The Penrose inequality for asymptotically locally hyperbolic spaces with nonpositive mass

Transcription

1 The Penrose inequality for asymptotically locally hyperbolic spaces with nonpositive mass Dan A. Lee André Neves September 18, 2014 Abstract In the asymptotically locally hyperbolic setting it is possible to have metrics with scalar curvature 6 and negative mass when the genus of the conformal boundary at infinity is positive. Using inverse mean curvature flow, we prove a Penrose inequality for these negative mass metrics. The motivation comes from a previous result of P. Chruściel and W. Simon, which states that the Penrose inequality we prove implies a static uniqueness theorem for negative mass Kottler metrics. 1 Introduction The Penrose inequality for asymptotically flat 3-manifolds M with mass m and nonnegative scalar curvature states that is an outermost minimal surface (i.e., there are no compact minimal surfaces from infinity), then r A m 16, where A is the area G. Huisken and T. Ilmanen [14] first proved this inequality for A equal to the largest area of a connected component H. Bray [3] later proved the more general inequality described above using a di erent method, and this result was later extended to dimensions less than 8 by Bray and the first author [4]. We are interested in an analog of this theorem for a class of asymptotically locally hyperbolic manifolds which we now define. 1

2 Definition 1.1. We say that a C i Riemannian metric g on a smooth manifold M 3 is C i asymptotically locally hyperbolic 1 if there exists a compact set K M and a constant curvature surface (, ĝ), called the conformal infinity of (M,g), such that M r K is di eomorphic to (1, 1) withthe metric satisfying where g =(k + 2 ) 1 d ĝ + 1 h + Q, k is the constant curvature of (, ĝ); is the coordinate on (1, 1); h is a C i symmetric two-tensor on. function µ = 3 4 tr ĝ h; We define the mass aspect Q is a C i symmetric two-tensor on M such that Q b + rq b + + i r i Q b = o( 3 ), where b is the hyperbolic metric (k + 2 ) 1 d ĝ and r are derivatives taken with respect to b. We will use the notation Q = o i ( 3 ) as a convenient abbreviation. For the sake of convenience, we assume that k is 1, 0, or 1, and in the case k = 0, we further assume that ĝ =4. (These assumptions simply serve the purpose of normalization.) Finally, we define the mass to be m = µdĝ = 1 µdĝ, ĝ and we also define m =supµ. An important class of asymptotically locally hyperbolic manifolds is given by the Kottler metrics (see [9] for instance), which are static metrics with cosmological constant = 3. 1 This definition requires M to have only one end. Although the definition could be altered to allow for multiple ends, our main interest is in the region outside outermost minimal surfaces, whose components necessarily only have one end. 2

3 Definition 1.2. Let (, ĝ) be a surface with constant curvature k equal to 1, 0, or 1, with area equal to 4 in the k = 0 case. Let m 2 R be large enough so that the function r 2 + k 2m r := V (r)2 has a positive zero. Let r m be the largest zero of V 2, and define the metric g = V 2 dr 2 + r 2 ĝ on (r m, 1). Define (M,g) to be the metric completion of this Riemannian manifold. We say that (M,g) is a Kottler space with conformal infinity (, ĝ) and mass m. Remark. The Kottler metrics have scalar curvature R = 6. The most familiar situation is when is a sphere (k = 1), in which case the metric is also called an Schwarzschild anti-de Sitter metric in the literature. In order for V 2 to have a positive zero, we must have m>0 when k =0or k =1. However, when k = 1, we need only have m 1 3 p 3. Therefore we define the critical mass m crit := 0 when k =0or k =1, and m crit := 1 3 p 3 when k = 1. As long m>m crit, we obtain M =[r m, 1) = {r m } as an outermost minimal surface boundary. One can see that these metrics are asymptotically locally hyperbolic by performing a substitution, in which case one has h = 2 3 mĝ. We briefly describe the critical cases when m = m crit. When k =1 and m = 0, g is just the hyperbolic metric, and when k = 0 and m =0, the metric g can be written as dt 2 + e 2t ĝ on R after a coordinate change. When k = 1 and m = m crit, the metric g on (r m, 1) is a two-ended complete Riemannian manifold, with one end asymptotically locally hyperbolic and the other end asymptotic to the cylindrical metric dt ĝ on R. We can now state our main theorem. Theorem 1.3 (Penrose Inequality for nonpositive mass). Let (M 3,g) be a C 2 asymptotically locally hyperbolic manifold with m apple 0 and conformal infinity (, ĝ), whose genus is g. 3

4 Assume that R is an outermost minimal surface, and there is a boundary g M of genus g. Then m 1 r A 1 g + A, (1) 16 4 where A is the area g M,and = (max{1, g 1}) 3/2 is a topological constant. Furthermore, equality occurs if and only if (M,g) is isometric to the Kottler space with infinity (, ĝ) and mass m. If one replaces m by m and removes the m apple 0 condition in our theorem, 2 one obtains the natural analog of Huisken and Ilmanen s Penrose inequality [14] in the asymptotically locally hyperbolic setting. Versions of this statement have been conjectured by P. Chruściel and W. Simon [9, Section VI], and in the g = 0 case, by X. Wang [23, Section 1]. The graph case of the conjecture has recently been established by L. de Lima and F. Girão when m 0 [10]. See also a related volume comparison result by S. Brendle and O. Chodosh [5]. Corollary 1.4. Let (M 3,g) be a C 2 asymptotically locally hyperbolic manifold with conformal infinity of genus g. Assume that R is an outermost minimal surface, and there is a boundary g M of genus g. Then m >0 for g =0or 1, 1 m > 3 p for g > 1. 3 Proof. If g = 0 or g = 1 and m apple 0 we obtain at once from Theorem 1.3 g M = 0, which is a contradiction. If g > 1, then it is an easily verifiable fact that r x 1 g + x (g 1) 3/ p for all x This inequality and Theorem 1.3 imply at once that m 3 p. If equality 3 holds then we are in the equality case of Theorem 1.3 and so M must be 1 isometric to a critical Kottler space of mass 3 p. But this is impossible 3 because a critical Kottler space has no compact minimal surfaces. (They are foliated by strictly mean convex surfaces.) 2 Although we are unable to replace m by m in general, we do obtain a slightly stronger inequality than (1). See Lemma

