Examensarbete i matematik, 15 hp
Handledare och examinator: Ernst Dieterich
Juni 2010
Department of Mathematics
Uppsala University
Triality
Rune Suhr
The real numbers are the dependable breadwinner of the family, the complete ordered eld we all rely on. The complex numbers are a slightly ashier but still respectable younger brother: not ordered, but algebraically complete. The quaternions, being noncommutative, are the eccentric cousin who is shunned at important family gatherings. But the octonions are the crazy old uncle
nobody lets out of the attic: they are nonassociative. - John C. Baez.
Abstract. In 1925, Élie Cartan discovered what nowadays is called the tri- ality principle. It asserts that for each special orthogonal operator γ ∈ SO(8) there exist α, β ∈ SO(8) such that α(x)β(y) = γ(xy) for all x, y ∈ R8, where the products involved in this identity are given by the octonion algebra struc- ture on R8. A proof of this is given and a recent application of the principle in the classi cation of 8 dimensional absolute valued algebras is reported on.
Thanks goes to my advisor Ernst Dieterich, for dedicating much more of his precious time than can or should be expected, to explain various concepts and the- orems that I needed to understand the sources from which I've drawn the majority of the contents of this thesis.
1. Inverse Loops.
1.1. Basic structure.
De nition 1. An inverse loop, is a pair (L, ·), L being a set, and · a function L× L → L, (x, y) 7→ xy, such that the following two conditions hold:
(1) There is an identity element 1∈ L: 1x=x1=x,
(2) For all x∈L there is a y∈L such that (zx)y=z=y(xz), for all z∈L.
Thus neither commutativity nor associativity of the operation is assumed. Condi- tion 1 implies that L ̸=∅. Following convention, we shall often confuse L the set with (L, ·) - the structure on L.
Remark 2. The identity element is unique, by the standard proof: 1'=1'·1 = 1.
Proposition 3. The inverse is unique:
Proof. Assume x ∈ L has inverses y,z∈L, ie. : xy = 1 = yx, and xz = 1 = zx. By condition 2 in de nition 1, we have that:
y=y1=y(xz)=(yx)z=1z=z.
Remark 4. Uniqueness being indicated, we will denote the inverse by x−1. It follows from uniqueness that (x−1)−1= x.
Proposition 5. The relation (xy)z=1 is equivalent to x(yz)=1.
Proof. From xy=z, x,y,z∈L, it follows on the one hand that:
y = x−1z⇔ yz−1= x⇔ (yz−1)x = 1, and on the other hand that
z−1= y−1x−1⇔ z−1x = y−1⇔ y(z−1x) = 1.
1
Remark 6. We omit parenthesis, merely writing xyz=1, when the above relation holds.
1.2. Isotopies.
De nition 7. An isotopy of an inverse loop L, is a triple of invertible maps α, β, γ: L →L, such that
α(x)· β(y) = γ(x · y), for all x, y ∈ L.
We write (α, β | γ) for this isotopy relation. If xyz=1 implies that α(x)β(y)γ(z) = 1, we write (α, β, γ).
Remark 8. We call (α, β | γ) the duplex form of an isotopy, and (α, β, γ) the triplex form of an isotopy.
Proposition 9. (α, β | γ)⇔ (α, β, ιγι), where ι is the inverse map: x 7−→ x−1. Proof. xyz = 1 ⇔ xy = z−1 ⇒ α(x)β(y) = γ(z−1) ⇔ α(x)β(y)(γ(z−1))−1 = 1, and we write (α, β, ιγι) for the isotopy in triplex form equivalent to the duplex form isotopy (α, β | γ).
γ(z−1) is well de ned, since z−1 ∈ L and therefore ι(γ(ι)) : x 7−→ (γ(x−1))−1 is also a bijection, by the uniqueness of the inverse. Let me now state the rst theorem of this section, so that the discussion leading to its proof will be clearer in purpose.
Theorem 10. Let I(L) be the set of isotopies on an inverse loop L. Then I(L) is a group under component wise composition.
We have considered an isotopy to be a relation on triplets of invertible maps on an inverse loop L. The theorem states that we can view isotopies as objects with a group structure. We shall use the same notation to denote the isotopy as an object as we use for denoting the relation on the triplets of maps, namely (α, β | γ).
De nition 11. Let (α, β | γ) be an isotopy. Then each of the invertible maps in the triple α, β, γ are called monotopies. We denote the set of all monotopies on L by M(L) := {α | α is a monotopy on L}.
In other words, α is a member of M(L) if and only if there exists invertible maps β and γ such that (α, β | γ) is an isotopy.
Remark 12. Analogously, a map φ is an endomorphism on a group G, if and only if (φ, φ | φ) is an isotopy on G.
De nition 13. The product of two isotopies is their component wise composition:
(α, β| γ)·(α′, β′| γ′) := (α(α′), β(β′)| γ(γ′)).
Lemma 14. The product of two isotopies is an isotopy.
