate bilinear form.

Now let λ : V −→ V ∗, v 7→ λv = 〈v, ·〉. We see that λ is linear since,λv+w = 〈v+w, ·〉 = 〈v, ·〉+ 〈w, ·〉 = λv +λw and λcv = 〈cv, ·〉 = c〈v, ·〉 = cλv forall v, w ∈ V and c ∈ k. Now look at the kernel of λ,

kerλ = {v ∈ V | λv = 0}= {v ∈ V | λv(w) = 0, ∀w ∈ V }= {v ∈ V | 〈v, w〉 = 0, ∀w ∈ V }= {0}.

So λ is injective. It is also surjective since dimV ∗ = dimV holds when V isfinite-dimensional. Hence λ is an isomorphism.

Theorem 2.6. If V is a vector space over k, 〈 , 〉 a bilinear form on V andU ⊆ V a finite-dimensional non-singular subspace, then

V = U ⊕ U⊥.

If moreover 〈 , 〉 is non-degenerate, then U⊥ is non-singular.

Proof. Assume that U ⊆ V is non-singular and that dimU

which concludes the proof of the first statement.Now assume that 〈 , 〉 is non-degenerate, that U⊥ is singular and that

x ∈ U⊥ ∩ (U⊥)⊥. If we can prove that x = 0 then we are done. Let y ∈ V .Then we can write y = u+ u′ where u ∈ U and u′ ∈ U⊥ since V = U ⊕ U⊥ bythe first statement of this theorem. By our assumption x ∈ U⊥ and x ∈ (U⊥)⊥so 〈x, u〉 = 0 and 〈x, u′〉 = 0 respectively. Adding these we get 〈x, u〉+ 〈x, u′〉 =〈x, u+ u′〉 = 0. This means that we have 〈x, y〉 = 0 for all y ∈ V which impliesthat x = 0 since 〈 , 〉 is non-degenerate.

Corollary 2.7. If 〈 , 〉 : V ×V −→ k is a bilinear form on a vector space V andU ⊆ V a finite-dimensional non-singular subspace, then

dimV = dimU + dimU⊥ (2.8)

Proof. Follows from Theorem 2.6.

Corollary 2.8. If 〈 , 〉 : V × V −→ k is a non-degenerate bilinear form and Ua finite-dimensional non-singular subspace of V , then

(U⊥)⊥ = U (2.9)

Proof. We already know from Proposition 2.2 that U ⊆ (U⊥)⊥ so we only needto prove the inclusion (U⊥)⊥ ⊆ U . Let x ∈ (U⊥)⊥. From Theorem 2.6 we knowthat V = U ⊕ U⊥ so we can write x = u + u′ where u ∈ U and u′ ∈ U⊥. Butu ∈ (U⊥)⊥ since u ∈ U which means that x − u ∈ (U⊥)⊥. On the other handx − u = u′ ∈ U⊥ so x − u ∈ U⊥ ∩ (U⊥)⊥ and hence x − u = 0. Moving the uto the other side gives x = u so x ∈ U .

2.2 Algebras

Definition An algebra A over a field k is a vector space over k together witha bilinear multiplication A × A −→ A, (x, y) 7→ xy. It follows that the mapsLa : A −→ k, x 7→ ax and Ra : A −→ k, x 7→ xa are linear for all a ∈ A.

Reminder A multiplication A×A −→ A, (x, y) 7→ xy is bilinear if the followingaxioms are satisfied.

1. x(y + z) = xy + xz for all x, y, z ∈ A.

2. (x+ y)z = xz + yz for all x, y, z ∈ A.

3. λ(xy) = (λx)y = x(λy) for all x, y ∈ A and λ ∈ k.

They are called right additivity, left additivity and homogenity, respectively.

Definition The dimension of an algebra A is the dimension of its underlyingvector space.

Here follows a list of some various types of algebras

Division Algebra A non-zero algebra in which the maps La and Ra are bi-jective for all a 6= 0. Division by non-zero elements is possible in divisionalgebras, hence its name. A commutative and associative division algebrawith unity forms a field together with vector addition and multiplication.

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Quadratic Algebra An algebra with unity e in which the square of an elementx is contained in the span of e and x, i.e

x2 = αe+ βx for some α, β ∈ k (2.10)

Alternative Algebra An algebra A is called alternative if for all x, y ∈ A

x(xy) = (xx)y

x(yy) = (xy)y

x(yx) = (xy)x.

In other words, the subalgebra generated by x and y is associative.

3 Composition Algebras

In this section we will speak of an algebraic objects known as compositionalgebras. These were introduced by Nathan Jacobson as a means to proveHurwitz Theorem for any field of characteristic not two.

The way this is presented in follows closely to that in [5].

3.1 Definition and Basic Properties

Definition A composition algebra C over a field k is a pair (C,N) where C isa non-zero algebra with identity e and N : C −→ k a non-degenerate quadraticform that satisfies

N(xy) = N(x)N(y) ∀x, y ∈ C. (3.1)

The equation (3.1) is what relates the composition algebras to the quadraticforms.

Composition algebras are sometimes called normed algebras. The quadraticform N is called the norm and the associated bilinear form 〈 , 〉N is called theinner product . Note that it is not neccesarily an inner product in the usualsense, it doesn’t have to be positive definite.

Definition A composition subalgebra D of a composition algebra C is a non-singular subspace D of C that is closed under multiplication and contains theidentity e.

