I guess this question isn't actually specific to the Ricci tensor, but to rank 2 symmetric tensors.(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]R_{mn} =

\begin{pmatrix}

a &e &f &g\\

e &b &h &i\\

f &h &c &j\\

g &i &j &d

\end{pmatrix}[/tex]

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?

[tex]R_{MN} =

\begin{pmatrix}

A &0 &0 &0\\

0 &B &0 &0\\

0 &0 &C &0\\

0 &0 &0 &D

\end{pmatrix}[/tex]

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?

--------

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?

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# Diagonal Ricci tensors

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