5 While the assumption of a boundary component with genus g is not desirable, we note that there exist examples (the AdS solitons due to G. Horowitz and R. Myers [13]) of asymptotically locally hyperbolic manifolds with g = 1, empty boundary, and negative mass. We note that the inverse mean curvature flow technique allows us to a give a new proof of a weakened version of the positive mass theorem for asymptotically hyperbolic manifolds, the full version of which was proved by Chruściel and M. Herzlich [8] and by X. Wang [23], and for the case of a manifold with minimal boundary, by V. Bonini and J. Qing [1]. Theorem 1.5 (Positive m theorem for asymptotically hyperbolic manifolds). Let (M 3,g) be a complete C 2 asymptotically hyperbolic 3 manifold (with or without a minimal boundary) with scalar curvature R 6. Then m 0. Moreover, if m =0, then (M,g) must be hyperbolic space. We prove Theorem 1.3 following the inverse mean curvature flow theory developed by Huisken and Ilmanen in [14]. The general idea is to g M outward with speed inversely proportional to the mean curvature and obtain a (weak) flow of surfaces ( ) t 0 (where 0 g M). To each compact surface one considers its Hawking mass to be m H ( ) := r 1 g 16 1 (H 2 4), (2) 16 where denotes the area of, and H is its mean curvature. Observe that our notation m H ( ) leaves out the dependence on g. The key property of inverse mean curvature flow is that m H ( ) is non-decreasing in time. Therefore, since 1 m H ( 0 ) coincides with the right-hand side of the inequality in Theorem 1.3, the desired result follows if 1 lim mh ( ) apple m. t!1 In the asymptotically locally hyperbolic setting this inequality is subtle, and in fact the second author constructed well-behaved examples (with g = 0) where the above inequality does not hold [19]. The central observation of this paper is that this inequality holds if m apple 0. The di erence between our result here and the one in [19] can be traced to the fact that if the mass aspect is positive, then a desired inequality goes in the wrong direction, whereas for nonpositive mass aspect, it goes in the right direction. See the 3 Here we define this to mean asymptotically locally hyperbolic with infinity equal to the round sphere. 5

6 end of the proof of Lemma 3.13 to see the exact place where the condition m apple 0isused. Acknowledgements: The authors would like to thank Piotr Chruściel and Walter Simon for bringing this problem to our attention and for their interest in this work, as well as Richard Schoen for some helpful conversations. We also thank the referees for their careful reading of the article and for pointing us to relevant references. 2 An application to static uniqueness Chruściel and Simon proved in [9] that Theorem 1.3 implies a uniqueness theorem for static metrics with negative mass that we now explain. Definition 2.1. We say that (M 3,g,V)isacomplete vacuum static data set with cosmological constant = 3 if and only if M is a smooth manifold (possibly with boundary) equipped with a complete C 2 Riemannian metric g and a nonnegative C 2 function V such = {V =0} and gv = V (3) Ric(g) = 1 V Hess g V + g. (4) If (M 3,g,V) is a vacuum static data set as above, then the Lorentzian metric h = V 2 dt 2 + g on R M is a solution to Einstein s equations with cosmological constant. In the = 0 case a 1987 result of G. Bunting and A. Masood-ul-Alam [6] shows that Schwarzschild spaces are the only asymptotically flat vacuum static data sets (with the case of connected boundary treated in much earlier work of W. Israel [15] and H. Müller zum Hagen, D. Robinson, and H. Seifert in 1973 [18]). We are interested in a similar characterization of the Kottler metrics defined in the previous section, which are known to be vacuum static data sets with cosmological constant = 3. A static uniqueness theorem for the hyperbolic space was proved in work of Boucher-Gibbons-Horowitz [2], J. Qing [22], and X. Wang [24]. The following definition is equivalent to the one given in [9, Section III.A]. Definition 2.2. We say that a complete vacuum static data set (M 3,g,V) with cosmological constant = 3isC i conformally compactifiable if g and V are C i and there exists a smooth compact manifold M 0 with boundary and a C i+1 embedding of M into M 0 such that M 0 = M [, 6

7 the function V 1 extends to a C i function up to hat vanishes at, and d(v 1 ) 6= 0 at ; and the formula ĝ = V 2 g defines a Riemannian metric near. If (M,g,V ) is a complete vacuum static data set, the fact that Hess g V =0 implies that rv is constant on each component and we call this constant the surface gravity apple of that component. The surface gravity is strictly positive for the following reason: Given p let be the unit speed geodesic that starts at p perpendicular and set f(t) =V ( (t)). From the static equation we see the existence of c>0and t 0 > 0 so that f 00 apple cf for all 0 apple t apple t 0. Standard o.d.e. comparison shows that if f 0 (0) = f(0) = 0 then f(t) apple 0 for all 0 apple t apple t 0, which is impossible because V is strictly positive on the interior of M. The (non-critical) Kottler metrics have exactly one component and, for fixed k, there is a bijection between possible surface gravities in 1 (0, 1) and possible masses in ( 3 p, 1) whenk = 1, or in (0, 1) whenk 3 is 0 or 1. Therefore, for fixed k we can define a bijection m(apple) according to the fixed relationship between mass and surface gravity for Kottler metrics whose infinities have curvature equal to k. For this bijection, when k = 1 1 one has lim apple!0 m(apple) = 3 p and m(1) = 0 (see [9, Section II] for details). 3 We can now state a static uniqueness theorem for Kottler metrics of negative mass, which will follow from Theorem 1.3 combined with some of the results of [9] that we will describe later. Theorem 2.3 (Static uniqueness with nonpositive mass). Let (M 3,g,V) be a complete vacuum static data set with cosmological constant = 3, and assume that it is C 5 conformally compactifiable with conformal infinity (, ĝ) of constant curvature k = 1. Assume that there is a component such that is homeomorphic to and has the largest surface gravity apple of any component. If m(apple) apple 0 (or equivalently apple apple 1), then (M,g) must be isometric to the Kottler metric with infinity (, ĝ) (as in Definition 1.2) and mass m(apple), while V is equal to the usual static potential of the Kottler metric, up to a constant multiple. Before we present the proof some comments are in order. It follows from Lemma 2.6 and the proof of Lemma 3.3 that no component can ever have larger genus than. It would be nice to also rule out the possibility has components of strictly smaller genus. 7

