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The curvature tensor, $R_{ab}{}^c{}_d$, can be obtained from a connection which not necessarily is a metric connection.

By construction it is antisymmetric in the first two indices, since roughly speaking $[\nabla_a , \nabla_b] \simeq R_{ab}{}^\bullet{}_\bullet$. I'm assuming the connection has vanishing torsion.

Question

Under what conditions $R_{ab}{}^c{}_c$ vanishes for a general curvature?

For example:

  • If $\Gamma_{b}{}^c{}_c = 0$ the trace of $R$ vanishes.
  • If $\partial_a \Gamma_{b}{}^c{}_c = 0$ the trace of $R$ vanishes.

Is there something more general? and Could that be interpreted as a gauge fixing?

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  • $\begingroup$ I think you should look at mathoverflow.net/questions/69374/… $\endgroup$
    – Ben McKay
    Commented Jul 31, 2015 at 20:20
  • $\begingroup$ Thank you @BenMcKay, it is an interesting post. However, the trace for constructing the Ricci tensor is different than the one I mention. Additionally, the trace in the last two indices vanishes for Riemannian manifolds. $\endgroup$
    – Dox
    Commented Aug 1, 2015 at 2:15
  • $\begingroup$ Ben, this doesn't look like Ricci to me. The result is a skew-symmetric tensor and vanishes if the connection is the Levi-Civita connection. $\endgroup$
    – Deane Yang
    Commented Aug 1, 2015 at 2:19
  • $\begingroup$ Do you have any thoughts about what kind of condition you're looking for? $\endgroup$
    – Deane Yang
    Commented Aug 1, 2015 at 4:52

1 Answer 1

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Part of the confusion is that your positional notational convention is not the standard one. In most books, what you are writing as $R_{ab}{^c}_d$ would be written as $R{^c}_{dab}=-R{^c}_{dba}$ (though the letters used would probably be different).

That said, the condition $R{^c}_{cab}=0$ is the condition that the connection (locally) admit a parallel volume form. In other words, $R{^c}_{cab}$ represents a $2$-form that is the curvature of the induced connection on the line bundle $\Lambda^n(T^*M)$ (where $n$ is the dimension of the manifold). In particular, it is the exterior derivative of the $1$-form that you are writing as $\Gamma{_b}^{c}{_c}$. This $1$-form need not be zero (which would be $\Gamma{_b}^{c}{_c}=0$) nor need its coefficients be constant (which would be $\partial_a\Gamma{_b}^{c}{_c}=0$). It just needs to be closed in order to have your condition on the curvature.

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  • $\begingroup$ Thank you for the response, I'm getting the picture here! First, I apologize for the indices confusion (a personal academic deformation). It'd be great if you could confirm if I'm right in the following: (a) The fact that the connection admits a parallel volume form is equivalent to the equiaffine condition?, and (b) If a connection is closed, is locally pure gauge. What happens globally? Is it possible to define a cohomology of these connections? $\endgroup$
    – Dox
    Commented Aug 1, 2015 at 12:04
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    $\begingroup$ Yes, the condition is equivalent to 'locally equi-affine'. I don't know what you mean by a connection being 'closed', nor do I know what 'locally pure gauge' means. Instead of 'cohomology', you might be thinking of 'holonomy', which is the global thing that is left to measure when a connection is locally flat (i.e., has vanishing curvature). $\endgroup$
    – Robert Bryant
    Commented Aug 1, 2015 at 12:49
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    $\begingroup$ Maybe it means that in some local coordinates the connection one-form is closed?! (could be component-wise?!) and pure gauge is basically that this one form is exact (it is the differential of a local section of the Endomorphism bundle of the tangent bundle?!)... I guess it's kind of like line-bundles in electrodynamics (the connection one-form there is the electromagnetic potential). Hence the "cohomology" reference... but I am guessing... $\endgroup$
    – Futurologist
    Commented Aug 3, 2015 at 0:48
  • $\begingroup$ @Futurologist That is the spirit of my comment! $\endgroup$
    – Dox
    Commented Aug 3, 2015 at 10:21

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