Monotonicity of the Yamabe invariant under connect sum over the boundary

Transcription

1 Ann Glob Anal Geom (2009) 35: DOI /s ORIGINAL PAPER Monotonicity of the Yamabe invariant under connect sum over the boundary Fernando Schwartz Received: 6 June 2008 / Accepted: 18 August 2008 / Published online: 5 September 2008 SpringerScience+BusinessMediaB.V.2008 Abstract We show that the Yamabe invariant of manifolds with boundary satisfies a monotonicity property with respect to connected sums along the boundary, similar to the one in the closed case. A consequence of our result is that handlebodies have maximal invariant. Keywords Manifold with boundary Positive scalar curvature Yamabe invariant 1 Introduction, main results, and sketch of proof The Yamabe invariant of a closed manifold is a smooth invariant. It is a minimax of the total scalar curvature functional over the space of unit volume metrics on the manifold. More precisely, one computes the infimum of the functional over a fixed conformal class, and then the supremum of the infimums is computed over the set of all conformal classes. The Yamabe invariant has been extensively studied. It is known that whenever the minimax is achieved the manifold admits an Einstein metric, which is an important question in geometry. When the invariant is positive it is very difficult to determine whether it is achieved or not. Furthermore, very few invariants have been computed in the positive case. The analysis of the Yamabe invariant can also be carried out in the realm of manifolds with boundary. The corresponding quantity to study is a quotient of the energy, which in this case is the total scalar curvature plus the total mean curvature of the boundary, divided by a normalization term. Since one can choose to normalize with respect to the total volume of the manifold or the total area of its boundary, two possible definitions for the invariant appear. F. Schwartz (B) Mathematics Department, Duke University, Durham, NC 27710, USA fernando@math.duke.edu Present Address: F. Schwartz Mathematics Institute, Zeeman Building, University of Warwick, Coventry CV4 7AL, UK

2 116 Ann Glob Anal Geom (2009) 35: Let (M n, g) be a compact manifold with boundary. Proceeding in analogy with the closed case, we define the (conformally invariant) Yamabe constant of (M, g) by E( g) Y λ (M, [g]) = inf g [g] N λ ( g), (1) where E( g) is the energy of the metric g and λ {0, 1} determines whether the normalization N λ is with respect to the volume of g (λ = 1) or the area of the boundary of g (λ = 0). It is a well known fact that Y λ (M n, [g]) Y λ (S n +, [g 0]), whereg 0 denotes the round metric on the hemisphere. Existence of minimizers of the above quotient have been studied by Escobar in [5,6], where he proved that solutions exist in most cases (See also [4,9,10]). A standard argument shows that, when writing Eq. 1 in terms of conformal factors of conformally related metrics (see Eqs. 5, 6), solutions of the Euler-Lagrange equations correspond to conformally related metrics that solve the Yamabe problem on (M, g). This is, conformal metrics with constant scalar curvature and minimal boundary (in the case λ = 1), and scalar flat metrics with boundary having constant mean curvature (whenever λ = 0). The Yamabe invariant of a smooth manifold with boundary is defined as σ λ (M) = sup Y λ (M, [g]), (2) [g] C where C is the set of all smooth conformal classes of metrics on M. From the considerations above it follows that < σ λ (M n ) σ λ (S n + ). In this paper we prove the following result. Theorem 1.1 Let M 1, M 2 be smooth n-manifolds with boundary, n 3.LetM 1 #M 2 denote their connected sum along the boundary. Then, for λ {0, 1}, σ λ (M 1 #M 2 ) σ λ (M 1 M 2 ), (3) where M 1 M 2 denotes the disjoint union of M 1, M 2 and { σ λ (M 1 M 2 ) = ( σ λ (M 1 ) n 2 + σ λ (M 2 ) n 2 ) n 2 σ λ (M 1 ), σ λ (M 2 ) 0, min{σ λ (M 1 ), σ λ (M 2 )} otherwise. (4) Our result is an extension, to manifolds with boundary, of Kobayashi s classical theorem about monotonicity of the Yamabe invariant over connect sums of closed manifolds [8]. The proof of our result parallels his proof with the added boundary ingredients, plus some approximation results that are adaptations of some found by Akutagawa and Botvinnik [1]. An interesting consequence of our result is the following corollary. Corollary 1.2 Handlebodies have maximal invariant. This is, if H n is the result of attaching a finite number of handles to the n-ball, then σ λ (H) = σ λ (S+ n ), λ {0, 1}. Sketch of the Proof of Theorem 1.1 The proof of the theorem is divided into two cases: (a) when the Yamabe invariant of both M 1 and M 2 is positive, and (b) when one of the invariants is nonpositive. The idea behind the proof of (a) is to show that if both M 1, M 2 have positive metrics, then after gluing a long enough handle over the boundary of M 1 M 2 one obtains a manifold whose Yamabe constant is larger than the Yamabe constant of M 1 M 2 minus a small error. Therefore, the invariant of M 1 #M 2 is no less than the invariant of M 1 M 2. To prove (b) we endow M 1 and M 2 with some particular negative metrics, which

