Intrinsic flat stability of the positive mass theorem for graphical hypersurfaces of Euclidean space

Theorem A.1.

Fix a precompact n-dimensional integral current space (X,d0,T) without boundary (e.g. ∂⁡T=0) and fix λ>0. Suppose that dj are metrics on X such that

Then there exists a subsequence, also denoted dj, and a length metric d∞ satisfying (A.1) such that dj converges uniformly to d∞

Furthermore,

and

In particular, (X,d∞,T) is an integral current space and set⁡(T)=X so there are no disappearing sequences of points xj∈(X,dj).

In fact, we have

and

Lemma A.2.

Under the hypothesis of Theorem A.1, there exists a subsequence, also denoted dj, and a length metric d∞ satisfying (A.1) such that dj converges uniformly to d∞,

and (X,d∞,T) is an integral current space.

Proof.

Observe that the functions dj may be extended to the metric completion

By (A.1) they are equicontinuous and so by the Arzela–Ascoli Theorem they have a subsequence converging uniformly to a function

Taking the limit of (A.1) we see that d∞ satisfies (A.1) as well. In particular, d∞ is a metric on X.

Furthermore, the Ambrosio–Kirchheim mass measure defined using d0 and defined using dj may be related as follows:

Recall that X=set0⁡(T)⊂X¯ because (X,d0,T) is an integral current space (by the definition of integral current space). In general the set of positive density also depends upon the metric just as the mass measure done. Here we have

and so (X,dj,T) is also an integral current space. This is true for all j=1,2,…,∞. ∎

Lemma A.3.

Given two metric spaces (X,dj) and (X,d∞) there exists a common metric space

where

with a metric dj′ on Zj such that

and

Thus we have metric isometric embeddings φj:(X,dj)→(Zj,dj′), φj′:(X,d∞)→(Zj,dj′) such that

In addition, if d0,dj satisfy (A.1), then

More precisely, we define dj′ by

where

Proof of Theorem A.1.

First apply Lemmas A.2 and A.3 and take the metric completion of Zj if it is not yet complete. Observe that

Thus

and we have (A.5) which implies (A.3).

To obtain (A.6), we take Bj=Iϵ×T to be the product integral current on Zj=Iϵ×X, where Iϵ=[-ϵj,ϵj] (see [21] for the precise definition of such products of intervals with currents). When T is just integration over a smooth manifold M, then Iϵ×T is just integration over Iϵ×M.

In [21] it is proven that

Since ∂⁡T=0, we have

Then by the definition of the intrinsic flat distance,

So we need only estimate the mass of Bj.

In [21] it is shown that

when the distance Dj is the isometric product metric on Zj defined with d0:

Since

We have

Thus we have (A.6) which implies (A.4).

This completes the proof of Theorem A.1. ∎

Proof.

First note that

Observe that dj′ is immediately symmetric and nonnegative. It is positive definite because

and clearly d-+⁢(z1,z2),d+-⁢(z1,z2)>2⁢ϵj for distinct z1,z2.

Before proving the triangle inequality, we apply (A.7) to prove (A.8):

Naturally, (A.9) follows in a similar way.

It suffices now to prove the triangle inequality. In (A.13)–(A.16) we prove the triangle inequality in the case where

and

Observe that

and similarly

Clearly,

so

Immediately by the definition we have

Thus we have shown the triangle inequality holds as long as (A.11)–(A.12) hold.

We need only prove the triangle inequality for all the five cases where

The rest of the cases will follow by symmetry in the definitions of d-+ and d+- and in swapping of the points z1,z2 with z3,z2. We have

Below we will use the following inequality, which follows from (A.7),

So

and

Thus dj′ is a metric. ∎

Example A.4.

Let X={0,12,14,…}∪{1}. Let

Let F:X→X be the identity map on X∖{0,1} and F⁢(0)=1 and F⁢(1)=0. Let

Both (X,dj) and (X,d∞) are complete but with different limits for the sequence {12,14,…}:

Observe that ϵj=1 because

So

Take the sequence of points zi=(0,1i). This sequence is Cauchy in Zj because

Assume on the contrary that this sequence of points converges to a point

Observe that for any z∈Zj,

Therefore, for i sufficiently large,

Thus p∞ is the limit of the sequence {1i} with respect to both metrics dj,d∞, which is a contradiction. Thus Zj is not complete.

The metric completion of Zj is

where (-1,p∞)∼(-1,0) and (1,p∞)∼(1,1). For ti∈[-1,1] and pi∈X we have

Note that with this distance

and that is why (-1,p∞)∼(-1,0) and (1,p∞)∼(1,1).