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?
 A: 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.
A: I think I see another argument, but I'm not sure it is valid, because everyone is using different terminology and I'm a bit confused by the profusion of terms.
I know "geodesic coordinates" as coordinates in which the Christoffel symbols of the second kind vanish at a point.  Their derivatives do not vanish here.  Various people talk about "normal coordinates" as a choice of coordinate system where the first derivatives of the metric vanish at a point, which also means that the Christoffel symbols vanish there.  Side question: are "geodesic coordinates" and "normal coordinates" two names for the same thing?
In that case, we can switch to normal coordinates and now at the point above where I had
$\frac{1}{2} \left [\frac{\partial g^{il}}{\partial x^j}[ik,l] -  \frac{\partial g^{il}}{\partial x^k}[ij,l] +g^{il} \left (\frac{\partial}{\partial x^j}[ik,l] - \frac{\partial}{\partial x^k}[ij,l]\right)\right ]$
at the special point in normal coordinates the first two terms vanish because the derivatives of the metric vanish.  The last two terms (as I've already said) have pairs of terms which cancel when we expand them out fully.  So, at the special point in our normal coordinates the identity is true.  But this is a proper tensor identity so it is true in all coordinate systems.  Further, our choice of where to make the derivatives of $g_{ij}$ vanish is arbitrary in our transformation to normal coordinates, so this must be true everywhere.
This feels hand-wavey, but my understanding is that arguments like this are precisely whey geodesic coordinates are useful.  Is my reasoning sound?
A: I've struggled with a very similar problem for a few weeks: #244 from Pavel Grinfeld's Introduction to Tensor Analysis...
$R^\alpha_{\cdot\alpha\gamma\delta} = 0$
The solution from a graduate student's solution manual online showed (note that he changed index variables) (pencil writing mine):

I couldn't figure out how to show that the seemingly self-evident difference of the Christoffel partial derivatives equals zero.  I tried many things, mostly writing the Christoffel symbols as dotted vectors. (Note: Grinfeld uses $S$ as the metric tensor variable for the surface space.)
For example,
$\Gamma^\alpha _{\delta\alpha} = \boldsymbol{S}^\alpha \cdot \frac{\partial \boldsymbol{S}_\delta}{\partial S^\alpha}$
then differentiating the corresponding terms for each Christoffel symbol using the product rule. Second order differentials canceled, but I kept ending up with equations like
$R^\alpha_{\cdot\alpha\gamma\delta} = \frac{\partial \boldsymbol{S}_\alpha}{\partial S^\gamma} \cdot \frac{\partial \boldsymbol{S}_\delta}{\partial S^\alpha} - \frac{\partial \boldsymbol{S}_\alpha}{\partial S^\delta} \cdot \frac{\partial \boldsymbol{S}_\gamma}{\partial S^\alpha}  \quad(*)$
This has nice symmetries, but it wasn't evident with my novice familiarity (or my primitive fiddling) that it equals zero.
Today, I went back in the book and tried extending a relationship which he had basically derived for the ambient space (and which I also saw in Schaum's Tensor Calculus) to the surface space:
$\Gamma^\alpha _{\alpha\beta} = \frac{\partial}{\partial S^\beta}(\frac{1}{2}ln{\vert S \vert})$
where $\vert S \vert$ is the determinant of the metric tensor.
Doing this led to
$R^\alpha_{\cdot\alpha\gamma\delta} = \frac{1}{2}[\frac{\partial^2}{\partial S^\gamma \partial S^\delta}(ln{\vert S \vert})-\frac{\partial^2}{\partial S^\delta \partial S^\gamma}(ln{\vert S \vert})]=0$
This doesn't show me the inner workings of why my equation $(*)$ equals zero, but it at least let me demonstrate the identity.