3 Ann Glob Anal Geom (2009) 35: can be done by a classic result of Kazdan and Warner. Using some approximation lemmas (found in the appendix) we prove that the Yamabe constant of the connect sum is bounded below by the appropriate quantities in terms of the Yamabe constants of M 1 and M 2. 2Preliminaries Let (M n, g) be a Riemannian manifold of dimension n 3. The energy of u C (M) with respect to g is defined as E g (u) = M ( g u 2 + n 2 ) 4(n 1) R gu 2 dv g + n 2 h g u 2 da g. (5) 2 M Using the transformation laws for the scalar curvature and the mean curvature, it is standard to note that the energy of u is the total scalar curvature plus the total mean curvature of the conformally related metric g = u 4/(n 2) g,i.e.,e g (u) = E( g) = M R gdv + n 2 2 M h gda. For λ {0, 1}, the normalization factor is just ( N g,λ (u) = λ u n 2 2n dv g + (1 λ) M M ) n u 2(n 1) n 1 n 2 da g, (6) It follows that N g,λ (u) = N λ ( g) = λvol(m, g)+(1 λ)(area( M, g) n/(n 1) ), as desired. This way, we get that Y λ (M, [g]) = inf { Eg (u) inf } { N g,0 (u) : u 0 on M } Eg (u) N g,1 (u) : u 0 on M if λ = 0, if λ = 1. Outline of the Proof of Theorem 1.1 The simple argument below shows how Eq. 4 of the theorem is satisfied. The proof of Eq. 3 is split into two cases. In Sect. 3 we deal with the case when both σ λ (M 1 ) and σ λ (M 2 ) are positive, where we prove that Eq. 3 follows from the (more general) Handle Monotonicity Theorem (Theorem 3.1). In Sect. 4 we study the case when at least one of σ λ (M 1 ), σ λ (M 2 ) is nonpositive. In Sects. 3 and 4 we use a series of approximation arguments. These are modifications of the approximation lemmas that appear in Kobayashi s original paper [8]andAkutagawa and Botvinnik s paper [1]. We include their proofs, for sake of completeness, in the Appendix. Lemma 2.1 Let λ {0, 1} and (M 1, g 1 ), (M 2, g 2 ) be Riemannian manifolds with boundary. The following holds: Y λ (M 1 M 2, g 1 g 2 ) { = ( Y λ (M 1, g 1 ) n 2 + Y λ (M 2, g 2 ) n 2 ) n 2 Y λ (M 1, g 1 ), Y λ (M 2, g 2 ) 0, min{y λ (M 1, g 1 ), Y λ (M 2, g 2 )} otherwise.

4 118 Ann Glob Anal Geom (2009) 35: Proof Write E 1 (u), V 1 (u) and E 2 (u), V 2 (u) for the expressions in Eqs. 5 and 6 integrated over M 1, M 1 and M 2, M 2, respectively. We have that Y λ (M 1 M 2, g 1 g 2 ) = inf { E 1 (u) + E 2 (u) : u C 1 ( M) with V 1 (u) + V 2 (u) = 1 } { } = inf V 1 (u) n 2 E 1 (u) n + V V 1 (u) n 2 2 (u) n 2 E 1 (u) n : V n V 1 (u) n 2 1 (u) + V 2 (u) = 1 n } = inf {α n 2 n Yλ (M 1, g 1 ) + (1 α) n 2 n Yλ (M 2, g 2 ) : 0 α 1 = { ( Y λ (M 1, g 1 ) n 2 + Y λ (M 2, g 2 ) n 2 ) n 2 Y λ (M 1, g 1 ), Y λ (M 2, g 2 ) 0, min{y λ (M 1, g 1 ), Y λ (M 2, g 2 )} otherwise. As a direct consequence of the lemma is the proof of Eq. 4. Corollary 2.2 Let M1 n, Mn 2, n 3 be Riemannian manifolds with boundary. The following holds: σ λ (M 1 M 2 ) = { ( σ λ (M 1 ) n 2 + σ λ (M 2 ) n 2 ) n 2 σ λ (M 1 ), σ λ (M 2 ) 0, min{σ λ (M 1 ), σ λ (M 2 )} otherwise. 3Proofofthepositivecase The proof of Theorem 1.1 is split in two cases: the positive case, which we present here, and the nonpositive case (where at least one of σ λ (M 1 ), σ λ (M 2 ) is nonpositive), which we prove in Sect. 4 below. Throughout this section we assume that M 1, M 2 are n-dimensional, compact manifolds with boundary (n 3) with positive Yamabe invariant. We actually prove here a more general theorem, valid only for the positive case. We show that the Yamabe invariant is monotonic when attaching a handle over the boundary. In particular, from this it follows that all handlebodies have maximal Yamabe invariant. Theorem 3.1 (Handle monotonicity) Let M be a compact manifold with boundary and positive Yamabe invariant (σ λ (M) >0). Denote by M the manifold obtained by attaching a handle to the boundary of M. Then σ λ ( M) σ λ (M). The handle monotonicity theorem has two immediate consequences. Proof of Theorem 1.1 in the Positive Case Assuming that σ λ (M 1 ), σ λ (M 2 ) > 0, apply Theorem 3.1 to M 1 M 2 when attaching a handle between a boundary component of M 1 and a boundary component of M 2.TogetherwithCorollary2.2 this gives σ λ (M 1 #M 2 ) σ λ (M 1 M 2 ) = min{σ λ (M 1 ), σ λ (M 2 )}. Proof of Corollary 1.2 Use the Handle monotonicity theorem on B n, taking into consideration that σ λ ( B n ) = σ λ (S+ n ) is maximal.