Examples The set of real numbers R and the set of complex numbers C are realcomposition algebras with the quadratic forms NR(x) = x

2 and NC(x + iy) =x2 + y2 respectively.

A slightly more advanced example is the algebra Z2 ⊕Z2 over Z2 with mul-tiplication (x1, y1)(x2, y2) = (x1x2, y1y2), quadratic form N((x, y)) = xy andidentity e = (1, 1).

Proposition 3.1. Let C be any composition algebra. Then the identity e sat-isfies

N(e) = 1. (3.2)

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Proof.N(e) = N(ee) = N(e)N(e) =⇒ N(e) = 1 or N(e) = 0

Assume that N(e) = 0 then N(x) = N(x)N(e) = 0 for all x ∈ C so

〈e, x〉 = 0 ∀x ∈ C

which, since 〈 , 〉 is non-degenerate, implies that e = 0. But a compositionalgebra is by definition non-zero so we have a contradiction. Hence N(e) =1.

Definition An element x of C is said to have an inverse if there is some y ∈ Csuch that xy = yx = e.

Remark Nothing is yet said about the uniqueness of inverses. Until the unique-ness is proven we have to include the possibility of mulitple inverses.

Proposition 3.2. If x is an element of a composition algebra and x has aninverse x−1 then

N(x−1) = N(x)−1.

In particular, N(x) and N(x−1) are both non-zero.

Proof. Assume x is an invertible element in a composition algebra C. Then thefollowing holds

1 = N(e) = N(xx−1) = N(x)N(x−1).

Hence, N(x−1) = N(x)−1.

We will see later that the condition N(x) 6= 0 is in fact sufficient for x to havean inverse and that the inverse is, in fact, unique.

We now continue with some more technical identities that will prove veryuseful.

Proposition 3.3. In every composition algebra C the following identities holdfor all x, x1, x2, y, y1, y2 ∈ C

〈x1y, x2y〉 = 〈x1, x2〉N(y) (3.3)〈xy1, xy2〉 = N(x)〈y1, y2〉 (3.4)

〈x1y1, x2y2〉+ 〈x1y2, x2y1〉 = 〈x1, x2〉〈y1, y2〉. (3.5)

Proof. Regarding (3.3) we have for all x1, x2, y ∈ C

〈x1y, x2y〉 = N(x1y + x2y)−N(x1y)−N(x2y)= N((x1 + x2)y)−N(x1y)−N(x2y)= N(x1 + x2)N(y)−N(x1)N(y)−N(x2)N(y)= (N(x1 + x2)−N(x1)−N(x2))N(y)= 〈x1, x2〉N(y).

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The identity (3.4) is proved in the same way as (3.3). The last one is proved byusing (3.3). On one hand we have

〈x1(y1 + y2), x2(y1 + y2)〉 = 〈x1y1 + x1y2, x2y1 + x2y2〉= 〈x1y1, x2y1 + x2y2〉+ 〈x1y2, x2y1 + x2y2〉= 〈x1y1, x2y1〉+ 〈x1y1, x2y2〉+ 〈x1y2, x2y1〉+〈x1y2, x2y2〉

= 〈x1y1, x2y2〉+ 〈x1y2, x2y1〉+ 〈x1, x2〉N(y1) +〈x1, x2〉N(y2)

and on the other we have

〈x1, x2〉N(y1 + y2) = 〈x1, x2〉(〈y1, y2〉+N(y1) +N(y2))= 〈x1, x2〉〈y1, y2〉+ 〈x1, x2〉(N(y1) +N(y2)).

By (3.3), the starting terms of both chains coincide. Subtracting 〈x1, x2〉(N(y1)+N(y2)) from the end terms we obtain (3.5).

Corollary 3.4. Let C be a composition algebra and let x, y ∈ C with 〈x, y〉 = 0.Then for all x1, y1 ∈ C,

〈x1x, y1y〉 = −〈x1y, y1x〉. (3.6)

Proof. Follows directly from (3.5).

Proposition 3.5. Let C be a composition algebra and x an element in C. Thenthe following identity holds

x2 − 〈x, e〉x+N(x)e = 0. (3.7)

Proof. Let x ∈ C. Then create the inner product of the left hand side of (3.7)with an arbitrary y.

〈x2 − 〈x, e〉x+N(x)e, y〉 = 〈x2, y〉 − 〈x, e〉〈x, y〉+N(x)〈e, y〉= 〈x2, y〉 − 〈x, e〉〈x, y〉+ 〈xe, xy〉= 0

The equalities are valid since N(x)〈e, y〉 = 〈xe, xy〉 by (3.4) and 〈x2, y〉 +〈xe, xy〉 = 〈x, e〉〈x, y〉 by (3.5). Using Proposition 2.3 we obtain (3.7).

Corollary 3.6. If (C,M) and (C,N) are composition algebras, then M = N .

Proof. For every x ∈ C we have by (3.7)

x2 = 〈x, e〉Mx−M(x)ex2 = 〈x, e〉Nx−N(x)e.

So〈x, e〉Mx−M(x)e = 〈x, e〉Nx−N(x)e. (3.8)

Now we have two cases, e and x are either linearly independent or linearlydependent.

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First case If e and x are linearly independent then −M(x) = −N(x) by (3.8)so M(x) = N(x).

Second case If e and x are linearly dependent then x = λe for some scalar λ.Thus we obtain

M(x) = λ2M(e) = λ2 = λ2N(e) = N(x).

Both cases lead to M(x) = N(x) which concludes the proof.

