# Diagonal Ricci tensors

1. Apr 29, 2010

### JustinLevy

I guess this question isn't actually specific to the Ricci tensor, but to rank 2 symmetric tensors.
If one is free to choose any local inertial coordinate system, what is the simpliest form we are garaunteed to be able to write any tensor components of a rank 2 symmetric tensor?

If we start from general components in 4 dimensional spacetime
$$R_{mn} = \begin{pmatrix} a &e &f &g\\ e &b &h &i\\ f &h &c &j\\ g &i &j &d \end{pmatrix}$$
in some local inertial coordinate system. This can general form can be specified with 10 free terms. We have three boosts and three spatial rotations that can be applied, and still be in a local inertial coordinate system. Which naively suggests we can eliminate 6 terms.

So can we always diagonalize R_{ab} by proper local inertial frame choice?
$$R_{MN} = \begin{pmatrix} A &0 &0 &0\\ 0 &B &0 &0\\ 0 &0 &C &0\\ 0 &0 &0 &D \end{pmatrix}$$
I'm not sure if the 6 transformations I mentioned above are "independent enough" to actually provide enough constraint to do this. Can someone comment on whether this is actually always possible?

In higher dimension N, there would be N(N+1)/2 unique terms in the symmetric tensor and we'd have N-1 boosts and N-1 spatial rotations. Which leaves:
N(N+1)/2 - 2(N-1) terms
which only equals N for N=4.

So there's something special about N=4 spacetime dimensions? We can't diagonalize it in higher dimensions?

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For a similar question. What about rank 2 anti-symmetric tensors? They seem to have only 6 free terms, so with 6 allowed transformations it naively suggests we could make it whatever we want, which is clearly wrong. Are some of the transformations no longer "independent" now or something?

2. Apr 30, 2010

### Phrak

I'd always assumed that, independent of dimension, a symmetric tensor could be diagonalized. However I haven't proved it. To decide if all symmetric tensors in any finite dimension can be diagonalized, I'd suggest solving for the eigenvalues of the matrix for arbitrary N. If it can be done, they are diagonalizable.

There are N(N-1)/2 rotations resulting from N dimensions taken two at a time. If, as you assume, each rotation is a coordinate dependent symmetry this will always leave N independent tensor elements in N dimensions. N=4 is not unique in this respect, if that's what you mean.

3. Apr 30, 2010

### JustinLevy

Doh! You are right. I didn't consider all the possible rotations. This simplifies things a lot.

Hmm... if we think along those lines, because it is symmetric and real valued, the matrix representing it in a coordinate system would be Hermitian and thus can be diagonalized (with real valued eigenvalues). So I think that fills in the gap. So yes, this is indeed always possible.

Thanks!

4. Apr 30, 2010

### George Jones

Staff Emeritus
No, using reasonable definitions of "spatial rotation" and "boost", there are $N - 1$ spatial rotations and $(N - 1)(N - 2)/2$ boosts. When $N = 4$, $(N - 1)(N - 2)/2 = N - 1$, so, in this special case, the number of boosts is $N - 1$.

The number of spatial rotations plus the number of boosts is

[tex]N - 1 + (N - 1)(N - 2)/2 = N(N - 1)/2,[/itex]

which is the number given by Phrak.

5. Apr 30, 2010

### JustinLevy

Yes, I mentioned I noticed that above. Thanks for adding clarification though.

So I think I understand the rank 2 symmetric tensor case now. How about a rank 2 anti-symmetric tensor?

This case seems much stranger because we have as operations available as there are free parameters to specify the components of a general rank 2 anti-symmetric tensor. But we clearly aren't completely free to set these components, as any of the unitary coordinate transformations will preserve the determinant. So what is going on here? Are some of the coordinates rotations somehow "effectively not independent" when dealing with anti-symmetric tensors?