5 Ann Glob Anal Geom (2009) 35: We now give the proof of the main theorem of this section. Proof of Theorem 3.1 Fix ɛ > 0small.Letg be a metric on M so that 0 < Y λ (M) ɛ < Y λ (M, g ). Pick p 1, p 2 M.ByLemmaA.8 there is a metric g close to g so that in a neighborhood of p 1 and p 2, g is isometric to a fixed neighborhood B λ (q, ɛ ) of q Sλ n,forsomeɛ > 0. The Lemma also gives that R g 0 on M. There exists a function λ C (M {p 1, p 2 }) which is 1 outside B λ (q, ɛ /2) and such that the metric g = e λ g is isometric to the half infinite cylinder [0, ) S+ n 1 around the removed points. For convenience we write (M {p 1, p 2 }, g) =[0, ) S n 1 + ( M, g) [0, ) S n 1 +, where M is the complement of the two cylinders. We glue ( M, g) with [0, l] S+ n 1 to get a smooth Riemannian manifold ( M, g l ). This is, M is the result of attaching a handle on M connecting neighborhoods of p 1 and p 2.We have the follwing decomposition: ( M, g l ) = ( M, g) [0, l] S n 1 +. (7) Since Y λ ( M, g l ) = inf u M ( u 2 + ( λ 2n M u n 2 dv gl + (1 λ) 4(n 1) n 2 R g l u 2 )dv gl + n 2 2 ( M M h g l u 2 da gl ) n 2(n 1) n 1 u n 2 da gl ) n 2 n we can take a positive function u l C ( M) such that ( u l 2 + R gl ul 2 )dv g l + n 2 h gl ul 2 M 2 dσ l < Y λ ( M, g l ) + 1 M 1 + l (8) and ( λ u l n 2 2n dv gl + (1 λ) M M ) n u l 2(n 1) n 1 n 2 da gl = 1. (9) Lemma 3.2 There exists a section, say {t l } S+ n 1 in the cylindrical part of M (see Eq.7) such that {t l } S n 1 + where A is a constant independent of l. Proof Let ( u l 2 + ul 2 )dv l + u 2 {t l } S n 2 l da l < A l, A = min{0, min h g (x)}area( M, g) n 1 1 min{0, min R g (x)}vol( M, g) n 2. x M x M

6 120 Ann Glob Anal Geom (2009) 35: Using Hölder s inequality on (8) andthefactthatr gl 0 outside the cylindrical part of M we get ( u l 2 + n 1 ) dv [0,l] S n 1 + 4(n 2) (n 1)(n 2)u2 l +(n 2)(n 3) u 2 [0,l] S n 2 l da < Y λ ( M, g l ) l + A. This way, there is a t l [0, l] such that ( u l 2 + n 1 {t l } S+ n 1 4(n 2) (n 1)(n 2)u2 l ( < Y λ ( M, g l ) + 1 ) /l. 1 + l + A ) dv + (n 2)(n 3) u 2 {t l } S n 2 l da This proves Lemma 3.2. We continue with the proof of Theorem 3.1 as follows: cut off M on the section {t l } S+ n 1 and attach two half-infinite cylinders to it, so (M {p 1, p 2 }, g) reappears. This time we write it as (M {p 1, p 2 }, g) =[0, ) S n 1 + ( M {t l } S n 1 +, g l) [0, ) S n 1 +. We think of u l as defined on M {t l } S n 1 + and extend it to the whole space M {p 1, p 2 } as follows: Let U l be a Lipschitz function on M {p 1, p 2 } such that and U l (t, x) = where ũ l = u l {tl } S+ n 1 M {p 1,p 2 } U l = u l on M {t l } S n 1 + { (1 t)ũl (x) for (t, x) [0, 1] S n for (t, x) [1, ) S n 1 + C (S+ n 1 ).Nowitiseasytoseefrom(8) and Lemma 3.2 that ( U l 2 + n 1 4(n 2) R gu l 2 )dv g + n 2 2 where B is a constant independent of l.fromeq.9 it follows that ( λ U l n 2 2n dv gl + (1 λ) M M Therefore, we have M {p inf 1,p 2 } ( U 2 + U (λ 2n M {p1,p2 } U n 2 dv g + (1 λ)( M σ λ ( M), M h g U 2 l da g < Y λ ( M, g l ) + B l, ) n U l 2(n 1) n 1 n 2 da gl > 1. 4(n 2) n 1 R gu 2 )dv g + n 2 2 M h gu 2 da g 2(n 1) U n 2 da g ) n 1 n ) (n 2) n (10)

7 Ann Glob Anal Geom (2009) 35: where the infimum is taken over all non-negative Lipschitz functions U with compact support. It follows from the choice of the metric g that the left side of Eq. 10 is equal to Y λ (M, g).sinceɛ was arbitrary we conclude σ λ (M) σ λ ( M),whichcompletestheproofof Theorem Proofofthenonpositivecase In this section we give a proof of Theorem 1.1 whenever one of σ λ (M 1 ), σ λ (M 2 ) is nonpositive. We first prove some lemmas. Lemma 4.1 Let (M n, g ), n 3 be a compact manifold with boundary. Suppose that < Y λ (M, g ) 0. Then the following holds: (λ = 1) The scalar curvature R g of a metric g [g ] with h g 0 satisfies n 2 4(n 1) (min R g)vol(m, g) 2/n Y 1 (M, g), and equality above implies that R g is a constant. (λ = 0) The mean curvature h g of a metric g [g ] with R g 0 satisfies n 2 (min h g )Area( M, g) 1/(n 1) Y 0 (M, g ), 2 where equality above implies that h g is a constant. Proof We first prove the case λ = 1. Let g [g ] with h g 0. A standard argument (c.f. [3]) shows that, since h g 0 we must have min R g 0, or else Y (M, [g ])>0. This way, Y 1 (M, [g M ]) = inf ( u 2 + 4(n 1) n 2 R gu 2 )dv g + n 2 2 M h gu 2 dσ u 0 in M ( M u2n/(n 2) dσ ) (n 2)/n n 2 4(n 1) inf u 0 in M (min R g ) M u2 dv g ( M u2n/(n 2) dσ ) (n 2)/n n 2 4(n 1) (min R g)vol(m, g) 2/n. The last line holds by Hölder s inequality. If this inequality were an equality, take a sequence {u n } so that (E g (u n )/N g,1 (u n )) Y 1 (M, [g ]).Then{u n } also converges to the infimum of M n 2 4(n 1) R gu 2 dv g (,andso u M u2n/(n 2) dσ ) (n 2)/n n 2 0. We deduce that min R g = ( M R g)/vol(m, g), which in turn implies that R g is constant. Proof of the case λ = 0. Let g [g ] with R g 0. A similar argument as the one above gives that min h g 0. This way, Y 0 (M, [g]) = inf u 0 on M n 2 2 M ( u 2 + inf u 0 on M 4(n 1) n 2 R gu 2 )dv g + n 2 2 M h gu 2 dσ ( M u2(n 1)/(n 2) dσ ) (n 2)/(n 1) (min h g ) M u2 dσ ( M u2(n 1)/(n 2) dσ ) (n 2)/(n 1) n 2 (min h g )Area( M, g) 1/(n 1). 2

