Geodesic Distance & Maximally Symmetric Spacetimes: Why Does it Matter?

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highflyyer
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Any physical quantity ##K(t,x,x')## on a maximally symmetric spacetime only depends on the geodesic distance between the points ##x## and ##x'##.

Why is this so?

N.B.:

This statement is different from the statement that

The geodesic distance on any spacetime is invariant under an arbitrary coordinate transformation of that spacetime.
 
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I haven't found a proof of it. I read this in a paper.

This is my understanding of the problem.

The Euclidean plane is a maximally symmetric space with ##3## translation symmetries and ##3## rotation symmetries. Any physical quantity ##K(x,y)## on the Euclidean plane, where ##x## and ##y## are two arbitrary spacetime points, is constrained by the symmetries of the spacetime to depend only on ##(x-y)^{2}##. This is because ##(x-y)## is translation invariant and ##(x-y)^{2}## is rotation invariant. Therefore, the physical quantity ##K(x,y)## depends on the Galilean-invariant geodesic distance ##(x-y)^{2}##.
 
But I am not sure how the dependence changes if we have a Euclidean disk (that is, a plane with a boundary).

My intuition is that the ##K(x,y)## now depends not only on the spacetime points ##x## and ##y##, but also on the 'border.' The dependence is such that ##K(x,y)## for the Euclidean disk tends to ##(x-y)^2## as the 'border' tends to infinity.

But I am not able to carry my intuition any further and write down an explicit form for the dependence of ##K(x,y)## for the Euclidean disk.

It would be really helpful if you share some thoughts here.
 
highflyyer said:
I am not sure how the dependence changes if we have a Euclidean disk (that is, a plane with a boundary).

A plane with a boundary is not maximally symmetric.