The identity (3.7) can be interpreted (weaker) as follows.

Corollary 3.7. Every composition algebra is a quadratic algebra.

Proof. Let C be a composition algebra. Then for all x ∈ C, x2 = 〈x, e〉x−N(x)eby (3.7). So x2 ∈ span{e, x}.

This gives us some geometric understanding of the multiplication in a com-position algebra.

3.2 Purely Imaginary Elements and Conjugation

We will now generalize the notion of imaginary complex numbers that shows upin many parts of mathematics.

Definition The set of all purely imaginary elements of a composition algebraC over k is the set

Im C = {x ∈ C | x2 ∈ ke, x /∈ ke \ {0}}. (3.9)

Proposition 3.8. Let C be a composition algebra over a field k with char k 6= 2.Then Im C is a non-singular subspace of C, Im C = (ke)⊥ and

C = Im C ⊕ ke.

Proof. To prove Proposition 3.8 it is enough to show that ke is a non-singularsubspace and that Im C = (ke)⊥. The rest will follow from Theorem 2.6 sincedim ke = 1.

Assume that ke is singular. Then x ∈ (ke)⊥ ∩ ke for some non-zero x ∈ C.So x = λe, λ 6= 0 and 〈x, e〉 = 0. But this is a contradiction since 0 = 〈λe, e〉 =λ〈e, e〉 = λ(N(2e)− 2N(e)) = 2λ when char k 6= 2. Hence ke is non-singular.

To show Im C = (ke)⊥ we first show Im C ⊆ (ke)⊥. Let x ∈ Im (C) \ {0}then e and x are linearly independent and x2 = λe for some λ. The identity (3.7)on the other hand gives us x2 = 〈x, e〉x −N(x)e. Combining these two we getN(x) = −λ and 〈x, e〉 = 0. So x ∈ (ke)⊥. Now assume that x ∈ (ke)⊥. Then〈x, e〉 = 0 and from (3.7) we have, once again, x2 = 〈x, e〉x−N(x)e = −N(x)e.Also, since x ⊥ ke, we have x /∈ ke \ {0} because ke is non-singular. Hencex ∈ Im C. That concludes the proof that Im C = (ke)⊥.

By Theorem 2.6, Im C is a non-singular subspace of C and C = Im C ⊕ ke,a direct sum decomposition.

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The above proposition has a generalization called Frobenius Lemma ([1,p. 227]) that states that all real quadratic algebras A can be written as A =Re⊕ Im A. That proof can easily be extended to all quadratic algebras over afield with characteristic not equal to two.

Hopefully, Proposition 3.8 gives the reader some more understanding of whatIm C looks like. For example, Im R = {0}, R = R ⊕ {0} and Im C = Ri, C =Ri⊕R. In Z2⊕Z2, however, we have Im (Z2⊕Z2) = {0} and Z2e = {0, (1, 1)}.It follows that Z2 ⊕ Z2 6= Im (Z2 ⊕ Z2)⊕ Z2e.

Definition Let C be a composition k-algebra. The conjugation map ¯ : C −→ C,x 7→ x is defined by

x = 〈x, e〉e− x. (3.10)

This can also be seen as −s(x) where s(x) is the reflection of x in (ke)⊥ (or Im Cif char k 6= 2). For C = C the definition coincides with the normal understandingof conjugates. In R we have x = x and in C we have x+ iy = x−iy. Computingthe conjugate for each element in Z2 ⊕ Z2 one gets 0 = 0, e = e, (0, 1) = (1, 0)and (1, 0) = (0, 1).

Many of the usual properties of conjugates hold with the above definition ofconjugation for every composition algebra.

Proposition 3.9. Let C be a composition algebra. Then conjugation is a linearmap and x = x.

Proof. The proof that conjugation is linear follows from the definition since 〈 , 〉is a bilinear form;

x+ y = 〈x+ y, e〉e− (x+ y) = 〈x, e〉e− x+ 〈y, e〉e− y = x+ y,

and similarily for multiplication with scalars. The last part follows from thecomputation

x = 〈x, e〉e− x= 〈〈x, e〉e− x, e〉e− 〈x, e〉e+ x= 〈x, e〉〈e, e〉e− 〈x, e〉e− 〈x, e〉e+ x= 2〈x, e〉e− 2〈x, e〉e+ x= x.

since 〈e, e〉 = 2N(e) = 2 by Proposition 2.1 and 3.1.

This is a good start. But one might ask what other properties hold. Forexample, in C (with the ordinary conjugate) we have xx = |x|2, xy = xy and|x| = |x|. It turns out that these also hold in general composition algebras, withslight modifications.

Proposition 3.10. In every composition algebra C the following holds for allx, y ∈ C.

(i) xx = xx = N(x)e

(ii) xy = yx

(iii) N(x) = N(x)

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(iv) 〈x, y〉 = 〈x, y〉

Proof. (i) Follows from (3.7) in the following way

N(x)e = 〈x, e〉x− x2

= (〈x, e〉e− x)x= xx.

The other way round, N(x)e = xx is done in the same way. Rememberthat λx = (λx)e = x(λe) for all scalars λ.

