Let $M$ be a smooth manifold with non-empty boundary.

Let $g$ be a smooth Riemannian metric on $M$. Is the following true?

For every $\epsilon >0$ there exist a Riemannian metric $g_{\epsilon}$ with non-positive scalar curvature such that $\|g-g_{\epsilon}\|_{C^0}<\epsilon$?

(Here $\|\cdot\|_{C^0}$ is the $C^0$ norm w.r.t $g$, but I guess one can use $g_{\epsilon}$ as well).

I ask whether the metrics of non-positive scalar curvature are dense in the space of metrics.

On other hand, I heard that the space of metrics with non-negative scalar curvature is closed in the uniform topology.

Where can I find references for these two facts?

(I am not sure if the boundary creates problems. I might have heard this about closed manifolds).

If it matters, I am fine with assuming that $M$ is contractible.

  • $\begingroup$ regarding "closed manifolds" does not it contradicts to Gauss Bonnete theorem? $\endgroup$ – Ali Taghavi Sep 3 '19 at 13:07
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    $\begingroup$ Yes, I think you are right. (I also wondered about that). $\endgroup$ – Asaf Shachar Sep 3 '19 at 13:10
  • $\begingroup$ Is there a kind of Gauss Bonnete theorem for manifolds with boundary? $\endgroup$ – Ali Taghavi Sep 3 '19 at 13:10
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    $\begingroup$ I think there is a chance that what I heard applies only to dimension $\ge 3$ (or even just to $3$-manifolds, I am not sure.). I think that the higher dimensional versions of the Gauss-Bonnet are less restrictive that the $2$-dimensional version. $\endgroup$ – Asaf Shachar Sep 3 '19 at 13:26
  • $\begingroup$ yes I see. BTW as another approximation question(with a geometric nature but not related to your question) I was interested in this question: Is a generic non vanishing vector field $X$ a geodesible vector field?That is we have a Riemannian metric s.t the orbits of X are unparametrized geodesics. $\endgroup$ – Ali Taghavi Sep 3 '19 at 13:51

Jochen Lohkamp answers your first question in "Curvature h-principles", Ann. of Math, vol 142, p 457–498. If $\dim M\ge 3$, then the space of negative scalar curvature metrics is dense in the $C^0$ topology. More remarkably, he even shows this for negative Ricci curvature metrics.


About the $C^0$-closure of metrics with nonnegative scalar curvature, the results actually holds without any completeness assumptions and thus can be applied locally, so the fact that your manifold has boundary doesn't make any difference here. There are now two proofs available :

The theorem was first proved by Gromov in Dirac and Plateau Billiards in Domains with Corners, it proceeds by showing that non-negative scalar curvature is equivalent to the non-existence of cubical domains with :

  • $(1)$ mean convex faces.
  • $(2)$ angles between adjacent faces less than $\pi/2$.

This is accomplished by gluing several (regflected) copies of such a cube and smoothing the resulting metric to get a metric with positive scalar curvature on the $n$-torus, which do not exist thanks to previous work (Gromov-Lawson, or alternatively Schoen-Yau in dimension $n\leq 7$).

To prove $C^0$ stability from this, one consider a sequence $g_i\xrightarrow{C^0} g$, and assume that $g$ has negative scalar curvature at a point $x$. Around such an $x$ one can build a small cubical domain $C$ with the two properties above, the hard work is then to show that for $i$ big enough, one can slightly deform $C$ to get a cubical domain $C_i$ which has the same properties $(1)$ and $(2)$ with respect to $g_i$, which is hard because the positive mean curvature condition at first glance looks like a $C^2$ condition on the metric.

The second proof is (for me) much simpler and uses Ricci flow, it was given by Bamler in A Ricci flow proof of a result by Gromov on lower bounds for scalar curvature.


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