Lie Groups lecture 2

This lecture was given on september 1st 2009.

This lecture concerned important examples of matrix lie groups: GL(n, k), SL(n,k), O(n), SO(n), U(n), SU(n), SP(n), O(n,m), SO(n,m).  Most of the material is taken from chapter 1 of Brian Hall’s book.  Also a classification of bilinear forms was presented.

The notation is that O(B) is the group of invertible matrices that preserve B: B(Ax,Ay) = B(x,y).  In general, it suffices to study symmetric and anti symmetric bilinear forms because it can be written as a sum of symmetric and antisymmetric forms

B = \frac{B + B^\intercal}{2} + \frac{B - B^\intercal}{2}=: B' + B''

O(B) = O(B') \cap O(B'')

Symmetric Bilinear Forms

Assume B is symmetric.  Then its normal so the spectral thm says there exists a basis under which its orthogonal.  Its also Hermitian so its eigenvalues are real.  So after scaling can assume B = I_p\oplus -I_n \oplus 0_m where $p + n + m = N$ is the dimension of the vector space.  Let this basis be given by v_1, ...., v_N.  Define

m = \dim V_0, V_0 = \{x \in \mathbb{R}^N : B(x,\mathbb{R}^N) = 0 \}

V_\pm = span \{v_i : \pm B(v_i, v_i) > 0\}

The inertial thm says if W: w_1, ...., w_N is another basis with W_0 = V_0 then \dim W_\pm = \dim V_\pm.  If not, say \dim W_+ > \dim V_+ then \dim W_+ + \dim(V_0 + V_-) > N so there subspaces have to intersect nontrivially giving a v such that B(v,v) > 0 and B(v,v) \le 0; gives contradiction.

Anti Symmetric Bilinear forms

If B is antisymmetric (B(x,y) = -B(y,x) ) then define V_0 \subset k^n as B(V_0, k^n) = 0.  Then k^n can be decomposted k^n = V_0\oplus W.  By retricting to W can assume B is nondegernate.  Choose x \in W arbirary, by assumption there exists y s.t. B(x,y) \ne 0.  Let V_1 = span(x,y) and decompose W = V_1 \oplus V_1^{\perp} where V_1^{\perp} = \{ w | B(w,V_1) = 0\}.  Repeating the argument (and noting that B is nondegenerate) it follows that W = V_1 \oplus ... \oplus V_m where each V_i are two dimensional and (afer scaling) B|_{V_i} = \left(\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right) =: H.  It follows that W is even dimensional and the contruction gives a matrix for B which is a direct sum of H‘s.

Its a fac that B also has a matrix of the form J:= \left(\begin{smallmatrix} 0 & I \\ -I & 0 \end{smallmatrix}\right).

It can be checked that GL(n, \mathbb{C}) \subset GL(2n, \mathbb{R}) by identitifying it with the \{ A : AJ = JA\}.  If \mathbb{C}^n \ni x,y = (x_1, ...., x_n), (y_1, ..., y_n) are decomposed as x_k = u_k + iv_k and y_k =w_k +iz_k taking the usual norm on \mathbb{C}^n gives

\langle x,y \rangle = \sum_k u_kw_k + v_kz_k + i\sum_k u_kz_k - v_kw_k

Which separate it into a symmetric and antisymmetric part.  Consequently U(n) = \{A: AJ = JA\} \cap O(2n, \mathbb{R}) \cap SP(n, \mathbb{R}) \subset GL(2n, \mathbb{R}).

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