Proof. Let (α, β | γ), (α′, β′ | γ′)∈ I(L). Then for all x, y ∈ L we have that α(α′(x))β(β′(y)) = γ(α′(x)β′(y)) = γ(γ′(xy)),
where the last equality follows from (α′, β′| γ′)being an isotopy. Hence
(αα′, ββ′ | γγ′) := (α(α′), β(β′)| γ(γ′)), is an isotopy. Lemma 15. Composition of isotopies is associative.
Proof. Exercise. Lemma 16. The component wise inverse of an isotopy I is the inverse of I in I(L), i.e. (α, β | γ)(α−1, β−1| γ−1) = (1L, 1L| 1L) = 1I(L).
Proof. xy = α(α−1(x))β(β−1(y)) = γ(α−1(x)β−1(y)), implies that γ−1(xy) = α−1(x)β−1(y), and so (α−1, β−1 | γ−1)is indeed an isotopy.
That (α, β | γ)(α−1, β−1 | γ−1) = (1L, 1L | 1L) holds is obvious, as is the last
equality of the lemma.
And now we can prove our theorem 10:
Proof. Lemma 14 establishes the closure of I(L) under the de ned product, lemma 15 states that the product is associative, lemma 16 yields inverses and discloses the
nature of the unit element.
1.3. Companions. If γ is a monotopy, we have by de nition that there exists α, β∈ M(L) such that xy = z implies that α(x)β(y) = γ(z), and in particular that
α(z)β(1) = γ(z)⇔α(z) = γ(z)(β(z))−1 = γ(z)b,
for some b ∈ L, and likewise β(z) = aγ(z), for some a ∈ L.
Proposition 17. γ ∈ M(L) if and only if there exists a, b in L, such that γ(xy) = (γ(x)b)(aγ(y)).
Proof. We have already established ” ⇒ ”. ” ⇐ ” follows by taking α = γRb
and β = γLa, where left and right multiplication by a xed element de nes the functions Rb: L→ L, x 7→ xb, and La: L→ L, x 7→ ax. De nition 18. A companion is one of either a or b in proposition 15.
So proposition 15 asserts that the existence of companions is a necessary and suf-
cient condition for a bijective function γ : L→ L to be a monotopy.
Note that a group endomorphism has a=1=b as companions, so that monotopies are a natural generalization of endomorphisms.
Lemma 19. Ba is well de ned, i.e. Ba(x) := Ra(La(x)) = La(Ra(x))is indepen- dent of the order or multiplication, and we can omit parenthesis: a(xa) = (ax)a =:
axa.
Proof. Let (α, β | γ)−1 and (ιαι, γ | β) be two isotopies on L. Then (α, β| γ)−1· (ιαι, γ | β) = (α−1ιαι, β−1γ| γ−1β) := h, is an isotopy.
Noting thatγ−1β = La and so β−1γ = (La)−1= La−1, and so h =
= (α−1ιαι, La−1 | La).Applying h to xa yields: (α−1ιαι(x).a−1a = α−1ιαι(x) = a(x)and so a(xa).a−1y = a(xy)and in particular a(xa).a−1= ax, hence a(xa) =
(ax)a.
Lemma 20. If (α, β | γ) is an isotopy, then so are (ιαι, γ | β), (ιγι, α | ιβι), (ιβι, ιαι| ιγι), (β, ιγι | ιαι), (γ, ιβι | α).
Proof. If xy=z, then the following are equivalent expressions:
xy = z⇔ x = zy−1⇔
z−1x = y−1⇔ z−1= y−1x−1⇔ yz−1 = x−1⇔ y = x−1z
It follows that if (α, β | γ) is an isotopy then these relations will also be equivalent:
α(x)β(y) = γ(z)⇔ γ(x) = α(z)β(y−1)⇔ α(z−1)β(x) = γ(y−1)⇔ γ(z−1) = α(y−1)β(x−1)⇔ α(y)β(z−1) = γ(x−1)⇔ γ(y) = α(x−1)β(z).
Again moving around the factors, restoring the order of x, y and z, will yield the following equivalences:
α(x)β(y) = γ(z)⇔ γ(x)β(y−1)−1 = α(z)⇔ β(x)γ(y−1)−1 = α(z−1)−1⇔ β(x−1)−1α(y−1)−1= γ(z−1)−1 ⇔ γ(x−1)−1α(y) = β(z−1)−1⇔ α(x−1)−1γ(y) = β(z).
Reading the duplex form of isotopy relations implied by the above set of equiva- lences, we get:
(α, β| γ) ⇔ (γ, ιβι| α) ⇔ (β, ιγι| ιαι) ⇔ (ιβι, ιαι| ιγι) ⇔ (ιγι, α| ιβι) ⇔ (ιαι, γ| β).
If we transform the above set of equivalences to the triplex form and replace γ by ιγι- to make the symmetry more obvious - we get the following six isotopies which will come in handy later:
(α, β, γ)⇔ (ιγι, ιβι, ιαι)⇔ (β, γ, α)⇔ (ιβι, ιαι, ιγι)⇔ (γ, α, β)⇔
(ιαι, ιγι, ιβι).
Theorem 21. If a = α(1) for some monotopy α ∈ M(L) we have that Ma(L) :=
{La, La−1, Ra, Ra−1, Ba, Ba−1} are monotopies since (La, Ra | Ba), (Ba, La−1 | La), (Ra, Ba−1 | Ra−1), (La−1, Ra−1 | Ba−1), (Ba−1, La | La−1), (Ra−1, Ba | Ra) are isotopies.
