# Homework Help: Lie Groups

1. Feb 23, 2010

### latentcorpse

Consider R^4 with nondegenerate inner product $\left( x,y \right)= x^T \eta y$ where $\eta=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)$
Let $e_1,e_2,e_3,e_4$ be the basis of R^4 with $(e_i,e_i)=\begin{cases} 1 \quad i=1,2,3 \\ -1 \quad i=4 \end{cases}$
Let $P \subset Aff(\mathbb{R}^4)$ be the subset of the affine group that preserves $\delta(x,y)=(x-y)^T \eta (x-y)$

we have to show that P is a linear group and determine its lie algebra $\mathfrak{p}$ as a Lie subalgebra of $gl(5,\mathbb{R})$ and then find the dimension of $\mathfrak{p}$.

to show its a linear group:

i guess i need to establish that its a subset of invertible matrices (these will be 2x2 as then it would be isomorphic to R^4 i think) but i dont really know how to show this explicitly
then given A in P
$\delta(Ax,Ay)=(Ax-Ay)^T \eta (Ax-Ay) = (x-y)^T A^T \eta A (x-y) = delta(x,y)$ iff $A^T \eta A = \eta$. (this is what elements of P must satisfy)

then given a particular A in P, consider A^{-1}

$\delta(A^{-1}x-A^{-1}y)^T \eta (A^{-1}x-A^{-1}y)=(x-y)^T(A^{-1})^T \eta A^{-1} (x-y)$
but $(A^{-1})^T \eta A^{-1}=(A^T)^{-1} \eta A^{-1} = A^{T}^{-1} \eta^{-1} A^{-1}=A^T \eta A = \eta$where we used the property $\eta^{-1}=\eta$
this means that $A^{-1} \in P$
similarly if A,B are in P

$\delta( ABx,ABy ) = (x-y)^T B^T A^T \eta AB (x-y) = (x-y)^T B^T \eta B (x-y) = (x-y)^T \eta (x-y) \Rightarrow AB \in P$
therefore P is a linear group. is this correct?

then for the next bit i said let $A= exp ( \tau X) \in P$
we know $A^T \eta A = \eta$
therefore $exp ( \tau X^T ) \eta exp( \tau X ) = \eta$

differentiating wrt tau and evaluating at tau=0 we get
$X=- \eta^{-1} X^T \eta$
which implies
$\mathfrak{p} := \{ X \in Mat ( 2 , \mathbb{R} ) | X=-\eta^{-1} X^T \eta \}$
is this right? i guess this is determining p but how do i determine it as a lie subalgebra of gl(5,R) - i dont even know what that means tbh!
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