(ii) We first observe that

xy = 〈y, e〉x+ 〈x, e〉y − 〈x, y〉e− yx (3.11)

since

xy = (x+ y)2 − x2 − y2 − yx= 〈x+ y, e〉(x+ y)−N(x+ y)e− 〈x, e〉x+

N(x)e− 〈y, e〉y +N(y)e− yx= 〈x, e〉y + 〈y, e〉x− (N(x+ y)−N(x)−N(y))e− yx= 〈y, e〉x+ 〈x, e〉y − 〈x, y〉e− yx

by (3.7). By the definition of conjugates, (3.11) and (3.5) we have

yx = (〈y, e〉e− y)(〈x, e〉e− x)= 〈x, e〉〈y, e〉e− 〈x, e〉y − 〈y, e〉x+ yx= 〈x, e〉〈y, e〉e− 〈x, y〉e− xy= 〈xy, e〉e− xy= xy.

(iii) From the previous proposition and part (i);

N(x)e = xx = xx = N(x)e

so N(x) = N(x).

(iv) Using part (iii) and the linearity of conjugation

〈x, y〉 = N(x+ y)−N(x)−N(y)= N(x+ y)−N(x)−N(y)= N(x+ y)−N(x)−N(y)= 〈x, y〉.

We saw earlier in Proposition 3.2 that if x has an inverse x−1 then N(x−1) =N(x)−1. We will now show something stronger using the newly establishedproperties of conjugation.

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Proposition 3.11. In every composition algebra C and for every x ∈ C thefollowing are equivalent

(i) x has an inverse.

(ii) N(x) 6= 0.

In that case, N(x)−1x is inverse to x.

Proof. (i)⇒ (ii) See Proposition 3.2.

(ii)⇒ (i) Assume N(x) 6= 0. Let y = N(x)−1x, then

xy = x(N(x)−1x) = N(x)−1(xx) = N(x)−1N(x)e = e

The equality yx = e is similar which concludes that xy = yx = e.

3.3 Associativity Properties

So far we have not established any facts about associativity in compositionalgebras. It turns out that they need not be associative but they satisfy someother, although weaker, properties. Unfortunately, these properties are notobvious and need some technical propositions first.

Proposition 3.12. Let C be a composition algebra and let x, y, z ∈ C, then

〈xy, z〉 = 〈y, xz〉 (3.12)〈xy, z〉 = 〈x, zy〉 (3.13)〈xy, z〉 = 〈yz, x〉 (3.14)

Proof.

〈y, xz〉 = 〈y, (〈x, e〉e− x)z〉= 〈y, 〈x, e〉z〉 − 〈y, xz〉= 〈x, e〉〈y, z〉 − 〈y, xz〉= 〈xy, z〉+ 〈xz, y〉 − 〈y, xz〉= 〈xy, z〉

The only non-trivial step follows from (3.5);

〈x, e〉〈y, z〉 = 〈xy, z〉+ 〈xz, y〉

To prove (3.13) and (3.14) we use part (ii) and (iv) of Proposition 3.10 and(3.12).

〈xy, z〉 = 〈y, xz〉= 〈y, zx〉= 〈zy, x〉= 〈x, zy〉

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proves (3.13) and

〈xy, z〉 = 〈z, xy〉= 〈z, xy〉= 〈z, yx〉= 〈yz, x〉

proves (3.14)

Remark Recall that the adjoint of a linear operator f on an inner productspace V is the unique linear operator f∗ on V such that 〈f(x), y〉 = 〈x, f∗(y)〉for all x, y ∈ V . Using that definition we can define the notion of adjointsin composition algebras, which gives us a nice way of interpreting (3.12) and(3.13);

L∗x = Lx

R∗y = Ry

since 〈Lx(y), z〉 = 〈xy, z〉 = 〈y, xz〉 = 〈y, Lx(z)〉 and similarily for Ry.

We can now prove a slight generalization to the identity xx = N(x)e.

Proposition 3.13. If C is a composition algebra and x, y ∈ C then the followingholds

x(xy) = N(x)y (3.15)

(xy)y = N(y)x (3.16)

Proof. Using Proposition 2.3 we can show that if 〈x(xy), z〉 = 〈N(x)y, z〉 for allz ∈ C then (3.15) will hold. That is done using (3.12), (3.4) and part (iii) ofProposition 3.10 in the following way:

〈x(xy), z〉 = 〈xy, xz〉= N(x)〈y, z〉= N(x)〈y, z〉= 〈N(x)y, z〉.

Now using part (ii) of Proposition 3.10 and conjugating both sides in (3.15)we get

x(xy) = (xy)x = (yx)x

andN(x)y = N(x)y.

Combining the above expressions proves (3.16).

We are now ready to state our first associativity related result.

Corollary 3.14. For all x and y in a composition algebra C the following holds

x(xy) = (xx)y (3.17)

x(yy) = (xy)y (3.18)

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Proof. Follows immediately from (3.15), (3.16) and Proposition 3.10 (i).

Corollary 3.15. Let C be a composition algebra. Then every element x ∈ Cthat satisfies N(x) 6= 0 has a unique inverse N(x)−1x.

Proof. Let x ∈ C and N(x) 6= 0. Then x has an inverse y = N(x)−1x byProposition 3.11. Assume that there is another element z such that xz = zx = e.Then

y = y(xz)

= N(x)−1x(xz)

= N(x)−1(xx)z

= N(x)−1N(x)z

= z

by Corollary 3.14.

Proposition 3.16. In every composition algebra C the Moufang identities holdfor all x, y, a ∈ C.

