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Why are SOnR and SLnR Lie Groups?

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove SOnR and SLnR are Lie groups, and determine their dimensions.

    SOnR = {nxn real hermitian matrices and determinant > 0}
    SLnR = {nxn real matrices with determinant 1}

    3. The attempt at a solution

    We can see that SLnR is level set at zero of the graph of a smooth function, namely the function f:Rm-1 -> R, x -> det[x] - 1, which is smooth because it is just a polynomial in the coefficients of elements of SLnR.
    We know that the graph of a smooth function is a smooth manifold, and by the implicit function theorem, so is a level set of a smooth function, and so is SLnR, in this case of dimension n^2 -1.

    The above is a very short version of what took the better part of several afternoons.

    I still have problems showing one thing:

    I want to make a similar map for SOnR, to show that this is too a smooth manifold. What is such map? I imagine it being slightly more complicated, since I looked up its dimension to be n(n-1)/2. I figured SOnR is also the set {nxn matrices with determinant +1 or -1}, and hence be the union of the previous level set, with som other graph, but this would never yiel a dimension of n(n-1)/2

    I feel quite lost, I hope someone can help me out.
     
    Last edited: Oct 7, 2008
  2. jcsd
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