On the Riemannian Penrose inequality in dimensions less than 8
The Positive Mass Theorem states that a complete asymptotically flat manifold of nonnegative scalar curvature has nonnegative mass. The Riemannian Penrose inequality provides a sharp lower bound for the mass when black holes are present. More precisely, this lower bound is given in terms of the area of an outermost minimal surface, and equality is achieved only for Schwarzschild metrics. The Riemannian Penrose inequality was first proved in three dimensions in 1997 by G. Huisken and T. Ilmanen for the case of a single black hole [HI01]. In 1999, H. Bray extended this result to the general case of multiple black holes using a different technique [Bra01]. In this paper we extend the technique of [Bra01] to dimensions less than 8.
The Penrose Conjecture is a longstanding conjecture in general relativity that provides a lower bound for the mass of an asymptotically flat spacelike slice of spacetime, in terms of the area of the black holes in the spacelike slice. Penrose originally formulated the conjecture as a test for the far more ambitious idea of cosmic censorship. In the case where the asymptotically flat spacelike slice is time-symmetric, the Penrose Conjecture reduces to a statement in Riemannian geometry, which we call the Riemannian Penrose inequality. In this paper we will restrict our attention to the Riemannian Penrose inequality. For more background on the general Penrose Conjecture, as well as some physical motivation, see [Bra01, Section 1] and references cited therein.
The Riemannian Penrose inequality was first proved in three dimensions in 1997 by G. Huisken and T. Ilmanen for the case of a single black hole [HI01]. In 1999, H. Bray extended this result to the general case of multiple black holes using a different technique [Bra01]. Before we state this theorem, let us review some definitions.
Let . A Riemannian manifold is said to be asymptotically flat111Note that there are various inequivalent definitions of asymptotic flatness in the literature, but they are all similar in spirit. This one is taken from [Sch89, Section 4]. if there is a compact set such that is a disjoint union of ends, , such that each end is diffeomorphic to , and in each of these coordinate charts, the metric satisfies
for some and some , where the commas denote partial derivatives in the coordinate chart, and is the scalar curvature of .
In this case, in each end , the limit
exists (see, e.g. [Sch89, Section 4]), where is the area of the standard unit -sphere, is the coordinate sphere in of radius , is its outward unit normal, and is the Euclidean area element on . We call the quantity , first considered by Arnowitt, Deser, and Misner (see, e.g. [ADM61]), the ADM mass of the end , or when the context is clear, we simply call it the mass, . (Under an additional assumption on the Ricci curvature, R. Bartnik showed that the ADM mass is a Riemannian invariant, independent of choice of asymptotically flat coordinates [Bar86].)
The Riemannian Penrose inequality may be thought of as a refinement of the celebrated Positive Mass Theorem when black holes are present. Indeed, we will need to use the Positive Mass Theorem for our proof.
Theorem 1.1 ((Riemannian) Positive Mass Theorem).
Let be a complete asymptotically flat manifold with nonnegative scalar curvature. If or if is spin, then the mass of each end is nonnegative. Moreover, if any of the ends has zero mass, then is isometric to Euclidean space.
The case was proved by R. Schoen and S.-T. Yau using minimal surface techniques [SY79] (see also [Sch89, Section 4]), and soon later E. Witten proved the spin case using a Bochner-type argument [Wit81] (see also [Bar86]).
Now fix a particular end of . Define to be the collection of hypersurfaces that are smooth compact boundaries of open sets in containing all of the other ends. Then each hypersurface in defines a meaningful outside and inside.
A horizon in is a minimal hypersurface in . A horizon is outer minimizing if its area minimizes area among all hypersurfaces in enclosing .
The (Riemannian) Schwarzschild manifold of dimension and mass is equipped with the metric
Given a mass , we also define the Schwarzschild radius of the mass to be
Note that in a Schwarzschild manifold, the coordinate sphere of radius is the unique outer minimizing horizon, and its area satisfies the equation
We can now state the main result of [Bra01].
Theorem 1.2 (Riemannian Penrose inequality in three dimensions).