8 122 Ann Glob Anal Geom (2009) 35: The last line is obtained by Hölder s inequality. The case of equality is similar to the cases of equality when λ = 1. The following is a well-known result of Escobar [5,6] which parallels that of Aubin [2] for the closed case. Theorem 4.2 Let (M n, g), n 3 be a compact manifold with boundary. If < Y λ (M, [g]) <Y λ (B n, [g 0 ]), then the infimum is achieved at a Yamabe metric. This is, (λ = 1) There exists a function u C (M) with E g (u)/n g,1 (u) = Y 1 (M, [g]), andthe metric g = u 4/(n 2) ghaszeromeancurvatureh g 0 and constant scalar curvature R g = Y 1 (M, [g])vol(m, g) 2/n. (λ = 0) There exists a function u C (M) with E g (u)/n g,0 (u) = Y 0 (M, [g]), so that the metric g = u 4/(n 2) ghaszeroscalarcurvaturer g 0 and constant mean curvature h g = Y 0 (M, [g])area( M, g) 1/(n 1). A direct consequence of the above theorem together with Lemma 4.1 is the following result. Corollary 4.3 Consider (M, g ) with < Y λ (M, [g ]) 0. Then (λ = 1) (λ = 0) max (min R g)vol(m, g) 2/n = Y 1 (M, [g ]). {g [g ]:h g 0} In particular, if σ 1 (M) 0, wegetthat sup (min R g )Vol(M, g) 2/n = Y 1 (M). {g : g [g ], h g 0, [g ] C} max (min h g)area( M, g) 1/(n 1) = Y 0 (M, [g ]). {g [g ]:R g 0} In particular, if σ 0 (M) 0, wegetthat sup (min h g )Area( M, g) 1/(n 1) = Y 0 (M). {g : g [g ], R g 0, [g ] C} Proof of Theorem 1.1 in the Nonpositive Case We first prove the case λ = 1. For a real number a R, leta denote the negative part of a. Thisisa = max{ a, 0}. By Kazdan and Warner s Theorem [7] therearemetricsg 1, g 2 on M 1, M 2, respectively, such that min R g1 = min R g2 = ((Y 1 (M 1 ) ) n/2 + (Y 1 (M 2 ) ) n/2 + 2ɛ) 2/n, R g1 (p 1 ) = R g2 (p 2 ) = n(n 1) at some points p i M i, h gi 0 on M i, (Y Vol(M i, g i ) = 1 (M i ) ) n/2 +ɛ for i = 1, 2 (Y 1 (M 1 ) ) n/2 +(Y 1 (M 2 ) ) n/2 +2ɛ where ɛ > 0isanarbitrarynumber.ByLemmaA.10 there exists a metric g of M 1 #M 2 such that h g 0, min R g = ((Y 1 (M 1 ) ) n/2 + (Y (M 2 ) ) n/2 + 2ɛ) 2/n,

9 Ann Glob Anal Geom (2009) 35: and Hence, from Corollary 4.3 we get Vol(M 1 #M 2, g) <1 + ɛ. Y 1 (M 1 #M 2 ) ((Y 1 (M 1 ) ) n/2 + (Y 1 (M 2 ) ) n/2 ) 2/n, which ends the proof of the first case. Proof of the case λ = 0. Let g i be metrics on M i, i = 1, 2 such that min h g1 = min h g2 = ((Y (M 1 ) ) n 1 + (Y (M 2 ) ) n 1 + 2ɛ) 1/(n 1), R gi 0 in M i, h gi (p i ) 0, R gi (p i ) = n(n 1), at some points p i M i, i = 1, 2 (Y (M Area( M i, g i ) = i ) ) n 1 +ɛ for i = 1, 2 (Y (M 1 ) ) n 1 +(Y (M 2 ) ) n 1 +2ɛ where ɛ is an arbitrary positive number. By Lemma A.10 there exists a metric g on M 1 #M 2 such that and R g 0, min h g = ((Y 0 (M 1 ) ) n 1 + (Y 0 (M 2 ) ) n 1 + 2ɛ) 1/(n 1) Hence, from Corollary 4.3 we get Area(M 1 #M 2, g) <1 + ɛ. Y 0 (M 1 #M 2 ) ((Y 0 (M 1 ) ) n 1 + (Y 0 (M 2 ) ) n 1 ) 1/(n 1). This ends the proof of the nonpositive case of Theorem 1.1. Acknowledgements This work was partially supported by NSF grant # DMS Appendix A: Approximation lemmas This section consists of several approximation lemmas. The results are, essentially, adaptations of those found in Kobayashi [8] and Akutagawa and Botvinnik [1] to fit our scenario. We include the more relevant proofs here. Lemma 4.4 For any δ > 0 there exists a smooth non-negative function 0 w δ 1 and a positive constant ɛ(δ), 0 < ɛ(δ) < δ, such that (i) w δ (t) [0,ɛ(δ)] 1, w δ (t) [δ, ) 0, (ii) tẇ δ (t) < δ for t 0, (iii) t 2 ẅ δ (t) < δ for t 0. Proof Elementary. Lemma 4.5 Let ḡ, g be two Riemannian metrics on M and h = g ḡ. Then R g Rḡ = Pḡ(h) + Qḡ(h), where Pḡ(h) = ḡ(trḡh) + i j h ij h, Ricḡ ḡ, Qḡ(h) c( h 2 q 3 + h h 2 q 2 + ( h h 2 + Ricḡ h 2 )q). Here, c = c(n) >0 is a constant that depends only on the dimension n of M, and q is a non-negative smooth function that satisfies q g ḡ.

