general relativity - What is the geometrical interpretation of Ricci tensor?

In differential geometry and general relativity space is said to be flat if the Riemann tensor $R=0$. If the Ricci tensor on manifold $M$ is zero, it doesn't mean that the manifold itself is flat. So what's the geometrical meaning of Ricci tensor since it's been defined with the Riemann tensor as

$$\mathrm{Ric}_{ij}=\sum_a R^a_{iaj}?$$

Answer

The local geometric structure of a pseudo-Riemannian manifiold $M$ is completely described by the Riemann tensor $R_{abcd}$. The local structure of a manifold is affected by two possible sources

1. 
   

   Matter sources in $M$: The matter distribution on a manifold is described by the stress tensor $T_{ab}$. By Einstein's equations, this can be related to the Ricci tensor (which is the trace of the Riemann tensor = $R_{ab} = R^c{}_{acb}$.

   $$R_{ab} = 8 \pi G \left( T_{ab} + \frac{g_{ab} T}{2-d} \right)$$
   
   


2. 
   

   Gravitational waves on $M$. This is described by the Weyl tensor $C_{abcd}$ which is the trace-free part of the Riemann tensor.

   
   
   

Thus, the local structure of $M$ is completely described by two tensors

1. 
   

   $R_{ab}$: This is related to the matter distribution. If one includes a cosmological constant, this tensor comprises the information of both matter and curvature due to the cosmological constant.

   
   


2. 
   

   $C_{abcd}$: This describes gravitational waves in $M$. A study of Weyl tensor is required when describing quantum gravity theories.