Let X ad Y be two vectors in R^{4}, and define the inner product of X and Y as:

(X*Y) = g_{ik}X_{i}Y_{k} (summation convention for repeated indicies)

Then we consider the 4x4 matrix g whose components are g_{ik}. I am of course interested in the case that g is NOT positive definite, because this is the situation when g represents the gravitational field in general relativity.

Let A be a 4x4 matrix which satisfies (X*Y) = (AX*AY), then I say that A is an element O(g), the orthogonal group determined by g.

I am interested in finding any sort of formula which relates the lie algebra of O(g) to the metric g.

In a previous question, it was suggested that I diagonalize the matrix g using the theorem on diagonalizing positive definite matrices. This method works nicely and gives a simple solution for the lie algebra in terms of the transformation matrix which diagonalizes g, but only when g is positive definite.

Can I still diagonalize my non positive definite g by finding the roots of the characteristic polynomial? I believe I must first somehow restrict the set of vectors I allow the inner product to work on, to avoid the case (X*Y) = 0. Nevertheless, for arguments sake let's assume that I can diagonalize g.

Let B be the transformation matrix, then I assume that I can write:

g = B^{-1}ηB, where η is the identity matrix of signature (1,3), i.e. the metric of flat space time.

We can characterize the elements of O(g) by realizing that our inner product can be written as:

(X*Y) = X^{T}gY

It's clear that if A is an element of O(g), then

(AX*AY) = X^{T}A^{T}gAY = X^{T}gY

Hence, A^{T}gA = g

This formula can be written as gA^{-1}g = A^{T}, since g = g^{-1} explicitly.

Now applying the transformation matrix B:

gA^{-1}g = B^{-1}ηBA^{-1}B^{-1}ηB = A^{T}, which I rearrange as:

ηBA^{-1}B^{-1}η = BA^{T}B^{-1}

If by some chance (BA^{-1}B^{-1})^{-1} = (BA^{T}B^{-1})^{T}, then I can immediately conclude that A is in O(g) so long as BA^{-1}B^{-1} is in O(1,3) (the group which preserves the metric η). From this step is it quite straightforward to compute the lie algebra, by taking advantage of the formula Be^{X}B^{-1} = e^{BXB-1}.

I am concerned about several steps of this procedure:

1) Is it legal to diagonalize g? I believe I need exclude any combination of vectors X, Y for which (X*Y) = 0. Since the squared norm is (X*X), then this amounts to disregarding vectors which lie along geodesic paths. Then I can deal separately for the case (X*X) > 0 and (X*X) < 0.

2) How am I supposed to deal with the condition that (BA^{-1}B^{-1})^{-1} = (BA^{T}B^{-1})^{T}? This condition seems quite restrictive.

not$P^{-1}gP$. – Deane Yang Jul 11 '10 at 19:38