(ax)(ya) = a((xy)a) (3.19)

a(x(ay)) = (a(xa))y (3.20)

x(a(ya)) = ((xa)y)a (3.21)

Proof. Once again we use Proposition 2.3 to show an identity. The reader shouldat this point be familiar with the tools used so we won’t write every identityand proposition used, which will increase readability. We will point out, howeverthat (3.5) is used twice.

〈(ax)(ya), z〉 = 〈ya, (ax)z〉= 〈ya, (xa)z〉= 〈y, xa〉〈a, z〉 − 〈yz, (xa)a〉= 〈xy, a〉〈a, z〉 − 〈yz,N(a)x〉= 〈xy, a〉〈a, z〉 −N(a)〈yz, x〉= 〈xy, a〉〈a, z〉 −N(a)〈xy, z〉= 〈xy, a〉〈a, z〉 −N(a)〈(xy)z, e〉= 〈xy, a〉〈a, z〉 − 〈(xy)z, aa〉= 〈(xy)a, az〉= 〈a((xy)a), z〉

This proves (3.19). To prove (3.20) we use a similar chain of equalities where

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we use (3.19) once.

〈a(x(ay)), z〉 = 〈x(ay), az〉= 〈x, (az)(ay)〉= 〈x, (az)(ay)〉= 〈x, (ay)(za)〉= 〈x, a((yz)a)〉= 〈x, a((yz)a)〉= 〈x, ((yz)a)a〉= 〈x, (a(zy))a〉= 〈xa, a(zy)〉= 〈a(xa), zy〉= 〈(a(xa))y, z〉

To prove the last equality we start with the second and conjugate it. Theleft hand side becomes

a(x(ay)) = (x(ay))a = ((ya)x)a

and the right hand side, with the help of (3.19) since (ax)a = (ax)(ea) =a((xe)a) = a(xa),

(a(xa))y = y(a(xa)) = y((ax)a) = y(a(xa)).

Substituting y, a and x for x′, a′ and y′ respectively (for readability) and com-bining the above equalities we get

x′(a′(y′a′)) = ((x′a′)y′)a′

which proves (3.21).

Proposition 3.17. Let C be a composition algebra. Then C is alternative.

Proof. To prove that C is alternative we have to prove the following equalities

x(yx) = (xy)x (3.22)

(xx)y = x(xy) (3.23)

(xy)y = x(yy). (3.24)

The first follows from (3.19)

(xy)x = (xy)(ex) = x((ye)x) = x(yx).

The second follows if we start with (3.17) and expand both sides, starting withthe left

x(xy) = x((〈x, e〉e− x)y)= x(〈x, e〉y − xy)= 〈x, e〉xy − x(xy).

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The right hand side is

(xx)y = (x(〈x, e〉e− x))y= (〈x, e〉x− xx)y= 〈x, e〉xy − (xx)y.

Combining the two chains proves (3.23).Proving (3.24) is similar.

Summarizing everything done so far we get the following theorem.

Theorem 3.18. Let C be a composition algebra. Then C is a quadratic, alter-native algebra that satisfies the Moufang identities.

Even if this says something about the elements it still doesn’t say anythingabout the structure of the algebra. We are still missing one vital part; a con-struction method called doubling that will prove very useful.

3.4 Doubling

Lemma 3.19. Let C be a composition algebra. If D ( C is a finite-dimensionalproper non-singular subspace of C, then there is an element a ∈ D⊥ \ {0} suchthat N(a) 6= 0.

Proof. Assume that N(x) = 0 for all x ∈ D⊥. Theorem 2.6 tells us thatC = D⊕D⊥ where D⊥ is non-singular. We also know that D 6= C so D⊥ 6= {0}

Let a be any non-zero element in D⊥. Then,

〈a, x〉 = N(a+ x)−N(a)−N(x) = 0− 0− 0 = 0, ∀x ∈ D⊥

and since D⊥ is non-singular, a = 0 which is a contradiction.

If D1 ( C is a finite-dimensional proper composition subalgebra, then thelemma makes sure that we can create the following subalgebra of C

D2 = D1 ⊕D1a

where D1 is a proper composition subalgebra of C and a is a non-zero elementin D⊥ such that N(a) 6= 0.

Proposition 3.20 (The doubling construction). If D1 is a finite-dimensionalproper composition subalgebra of a composition algebra C, a ∈ D⊥ \ {0} andN(a) 6= 0 then the subspace

D2 = D1 +D1a (3.25)

of C is a composition subalgebra of C with dimD2 = 2 dimD1 and the sumbeing a direct sum decomposition of D2. Multiplication in D2 then is

(x+ ya)(z + wa) = (xz −N(a)wy) + (wx+ yz)a. (3.26)

Proof. There is quite a lot of information in this proposition but the proofessentially boils down to the following two parts.

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(i) By showing D1a ⊆ D⊥1 we will show that the sum is direct.

(ii) By showing (3.26) we will show that D2 is closed under multiplication andnon-singular.

Proof of (i). We know that a ∈ D⊥1 so we need to show that xa ⊥ D1 for allx ∈ D1. Therefore, let x ∈ D1 and look at the inner product 〈xa, y〉 where y ∈D1. Identity (3.12) gives us 〈xa, y〉 = 〈a, xy〉. Since D1 is a composition algebrait must be closed under conjugation and multiplication so xy ∈ D1. So, sincea ⊥ xy we must have xa ⊥ y. This concludes the proof that D1a ⊆ D⊥1 .