8 The Horowitz-Myers AdS solitons [13] described in the Introduction do not only have have negative mass, no boundary, and k = 0, but they are also static. A static uniqueness theorem for these examples was proved by G. Galloway, S. Surya, and E. Woolgar [12]. It would be desirable to remove the condition on m(apple). Chruściel and Simon raise the question of whether the requirement of constant curvature at conformal infinity is necessary [9, Section III.A]. 2.1 Proof of Theorem 2.3 The following proposition is a consequence of Theorem I.1 and Proposition III.7 of [9], together with a coordinate change. Proposition 2.4. Let (M 3,g,V) be a C i+3 conformally compactifiable complete vacuum static data set with cosmological constant = 3. Further assume that the induced metric ĝ on (as defined in the previous section) has locally constant Gauss curvature k equal to 1, 0, or 1. Then is connected, (M,g) is C i asymptotically locally hyperbolic (as defined in the Introduction) with conformal infinity (, ĝ), and V 2 = 2 + k 4 3 µ 1, where can be used as the radial coordinate in the definition of asymptotically locally hyperbolic, and µ is the mass aspect. In particular, static data sets with = and m. 3 have a well-defined mass m Proof. Theorem I.1 of [9] states that the conformal boundary mustbe connected. Proposition III.7 of [9] (together with the assumption of C i+3 smoothness of the compactified metric) shows that if we choose a radial coordinate r so that V 2 = r 2 + k, then the metric takes the form g = r 2 + k 2 r 1 1 dr 2 + r 2 ĝ + r 1 h 0 + o i (r 3 ), where is a function with the property that 1 := lim r!1 = 1 2 tr ĝ h 0. In order to show that g is asymptotically locally hyperbolic, we perform the coordinate change r 2 = We analyze the first term in the expression for g. Note that r 2 +k 2 r 1 = 2 +k r 1 = 2 +k (lower order terms), 8

9 and dr 2 = dr 2 d 2 = 1+ 4 d d 2 + (lower order terms). Putting it together, we obtain the asymptotically local hyperbolic form g =( 2 + k) 1 d ĝ + 1 (h ĝ)+o i ( 3 ). Then the mass aspect function is µ = 3 4 tr ĝ(h ĝ) = Thus V 2 = r 2 + k = 2 + k = 2 + k 4 3 µ 1. The key theorem of [9] for the purposes of this article is Theorem I.5. However, since we require a slightly di erent statement, we provide a complete proof in Section 4 using their method. Theorem 2.5 (Chruściel-Simon). Let (M 3,g,V) be a C 3 conformally compactifiable complete vacuum static data set with cosmological constant = 3, conformalinfinity(, ĝ) of constant curvature k = 1, and@m 6= ;. apple M denote the boundary component with the largest surface gravity apple and suppose m 0 := m(apple) apple 0 (i.e., 0 <appleapple 1). If (M 0,g 0,V 0 ) denotes the Kottler space with infinity (, ĝ) and mass m 0 (i.e., the one with surface gravity apple), then (@ apple apple M (@M 0 0 g0 and m apple m 0, 0 g0 is the area with respect to g 0. In the case apple M has the same genus as, this theorem provides a simple comparison between the masses and boundary areas of a vacuum static data set and its so-called reference solution (M 0,g 0,V 0 ). Lemma 2.6. Let (M 3,g,V) be a C 2 asymptotically locally hyperbolic, complete vacuum static data set with cosmological constant = 3. If@M 6= ;, is an outermost minimal surface. In fact, there are no compact minimal surfaces in the interior of M. This result is somewhat of a converse to a result of P. Miao [17]. (See also [7].) 9

10 Proof. First note that the static equations imply is totally geodesic, so we need only show that there are no other compact minimal surfaces. Consider the trapped region K of M, which is the union of all compact minimal surfaces in M, together with all regions of M that are bounded by these minimal surfaces. The boundary of the trapped must itself be a smooth compact minimal surface. (See the proof of Lemma 4.1(i) of [14].) Following [14], we define the exterior region M 0 of M to be the metric completion of M r K. Thus (M 0,g,V) is a vacuum static data set, except for the requirement that {x 2 M 0 V (x) =0} 0. The exterior region M 0 has a strictly outward minimizing minimal boundary and no interior compact minimal surfaces (see [14]), where strictly outward minimizing means 0 is strictly less than the area of any other surface that encloses it. Suppose that M has a compact minimal surface other Since V > 0 away it follows from the definition of M 0 that V does not vanish identically 0. We consider the outward normal flow of surfaces ( ) t 0 with initial condition 0 0 that flows with speed V. Since V 0 does not vanish 0, this flow is nontrivial. According to the formula for the variation of mean curvature, we see that the mean curvature of evolves = V (Ric(, )+ A t 2 )V = ( gv r r V + hh, rv i) (r r V 3V + A t 2 V ) = hh, rv i A t 2 V, where is the outward unit normal, and A t is the second fundamental form. Since V 0, it follows that must have H apple 0 for all small t. By the first variation of area formula, must have area less than or equal to that of 0 0. But this contradicts the strictly outward minimizing property 0. We can now prove Theorem 2.3 following the description in [9]. Let (M,g,V ) be as in the statement of the theorem. In particular, all of the hypotheses of Chruściel-Simon s Theorem (Theorem 2.5) are satisfied and so m apple 0. Hence Proposition 2.4 and Lemma 2.6 imply that all of the hypotheses of our Penrose inequality (Theorem 1.3) are also satisfied. Since we are assuming that is homeomorphic to and is also the component with largest surface gravity, Theorem 2.5 tells us that A A 0 and m apple m 0, 10

11 where A is the area of and A 0 is the area 0 in the reference solution. Moreover, since the curvature k of ĝ is equal to 1 our Penrose inequality (Theorem 1.3) tells us that r A m(g 1) 3/2 1 g + A. (5) 16 4 It is convenient to define constants s s A A 0 r := and r 0 := 4 (g 1) 4 (g 1), so that we have r r 0. Inequality (5) then becomes m 1 2 r( 1+r2 ) =) 2 m + r r 3 0. Meanwhile, on the reference space we know that r 0 is the largest root of 2m 0 + r 0 r 3 0 =0. and so r 0 1 p3 using elementary reasoning. Thus 0 apple 2 m + r r 3 =(2 m + r r 3 ) (2m 0 + r 0 r 3 0) = 2( m m 0 )+(r r 0 ) (r 3 r 3 0) = 2( m m 0 )+(r r 0 )[1 (r 2 + rr 0 + r 2 0)] apple 2( m m 0 )+(r r 0 )(1 3r 2 0) apple 0. Therefore all of the inequalities must be equalities and so it follows from the rigidity part of Theorem 1.3 that (M,g) is isometric to the Kottler space with infinity and mass m 0. It is simple to check that any two static potentials on the Kottler space (M,g) must be proportional and so V is the usual static potential, up to a constant multiple. 3 Proof of Theorem 1.3 Assume that (M 3,g)beaC 2 asymptotically locally hyperbolic manifold with R 6 such is an outermost minimal surface. In order to establish existence of the weak inverse mean curvature flow, we first need to find a weak subsolution. 11

