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What's a simple example of metric on $R^{3}$ which has negative scalar curvature inside of a (limited) set N, and is equal to the standard (Euclidean) metric outside?

Basically, I am asking for a deformation of $R^{3}$ which decreases curvature.

Remarks: Such a thing wouldn't be possible in $R^{2}$ by the Gauss-Bonnet theorem.
In 3 or more dimensions it is possible by a theorem of Lohkamp stated here (Theorem 1.6).

The same problem, with positive instead of negative curvature, has no solution by the Positive Energy Theorem (Theorem 1.2 in the link above).

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  • $\begingroup$ What do you mean by the "standard Euclidean metric outside"? Do you mean that after removing your $N$ there is an isometry to the standard $\mathbb R^3$ minus a compact set? BTW, in dimension 2 any smooh nonpositive function on $\mathbb R^2$ can be realized as the scalar (or equaivalently sectional) curvature of a complete Riemannian metric. One simply solves the Jacobi equation $f_{xx}+Kf=0$ where $K$ is the sectional curvature, and then the metric is $dx^2+f^2dy^2$. See page 217 of [Kazdan-Warner, "Curvature Functions for Open 2-Manifolds", Annals of Math. 99, No. 2, (1974), pp. 203-219]. $\endgroup$
    – Igor Belegradek
    Commented Jun 15, 2013 at 1:48
  • $\begingroup$ @Igor By "standard Euclidean metric outside" I mean it IS the standard metric in each point outside N, not just up to isometry. Your construction is of little use here, because it won't generally produce standard flat space outside of a compact set, not even up to isometry, unless $\int f dA=0$, which implies $f$ has positive AND negative values (or is constantly zero). This can be seen by Gauss-Bonnet, as follows. Suppose the space to be standard $R^{2}$ outside N. Take a polygon enclosing N: its angular excess is 0, so $\int k dA=0$ inside the polygon, where $k$ is the curvature. $\endgroup$
    – delenda
    Commented Jun 15, 2013 at 3:01
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    $\begingroup$ I think the paper mathnet.or.kr/mathnet/thesis_file/BKMS-49-3-581-588.pdf in [Bull. Korean Math. Soc. 49 (2012), No. 3, pp. 581–588] does exactly what you want. $\endgroup$
    – Igor Belegradek
    Commented Jun 15, 2013 at 3:19
  • $\begingroup$ @Igor Yes! Section 2 does exactly what I asked for. Also, through the References, I found jstor.org.sci-hub.org/stable/2118620 (Section 3) with a different solution. Both are quite complicated, though! I wish there was a simpler example. I'll work my way through these papers, anyway. Thanks. $\endgroup$
    – delenda
    Commented Jun 15, 2013 at 4:18

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