# The 3D equivalent to Quaternion?

1. Jul 23, 2004

### Aidman

Hi all,

Was wondering if the 3-dimensional equivalent to Quaternion has a name? And why does it seem like (at least for me) that only the groups, who’s number of values it holds is 2^n (where n is a integer value), are more intensively used compared to those who’s value count is not 2^n? I am for instance talking about complex numbers, quaternion and octonion. Is it just me or is there a reason for why these are more popular then groups who for example uses the imaginary numbers i and j? Please note that I am just asking of curiosity.

EDIT: sry if this all sounds foolish or unnecessary

Last edited: Jul 23, 2004
2. Jul 23, 2004

Staff Emeritus
In the middle 19th century, people had been looking for a three component "hypernumber" that would do for three dimensional space what complex two component numbers did for the plane. Nobody succeeded. Hamilton realized that the algbra that described three dimensional rotations was not three component but four component. His quaternions are actually the algebraic decriptors of three dimensional motions. They obey all the algebraic laws except commutation, but that is all right, because the rotations in three space are not commutative.

You are perfectly right about the powers of two. Google on "Clifford Algebras" to see the reasons. The next algebra up from the quaternions is that of the octonions. Eight components. But there is a theorem that there are no new ones after that, and the octonions in addition to being not commutative, are also not associative.

3. Jul 23, 2004

### fourier jr

AKA "Cayley numbers", and it was Frobenius who proved that there are only the complex numbers (which aren't ordered), quaternions (which aren't communative), and the Cayley numbers (which aren't even associative)

4. Aug 12, 2004

### mathwonk

There is a cute way to look at this problem using topology (and related to clifford algebras probably).

Note that the complex number multiplication on R^2 lets us define for each unit vector v in R^2 a perpendicular one iv, and hence a non zero vector at v, tangent to the unit circle, which is also viewed as P^1, the projective line.

Thus the existence of this multiplication implies that P^1 has one field of non zero tangent vectors, i.e. a trivial tangent bundle.

In general a division algebra structure R^n x R^n -->R^n, implies that P^(n-1) has (n-1) independent fields of non zero tangent vectors, hence a trivial tangent bundle.

Now there is a little gadget for confirming triviality of a tangent bundle called a stiefel whitney class w. These can actually be computed for projective spaces and one can show that w(P^r) is only zero when r +1 = 2^s, for some s.

It follows immediately that the only possible division algebras have dimension 2^s.

That the only actual ones have dimensions 1,2,4,8, is harder I guess.