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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?

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  • $\begingroup$ I am permanently confused by the tensor notation, but maybe it possible to use antisymmetry in those coordinates? Does $R_{ijk}^s = - R_{sjk}^i$? If so, that would show it ... right? $\endgroup$ – Lorenzo Aug 18 '15 at 0:59
  • $\begingroup$ Yes, $R^s_{ijk} = - R^i_{sjk}$, but proving that is completely equivalent to proving what I'm trying to show. So this is part of why I am confident that the thing I'm trying to show must be true. But that doesn't help me prove it. $\endgroup$ – gleedadswell Aug 18 '15 at 1:18
  • $\begingroup$ In other words, I'd also like to prove that $R^s_{ijk} = -R^i_{sjk}$, and haven't been able to prove that either. Proving either one immediately makes the other proof easy. But I've been unable to do either proof. $\endgroup$ – gleedadswell Aug 18 '15 at 1:21
  • $\begingroup$ Oops! And actually, that's not true. It is true that $R_{pijk} = - R_{ipjk}$ where $R_{pijk} = g_{pr}R^r_{ijk}$. That's not quite the same as $R^s_{ijk} = - R^i_{sjk}$. $\endgroup$ – gleedadswell Aug 18 '15 at 1:24
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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.

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    $\begingroup$ Let me see if I understand your argument. The derivative of the inverse you mean is $(d/dt)A^{-1} = -A^{-1}(dA/dt) A^{-1} $, correct? So now I see how (using commas for partial w.r.t. x_j or x_k) $g^{il}_{,j} g_{il,k} = Tr[G^{-1}(dG/dx_j)G^{-1}(dG/dx_k)]$ and now clearly switching the derivatives w.r.t. $x_k$ and $x_j$ between $g^{il}$ and $g_{il}$ makes no difference since these are symmetric matrices. Thanks! Does anyone else see another answer more in terms of standard identities on the metric? $\endgroup$ – gleedadswell Aug 18 '15 at 1:58
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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?

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  • $\begingroup$ The argument I presented above also leads quickly to the identity $R_{pijk} = \partial_j \partial_i g_{pk} + \partial_k \partial_p g_{ij} - \partial_j \partial_p g_{ik} - \partial_k \partial_i g_{pj} +g_{sr}(\Gamma^s_{ij} \Gamma^r_{pk} - \Gamma^s_{pj} \Gamma^r_{ik})$ which is a well known identity. So I'm becoming confident that the argument may be valid. $\endgroup$ – gleedadswell Aug 18 '15 at 2:57
  • $\begingroup$ The normal coordinate system is defined in a neighborhood of a point where the torsion is zero (symmetric connection) If, in addition, the connection is the Levi-Civita one (zero torsion everywhere) you can make it so that, at that point, the partial derivatives of the metric vanish. But these properties do not hold in the whole manifold with the same chart... $\endgroup$ – bartgol Aug 18 '15 at 13:13

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