5
$\begingroup$

Is there a concrete example of a $4$ tensor $R_{ijkl}$ with the same symmetries as the Riemannian curvature tensor, i.e. \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*} for which there is no metric for which it is the Riemannian curvature tensor?

The existence of such a curvature was already shown by Robert Bryant, however, I'm looking for a concrete example.

Improve this question
$\endgroup$
8
  • 4
    $\begingroup$ Do you want a local example or is a global example enough? A trivial global example would be the zero tensor on a sphere $S^n$ for $n\geq 2$ (or any other manifold that can't admit a flat metric). $\endgroup$
    – RBega2
    Jan 11, 2023 at 12:00
  • $\begingroup$ Choose a tensor field that does not meet the differential Bianchi identity. $\endgroup$
    – Anton Petrunin
    Jan 11, 2023 at 15:06
  • 1
    $\begingroup$ @AntonPetrunin, is that enough? I know it has to hold at a point if the coordinates are such that the Christoffel symbols vanish. But otherwise it does not have to hold. But I guess you can say that there do not exist any coordinates for which the Boanchi identity holds at the given point. $\endgroup$
    – Deane Yang
    Jan 11, 2023 at 16:53
  • 1
    $\begingroup$ @DeaneYang Do you know what happens when you take the Berger spheres and shrink the fiber direction to zero? In the limit you get a torsion free connection and Riemann curvature like tensor, but naively it is not associated to a Riemannian metric anymore (as the natural candidate is not positive definite any longer). Unfortunately, I don't have a good explanation for why there isn't some other more exotic metric that does have this tensor field as its curvature. $\endgroup$
    – RBega2
    Jan 11, 2023 at 20:45
  • 3
    $\begingroup$ @RBega2 The curvature tensor will be the same as in $\mathbb{S}^2_{1/2}\times\mathbb{R}$ $\endgroup$
    – Anton Petrunin
    Jan 11, 2023 at 21:29

2 Answers 2

Reset to default
6
$\begingroup$

A simple example (which just uses Deane Yang/Robert Bryant's idea) is to consider any space of dimension at least three and consider the tensor field $$ R_{ijkl} = f(x)(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$$ where $f(x)$ is your favorite function which changes sign and whose derivative is non-vanishing when $f=0$ (thanks to Robert Bryant for the correction). When $f=0$, this curvature tensor has constant sectional curvature $0$, no matter what metric we choose.

However, we can now apply the proof of Schur's lemma (i.e., trace the second Bianchi identity twice) to see that when $f$ vanishes, we have that $$dR=\frac{n}{2}dR$$ where $R$ is the scalar curvature. As such, the differential of the scalar curvature must vanish when $f=0$. However, no matter which metric we pick, there is no way to make this happen if the differential of $f$ is nonzero as it goes from being positive to negative.

Edit: My original answer had a mistake which was pointed out in the comments. Here is a revised version which uses the same idea which should (hopefully) work.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nice and simple. $\endgroup$
    – Deane Yang
    Jan 12, 2023 at 2:38
  • 5
    $\begingroup$ Actually, this doesn't quite work. There are non constant functions $h(x)$ such that $g = h(x)\,\bigl (\mathrm{d}x^1)^2 + \cdots + (\mathrm{d}x^n)^2\bigr)$ has constant sectional curvature. For example, $h(x) = 4/(1+|x|^2)^2$. Won't the Riemann curvature tensor of $g$ have the above form for a nonconstant $f(x)$? $\endgroup$
    – Robert Bryant
    Jan 12, 2023 at 6:37
  • 4
    $\begingroup$ Moreover, in dimension $3$, if $f(x)$ is nonvanishing, then $R_{ijkl}=f(x)(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$ is locally the Riemann curvature tensor of some metric $g$. I proved this when $f$ is real-analytic (and nonvanishing) and later DeTurck and Yang proved this when $f$ is smooth (and nonvanishing). $\endgroup$
    – Robert Bryant
    Jan 12, 2023 at 7:31
  • 1
    $\begingroup$ Ah. Thanks for the correction. I guess to fix this we can add the assumption that f changes sign, right? That will definitely contradict Schur’s lemma then. $\endgroup$
    – Gabe K
    Jan 12, 2023 at 9:44
6
$\begingroup$

@AntonPetrunin's comment points to, I think, another way to describe the counterexample given by Robert Bryant in his answer.

Consider a curvature-like tensor $$ R_{ijkl}(dy^i\wedge dy^j)(dy^k\wedge dy^l) $$ where $(y^1, \dots, y^n)$ are coordinates in a neighborhood of $0$. If $R$ is the curvature tensor of a Riemanian metric, then there exists a change of coordinates $y=\phi(x)$, such that $\phi(0) = 0$, $\partial_i\phi^j(0) = \delta_i^j$, and, with respect to the coordinates $x= (x^1, \dots, x^n)$, the Christoffel symbols vanish at $0$. It follows by the second Bianchi identity that at the point $0$, $$ \partial_mR_{ijkl}-\partial_lR_{ijkm} = 0. $$

Now consider a curvature-like tensor $R$ in a neighborhood of $0$ such that $R(0) = 0$ but for some choice of $i,j,k,l,m$, $$ \partial_mR_{ijkl} - \partial_lR_{ijkm} \ne 0. $$ You can now verify that this inequality will still hold at $0$ for any change of coordinates. Therefore, this tensor cannot be the curvature tensor of a Riemannian metric.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Your example is basically what I gave in my answer that the OP linked. It may be that the OP will not consider your class of examples sufficiently 'explicit' either. $\endgroup$
    – Robert Bryant
    Jan 11, 2023 at 22:42
  • $\begingroup$ @RobertBryant, sorry about that. I'm editing the answer to attribute this to you properly. $\endgroup$
    – Deane Yang
    Jan 11, 2023 at 23:49
  • 4
    $\begingroup$ Oh, I wasn't worried about 'attribution'. I was just pointing out that it's not as 'concrete' as the OP might want. What I was thinking was that the OP might just want a 'concrete' answer, such as $R = x^3\,(\mathrm{d}x^1\wedge\mathrm{d}x^2)\otimes (\mathrm{d}x^1\wedge\mathrm{d}x^2)$ on $\mathbb{R}^3$. $\endgroup$
    – Robert Bryant
    Jan 12, 2023 at 6:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.