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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*}

Now suppose that we start with some covariant $4$-tensor field $R$ that satisfies these equations, would we always be able to find a metric globally such that its Riemannian curvature tensor is $R$?

If this were possible, would there be a formula to obtain the metric from the curvature tensor?

If this were not possible, what would be an example of a ‘curvature’ tensor that does not have an associated metric?

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.

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.

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*}

Now suppose that we start with some covariant $4$-tensor field $R$ that satisfies these equations, would we always be able to find a metric globally such that its Riemannian curvature tensor is $R$?

If this were possible, would there be a formula to obtain the metric from the curvature tensor?

If this were not possible, what would be an example of a ‘curvature’ tensor that does not have an associated metric?

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

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.