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The Ricci tensor is defined as the contraction of the Riemann tensor in its upper and the second lower index. I was wondering why it is defined this way.

What happens if the Ricci tensor is defined as a different contraction of the Riemann tensor? Would it satisfy Einstein equations? Does the usual definition have any physical or geometrical meaning?

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    $\begingroup$ There are not many possible choices, because of the symmetry properties of $R_{abcd}$, i.e, $R_{abcd}=-R_{bacd}$, $R_{abcd}=-R_{abdc}$ and $R_{abcd}=-R_{cdab}$. If you try to contract the first and second indices, for example, you would get $g^{ab}R_{abcd}=0$, and likewise with the other combinations. The only non-zero one are $R^a_{bad}$ and $R^a_{bca}=-R^a_{bac}$. $\endgroup$ – AccidentalFourierTransform Feb 14 '16 at 15:20
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There are only six possible contractions, each of which can be simplified using the symmetries of the Riemann tensor:

$R^{\mu}_{\enspace\mu\lambda\sigma}=0$ because $R_{\kappa\nu\lambda\sigma}=-R_{\nu\kappa\lambda\sigma}$.

$R^{\mu}_{\enspace\nu\mu\sigma}=\text{Ric}_{\nu\sigma}$ is the usual definition.

$R^{\mu}_{\enspace\nu\lambda\mu}=-\text{Ric}_{\nu\lambda}$ because $R_{\kappa\nu\lambda\sigma}=-R_{\kappa\nu\sigma\lambda}$.

$R^{\enspace\mu}_{\kappa\enspace\mu\sigma}=-\text{Ric}_{\kappa\lambda}$ because $R_{\kappa\nu\lambda\sigma}=R_{\nu\kappa\lambda\sigma}$.

$R^{\enspace\mu}_{\kappa\enspace\lambda\mu}=\text{Ric}_{\kappa\lambda}$ because $R_{\kappa\nu\lambda\sigma}=R_{\nu\kappa\sigma\lambda}$.

$R^{\enspace\enspace\mu}_{\kappa\lambda\enspace\mu}=0$ because $R_{\kappa\nu\lambda\sigma}=R_{\kappa\nu\sigma\lambda}$.

So all the possible contractions result either in the Ricci tensor, its negation, or zero. Thus the Ricci tensor is unique: it (or its negation) is the only non-zero order 2 tensor you can make by contracting the Riemann tensor.

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The Ricci tensor's geometric interpretation is that it describes how open balls on a manifold behave i.e. how the radius of an open ball behaves. This is naturally a symmetric quantity since the manifold is constructed in a way in which you have a freedom of choice of what basis to choose. The symmetries of the Riemann tensor are $$R_{abcd} = R_{cdab} = - R_{cdba} = R_{cdba}.$$ SO it is obvious from the symmetric properties that you have to contract either the first and third or second and fourth indices with each other. And now you choose to define a symmetric contraction of the Riemann tensor as $$R_{bd} = R_{db} = R^a_{\ bad}. $$

It's a choice based on symmetry.

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The Ricci curvature tensor is a rank 2, symmetric tensor that arises naturally in pseudo-Riemannian geometry. Let (M,gij) be a smooth, n-dimensional pseudo-Riemannian manifold, and let Rijkl denote the corresponding Riemann curvature tensor. The Ricci tensor Rij is commonly defined as the following contraction of the full curvature tensor:

Rij=Rk.ikj

The index symmetry of Rij, so defined, follows from the symmetry properties of the Riemann curvature. To wit,

Rij=Rk=ikjRki=jkRk=jkiRji.

It is also convenient to regard the Ricci tensor as a symmetric bilinear form. To that end for vector-fields X,Ywe will write

Ric(X,Y)=XiYjRij.

source : http://www.studygtu.com/2016/02/what-is-ricci-tensone.html

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    $\begingroup$ Hi mitesh, we do have MathJax enabled on this site so that you can write nice math equations, rather than incomprehensible text lines. $\endgroup$ – Kyle Kanos Feb 15 '16 at 13:24

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