International Congress of Mathematicians

Black Holes and the Penrose Inequality in General Relativity 267surface or that (M 3 ,g 0 ) have nonnegative scalar curvature. Hence, we quickly seethat equation 19 is a fairly deep conjecture which says something quite interestingabout manifold with nonnegative scalar curvature. Well, the Riemannian positivemass theorem is also a deep conjecture which says something quite interesting aboutmanifolds with nonnegative scalar curvature. Hence, it is natural to try to use theRiemannian positive mass theorem to prove equation 19.Thus, we want to create a manifold whose total mass depends on c fromequation 17. The idea is to use a reflection trick similar to one used by Bunting andMasood-ul-Alani for another purpose in [11]. First, remove the region of M 3 insideS(0) and then reflect the remainder of (M 3 ,g 0 ) through S(0). Define the resultingRiemannian manifold to be (M 3 ,g 0 ) which has two asymptotically flat ends since(M 3 ,g 0 ) has exactly one asymptotically flat end not contained by S(0). Note that(M 3 ,g 0 ) has nonnegative scalar curvature everywhere except on S(0) where themetric has corners. In fact, the fact that S(0) has zero mean curvature (since itis a minimal surface) implies that (M 3 ,g 0 ) has distributional nonnegative scalarcurvature everywhere, even on S(0). This notion is made rigorous in [6]. Thus wehave used the fact that S(0) is minimal in a critical way.Recall from equation 13 that vo(x) was defined to be the harmonic functionequal to zero on S(0) which goes to — 1 at infinity. We want to reflect vo(x) to bedefined on all of (M 3 ,g 0 ). The trick here is to define vo(x) on (M 3 ,g 0 ) to be theharmonic function which goes to —1 at infinity in the original end and goes to 1 atinfinity in the reflect end. By symmetry, VQ(X) equals 0 on S(0) and so agrees withits original definition on (M 3 ,g 0 ).The next step is to compactify one end of (M 3 , g 0 ). By the maximum principle,we know that VQ(X) > —1 and c > 0, so the new Riemannian manifold (M 3 , (vo(x) +l) 4 9o) does the job quite nicely and compactifies the original end to a point. Infact, the compactified point at infinity and the metric there can be filled in smoothly(using the fact that (M 3 ,g 0 ) is harmonically flat). It then follows from equation 15that this new compactified manifold has nonnegative scalar curvature since vo(x) +1is harmonic.The last step is simply to apply the Riemannian positive mass theorem to(M 3 , (vo(x) + l) 4 go)- R is not surprising that the total mass rô(0) of this manifoldinvolves c, but it is quite lucky that direct calculation yieldsm(0) =-4(c-m(0)), (20)which must be positive by the Riemannian positive mass theorem. Thus, we havethatro'(0) = 2(c-m(0)) = -^m(O) < 0. (21)2.4. The harmonic conformai class of a metricAs a final topic which is also of independent interest, we define a new equivalenceclass and partial ordering of conformai metrics. These new objects providea natural motivation for studying conformai flows of metrics to try to prove the