# Question on N-dimensional Lie Groups

1. Aug 22, 2010

### 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.

2. Aug 22, 2010

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).)

3. Aug 22, 2010

### 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.

Last edited: Aug 22, 2010
4. Aug 22, 2010

### antibrane

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