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Suppose that I want to check if a given metric is singular or not. I'm interested in curvature singularities, not coordinate singularities, so I can look to scalars made with Ricci, Riemann and Weyl Tensor.

If I found that one of this scalar is divergent somewhere, then I'm done. My problem is the opposite, suppose that I don't find singularities after checking some invariants. How can I be sure that the space is non singular? Rephrased: Is there a COMPLETE basis of scalar curvature invariants in general relativity? Let's say in $D=4$ for concreteness. The vacuum case in particular.

I heard somewhere that in vacuum and in $D=4$ is enough to restrict to: $R$, $R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} $, $R_{\mu \nu } R^{\mu \nu } $, ${{}^\star R}_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} $, $ {{}^\star R}^{\star}_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} $. Is this true?

References are welcome.

EDIT: To be more precise, referring only to curvature singularities (I know that there are other way to characterize a singularity like explicitly working with geodesics) is there a minimum number of invariants to check, in order to conclude that the metric is free of curvature divergencies?

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  • $\begingroup$ Indeed, XD. Anyway now it's even more clear. $\endgroup$ – Rexcirus Apr 12 '16 at 15:00
  • $\begingroup$ The paper here goes in to a lot of detail (much more than I can digest myself!), but I think it goes a long way to answering your question. arxiv.org/abs/gr-qc/0302095 $\endgroup$ – m4r35n357 Apr 12 '16 at 18:37
  • $\begingroup$ For just the Ricci tensor part, I think you would need $4$. You could think of these as the four eigenvalues of $R^\mu_\nu$, which can be expressed in terms of $R$, $R_{\mu\nu}R^{\mu\nu}$, $R^\mu_\alpha R^\alpha_\beta R^\beta_\mu$, and $R^\mu_\alpha R^\alpha_\beta R^\beta_\nu R^\nu_\mu$. For the Weyl tensor, my guess would be you need 5. This comes from viewing $C^{\alpha \beta}_{\mu\nu}$ as a $6\times 6$ tracefree matrix, so there are 6 eigenvalues subject to the constraint that they sum to zero, $6-1=5$. $\endgroup$ – asperanz Apr 12 '16 at 18:46
  • $\begingroup$ The other possibility is that you need 20 total invariants, since the Riemann tensor has 20 independent components in $D=4$. $\endgroup$ – asperanz Apr 12 '16 at 18:47
  • $\begingroup$ Also note that $^*R_{\mu\nu\rho\sigma}^*$ is not independent of $R_{\mu\nu\rho\sigma}$ in $D=4$. $\endgroup$ – asperanz Apr 12 '16 at 18:48
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To be more precise, referring only to curvature singularities (I know that there are other way to characterize a singularity like explicitly working with geodesics) is there a minimum number of invariants to check, in order to conclude that the metric is free of curvature divergencies?

No. There is a broad class of spacetimes, called vanishing scalar invariant (VSI) spacetimes, for which every scalar invariant vanishes. In particular, a gravitational plane wave is VSI. (This follows from the fact that the wave can experience Doppler shifts, but a scalar has to stay the same under a boost.) You can have gravitational waves that are singular in the sense of geodesic incompleteness (such as Penrose-Hawking "thunderbolts"), so no investigation of any number of curvature scalars will suffice to prove that there is no singularity.

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DISCLAIMER: I have never seen the below conlcusion made explicitly, but it seems to me to be a straight-forward consequence of established theory. I cannot find any error in my thinking, whence I will post it and await your judgement.

The equivalence theorem, proved e.g. in Cartan's Leçons sur la géométrie des espaces de Riemann, states that in a rigid frame a finite number of the covariant derivatives of the Riemann tensor is sufficient to completely classify the local geometry of a semi-Riemannian manifold. This theorem is the basis of the Cartan-Karlhede Algorithm, and a more formal statement can be found in English in Karlhede's paper here (although note that he references his own original USIP report, which I can't seem to find anywhere online; maybe I just do not know where to look). The maximum number of covariant derivatives needed in four dimensions is seven, but many solutions require fewer, and the components of any higher covariant derivatives are functionally dependent on previous components.

Thus consider a rigid frame. If the squares of all the required covariant derivatives are finite, e.g. $$ |R_{ijk\ell;m_1\ldots m_{r}}R^{ijk\ell;m_1\ldots m_{r}}| < \infty, $$ then the covariant derivatives are finite, so by functional dependence all covariant derivatives are finite. Additionally, in a rigid frame, all contractions of the Riemann tensor are finite if all components are, as are all duals. Therefore it seems to me that a finite number of covariant derivatives of the Riemann tensor is sufficient in a rigid frame.

Now, in a given (well-defined) rigid frame, i.e. such that the frame vectors $e_i$ are finite and smooth, the covariant derivatives are finite as long as the components $R_{ijk\ell}$ and the Ricci rotation coefficients $\gamma^i{}_{jk}$ are also all finite and smooth ($C^\infty$). This matches well with established theory in fixed frame, see e.g. this paper.

In conclusion: under restriction to a well-defined rigid frame it is sufficient to show that the Riemann tensor components and tbe Ricci rotation coefficients are all finite and smooth.

In a non-rigid frame I do not think a similar statement can be made since there are cases (e.g. gravitational wave solutions) where an infinite number of covariant derivatives are needed to give a complete local description of the geometry.

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