Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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
2  
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
3  
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
2  
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

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
    
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 $O(1,3)$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.