4
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • $\begingroup$ I think you should look at mathoverflow.net/questions/69374/… $\endgroup$
    – Ben McKay
    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
    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
    Aug 1, 2015 at 2:19
  • $\begingroup$ Do you have any thoughts about what kind of condition you're looking for? $\endgroup$
    – Deane Yang
    Aug 1, 2015 at 4:52

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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
    Aug 1, 2015 at 12:04
  • 3
    $\begingroup$ 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
    Aug 1, 2015 at 12:49
  • 2
    $\begingroup$ 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
    Aug 3, 2015 at 0:48
  • $\begingroup$ @Futurologist That is the spirit of my comment! $\endgroup$
    – Dox
    Aug 3, 2015 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.