Proof. La takes 1 to a, and is a monotopy, since (La, Ra | Ba) is an isotopy, by lemma 17. By lemma 20 we have the six isotopies, where we can write ιLaι = Ra−1, ιRaι = La−1 and ιBaι = Ba−1. By de nition the members of Ma(L) are
monotopies.
Theorem 22. The action of M(L) on L, M×L→ L,α·x 7→ α(x) is transitive if and only if (zx)(yz) = z(xy)z holds for all x, y ∈ L. (zx)(yz) = z(xy)z is also known as the Moufang identity, and inverse loops in which it holds are called Moufang loops.
Proof. Assume (zx)(yz) = z(xy)z, for all x, y ϵL. But (zx)(yz) = Lz(z)Rz(x)and (z(xy)z = Bz(xy),
hence (Lz, Rz | Bz) for all z ∈ L, and so Lz, Rz and Bz are monotopies for all z∈ L.
So given x, y ϵL, we have that Lyx−1(x) = (yx−1)x = y(x−1x) = yand likewise for R, and the action is transitive.
Conversely assume the action M × L → L is transitive, i.e. for all x, y ∈ L there exists α ∈ M(L) such that α(x) = y, which is equivalent to there being only one orbit of the action on M(L) on L.
In particular for all z ∈ L there exists α ∈ M such that α(1) = z. By theorem 19 Lzis a monotopy, taking 1 to z, and we have the isotopy (Lz, Rz| Bz)for all z ∈ L, and so Lz(x)Rz(x) = Bz(x),i.e. (zx)(yz) = z(xy)z.
2. composition Algebras, Absolute Valued Algebras and the octonions.
In this section we will de ne absolute valued algebras and composition algebras, derive some basic consequences and describe some features of their intersection.
De nition 23. Let k be a eld of characteristic not 2. A k-algebra is vector space Aover k that is equipped with a k-bilinear multiplication A × A → A, (x, y) 7→ xy.
We restrict the choice of eld to be able to de ne a quadratic form on the algebra.
De nition 24. A composition algebra A = (A, m) is a non-zero k-algebra endowed with a quadratic form m : A → k that is multiplicative and non-degenerate, i.e.
m(xy) = m(x)m(y), and < x, y >:= 12(m(x + y)− m(x) − m(y)) is non-degenerate:
< x, y >= 0for all y∈ A if and only if x = 0.
We'll now deduce some consequences the composition law, m(xy) = m(x)m(y).
Proposition 25. < xy, xz >= m(x) < y, z > and < xz, yz >=< x, y > m(z).
(The scaling laws).
Proof. Replacing y by y + z in the the composition law, gives:
m(xy + xz) = 2 < xy, xz > +m(xy) + m(xz), but
m(xy + xz) = m(x(y + z)) = m(x)m(y + z) = m(x)(2 < y, z > +m(y) + m(z)),
so we we cancel the terms m(xy) and m(xz) using the composition law, divide by two and we have our equality. < xz, yz >=< x, y > m(z) is proved in a similar
way.
Proposition 26. < xy, uz >= 2 < x, u >< y, z > − < xz, uy >. (The Exchange Law).
Proof. replacing x by x + u in < xy, xz >= m(x) < y, z >, gives
< (x + u)y, (x + u)z >=< xy + uy, xz + uz >=< xy, xz > + < xy, uz >
+ < uy, xz > + < uy, uz >, which by the scaling laws is (m(x) + 2 < x, u > +m(u)) < y, z >.
We cancel terms and rearrange.
De nition 27. conjugation of an element x, denoted x∗, is de ned as x∗ := 2 <
x, 1 >−x.
Proposition 28. < xy, z >=< y, x∗z >, and < xy, z >=< x, zy∗>. (The Braid Laws).
Proof. Put u = 1 in the exchange law and we get
2 < x, 1 >< y, z >− < xz, y >=< y, (2 < x, 1 > −x)z >.
The left hand side is equal to < xy , z >, and the right hand side is by de nition equal to < y, x∗z >. The other case is similar. Proposition 29. x∗∗:= (x∗)∗= x. (Biconjugation)
Proof. Setting y = 1 and z = t and applying the braid law twice we have:
< x, t >=< x1, t >=< 1, x∗t >=< (x∗)∗1, t >=< x∗∗, t >, for all t. Proposition 30. (xy)∗= y∗x∗. (Product Conjugation)
Proof. Repeated use of biconjugation gives that:
< y∗x∗, t >=< x∗, yt >=< x∗t∗, y >=< t∗, xy >=< t∗, (xy)1 >=
=< t∗(xy)∗, 1 >=< (xy)∗, t >.
Proposition 31. De ne x−1 = x/m(x), for x ̸= 0. Then x∗(xy) = m(x)y = (yx)x∗, or equivalently x−1(xy) = y = (yx)x−1
Proof. < x∗(xy), t >=< xy, xt >= m(x) < y, t >=< m(x)y, t >. Theorem 32. The moufang laws hold: (xy)(zx) = (x(yz))x = x((yz)x) for all x, y, z in a composition algebra.