12 Lemma 3.1. Let (M 3,g) be a C 2 asymptotically locally hyperbolic metric with radial coordinate as in the definition of asymptotically locally hyperbolic. There exists r 0 so that for all r r 0, = { =(r + 1)e t/2 1} and + t = { =(r 1)e t/2 +1}, t 0 are, respectively, subsolutions and supersolutions for inverse mean curvature flow with initial condition { = r}. Proof. We will prove that + t is a supersolution. (The proof for is similar.) Using the asymptotics of g, one can see that the inverse mean curvature of the constant surface in (M,g) is H 1 = + O( 2 qk 2 ). + 2 Since that + t is just the constant surface with =(r 1)e t/2 + 1, we see that the speed of the flow is just d g = 1 2 (r 1)et/2 [(k+ 2 ) 1/2 +O( 3 )] = q 2 1 +O( k qk 2 ). + 2 Clearly, for su ciently large, this speed is less than H 1, showing that + t + t is a supersolution. Given Lemma 3.1, we may now apply Huisken and Ilmanen s Weak Existence Theorem 3.1 of [14] to find a weak solution u for inverse mean curvature flow with initial More precisely, u is a proper, locally Lipschitz nonnegative function u defined on M with u = 0 that satisfies a certain variational property (defined on page 365 of [14]). The surfaces <t} are C 1, and strictly outward minimizing, 4 as defined in the proof of Lemma 2.6. In particular, each is mean convex. There are only countably many jump times, that is, values of t for which <t} does not equal + t apple t}). In a nonrigorous sense, may be regarded as flowing by smooth inverse mean curvature flow, except when it ceases to be strictly outward minimizing, at which time it jumps to a strictly outward minimizing surface + t of equal area. In is not connected, Huisken and Ilmanen explained how one can single out a (spherical) component 0 as the initial surface while 4 Huisken and Ilmanen instead describe the region enclosed by as a strictly minimizing hull [14]. 12

13 treating the other components as obstacles. See Section 6 of [14] for details. Essentially, we arbitrarily fill in all other components to obtain a new space M and then run the weak inverse mean curvature flow in M with initial condition 0, except that whenever the surface is about to enter the filled-in region, we jump to a connected strictly outward minimizing surface F enclosing both and one or more of the filled-in regions. We then restart the flow with initial condition F. There is another important alteration introduced by Huisken and Ilmanen [14, Section 4]. We consider the exterior region M 0 of M, as defined in the proof of Lemma 2.6. was the outermost minimal surface of M, it follows is still part of the boundary of M 0,butnowthere might be more minimal boundary components. The exterior region M 0 is an improvement over M because it is completely free of compact minimal surfaces in its interior. We will actually run the weak inverse mean curvature flow in the exterior region of M rather than in M itself. So for our proof of Theorem 1.3, we may assume without loss of generality that M is an exterior region. 3.1 Monotonicity of inverse mean curvature flow Fix an integer g and recall our definition of the Hawking mass of a surface in (M,g) tobe r 1 m H ( ) := 1 g (H 2 4) The proof of the Geroch Monotonicity Formula 5.8 in [14] adapts straightforwardly to the locally hyperbolic setting to show the following: Theorem 3.2 (Huisken-Ilmanen). Let (M 3,g) be a complete C 2 asymptotically locally hyperbolic manifold with outermost minimal boundary, and let 0 be one of its boundary components. Let be a weak solution to inverse mean curvature flow (possibly with obstacles, as described above), with initial surface 0. Then for 0 apple <, if there are no obstacles between and, then m H ( ) m H ( ) 1 3/2 (16 ) "8 ( 1/2 ( )) 2 + 2(R + 6) + Å 2 +4H 2 rh 2 13 # dt, (6)

14 where := 2 form. 2g, and Å is the trace-free part of the second fundamental To make use of this theorem we need the following lemma controlling the topology of. Lemma 3.3. Let (M 3,g) be a complete C 2 asymptotically locally hyperbolic manifold, which is an exterior region. g M be a component with genus g, and let be a weak solution to inverse mean curvature flow (possibly with obstacles), with initial surface 0 g M. For all t, the surface is connected and has genus at least g. In particular, ( ). Proof. Let M be the manifold M with the obstacles filled in. Let u be the function defining the weak flow (possibly with obstacles). For each t>0, let M t be the closure of {x 2 M u(x) <t}, so M t = g M. We claim that M t is connected. If it were not connected, one of the components of Mt would be disjoint g M. By the variational property that characterizes u, one can deduce that u must be constant over. (See the proof of [14, Connectedness Lemma 4.2(i)].) Since M is connected, must meet, and thus u = t on, which is a contradiction to the definition of Mt. Since M t is connected, it follows that M t := M t \ M is connected. Note t = g M 1 M [ [@ k M,where@ 1 M,...,@ k M is some labeling of the other components that touch M t. The rest of the proof does not use inverse mean curvature flow. It is essentially a topological argument that relies only on the following facts about : There exists a connected manifold M t whose boundary is g 1 M [ [@ k M, is mean convex, g M,@ 1 M,...@ k M are the only compact minimal surfaces in M t. This last part is where we use the assumption that M is an exterior region. Let 1,..., ` be the connected components of. We minimize area in the isotopy class of 1 in M t. Note that Theorem 1 of Meeks, Simon, and Yau applies because M t has mean convex boundary, as explained in Section 6 of [16]. According to Theorem 1 and Remark 3.27 of [16], there exists some surface 1 obtained from 1 via isotopy and a series of -reductions such that each component of 1 is a parallel surface of a connected minimal surface, except for one component that may be taken to have arbitrarily small area. Recall that a -reduction is a surgery procedure that deletes an annulus and replaces it with two disks in such a way that the annulus and two disks bound a ball in M t. (See [16, Section 3] for details.) Note that -reduction preserves homology class. Since the only compact minimal surfaces in M t g M,@ 1 M,...@ k M, and because a surface of 14

