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Let $(M^n,g)$ be an asymptotically flat manifold of decaying-order $\tau>\frac{n-2}{2}$, the positive mass theorem states that if the scalar curvature $S_g$ is non-negative, then the ADM mass $m_g$ must be non-negative too. Moreover if $m_g = 0$ then $(M^n,g)$ is isometric to the Euclidean space $(\mathbb{R}^n,\delta)$.

I wonder if we can find a pair $(M^n,g)$ such that the scalar curvature is negative somewhere and yet $m_g\geq 0$. Or even if we can find a manifold with striclty negative scalar curvature everywhere and non negative mass.

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  • $\begingroup$ Probably there's something I'm not understanding here, but isn't the first conjecture trivially true? If we have one asymptotically flat manifold that has negative scalar curvature somewhere and ADM mass $m_1<0$, then we can construct some other asymptotically flat manifold that has positive ADM mass $m_2>|m_1|$, e.g., a Schwarzschild spacetime with a large mass. Then these two spacetimes can surely be "joined" side by side in some sense to make one spacetime, since both are Minkowski at large distances. $\endgroup$
    – user21349
    Nov 14, 2018 at 17:05
  • $\begingroup$ I didn't know this. Does the scalar curvature of the second manifold you mention change? Where can I read about this? $\endgroup$
    – Overflowian
    Nov 14, 2018 at 17:38
  • $\begingroup$ Does the scalar curvature of the second manifold you mention change? The reason I put "joined" in scare quotes is that in principle, GR is nonlinear, so you can't superpose solutions, glue them together at a boundary, etc. However, the notion of an asymptotically flat spacetime is that we're describing an isolated object that is very far away from anything else. Therefore it should be possible to make the glue job match up as well as desired (metric and all its derivatives have discontinuities less than $\epsilon$), simply by doing the surgery sufficiently far away from the objects. $\endgroup$
    – user21349
    Nov 14, 2018 at 20:31

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