Let be a complete asymptotically flat 3-manifold with nonnegative scalar curvature. Fix one end. Let be the mass of that end, and let be the area of an outer minimizing horizon (with one or more components). Then
with equality if and only if the part of outside the horizon is isometric to a Riemannian Schwarzschild manifold outside its unique outer minimizing horizon.
Even though the original motivation from general relativity may have been specific to three dimensions, because of string theory there is a great deal of interest in higher dimensional black holes. More importantly, from a purely geometric perspective, there appears to be nothing inherently three-dimensional about the Riemannian Penrose inequality, so it is natural to wonder whether the result holds in higher dimensions. The goal of this paper is to prove the following generalization.
Theorem 1.3 (Riemannian Penrose inequality in dimensions less than 8).
Let be a complete asymptotically flat manifold with nonnegative scalar curvature, where . Fix one end. Let be the mass of that end, and let be the area of an outer minimizing horizon (with one or more components). Let be the area of the standard unit -sphere. Then
with equality if and only if the part of outside the horizon is isometric to a Riemannian Schwarzschild manifold outside its unique outer minimizing horizon.
Our proof is limited to dimensions less than 8 for the same reason that Schoen and Yau’s proof is limited; we need to use regularity of minimal hypersurfaces.
The geometry inside the horizon plays no role at all in the proof of the theorem. Accordingly, our objective is to prove the following theorem.
Let be a complete one-ended asymptotically flat manifold with boundary, where . If has nonnegative scalar curvature, and if the boundary is an outer minimizing horizon (with one or more components) with total area , then
with equality if and only if is isometric to a Riemannian Schwarzschild manifold outside its unique outer minimizing horizon.
We will prove this theorem using the first author’s conformal flow method [Bra01]. One might wonder whether Huisken and Ilmanen’s inverse mean curvature flow method [HI01] could also be used for this purpose. Unfortunately, since the Gauss-Bonnet Theorem lies at the heart of that method, it would require major new insights to adapt it higher dimensions.
2 Overview of proof
The vast majority of the first author’s proof of the Riemannian Penrose inequality in dimension three applies to dimensions less than 8 [Bra01]. In this section we will review the main features of the proof and describe the parts that require modification. Since a large portion of our proof is actually contained in [Bra01], we will try to maintain consistent notation. The main technical tool that we will employ is the conformal flow.
Let be a manifold with a distinguished end. Let be a family of metrics on , and let be a family of hypersurfaces in such that is Lipschitz in , in , and smooth in outside . We say that is a conformal flow if and only if the following conditions hold for each :
outside is a complete asymptotically flat manifold with boundary, and it has nonnegative scalar curvature.
is an outer minimizing horizon in .
, where inside , and outside , is the unique solution to the Dirichlet problem
The formulation of the conformal flow in [Bra01] is slightly different but defines the same flow. Instead of using the last item in the above definition, we could set and demand that
where inside , and outside , is the unique solution to the Dirichlet problem
The fact that these two formulations are equivalent follows from the following simple lemma, which we will use repeatedly.
If and are smooth metrics and is a smooth function such that
then for any smooth function ,
Given initial data satisfying the first two properties of the conformal flow described above, with , there exists a conformal flow for all . Moreover,
For all , encloses without touching it.
can “jump” at most countably many times. At these jump times, we write and to denote the hypersurface “before” and “after” it jumps, respectively.222See [Bra01, Section 4] for a precise statement. With the definition of the conformal flow given above, at a jump time, could lie somewhere between and . However, in the construction of the conformal flow in [Bra01, Section 4], is the outermost horizon in containing , and consequently, we have for .
The proof of this theorem in [Bra01, Theorem 2] is unchanged in higher dimensions, as long as . The basic idea behind the proof is to use a discrete time approximation, and then take the limit as the length of the discrete time intervals approaches zero. The hypothesis is required in the proof in order to find smooth outermost minimal area enclosures. The only other place that this dimensional restriction will be used again is when we invoke the Positive Mass Theorem.
To prove our main theorem (Theorem 1.4), we will use the hypotheses of the theorem as initial data for the conformal flow and prove that the conformal flow has the following properties:
The area of in is constant in . Call it .
