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So we know that given a Riemannian manifold $(M,g)$ we can calculate its Riemannian curvature $R$. This covariant $4$-tensor field then satisfies some important symmetries \begin{gather*} R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\ R_{ijkl} + R_{iklj} + R_{iljk} = 0. \end{gather*}

The main question of the global existence of a metric for a prescribed curvature was already explained by Robert Bryant (thank you to @Deane Yang for the summary). However, this answer only describes why such a curvature exists. Hence, I'd like to know whether there is a concrete example of a curvature that is not generated by a metric.

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    $\begingroup$ This was answered already. The short answer is you can do it at a point but not globally in general once the dimension is 3 or more. $\endgroup$
    – RBega2
    Commented Jan 9, 2023 at 22:37
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    $\begingroup$ I suggest you post a new question specifically about a concrete example. I won't be of much help with this, but maybe someone else can be. $\endgroup$
    – Deane Yang
    Commented Jan 10, 2023 at 21:28

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Here's a quick summary. The answers provided in the link cited by @RBega2 have more details.

  1. Given a curvature-like tensor $R$ at a point, there always exists a metric whose curvature tensor at that point is equal to $R$.
  2. In general, this is system of nonlinear second order PDEs, where the unknown functions are the components of the metric tensor and there is an equation for each component of the curvature tensor.
  3. In dimensions 4 and higher, there are more equations than unknown functions and therefore there are no solutions unless further conditions are imposed on the curvature-like tensor. These conditions are poorly understood.
  4. In dimension 3, the number of equations is equal to the number of unknown functions. Again, the necessary and sufficient conditions are not known. However, as Robert Bryant explains in his answer, he proved that if the curvature-like tensor satisfies a nondegeneracy condition and is real analytic, then there is a local real analytic solution. Later, DeTurck and I extended this to the smooth category.
  5. As far as I know, nothing is known about the global problem. This system of PDEs is quite nasty. A fundamentally new advance in how to solve system of PDEs like this is needed.
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  • $\begingroup$ Could the special case of dimension 4 be solved requiring the tangent space at any point be isomorphic to the quaternions skew field and the conditions stated in math.stackexchange.com/questions/821881/… be fulfilled? $\endgroup$
    – Sylvain JULIEN
    Commented Jan 10, 2023 at 9:21
  • $\begingroup$ Thank you very much for the answer! The one question I'm left with is whether there is some example of a curvature that is not generated by a metric. The linked answer only gives an argument for the existence (it does not construct one as far as I understand). Is this due to the fact that the conditions are poorly understood for when a solution would exist? $\endgroup$
    – Matthew Lou
    Commented Jan 10, 2023 at 10:01
  • $\begingroup$ I edited the question to be more about a concrete example rather than a general proof, I hope that is okay with you. If not, please let me know. $\endgroup$
    – Matthew Lou
    Commented Jan 10, 2023 at 13:50
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    $\begingroup$ In terms of a concrete example where there is no solution, you can do this using only Taylor polynomials of the curvature-like tensor and the metric at a single point. The analysis described by Robert yields explicit equations and/or inequalities that the curvature-like tensor has satisfy in order for it to be a curvature tensor. In principle, it is then easy to find a counterexample that does not satisfy these necessary conditions. In practice, this can be a bit painful to do $\endgroup$
    – Deane Yang
    Commented Jan 10, 2023 at 19:28

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