# Why are SOnR and SLnR Lie Groups?

1. Oct 7, 2008

### jacobrhcp

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