The mass of , which we will call , is nonincreasing.
With the right choice of end coordinates, the metric outside converges to a Schwarzschild metric.
The area of the horizon in this Schwarzschild manifold is greater than or equal to .
Once we have established these properties, the main theorem follows immediately.333Except for the case of equality, which requires an additional simple argument.
The area of in is constant in .
The proof of this lemma in [Bra01, Section 5] is unchanged in higher dimensions. The basic idea behind the proof is that the rate of change of the area has a contribution from changing while leaving fixed and a contribution from changing while leaving fixed. The first contribution is zero because is minimal, and the second contribution is zero because the metric is not changing at . (Specifically, at .) However, the proof is more subtle than this because can jump. See [Bra01, Section 5] for details.
In order to simplify the rest of our arguments, we use a tool called harmonic flatness.
A Riemannian manifold is said to be harmonically flat at infinity if there is a compact set such that is the disjoint union of ends, , such that each end is diffeomorphic to , and in each of these coordinate charts, there is a (Euclidean) harmonic function such that
In other words, each end is conformally flat with a harmonic conformal factor.
Note that a harmonically flat end necessarily has zero scalar curvature. Expanding in spherical harmonics in a particular end , we see that
for some constants and . Clearly, a manifold that is harmonically flat at infinity is asymptotically flat.444However, when , it is necessary to change the distinguished coordinate chart. A simple computation shows the following:
In the situation described above, the mass of the end is equal to .
In order to prove our main theorem (Theorem 1.4), we may assume without loss of generality that is harmonically flat at infinity.
From now on we will always work in the situation of initial data that is harmonically flat at infinity, and then evolved by the conformal flow. From Lemma 2.1 it is clear that the conformal flow preserves harmonic flatness outside .
We now consider a third formulation of the conformal flow. By the harmonic flatness assumption, we know that for some harmonic function on the exterior region for some . (We adopt the shorthand notation and .) We choose end coordinates so that , and consequently, we are not allowed to choose the constant arbitrarily. Now extend to a positive function on all of and define the metric by
Note that for , . We can now reformulate the conformal flow by setting
and demanding that
where inside , and outside , is the unique solution to the Dirichlet problem555We know that there is a unique solution because this formulation is equivalent to the previous ones.
Note that in the region outside with , both and are (Euclidean) harmonic functions. We summarize the relationships between the three formulations of the conformal flow:
The mass, , is nonincreasing.
The proof of this lemma in [Bra01, Sections 6 and 7] is also unchanged in higher dimensions. However, the main idea used in the proof is central to this paper, so we will summarize the basic argument.
For each time , consider the two-ended manifold obtained by reflecting the manifold through . Let be the -harmonic function that approaches at one end and at the other end. We can use to conformally close the -end by considering the metric on . The result is a new one-ended manifold with nonnegative scalar curvature.666One can show that the singularity at is removeable. Similarly, if is a jump time, then we can construct by first reflecting through . Lemma 2.6 will follow from the following key lemma.
Let be the mass of . If is not a jump time, then
If is a jump time, let be the mass of . Then
where denotes the right and left side limits of .
Lemma 2.6 follows from this lemma because the Positive Mass Theorem tells us that (and that ). However, there is a technical point to deal with here: Since the metric is not smooth along where the gluing took place, the standard version of the Positive Mass Theorem does not immediately apply. However, since was obtained by reflecting through a minimal surface, one can show that is a limit of smooth manifolds with nonnegative scalar curvature, and we still have the desired result. This argument was carried out in [Bra01, Section 6] and described in further detail in [Mia02].