Proof. We have the following equalities:
< (xy)(zx), t >=< xy, t(x∗z∗) >=
= 2 < x, t >< y, x∗z∗>− < x(x∗z∗), ty >=
= 2 < x, t >< yz, x∗>− < x∗z∗, x∗(ty) >=
= 2 < yz, x∗>< x, t >−m(x) < z∗y∗, t >=
= 2 < x, (yz)∗>< x, t >−m(x) < (yz)∗, t >.
So (xy)(zx) = 2 < x, (yz)∗ > x− m(x)(yz)∗ is a function of x and yz only. We can therefore replace y and z by any two other elements with the same product, so deduce that:
(xy)(zx) = (x(yz))1x = (x(yz))x, and (xy)(zx) = x1((yz)x) = x((yz)x). Corollary 33. Bimultiplication is well de ned in a composition algebra:
Proof. (xy)x = (xy)(1x) = (x1)(yx) = x(yx).
Remark 34. Thus we have proved that unital composition algebras are special cases of Moufang Loops.
De nition 35. An absolute valued algebra A, is a non-zero real algebra and a multiplicative norm ∥ · ∥: A → R≥0 i.e. for all x, y∈ A, ∥ xy ∥=∥ x ∥∥ y ∥, where the last juxtaposition is multiplication in R. We also say that the norm respects the multiplication on A.
Proposition 36. ∥ xy ∥=∥ x ∥∥ y ∥ implies that A has no zero dividers.
Proof. For otherwise there would be non-zero x and y in A such that 0 =∥ xy ∥=
=∥ x ∥∥ y ∦= 0.
Proposition 37. If the dimension of A is nite, then A contains no zero dividers if and only if A is a division algebra, i.e. La and Ra are bijective for all a̸= 0.
Proof. The lack of zero dividers in A implies that Laand Raare injective and linear algebra teaches that linear operators on nite vectorspaces are surjective if they are
injective.
Theorem 38. If A is a nite dimensional absolute valued algebra then A is a composition algebra.
Proof. We need a quadratic form that is multiplicative and non-degenerate. m : A → R. Our candidate is m(x) :=∥ x ∥2. It is multiplicative since m(xy) =∥
xy ∥2= (∥ x ∥∥ y ∥)2 =∥ x ∥2∥ y ∥2= m(x)m(y). It is non degenerate since if
< x, y >= 0for all y ∈ A then in particular 0 =< x, x >= m(x) =∥ x ∥2 which implies that x = 0.
However it is not in general a qudratic form since the inner product it induces
< x, y >:= 12(m(x + y)− m(x) − m(y)) need not be bilinear. However, Urbanik and Wright [7] proved that if A is unital then A is nite dimensional and then the
norm induces a quadratic form [1].
2.1. The octonion algebra.
De nition 39. The real octonion algebra O is the vectorspace R8 with a bilin- ear multiplication O × O : (x, y) 7→ xy, given by subjecting the basis vectors (1, i0, i1, i2, i3, i4, i5, i6) := (e1, e2, e3, e4, e5, e6, e7, e8)to the following relations:
i2n=−1 and
in+1in+2= in+4=−in+2in+1
in+2in+4= in+1=−in+4in+2
in+4in+1= in+2=−in+1in+4, where the subscripts run modulo 7.
Each octonion o ∈ O is thus on the form:
o = x∞+ x0i0+ x1i1+ x2i2+ x3i3+ x4i4+ x5i5+ x6i6, xt∈ R.
The standard norm on R8makes the octonion algebra an 8 dimensional absolute valued algebra.
3. Triality
Proposition 40. Every element α of O(n) that xes a k-dimensional subspace can be written as a product of at most n − k re ections.
Proof. Take a vector v not xed by α, i.e. α(v) = w ̸= v. The re ection in v − w, call it σ, restores w to v and xes any vector u in U, the subspace xed by α. To see this rst note that since α ϵ O(n) we have that ∥ v ∥=∥ α(v) ∥=∥ w ∥. Hence the parallelogram spanned by v and w is a rhombus and so the diagonals v + w and v − w intersect at right angles, indeed:
< v + w, v− w >=< v, v > − < v, w > + < w, v > − < w, w >=
=< v, v >− < w, w >=< v, v > − < α(v), α(v) >= 0.
We can thus make a unique orthogonal decomposition of v and w, namely:
v = 12((v + w) + (v− w)) and w = 12((v + w)− (v − w)). The re ection σ in (v− w) will x (v + w) since it's orthogonal to (v − w) and σ(v − w) = −(v − w).
It follows that σ(w) = v.
If u is xed by α then < u, v >=< α(u), α(v) >=< u, w > which implies that
< u, v− w >= 0 and u is thus orthogonal to v − w and hence xed by the re ection so we have that α(u) = u implies σ(u) = u. But we also have that σα(v) = σ(w) = v and so σα(x) = x, for all x in U + Rv. But U ( U + Rv since v was not in U, and so we procede by induction.
