Question on N-dimensional Lie Groups

[antibrane]

I'm currently learning Lie groups/algebras and I am trying to find the infinitesimal generators of the special orthogonal group SO(n). It is the n-dimensions that throws me off. I know that the answer is $n(n-1)/2$ generators of the form,

$$X_{\rho,\sigma}=-i\left(x_{\rho}\frac{\partial}{\partial x_{\sigma}}-x_{\sigma}\frac{\partial}{\partial x_{\rho}}\right)$$

where $1\leq x_{\rho}\leq n$ such that $\sigma > \rho$, but how do I get this? The only way I could think of is somehow trying to find the n-dimensional rotation matrix in general and then going from there (I have actually tried this and it gets me something similar)--there must be a simpler way though.

I also run into this problem in other areas of the book I am using. For example attempting to establish that the Lie algebra sl(n,C) is an ideal of gl(n,C). I guess I am looking for some advice on how to compute these algebras in n-dimensions. Thanks in advance for any help/suggestions, and let me know if I can make anything more clear.

 

[adriank]

Can't you prove that sl(n) is an ideal of gl(n) directly? You know that sl(n) is the subset of gl(n) consisting of matrices of trace zero. So prove that for A in sl(n) and B in gl(n), [A,B] is in sl(n). Use the fact that tr(AB) = tr(BA). (You could also note that the trace is a Lie algebra homomorphism gl(n) -> C with kernel sl(n).)

 

[Simon_Tyler]

Adriank is correct about the approach to sl(n) being an ideal of gl(n).

As for the dimensions of so(n), the representation you give for the generators is clearly antisymmetric in $$\rho$$ and $$\sigma$$. This means you can group your generators $$X_{\rho \sigma}$$ into a matrix where $$X_{\sigma\rho}=-X_{\rho \sigma}$$, ie a $$n\times n$$ antisymmetric matrix. The only independent elements in such a matrix are (say) the upper right triangle. This has $$1 + 2 + \ldots + n-1 = \tfrac12 n(n-1)$$ elements.

 

[antibrane]

thanks very much to both of you. the answers to both questions actually seem quite obvious after the fact. hah