- #1

- 894

- 4

Show that the number of group parameters for [tex] O(N) [/tex] and [tex] SO(N) [/tex] are [tex] \frac{N(N-1)}{2} [/tex]

Does this make sense as the solution:

We have [tex] n^2 [/tex] parameters. For [tex] O(N) [/tex] we have the condition that [tex] O^T O = 1[/tex]. This condition places [tex] \frac{n(n+1)}{2} [/tex] constraints on the parameters. This leaves [tex] n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2} [/tex]

Since [tex] O(N) [/tex] has [tex] detO(N) = \pm 1[/tex] and [tex] SO(N) [/tex] has [tex] detSO(N)=+1[/tex], simply picking the sign of the determinent doesn't force any additional constraint on the parameters.

I can only show this statement: "This condition places [tex] \frac{n(n+1)}{2} [/tex] constraints on the parameters." for specific examples, like N=2 and 3. I can see how it is extended to higher orders, but I cannot prove it in the general case for N. Any ideas? Also, any good books on Group Theory and Lie Algebras for Physicists?

Thanks,

Ryan