If dim(U) = n then α = 1Rn=∏
i ϵØσi. Assume that if α xes U pointwise and that dim(U) = k ≥ 1 then α = σl...σ1, l ≤ n − k. Now if α xes T where dim(T ) = k − 1, then σvα xes the subspace T + Rv and
dim(T + Rv) = dim(T ) + 1 = k, and by assumption we must have that σvα = σl+1σl...σ1, where σv= σl+1, and l + 1 = n − k + 1 = n − (k − 1).
Theorem 41. If α ϵ SO(8), then α is a product of an even number of re ections.
Proof. Re ections α have det(α) = −1, SO(n) is the subgroup of maps β with det(β) = 1 and determinants are multiplicative, i.e. det(αβ) = det(α)det(β). It follows from proposition 40 that all elements of SO(n) must be the product of an
even number of re ections.
The octonions are strongly non-associative in the sense that
Proposition 42. If x(ry) = (xr)y for all octonions x, y and r, then r is real.
Proof. (in+1in)in+2)=−in+3in+2= in+5and in+1(inin+2)) = in+1in+6=−in+5̸=
in+5. Thus for r to associate with all octonions r must have all imaginary coe cients
equal to zero so r is real.
Proposition 43. If a, b is any pair of companions for the monotopy γ then any other pair has the form ar, r−1b, where r is real.
Proof. The condition that A, B should be a second pair of companions is the identity (γ(x)a)(bγ(y)) = (γ(x)A)(Bγ(y)). If we write γ(x) = X and γ(y) = Y then the identity becomes (Xa)(bY ) = (XA)(BY ) for all X, Y. We can choose r such that (Rr is transitive) A = ar, and since γ is bijective, there is x and y such that X = a−1, Y = 1, which gives b = rB, so B = r−1b, and the identity is now that (Xa)(bY ) = (X(ar))((r−1b)Y ), for all X, Y. Set Y = (r−1b)−1= b−1r, shows that (Xa)r = X(ar), and setting X = (ar)−1= r−1a−1shows that r−1(bY ) = (r−1b)Y and we get the identity (Xa)(bY ) = ((Xa)r)(r−1(bY )), for all X, Y. Finally set Xa = xand bY = rz and we get the identity on the desired form x(rz) = (xr)z,
which by proposition 42 implies that r is real.
Theorem 44. Let A be a real division algebra with 2 ≤ dim(A) < ∞. Let x and y be elements in A \ {0}. Then det(Lx)and det(Ly) have the same sign.
Proof. Consider the sign map: R \ {0} → C2, sign(x) = |x|x as having values in the cyclic group C2={1, −1}. We de ne the generalized sign map
s: GLR(A)→ C2, s(x) = sign(det(x)) det(x),
which is de ned for every nite-dimensional real vector space A. The maps:
L : A\ {0} → GLR(A), a 7→ La and R : A \ {0} → GLR(A), a 7→ Ra
yield maps l and r when composed with the generalized sign map:
l : A\ {0} → C2, l(a) = s(La)and r : A \ {0} → C2, r(a) = s(Ra).
If A is either C, H or O then both l and r are constant. For we equip A \ {0} with the standard Euclidean topology of Rdim(A) and C2 with the discrete topology.
A\ {0} is connected, since the dimension is ≥ 2, and l and r are continuous - since they are the composition of continuous functions - hence their image is connected
and they must therefore be constant.
Proposition 45. Let A be an absolute valued algebra. If ∥ x ∥= 1, then Lx is an orthogonal map.
Proof. We have the equality:
< v, v >=∥ v ∥2=∥ x ∥2∥ v ∥2=∥ xv ∥2=∥ Lxv∥2=< Lxv, Lxv >. Corollary 46. det(Lx) = 1for all unit octonions.
Proof. det(L1) = 1, and if ∥ x ∥= 1 then x is in O(8), and so it must be in
SO(8).
Proposition 47. Bxis a scalar multiple of ref(1)·ref(x), where ref(x) is the re ec- tion in the vector x, i.e. in the hyperplane of which x is orthogonal.
Proof. In theorem 32 we derived that Bx(yz) = 2 < x, (yz)∗ > x− m(x)(yz)∗. Comparing this with the formula for the re ection: ref(x) : t 7→ t−2 < x, t > x/ <
x, x >, we get that − < x, x > refx((yz)∗) =− < x, x > (yz)∗+ 2 < x, (yz)∗ >
x = (xy)(zx) = Bx(yz). So < x, x > ref1refx(yz) = Bx(yz). This is so since for real r, rz∗= a− b0i0− ... − b6i6 which implies that −rz∗ =−a + b0i0+ ... + b6i6
which implies that ref1(−rz∗) = a + b0i0+ ... + b6i6= rz. Of course real numbers
commutes with re ections.
Lemma 48. The operations ref(1)ref(a) and ref(a)ref(1) are bimultiplications by unit octonions.
Proof. Re ections are naturally unaected if we rescale a to have norm 1. But by proposition we have that the rst one is nothing but bimultiplication with a: ref(1)ref(a) = Ba. The operations are obviously inverses of each other, ref (a)ref (1)◦ ref(1)ref(a) = 1, and the inverse of Ba is of course given by (Ba)−1 = Ba−1 and so the second operation is also bimultiplication by a unit
octonion.