10 124 Ann Glob Anal Geom (2009) 35: Proof Astraightforwardcalculationofwritingoutthescalarcurvatureintermsofthemetric and its derivatives that can be found in [8]. Lemma 4.6 (Fermi coordinates) Let (M n, g), n 3 be a compact Riemannian manifold with boundary and o M. There exists Fermi coordinates around o. This is, coordinates (r, x) around o so that γ x (r) = (r, x) is a geodesic and the set {(0, x) : x} corresponds to Mnearo.Wehave (i) g 00 = g( r, r ) = 1, g 00 = r g 00 = 0, (ii) g 0i = 0, g 0i = r g 0i = 0 for i > 0, (iii) g ij = (g M ) ij, g ij = 2A ij, i, j > 0. Proposition A.4 Let o M and ḡ, g be metrics on M such that Rḡ(o) = R g (o) and jo 1ḡ = j o 1 g (i.e.,ḡand g coincide up to first order derivatives on o). Consider Fermi coordinates around o as in Lemma 4.6. Thenthefamilyofmetrics has the following properties: (i) g δ =ḡ outside B δ (o), (ii) g δ = gonb ɛ(δ) (o), (iii) h gδ = h g on B ɛ(δ) (o), (iv) g δ ginc 1 (M) as δ 0, (v) R gδ Rḡ in C 0 (M) as δ 0, (vi) h gδ h g in C 0 ( M) as δ 0. g δ =ḡ + w δ (r)( g ḡ) Proof The proof of this proposition is similar to Kobayashi s Theorem (Lemma 3.2 of [8]). We proceed as follows: (i) and (ii) are direct and (iii) follows from (ii). Toprove(iv) consider the function w δ from Lemma 4.4.Notethatw δ is supported inside [0, δ].thisway g δ ḡ = w δ (r)( g ḡ) = O(r 2 ), and so g δ ḡ in C 0 (M). TogettheC 1 (M) convergence we see that ( g δ ḡ) = ẇ δ (r)( g ḡ) + w δ (r) ( g ḡ). Since g ḡ = O(r 2 ) and ( g ḡ) = O(r),Lemma4.4 implies g δ ḡ ẇ δ (r)r O(r 2 ) r + w δ (r)o(r) δo(δ) + O(δ), whichshows g δ ḡ in C 0 (M), as desired. For (v) we use Lemma 4.5 twice to get R gδ Rḡ = Pḡ(w δ (r)( g ḡ)) + Qḡ(w δ (r)( g ḡ)), 0 = R g + Rḡ + Pḡ( g ḡ) + Qḡ( g ḡ). Multiplying the second line above by w δ (r) and adding it to the first one we get R gδ Rḡ Pḡ(w δ (r)( g ḡ)) w δ (r)pḡ( g ḡ) + Qḡ(w δ (r)( g ḡ)) For convenience put + w δ (r)qḡ( g ḡ) + w δ (r)(rḡ R g ). T 1 = Pḡ(w δ (r)( g ḡ)) w δ (r)pḡ( g ḡ), T 2 = Qḡ(w δ (r)( g ḡ)), T 3 = w δ (r)qḡ( g ḡ), T 4 = w δ (r)(rḡ R g ).

