Let X ad Y be two vectors in R4, and define the inner product of X and Y as:


(X\*Y) = gikXiYk (summation convention for repeated indicies)


Then we consider the 4x4 matrix g whose components are gik. 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-1ηB, 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) = XTgY


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


(AX\*AY) = XTATgAY = XTgY


Hence, ATgA = g


This formula can be written as gA-1g = AT, since g = g-1 explicitly.


Now applying the transformation matrix B:


gA-1g = B-1ηBA-1B-1ηB = AT, which I rearrange as:


ηBA-1B-1η = BATB-1


If by some chance (BA-1B-1)-1 = (BATB-1)T, then I can immediately conclude that A is in O(g) so long as BA-1B-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 BeXB-1 = eBXB-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-1B-1)-1 = (BATB-1)T? This condition seems quite restrictive.



The lie algebra of the orthogonal group of an arbitrary space time metric
You don't need to diagonalize. You are looking at the group of those $A$ such that $A g A^T = g$. Putting $A=1+ \epsilon B$ for some small $\epsilon$, you want $(1+\epsilon B)g(1+\epsilon B^T) = g$ or $\epsilon(Bg+g B^T) = O(\epsilon^2)$. So the Lie algebra you want is $\{ B : Bg+gB^T=0 \}$. Since $g=g^T$, this can also be stated as the set of $B$ such that $Bg$ is skew-symmetric.


