# Lie Groups Lecture 4

This lecture ( lie4 ) was given on 9/8/09

### Excerpts

Let $G,H$ be matrix lie groups, a morphism between them is a continous group homomoprhism; if $\phi \colon G \to H$ is a morphism then $\phi(G) \subset H$ need not be a matrix lie subgroup.  Consider the map $\mathbb{R} \to \mathbb{C}/\mathbb{Z}^2$ defined by $r \mapsto$latex r\exp(i \alpha)\$ for some irrational angle $\alpha$, the image is not a closed subgroup.

Polar decomposition for $SL(n, \mathbb{C})$ says any element in the special linear group can be written as a product $A = UP$ where $U$ is unitary of determinant 1, and $P \in \mathscr{P}_\mathbb{C}^1$; the set of positive Hermetian matrices of determinant 1

As an application,  let $B \in \mathscr{P}_\mathbb{C}^1$.  Then $B = \bigl( \begin{smallmatrix} a & b \\ \bar b & d \end{smallmatrix}\bigr)$ and $a,d$ are real and $ad + |b|^2 = 1$.  Parametrize this as $a = x_1 + x_4, b = x_2 + ix_3, d = x_4 - x_1$.  This gives a vector space isomorphism between $\mathbb{R}^4$ and $\mathscr{P}_\mathbb{C}^1$.  Also the det on $\mathscr{P}_\mathbb{C}^1$ gives minus the form associated to $O(3;1)$, namely $-(x_1^2 + x_2^2 + x_3^2 - x^4$.  It can be check (using polar decomposition) that conjugation by an element of $SL(2,\mathbb{C})$ preserves $\mathscr{P}_\mathbb{C}^1$, this gives a homomorphism $SL(2, \mathbb{C}) \to O(3,1; \mathbb{R})$ with kernel being plus and minus the identity.

A lie group that is not a matrix lie group.  $SL(n, \mathbb{Z})$ acts on the n torus $\mathbb{R}^n/\mathbb{Z}^n$.  It turns our for $n \ge 3$ the semi direct product

$T^n \rtimes SL(n,\mathbb{Z})$

is a lie group that is not a matrix lie group.