De nition 49. (α, β | γ) is called a special orthogonal isotopy if α, β, γ are all in SO(8).
Proposition 50. The special orthogonal isotopies form a subgroup of the group of all isotopies.
Proof. SO(8) is a group, so the component wise composition of isotopies is closed
in the set of special orthogonal isotopies.
Remark 51. The group of special orthogonal isotopies is also called the spin group Spin8.
Theorem 52. If γ is any element in SO(8) then there exists α, β in SO(8) for which (α, β | γ) is an isotopy which implies that γ is a monotopy. Moreover, α and β are uniquely determined by γ up to sign, the only other pair being −α and −β.
Proof. We can write γ as the product of an even number of re ections, γ = ref (a1)ref (b1)ref (a2)ref (b2)...ref (a2n)ref (b2n). But
ref (ai)ref (bi) = ref (ai)ref (1)ref (1)ref (bi)
and the left hand is the product of two bimultiplications of unit octonions, so γ is the product of 4n unit bimultiplications, Bc1Bc2...Bc4n. But then
(α, β| γ) = (Lc1Lc2...Lc4n, Rc1Rc2...Rc4n | Bc1Bc2...Bc4n)
is the desired isotopy, since c1c2...c4n are unit octonions α and β are in SO(8).
Companions being unique up to scalar multiplication α, β are unique up to sign, as the only non-trivial scalar that preserves SO(8) membership is -1, for det(λx) = λndet(x)and −La= L−a, so if Lais in SO(8) then −La is in SO(8). Therefore we also have the isotopy (−α, −β | γ), where −α = −Lc1Lc2...Lc4n = L−1Lc1Lc2...Lc4n.
Remark 53. The the theorem states that Spin8 forms a 2-to-1 cover of SO(8), a 2-to-1 cover meaning a surjective homomorphism where the pre-image of each element in the codomain is a set containing exactly two elements.
Lemma 54. Conjugate elements in SO(8) x subspaces of the same dimension.
Proof. Given α, β in SO(n) such that α = γβγ−1 for some γ in SO(n), consider F ix(α) :={x ∈ Rn| α(x) = x} = ker(α − 1n),
where the last equality, apart from being obvious, shows that F ix(α) is a subspace.
Let F ix(β) = {x ∈ Rn| β(x) = x} and our goal is to show that dim(F ix(α)) = dim(F ix(β)).
But x ∈ F ix(α) gives that α(x) = x which is equivalent to γβγ−1(x) = x, if and only if βγ−1(x) = γ−1(x), if and only if γ−1(x)∈ F ix(β). Let γ−1(x) = y and the expression is that y ∈ F ix(β) if and only if γ(x) ∈ F ix(α), thus F ix(α) = γ(F ix(β).
But γ is a bijection, therefore dim(F ix(α)) = dim(F ix(β)). Lemma 55. Bi0 xes a space of dimension ≥ 6. Li0 is x-point free.
Proof. We study the multiplication table:
i0i0=−1.
i0i0i0=−i0
i0i1i0= i3i0= i1
i0i2i0= i6i0= i2
i0i3i0= i0i1= i3
i0i4i0= i5i0= i4
i0i5i0= i0i4= i5
i0i6i0= i0i2= i6,
and the rst statement of the theorem follows. For the second statement observe that Li0(x) = i0x. So x ∈ F ix(Li0) implies that i0x = x = 1x if and only if (i0− 1)x = 0 but i0− 1 ̸= 0 and O has no zero-divisors. This implies that x = 0,
therefore dim(F ix(Li0)) = 0.
Theorem 56. The spin group Spin8has an outer automorphism of order 3, namely (α, β, γ)7→ (β, γ, α).
Proof. By lemma 20 the image of an isotopy is an isotopy, and it obviously respects composition, so this is an automorphism and it obviously has order 3. To see that it is an outer automorphism (i.e. not a conjugation x 7→ y−1xy) consider the isotopy (Li0, Ri0, Bi0) which is sent to (Ri0, Bi0, Li0). The two are not conjugates since
Li0 has no xpoint and Bi0 xes a 6-space.
Remark 57. SO(8) does not have this triality because γ determines α and β only up to sign.
4. Trialy and an isomorphism conditions for absolute valued algebras.
We start by stating a classic result by Albert [5].
Theorem 58. Let Af denote the category of nite dimensional absolute valued algebras. Let O1(V )⊆ O(V ) be the set of orthogonal maps xing the idenity 1V of the vector space V . Let A be either C, H or O. Then {Af,g| f, g ϵ O1(A)}, is dense in Af, i.e. for every A in Af there is f and g in O1 such that Af,g≃ A.
De nition 59. If A is in {C, H, O} and f, g in O1(A) then Af g=A as a vector space and multiplication is de ned as x ◦ y = f(x)g(y), where the juxtaposition is multiplication in A and the norm is the norm of A.
Proposition 60. If f and g are in O(A) then Af,g is an absolute valued algebra.