11 Ann Glob Anal Geom (2009) 35: Using Lemmas 4.4 and 4.5 and the fact that jo 1 g = j o 1 ḡ we get: T 1 = Pḡ(w δ (r)( g ḡ)) w δ (r)pḡ( g ḡ) = ḡ(trḡ(w δ (r)( g ḡ))) + w δ (r) ḡ(trḡ( g ḡ)) + i j (w δ (r)( g ij ḡ ij )) w δ (r) i j ( g ij ḡ ij ) w δ (r)( g ḡ), Ricḡ ḡ + w δ (r) g ḡ, Ricḡ ḡ c 1 ( ẅ δ (r) g ḡ + ẇ δ (r) ( g ḡ) ) = ẅ δ (r)r 2 O(r 2 ) = O(δ), r 2 + ẇ δ (r)r O(r) r where c 1 > 0 above is a constant independent of δ. For the second term we get T 2 = Qḡ(w δ (r)( g ḡ)) c(n)( (w δ (r)( g ḡ)) 2 q 3 + (w δ (r)( g ḡ)) (w δ (r)( g ḡ)) 2 q 2 + ( (w δ (r)( g ḡ)) (w δ (r)( g ḡ)) 2 + Ricḡ (w δ (r)( g ḡ)) 2 )q). Notice that there are only first order derivatives in the above estimate. It is easy to see that T 2 = O(δ 2 ).ForT 3 we get T 3 = w δ (r)qḡ( g ḡ) c(n) w δ (r) ( ( g ḡ) 2 q 3 + g ḡ ( g ḡ) 2 q 2 +( g ḡ ( g ḡ) 2 + Ricḡ g ḡ 2 )q) = O(δ 2 ). To estimate T 4 we see that T 4 = w δ (r)(rḡ R g ) =O(δ) since Rḡ(o) = R g (o).thiswayt 1 + T 2 + T 3 + T 4 = O(δ) and (iv) is proved. For (vi) we use Lemma 4.6 (iii). We have that r ( g δ ) ij = 2Ã δ ij,so h gδ = (1/2)Tr gδ (Ã δ ) = (1/2)Tr gδ ( r ( g δ ) ij ). Since g δ ḡ in C 1 (M), r ( g δ ) r ḡ in C 0 (M).Hence,h gδ hḡ in C 0 ( M), as desired. This ends the proof. Theorem A.5 Let M be a compact manifold with boundary, ɛ 0 > 0, o Mandḡametric on M. Put g ḡ M and let Aḡ be the second fundamental form of ḡon M. Then there exists a family of metrics g δ on M such that (i) g δ =ḡinm\ B δ (o), (ii) h gδ = hḡ on B ɛ0 (o, ḡ), (iii) g δ is conformally equivalent to the metric (g 2rAḡ) + dr 2 on B ɛ(δ) (o, ḡ). (iv) g δ ḡinc 1 (M) as δ 0, (v) R gδ Rḡ in C 0 (M) δ 0, (vi) h gδ hḡ in C 0 ( M) as δ 0, Proof Consider Fermi coordinates (x, r) near o. By Lemma 4.6 we get the following expansion of the metric ḡ near o: ḡ(x, r) = (g ij (x) 2rA ij (x) + O(r 2 ))dx i dx j + i O(r 2 )drdx i + dr 2,

12 126 Ann Glob Anal Geom (2009) 35: where ḡ M = g and A ij denotes the second fundamental form of M with respect to ḡ. Consider the metric which has j 1 M ĝ = j 1 M ḡ.nowput where ĝ(x, r) = (g ij (x) 2rA ij (x))dx i dx j + dr 2 g(x, r) = u(x, r) 4/(n 2) ĝ, (11) u(x, t) = r 2 φ 2 (x), φ(x) = n 2 4(n 1) (R ḡ Rĝ). (12) Then u M = u(x, 0) = 0, r u(x, 0) = 0andsoon M, i u = 0, i j u = 0and r u = 0. This way j 1 o g = j 1 o ḡ. The well-known transformation law for the scalar curvature under conformal deformations of the metric gives that ĝu = n 2 4(n 1) (R gu (n+2)/(n 2) Rĝu). In our case u(x, 0) 1 so On the other hand, also on M we have ĝu = α α u ĝu = n 2 4(n 1) (R g Rĝ) on M. (13) =ĝ αβ ( α β u ˆƔ γ αβ γ u) = 2 r u + gij i j u ˆƔ 0 00 r u ˆƔ i 00 i u g ij ( ˆƔ 0 ij r u + ˆƔ k ij ku) = 2 r u. From this last line and Eq. 13 we get that, on M, n 2 4(n 1) (R g Rĝ) = ĝu = r 2 n 2 u = φ(x) = 4(n 1) (R ḡ Rĝ). This way R g = Rḡ on M. Now apply Proposition A.4 to ḡ, g to get a family g δ that satisfies (i) (vi). Theorem A.6 Let M be a compact manifold with boundary, o Mandɛ 0 > 0. Letḡ be a metric on M with hḡ = 0 on B ɛ0 (o). Write g ḡ M,andputAḡ to be the second fundamental form of M.Thenthereexistsafamilyofmetrics g δ such that for δ small enough (compared to ɛ 0 ) the following holds: (i) g δ ḡonm\ B δ (o), (ii) g δ is conformally equivalent to the metric g + dr 2 on B ɛ(δ) (o, ḡ), (iii) g δ ḡinc 0 (M) as δ 0, (iv) R gδ Rḡ in C 0 (M) δ 0, (v) h gδ hḡ in C 0 ( M) as δ 0. Proof The proof of this theorem is similar to Akutagawa and Botvinnik s Approximation Trick (Theorem 4.6 of [1]).

