Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on N-dimensional Lie Groups

  1. Aug 22, 2010 #1
    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 [itex]n(n-1)/2[/itex] generators of the form,

    [tex]X_{\rho,\sigma}=-i\left(x_{\rho}\frac{\partial}{\partial x_{\sigma}}-x_{\sigma}\frac{\partial}{\partial x_{\rho}}\right)[/tex]

    where [itex]1\leq x_{\rho}\leq n[/itex] such that [itex]\sigma > \rho[/itex], 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. jcsd
  3. Aug 22, 2010 #2
    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).)
  4. Aug 22, 2010 #3
    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 [tex]\rho[/tex] and [tex]\sigma[/tex]. This means you can group your generators [tex]X_{\rho \sigma}[/tex] into a matrix where [tex]X_{\sigma\rho}=-X_{\rho \sigma}[/tex], ie a [tex]n\times n[/tex] antisymmetric matrix. The only independent elements in such a matrix are (say) the upper right triangle. This has [tex]1 + 2 + \ldots + n-1 = \tfrac12 n(n-1)[/tex] elements.
    Last edited: Aug 22, 2010
  5. Aug 22, 2010 #4
    thanks very much to both of you. the answers to both questions actually seem quite obvious after the fact. hah
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook