Q curvature on a class of manifolds with dimension at least 5
aa r X i v : . [ m a t h . DG ] O c t Q CURVATURE ON A CLASS OF MANIFOLDS WITHDIMENSION AT LEAST FENGBO HANG AND PAUL C. YANG
Abstract.
For a smooth compact Riemannian manifold with positive Yam-abe invariant, positive Q curvature and dimension at least 5, we prove theexistence of a conformal metric with constant Q curvature. Our approach isbased on the study of extremal problem for a new functional involving thePaneitz operator. Introduction
Recall the definition of the 4th order Paneitz operator and its associated Q curvature [B, P]: when ( M, g ) is a smooth compact n dimensional Riemannianmanifold with n ≥
3, the Q curvature is given by Q = −
12 ( n −
1) ∆ R − n − | Rc | + n − n + 16 n −
168 ( n − ( n − R (1.1)= − ∆ J − | A | + n J . Here R is the scalar curvature, Rc is the Ricci tensor and J = R n − , A = 1 n − Rc − Jg ) . (1.2)The Paneitz operator is given by P ϕ (1.3)= ∆ ϕ + 4 n − Rc ( ∇ ϕ, e i ) e i ) − n − n + 82 ( n −
1) ( n −
2) div ( R ∇ ϕ ) + n − Qϕ = ∆ ϕ + div (4 A ( ∇ ϕ, e i ) e i − ( n − J ∇ ϕ ) + n − Qϕ.
Here e , · · · , e n is a local orthonormal frame with respect to g . When n = 4, undera conformal change of the metric, the operator satisfies P ρ n − g ϕ = ρ − n +4 n − P g ( ρϕ ) . (1.4)This is similar to the conformal Laplacian operator, which appears naturally whenconsidering transformation law of the scalar curvature under conformal change ofmetric in dimension greater than 2 ([LP]). As a consequence we have P ρ n − g ϕ · ψdµ ρ n − g = P g ( ρϕ ) · ρψdµ g . (1.5)Here µ g is the measure associated with metric g . The research of Yang is supported by NSF grant 1104536.
In dimension 4, the Paneitz operator is given by
P ϕ = ∆ ϕ + 2 div ( Rc ( ∇ ϕ, e i ) e i ) −
23 div ( R ∇ ϕ ) , (1.6)and its conformal covariance property takes the form P e w g ϕ = e − w P g ϕ. (1.7)Following the basic work [CGY] in dimension 4 on the 4th order Q curvature equa-tion, there has been several studies on this equation in dimension 3 by [HY1, XY2,YZ], and in dimensions greater than 4 by [DHL, HeR1, HeR2, HuR, QR1, QR2].While it is important to determine conditions under which the Paneitz operatoris positive, we discover that it is sufficient for our purpose in this article to deter-mine when its Green’s function is positive. This is a property that is conformallyinvariant: observe that by (1.4),ker P g = 0 ⇔ ker P ρ n − g = 0 , (1.8)and under this assumption, the Green’s functions G P satisfy the transformationlaw G P,ρ n − g ( p, q ) = ρ ( p ) − ρ ( q ) − G P,g ( p, q ) . (1.9)In analogy with the preliminary study of the classical Yamabe problem ([LP]),the first question would be whether one can find a conformal invariant condition forthe existence of a conformal metric with positive Q curvature. In the case Yamabeinvariant Y ( g ) is positive, the existence of a conformal metric with positive Q curvature is equivalent to the requirements that ker P = 0 and the Green’s function G P > Q curvature metric in a con-formal class, in the same spirit as Yamabe problem. The main aim of the presentarticle is to prove the following Theorem 1.1.
Let ( M, g ) be a smooth compact n dimensional Riemannian man-ifold with n ≥ , Y ( g ) > , Q ≥ and not identically zero, then ker P = 0 , theGreen’s function of P is positive and there exists a conformal metric e g with e Q = 1 . Remark 1.1.
Let ( M n , g ) be a smooth compact Riemannian manifold with n ≥ , Y ( g ) > . Denote L = − n − n − ∆ + R as the conformal Laplacian operator and for p ∈ M , G L,p as the Green’s function of L with pole at p . Define Γ ( p, q ) = 2 n − n − n − n − ( n − n − n − ( n − − ω − n − n G L ( p, q ) n − n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Rc G n − L,p g (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) g ( q ) . Here ω n is the volume of unit ball in R n , G L ( p, q ) = G L,p ( q ) . The associatedintegral operator T Γ is given by T Γ ( ϕ ) ( p ) = Z M Γ ( p, q ) ϕ ( q ) dµ ( q ) for any nice function ϕ on M . In [HY5] , it is shown that the spectrum σ ( T Γ ) andspectral radius r σ ( T Γ ) are conformal invariants, moreover the following statementsare equivalent: (1) there exists a conformal metric e g with e Q > . CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 3 (2) ker P = 0 and the Green’s function of Paneitz operator G P ( p, q ) > for p = q . (3) ker P = 0 and there exists p ∈ M such that G P ( p, q ) > for q = p . (4) r σ ( T Γ ) < .Under the assumption Q ≥ and not identically zero, we have r σ ( T Γ ) < . The fundamental difficulty of the lack of maximum principle in this 4th orderequation has recently been overcome by the work in [GM]. Following this devel-opment, similar results in dimension 3 were proved in [HY3, HY4] (see also closelyrelated [HY2]). Dimension 4 case does not suffer from this difficulty and was treatedin many articles like [CY, DM, FR] and so on. For a locally conformally flat mani-fold with positive Yamabe invariant and Poincare exponent less than n − (see [SY]),Theorem 1.1 was proved in [QR2] by apriori estimates and connecting the equationto Yamabe equation through a path of integral equations. Under the slightly morestringent conditions R >
Q >
0, Theorem 1.1 was proved in [GM] throughthe study of a non-local flow. Here we will derive Theorem 1.1 by maximizing afunctional (see (1.16) and (2.2)) involving the Paneitz operator (see Theorem 1.3for more details).For u, v ∈ C ∞ ( M ), we denote the quadratic form associated to P as E ( u, v ) (1.10)= Z M P u · vdµ = Z M (cid:18) ∆ u ∆ v − n − Rc ( ∇ u, ∇ v ) + n − n + 82 ( n −
1) ( n − R ∇ u · ∇ v + n − Quv (cid:19) dµ = Z M (cid:18) ∆ u ∆ v − A ( ∇ u, ∇ v ) + ( n − J ∇ u · ∇ v + n − Quv (cid:19) dµ, and E ( u ) = E ( u, u ) . (1.11)By the integration by parts formula in (1.10) we know that E ( u, v ) extends con-tinuously to u, v ∈ H ( M ).To find the metric e g in Theorem 1.1, we write e g = ρ n − g , then the equation e Q = 1 becomes P g ρ = n − ρ n +4 n − , ρ ∈ C ∞ ( M ) , ρ > . (1.12)Let Y ( g ) = inf u ∈ H ( M ) \{ } E ( u ) k u k L nn − , (1.13)then Y (cid:16) τ n − g (cid:17) = Y ( g ) for any positive smooth function τ . Hence Y ( g ) is aconformal invariant. If ( M, g ) is not locally conformally flat and n ≥
8, or (
M, g )is locally conformally flat with Y ( g ) >
0, ker P = 0 and the Green’s function of P , G P >
0, or n = 5 , , Y ( g ) >
0, ker P = 0 and G P >
0, one can show Y ( g )is achieved (see [ER, GM, R]), but in general it is difficult to know whether theminimizer is positive. Under the additional assumption Y ( g ) > G P >
0, it
FENGBO HANG AND PAUL C. YANG was observed in [R] that the minimizer cannot change sign. Combining this withthe positivity criterion of Green’s function in [HY4], we arrive at
Theorem 1.2.
Let ( M, g ) be a smooth compact n dimensional Riemannian mani-fold with n ≥ , Y ( g ) > , Y ( g ) > , Q ≥ and not identically zero, then (1) Y ( g ) ≤ Y ( S n ) , and equality holds if and only if ( M, g ) is conformallydiffeomorphic to the standard sphere. (2) Y ( g ) is always achieved. Any minimizer must be smooth and cannotchange sign. In particular we can find a constant Q curvature metric inthe conformal class. (3) If ( M, g ) is not conformally diffeomorphic to the standard sphere, then theset of all minimizers u for Y ( g ) , after normalizing with k u k L nn − = 1 , iscompact in C ∞ topology. Note the positivity of Y ( g ) is equivalent to the positivity of Paneitz operator P . There are several criterion for the positivity of P (see [CHY, Theorem 1.6]and [GM, XY1]). On the other hand, in a recent preprint [GHL], it is proved thatif ( M, g ) is a smooth compact Riemannian manifold with dimension n ≥
6, and Y ( g ) > Y ( g ) >
0, then we can find a conformal metric e g with e R > e Q >
0. In particular, it follows from [GM] that any conformal metric with constant Q curvature must have positive scalar curvature. Similar statement for n = 5 islikely to be true but could not be justified due to the approach there.In general it is not known whether Y ( g ) > , Q ≥ Y ( g ) >
0. To get around this difficulty when proving Theorem 1.1we note that by [HY4, Proposition 1.1] if Y ( g ) > Q ≥ P = 0, and the Green’s function of P , G P >
0. Hence we can define anintegral operator (the inverse of P ) as G P f ( p ) = Z M G P ( p, q ) f ( q ) dµ ( q ) . (1.14)If we denote f = ρ n +4 n − , then equation (1.12) becomes G P f = 2 n − f n − n +4 , f ∈ C ∞ ( M ) , f > . (1.15)Let Θ ( g ) = sup f ∈ L nn +4 ( M ) \{ } R M G P f · f dµ k f k L nn +4 (1.16)= sup f ∈ L nn +4 ( M ) \{ } R M × M G P ( p, q ) f ( p ) f ( q ) dµ ( p ) dµ ( q ) k f k L nn +4 . It follows from the classical Hardy-Littlewood-Sobolev inequality ([St]) that Θ ( g )is always finite. Moreover it follows from (1.9) that for a positive smooth function ρ , Θ (cid:16) ρ n − g (cid:17) = Θ ( g ) i.e. Θ ( g ) is a conformal invariant. If Θ ( g ) is achieved bya maximizer f , using the fact that G P >
0, we easily deduce that f cannot change CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 5 sign. Θ ( g ) has a nice geometric description (see Lemma 2.1):Θ ( g ) = 2 n − R M e Qd e µ (cid:13)(cid:13)(cid:13) e Q (cid:13)(cid:13)(cid:13) L nn +4 ( M,d e µ ) : e g ∈ [ g ] (1.17)Here [ g ] denotes the conformal class of g i.e.[ g ] = (cid:8) ρ g : ρ ∈ C ∞ ( M ) , ρ > (cid:9) . (1.18) Theorem 1.3.