Proof. The vectorspace is left intact and all that needs to be proved is that the norm respects the multiplication. This is established by the following calculation:
∥ x ◦ y ∥=∥ f(x)g(y) ∥=∥ f(x) ∥∥ g(y) ∥=∥ x ∥∥ y ∥.
De nition 61. Let A be an absolue valued algebra with 2 ≤ dim(A) ≤ ∞, then the the double sign of A is (l(A), r(A)) ∈ C2× C2, where l(A) := l(a) for some a∈ A \ {0}.
Remark 62. l(A) := l(a), a ∈ A \ {0} is well de ned, since l(a) has been shown to be constant.
Theorem 63. If two nite dimensional absolue valued algebras have dierent dou- ble signs they are not isomorphic.
Proof. We will show that φ : A ≃ B ⇒ (l(A), r(A)) = (l(B), r(B)).
But l(A) := l(a) = sign(det(La))for some a ∈ A \ {0}. Chose some such a and set b = φ(a). Then
Lb(φ(x)) = by = φ(a)φ(x) = φ(ax) = φLa(x), x = φ−1(y)∈ A. So Lbφ = φLa. Therefore chose an arbitrary basis for A: a = (a1, ..., an) and let b = φ(a) = (φ(a1), ..., φ(an)) be a basis for B. Then mat(La)a = mat(Lb)φ(a). It follows in particular that sign(det(mat(La)a)) = sign(det(mat(Lb)φ(a))). Proposition 64. Let ψ be an isomorphism of nite dimensional absolute valued algebras A and B. Then ψ is a proper orthogonal map.
Proof. For linear maps f we have that < v, v′ >=< f (v), f (v′) if and only if
∥ v ∥=∥ f(v) ∥. But isomorphisms of nite dimensional absolute valued algebras respect the norm [2], i.e. ∥ v ∥=∥ ψ(v) ∥ and for nite dimensional absolute valued algebras
< v, w >:= 12(∥ v + w ∥ − ∥ v ∥ − ∥ w ∥),
de nes a scalar product [1]. Thus if ψ is an isomorphism, ψ is orthogonal.
But if A and B are isomorphic they have the same double sign (l(A), r(A)) = (l(B), r(B))and to preserve that, ψ must be proper orthogonal. We will now prove two lemmas, masqerading as propositions, before proving our
nal theorem.
Proposition 65. Let ψ : A → B be an isomorphism of nite dimensional absolute valued algebras and let f, g : A → A be two linear isometries. Then
ψf ψ−1, ψgψ−1: B→ B, are linear isometries and ψ : Af,g→ Bψf ψ−1,ψgψ−1, is an isomorphism.
In particular we have that:
(1) ψ : Af,g→ A is an isomorphism if and only if ψ : A → Aψf−1ψ−1,ψg−1ψ−1
is an isomorphism, where Af,gis unital with unity ψ−1(1).
(2) If this unity is xed by f and g, then ψf−1ψ−1 and ψg−1ψ−1 x 1.
Proof. By Rodríguez [6] we have that any isomorphism of nite dimensional abso- lute valued algebras is linearly isometric. So ψfψ−1 and ψgψ−1 are linear isome- tries. That they are from B to B is clear.
That Af,g≃ Bψf ψ−1,ψgψ−1 is clear since A ≃ B and we can therefore consider the multiplication in B to be of the form
x◦ y = ψ1Aψ−1(x)ψ1Aψ−1(y).
Then clearly Af,g≃ Bψf ψ−1,ψgψ−1, since
ψ(x◦ y) = ψ(f(x)g(y)) = ψ(f(x))ψ(g(y)) = x ⋄ y,
where ◦ is multiplication in Af,g, ⋄ is multiplication in Bψf ψ−1,ψgψ−1 and the second equality holds since ψ is an isomorphism on A.
(1) follows by taking A = Af,g, f = f−1 and g = g−1 in the main statement of the theorem.
(2) is obvious.
Proposition 66. Let f, g be two linear isometries of an Euclidean space A. Then Af,g is isomorphic to A if and only if f = Ra and g = Lb for suitable norm-one a, b of A. If moreover the linear isometries f, g x 1, then f = g = 1A.
Proof. Let ψ : A → Af,g be an isomorphism. Then for every x ∈ A we have ψ(x) = ψ(x· 1) = f(ψ(x))g(ψ(1)) and ψ(x) = ψ(1 · x) = f(ψ(1))g(ψ(x)). This shows that f = Ra and g = Lb where a = (g(ψ(1)))−1 and b = (f(ψ(1)))−1. If f and g x 1 then a = Ra(1) = f (1) = 1 = g(1) = Lb(1) = b.
For the converse observe that Af,g is unital with unity b−1a−1: x(b−1a−1) = xa(b(b−1a−1)) = xa(a−1) = x, and
(b−1a−1)x = ((b−1a−1)a)bx = (b−1)bx = x,
and by Albert the only unital nite dimensional algebras are the classic ones. Theorem 67. All nite dimensional absolute valued algebras are isomorphic to one of the form Af,g where f, g are two linear isometries xing 1 and A is R, C, H or O. Except in the case of A = R these algebras fall into four disjoint double-sign classes that corresponds to the following cases:
(1) f, g ∈ SO(n), (1, 1) ∈ C2× C2,
(2) f ∈ SO(n), g ∈ O(n) − SO(n), (1, -1) ∈ C2× C2, (3) g ∈ SO(n), f ∈ O(n) − SO(n), (-1, 1) ∈ C2× C2, (4) f, g ∈ O(n) − SO(n), (-1, -1) ∈ C2× C2.
