# Why is a flat, asymptotically flat 3-manifold isometric to $\mathbb{R}^3$?

I'm currently reading Schoen & Yau's 1979 proof of the positive-mass theorem and arrived at the very last sentence on the very last page (modulo the appendix) where they say:

Hence we conclude that $$\mathop{Ric} = 0$$ and because we are working in dimension three, $$ds²$$ is flat. This completes the proof of Theorem 2.

Theorem 2, however, made the claim that the manifold-with-boundary $$N$$ is (globally) isometric to $$\mathbb{R}^3$$ and I haven't been able to figure out in detail why local isometry to $$\mathbb{R}^3$$ (away from the boundary, of course) implies global isometry in this case. (My gut tells me that I'm missing something very elementary about what 3-manifolds of the given kind must look like, so I hope you'll excuse if this is a very simple question.)

For definiteness, let me restate the situation and the claim in detail:

Let $$N$$ be a connected¹, complete², asymptotically flat Riemannian manifold-with-boundary. Here, asymptotic flatness means that (1) there is a compact set $$K \subset N$$ such that the open set $$N \setminus K$$ consists of finitely many connected components ("ends") $$N_k$$ each of which is diffeomorphic to some $$\mathbb{R^3} \setminus (\text{closed ball})$$, and (2) the boundary of $$N$$ has mean curvature $$H < 0$$ with respect to the outward-pointing normal $$n$$. (Here, $$H$$ is defined as $$H := \mathop{tr}_g II$$ and $$II(v, w) := \langle \nabla_v w, n \rangle$$ is the 2nd fundamental form.)

Claim: Suppose $$N$$ is flat and only has one end (called $$N_k$$ in the theorem). Then $$N$$ is isometrically isomorphic to $$\mathbb{R}^3$$.

¹, ²: These requirements are not explicitly mentioned by Schoen & Yau but seem natural and, in fact, are necessary for some of the proofs in the paper to work. In particular, without them the claim would be false right away. I hope I'm not missing any further implicit assumptions on $$N$$.

My intuition is that the requirement on the boundary's mean curvature prevents it from bounding any "holes" in $$N$$. (In particular, K cannot merely be the boundary of $$N_k \subset \mathbb{R}^3$$, i.e. a sphere.) Moreover, the completeness of $$N$$ prevents us from choosing $$K$$ to be e.g. the empty set.

Finally, the fact that $$N = K \cup N_k$$ and that $$N$$ is flat everywhere should prevent $$K$$ from including something like a flat 3-torus. After all, topologically, there would be nothing preventing me from e.g. gluing $$N_k$$ to $$T^3$$. So I suspect that it must be the flatness forbidding this, in the sense that I cannot actually choose the throat between $$N_k$$ and $$T^3$$ to be flat. But I'm having trouble making this precise and, in particular, generalizing this argument to any other $$K$$ that is not a ball.

• Do you know Cartan-Hadamard theorem? What does it say in the case of a complete simply-connected flat manifold? – Moishe Kohan Feb 27 at 23:04
• Well, yes, in the case you describe, we'd immediately get that the manifold is diffeomorphic to $\mathbb{R}^3$. But in order to apply this to the present case, we'd first of all would have to argue that $∂N$ is empty (which would give us that the universal cover is diffeomorphic to $\mathbb{R}^3$) and, later, that $N$ is simply connected, so that $N$ is its own universal cover and thus diffeomorphic to $\mathbb{R}^3$. By pulling the metric on $N$ back to $\mathbb{R}^3$, we'd then, by construction, get an isometric isomorphy to $\mathbb{R}^3$ with the flat metric. But my problem here is: … – balu Feb 28 at 7:14
• 1) How would one show $∂N = \emptyset$? 2) Showing simple-connectedness already amounts to showing that we can't glue Nk to something like $T^3$. So unless I'm missing something (like a generalization of Cartan-Hadamard to manifolds-with-boundary), we're not really gaining anything here. (Ok, admittedly, once we've proved that the boundary is indeed empty, we only have to show that $\pi_1(N)=0$ and then Cartan-Hadamard basically gives us $\pi_k(N)=0$ for $k > 1$ for free, so that'd be something.) – balu Feb 28 at 7:15
• Re 1): I can think of (non-simply connected) cases where the boundary, a priori, would have negative mean curvature as required but where it's the gluing construction that can probably not be made flat. For instance, imagine gluing the end $N_k$ to the 3-dimensional solid torus along some ball contained in the torus. – balu Feb 28 at 7:40
• As an addendum to my earlier statement that pulling back the flat metric from $N$ to $\mathbb{R}^3$ via the diffeomorphism would give an isometry to $\mathbb{R}^3$ with the flat (standard!) metric: It seems one still has to do a bit of work there to show that $\mathbb{R}^3$ with the pulled-back flat metric is indeed standard Euclidean space, see math.stackexchange.com/questions/2229227/… & math.stackexchange.com/questions/352135/… – balu Feb 28 at 8:00

This is only a partial answer as of now but in the case of a manifold $$N$$ without boundary, a comparison result like the Bishop-Gromov inequality (see this PDF for a good introduction) can be used together with asymptotic flatness to show that every geodesic ball in $$N$$ is isometric to a ball in $$\mathbb{R}³$$ and these isometries can be glued together to an isometry between $$N$$ and $$\mathbb{R}³$$.