Assume ( M, g ) is a smooth compact n dimensional Riemannianmanifold with n ≥ , Y ( g ) > , Q ≥ and not identically zero, then (1) Θ ( g ) ≥ Θ ( S n ) , here S n has the standard metric. Θ ( g ) = Θ ( S n ) ifand only if ( M, g ) is conformally diffeomorphic to the standard sphere. (2) Θ ( g ) is always achieved. Any maximizer f must be smooth and cannotchange sign. If f > , then after scaling we have G P f = n − f n − n +4 i.e. Q f n +4 g = 1 . (3) If ( M, g ) is not conformally diffeomorphic to the standard sphere, then theset of all maximizers f for Θ ( g ) , after normalizing with k f k L nn +4 = 1 , iscompact in the C ∞ topology. It is worthwhile to note the similarity of Theorem 1.2 and 1.3 to classical Yam-abe problem ([LP, S]) and the integral equation considered in [HWY1, HWY2].Indeed, the formulation of our approach follows that of [HWY2]. Integral equationformulation of the Q curvature equation was used in [QR2]. A similar functionalfor the conformal Laplacian operator, Θ (see (4.8)) is also considered in [DoZ]. InSection 2 below we will first give other expressions for Θ ( g ) and discuss its rela-tion with Y ( g ), then we will derive the concentration compactness principle forthe extremal problem of Θ ( g ) and find the asymptotic expansion formula for theGreen’s function of Paneitz operator. In Section 3 we will show that maximizersalways exist and that they are smooth. In particular Theorem 1.3 will follow. Atlast, in Section 4 we will prove Theorem 1.2. Moreover we will show the approachto Theorem 1.3 gives another way to find constant scalar curvature metrics in aconformal class.The authors would like to thank Gursky and Malchiodi for making their workavailable. We would also like to thank the referee for his/her careful reading of thearticle and many comments which improve the presentation of the paper.2. Some preparations
The conformal invariants Y ( g ) , Y +4 ( g ) and Θ ( g ) . Throughout this sub-section we will assume (
M, g ) is a smooth compact n dimensional Riemannianmanifold with n ≥
5. Recall that Y ( g ) = inf u ∈ H ( M ) \{ } E ( u ) k u k L nn − = inf u ∈ C ∞ ( M ) \{ } R M P u · udµ k u k L nn − . (2.1) FENGBO HANG AND PAUL C. YANG
If in addition Y ( g ) > Q ≥ ( g ) = sup f ∈ L nn +4 ( M ) \{ } R M G P f · f dµ k f k L nn +4 (2.2)= sup u ∈ W , nn +4 ( M ) \{ } R M P u · udµ k P u k L nn +4 . The second equality in (2.2) will be very useful for us later on because the expressionis local. It will facilitate our calculations in estimating Θ ( g ). Θ ( g ) also has ageometric description. Lemma 2.1. If n ≥ , Y ( g ) > , Q ≥ and not identically zero, then Θ ( g ) = 2 n − R M e Qd e µ (cid:13)(cid:13)(cid:13) e Q (cid:13)(cid:13)(cid:13) L nn +4 ( M,d e µ ) : e g ∈ [ g ] . (2.3) Proof.
Note that 2 n − R M e Qd e µ (cid:13)(cid:13)(cid:13) e Q (cid:13)(cid:13)(cid:13) L nn +4 ( M,d e µ ) : e g ∈ [ g ] = sup R M P u · udµ k P u k L nn +4 : u ∈ C ∞ ( M ) , u > ≤ Θ ( g ) . On the other hand, by the positivity of G P we haveΘ ( g )= sup R M G P f · f dµ k f k L nn +4 : f ∈ L nn +4 ( M ) \ { } , f ≥ = sup R M G P f · f dµ k f k L nn +4 : f ∈ C ∞ ( M ) \ { } , f ≥ = sup R M P u · udµ k P u k L nn +4 : u ∈ C ∞ ( M ) \ { } , P u ≥ ≤ sup R M P u · udµ k P u k L nn +4 : u ∈ C ∞ ( M ) , u > = 2 n − R M e Qd e µ (cid:13)(cid:13)(cid:13) e Q (cid:13)(cid:13)(cid:13) L nn +4 ( M,d e µ ) : e g ∈ [ g ] . In between we have used the fact for smooth function u , P u ≥ u not identi-cally zero implies u > CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 7
To better understand the relation between Y ( g ) and Θ ( g ), we define Y +4 ( g ) = inf R M P u · udµ k u k L nn − : u ∈ C ∞ ( M ) , u > (2.4)= n −
42 inf ( R M e Qd e µ ( e µ ( M )) n − n : e g ∈ [ g ] ) . Clearly we have Y ( g ) ≤ Y +4 ( g ) . (2.5) Lemma 2.2. If n ≥ , Y ( g ) > , Q ≥ and not identically zero, then Y +4 ( g ) Θ ( g ) ≤ . (2.6) Moreover if Y +4 ( g ) is achieved, then Y +4 ( g ) Θ ( g ) = 1 and Θ ( g ) must be achievedtoo.Proof. It is clear that Θ ( g ) >
0. To prove the inequality we only need to dealwith the case Y +4 ( g ) >
0. Under this assumption for u ∈ C ∞ ( M ) , u >
0, we have R M P u · udµ >
0. By Holder’s inequality we have (cid:0)R M P u · udµ (cid:1) k u k L nn − k P u k L nn +4 ≤ . It follows that Y +4 ( g ) R M P u · udµ k P u k L nn +4 ≤ . By the proof of Lemma 2.1 we haveΘ ( g ) = sup R M P v · vdµ k P v k L nn +4 : v ∈ C ∞ ( M ) , v > , hence Y +4 ( g ) Θ ( g ) ≤ Y +4 ( g ) is achieved, say at u ∈ C ∞ ( M ) , u >
0, then
P u = κu n +4 n − for some constant κ . Since G P >
0, we see that κ >
0. HenceΘ ( g ) ≥ R M P u · udµ k P u k L nn +4 = 1 κ k u k − n − L nn +4 = 1 Y +4 ( g ) ≥ Θ ( g ) . Hence all the inequalities are equalities. Θ ( g ) = Y +4 ( g ) and it is achieved at u too. Remark 2.1.
Assume Y +4 ( g ) Θ ( g ) = 1 . Later we will show that Θ ( g ) is alwaysachieved by positive smooth functions i.e. Θ ( g ) = R M G P f · f dµ k f k L nn +4 = R M P v · vdµ k P v k L nn +4 , here f ∈ C ∞ ( M ) , f > , v = G P f . Hence v ∈ C ∞ ( M ) , v > and P v = κv n +4 n − FENGBO HANG AND PAUL C. YANG for some constant κ . Using G P > we see that κ > . On the other hand Θ ( g ) = R M P v · vdµ k P v k L nn +4 = κ − k v k − n − L nn − . Hence Y +4 ( g ) = κ k v k n − L nn − = R M P v · vdµ k v k L nn − . In other words, positive maximizers for Θ ( g ) are also minimizers for Y +4 ( g ) . The sphere S n . On S n ( n ≥
5) with standard metric we have Q = n ( n + 2) ( n − P u = ∆ u − n − n −
42 ∆ u + n ( n + 2) ( n −
2) ( n − u. (2.8)Let N be the north pole and π N : S n \ { N } → R n be the stereographic projection.Using x = π N as the coordinate, then the Green’s function of P with pole at N isgiven by G P,N = 1 n ( n −
2) ( n −
4) 2 n − ω n (cid:16) | x | + 1 (cid:17) n − . (2.9)Here ω n is the volume of the unit ball in R n i.e. ω n = π n Γ (cid:0) n + 1 (cid:1) , (2.10)Γ is the Gamma function given byΓ ( α ) = Z ∞ e − t t α − dt for α > . (2.11)From [CnLO, Li] we know Y ( S n ) = inf u ∈ C ∞ c ( R n ) \{ } k ∆ u k L ( R n ) k u k L nn − ( R n ) (2.12)= k ∆ u k L ( R n ) k u k L nn − ( R n ) = n ( n + 2) ( n −
2) ( n − n π n +1) n Γ (cid:0) n +12 (cid:1) n = Y +4 ( S n ) . Here u ( x ) = (cid:16) | x | + 1 (cid:17) − n − . (2.13)For λ >
0, let u λ ( x ) = λ − n − u (cid:16) xλ (cid:17) = λ | x | + λ ! n − , (2.14)then ∆ u λ = n ( n + 2) ( n −
2) ( n − u n +4 n − λ . (2.15) CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 9
On the other hand it follows from [CnLO, Li] thatΘ ( S n ) (2.16)= 12 n ( n −
2) ( n − ω n sup f ∈ L ( R n ) \{ } R R n × R n f ( x ) f ( y ) | x − y | n − dxdy k f k L nn +4 ( R n ) = sup u ∈ C ∞ c ( R n ) \{ } R R n (∆ u ) dx k ∆ u k L nn +4 ( R n ) = 12 n ( n −
2) ( n − ω n R R n × R n f ( x ) f ( y ) | x − y | n − dxdy k f k L nn +4 ( R n ) = 1 Y ( S n ) . Here f ( x ) = (cid:16) | x | + 1 (cid:17) − n +42 . (2.17)For λ >
0, let f λ ( x ) = λ − n +42 f (cid:16) xλ (cid:17) = λ | x | + λ ! n +42 , (2.18)then ∆ u λ = n ( n + 2) ( n −
2) ( n − f λ . (2.19)2.3. Concentration compactness principle.
Here we apply the concentrationcompactness principle in [Ln] to extremal problem (2.2). To achieve this goal westart with an almost sharp Sobolev inequality. Recall by (2.16) for u ∈ C ∞ c ( R n ) , Z R n (∆ u ) dx ≤ Θ ( S n ) (cid:13)(cid:13) ∆ u (cid:13)(cid:13) L nn +4 ( R n ) . (2.20) Lemma 2.3.
Assume M is a smooth compact Riemannian manifold with dimen-sion n ≥ . Then for any ε > , we have k ∆ u k L ( M ) ≤ (Θ ( S n ) + ε ) k P u k L nn +4 ( M ) + C ( ε ) k u k L nn +4 ( M ) (2.21) for all u ∈ W , nn +4 ( M ) . The passage from (2.20) to (2.21) is standard and we refer the readers to [DHL,He] for further details. The above almost sharp Sobolev inequality can be usedto prove the following concentration compactness lemma. We refer the readers to[He, Ln] for the now standard argument.
Lemma 2.4.
Let M be a smooth compact Riemannian manifold with dimension n ≥ , ker P = 0 , f i ∈ L nn +4 ( M ) such that f i ⇀ f weakly in L nn +4 . Let u i , u ∈ W , nn +4 ( M ) such that P u i = f i , P u = f . Assume | f i | nn +4 dµ ⇀ σ in M ( M ) and | ∆ u i | dµ ⇀ ν in M ( M ) , here M ( M ) is the space of all Radon measures on M . Then there exists countablymany points p i ∈ M such that σ ≥ | f | nn +4 dµ + X i σ i δ p i and ν = | ∆ u | dµ + X i ν i δ p i , here σ i = σ ( { p i } ) , ν i = ν ( { p i } ) . Moreover ν i ≤ Θ ( S n ) σ n +4 n i . Now we are ready to derive a criterion for the existence of maximizers. Suchkind of criterion is an analog statement for those of Yamabe problems ([LP]).
Proposition 2.1.
Assume ( M, g ) is a smooth compact n dimensional Riemannianmanifold with n ≥ , ker P = 0 . Let Θ ( g ) = sup f ∈ L nn +4 ( M ) \{ } R M G P f · f dµ k f k L nn +4 . If Θ ( g ) > Θ ( S n ) and f i ∈ L nn +4 satisfies k f i k L nn +4 = 1 , R M G P f i · f i dµ → Θ ( g ) ,then after passing to a subsequence, we can find a f ∈ L nn +4 such that f i → f in L nn +4 . In particular, k f k L nn +4 = 1 and R M G P f · f dµ = Θ ( g ) , f is a maximizerfor Θ ( g ) .Proof. After passing to a subsequence we can assume f i ⇀ f weakly in L nn +4 . Let u i , u ∈ W , nn +4 such that P u i = f i , P u = f . Then u i ⇀ u weakly in W , nn +4 , u i → u in W , nn +4 and u i → u in W , . After passing to another subsequence wehave | f i | nn +4 dµ ⇀ dσ and (∆ u i ) dµ ⇀ dν in M ( M ) , moreover it follows from Lemma 2.4 that σ ≥ | f | nn +4 dµ + X i σ i δ p i , ν = (∆ u ) dµ + X i ν i δ p i , here σ i = σ ( { p i } ) , ν i = ν ( { p i } ) and ν i ≤ Θ ( S n ) σ n +4 n i . It follows that σ ( M ) = 1 and Z M G P f i · f i dµ = Z M u i P u i dµ = E ( u i )= Z M (cid:18) (∆ u i ) − A ( ∇ u i , ∇ u i ) + ( n − J |∇ u i | + n − Qu i (cid:19) dµ → E ( u ) + X i ν i . CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 11
Hence Θ ( g ) = E ( u ) + X i ν i ≤ Θ ( g ) k f k L nn +4 + Θ ( S n ) X i σ n +4 n i ≤ Θ ( g ) "(cid:18) k f k nn +4 L nn +4 (cid:19) n +4 n + X i σ n +4 n i ≤ Θ ( g ) k f k nn +4 L nn +4 + X i σ i ! n +4 n ≤ Θ ( g ) . Hence all inequalities become equalities. In particular, σ i = 0, ν i = 0, k f k L nn +4 = 1.Hence f i → f in L nn +4 , E ( u ) = R M G P f · f dµ = Θ ( g ).2.4. Expansion of Green’s function of the Paneitz operator.
In [LP], theexpansion formula of Green’s function of conformal Laplacian operator plays animportant role. Here we determine the expansion formulas for Green’s function ofPaneitz operator. These formulas will be crucial in the choice of test function insection 3.We use the same strategy as [LP, section 6], but since we need to take intoaccount lower order terms, some efforts are needed in doing the algebra. Let usintroduce some notation. For m ∈ Z + , let P m = { homogeneous degree m polynomials on R n } , (2.22)and H m = { harmonic degree m homogeneous polynomials } . (2.23)Let f be a function defined on a neighborhood of 0 except at 0, namely U \ { } ,m be nonnegative integer, and θ ∈ R . Then we write f = O ( m ) (cid:0) r θ (cid:1) as r → f ∈ C m ( U \ { } ) and ∂ i ··· i k f ( x ) = O (cid:0) r θ − k (cid:1) as r → k = 0 , , · · · , m . Here r = | x | .Another useful notation is as follows. Let f be a function defined on a neighbor-hood of 0, namely U , m and k be nonnegative integers. Then we write f = O m (cid:0) r k (cid:1) if f ∈ C m ( U ) and f ( x ) = O (cid:0) r k (cid:1) as r →
0. Similarly we write f = O ∞ (cid:0) r k (cid:1) if f ∈ C ∞ ( U ) and f ( x ) = O (cid:0) r k (cid:1) as r → M be a smooth compact manifold with a conformal class of Riemannianmetrics. For a point p ∈ M , choose a conformal normal coordinate (see [LP]) at p , x , · · · , x n . Let the metric g = g ij dx i dx j . Then we have J ( p ) = 0 , J i ( p ) = 0 , ∆ J ( p ) = − | W ( p ) |
12 ( n − , (2.25) A ij ( p ) = 0 , A ijk ( p ) x i x j x k = 0 , (2.26)and A ijkl ( p ) x i x j x k x l = −
29 ( n − X kl ( W ikjl ( p ) x i x j ) − r n − J ij ( p ) x i x j . (2.27) Here A ijk and A ijkl are covariant derivatives of the Schouten tensor A (see (1.2)). Proposition 2.2.
Assume n ≥ and ker P = 0 . Then in conformal normalcoordinate at p , we have the following statements: • If the original conformal class is conformal flat in a neighborhood of p , thenwe may choose g such that it is flat near p , and n (2 − n ) (4 − n ) ω n G P,p = r − n + O ∞ (1) . (2.28) • If n is odd, then n (2 − n ) (4 − n ) ω n G P,p = r − n n X i =4 ψ i ! + O (1) . (2.29) Here ψ i ∈ P i . • If n is even and larger than or equal to , then n (2 − n ) (4 − n ) ω n G P,p (2.30)= r − n n X i =4 ψ i ! + r − n log r n X i = n − ψ ′ i + r − n log r n X i = n − ψ ′′ i + r − n log r · ψ ′′′ n + O (1) . Here ψ i , ψ ′ i , ψ ′′ i , ψ ′′′ i ∈ P i . • If n = 6 , then ω G P,p = r − (1 + ψ + ψ + ψ ) + r − log r (cid:0) ψ ′ + ψ ′ + ψ ′ (cid:1) (2.31)+ r − log r · ψ ′′ + O (1) . Here ψ i , ψ ′ i , ψ ′′ i ∈ P i .As a consequence, we have • If n = 5 , , or M is conformal flat near p , then n (2 − n ) (4 − n ) ω n G P,p = r − n + A + O (4) ( r ) . (2.32) Here A is a constant. • If n = 8 , then ω G P,p = r − − | W ( p ) | r + O (4) (1) . (2.33) • If n ≥ , then n (2 − n ) (4 − n ) ω n G P,p = r − n + r − n ψ + O (4) (cid:0) r − n (cid:1) , (2.34) CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 13 here ψ ∈ P and in fact ψ (2.35)= 140 ( n − X kl ( W ikjl ( p ) x i x j ) − r n + 4) X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) + | W ( p ) | n + 2) ( n + 4) r + r
48 ( n −
49 ( n + 4) X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) − n − J ij ( p ) x i x j − (cid:0) n + 6 n − (cid:1) | W ( p ) | n ( n + 4) ( n − r + r · ( n − (cid:0) n − n − (cid:1) | W ( p ) | n ( n + 2) ( n −
1) ( n −
6) ( n − . The terms in the square brackets are harmonic polynomials.
To derive these expansions, we need some algebraic preparations. Note that P m has the following decomposition (see [S]) P m = [ m ] M k =0 (cid:0) r k H m − k (cid:1) . (2.36)Under this decomposition, we have (cid:0) r ∆ (cid:1)(cid:12)(cid:12) r k H m − k = 2 k (2 m − k + n −
2) for k = 0 , , , · · · , h m i . (2.37)Here ∆ denotes the Laplace operator with respect to the Euclidean metric.For α ∈ R , let A α = r ∆ + 2 αr∂ r + α ( α + n − , (2.38)and B α = ∂∂α A α = 2 r∂ r + (2 α + n − , (2.39)then ∆ ( r α ϕ ) = r α − A α ϕ,A α (cid:0) r β ϕ (cid:1) = r β A α + β ϕ,A α ( ϕ log r ) = ( A α ϕ ) log r + B α ϕ,B α (cid:0) r β ϕ (cid:1) = r β B α + β ϕ,B α ( ϕ log r ) = ( B α ϕ ) log r + 2 ϕ. In addition, A α | P m = r ∆ + α (2 m + α + n − , (2.40) B α | P m = 2 m + 2 α + n − , (2.41)and A α | r k H m − k = ( α + 2 k ) (2 m − k + α + n −
2) (2.42)for k = 0 , , , · · · , (cid:2) m (cid:3) . In particular,( A − n A − n ) | r k H m − k (2.43)= (2 m − k ) (2 m − k + 2) (2 k + 2 − n ) (2 k + 4 − n ) , for k = 0 , , , · · · , (cid:2) m (cid:3) . Lemma 2.5.
For any real numbers α and β , and any nonnegative integer k , wehave B α (cid:16) ϕ log k r (cid:17) = B α ϕ · log k r + 2 kϕ log k − r,A α (cid:16) ϕ log k r (cid:17) = A α ϕ · log k r + kB α ϕ · log k − r + k ( k − ϕ log k − r, and A α A β (cid:16) ϕ log k r (cid:17) = A α A β ϕ · log k r + k ( A α B β ϕ + B α A β ϕ ) log k − r + k ( k −
1) ( A α ϕ + A β ϕ + B α B β ϕ ) log k − r + k ( k −
1) ( k −
2) ( B α ϕ + B β ϕ ) log k − r + k ( k −
1) ( k −
2) ( k − ϕ log k − r. Proof.
Observe ∂∂α B α ϕ = 2 ϕ, ∂ ∂α B α ϕ = 0 . Now since B α (cid:0) r β ϕ (cid:1) = r β B α + β ϕ , we know B α (cid:16) ϕ log k r (cid:17) = ∂ k ∂β k (cid:12)(cid:12)(cid:12)(cid:12) β =0 B α (cid:0) r β ϕ (cid:1) = ∂ k ∂β k (cid:12)(cid:12)(cid:12)(cid:12) β =0 (cid:0) r β B α + β ϕ (cid:1) = B α ϕ · log k r + 2 kϕ log k − r, here we have used the Newton-Lebniz formula. For the second equation, we startwith ∂∂α A α ϕ = B α ϕ, ∂ ∂α A α ϕ = 2 ϕ, ∂ ∂α A α ϕ = 0 , then A α (cid:16) ϕ log k r (cid:17) = ∂ k ∂β k (cid:12)(cid:12)(cid:12)(cid:12) β =0 A α (cid:0) r β ϕ (cid:1) = ∂ k ∂β k (cid:12)(cid:12)(cid:12)(cid:12) β =0 (cid:0) r β A α + β ϕ (cid:1) = A α ϕ · log k r + kB α ϕ · log k − r + k ( k − ϕ log k − r. Define an operator M g ϕ = 4 div ( A ( ∇ g ϕ, e i ) e i ) + (2 − n ) div ( J ∇ g ϕ ) . (2.44)The Paneitz operator can be written as P g ϕ = ∆ g ϕ + M g ϕ + n − Qϕ. (2.45)
CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 15
For any α ∈ R , define N α,g ϕ = r M g ϕ + 8 αr A ( r∂ r , ∇ g ϕ ) + 2 (2 − n ) αr J · r∂ r ϕ (2.46)+4 αr div ( A ( r∂ r , e i ) e i ) ϕ + (2 − n ) αr · r∂ r J · ϕ +4 α ( α − A ( r∂ r , r∂ r ) ϕ + (2 − n ) α ( α + n − r Jϕ, then M g ( r α ϕ ) = r α − N α,g ϕ. (2.47)At first, we claim that P g (cid:0) r − n (cid:1) = 2 n (2 − n ) (4 − n ) ω n δ p + f r − n , (2.48)with f = O ∞ (cid:0) r (cid:1) .Indeed, since r − n is radial, we have∆ g (cid:0) r − n (cid:1) = 2 n (2 − n ) (4 − n ) ω n δ p . (2.49)On the other hand, M g (cid:0) r − n (cid:1) = r − n N − n,g . In view of the facts div ( A ( r∂ r , e i ) e i )= ∂ k (cid:0) x i A ij g jk (cid:1) = g ij A ij + x i ∂ k A ij g jk + O ∞ (cid:0) r (cid:1) = J + x i A ijk ( p ) δ jk + O ∞ (cid:0) r (cid:1) = x i J i ( p ) + O ∞ (cid:0) r (cid:1) = O ∞ (cid:0) r (cid:1) , and A ( r∂ r , r∂ r ) = A ij x i x j = A ijk ( p ) x i x j x k + O ∞ (cid:0) r (cid:1) = O ∞ (cid:0) r (cid:1) , we see N − n,g ∈ O ∞ (cid:0) r (cid:1) , (2.48) follows.To continue, first we introduce a notation. For any α ∈ R , let A α,g = r ∆ g + 2 αr∂ r + α ( α + n − , (2.50)then ∆ g ( r α ϕ ) = r α − A α,g ϕ,A α,g (cid:0) r β ϕ (cid:1) = r β A α + β,g ϕ,A α,g ( ϕ log r ) = A α,g ϕ · log r + B α ϕ. Note that A α,g = A α + r (∆ g − ∆) = A α + r ∂ i (cid:0)(cid:0) g ij − δ ij (cid:1) ∂ j (cid:1) . (2.51)A straightforward computation shows P g ( r α ϕ ) = r α − ( A α − A α ϕ + K α ϕ ) , (2.52)where K α ϕ (2.53)= A α − (cid:0) r (∆ g − ∆) ϕ (cid:1) + r (∆ g − ∆) A α,g ϕ + N α,g ϕ + n − r Qϕ.
We easily see that for any nonnegative integer k , ϕ = O ∞ (cid:0) r k (cid:1) implies K α ϕ = O ∞ (cid:0) r k +2 (cid:1) .We also introduce the following two operators, K (1) α ϕ = ∂∂α K α ϕ (2.54)= B α − (cid:0) r (∆ g − ∆) ϕ (cid:1) + r (∆ g − ∆) B α ϕ +8 r A ( r∂ r , ∇ g ϕ ) + 2 (2 − n ) r J · r∂ r ϕ +4 r div ( A ( r∂ r , e i ) e i ) ϕ + (2 − n ) r · r∂ r J · ϕ +8 ( α − A ( r∂ r , r∂ r ) ϕ + (2 − n ) (2 α + n − r Jϕ, and K (2) α ϕ = ∂∂α K (1) α ϕ (2.55)= 4 r (∆ g − ∆) ϕ + 8 A ( r∂ r , r∂ r ) ϕ + 2 (2 − n ) r Jϕ = K (2) ϕ. Clearly, ϕ = O ∞ (cid:0) r k (cid:1) for some nonnegative integer would imply K (1) α ϕ, K (2) ϕ = O ∞ (cid:0) r k +2 (cid:1) . In addition, these operators satisfy the following K α (cid:0) r β ϕ (cid:1) = r β K α + β ϕ,K α ( ϕ log r ) = K α ϕ · log r + K (1) α ϕ,K (1) α (cid:0) r β ϕ (cid:1) = r β K (1) α + β ϕ,K (1) α ( ϕ log r ) = K (1) α ϕ · log r + K (2) α ϕ,K (2) (cid:0) r β ϕ (cid:1) = r β K (2) ϕ,K (2) ( ϕ log r ) = K (2) ϕ · log r. More generally, we have
Lemma 2.6.
For any nonnegative integer k , we have K (1) α (cid:16) ϕ log k r (cid:17) = K (1) α ϕ · log k r + kK (2) ϕ · log k − r,K α (cid:16) ϕ log k r (cid:17) = K α ϕ · log k r + kK (1) α ϕ · log k − r + k ( k − K (2) ϕ · log k − ϕ. This follows from the same proof of Lemma 2.4.
Case 2.1.