Since dimD1

To prove the final equality we observe that from (3.30) we get

va = va = −va = −av = av (3.31)

for all v ∈ D1 so we have

(ya)(wa) = (ay)(wa)

= a((yw)a)

= a(a(yw))

= a(a(wy))

= (aa)(wy)

= −(aa)(wy)= −N(a)(wy).

This concludes the proof of (3.26) which shows that D2 is closed undermultiplication. The only thing left is to show that D2 is non-singular. Lookingat the norm N on D2 we see that

N(x+ ya) = 〈x, ya〉+N(x) +N(ya) = N(x) +N(y)N(a).

Looking at the inner product we similarily get for x, y, v, w ∈ D1

〈x+ ya, v + wa〉 = 〈x, v〉+ 〈x,wa〉+ 〈ya, v〉+ 〈ya, wa〉= 〈x, v〉+ 〈ya,wa〉= 〈x, v〉+ 〈y, w〉N(a)

Let x + ya ∈ D2 and assume that 〈x + ya, v + wa〉 = 0 for all v, w ∈ D1.Then 〈x, v〉 + 〈y, w〉N(a) = 0 for all v, w ∈ D1. In particular we have w = 0which implies that 〈x, v〉 = 0 for all v ∈ D1. Since D1 is non-singular we havex = 0. With a similar argument we also have 〈y, w〉N(a) = 0 for all w ∈ D1which (since N(a) 6= 0) implies that y = 0. Together these two cases gives usx+ ya = 0 so D2 is non-singular.

To summarize the proof so far we know that D2 = D1⊕D1a is a direct sumdecomposition, that D2 is a non-singular subspace of C and it is closed undermultiplication. Hence D2 is a composition subalgebra.

We now conclude the proof by showing that dimD2 = 2 dimD1. LetRa : D1 −→ D1a be the map x 7→ xa. It is a linear map since multiplica-tion is bilinear. Moreover a has an inverse since N(a) 6= 0 so Ra has in-verse R−1a = Ra−1 which proves that Ra is bijective and hence an isomor-phism. It then follows that dimD1a = dimD1. From the direct sum we getdimD2 = dimD1 + dimD1a = 2 dimD1 which completes the proof.

Proposition 3.20 is a very powerful tool. It gives us a contructive methodof creating new composition subalgebras from already existing ones. However,it does not say anything about the properties of the new ones except that theyhave double dimension.

Proposition 3.21. Let C be a composition algebra and D a finite-dimensionalproper composition subalgebra. Then D is associative.

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Proof. Let a ∈ D⊥\{0} with N(a) 6= 0. The existence of such an a is guaranteedby Lemma 3.19. Using the equality N((x+ya)(z+wa)) = N(x+ya)N(z+wa)for x, y, z, w ∈ D we have

N((x+ ya)(z + wa)) = N((xz −N(a)wy) + (wx+ yz)a)= N(xz −N(a)wy) +N(a)N(wx+ yz)= 〈xz,−N(a)wy〉+N(xz) +N(−N(a)wy) +

N(a)(〈wx, yz〉+N(wx) +N(yz))

and

N(x+ ya)N(z + wa) = (N(x) +N(y)N(a))(N(z) +N(w)N(a))

= N(x)N(z) +N(x)N(w)N(a) +

N(y)N(z)N(a) +N(y)N(w)N(a)N(a)

= N(xz) +N(wx)N(a) +N(yz)N(a) +N(a)2N(wy)

by Proposition 3.20. Combining the above we get

〈wx, yz〉 = 〈xz,wy〉.

Since we, from (3.13) and Proposition 3.10, have

〈wx, yz〉 = 〈yz, wx〉= 〈(yz)x,w〉= 〈x(zy), w〉

and

〈xz,wy〉 = 〈(xz)y, w〉

we then have〈x(zy), w〉 = 〈(xz)y, w〉.

If we do a substitution for readability we see that

〈x′(y′z′), w′〉 = 〈(x′y′)z′, w′〉

holds for all x′, y′, z′, w′ ∈ D so x′(y′z′) = (x′y′)z′ for all x′, y′, z′ ∈ D whichshows that D is associative.

Proposition 3.22. Let C be a composition algebra and D1 a finite-dimensionalproper composition subalgebra. Also let a ∈ D⊥1 with N(a) 6= 0 and let D2 =D1 ⊕D1a. Then

D2 is associative ⇐⇒ D1 is assosiative and commutative.

Proof. D2 is a composition subalgebra by Proposition 3.20 with multiplicationas in (3.26).

“⇒” If D2 is associative then so is every subalgebra of D2. In particular,D1 is associative. Let x, y ∈ D1. Then (xy)a = x(ya) and from (3.27)x(ya) = (yx)a. Combining these we get

xy = xyaa−1 = ((xy)a)a−1 = ((yx)a)a−1 = yxaa−1 = yx.

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“⇐” Assume that D1 is both commutative and associative. Then, for allx1, x2, x3, y1, y2, y3 ∈ D1 we get3

(z1z2)z3 = (x1x2)x3 −N(a)((y2y1)x3 + y3(y2x1) + y3(y1x1)

)+(

y3(x1x2)−N(a)y3(y2y1) + (y2x1)x3 + (y1x2)x3)a

z1(z2z3) = x1(x2x3)−N(a)(x1(y3y2) + (y3x2)y1 + (y2x3)y1

)+(

(y3x2)x1 + (y2x3)x1 + y1(x3x2)−N(a)y1(y2y3))a

where zi = xi + yia, i = 1, 2, 3.

