2
$\begingroup$

Let $(M,g)$ be a closed Riemannian manifold, if $Scal^g>0$, we know that in a metric space of $M$, there is a neighborhood of $g$, such that all metrics in this neighborhood have the positive scalar curvature. In other words, we can choose a generic metric with P.S.C.-property.

Q If we replace the PSC with Non-negative scalar curvature, can we get a similar property? Is there any/some research about such a question?

$\endgroup$
  • 3
    $\begingroup$ Similar, maybe. Exactly the same, no. Just look at the two dimensional case with the flat metric on $\mathbb{T}^2$ and apply Gauss-Bonnet. $\endgroup$ – Willie Wong Oct 12 '18 at 15:12
  • 3
    $\begingroup$ Expanding on Willie's comment: there is work of Schoen--Yau and Gromov--Lawson on obstructions to the existence of a metric with positive scalar curvature. One consequence is that if a metric on $\mathbb{T}^n$ has nonnegative scalar curvature, then that metric is flat. So, at least in certain cases, "nonnegative scalar curvature" is a rigid property of a metric. $\endgroup$ – Jeffrey Case Oct 12 '18 at 17:38
3
$\begingroup$

In general, there's probably no hope of getting anything too similar for non-negative curvature. Willie Wong gave an excellent example of how the Gauss-Bonnet restricts the scalar curvature (a deformation of the flat metric on the torus must have negative scalar curvature unless it is totally flat). In fact, a small deformation of a manifold with non-negative sectional curvature might not even have non-negative scalar curvature. To see this, take a round 2-sphere and flatten its north pole until it becomes flat at that pole. This can be done so that the induced metric $g$ on this flattened sphere has positive sectional curvature everywhere except for the north pole. If we then consider a small conformal deformation of the metric $\tilde g= e^{2f} g$, then $\tilde g$ has negative scalar curvature at the north pole whenever $\triangle\left( e^{(n-2)f/2} \right) <0$ at the north pole. There is nothing special about conformal deformations here, they just allow for easy computation of the scalar curvature.

It's possible to extend this example to higher dimensions, but the moral of the story is that we can't expect arbitrary deformations of metrics at "flat" points to remain curvature-definite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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