# The Riemann-Einstein manifolds

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Let $(M,g)$ be a riemannian manifold with riemannian curvature $R$. Then the Riemann-Ricci curvature is:

$$RR(X,Y)=tr(R(X,e_i)R(Y,e_i))$$

A Riemann-Einstein metric is such that:

$$RR(X,Y)=\lambda g(X,Y)$$

with $\lambda$ a scalar.

Can we have spherical solutions of the Riemann-Einstein?

asked Jun 6, 2020

Doesn't the Schwarzschild-deSitter solution qualify? Or am I missing something?

## 1 Answer

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In a vacuum, the gravitational field equations take the form $$RR(X,Y) - \frac{1}{2}tr(RR(e,e))g(X,Y) + \Lambda g(X,Y) = 0$$
By taking the trace reverse they can be written as $$RR(X,Y)=\Lambda g(X,Y)$$
so any vacuum solution with a cosmological constant is a Riemann-Einstein metric. As user AzureSea states the deSitter-Schwartzchild solution is a spherically symmetric example.

answered Oct 16, 2020 by (45 points)

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