SinceD1 is associative and commutative we have that x1(x2x3) = (x1x2)x3and that terms containing a in the above are equal. To show that (z1z2)z3 =z1(z2z3) we are left with proving that

N(a)(y2y1)x3 = N(a)x1(y3y2)

N(a)y3(y2x1) = N(a)y3(y1x1)

N(a)(y3x2)y1 = N(a)(y2x3)y1.

Using (3.28) we see that for x, y, z ∈ D1

N(a)xyz = aaxyz = az((ya)x) = az((yx)a) = N(a)xyz

and similarily for y and z. So

N(a)xyz = N(a)xyz = N(a)xyz = N(a)xyz

which proves the three equalities above since D1 is both commutative andassociative.

4 The (1, 2, 4, 8)-Theorem

With the doubling method to back us up we are now ready to take on the bigtheorem of this paper.

Theorem 4.1 (Theorem 3.18 revisited). Let C be a composition algebra overa field k. Then C is a quadratic, alternative algebra that satisfies the Moufangidentities. Moreover, it must also satisfy one of the following;

dimC k commutative associative alternative1 char(k) 6= 2 Yes Yes Yes2 Any Yes Yes Yes4 Any No Yes Yes8 Any No No Yes

Figure 1: Characteristics of composition algebras

3We omit most of the multiplications for readability since they are not the most importantpart.

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Proof. Let C be a composition algebra. Then C is quadratic by Corollary3.7, alternative by Proposition 3.17 and it satisifes the Moufang identities byProposition 3.16.

1. Let’s start with any composition algebra C. Then, if char k 6= 2, we canfind a one-dimensional composition subalgebra D1 = ke. It is non-singularsince 〈λe, µe〉 = λµ〈e, e〉 = 2λµ 6= 0 if λ, µ 6= 0. However if char k = 2then 〈λe, µe〉 = 0 for all λ, µ ∈ k so ke is singular. Hence if char k = 2then dimC > 1.

2. If dimC > 1 then, in the case of char k 6= 2, Lemma 3.19 and Proposition3.20 gives us a two-dimensional subalgebra D2 = D1 ⊕ D1a for someelement a ∈ C. Since D1 is both associative and commutative D2 mustbe associative by Proposition 3.22. A simple computation shows that it isalso commutative. If char k = 2 we then use the following lemma.

Lemma 4.2. If C is a composition k-algebra with char k = 2, then thereis a ∈ C such that 〈e, a〉 6= 0. Also, D = ke ⊕ ka is a two-dimensionalassociative and commutative composition subalgebra.

Proof of Lemma 4.2. Assume that 〈e, a〉 = 0 for all a, then e = 0 since〈 , 〉 is non-degenerate on C. So 〈e, a〉 6= 0 for some a ∈ C. Note thata /∈ ke (otherwise 〈e, a〉 = 〈e, λe〉 = 0) so D must be two-dimensional. Itis non-singular since if 〈λe + µa, x〉 = 0 for all x ∈ D then in particular〈λe+µa, e〉 = 0 and 〈λe+µa, a〉 = 0. These in turn give µ = 0 and λ = 0respectively since 〈e, e〉 = 0 and 〈a, a〉 = 0 so λe + µa = 0. It is closedunder multiplication as the following computation shows

xy = (αe+ βa)(γe+ δa) = αγe+ (αδ + βγ)a+ βδa2

= αγe+ (αδ + βγ)a+ βδ(〈a, e〉a−N(a)e)= (αγ − βδN(a))e+ (αδ + βγ + βδ〈a, e〉)a.

That it is associative and commutative follows from two similar computa-tions.

By this lemma we create, in the case char k = 2, D2 = ke⊕ ka′ for somea′ ∈ C with 〈e, a′〉 6= 0. So now we have, no matter what the characteristic,a two-dimensional composition subalgebra D2 ⊆ C.

3. If we assume that dimC > 2 then we can perform the doubling anothertime and get a four-dimensional subalgebra D3 = D2 ⊕ D2b. It is alsoassociative by Prop. 3.22 but not commutative as the following shows.

Let y = a or y = a′ depending on the characteristic. If y = a then by thechoice of a we have 〈a, e〉 = 0, so y = −y 6= y. If y = a′ then assume thaty = y. We then have y = 〈y, e〉e − y which in turn yields 2y = 〈y, e〉e.So 〈y, e〉e = 0 =⇒ 〈y, e〉 = 0 which contradicts the choice of a′. In bothcases we have y 6= y. We will now show that D2b is non-singular. Assumethat for some xb ∈ D2b we have 〈xb, zb〉 = 0 for all zb ∈ D2b. Then〈xb, zb〉 = 〈x, z〉N(b) = 0 which implies that 〈x, z〉 = 0 for all z ∈ D2.Since D2 is non-singular must have x = 0 so xb = 0 and D2b is non-singular. Using the same method as in the proof of Lemma 3.19 we find

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an element x ∈ D2b such that N(x) 6= 0. We’re now finally able to showthat D3 is not commutative. Since x ∈ D2b we also have x ∈ D⊥2 soxy = −xy by (3.30). But xy = yx = −yx since x ⊥ e. Combining theequalities and remembering that we have y 6= y we get

xy = yx 6= yx.

4. If we assume that dimC > 4 we can perform the doubling one final timeto obtain an eight-dimensional subalgebra D4 = D3 ⊕ D3c. It is notcommutative since D3 ⊂ D4. It is not associative, by Proposition 3.22.Moreover, it is not a proper composition subalgebra of C by Proposition3.21. So C = D4 and the proof is complete.