The dimension n is odd. In this case, we claim that we may find a ψ = P ni =1 ψ i , with ψ i ∈ P i such that A − n A − n ψ + K − n ψ + f = O ∞ (cid:0) r n +1 (cid:1) . (2.56)Once this has been done, then we have r − n ( A − n A − n ψ + K − n ψ + f ) ∈ C α for any 0 < α < ψ ∈ C ,α such that P g ψ = − r − n ( A − n A − n ψ + K − n ψ + f ) . Then P g (cid:0) r − n (1 + ψ ) + ψ (cid:1) = 2 n (2 − n ) (4 − n ) ω n δ p . (2.57) CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 17
Hence the Green’s function satisfies2 n (2 − n ) (4 − n ) ω n G p = r − n (1 + ψ ) + ψ + O ∞ (1) . (2.58)To define ψ , · · · , ψ n , we let ψ = 0 , ψ = 0 and ψ = 0. One easily see f = A − n A − n ( ψ + ψ + ψ ) + K − n ( ψ + ψ + ψ ) + f (2.59)= f = O ∞ (cid:0) r (cid:1) . Assume we have found ψ , ψ , · · · , ψ k for 3 ≤ k ≤ n −
1, such that ψ i ∈ P i and f k = A − n A − n k X i =1 ψ i ! + K − n k X i =1 ψ i ! + f = O ∞ (cid:0) r k +1 (cid:1) , then we write f k = φ k +1 + O ∞ (cid:0) r k +2 (cid:1) , φ k +1 ∈ P k +1 . Since A − n A − n | r j H k +1 − j = (2 ( k + 1) − j ) (2 ( k + 1) − j + 2) (2 j + 2 − n ) (2 j + 4 − n ) = 0for j = 0 , , , · · · , (cid:2) k +12 (cid:3) , A − n A − n is invertible on P k +1 . We may find a unique ψ k +1 ∈ P k +1 , such that A − n A − n ψ k +1 + φ k +1 = 0 . (2.60)Then f k +1 = A − n A − n k +1 X i =1 ψ i ! + K − n k +1 X i =1 ψ i ! + f = f k + A − n A − n ψ k +1 + K − n ψ k +1 = O ∞ (cid:0) r k +2 (cid:1) . This finishes the induction process.
Case 2.2. n is even and larger than or equal to . In this case, we first set ψ = 0 , ψ = 0 and ψ = 0. Since A − n A − n is invertibleon P k for 0 ≤ k ≤ n −
5, by the same induction procedure as Case 2.1, we can find ψ , · · · , ψ n − such that ψ i ∈ P i and f n − = A − n A − n n − X i =1 ψ i ! + K − n n − X i =1 ψ i ! + f = O ∞ (cid:0) r n − (cid:1) . To continue, we write f n − = φ n − + O ∞ (cid:0) r n − (cid:1) , φ n − ∈ P n − . Let ψ (0) n − = α (0) n − + β (0) n − log r with α (0) n − , β (0) n − ∈ P n − , then A − n A − n ψ (0) n − = A − n A − n α (0) n − + ( A − n B − n + B − n A − n ) β (0) n − + A − n A − n β (0) n − · log r. Let β (0) n − ∈ r n − H , then since( A − n B − n + B − n A − n ) | r n − H = − n −
2) ( n − = 0 , and A − n A − n | r k H n − − k = (2 ( n − − k ) (2 ( n − − k + 2) (2 k + 2 − n ) (2 k + 4 − n ) = 0 , for 0 ≤ k ≤ n −
3, we may find a α (0) n − ∈ P n − and a β (0) n − ∈ r n − H such that A − n A − n ψ (0) n − + φ n − = 0 . This implies f n − = A − n A − n n − X i =1 ψ i + ψ (0) n − ! + K − n n − X i =1 ψ i + ψ (0) n − ! + f = f n − + A − n A − n ψ (0) n − + K − n ψ (0) n − = O ∞ (cid:0) r n − (cid:1) + O ∞ (cid:0) r n − (cid:1) log r. Next we write f n − = φ n − + O ∞ (cid:0) r n − (cid:1) log r + O ∞ (cid:0) r n − (cid:1) , φ n − ∈ P n − . Again by similar arguments, we can find ψ (0) n − ∈ P n − + r n − H log r such that A − n A − n ψ (0) n − + φ n − = 0 . Then f n − = A − n A − n n − X i =1 ψ i + ψ (0) n − + ψ (0) n − ! + K − n n − X i =1 ψ i + ψ (0) n − + ψ (0) n − ! + f = f n − + A − n A − n ψ (0) n − + K − n ψ (0) n − = O ∞ (cid:0) r n − (cid:1) log r + O ∞ (cid:0) r n − (cid:1) . We write f n − = φ (1) n − log r + O ∞ (cid:0) r n − (cid:1) + O ∞ (cid:0) r n − (cid:1) log r. Similar as before, we may find ψ (1) n − ∈ P n − log r + (cid:0) r n − H + r n − H (cid:1) log r such that A − n A − n ψ (1) n − + φ (1) n − log r ∈ P n − . Indeed, for ψ (1) n − = α (1) n − log r + β (1) n − log r , with α (1) n − , β (1) n − ∈ P n − , we have A − n A − n ψ (1) n − = (cid:16) A − n A − n α (1) n − + 2 ( A − n B − n + B − n A − n ) β (1) n − (cid:17) log r + A − n A − n β (1) n − · log r + P n − . Let β (1) n − ∈ r n − H + r n − H . Since2 ( A − n B − n + B − n A − n ) | r n − H = 4 n ( n − = 0 , A − n B − n + B − n A − n ) | r n − H = − n ( n + 2) = 0 , and A − n A − n | r k H n − − k = (2 ( n − − k ) (2 ( n − − k + 2) (2 k + 2 − n ) (2 k + 4 − n ) = 0 CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 19 for 0 ≤ k ≤ n −
3, we may find the above needed ψ (1) n − . Then f (1) n − = A − n A − n n − X i =1 ψ i + ψ (0) n − + ψ (0) n − + ψ (1) n − ! + K − n n − X i =1 ψ i + ψ (0) n − + ψ (0) n − + ψ (1) n − ! + f = f n − + A − n A − n ψ (1) n − + K − n ψ (1) n − = O ∞ (cid:0) r n − (cid:1) + O ∞ (cid:0) r n − (cid:1) log r + O ∞ ( r n ) log r. The next step is to remove the P n − term in O ∞ (cid:0) r n − (cid:1) , then the P n − log r termin O ∞ (cid:0) r n − (cid:1) log r and so on, until we reach O ∞ (cid:0) r n +1 (cid:1) log r + O ∞ (cid:0) r n +1 (cid:1) log r + O ∞ (cid:0) r n +1 (cid:1) + O ∞ (cid:0) r n +2 (cid:1) log r . That is, we find ψ (0) n − ∈ P n − + r n − H log r,ψ (0) n − ∈ P n − + r n − H log r,ψ (1) n − ∈ P n − log r + (cid:0) r n − H + r n − H (cid:1) log r,ψ (0) n − ∈ P n − + (cid:0) r n − H + r n − H (cid:1) log r,ψ (1) n − ∈ P n − log r + (cid:0) r n − H + r n − H (cid:1) log r,ψ (0) n − ∈ P n − + (cid:0) r n − H + r n − H (cid:1) log r,ψ (2) n ∈ P n log r + (cid:0) r n − H + r n − H (cid:1) log r,ψ (1) n ∈ P n log r + (cid:0) r n − H + r n − H (cid:1) log r, and ψ (0) n ∈ P n + (cid:0) r n − H + r n − H (cid:1) log r, such that f n = A − n A − n n − X i =1 ψ i + n X i = n − ψ (0) i + n X i = n − ψ (1) i + ψ (2) n ! + K − n n − X i =1 ψ i + n X i = n − ψ (0) i + n X i = n − ψ (1) i + ψ (2) n ! + f = O ∞ (cid:0) r n +1 (cid:1) log r + O ∞ (cid:0) r n +1 (cid:1) log r + O ∞ (cid:0) r n +1 (cid:1) + O ∞ (cid:0) r n +2 (cid:1) log r. Clearly r − n f n ∈ C α for any 0 < α <
1. This implies locally we may find ψ ∈ C ,α such that P g ψ = − r − n f n . Let ψ = n − X i =1 ψ i + n X i = n − ψ (0) i + n X i = n − ψ (1) i + ψ (2) n , then P g (cid:0) r − n (1 + ψ ) + ψ (cid:1) = 2 n (2 − n ) (4 − n ) ω n δ p on a small disk. Hence2 n (2 − n ) (4 − n ) ω n G p = r − n (1 + ψ ) + ψ + O ∞ (1) . Case 2.3. n = 6 . This case can be done similarly as for Case 2.2. That is, we can find ψ (0)4 ∈ P + (cid:0) r H + r H (cid:1) log r,ψ (0)5 ∈ P + (cid:0) r H + r H (cid:1) log r,ψ (1)6 ∈ P log r + (cid:0) r H + r H (cid:1) log r, and ψ (0)6 ∈ P + (cid:0) r H + r H (cid:1) log r, such that f = A − A − (cid:16) ψ (0)4 + ψ (0)5 + ψ (0)6 + ψ (1)6 (cid:17) + K − (cid:16) ψ (0)4 + ψ (0)5 + ψ (0)6 + ψ (1)6 (cid:17) + f = O ∞ (cid:0) r (cid:1) log r + O ∞ (cid:0) r (cid:1) + O ∞ (cid:0) r (cid:1) log r. The remaining argument can be done as before.
Case 2.4. M is conformal flat near p . In this case, we may take the metric g such that it is flat near p . This implies P g = ∆ , and hence P g (cid:0) r − n (cid:1) = 2 n (2 − n ) (4 − n ) ω n δ p . It follows that 2 n (2 − n ) (4 − n ) ω n G P,p = r − n + O ∞ (1) . Finally, to get the leading terms in the expansion for n ≥
8, by a direct compu-tation we have f = f = φ + O ∞ (cid:0) r (cid:1) , with φ ∈ P and φ (2.61)= − n − X kl ( W ikjl ( p ) x i x j ) + 2 ( n −
4) ( n − r J ij ( p ) x i x j + ( n − | W ( p ) |
24 ( n − r . From this, we can compute the leading terms of G P,p directly from the argumentsin Case 2.2. 3.