For ease of notation, let us assume that is not a jump time. (The proof for jump times is the same, but with superscripts everywhere.) By symmetry, we know that the function used in the construction of must be on one end (and on the end to be closed up). Therefore, in the one end of , for ,
We will now compute by expanding . For , is harmonic and thus we can expand it as
We know that and , so we can write
Therefore, in order to prove Theorem 1.4, the only part of [Bra01] that needs to be modified is the part that deals with the convergence to Schwarzschild. The basic idea here is that since is nonincreasing and bounded below by zero (by Positive Mass Theorem), we might hope that its derivative, , converges to zero. Indeed, that turns out to be the case (see Lemma 3.1). The equality case of the Positive Mass Theorem states that the only complete asymptotically flat manifold of nonnegative scalar curvature and zero mass is Euclidean space. Therefore we might also hope that since is converging to zero, must converge to the flat metric at infinity, in some sense. In order to establish this fact, we will need a strengthened version of the equality case of the Positive Mass Theorem (see Theorem 3.4), proved in a separate paper [Lee]. Then it is not hard to see that (with the right choice of end coordinates), must converge to a Schwarzschild metric outside .777This is a refined version of the fact that the only asymptotically flat manifolds that are scalar-flat and conformal to Euclidean are Schwarzschild manifolds.
In order to make this basic argument work, we need to control . (Specifically, we need Lemma 3.3.) In [Bra01], this control was obtained using curvature estimates by way of the Gauss-Bonnet Theorem, together with a Harnack-type inequality from [BI02] that was only applicable in three dimensions. It is this part of the proof that needs to be completely re-worked for application to higher dimensions. Even though our new proof is more general, it is actually more elementary and straightforward than the original proof. This content appears in Section 5 of this paper.
Section 3 of this paper serves as a replacement of Sections 8 through 12 in [Bra01], although there is a fair amount of overlap. To summarize the differences: First, using the three-dimensional curvature estimates described above, it was proved in [Bra01] that eventually encloses any bounded region, and consequently one can then assume that is an exterior domain of . It turns out that this simplification is not actually needed for our proof, but it means that we have to be a bit more careful than in [Bra01]. Second, the strengthened version of the equality case of the Positive Mass Theorem (Theorem 3.4) mentioned above was proved in [Bra01] using spinors. We need a different proof here since higher dimensional manifolds need not be spin; the proof is given in [Lee]. Third, with the benefit of hindsight we are able to simplify and streamline many aspects of the original proof.
3 Convergence to Schwarzschild
As mentioned earlier, we want to show that converges to zero as .
The quantity is nondecreasing in .
Recall that for , encloses . By the maximum principle and the definition of , it is evident that is nondecreasing in , for any fixed . Therefore is also nondecreasing in , for any fixed . Recall from the proof of Lemma 2.7 that
The claim follows.
For now assume that is smooth.
Differentiating the monotone quantity from the previous claim,
Since , we have
proving the claim. Since is a nonnegative function with finite integral and derivative bounded above, it follows that . Of course, is not necessarily smooth, but it is a simple exercise to show that the result still holds. ∎
Since is nonincreasing and bounded below by zero, it must have a limit.
Let . .
We postpone the proof of this lemma until the next section, so as not to interrupt the flow of the main argument.
Let , and choose a diffeomorphism . That is, we choose coordinates in for the end. Recall that since we chose the normalization , we cannot say that is harmonic in without losing generality. We can only say that is harmonic in for some possibly large .
We want to talk about convergence of our Riemannian manifold as , but we will see that, with respect to a fixed coordinate system at infinity, runs off to infinity. Consequently, the part of that we care about (the part outside ) disappears in the limit. Therefore we need to change our choice of coordinates as changes. One way to do this is to introduce a one-parameter group of diffeomorphisms. Choose a smooth vector field on such that
on , where is the radial coordinate on . (We extend inside so that it is smooth.) Let be the one-parameter group of diffeomorphisms of generated by .
Given our conformal flow , we define the normalized conformal flow by
Define new functions
and a new metric
Note that . Also note that inside , and outside , is the unique solution to the Dirichlet problem
Differentiating the definition of , we see that
For all and , we see that and
Since we are concerned with what happens for large , from now on we will always assume that .
Let outside of . Now define
on the exterior of . Observe that is isometric to , and consequently it has mass equal to . Note that for and outside with ,
As mentioned earlier, in order to make this argument work, we need to obtain control on . We postpone the proof of this lemma so as not to interrupt the flow of the main argument.