4.1 Proving Hurwitz Theorem

With all the theory developed in the framework of composition algebras it isnow trivial to prove Hurwitz Theorem for all fields with characteristic not equalto two4.

Theorem 4.3 (Hurwitz Theorem). Let k be a field with char k 6= 2. The onlyvalues of n ∈ N \ {0} for which (1.1) holds are n ∈ {1, 2, 4, 8} where xi, yj ∈ kand zi = zi(x, y) is bilinear.

Proof. Assume that (1.1) holds for some bilinear map z = z(x, y). Let N(x) =∑ni=1 x

2i . ThenN is clearly a non-degenerate quadratic form. Moreover, N(x)N(y) =

N(z(x, y)) by (1.1). The only thing missing before (kn, N) is a composition al-gebra is the existence of an identity element e.

Lemma 4.4. Let N : A −→ k be a non-degenerate quadratic form and A×A −→A, (x, y) 7→ xy a bilinear map such that

N(xy) = N(x)N(y) ∀x, y ∈ A.

Then there is a map ∗ : A×A −→ A, (x, y) 7→ x ∗ y such that

N(x ∗ y) = N(x)N(y) ∀x, y ∈ A

and there is an element e ∈ A such that

e ∗ x = x ∗ e = x ∀x ∈ A.

Proof of Lemma 4.4. Let v ∈ A such that N(v) 6= 0. It is guaranteed to existsince N is non-degenerate. Let u = N(v)−1v2. Then N(u) = 1 and N(ux) =N(xu) = N(x) for all x ∈ A. We also have 〈ux, uy〉 = N(ux + uy) −N(ux) −N(uy) = N(x+ y)−N(x)−N(y) = 〈x, y〉 so

〈Lu(x), Lu(y)〉 = 〈x, y〉 (4.1)4The case char k = 2 is uninteresting since (

∑xi)

2 = (∑

x2i ) in that case.

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and similarily for Ru. Let L∗u be the adjoint

5 of Lu. Then for all x, y ∈ A

〈x, y〉 = 〈Lu(x), Lu(y)〉 = 〈x, L∗u(Lu(y))〉.

Which means that y = L∗u(Lu(y)) for all y ∈ A. So L∗u = L−1u . Note thatN(L−1u (x)) = N(uL

−1u (x)) = N(x). Once again, a similar argument holds for

Ru.Now define the map ∗ : A×A −→ A as

x ∗ y = R−1u (x)L−1u (y).

Then N(x ∗ y) = N(R−1u (x))N(L−1u (y)) = N(x)N(y). We also have

u2 ∗ x = R−1u (u2)L−1u (x) = uL−1u (x) = xx ∗ u2 = R−1u (x)L−1u (u2) = R−1u (x)u = x

which means that u2 is an identity element relative to ∗.

Lemma 4.4 gives us a way to construct a new bilinear map z′ such that(kn, N) is a composition algebra with the bilinear multiplication being z′. The-orem 4.1 then implies that n must be 1, 2, 4 or 8 and we are done.

5We’ll leave it to the reader to prove that the adjoint is well-defined and unique on afinite-dimensional vector space together with a non-degenerate bilinear form.

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References

[1] H.-D. Ebbinghaus. Numbers, chapter 8. Springer Verlag, 1991.

[2] F. G. Frobenius. Über lineare substitutionen und bilineare formen. Journalfür die reine und angewandte Mathematik, 1878.

[3] N. Jacobson. Composition algebras and their automorphisms. Rendicontidel Circolo Matematico di Palermo, 1958.

[4] T. A. Springer. Oktaven, Jordan-Algebren und Ausnahmegruppen. Math.Institut der Universität Göttingen, 1963.

[5] T. A. Springer and F. D. Veldkamp. Octonions, Jordan Algebras, and Ex-ceptional Groups. Springer Verlag, 2000.

[6] M. Zorn. Theorie der alternativen ringe. Abhandlungen aus dem Mathema-tischen Seminar der Universität Hamburg, 1931.

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Appendix A Hurwitz Problem

The identity for n = 4 is

(x21 + x22 + x

23 + x

24)(y

21 + y

22 + y

23 + y

24) = z

21 + z

22 + z

23 + z

24 (A.1)

where

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

The identity for n = 8 is

(x21 + · · ·+ x28)(y21 + · · ·+ y28) = z21 + · · ·+ z28 (A.2)

where

z1 = x1y1 − x2y2 − x3y3 − x4y4 − x5y5 − x6y6 − x7y7 − x8y8z2 = x1y2 + x2y1 + x3y4 − x4y3 + x5y6 + x6y5 + x7y8 − x8y7z3 = x1y3 − x2y4 + x3y1 + x4y2 + x5y7 − x6y8 + x7y5 + x8y6z4 = x1y4 + x2y3 − x3y2 + x4y1 + x5y8 + x6y7 − x7y6 + x8y5z5 = x1y5 − x2y6 − x3y7 − x4y8 + x5y1 + x6y2 + x7y3 + x8y4z6 = x1y6 + x1y5 + x3y8 − x4y7 − x5y2 + x6y1 − x7y4 + x8y3z7 = x1y7 − x2y8 + x3y5 + x4y6 − x5y3 + x6y4 + x7y1 − x8y2z8 = x1y8 + x2y7 − x3y6 + x4y5 − x5y4 − x6y3 + x7y2 + x8y1.

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