Existence and regularity of maximizers
The main aim of this section is to show the strict inequality between Θ ( g )and Θ ( S n ) in the assumption of Proposition 2.1 is valid as long as ( M, g ) is notconformally equivalent to the standard sphere. As in the Yamabe problem case([LP]), this is achieved by a careful choice of test function.
Proposition 3.1.
Assume ( M, g ) is a smooth compact n dimensional Riemannianmanifold with n ≥ , Y ( g ) > , Q ≥ and not identically zero, then Θ ( g ) ≥ Θ ( S n ) (3.1) and equality holds if and only if ( M, g ) is conformally equivalent to the standardsphere. CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 21
Before we start the proof of Proposition 3.1, we list several basic identities whichwill facilitate the calculations. For b > − n and 2 a − b > n , Z R n | x | b (cid:16) | x | + 1 (cid:17) a dx = nω n (cid:0) b + n (cid:1) Γ (cid:0) a − b + n (cid:1) Γ ( a ) = π n Γ (cid:0) b + n (cid:1) Γ (cid:0) a − b + n (cid:1) Γ ( a ) Γ (cid:0) n (cid:1) . (3.2)If we fix an orthonormal frame at p , and let ∆ be the Euclidean Laplacian, then∆ X k,l ( W ikjl ( p ) x i x j ) (3.3)= 4 W ikjl ( p ) W ikml ( p ) x j x m + 4 W ikjl ( p ) W ilmk ( p ) x j x m = 2 X ikl ( W ikjl ( p ) x j + W iljk ( p ) x j ) = 2 X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) , and∆ X kl ( W ikjl ( p ) x i x j ) = 8 (cid:16) | W ( p ) | + W ikjl ( p ) W iljk ( p ) (cid:17) = 12 | W ( p ) | , (3.4)here we have used W ikjl ( p ) W iljk ( p ) = 12 | W ( p ) | , (3.5)which follows from the usual Bianchi identity. Hence X kl ( W ikjl ( p ) x i x j ) (3.6)= X kl ( W ikjl ( p ) x i x j ) − r n + 4 X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) + 32 ( n + 2) ( n + 4) | W ( p ) | r (cid:21) + r · n + 4 X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) − n ( n + 4) | W ( p ) | r (cid:21) + r · n ( n + 2) | W ( p ) | . The polynomials in the square brackets are harmonic. In particular, Z S n − X kl ( W ikjl ( p ) x i x j ) dS = 3 ω n n + 2) | W ( p ) | . (3.7)Due to the fact that in (2.2) the power nn +4 < < nn − , to control the error onannulus region, the choice of test functions for Θ ( g ) will be more delicate thanthose for the classical Yamabe problem or for Y ( g ) (see (1.13)) in the literature.In particular, dimension 8 and 9 has to be separated from dimensions 5 , , η ∈ C ∞ ( R , R ) such that η | ( −∞ , = 0 , η | (2 , ∞ ) = 1 and 0 ≤ η ≤
1. Denote η = 1 − η . For convenience we always denote H = 2 n (2 − n ) (4 − n ) ω n G P,p . (3.8) Let δ be a small positive number. For 0 < λ < δ , let u λ = λ | x | + λ ! n − (3.9)and β = λ n − r − n − u λ . (3.10)If we write φ ( x ) = | x | − n − (cid:16) | x | + 1 (cid:17) − n , then β = λ − n φ (cid:0) xλ (cid:1) . Case 3.1. M is conformally flat near p , n ≥ . In this case we can assume that the metric g is flat near p . Using the Euclideancoordinate at p , namely x , · · · , x n we have H = r − n + A + α. (3.11)Here A is a constant, α = O ∞ ( r ) is a biharmonic function (with respect to Eu-clidean metric).Define ϕ λ = ( u λ + η (cid:0) rδ (cid:1) β + λ n − A + λ n − α, on B δ ( p ) ,λ n − H, on M \ B δ ( p ) . (3.12)It is clear that ϕ λ ∈ C ∞ ( M ). Note that P ϕ λ (3.13)= n ( n + 2) ( n −
2) ( n − λ n +42 (cid:16) | x | + λ (cid:17) − n +42 , on B δ ( p ) ,O (cid:16) λ n (cid:17) , on B δ ( p ) \ B δ ( p ) , , on M \ B δ ( p ) . Hence Z M | P ϕ λ | nn +4 dµ (3.14)= ( n ( n + 2) ( n −
2) ( n − nn +4 Γ (cid:0) n (cid:1) π n ( n − O (cid:16) λ n n +4 (cid:17) . It follows that k P ϕ λ k L nn +4 (3.15)= ( n ( n + 2) ( n −
2) ( n − Γ (cid:0) n (cid:1) n +4 n π n +42 (( n − n +4 n + O (cid:16) λ n n +4 (cid:17) . On the other hand, Z M P ϕ λ · ϕ λ dµ (3.16)= n ( n + 2) ( n −
2) ( n −
4) Γ (cid:0) n (cid:1) π n ( n − n −
2) ( n − π n Γ (cid:0) n (cid:1) A λ n − + o (cid:0) λ n − (cid:1) . CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 23
Hence R M P ϕ λ · ϕ λ dµ k P ϕ λ k L nn +4 (3.17)= Θ ( S n ) + 4 (( n − n +4 n n ( n + 2) ( n −
2) ( n −
4) Γ (cid:0) n (cid:1) n +4 n π A λ n − + o (cid:0) λ n − (cid:1) . If (
M, g ) is not conformally diffeomorphic to the standard sphere, then it followsfrom the arguments in [HY4, section 6] that A >
0. Fix δ small and let λ ↓
0, wesee Θ ( g ) > Θ ( S n ). Case 3.2. n = 5 , , . In this case by a conformal change of the metric we can assume exp p preservesthe volume near p (note this is another way of saying we choose a conformal normalcoordinate, see [LP]). Using the normal coordinate at p , x , · · · , x n , we have H = r − n + A + α. (3.18)Here A is a constant and α = O (4) ( r ). Define ϕ λ = ( u λ + η (cid:0) rδ (cid:1) β + λ n − A + λ n − α, on B δ ( p ) ,λ n − H, on M \ B δ ( p ) . (3.19)then ϕ λ ∈ W , nn +4 ( M ). On B δ ( p ) \ { p } ,P ϕ λ (3.20)= P u λ − λ n − P (cid:0) r − n (cid:1) = ∆ u λ − A ( ∇ β, e i ) e i ) + ( n −
2) div ( J ∇ β ) − n − Qβ = n ( n + 2) ( n −
2) ( n − λ n +42 (cid:16) | x | + λ (cid:17) − n +42 + O (cid:16) λ n | x | − n (cid:17) . Here we will need to use (2.25) and (2.26). On B δ ( p ) \ B δ ( p ), P ϕ λ = − P (cid:16) η (cid:16) rδ (cid:17) β (cid:17) = O (cid:16) λ n (cid:17) (3.21)and on M \ B δ ( p ), P ϕ λ = 0. Hence Z M | P ϕ λ | nn +4 dµ = ( n ( n + 2) ( n −
2) ( n − nn +4 Γ (cid:0) n (cid:1) π n ( n − o (cid:0) λ n − (cid:1) , and k P ϕ λ k L nn +4 = ( n ( n + 2) ( n −
2) ( n − Γ (cid:0) n (cid:1) n +4 n π n +42 (( n − n +4 n + o (cid:0) λ n − (cid:1) . On the other hand, Z M P ϕ λ · ϕ λ dµ (3.22)= n ( n + 2) ( n −
2) ( n −
4) Γ (cid:0) n (cid:1) π n ( n − n −
2) ( n − π n Γ (cid:0) n (cid:1) A λ n − + o (cid:0) λ n − (cid:1) . Summing up we have R M P ϕ λ · ϕ λ dµ k P ϕ λ k L nn +4 (3.23)= Θ ( S n ) + 4 (( n − n +4 n n ( n + 2) ( n −
2) ( n −
4) Γ (cid:0) n (cid:1) n +4 n π A λ n − + o (cid:0) λ n − (cid:1) . By [HY4, section 6] we know when (
M, g ) is not conformally diffeomorphic to thestandard sphere, A is strictly positive. Letting λ ↓
