Lie Groups Lecture 4

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


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.


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