13 Ann Glob Anal Geom (2009) 35: From Eqs. 11 and 12 from the previous proof we may assume that ḡ = in Fermi coordinates near the point o.let ( ) 4 2 r 2 n 2 ( φ(x) (g(x) 2rAḡ(x)) + dr 2) G δ (x, r) = (g(x) 2r(1 w δ (r))aḡ(x)) + dr 2, and put g δ (x, r) = ( ) 4 2 r 2 n 2 φ δ (x, r) Gδ (x, r), φ δ (x, r) = φ(x) 3(n 2) 4(n 1) (2 w δ(r))w δ (r) Aḡ 2 g. We see that (i) and (ii) come from the construction. For (iii) notice that φ δ φ in C 0 (M) and G δ ḡ in C 0 (M), andso g δ ḡ. For (iv) a computation shows that near o we have Since hḡ = 0 on B ɛ0 (o) we get near o.thisway R gδ = R Gδ = R g + 3(1 w δ (r)) 2 Aḡ 2 g (1 w δ(r)) 2 h 2 ḡ (4ẇ δ (r) + 2rẅ δ (r))hḡ + O(δ). R Gδ = R g + 3(1 w δ (r)) 2 Aḡ 2 g + O(δ) ( ) n+2 ( 2 r 2 n 2 φ δ (x, r) +R Gδ (1 + 1 ) 2 r 2 φ δ (x, r)) ( 4(n 2) n 1 G δ ) 2 r 2 φ δ (x, r) ( ) = (1 + O(δ 2 4(n 2) )) n 1 φ + 3(2 w δ(r))w δ (r) Aḡ 2 g + R G δ + O(δ) = R g + 3(1 w δ (r)) 2 Aḡ 2 g + R ḡ R g 3 Aḡ 2 g + 3(2 wδ(r))w δ(r) Aḡ 2 g +O(δ) = Rḡ + O(δ), so (iv) follows. To prove (v) note that near o ( r 2 φ δ ) (x, 0) = 1, r ( r 2 φ δ ) (x, 0) = 0. The usual transformation law for the mean curvature under conformal deformations gives that r u = n 2 2 (h g δ u n/(n 2) h Gδ u). This way, h gδ = h Gδ near o. Sinceh Gδ = hḡ = 0 near o the result follows. Lemma A.7 Let (M, ḡ) be a manifold with boundary and o M such that, in Fermi coordinates (r, x), ḡ = g(x) + dr 2 near o, and so that Rḡ(o) = n(n 1). Lete S+ n be an

14 128 Ann Glob Anal Geom (2009) 35: equatorial point in the boundary of (S n +, g n),sothatinpolarnormalcoordinatesarounde, the round metric g n = dt 2 + sin 2 (t)g n 1.Here,g n 1 denotes the round metric on S n 1. Define g = dt 2 + sin 2 (t)g n 1 (y) around o M, where (t, y) are polar normal coordinates centered at o. Then: (i) Rḡ(o) = R g (o), (ii) j 1 o (ḡ) = j 1 o ( g). Proof Part (i) is direct since R g (o) = R gn (e) = n(n 1).For(ii)wenotethatallfirstorder partial derivatives of ḡ vanish at o since the boundary M has zero second fundamental form near o (see Lemma 4.6 about Fermi coordinates). The same thing holds at e, since the equator in S n is totally geodesic. Lemma A.8 Let (M, ḡ) be a manifold with Y λ (M, ḡ) >0, p 1, p 2 Mandδ > 0. Then there exist a metric g onmand0 < ɛ < δ such that (i) Y (M, [ḡ]) Y (M, [ g]) < δ, (ii) g Bɛ (p i ) around p i is isometric to a neighborhood of e S+ n inside Sn +,fori= 1, 2. (iii) R g 0 on M, (iv) h g = 0 outside a small neighborhood of p i. Proof We start by finding g.sincetheyamabeconstantof(m, ḡ) is positive we may assume that ḡ has positive scalar curvature and minimal boundary. Using Theorem A.6 we can approximate ḡ by a metric ĝ that is conformal to a product near each p i.lemmaa.7 together with Proposition A.4 imply that there exists a metric ǧ on M that is close to ĝ and satisfies (ii). Apply Proposition A.4 to ḡ and ǧ to produce a metric g that coincides with ǧ near p i and with equals ḡ elsewhere. Now g is close to the original ḡ so (i) follows. (ii) holds by construction. Since R g is close to Rḡ > 0, it is non-negative, so (iii) holds. (iv) holds since all constructions took place in a small neighborhood of p i and ḡ had minimal boundary. Let (S+ n, g n), n 3 be the Euclidean unit hemisphere and r the intrinsic distance relative to g n from a point e on the boundary, so that g n = dr 2 + sin 2 rg n 1,whereg n 1 is the standard metric on the unit n 1 hemisphere. For an interval I [0, π] denoted by A(I ) the region A(I ) ={x S+ n : r(x) I }. Lemma A.9 For any ɛ 1 > 0, 0 < ɛ 2 < π there exists a positive function f = f (r) on S n + such that: (i) R g n(n 1) < ɛ 1, where g = f 2 g n, (ii) h g = 0, (iii) Vol(S+ n, g ) 2Vol(S+ n, g n) < ɛ 1, (iv) Area( S+ n, g ) 2Area( S+ n, g n 1) < ɛ 1, (v) f (r) = 1 for r > ɛ 2 and (A([0, ɛ 2 )), g ) is isometric to (A((ɛ 2, π]), g ) = (A((ɛ 2, π]), g n ) for some ɛ 2 < ɛ 2, (vi) 0 < f (r) 1 and f (r) 2/ sin(r) for all r, where means d/dr. Proof This is similar to Lemma 3.1 of [8]. For (ii) note that h gn = 0, g = ( f (1 n)/2 ) 4/(n 1) g 0, so the transformation law for the mean curvature gives h g = 2 n 2 f n(n 1)/2(n 2) η f (1 n)/2.