0, we get Θ ( g ) > Θ ( S n ) inthis case. Case 3.3. ( M, g ) is not locally conformally flat and n = 8 . In this case we can choose p such that W ( p ) = 0. By a conformal change ofthe metric we can assume exp p preserves the volume near p . Using the normalcoordinate at p , x , · · · , x , we have H = r − − | W ( p ) | r + α. (3.24)Here α = O (4) (1). Define ϕ λ = ( u λ + η (cid:0) rδ (cid:1) β − | W ( p ) | λ log r + λ α, on B δ ( p ) ,λ H, on M \ B δ ( p ) . (3.25)Then ϕ ∈ W , ( M ). On B δ ( p ) \ { p } ,P ϕ λ = P u λ − λ P (cid:0) r − (cid:1) (3.26)= 1920 λ (cid:16) | x | + λ (cid:17) − − A ( ∇ β, e i ) e i ) + 6 div ( J ∇ β ) − Qβ = 1920 λ (cid:16) | x | + λ (cid:17) − + O ( β ) + O (cid:0) β ′ r (cid:1) + O (cid:0) β ′′ r (cid:1) . Here we have used (2.25) and (2.26). On B δ ( p ) \ B δ ( p ), P ϕ λ = − P (cid:16) η (cid:16) rδ (cid:17) β (cid:17) = O (cid:0) λ (cid:1) (3.27)and on M \ B δ ( p ), P ϕ λ = 0. Note that β = λ r − − λ (cid:0) r + λ (cid:1) − ,β ′ = − λ r − + 4 λ (cid:0) r + λ (cid:1) − r,β ′′ = 20 λ r − − λ (cid:0) r + λ (cid:1) − r + 4 λ (cid:0) r + λ (cid:1) − . Hence we have Z M | P ϕ λ | dµ = 1920 π
840 + O (cid:0) λ (cid:1) , (3.28) CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 25 and Z M P ϕ λ · ϕ λ dµ = 1920 π
840 + π | W ( p ) | λ log 1 λ + O (cid:0) λ (cid:1) . It follows that R M P ϕ λ · ϕ λ dµ k P ϕ λ k L = Θ (cid:0) S (cid:1) + 210 π | W ( p ) | λ log 1 λ + O (cid:0) λ (cid:1) . Hence Θ ( g ) > Θ (cid:0) S (cid:1) . Case 3.4. M is not conformally flat and n = 9 . In this case we can choose p such that W ( p ) = 0. By a conformal changeof metric we can assume exp p preserves the volume near p . Using the normalcoordinate at p , x , · · · , x , we have H = r − + r − ψ + α. (3.29)Here α = O (4) (1) and ψ (3.30)= 1280 X kl ( W ikjl ( p ) x i x j ) − r X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) + | W ( p ) | r + r X jkl ( W ijkl ( p ) x i + W ilkj ( p ) x i ) − J ij ( p ) x i x j − | W ( p ) | r (cid:21) + 8051368576 | W ( p ) | r . Define ϕ λ = ( u λ + η (cid:0) rδ (cid:1) β + λ r − ψ + λ α, on B δ ( p ) ,λ H, on M \ B δ ( p ) . (3.31)Then ϕ ∈ W , ( M ). On B δ ( p ) \ { p } ,P ϕ λ = P u λ − λ P (cid:0) r − (cid:1) (3.32)= 3465 λ (cid:16) | x | + λ (cid:17) − − A ( ∇ β, e i ) e i ) + 7 div ( J ∇ β ) − Qβ = 3465 λ (cid:16) | x | + λ (cid:17) − − (cid:18) β ′ r (cid:19) ′ A ijkl ( p ) x i x j x k x l r + 72 (cid:18) β ′ r (cid:19) ′ rJ ij ( p ) x i x j + 652 β ′ r J ij ( p ) x i x j − | W ( p ) | β + O ( βr ) + O (cid:0) β ′ r (cid:1) + O (cid:0) β ′′ r (cid:1) . On B δ ( p ) \ B δ ( p ), P ϕ λ = − P (cid:16) η (cid:16) rδ (cid:17) β (cid:17) = O (cid:16) λ (cid:17) (3.33) and on M \ B δ ( p ), P ϕ λ = 0. Note that β = λ r − − λ (cid:0) r + λ (cid:1) − ,β ′ = − λ r − + 5 λ (cid:0) r + λ (cid:1) − r,β ′ r = − λ r − + 5 λ (cid:0) r + λ (cid:1) − ,β ′′ = 30 λ r − − λ (cid:0) r + λ (cid:1) − r + 5 λ (cid:0) r + λ (cid:1) − , (cid:18) β ′ r (cid:19) ′ = 35 λ r − − λ (cid:0) r + λ (cid:1) − r. A straightforward calculation shows Z M | P ϕ λ | dµ (3.34)= 3465 π (cid:20) (cid:18) π − (cid:19) | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:21) , hence k P ϕ λ k L (3.35)= 3465 π (cid:20) (cid:18) π − (cid:19) | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:21) . On the other hand, Z M P ϕ λ · ϕ λ dµ (3.36)= 11552048 π (cid:20) (cid:18) π − (cid:19) | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:21) . Summing up we get R M P ϕ λ · ϕ λ dµ k P ϕ λ k L = Θ (cid:0) S (cid:1) (cid:18) | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:19) . (3.37)Hence we see that Θ ( g ) > Θ (cid:0) S (cid:1) . Case 3.5. M is not conformally flat and n ≥ . We can find a point p such that W ( p ) = 0. Let x , · · · , x n be conformal normalcoordinate at p , δ be a small fixed positive number, and ϕ λ = u λ ( x ) η (cid:18) | x | δ (cid:19) . (3.38)Then on B δ ( p ) \ B δ ( p ), P ϕ λ = O (cid:16) λ n − (cid:17) . (3.39) CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 27 On B δ ( p ) ,P ϕ λ (3.40)= n ( n + 2) ( n −
2) ( n − λ n +42 (cid:16) | x | + λ (cid:17) − n +42 −
49 ( n − λ n − (cid:16) | x | + λ (cid:17) − n X kl ( W ikjl ( p ) x i x j ) + n − λ n − (cid:16) | x | + λ (cid:17) − n (cid:16) n − | x | + (cid:0) n − (cid:1) λ (cid:17) J ij ( p ) x i x j + n −
424 ( n − λ n − | W ( p ) | (cid:16) | x | + λ (cid:17) − n − + O (cid:18) λ n − (cid:16) | x | + λ (cid:17) − n − | x | (cid:19) . Using the basic inequality (cid:12)(cid:12)(cid:12)(cid:12) | t | nn +4 − − nn + 4 t (cid:12)(cid:12)(cid:12)(cid:12) ≤ C | t | nn +4 (3.41)we see on B δ ( p ) , | P ϕ λ | nn +4 = ( n ( n + 2) ( n −
2) ( n − nn +4 λ n (cid:16) | x | + λ (cid:17) − n · − λ − (cid:16) | x | + λ (cid:17) ( n + 2) ( n + 4) ( n − X kl ( W ikjl ( p ) x i x j ) + λ − (cid:16) | x | + λ (cid:17) ( n + 2) ( n + 4) ( n − (cid:16) n − | x | + (cid:0) n − (cid:1) λ (cid:17) J ij ( p ) x i x j + λ − | W ( p ) |
12 ( n + 2) ( n + 4) ( n −
1) ( n − (cid:16) | x | + λ (cid:17) + O (cid:18) λ − (cid:16) | x | + λ (cid:17) | x | (cid:19) + O (cid:18) λ − nn +4 (cid:16) | x | + λ (cid:17) nn +4 (cid:19) + O (cid:18) λ − nn +4 (cid:16) | x | + λ (cid:17) nn +4 | x | nn +4 (cid:19)(cid:21) . A straightforward calculation shows Z M | P ϕ λ | nn +4 dµ (3.42)= ( n ( n + 2) ( n −
2) ( n − nn +4 π n Γ (cid:0) n (cid:1) ( n − · (cid:18) − n − n − n + 2) ( n + 4) ( n −
2) ( n −
6) ( n − | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:19) . Hence k P ϕ λ k L nn +4 (3.43)= ( n ( n + 2) ( n −
2) ( n − π n +42 Γ (cid:0) n (cid:1) n +4 n (( n − n +4 n · (cid:18) − n − n − n ( n + 2) ( n −
2) ( n −
6) ( n − | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:19) . On the other hand, Z M P ϕ λ · ϕ λ dµ (3.44)= n ( n + 2) ( n −
2) ( n − π n Γ (cid:0) n (cid:1) ( n − · (cid:18) − n − n − n ( n + 2) ( n −
2) ( n −
6) ( n − | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:19) . Summing up we get R M P ϕ λ · ϕ λ dµ k P ϕ λ k L nn +4 (3.45)= Θ ( S n ) (cid:18) n − n − n ( n + 2) ( n −
2) ( n −
6) ( n − | W ( p ) | λ + o (cid:0) λ (cid:1)(cid:19) . It follows that Θ ( g ) > Θ ( S n ).Next we turn to the regularity issue for maximizers of Θ ( g ) in (1.16). Assume f ∈ L nn +4 ( M ), f ≥ ( g ),then after scaling we have G P f = 2 n − f n − n +4 . (3.46)Note that this equation is critical in the sense that if we start with f ∈ L nn +4 anduse the equation, the usual bootstrap method simply ends with f ∈ L nn +4 again.Approaches in deriving further regularity for such kind of equations has been wellunderstood (see for example [DHL, ER, R, V] and so on). Here is a regularity resultparticularly tailored for our purpose, we refer the readers to [DHL, ER, R, V] fordetailed proofs. Lemma 3.1.
Assume ( M, g ) is a smooth compact n dimensional Riemannian man-ifold with n ≥ , Y ( g ) > , Q ≥ and not identically zero, f ∈ L nn +4 ( M ) , f ≥ and not identically zero, and it satisfies (3.46), then f ∈ C ∞ ( M ) , f > . Now we have all the ingredients to prove Theorem 1.3. Theorem 1.1 clearlyfollows from Theorem 1.3.
Proof of Theorem 1.3.
