contraction of the Riemann-Christoffel tensor I'm attempting to prove that a particular contraction of the Riemann-Christoffel tensor is zero.  I know that when the top and second of the bottom indices are contracted we get the Ricci tensor.  But when the top and first bottom indices are contracted it is apparently zero:
$R^i_{ijk} = 0$.
This contraction is
$\frac{\partial}{\partial x^j}\Gamma^i_{ik} - \frac{\partial}{\partial x^k}\Gamma^i_{ij} + \Gamma^p_{ik}\Gamma^i_{pj} - \Gamma^p_{ij}\Gamma^i_{pk}$
The final two terms clearly cancel simply by relabelling dummy indices.  We are left with
$\frac{1}{2} \left [\frac{\partial g^{il}}{\partial x^j}[ik,l] -  \frac{\partial g^{il}}{\partial x^k}[ij,l] +g^{il} \left (\frac{\partial}{\partial x^j}[ik,l] - \frac{\partial}{\partial x^k}[ij,l]\right)\right ]$
In the last term if we just expand the Christoffel symbols of the first kind out and carry the derivatives through then all terms pair up and cancel.  So, expanding the first two terms we are left with
$\frac{1}{2} \left [\frac{\partial g^{il}}{\partial x^j}\left ( \frac{\partial g_{il}}{\partial x_k} + \frac{\partial g_{kl}}{\partial x_i} - \frac{\partial g_{ik}}{\partial x_l}\right ) -  \frac{\partial g^{il}}{\partial x^k} \left ( \frac{\partial g_{il}}{\partial x_j} + \frac{\partial g_{jl}}{\partial x_i} - \frac{\partial g_{ij}}{\partial x_l} \right )\right ]$
Again, simply relabelling dummy indices makes the second and third term cancel and the fifth and sixth term cancel.  So we are left with just
$\frac{1}{2} \left [ \frac{\partial g^{il}}{\partial x^j}\frac{\partial g_{il}}{\partial x^k} - \frac{\partial g^{il}}{\partial x^k}\frac{\partial g_{il}}{\partial x^j} \right ]$
This is beautifully symmetric so it is hard to believe that it isn't zero.  But, frustratingly, I haven't been able to show that it is zero.  I'm sure there is some "obvious" reason but, not seeing it, I've been reduced to doing disgusting things like expanding the contravariant metric out in terms of the covariant metric and permutation symbols, which still hasn't worked.  Can someone point me at the "obvious" reason this is zero?

I think it's true. If $G$ is your metric tensor, then, using the formula for the derivative of the inverse, your expression is zero if $tr(G^{-1}G_{x_j}G^{-1}G_{x_k})=tr(G^{-1}G_{x_k}G^{-1}G_{x_j})$, where $G_{x_i}$ is the derivative of $G$ wrt $x_i$. Let $A=G^{-1}G_{x_k}$ and $B=G^{-1}G_{x_j}$. Since $G$ is symmetric, so are $A$ and $B$, and your expression is $tr(AB)-tr(BA)$. Therefore, for the property of the trace, you have your result.