15 Ann Glob Anal Geom (2009) 35: Since f is a function that depends only on r, the normal derivative η f (1 n)/2 = 0 by Gauss Lemma. This way h g = 0 as desired. The scalar curvature of g is given by 1 n(n 1) R g = 2 n f ( f + (n 1) f cot r) f 2 + f 2. We make the following change of variables: Put Then g = u 2 (dt 2 + g n ) and where = d/dt = (sin r)d/dr.weput so that from (16) weget We fix t 0 > max{0, log cot(ɛ 2 /2)}, andset cos r = tanh t, 0 < r < π, > t >. (14) u(t) = f (r(t)) cosh t. (15) 1 n(n 1) R g = 2 n uu (u ) 2 + n 2 u 2, (16) n B(t) = u n ((u ) 2 u 2 + 1), (17) B (t) = (u n ) (1 R g ). (18) n(n 1) u(t) = cosh t for t (, t 0 ], (19) hence, B(t) 0fort < t 0. We want to consider the solution u of (17) withasuitableb(t). First note that u(t) cosh t for t t 0, if B(t) 0 for t t 0, (20) which follows from a comparison argument. Let B(t) = 0 for t t 0 B(t) 0, 2δ B (t) 0 for t 0 t t B(t) = δ for t t t 1. If δ is sufficiently small, then (17)togetherwith(19)issolvableforu in the interval (, t 1 ) with arbitrary t 1 t 0 + 1, and u (t) >0, u (t) >0fort 0 t t Therefore, taking δ > 0 smaller we have from (18) R g n(n 1) 2(n 1) coshn+1 (t 0 + 1) sinh t 0 δ < ɛ 1, for t t We choose t 1 so that u (t 1 ) = 0andu (t) >0fort < t < t 1.ThiswayR g = n(n 1) for t t t 1.Fort t 1 we put B(t) = B(2t 1 t). Then u(t) = u(2t 1 t) for t t 1. (21) Thus R g n(n 1) < ɛ 1 for all t, and(v) follows from (19) and (21) via (14) and (15). As for the volume we have Vol(S n +, g ) = Vol(S n +, g n) + u(t1 ) 1 du u n u

16 130 Ann Glob Anal Geom (2009) 35: and we can see by a long and elementary calculation that the second term of the right hand side converges to Vol(S+ n, g n) as δ 0. Therefore, (iii) holds for δ small enough, and (iv) follows from an analogous computation. From (20), u(t) cosh t and hence u (t) sinh t from (17), because B(t) 0, which proves (vi). Lemma A.10 Let (M n 1, g 1), (M n 2, g 2), n 3 be two compact Riemannian manifolds with boundary and ɛ > 0. (λ = 1) Ifh gi 0, R gi (p i ) = n(n 1), i = 1, 2 at some points p i M i, then there exists ametricgonm 1 #M 2 so that: (1.a) h g 0 and Vol(M 1 #M 2, g) 2 i=1 Vol(M i, g i ) < ɛ. (1.b) There are isometric embeddings φ i : (M i B i, g i ) (M 1 #M 2, g), i = 1, 2, where B i are small balls around p i M i,and R g (x) n(n 1) < ɛ for x M 1 #M 2 (Im(φ 1 ) Im(φ 2 )). (λ = 1) If R gi 0, R gi (p i ) = n(n 1), h gi > 0 at some points p i M i, i = 1, 2 then there exists a metric g on M 1 #M 2 so that: (0.a) R g 0 and Area( (M 1 #M 2 ), g) 2 i=1 Area( M i, g i ) < ɛ. (0.b) There are isometric embeddings φ i : (M i B i, g i ) (M 1 #M 2, g), i = 1, 2, whereb i are small balls around p i M i,and h g (x) < ɛ for x (M 1 #M 2 ) (Im(φ 1 ) Im(φ 2 ))). Proof Proof of λ = 1. By Lemmas A.4, A.7,andA.9 we can take a metric ḡ i of M i that coincides with g i outside a small (half) ball B i containing p i and such that R gi (x) n(n 1) < ɛ for x B i, Vol(M i, ḡ i ) Vol(M i, g i ) ɛ/4 andb i contains a smaller ball B i such that (B i, ḡ i ) is isometric to a geodesic δ-(half) ball in the unit hemisphere and Vol(B i, ḡ i )<ɛ/4. From Lemma A.9 with ɛ 1, ɛ 2 smaller than ɛ and δ, respectively, we have a small piece (A[δ, δ], g ) for some δ < δ such that R g n(n 1) < ɛ, Vol(A[δ, δ], g )<ɛ/4 and aneighborhoodofeachoftheboundarycomponentsisisometrictoaneighborhoodofthe boundary of the δ-ball inside the hemisphere. This way, the manifold (M 1 #M 2, g) is obtained by putting together (M B i, ḡ 1), (M 2 B 2, ḡ 2) and (A([δ, δ]), g ). The proof of λ = 0issimilar:takeḡ i as above with R gi (x) n(n 1) < ɛ for x B i, Area( M i, ḡ i ) Area( M i, g i ) ɛ/4andb i contains a smaller ball B i such that (B i, ḡ i ) is isometric to a geodesic δ-(half) ball in the unit hemisphere and Area(( M i ) B i, ḡ i ) ɛ/4. From Lemma A.9 with ɛ 1, ɛ 2 smaller than ɛ and δ, respectively, we have a small piece (A[δ, δ], g ) for some δ < δ such that R g n(n 1) < ɛ, Area( A[δ, δ], g )<ɛ/4 and a neighborhood of each of the boundary components is isometric to a neighborhood of the boundary of the δ-ball inside the hemisphere. The manifold (M 1 #M 2, g) is obtained by putting together (M B i, ḡ 1), (M 2 B 2, ḡ 2) and (A([δ, δ]), g ). References 1. Akutagawa, K., Botvinnik, B.: The relative Yamabe invariant. Comm. Anal. Geom. 10(5), (2002) 2. Aubin, T.: Équations différentielles non linéaires et problème de Yamabe concernant la courbure scalaire. J. Math. Pures Appl. 55(3), (1976) 3. Aubin, T.: Some Nonlinear Problems in Riemannian Geometry. Springer-Verlag, Springer Monographs in Mathematics (1998) 4. Brendle, S.: A generalization of the Yamabe flow for manifolds with boundary. Asian J. Math. 6(4), (2002)

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