15 small enough area must be homologically trivial, it follows that [ 1 ]=[ 1 ]=n 0 [@ g M] kx n i [@ i M]inH 2 (M t, Z), (7) i=1 for some integers n i. Using the long exact sequence for the pair (M t,@m t ), we have exactness of H 3 (M t,@m t H2 (@M t, Z)! H 2 (M t, Z). Since M t is connected, ker i must be generated t ]= `X [ i ] [@ g M] i=1 kx [@ i M], i=1 where i are oriented using the outward normal in M as usual. Since equation (7) says that [ 1 P ] [@ g M] k i=1 n i[@ i M] 2 ker i, it follows that must be connected and hence equal to 1. In particular, [ 1 ]=[@ g M]+ kx [@ i M]inH 2 (M t, Z). i=1 Since each component of 1 is either isotopic to one g M,@ 1 M,...@ k M (with some orientation) or is null homologous, and since there are no relations among [@ g M], [@ 1 M],...[@ k M]inH 2 (M t, Z), the previous equation implies that at least one component of 1 is isotopic 1 gm. Finally, since -reduction can only reduce the total genus of all components of a surface, we know that has genus at least as large as that g M. Note that if we apply the reasoning in the proof of Lemma 3.3 above to the conformal infinity of M, we see that the genus of is at least as large as the genus of any component Compare this to [11], where it is shown that in a spacetime obeying topological censorship, the horizon must obey the same property. Corollary 3.4 (Geroch monotonicity). Let (M 3,g) be a complete C 2 asymptotically locally hyperbolic manifold, which is an exterior region, and assume R 6. g M be a component with genus g, and let be a weak solution to inverse mean curvature flow (possibly with obstacles), with initial surface 0 g M. Then the Hawking mass of is nondecreasing in t. 15

16 Proof. The result follows immediately from Theorem 3.2 and Lemma 3.3 in the absence of obstacles. Since there are only finitely many obstacles, all that is left to show is that the mass cannot drop when we jump over an obstacle. Let t be the first time that we have to jump over an obstacle, and let F denote the strictly outward minimizing surface that jumps to. Then we know from [14, Equation 6.1] that apple F and HF 2, H 2 where the second inequality essentially follows from the fact that the strictly minimizing hull of a surface should be minimal away from where it agrees with the original surface. In the case g < 2, these inequalities combine with nonnegativity of m H ( ) (from monotonicity in the absence of obstacles) to immediately show that m H ( ) apple m H (F ), just as in the asymptotically flat case [14, Section 6]. To handle the case g 2 we proceed as follows. m H (F ) m H ( )= F 1/2 m H(F ) F 1/2 1/2 m H( ) 1/2 =( F 1/2 1/2 ) m H( ) + F 1/2 1/2 F mh (F ) m H ( ) F 1/2 1/2 =( F 1/2 1/2 ) m H( ) 1/2 + F 1/2 (16 ) 3/2 (HF 2 4) + (H 2 t 4) F ( F 1/2 1/2 ) m H( ) 1/2 + F 1/2 (16 ) 3/2 (4 F 4 ) ( F 1/2 1/2 ) m H( ) 1/2 + 1/2 (16 ) 3/2 (4 F 4 ) =( F 1/2 1/2 mh ( ) ) 1/2 + 4(16 ) 3/2 1/2 ( F 1/2 + 1/2 ) ( F 1/2 1/2 mh ( ) ) 1/2 + 8(16 ) 3/2. Therefore it only remains to show that m H ( ) 8(16 ) 3/2 3/2. Note that for any A 0, r A 1 g + 1 g 1 3/ A. 3 16

17 Taking A g M and using monotonicity in the absence of obstacles, we then have g 1 3/2 m H ( ) m H (@ g M). 3 On the other hand, observe that a stable compact minimal surface of genus g in a 3-manifold with R 6 must have area at least 4 3 (g 1). (This follows from a standard computation using the second variation of area, see [21, Section 2]). Thus 8(16 ) 3/2 3/2 8(16 ) g M 3/2 8(16 ) 3/2 apple 4 3 (g 1) 3/2 which completes the proof. g 1 3/2 =, The long-time limit of inverse mean curvature flow First we compute the asymptotics of Ricci curvature for asymptotically locally hyperbolic manifolds. Proceeding as in Lemma 3.1 of [20], we deduce the following: Lemma 3.5. Let (M 3,g) be C 2 asymptotically locally hyperbolic, with radial r coordinate. If r,e 1,e 2 is an orthonormal frame at a point in M, then r Ric r, r = 2 2µ 3 + o( 3 ) r Ric(e i,e j )= 2 ij + O( 3 ) r Ric r,e i = o( 3 ) R = 6+o( 3 ). In order to make certain computations easier, we consider a conformal compactification g = 2 g of the exterior region of (M,g). Then if we set s = 1,wehave g = ds 2 +ĝ + Q, on the space (0,s 1 ) for small enough s 1,where Q + s r Q + s 2 r Q = O 2 (s 2 ). 17

18 Lemma 3.6. There is a constant C such that for su ciently large t, the coordinate on lies in ( 1 C et/2,ce t/2 ) and the s coordinate on must lie in ( 1 C e t/2,ce t/2 ). Proof. This follows immediately from the subsolutions and supersolutions of inverse mean curvature flow described in Lemma 3.1. We will use the following area bound repeatedly. Lemma 3.7. Let denote the surface endowed with the metric induced from g. Theareaof is uniformly bounded in time. Proof. In the absence of obstacles, this follows immediately from the fact that = 0 e t, the definition of g, and the previous Lemma. The presence of obstacles does not change the conclusion because can only jump a finite number of times. Lemma 3.8. (H 2 4) is uniformly bounded above and below. Proof. The upper bound follows easily from Corollary 3.4: Since r t 1 m H ( 0 ) apple m H ( )= 1 g (H 2 4), we have s! (H ) apple 16 1 g m H ( 0 ) = 16 (1 g)+o(e t/2 ). To prove the lower bound, we consider the Gauss-Codazzi equations in (M,g): 2 Å 2 +4K = H 2 +2R 4Ric(, ) (8) Performing the same computation in the compactified metric g and using the fact that integral of the left-hand side is conformally invariant, we see that (H 2 +2R 4Ric(, )) = ( H 2 +2 R 4Ric(, g )). By Lemma 3.5, we see that (H 2 4) = ( H 2 +2 R 4Ric(, g )) + O(e t/2 ) C + H2 C, (9) for some constant C independent of t, where we used the fact that the curvature of g is bounded. 18

