5

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.

Matthew Lou
  • 113
  • 7
  • 4
    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). – RBega2 Jan 11 '23 at 12:00
  • Choose a tensor field that does not meet the differential Bianchi identity. – Anton Petrunin Jan 11 '23 at 15:06
  • 1
    @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. – Deane Yang Jan 11 '23 at 16:53
  • @DeaneYang yes, you are right --- we do not have connection. – Anton Petrunin Jan 11 '23 at 18:59
  • I don't know any systematic way to find an explicit example, but @AntonPetrunin – Deane Yang Jan 11 '23 at 20:05
  • 1
    @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. – RBega2 Jan 11 '23 at 20:45
  • @RBega2, nice observation and question. Offhand, I don't know. – Deane Yang Jan 11 '23 at 20:47
  • 3
    @RBega2 The curvature tensor will be the same as in $\mathbb{S}^2_{1/2}\times\mathbb{R}$ – Anton Petrunin Jan 11 '23 at 21:29

2 Answers2

6

@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.

Deane Yang
  • 26,941
  • 2
    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. – Robert Bryant Jan 11 '23 at 22:42
  • @RobertBryant, sorry about that. I'm editing the answer to attribute this to you properly. – Deane Yang Jan 11 '23 at 23:49
  • 4
    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 $\mathbb{R}^3$. – Robert Bryant Jan 12 '23 at 06:27
6

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.

Gabe K
  • 5,374
  • Nice and simple. – Deane Yang Jan 12 '23 at 02:38
  • 5
    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 $f(x)$? – Robert Bryant Jan 12 '23 at 06:37
  • 4
    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). – Robert Bryant Jan 12 '23 at 07:31
  • 1
    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. – Gabe K Jan 12 '23 at 09:44