I've been asked to show that the Ricci tensor

$$ \mathrm{Ric}(X,Y) = \mathrm{tr}(Z \mapsto \mathrm R(Z,X)Y),$$

where $\mathrm R$ is the curvature endomorphism, does not completely determine the curvature tensor of an arbitrary Riemannian manifold.

In Lee's Riemannian Manifolds, Lemma 8.9, he shows that the sectional curvatures determine the curvature tensor $\mathrm{Rm}$ by assuming that two $4$-tensors $\mathrm R_1$ and $\mathrm R_2$ satisfying the Bianchi identity and the symmetries of the curvature tensor give rise to the same sectional curvatures, and then prove they are the same. I thought I might do something similar here by providing a pair of tensors satisfying the properties of curvature tensors, such that they induce the same Ricci tensors but different sectional curvatures. I do know though that sectional curvatures determine the Ricci tensor as well, so I'm not sure to which extent this approach can work.

Also in Lee's book we can see, in Lemmas 8.7 and 8.10 respectively, that the Ricci tensor does determine the curvature for $2$-dimensional manifolds and for manifolds of constant sectional curvature. I've seen a number of claims that this is also true when the manifold has dimension $3$, so whatever counter example one can find has to be in dimension at least $4$. In dimension $4$ I have been unsuccessful so far in constructing two tensors as described above.

It appears to be common knowledge that only the Ricci tensor is not enough to determine the curvature, but what is the main argument to prove that? Is my approach a good one or is there a simpler counter example going around?


The quick way is to count the number of independent components.

The Ricci curvature tensor is a symmetric $(0,2)$-tensor, so it has $\frac12n(n+1)$ independent components.

On the other hand, there are $\binom{n}{2}+3\binom{n}{3}+\frac{4!}{8}\frac23\binom{n}{4}=\frac1{12}n^2(n^2-1)$ (this uses symmetry/antisymmetry/Bianchi identities to count) independent components of $Rm$.

So for $n\geq 4$, there are insufficient information encoded in the Ricci curvature tensor to determine Riemann curvature tensor.

  • $\begingroup$ What do you mean by independent components? Nonzero coefficients in a specific basis? $\endgroup$ – Douglas Finamore Jun 23 at 18:35
  • 1
    $\begingroup$ At any point $p\in M$, without a specified Riemannian metric to start with, the all possible tensor $Rm(p)$ lives in a $\frac1{12}n^2(n^2-1)$-dimensional vector subspace of $T_p^{(0,4)}M$, and each is possible (i.e., there is a metric $g$ that gives a particular candidate $Rm(p)$). Similarly Ricci. $\endgroup$ – user10354138 Jun 23 at 18:43
  • $\begingroup$ I understand your argument, but I'm not sure how to calculate these dimensions or why all of these tensors occur as curvature tensors for some metric. Do you mind detailing a bit more or providing some reference for further reading? $\endgroup$ – Douglas Finamore Jun 23 at 19:06
  • $\begingroup$ Pick a coordinate patch about $p$, let $g_{ij}=\delta_{ij}+\text{quadratic }Q_{ij}(x)+o(x^2)$ and start differentiating. $\endgroup$ – user10354138 Jun 24 at 8:57

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