19 Lemma 3.9. We have r T s 2 = O(e t/2 ). Proof. r T s 2 = s t s = s( gs r r s + h H,@ s i) t apple s gs r r s + s H rs t = O(s 2 )+ H O(s) 1/2 apple O(e t )+O(e t/2 ) H2, where we used the Hölder inequality in the last line. The result now follows because inequality (9) states that apple C + (H 2 4), while Lemma 3.8 implies that the right-hand side is bounded. Lemma = + O(e t/2 ). Proof. Choose to be the inward pointing normal of.wewillfirstshow that apple + O(e t/2 ). Using the fact that s is approximately a distance function with respect to g, rs 2 =1+O(s 2 ), and therefore = ( rs 2 + O(s 2 )) = ( r T s 2 + r N s 2 )+O(e t ) = (s) 2 + O(e t/2 ), where is the normal vector to with respect to the g metric, and the last line follows from Lemma 3.9. So it su ces to estimate t (s) 2 in terms of. 19

20 We now divide into three parts: A t = {x 2 (s) apple e t/4 }, B t = {x 2 e t/4 < (s) apple 0} and C t = {x 2 0 < (s)}. Again, since s is approximately a distance function, (s) apple1+o(s 2 ), and therefore, using the definitions of B t and C t, we can estimate (s) 2 = (s) 2 + (s) 2 + (s) 2 A t B t C t apple (1 + O(e t )) A t + e t/2 B t +(1+O(e t )) (s) C t apple A t + (s)+o(e t/2 ), where we used the boundedness of to obtain the last line. If we take M t to be the region of (0,s 1 ) so t = {0} [,thenthe divergence theorem tells us that (s) = (s)+ s = + O(e t ). M t Thus {0} (s) 2 apple A t + + O(e t/2 ). (10) It remains to estimate A t. The mean curvature changes under under the conformal change g = s 2 g according to the formula H = s H +2 (s). Since evolves by inverse mean curvature flow, we know that H 0. Therefore s (s) H 2. So the definition of A t tells us that e t/4 (s) s H 2 on A t. Squaring this and integrating over A t gives e t/2 s apple A t At 2 H2 4 =) A t appleo(e t/2 ) H2. 20

21 As mentioned in the proof of Lemma 3.9, t H2 is bounded, and therefore this inequality combined with inequality (10) gives us (s) 2 apple + O(e t/2 ), completing the proof of one side of the inequality. The reverse inequality follows from the fact that is within error O(e t ) from the area of as measured in the product metric ds 2 +ĝ on (0,s 1 ), and the projection map of the product metric onto is areanonincreasing. Lemma There exists a sequence of times t i such that lim Å 2 =0. i!1 i Proof. We use formula (5.22) from [14]: For <,wehave H 2 H DH 2 H 2 +2 A 2 + 2Ric(, ) H 2 dt H 2 + (2 Å 2 + 2Ric(, )) Since d dt =, it follows that (H 2 4) (H 2 4) + The last term is easily bounded: 2(Ric(, ) + 2) dt = (2 Å 2 + 2(Ric(, ) + 2)) dt. O( 3 ) dt = O(e t/2 ) dt = O(e /2 ) Hence, 2 Å 2 dt apple (H 2 4) (H 2 4) + O(e /2 ). Since (H 2 4) is bounded above and below by Lemma 3.8, the integral on the left-hand side is bounded as!1, completing the proof. 21

22 Lemma Using the same sequence as in Lemma 3.11, we have (i )= ( ) for su ciently large i. Proof. Note that by Lemma 3.3, we already know that (i ) apple ( ), so we need to show that (i ) ( ) for all i large. The Gauss-Codazzi equations in the compactified metric tell us that 2 Ã 2 +4 K = H 2 +2 R 4 g Ric(, ), where is the normal vector to the surface. If we apply this to the surfaces {s} we obtain that 2 R 4Ric( g rs, rs) =4k, (11) lim s!0 where k is the constant Gaussian curvature of (, ĝ). Decompose = a rs + v, rs so that v is orthogonal to rs. Wehave v = r T s rs = r T s + O(s 2 ). Integrating Gauss-Codazzi equations on and recalling Lemma 3.9 we obtain 2 Ã 2 +8 ( )= H2 +2 R 4Ric(, g ) 2 R 4Ric(, g ) t = (2 R 4a 2 Ric( g rs, rs) 8Ric( g rs, v) 4Ric(v, g v)) + O(s 2 ) (2 R 4Ric( g rs, rs) C( r T s + r T s 2 )) + O(s 2 ) = (2 R 4Ric( g rs, rs)) + O(e t/4 ). Thus, using (11), Lemma 3.10, and Lemma 3.11, we obtain lim 8 ( i ) 4k =8 ( ). i!1 This implies ( i ) ( ) for all i su ciently large. Lemma If m apple 0, then lim m H( ) apple t!1! 3/2 3/2 µ 2/

23 Proof. Using Gauss-Codazzi equations, Gauss-Bonnet Theorem, and Lemma 3.5, we have r t m H ( )= 16 s = (16 ) 3 s = (16 ) 3 = s (16 ) 3 1 g 16 (1 g) 8 ( ) 8 ( ) 8 ( ) 1 (H 2 4) 16 ( 2R +4K t + 4Ric(, )+ Å 2 4) (4K t + 4(Ric(, ) + 2) + Å 2 + o( 3 )) Å 2 4 (Ric(, ) + 2) + o(e t/2 ) If we choose t to be one of the times from the sequence described in Lemma 3.12, we have m H ( ) apple 1 2 (4 ) 3/2p (Ric(, ) + 2) + o(1). (12) Decompose = a r r + v so that v is orthogonal to r. We have v = rt r while a 2 =1 v 2. It follows from Lemma 3.5 that Note that Ric(, ) =a 2 ( 2 2µ 3 )+ v 2 ( 2+O( 3 )) + o( 3 ) v 2 3 = = 2 2µ 3 + v 2 O( 3 )+o( 3 ). (13) r T 2 3 = r r T 2 s s 3 rs = r! 2 T s rs s = O(e t ), where we used Lemma 3.9 in the last line. Putting this together with formulas (12) and (13) we obtain m H ( ) apple (4 ) 3/2p µ 3 + o(1) 23

