Riemannian curvature of maximally symmetric spaces

[center o bass]

A maximally symmetric is a Riemannian n-dimensional manifold for which there is n/2 (n+1) linearly independent (as solutions) killing vectors. It is well known that in such a space
$$R_{abcd} \propto (g_{ab}g_{cd} - g_{ac}g_{bd}) .$$

How is this formula derived for a general maximally symmetric space?

 

[George Jones]

The answers to this, and to your question on Killing vectors in your previous thread, are given in the book "A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry" by Peter Szekeres,

https://www.amazon.com/dp/0521829607/?tag=pfamazon01-20

 

[lpetrich]

Steven Weinberg's Gravitation and Cosmology has a nice derivation. Here's a summary.

The Killing vector ξ is defined as satisfying these two equivalent equations:
(covariant derivative of ξ) + (reversed indices) = 0
Lie derivative by ξ of the metric = 0

Take the covariant derivative of this equation, rearrange indices, add and subtract, and use the Riemann curvature tensor R's definition to get
Second covariant derivative of ξ = R.ξ (very schematic)

The Riemann tensor is defined from
Commutator of covariant deriative of X = R.X (very schematic)
Commutator = (second derivative) - (reversed indices)

From the second derivative of ξ, one can find it at any point using only its value and its first derivatives' value at that point. That must be the antisymmetric part of its first derivative, to satisfy its defining equation. For n dimensions, that's n values of ξ and n(n-1)/2 values of its antisymmetrized first derivative, giving n(n+1)/2 possible Killing vectors. In practice, there may be fewer than that, so n(n+1)/2 is the theoretical maximum.

With the second covariant derivative of ξ, take another covariant derivative, and subtract out the last two indices reversed. Go from covariant-derivative commutators to the Riemann tensor as appropriate. One ends up with an equation that is
Lie derivative by ξ of R = 0

At some point, take ξ = 0 but with nonzero first derivatives. One gets R.(first derivatives of ξ) = 0 (very schematic)

Going through all the possible first derivatives of ξ, one finds the OP's result: ## R_{ijkl} = K (g_{ik}g_{jl} - g_{il}g_{jk} ) ##.

Contracting to get the Ricci tensor, one finds ## R_{ik} = (n-1)K g_{ik} ##, and contracting further to get the Ricci scalar, one finds ## R = n(n-1)K ##.

Since the Lie derivative of the Ricci scalar is (gradient of Ricci scalar).ξ, and since ξ can be arbitrary at a point, one finds that the Ricci scalar is constant, and thus that K is constant. The Riemann and Ricci tensors, however, are not constant in the ordinary sense, but covariantly constant, because the covariant derivative of them is zero.

 

[lpetrich]

One can get a metric of a maximally-symmetric space in an interesting way. Using its value of the Riemann tensor, one can show that a maximally-symmetric metric is conformally flat. That is, its metric = (conformal function) * (flat-space metric).

One can use a manifestly flat metric like a constant metric for the flat-space metric, and then find the Riemann tensor from the complete metric. One can then solve for that metric's conformal function.

Once one has that metric, one can then solve for all its Killing vectors.

 

[blue_raver22]

The formula for the Riemannian curvature in a general maximally symmetric space is derived from the definition of a maximally symmetric space and the properties of killing vectors.

First, let's define a maximally symmetric space as a Riemannian n-dimensional manifold for which there are n/2 (n+1) linearly independent killing vectors. This means that there are n/2 (n+1) vectors that satisfy the equation:

$$\mathcal{L}_Xg_{ab}=0$$

where $X$ is a killing vector and $\mathcal{L}_X$ is the Lie derivative. This equation essentially means that the metric tensor $g_{ab}$ is invariant under the action of the killing vector $X$.

Now, the Riemannian curvature tensor $R_{abcd}$ is a measure of the non-commutativity of the covariant derivatives $\nabla_a$ and $\nabla_b$. In a maximally symmetric space, the covariant derivatives commute, meaning that:

$$\nabla_a \nabla_b = \nabla_b \nabla_a$$

This can be shown by using the fact that the metric tensor $g_{ab}$ is invariant under the action of the killing vectors. Since the covariant derivatives commute, we can apply the Bianchi identity:

$$R_{abcd} = \nabla_a \nabla_b g_{cd} - \nabla_b \nabla_a g_{cd}$$

Using the invariance of the metric tensor, we can simplify this to:

$$R_{abcd} = \nabla_a \nabla_b g_{cd} - \nabla_a \nabla_b g_{cd} = 0$$

This shows that in a maximally symmetric space, the Riemannian curvature tensor is identically zero, except for a proportionality factor.

Next, we can use the properties of killing vectors to further simplify the expression for the Riemannian curvature. Since the killing vectors commute with the covariant derivatives, we can write:

$$R_{abcd} = \nabla_a \nabla_b g_{cd} - \nabla_a \nabla_b g_{cd} = \nabla_a (X_bg_{cd}) - \nabla_b (X_ag_{cd})$$

Using the