Real Division Algebras Let me rst put up some properties for an algebra: Denition (Algebra over R) together with a multiplication An algebra over R is a vector space A over R A A R which satises Bilinearly: for any x, y, z A and any λ R. (x + y) z = x z + y z (λx) z = λ(z y) Unit axiom: There is an element 1 A such that 1 x = x 1 = x for all x A. By this we dene the Real division algebras as follows Denition (Real Division Algebra) Let A be an algebra over R not containing only of it's zero-element. A is a real division algebra if for any a A and any non-zero b A there exists exactly on x and y in A such that a = bx and a = yb This denition does not assume that the algebra A is either commutative or associative. The problem of real division algebra is to show that there exist four and only four real division algebras. That is to show that R n is a real division algebra if and only if n = 1, 2, 4 or 8. I will now write up some properties for these four real divison algebras in the following example: Example It is assumed that the algebras have conjugation properties. The most simple real division algebra is the real numbers, R. It has the properties R is Nicely-Normed, i.e. for any a R there exists a conjugation to a denoted ā R such that a + ā R. 1
R is Associative, i.e. for any a, b, c R it is true that a(bc) = (ab)c. R is Commutative, i.e. for any a, b R it is true that ab = ba. R is Real, i.e. for any a R there exists a conjugation to a denoted ā R such that a = ā. For n = 2 we have R 2 = C where = means isomophic. The complex numbers are on the form a + bi. They have some of the same properties as R but not all of them. C has the properties that C is Nicely-Normed Associative Commutative not Real It can be prooved that C is not real by looking at the conjugation of a complex number. (a + bi) = a bi a + bi So therefore C is not real. For n = 4 we have R 4 = H. These are called the Quarternions and was found by the mathematician Hamilton in 1843. It was also Hamilton who discovered the complex numbers eigth years earlier. The Quarternions are denoted H after Hamilton. They are on the form a + bi + cj + dk where i 2 = j 2 = k 2 = ijk = 1 One can easily multiply two Quarternions with each other, if one uses the gure below. This we shall see later. The Quarternions has the properties that they are Nicely-Normed Associative not Commutative To show the last part, let's look at two simple Quarternions: i j = k but j i = k 2
so when you move clockwise in the gure you get a plus at multiplication, but when you go anticlockwise you get a minus. Therefore the Quarternions are not commutative. The last dimension of the real numbers that gives a real division algebra is n = 8. So R 8 = O are the Octonions. They are given on the form a0 +a 1 e 1 +...+a 7 e 7 where e 2 i = 1 and e i e j can be found from the Fano plane. Figur 1: Fano plane The octonions have the property that it is Nicely-Normed. But it is not associative since e 1 (e 2 e 3 ) = e 1 e 5 = e 6 but (e 1 e 2 )e 3 = e 4 e 3 = e 6 3
The problem is to show that these four real division algebras are the only ones there exists. To do this I have to bring in some geometric terms. So rst I will go through some denitions Denition (Vector eld) A vector eld is a continuous map s : S n 1 R n such that s(x) is a tangent vector to S n 1 at x. Figur 2: Example of a vector eld for n = 2 Denition (k-frame) A k-frame is a set of vector elds s 1,..., s k such that s 1 (x),..., s k (x) are linearly independent (orthogonal) for all x S n 1. Denition (Parallelizable) on s 1,..., s n 1 on S n 1 S n 1 is parallelizable if there is a (n 1)-frame Examples: S 1 is parallelizable (see gure) S 2 is not parallelizable. This was proven by the Hairy ball theorem, which says that one can not take a ball with hair and comb it. I will now put up an impordent result for the real division algebras Propersition If R n is a real division algebra then S n 1 is parallelizable. In the proof we will assume that x y = x y, but rst look at the following example 4
Figur 3: Example of a real division algebra being parallelizable for n = 2 Example s(x) = ix For n = 2 we will have to dene the map s : S 1 R 2 = C such that Proof In the general case for n dene s 0, s 1,..., s n 1 : S n 1 R n by s i (x) = e i x where 1 = e 0, e 1,..., e n 1 is the orthogonal basis for R n. We need to check that these vector elds denes a (n 1)-frame for S n 1. We use that two vectors v and w are orthogonal if v + w 2 = v 2 + w 2. So for i j we have s i (x) + s j (x) 2 = e i x + e j x 2 = (e i + e j )x 2 1 = e i + e j 2 x 2 2 = e i 2 x 2 + e j 2 x 2 1 = e i x 2 + e j x 2 = s i (x) 2 + s j (x) 2 where it is used that R n is normed at 1, and that the basis is chosen so that it is linearly independent in itself at 2. Now we have shown that s 1,..., s n 1 are linearly independent and that they therefore denes a (n 1)-frame. That it is a (n 1)-frame to S n 1 follows from the fact that s 0 (x) = 1 x is also linearly independent of all the s i 's where i 0. Hence if R n is a real division algebra S n 1 is parallelizable. 5
Now the problem can be reduced to proving that the only spheres S n 1 that are parallelizable are the ones where n = 1, 2, 4 and 8. It was proved independently by Milner and Bott in 1958 that S n 1 is parallelizable if and only if n {1, 2, 4, 8}. Four years later the mathematician Adams proved a theorem which gave another and easier way to solve the problem. He stated that the span of S n 1 could be found by calculating a number called the Hurwitz-Radon number. In other words Span(S n 1 ) = ρ(n) 1 where ρ(n) = 8a + 2 b and n = 2 4a+b k where k is an odd number and a 0 and 0 b 3 are integers. So S n 1 is parallelizable if and only if ρ(n) = n Example For n = 2: n = 2 = 2 4a+b k a = 0, b = 1, k = 1 ρ(2) = 8 0 + 2 1 = 2 Hence S 1 is parallelizable, which is no surprice since we have already seen this above. For n = 3: n = 3 = 2 4a+b k a = 0, b = 0, k = 3 ρ(3) = 8 0 + 2 0 = 1 3 Hence S 2 is not parallelizable, which we have also seen in the Hairy ball theorem. Proof We have to solve the equation ρ(n) = n. Hence we have When we look at the term 2 4a we get 8a + 2 b = 2 4a+b k 2 4a 2 4a+b k = 8a + 2 b 8a + 8 = 8(1 + a) where at least one of the inequities denoted by * has to be a strict inequality, since if the rst should be an equality then b = 0 but if the last one should be an 6
equality then b = 3, and b can only hold one value at the time. Therefore we get 2 4a 3 < 1 + a This means that a has to be 0 for all n or else the left hand side will be to bigg. By plugging a = 0 into ρ(n) = n we get 2 b = 2 b k k = 1 Since b {0, 1, 2, 3} we now get that since n = 2 b, n can only take the values 1,2,4 and 8. 7