# Locally Maximally Symmetric Spacetimes

HomerSimpson
Can one say that every general curved spacetime, locally is maximally symmetric?
I know that one can say that every general curved spacetime is locally flat (and therefore maximally symmetric with R=0), but I'm talking about a very high curvature spacetime, where still we can consider nonzero curvatures.

Mentor
Can one say that every general curved spacetime, locally is maximally symmetric?

What do you mean by "maximally symmetric"? Most general curved spacetimes have no symmetries at all (i.e., no Killing vector fields).

I know that one can say that every general curved spacetime is locally flat (and therefore maximally symmetric with R=0)

No, you can't say that. What you can say is that every general curved spacetime is locally Lorentz, meaning that at a given event, you can choose a coordinate chart that makes ##g_{ab} = \eta_{ab}## at that event, and makes all of the connection coefficients (first derivatives of ##g_{ab}##) vanish at that event. But that doesn't make ##R = 0##.

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HomerSimpson
Thanks for you replay Peter, but I'm still a little bit confused.

What do you mean by "maximally symmetric"? Most general curved spacetimes have no symmetries at all (i.e., no Killing vector fields).

I know it, but I meant "locally" not globally.

No, you can't say that. What you can say is that every general curved spacetime is locally Lorentz, meaning that at a given event, you can choose a coordinate chart that makes ##g_{ab} = \eta_{ab}## at that event, and makes all of the connection coefficients (first derivatives of ##g_{ab}##) vanish at that event. But that doesn't make ##R = 0##.

What you mention here, looks like the Riemann normal coordinate, is it?

But that doesn't make ##R = 0##.

Can we assume that R is almost constant as in maximally symmetric spacetimes and therefore $R_{ab}\propto Rg_{ab}$?

What I get from your answer, is we can not say that a general high curved spacetime, is "locally" maximally symmetric. Am I right?

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