# Completeness hypothesis in the positive mass theorem

I am trying to understand and further formalize Witten's proof of the positive mass theorem. Dan Lee, in his book "Geometric relativity" did a wonderful job with formalizing and carrying out the details of Parker and Taubes' work, which was already a formalization of Witten's work.

The statement of the theorem in his book is more or less the following:

Theorem: Let $$(N,g)$$ be a complete asymptotically euclidean spin $$n$$-manifold with nonnegative scalar curvature and $$n \geq 3$$. Suppose further that $$N$$ has a well defined ADM mass. Then the ADM mass of each end is nonnegative. Moreover, if the mass of any end is zero, then $$(N,g)$$ is globally isometric to euclidean space.

I do not particularly like the completeness hypothesis as in most cases of interest in physics, the manifold is not complete. Therefore I am wondering why the completeness hypothesis is necessary. The only place I can find in the proof in his book where the completeness hypothesis is used explicitly is for positive mass rigidity, that is to say to prove that if the mass of any end is zero, then $$(N,g)$$ is globally isometric to euclidean space.

The completeness hypothesis is almost never stated in other surveys. Parker and Lee, in their survey on the Yamabe problem state the theorem as follows:

Theorem: Let $$(N,g)$$ be an asymptotically flat Riemannian manifold of dimension $$n \geq 3$$ such that the ADM mass is well defined, and with nonnegative scalar curvature. Then its mass $$m(g)$$ is nonnegative, with $$m(g) = 0$$ if and only if $$(N, g)$$ is isometric to $$\mathbb{R}^n$$ with its Euclidean metric.

The positive mass rigidity part in this theorem is plainly false, as $$\mathbb{R}^n \setminus \{0\}$$ satisfies all hypotheses but is not isometric to $$\mathbb{R}^n$$, so for this part the completeness is necessary. However, without completeness it is possible to prove that the manifold has to be flat. Hence I think the following theorem is also true:

Theorem: Let $$(N,g)$$ be an asymptotically euclidean spin $$n$$-manifold with nonnegative scalar curvature and $$n \geq 3$$. Suppose further that $$N$$ has a well defined ADM mass. Then the ADM mass of each end is nonnegative. Moreover, if the mass of any end is zero, then $$(N,g)$$ is flat.

Can someone confirm this?

Completeness is necessary. Otherwise you can just take a maximal spatial slice of the negative-mass Schwarzschild solution (i.e. a constant $$t$$ slice in Boyer-Lindquist coordinates) and it has vanishing scalar curvature with $$m < 0$$. There are however no complete maximal spatial sections of the negative-mass Schwarzschild solution.

For Lee and Parker, the implicit assumption of completeness is harmless: for applications to the Yamabe problem your asymptotically flat manifold comes from stereographic projection of a compact (closed) manifold about a point, so is always complete.

In the Witten proof included in the Appendix to Lee and Parker, completeness is used in the integration by parts argument equation (A.5). Here $$N_R$$ is assumed to be a compact manifold with boundary $$S_R$$. If your original manifold were not complete, but asymptotically flat, then for sufficiently large $$R$$ there must be another component of the boundary $$N_R$$ corresponding to the ends of the terminating geodesics.

Finally, if you make your definitions carefully, then incomplete manifolds are also ruled out of the Lee and Parker statement.

Recall that their definition of Asymptotically Flat manifold starts with a Riemannian manifold $$N$$ which is decomposed into some non-compact ends each of which are asymptotically flat in the usual sense, plus a compact portion.

If you start with an incomplete Riemannian manifold (here we are careful in having "manifolds with boundary" being not "manifolds"), and take away the asymptotically flat ends, you will necessarily still be left with something non-compact due to the incompleteness. (Incompleteness will always leave the manifold [which by definition cannot include a boundary] non-compact [Cauchy sequence alone the incomplete geodesic], and finite-length means it cannot be coming from an AF end.) So "complete" can be regarded as implicit.

• (I would also like to dispute the statement that "in most cases of interest in physics, the manifold is not complete." In physics, the space-time manifold may not be complete, but the spatial sections generally are. – Willie Wong Apr 3 at 20:08
• Thank you very much for your answer. When I said that in most cases of interest in physics, the manifold is not complete I was thinking primarily to a maximal spatial slice of the positive mass Schwarzschild solution. As your counterexample satisfies also the hypothesis for Schoen and Yau's theorem I guess that in such case the positive mass theorem cannot simply be used? – Nicolò Cavalleri Apr 6 at 0:53
• A maximal spatial slice of the maximally extended positive mass Schwarzschild solution passes through the bifurcate sphere on the event horizon and continues to the "other" domain of outer communications. This is a complete Riemannian manifold with two asymptotically flat ends. It shrinks to a throat at the bifurcate sphere, and gives the usual popular-science illustration of what a blackhole/wormhole looks like. – Willie Wong Apr 6 at 17:27