Why are SOnR and SLnR Lie Groups?

[jacobrhcp]

Homework Statement

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}

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.

 

[mighty2000]

To prove that SOnR is a Lie group, we need to show that it is a smooth manifold and that the group operations (multiplication and inversion) are smooth maps.

To show that SOnR is a smooth manifold, we can use a similar approach as for SLnR. We can define the map f: Rm-1 -> R, x -> det[x] - 1, where x is a matrix in SOnR. This function is smooth because it is just a polynomial in the coefficients of the elements of SOnR. The level set of this function will be SOnR, and by the implicit function theorem, SOnR will be a smooth manifold of dimension n(n-1)/2.

To show that the group operations are smooth maps, we can use the fact that the group operations (multiplication and inversion) are just matrix operations, which are known to be smooth. Therefore, the group operations on SOnR will also be smooth.

To determine the dimensions of SOnR and SLnR, we can use the fact that the dimension of a Lie group is equal to the dimension of its Lie algebra. The Lie algebra of SOnR is the set of skew-symmetric matrices, which has dimension n(n-1)/2. The Lie algebra of SLnR is the set of traceless matrices, which has dimension n^2 - 1. Therefore, the dimensions of SOnR and SLnR are n(n-1)/2 and n^2 - 1, respectively.