dimension of sl(2,H)

[BruceG]

dimension of sl(2,R) = 1*(2*2-1) = 3, is isomorphic to so(2,1) : 2+1 = 3
dimension of sl(2,C) = 2*(2*2-1) = 6, is isomorphic to so(3,1) : 3+2+1 = 6
dimension of sl(2,H) = 15, is isomorphic to so(5,1) : 5+4+3+2+1 = 15
dimension of sl(2,O) = 45, is isomorphic to so(9,1) : 9+8+7+6+5+4+3+2+1 = 45

How do you prove the 15 and 45?
Naively I would expect 4*(2*2-1)=12 and 8*(2*2-1)=24.

[arkajad]

The tricky part is the definition of sl(2,H) and sl(2,O). John Baez discusses it (but not in all the details) here (http://math.ucr.edu/home/baez/octonions/node11.html).

[BruceG]

OK, ta, I got it.

So with R and C the set of traceless matrices are closed (if a,b are traceless then so is [a,b]). So once you've counted the traceless matrices you've got the whole algebra.

H and O are not commutative, so if you start with the set of traceless matrices, then to close off the algebra you have to add in some matrices of the form [a,b] which are not traceless.

Hence 15 > 12
and 45 > 24

This then makes this result all the more fascinating: take any normed division algebra K and generate sl(2,K) according to the above procedure and you magically end up with something isomorphic to the Lorentz algebra so(dim(K)+1,1).

There must be a deeper reasoning behind all this: I'll read more of Baez to see what he has to say, but any other references on this would be appreciated.