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Let $\text{Riem}(g)$ be defined as the function which takes the metric and gives the $(0,4)$ Riemann tensor. We have the following implication: \begin{equation} R=\text{Riem}(g)\implies \begin{cases} R_{(ab)cd}=0 \\\\ R_{ab(cd)}=0 \\\\ R_{abcd}-R_{cdab}=0 \\\\ R_{a[bcd]}=0 \\\\ R_{ab[cd;e]}=0 \end{cases} \end{equation} where the final identity is to be understood as a relation involving $R$ and $g$.

My question is if the reverse implication is true: i.e. if given an $R$ and $g$ satisfying the 5 equations on the RHS, does that imply $R=\text{Riem}(g)$? If not, what identities do we have to append to the RHS to make the reverse implication true?

(I want to ask this question because it is similar in spirit to \begin{equation} \Gamma=\text{LC}(g)\Longleftrightarrow \begin{cases} \Gamma_{[ij]}^k=0 \\\\ \nabla g=0 \end{cases} \end{equation} where $\text{LC}(g)$ is the function that takes the metric and gives the Levi-Civita connection.)

dennis
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1 Answers1

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The answer is no, for the obvious reason that the zero tensor satisfies all the identities you want, but is rarely the curvature tensor of any metric on $M$. More generally, any multiple of the curvature tensor will satisfy these identities.

The first four identities are satisfied by many tensors, which are called curvature tensors on $M$. They have a nice description in terms of the irreducible representations of the orthogonal group (once a Riemannian metric $g$ is fixed). I don't know if curvature tensors satisfying the fifth identity (the differential Bianchi identity) have been studied in depth though.

Didier
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