If (
M, g ) is conformal equivalent to the standard sphere,then everything follows from discussions in Section 2.2. From now on we assumethat (
M, g ) is not conformally equivalent to the standard sphere. By Proposition3.1 we know that Θ ( g ) > Θ ( S n ). [HY4, Proposition 1.1] tells us ker P = 0 and CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 29 G P >
0. By Proposition 2.1 we know the set M = (cid:26) f ∈ L nn +4 ( M ) : k f k L nn +4 ( M ) = 1 , Z M G P f · f dµ = Θ ( g ) (cid:27) is nonempty and compact in L nn +4 ( M ). If f ∈ M , we can assume f + = 0, then f − must be equal to zero. IndeedΘ ( g )= Z M G P f · f dµ = Z M (cid:0) G P f + · f + − G P f + · f − + G P f − · f − (cid:1) dµ ≤ Z M G P | f | · | f | dµ ≤ Θ ( g ) . Hence R M G P f + · f − dµ = 0. Using the fact that G P > f + = 0, we see f − = 0. In another word, f does not change sign. It follows from Lemma 3.1that f ∈ C ∞ ( M ) and f >
0. Moreover the compactness of M under C ∞ ( M )topology follows from its compactness in L nn +4 ( M ) and the proofs of Lemma 3.1in [DHL, ER, R, V]. 4. Some discussions
Here we turn to the variational problem (1.13).
Proposition 4.1.
Let ( M, g ) be a smooth compact n dimensional Riemannianmanifold with n ≥ , Y ( g ) > , Q ≥ and not identically zero, then (1) Y ( g ) ≤ Y ( S n ) , here S n has the standard metric. Y ( g ) = Y ( S n ) if andonly if ( M, g ) is conformally diffeomorphic to the standard sphere. (2) Y ( g ) is always achieved. Let M P = (cid:26) u ∈ H ( M ) : k u k L nn − ( M ) = 1 and E ( u ) = Y ( g ) (cid:27) , (4.1) then M P is not empty. For any α ∈ (0 , , M P ⊂ C ,α ( M ) and when ( M, g ) is not conformally diffeomorphic to the standard sphere, M P is com-pact under C ,α topology. We start with the following standard fact (see [DHL, He]).
Lemma 4.1.
Let M P = (cid:26) u ∈ H ( M ) : k u k L nn − ( M ) = 1 and E ( u ) = Y ( g ) (cid:27) . If Y ( g ) < Y ( S n ) , then M P is nonempty. Moreover for any α ∈ (0 , , M P ⊂ C ,α ( M ) and it is compact in C ,α topology.Proof of Proposition 4.1. If (
M, g ) is conformal equivalent to the standard sphere,then the conclusion follows from discussions in Section 2.2. Assume (
M, g ) is notconformal equivalent to the standard sphere, then it follows from Lemma 2.2 andProposition 3.1 that Y ( g ) ≤ ( g ) < ( S n ) = Y ( S n ) . Here we want to point out that the fact Y ( g ) < Y ( S n ) can be verified, with thehelp of positive mass theorem for Paneitz operator ([HuR, GM, HY4]), by choosinga particular test function in (1.13) (see [ER, R, GM]). In fact the correspondingcalculation is easier than what we have in the proof of Proposition 3.1, but thestatement in Proposition 3.1 is stronger. By Lemma 4.1, we know M P is nonemptyand M P ⊂ C ,α ( M ) and it is compact in C ,α ( M ) for any α ∈ (0 , Proposition 4.2.
Let ( M, g ) be a smooth compact n dimensional Riemannianmanifold with n ≥ , Y ( g ) > , Y ( g ) > , Q ≥ and not identically zero. Denote M P = (cid:26) u ∈ H ( M ) : k u k L nn − ( M ) = 1 and E ( u ) = Y ( g ) (cid:27) and M Θ = u ∈ W , nn +4 ( M ) : k u k L nn − ( M ) = 1 and E ( u ) k P u k L nn +4 = Θ ( g ) . then (1) M P ⊂ C ∞ ( M ) and for any u ∈ M P , either u > or − u > . (2) Y ( g ) Θ ( g ) = 1 . (3) M P = M Θ .Proof. By Proposition 4.1 we know M P is nonempty and for any α ∈ (0 , M P ⊂ C ,α ( M ). By [HY4, Proposition 1.1] we know G P >
0. Assume u ∈ M P , withoutlosing of generality we can assume u + = 0. Now we will use an observation in [R]to show u >
0. In fact u satisfies k u k L nn − = 1 and P u = Y ( g ) | u | n − u. Let v = G P ( | P u | ), then v ∈ C ,α ( M ), v > | u | ≤ v . We have Y ( g ) ≤ E ( v ) k v k L nn − = Y ( g ) R M | u | n +4 n − vdµ k v k L nn − ≤ Y ( g ) k v k − L nn − ≤ Y ( g ) . Hence all the inequalities become equalities. In particular k v k L nn − = 1 = k u k L nn − .Since v ≥ | u | , we see v = | u | . This together with u + = 0 implies u = v >
0. Stan-dard bootstrap method shows u ∈ C ∞ ( M ). Hence M P ⊂ C ∞ ( M ), moreover when( M, g ) is not conformally diffeomorphic to the standard sphere, M P is compact in C ∞ ( M ).For u ∈ M P , we can assume u >
0, then k u k L nn − = 1 and P u = Y ( g ) u n +4 n − . It follows that from this equation and Lemma 2.2 thatΘ ( g ) ≥ E ( u ) k P u k L nn +4 = 1 Y ( g ) ≥ Θ ( g ) . Hence Y ( g ) Θ ( g ) = 1 and u ∈ M Θ .On the other hand, if u ∈ M Θ , let f = P u , thenΘ ( g ) = R M P u · udµ k P u k L nn +4 = R M G P f · f dµ k f k L nn +4 . CURVATURE ON A CLASS OF MANIFOLDS WITH DIMENSION AT LEAST 5 31
Hence it follows from Theorem 1.3 that f ∈ C ∞ ( M ) and either f > − f > f >
0, then u = G P f ∈ C ∞ ( M ), u > P u = κu n +4 n − for some positive constant κ . Using k u k L nn − = 1 we seeΘ ( g ) = E ( u ) k P u k L nn +4 = 1 κ , and hence κ = Y ( g ). It follows that E ( u ) = Y ( g ) and hence u ∈ M P . Summingup we see M P = M Θ .Now we are ready to prove Theorem 1.2. Proof of Theorem 1.2.
It is clear Theorem 1.2 follows from Proposition 4.1 and 4.2.The compactness of M P in C ∞ topology was shown in the proof of Proposition4.2.At last we will show the approach to the Q curvature equation in Theorem 1.3gives another way to find constant scalar curvature metric in a conformal class withpositive Yamabe invariant. Here we always assume ( M, g ) is a smooth compact n dimensional Riemannian manifold with n ≥ Y ( g ) >
0. To find a conformalmetric with scalar curvature 1 is the same as solving Lρ = ρ n +2 n − , ρ ∈ C ∞ ( M ) , ρ > . (4.2)Here L is the conformal Laplacian operator. For any u ∈ C ∞ ( M ) we write E ( u ) = Z M Lu · udµ (4.3)= Z M (cid:18) n − n − |∇ u | + Ru (cid:19) dµ. Clearly E ( u ) extends continuously to u ∈ H ( M ). To solve (4.2), people considerthe variational problem (see [LP]) Y ( g ) = inf u ∈ H ( M ) \{ } E ( u ) k u k L nn − . (4.4)Denote M L = n u ∈ H ( M ) : k u k L nn − = 1 and E ( u ) = Y ( g ) o , (4.5)then it is well known that M L is always nonempty, M L ⊂ C ∞ ( M ) and for any u ∈ M L , either u > − u >
0. If u >
0, then after scaling u solves (4.2).Moreover when ( M, g ) is not conformally diffeomorphic to the standard sphere, wehave Y ( g ) < Y ( S n ) and M L is compact in C ∞ topology (see [LP, S]).Now we turn to another approach to solve (4.2). Since Y ( g ) >
0, we know theGreen’s function of L exists and it is always positive. We can define an operator( G L f ) ( p ) = Z M G L ( p, q ) f ( q ) dµ ( q ) . (4.6)Let f = ρ n +2 n − , then (4.2) becomes G L f = f n +2 n − , f ∈ C ∞ ( M ) , f > . (4.7) LetΘ ( g ) = sup f ∈ L nn +2 ( M ) \{ } R M G L f · f dµ k f k L nn +2 = sup u ∈ W , nn +2 ( M ) \{ } R M Lu · udµ k Lu k L nn +2 . (4.8)Note that this functional is considered in [DoZ]. Same argument as in the proof ofLemma 2.1 shows Θ ( g ) = sup e g ∈ [ g ] R M e Rd e µ (cid:13)(cid:13)(cid:13) e R (cid:13)(cid:13)(cid:13) L nn +2 ( M,d e µ ) . (4.9)With the solution to Yamabe problem ([LP, S]) we can deduce Lemma 4.2.
Let ( M, g ) be a smooth compact n dimensional Riemannian manifoldwith n ≥ , Y ( g ) > . Denote M Θ = u ∈ W , nn +2 ( M ) : k u k L nn − ( M ) = 1 and E ( u ) k Lu k L nn +2 = Θ ( g ) . Then (1) Y ( g ) Θ ( g ) = 1 . (2) M L = M Θ . Since the proof is essentially the same as the one for Proposition 4.2, we omit ithere. Roughly speaking Lemma 4.2 tells us the maximization problem for Θ ( g ) willnot produce new constant scalar curvature metrics other than those by minimizingproblem for Y ( g ). However, without using the solution to Yamabe problem , wecan use the same argument as for Theorem 1.3 to show Θ ( g ) ≥ Θ ( S n ), withequality holds if and only if ( M, g ) is conformally diffeomorphic to the standardsphere (here one needs to use the positive mass theorem); M Θ is always nonempty, M Θ ⊂ C ∞ ( M ) and any u ∈ M Θ must be either positive or negative; M Θ iscompact in C ∞ ( M ) when ( M, g ) is not conformally diffeomorphic to the standardsphere. In particular, this gives another way to solve (4.2). The details are left tointerested readers.
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