# The lie algebra of the orthogonal group of an arbitrary space time metric

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.

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If you are talking about relativity, then you mean that $g$ is a quadratic form of signature $(3,1)$. Then up to a change of coordinates, you can assume it is the usual Lonretz form. Its group is called $O(3,1)$ and any textbook on bilinear algebra shall contain everything you want. This is fairly classical and does not fit MO as far as I understand it. –  Benoît Kloeckner Jul 11 '10 at 18:59
Another point: a change of coordinates P acts on g by $P^t g P$, not by $P^{-1} g P$ since we are talking of bilinear forms, not of endomorphisms. –  Benoît Kloeckner Jul 11 '10 at 19:01
I am well aware of the lie algebra of the lorentz group, naturally. Is this the reference you are referring me to? I believe my group O(g) is quite distinct from O(3,1), since O(3,1) is a special case of O(g) when g is the flat space time metric. For example, the spin group and lorentz group have different lie algebras, and both are derived by substiting a specific g into the formulas in my post. –  Matt Jul 11 '10 at 19:03
To repeat in more detail what Benoît has already tried to explain: Given any non-degenerate symmetric matrix $g$, there exists an invertible matrix $P$ such that $P^tgP$ is diagonal and contains only $1$ or $-1$ along the diagonal. If there are $p$ $1$'s and $q$ $-1$'s, then we say that it has signature $(p,q)$. In other words, using a change of co-ordinates you can always assume that $g$ is diagonal with only ones or negative-ones along the diagonal. In particular, if $g$ is signature $(3,1)$, then the Lie group and algebra of $g$ is isomorphic to $O(3,1)$ and $o(3,1)$. –  Deane Yang Jul 11 '10 at 19:17
No one is claiming that in general $P^{-1} = P^t$. What Benoît is trying to explain is that if you change co-ordinates using a linear transformation $P$, then the metric in the new co-ordinates is given by $P^tgP$ and not $P^{-1}gP$. –  Deane Yang Jul 11 '10 at 19:38

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.

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I really like this solution, thanks! –  Matt Jul 14 '10 at 16:11

(1) There is no difficulty in diagonalizing the quadratic form $g$, regardless of its signature. However, you must be careful: either $g = B^{-1} \eta B$, where $B$ is an orthogonal matrix and $\eta$ is an arbitrary diagonal matrix of signature (3,1), or $g = B^T \eta B$, where $B$ may not be orthogonal, but $\eta$ is the signature-(3,1) identity matrix.

(2) I am somewhat puzzled by your assertion that $g = g^{-1}$. This will not be true for most metrics (although it might be true for the case you care about).

(3) In any case, here's how I think the calculation ought to go. From the formula $A^TgA=g$, we get $$gA^{-1} g^{-1} = A^T,$$ which, after substituting in $g = B^T \eta B$, becomes $$B^T \eta B A^{-1} B^{-1} \eta B^{-T} = A^T,$$ which can be rearranged as $$\eta B A^{-1} B^{-1} \eta = B^{-T} A^T B^T,$$ or $$\eta (B A B^{-1})^{-1} \eta = (BAB^{-1})^T.$$ As you can see, $A$ is an element of $O(g)$ if and only if $BAB^{-1}$ lies in $O(1,3)$

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I am a little concerned about transforming g to the identity matrix with signature (3,1). How can we be sure that the eigenvalue's of g are all 1 or -1? –  Matt Jul 11 '10 at 20:14
If you view $g$ as a symmetric matrix, its eigenvalues are not necessarily all $1$ or $-1$. See, however, the comments below your question. –  Deane Yang Jul 11 '10 at 20:28