24 1/2 =(4 ) 3/2 s 2 µs + o(1). (14) We use the Hölder inequality with p = 3 and q =3/2 to see that µ 2/3 = s 2/3 (µs) 2/3 1/3 2/3 apple s 2 µ s. Taking the 3/2 power of both sides, and using the fact that µ apple 0, we have 3/2 1/2. µ 2/3 s 2 µs Combining this with inequality (14) to conclude that 1 3/2 m H ( ) apple µ 2/3 + O(e t/2 ), 4 for any sequence of t s as in the previous lemma. Now the result follows from taking the limit as this sequence approaches infinity and applying monotonicity of m H ( ) and Lemma Proof of Theorem 1.3. Recalling that ĝ has constant curvature 1, 0, or 1 on and our normalizaton of area in the k = 0 case, we have =4 max(1, g 1). Using Geroch Monotonicity (Corollary 3.4) and the previous lemma, we can now conclude! 3/2 3/2 m H (@ g M) apple lim m H ( ) apple µ 2/3 t!1 4 apple m(max{1, g 1}) 3/2, where we used the fact that µ apple m apple 0. All that remains is to prove the rigidity. We assume all of the hypotheses of Theorem 1.3, as well as equality in the Penrose inequality. By Corollary 3.4, the Hawking mass m H ( ) must be constant, and moreover, by equation (6), 2(R + 6) + Å 2 +4H 2 rh 2 =0. (15) 24

25 As argued on page 422 of [14], it follows that H is a positive constant on each, each is smooth, and there are no jump times. By the Smooth Start Lemma 2.4 of [14], is a classical solution of inverse mean curvature flow foliating the manifold M. This allows us to think of inverse mean curvature flow as defining a di eomorphism from [0, g M to M. (In must be connected.) We can write the metric as g = H 2 dt 2 + g t. Note that equation (15) implies that R = 6 everywhere, and Å = 0 on each. In particular, A = H 2 g and t g t = 2 H A = g. Hence g = H 2 dt 2 + e t g 0. (16) Recall that dh dt = H 1 ( A 2 +Ric(, ))H 1, which implies Ric(, ) is constant on each and so, by the Gauss-Codazzi equations, so is the Gauss curvature K. In particular, 0 has constant curvature. From here it is clear that (M,g) must be the Kottler space, since Kottler spaces are the unique asymptotically locally hyperbolic manifolds with R = 6 of the form g = f(r) 2 dr 2 + r 2 g 0,whereg 0 is a constant curvature metric on 0. Proof of Theorem 1.5. Assume the hypotheses of Theorem 1.5. If there exists at least one compact minimal surface in M, then the result follows from Corollary 1.4 applied to the exterior region of M. So let us assume that M contains no compact minimal surfaces, and suppose that m <0. In particular, the mass aspect µ is everywhere negative. We can consider a weak inverse mean curvature flow whose initial condition is e ectively a point, just as Huisken and Ilmanen did in [14, Section 8]. Then the same arguments used to prove Theorem 1.3 show that 0=lim t!0 m H ( ) apple lim t!1 m H ( ) apple m, where the Hawking mass here has g = 0. Rigidity follows according to an argument similar to the one used in Theorem 1.3 above. 4 The results of Chruściel-Simon: Proof of Theorem 2.5 Assume the hypotheses of Theorem 2.5. In particular, we have a static data set (M,g,V ) and a reference Kottler space (M 0,g 0,V 0 ) that has the same 25

26 surface gravity. We also have the important assumption that m 0 apple 0. We define W = rv 2, and also a reference function W 0 on M as follows: On the reference space M 0, rv 0 2 is constant on each level set of V 0 and, because of this, one may regard it as a function composed with V 0. That is, there exists a single-variable function! with the property that rv 0 2 =!(V 0 ) as functions on M 0. We define the function W 0 on M to be the function W 0 =!(V ). q Recall the formula V 0 = r 2 + k 2m r for an appropriate coordinate r on M 0. Inverting this formula, we may think of r as some function of V 0. In a manner similar to the way we defined W 0, we can define a reference function r on M by composing this function with V. (It might be logical to call this function r 0, but that is unnecessary because there is no ambiguity here.) Lemma 4.1. Under the hypotheses of Theorem 2.5 and using the notation introduced above, if we consider an open set in M on which W does not vanish, then on that open set W W 0 satisfies the elliptic inequality (W W 0 )+h,r(w W 0 )i + (W W 0 ) 0, where is a smooth vector field and is a smooth function whose sign is the opposite of m 0. This is the critical ingredient of the proof of Theorem 2.5 and it corresponds to equation (VII.15) in [9]. Proof. Since rv 6= 0, the level sets of V are smooth surfaces which we denote V. We introduce the notation Note V V = rv rv 2 = rv W. has the nice property that if f is a single-variable function, Recall the static V (f(v )) = f 0 (V ). gv =3V and Ric(g) = 1 V Hess(V ) 3g. 26

27 Also note that rw = 2 Hess V (rv, ) and thus V W = 2 Hess V rv, rv = 2 Hess V (rv,rv ). rv W Choose an orthonormal frame e 1,e 2,e 3 such that e 3 = rv rv. Starting with the Bochner formula and the static equations above, we have W =2 Hess V 2 + 2Ric(rV,rV )+2hr( V ), rv i 3X =2 Hess V (e i,e j ) V Hess V (rv,rv ) i,j=1 =2 Hess V (e 3,e 3 ) i,j=1 =2 Hess V 2X Hess V (e 3,e i ) 2 i=1 2X Hess V (e i,e j ) V Hess V (rv,rv ) +2 rv 2 A V rv rv, rv rv + 4 rv 2 2 rv 2 V Hess V 2X Hess V (rv,e i ) 2 i=1 rv rv, rv rv = 1 2 (@ V W ) W rt W 2 +2W ( Å V H2 V )+ W V W = 1 2 (@ V W ) 2 + WH 2 V + W V W + 1 W rt W 2 +2W Å V 2 = 1 2 (@ V W ) 2 + [Hess V (e 1,e 1 ) + Hess V (e 2,e 2 )] 2 + W V W + 1 W rt W 2 +2W Å V 2 = 1 2 (@ V W ) 2 +[ V Hess V (e 3,e 3 )] 2 + W V W + 1 W rt W 2 +2W Å V 2 = 1 2 (@ V W ) 2 +[3V Hess V (e 3,e 3 )] 2 + W V W + 1 W rt W 2 +2W Å V 2 = 1 2 (@ V W ) 2 +[3V 1 V W ] 2 + W V W + 1 W rt W 2 +2W Å V 2 = 3 4 (@ V W ) 2 3V@ V W +9V 2 + W V W + 1 W rt W 2 +2W Å V 2 27