There exists some such that is always enclosed by the coordinate sphere of radius .
As mentioned earlier, we will need to use a strengthened version of the equality case of the Positive Mass Theorem. Essentially, we want to say that a sequence of asymptotically flat manifolds of nonnegative scalar curvature becomes flatter as the mass approaches zero. The proof of this theorem is the subject of a separate paper [Lee].
Let be any smooth manifold on which the Positive Mass Theorem holds. Let be a coordinate chart for one of the ends of . Let be a complete asymptotically flat metric of nonnegative scalar curvature on , and suppose that
in , where is a positive (Euclidean) harmonic function on with .
Then for any , there exists such that if the mass of in this end is less then , then
where is a universal constant depending only on . The constant depends only on and . In particular, it does not depend on the topology of .
Now observe that in , is harmonic on , and . Also observe that since is isometric to , it is a limit of metrics that extend to complete metrics of nonnegative scalar curvature [Mia02]. In short, we may apply the previous theorem to in order to conclude that uniformly for .
The following limits hold uniformly over all .
Let . By the previous theorem and the following discussion, the third equality in the statement of the lemma follows immediately. In other words, we know that for large enough , is small enough so that
for large enough . Since is harmonic ans has no constant or terms in its expansion, it follows from the maximum principle and a gradient estimate that for all ,
for some constant independent of . Analyzing equation (3), we conclude that for large enough ,
The first equation in the statement of the lemma now follows from the definition of and the fact that . (The second equation in the statement of the lemma follows from the other two.) ∎
For , , let denote the -neighborhood of , that is, the set of points that are distance less than away from . For all , there exists some large such that
where is the sphere of radius in .888Actually, it is only necessary to show that lies within the sphere of radius .
Using maximum principle arguments and Lemma 3.3, one can prove uniform upper and lower bounds on on .
Since the area of with respect to is constant, and since there is a uniform lower bound on , we have a uniform upper bound on the Euclidean area of . Therefore we can show that for some sequence , the part of in weakly converges to some . But moreover, using the uniform bounds on , one can argue the stronger statement that converges to in Hausdorff distance. (See [Bra01, Section 12 and Appendix E] for details. One can also argue directly using Lemma 5.1.)
Since the ’s are harmonic and uniformly bounded, we can choose a subsequence such that converges uniformly on compact subsets of the exterior of in . Since the limit must be a harmonic function, it follows from the previous lemma that the limit is . More precisely, given , for large enough , we know that and that for all outside . Our goal is to show that is just the sphere of radius , and then the result will follow from the Hausdorff convergence.
Suppose that part of lies inside the sphere of radius . Then we can find some and some point such that is outside and yet , which is a contradiction.
We now come to the critical part of the proof. Suppose that part of lies outside the sphere of radius . Then for some and some , the ball lies completely outside the sphere of radius . The basic intuitive argument is as follows: We know that is zero at , but in we know that is significantly smaller than zero. The only way this can happen is if the gradient of is blowing up. In fact, we show that it blows up badly enough that the energy of blows up, which is a contradiction since we have a bound on energy (described below).
Consider the unique harmonic function that approaches at infinity and is zero at the sphere . Since is contained in , we can deduce from the energy-minimizing property of harmonic functions that the energy of in the exterior of is less than the energy of in the exterior of , namely . Let be the region outside , let , and let be the area form of . (Note that we suppress the dependence on in the notation.) Then by the co-area formula and the Hölder inequality,
Let . Then and we have
On the other hand, we know that for some nonzero constant , we have in . Now let , and choose large enough so that for all lying outside . Therefore
and it follows that
Since , we can choose large enough so that on a set of measure at least . We also know that for , , and consequently these ’s are Hausdorff converging to . In particular for , is uniformly bounded below by some constant . Plugging this into our energy bound (4), we see that
which contradicts equation (5).
The main result follows easily from this lemma. Let . Since is outer minimizing with respect to , we see that is less than or equal to the area of the sphere of radius with respect to . Also, the argument in the previous lemma shows that converges to uniformly on . So for large enough , we have