- #1

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**Context: Deriving the maximally symmetric- isotropic and homogenous- spatial metric**

I've seen a fair few sources state that the Rienamm tensor associated with the metric should take the form:

* ##R_{abcd}=K(g_{ac}g_{bd}-g_{ad}g_{bc})##

The arguing being that a maximally symmetric space has constant curvature and so Riemann Tensor can only depend upon the

**metric a**nd the

**k**which specifies the curvature. This combined with respecting the anti-symmetries of the Riemann tensor leads to form *.

**My question:**

I assume it should be obvious, since most sources simply state it without any justifcation, but, I don't understand how the requirement of constant curvature forces the Riemann Tensor to only depend on the metric ?

**Here's what I do know:**

*The Riemann tensor arises from quantifying the extent to which the second covariant deriviatives fail to commute. It has a associate Ricci vector and scalar. And, the Ricci scalar is the curvature scalar of the manifold. I know that taking traces we can 'reduce' the Riemann tensor to get the Ricci scalar.*

Actually, I may partly have an idea why: to get the Ricci scalar from the Riemann tensor (in the form of 4 lower indices) we need to multiply by 2 metrics with raised indicies, so if there are only metrics on the RHS of *, we get traces of the metric with itself - so just 3 in space - and so we have got R=constant as desired.

But this idea doesn't seem complete to me - it doesnt say why we can't include other tensors - w

Also, I'm not looking for a mathematical proof, I have found a couple of sources for these, using killing vectors, I'm just looking for a logical explanation.

But this idea doesn't seem complete to me - it doesnt say why we can't include other tensors - w

**ould this be along the right lines anyway or not?**Also, I'm not looking for a mathematical proof, I have found a couple of sources for these, using killing vectors, I'm just looking for a logical explanation.

**Thanks very much for your help !**