28 Meanwhile, W 0 =div(r(!(v ))) = div(! 0 (V )rv ) =! 00 (V )W +! 0 (V ) V =! 00 (V )W +3! 0 (V )V. Notice that both of the computations above provide a way to compute 0(!(V 0 )) in the reference space M 0. Since!(V 0 ) is constant on level surfaces of V 0 and since those level surfaces are umbilic, the first computation yields 0(!(V 0 )) = 3 V 0 (!(V 0 )) 2 3V@ V0 (!(V 0 )) + 9V 2 0 +!(V 0) V V0 (!(V 0 )) = 3 4!0 (V 0 ) 2 3V! 0 (V 0 )+9V 2 0 +!(V 0) V 0! 0 (V 0 ). Meanwhile, if we do the second computation in the reference space, we find 0(!(V 0 )) =! 00 (V 0 )!(V 0 )+3! 0 (V 0 )V 0. Equating the right-hand sides of the two equations above, we actually obtain a di erential equation for the single-variable function!, which means that we can replace the V 0 by V to obtain the following equation on the original space M:! 00 (V )!(V )+3! 0 (V )V = 3 4!0 (V ) 2 3V! 0 (V )+9V 2 +!(V ) V!0 (V )! 00 (V )W 0 +3! 0 (V )V = 3 V W 0 2 3V@ V W 0 +9V 2 + W 0 V W 0. Picking up from our expression for W 0 above, we now have W 0 =! 00 (V )W +3! 0 (V )V =! 00 (V )W 0 +3! 0 (V )V +! 00 (V )(W W 0 ) = 3 V W 0 2 3V@ V W 0 +9V 2 + W 0 V W 0 +! 00 (V )(W W 0 ). Now we can subtract this expression from our expression for for the last two nonnegative terms which we ignore) to find W (except 3 (W W 0 ) 4 (@ V W ) 2 3V@ V W +9V 2 + W V W 3 V W 0 2 3V@ V W 0 +9V 2 + W 0 V W 0! 00 (V )(W W 0 ) = 3 V (W + W 0 )@ V (W W 0 ) 3V@ V (W W 0 ) 28

29 + W V W W V W 0 + W W V W 0 0 V W 0! 00 (V )(W W 0 ) = 3 V (W + W 0 ) 3V + W (W W 0 )+(W W 0 )!0 (V ) V! 00 (V )(W W 0 ) = 3 V (W + W 0 ) 3V + W (W W 0 )+[!0 (V ) V! 00 (V )](W W 0 ). It only remains to compute the sign of the zero order coe the formula V 0 = r 2 + k 2m 1/2 0 r cient. Using the in M 0, we can perform simple computations using the chain rule to show that!(v )= r + m 0 2. r 2 and p dv!(v ) dr =. V Therefore! 0 2m 0 (V )=2V 1 r 3, and Thus! 00 (V )=2V 1 =!0 (V ) V 2m 0 r 3 +2V 6m 0 dr r 4 dv + 12V 2 p!(v ) m 0 r 4.! 0 (V ) V! 00 (V )= 12V 2 p W0 m 0 r 4. So we see that as long as m 0 apple 0, the coe cient of W W 0 is nonnegative. Corollary 4.2. Under the hypotheses of Theorem 2.5 and using the notation introduced above, we have W apple W 0 on M. Proof. By the definition of W 0, we know that W = W 0 apple M, and W apple W 0 on all apple M was assumed to have the largest surface gravity of any boundary component. We also know from Proposition 2.4 that W W 0 = 0 at the conformal infinity. (We will see a more detailed calculation below.) Suppose that W>W 0 somewhere in M. SinceW apple W 0 at the boundary and also at infinity, the function W W 0 must achieve 29

30 its positive maximum value at some point p in the interior of M. Since W (p) >W 0 (p) > 0, we see that the previous lemma applies to some open set U containing p. By the maximum principle for the elliptic inequality given by the Lemma, W W 0 cannot have a local maximum in U, whichis a contradiction. Proof of Theorem 2.5. We claim that for any point p apple M, where is the Gaussian curvature apple M at p, is the constant Gaussian curvature of 0 in the reference space. Integrating this inequality and using the Gauss-Bonnet Theorem, the claim clearly implies that (@ apple M) (@M 0 apple apple M 0, so we now focus on proving the claim. For the following, we use the same notation as in the proof of Lemma 4.1. Since W 6= 0 the vector V is well-defined apple M, and we can consider the flow ' v generated V apple M. Choose a point p apple M. Applying Taylor s Theorem to the function W restricted to the flow line starting at p, we have W (' v (p)) = W (p)+[(@ V W )(p)]v +[(@ 2 V W )(p)]v 2 + O(v 3 ). We know that W (p) =apple 2,whereapple is the surface gravity apple M. V W = 2 Hess V (rv,@ V ) = 2 Hess V (e 3,e 3 ) = V (2Ric(e 3,e 3 ) + 6) = 0, where the last identity follows because V vanishes Following same reason and using the Gauss-Codazzi equations plus the fact is totally geodesic, we 2 V W = (2Ric(e 3,e 3 ) + 6) + V@ V (2Ric(e 3,e 3 ) + 6) Thus = 2Ric p (e 3,e 3 )+6=R W (' v (p)) = apple O(v 3 ). Performing the same computation in the reference solution, we obtain W 0 (' v (p)) = apple 2 v 2 + O(v 3 ). 30

31 The claim now follows from Corollary 4.2. We now focus on proving the mass inequality µ apple m 0. Proposition 2.4 that we have V 2 = 2 + k 4 3 µ 1 + o 1 ( 1 ), Recall from where is a coordinate as in the definition of asymptotically locally hyperbolic. Di erentiating this, we find 2V rv =( µ 2 )r + o( 1 ). Taking the norm-square of both sides, and using the asymptotically locally hyperbolic property, Then V 2 W =( µ 2 ) 2 r 2 + o( ) W = =( µ 1 )( 2 + k)+o( ) = 4 + k µ + o( ). 4 + k µ + o( ) 2 + k 4 3 µ 1 + o( 1 ) = 2 1+k µ 3 1+k µ 3 + o( 1 ) = µ 1 + o( 1 ). Recall that we also have r 2 + k 2m 0 = V 2 = 2 + r k 4 3 µ 1 + o 1 ( 1 ), by definition of the function r. In particular, r 1 = 1 + o( 1 ). By changing variables from r to, wehave W 0 = r + m 0 2 = r 2 r 2 +2m 0 r 1 + O(r 4 ) =(r 2 + k 2m 0 r 1 ) k +4m0 r 1 + O(r 4 ) = µ 1 +4m o( 1 ). Comparing these asymptotic expansions for W and W 0 and using Corollary 4.2, we see that as!1,wehave from which the result follows. 8 3 µ apple 4 3 µ +4m 0, 31

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