SL(n;R) is connected
The special feature is that although not every element of is in the image of the exponential map of , is still a connected Lie group. The proof below is a guideline and should be clarified a bit more at some points, but this was done intentionally.
To illustrate the point, first we show
, but it is in .
We have with .
The keyword here is polar decomposition. We notice that is symmetric and positive definite, since , with the standard inner product on . Hence, we can write , with and . is well defined, since any real symmetric, positive definite matrix is diagonalizable. It’s easy to check that , hence . We had and , hence and so . Since the choice of positive root is unique, and are unique. Moreover, is exactly generated by the set and , the set of real symmetric matrices of determinant , by , we have the wanted statement: . ∎
This is now clear from the fact that both and are connected and so , a fact easily checked by taking the determinant. So is path-connected, hence connected. ∎
|Title||SL(n;R) is connected|
|Date of creation||2013-03-22 18:52:05|
|Last modified on||2013-03-22 18:52:05|
|Last modified by||Stephaninos (23208)|