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Why the Ricci tensor is defined the contraction of first and third indices of Riemann tensor? I guess it is more natural to define it as to contract the first and the second indices? Since from Wiki : $$R_{\sigma \mu \nu}^\rho = dx^\rho(R(\partial_\mu,\partial_\nu)\partial_\sigma)$$ To do a contraction, I guess it should be done on $\rho $ and $\sigma$ so that
$$R_{\mu \nu} = R_{\sigma \mu \nu}^ \sigma$$

but why the definition is $$R_{\mu \nu} = R_{\mu \sigma \nu}^ \sigma$$

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The symmetries of the Riemann tensor imply that $R^\sigma{}_{\sigma \mu \nu} = 0$. In fact the Ricci curvature is the only non-zero contraction of the Riemann tensor.

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  • $\begingroup$ thanks. the (anti)symmetry of third and fourth indices are easily seen, but how can one see the symmetry of first and second indices since one index is contravariant and the other is covariant? $\endgroup$ – ahala May 27 '14 at 18:54
  • $\begingroup$ @ahala: you can see this antisymmetry by applying the metric-compatibility of the connection to $R_{ijkl} = g(\partial_i, [\nabla_k, \nabla_l] \partial_j)$ a few times - move the derivatives off $\partial_j$ and on to $\partial_i$. The contraction is then $R^i{}_{ikl} = g^{ij} R_{ijkl}$ which vanishes since $g$ is symmetric while $R$ is antisymmetric. $\endgroup$ – Anthony Carapetis May 28 '14 at 0:51
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I know you're familiar with geometric algebra, ahala. The Riemann tensor can be understood as a bivector-valued function of a bivector:

$$R(a \wedge b) = c \wedge d$$

for some $a, b, c, d$. Components of the Riemann tensor are evaluated through inner products of bivectors. For instance,

$$R_{\alpha \beta \gamma \delta} a^\alpha b^\beta c^\gamma d^\delta = R(a \wedge b) \cdot (d \wedge c)$$

A contraction over the first two indices is nonsensical. It would be an expression of the form

$$R( \frac{\partial}{\partial a} \wedge a) \cdot (d \wedge c)$$

where $\partial/\partial a$ should be understood as a vectorial derivative (akin to the typical $\nabla$ in vector calculus, except with respect to $a$ as its "position" variable). But $\frac{\partial}{\partial a} \wedge a = 0$, so this can only be zero.

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  • $\begingroup$ Thanks. What if the first index is contravariant? $e^i \wedge e_i = 0$? $\endgroup$ – ahala May 27 '14 at 4:14
  • $\begingroup$ That's why I used a derivative $\partial/\partial a$ there. The expression $\frac{\partial}{\partial a} \wedge a = \sum_i e^i \wedge e_i = 0$, in any basis. $\endgroup$ – Muphrid May 27 '14 at 4:18
  • $\begingroup$ Thanks Muphrid. How can I see $\sum e^i \wedge e_i = 0$ if ${e_i}$ is not necessarily orthogonal? Is every term zero or it has to sum over $i$? $\endgroup$ – ahala May 27 '14 at 4:44
  • $\begingroup$ I think you're gonna have to rely on the basis independence of the final answer on this one. I mean, this can still be interpreted as a curl, with $a$ being the "position" vector in a flat vector space. It seems natural to say that the position vector has no curl. This could probably be done in a basis independent fashion by writing that curl in terms of the limit of a surface integral as the enclosed volume goes to zero. $\endgroup$ – Muphrid May 27 '14 at 4:56

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