Theorem 68. The full subcategory O = {Of,g| f, g ∈ O1(O)} of A8, the category of 8 dimensional absolute valued algebras, is dense, and is a grupoid.
If ψ : Of,g → Of′,g′ is an isomorphism in O, then ψ ∈ SO(8). Conversely, let Of,gand Of′,g′ be objects in O and let ψ ∈ SO(8) with triality correspondents ψ1, ψ2∈ SO(8). Then ψ : Of,g→ Of′,g′ is an isomorphism in O if and only if
f′= Rψ2(1)−1ψf ψ−1 and g′= Lψ1(1)−1ψgψ−1.
Proof. It has already been established that ψ is in SO(8). O is dense by Albert's theorem. It is a grupoid since all morphisms are isomorphisms (excluding the 0 morphisms from the category). This is so because morphisms in A8 respect the multiplication. So if f : A → B, is a morphism between two objects in A8 then injectivity implies surjectivity, since A and B are vectorspaces with the same nite dimension. To show that f is injective let x ∈ ker(f). Then for all y in A we have that f(xy) = f(x)f(y) = 0f(y) = 0. If x ̸= 0 then Lx has been shown to be bijective, thus every z ∈ A is on the form z = xy for some y ∈ A, this implies that f = 0, which we had already excluded. Thus x ∈ ker(f) implies that x = 0.
Assume that ψ : Of,g → Of′,g′ is an isomorphism. Take Of,g = A and Of′,g′ = B in proposition 66. It follows that ψ : O → Of′ψf−1ψ−1,g′ψg−1ψ−1 is also an isomorphism. By proposition 67 we have that
(f′ψf−1ψ−1, g′ψg−1ψ−1) = (Ra, Lb), where
(a, b) = ((g′ψg−1ψ−1ψ(1))−1, (f′ψf−1ψ−1ψ(1))−1) =
= ((g′ψg−1(1))−1, (f′ψf−1(1))−1), which implies that
(f′, g′) = (Raψf ψ−1, Lbψgψ−1). But we also have that
ψ(xy) = ψ(x)◦ ψ(y) = f′ψf−1ψ−1ψ(x)g′ψg−1ψ−1ψ(y) = f′ψf−1(x)g′ψg−1(y), for all x, y ∈ O.
So (f′ψf−1, g′ψg−1) = (ψ1, ψ2)both of which are clearly in SO(8). And the rst equivalence statement follows.
Conversely, assume that ψ ∈ SO(8) with ψ1, ψ2 ∈ SO(8), the triality correspon- dents, let f and g be given and set (f′, g′) = (Rψ2(1)−1ψf ψ−1, Lψ1(1)−1ψgψ−1). But we have that ψ(y) = ψ1(1)ψ2(y)if and only if (Lψ1(1))−1ψ(y) = ψ2(y) if and only if Lψ1(1)−1ψ = ψ2, and likewise we have that Rψ2(1)−1ψ = ψ1. But
ψ(x◦ y) = ψ(f(x)g(y)) = ψ1(f (x))ψ2(g(y)) = Rψ2(1)−1ψ(f (x))Lψ1(1)−1ψ(g(y)), and
ψ(x)⋄ ψ(y) = f′ψ(x)g′ψ(y) = Rψ2(1)−1ψf ψ−1ψ(x)Lψ1(1)−1ψgψ−1ψ(y) =
= Rψ2(1)−1ψf (x)Lψ1(1)−1ψg(y).
So ψ(x ◦ y) = ψ(x) ⋄ ψ(y) and ψ is an algebra morphism. It is bijective on the vectorspace and is therefore an algebra isomorphism.
References
[1] Lemma 2.2, M.L. El-Mallah, On nite dimensional absolute valued algebras satisfying (x,x,x)
= 0, Arch. Math. 49 (1987), 16-22.
[2] Lemma 2.1, J.A. Cuenca Mira and A. Rodríguez, Absolute values on H*-algebras, Comm.
Algebra 23 (1995), 2595-2610.
[3] T.A Springer, F.D. Veldkamp, Octonions, Jordan Algebras and Exceptional Groups, Springer Monographs in Mathematics (2000).
[4] J. H. Conway, D. A. Smith, On Quarternions and Octonions: Their Geometry, Arithmetic, and Symmetry, A K Peters, Ltd. (2003).
[5] A.A. Albert, Absolue Valued Algebras, Ann. of Math. (2) 48 (1947), 495-501.
[6] A. Rodríguez, Absolute-valued algebras, and absolute valuable Banach spaces, Proposition 1.1 p. 101, Advanced courses of mathematical analysis I, 99-155, World Sci. Publ., Hackensack, NJ, 2004.
[7] K. Urbanik, Fred B. Wright. Absolute-valued algebras. Proc. Amer. Math. Soc., 11:861-866, 1960.
