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.

 

[PeterDonis]

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 $R = 0$.

 

[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?

 

[Bill_K]

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

That's right. And you can strike the word "high". A general curved spacetime is not locally maximally symmetric.

Since the Riemann tensor has dimension L^-2, curvature only sets a length scale, and you can only say it's large or small in comparison to some other length.

 

[sardar]

I must clarify that the concept of "maximally symmetric" spacetimes is not a universally accepted term in the field of general relativity. However, I will address your question in the context of locally symmetric spacetimes.

In general, a spacetime is said to be locally symmetric if it possesses a continuous group of isometries, meaning that the metric remains invariant under certain transformations. This group of isometries can be used to classify spacetimes into different types, such as flat, anti-de Sitter, or de Sitter spacetimes.

Now, to answer your question, it is not accurate to say that every general curved spacetime is locally maximally symmetric. This is because the existence of non-zero curvature does not necessarily imply the presence of isometries. In fact, most general curved spacetimes do not possess a continuous group of isometries and therefore cannot be classified as locally symmetric.

However, it is true that every general curved spacetime is locally flat, meaning that in a small region of the spacetime, the curvature is negligible and the spacetime can be approximated by a flat spacetime. This is known as the equivalence principle and is a fundamental concept in general relativity.

In summary, while it is not accurate to say that every general curved spacetime is locally maximally symmetric, it is true that every general curved spacetime is locally flat. I hope this provides a clear understanding of the concept of locally symmetric spacetimes.