Sign of Green's function of Paneitz operators and the Q curvature
aa r X i v : . [ m a t h . DG ] M a y SIGN OF GREEN’S FUNCTION OF PANEITZ OPERATORSAND THE Q CURVATURE
FENGBO HANG AND PAUL C. YANG
Abstract.
In a conformal class of metrics with positive Yamabe invariant,we derive a necessary and sufficient condition for the existence of metrics withpositive Q curvature. The condition is conformally invariant. We also provesome inequalities between the Green’s functions of the conformal Laplacianoperator and the Paneitz operator. Introduction
Since the fundamental work [CGY] in dimension 4, the Paneitz operator and as-sociated Q curvature in dimension other than 4 (see [B, P]) attracts much attention(see [DHL, GM, HY1, HeR1, HeR2, HuR, QR] etc and the references therein). Let( M, g ) be a smooth compact n dimensional Riemannian manifold. For 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)Here R is the scalar curvature, Rc is the Ricci tensor. The Paneitz operator is givenby P ϕ (1.2)= ∆ ϕ + 4 n − Rc ( ∇ ϕ, e i ) e i ) − n − n + 82 ( n −
1) ( n −
2) div ( R ∇ ϕ ) + n − Qϕ.
Here e , · · · , e n is a local orthonormal frame with respect to g . For n = 4, underconformal transformation of the metric, the operator satisfies P ρ n − g ϕ = ρ − n +4 n − P g ( ρϕ ) . (1.3)Note this is similar to the conformal Laplacian operator, which appears naturallywhen considering transformation law of the scalar curvature under conformal changeof metric ([LP]). As a consequence we know P ρ n − g ϕ · ψdµ ρ n − g = P g ( ρϕ ) · ρψdµ g . (1.4)Here µ g is the measure associated with metric g . Moreoverker P g = 0 ⇔ ker P ρ n − g = 0 , (1.5)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 ) . (1.6) Mathematics Subject Classification.
For u, v ∈ C ∞ ( M ), we denote the quadratic form associated with P as E ( u, v ) (1.7)= 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 and E ( u ) = E ( u, u ) . (1.8)By the integration by parts formula in (1.7) we know E ( u, v ) also makes sense for u, v ∈ H ( M ).To continue we recall (see [LP]) for n ≥
3, on a smooth compact Riemannianmanifold ( M n , g ), the conformal Laplacian operator is given by L g ϕ = − n − n − ϕ + Rϕ. (1.9)The Yamabe invariant is defined as Y ( g ) (1.10)= inf ( R M e Rd e µ ( e µ ( M )) n − n : e g = ρ g for some positive smooth function ρ ) = inf R M L g ϕ · ϕdµ k ϕ k L nn − : ϕ is a nonzero smooth function on M . A basic but useful fact about the scalar curvature is Y ( g ) > ⇔ λ ( L g ) > ⇔ g is conformal to a metric with scalar curvature > > ” by ”= ”or ” < ”. Here λ ( L g ) is the first eigenvalue of L g . It is clear Y ( g ) is a conformalinvariant, on the other hand the sign of λ ( L g ) is also conformally invariant. Themain reason that (1.11) holds is based on the fact the first eigenfunction of a secondorder symmetric differential operator does not change sign. Unfortunately suchkind of property is known to be not valid for higher order operators. The followingquestion keeps puzzling people from the beginning of research on Q curvature indimension other than 4, namely: can we find a conformal invariant condition whichis equivalent to the existence of positive Q curvature in the conformal class (in thesame spirit as (1.11))? Here we give an answer to this question for conformal classwith positive Yamabe invariant. Theorem 1.1.
Let n > and ( M n , g ) be a smooth compact Riemannian manifoldwith Yamabe invariant Y ( g ) > , then the following statements are equivalent (1) there exists a positive smooth function ρ with Q ρ g > . (2) ker P g = 0 and the Green’s function G P ( p, q ) > for any p, q ∈ M, p = q . (3) ker P g = 0 and there exists a p ∈ M such that G P ( p, q ) > for q ∈ M \ { p } . IGN OF GREEN’S FUNCTION AND THE Q CURVATURE 3
Along the way we also find the following comparison inequality between Green’sfunction of L and P . Proposition 1.1.
Assume n > , ( M n , g ) is a smooth compact Riemannian man-ifold with Y ( g ) > , Q ≥ and not identically zero, then ker P = 0 and G n − n − L ≤ c n G P . (1.12) Here c n = 2 − n − n − n n − ( n − − n − n − ( n −
2) ( n − ω n − n , (1.13) ω n is the volume of unit ball in R n . Moreover if G n − n − L ( p, q ) = c n G P ( p, q ) for some p = q , then ( M, g ) is conformal diffeomorphic to the standard sphere. In dimension 3 we have
Theorem 1.2.
Let ( M, g ) be a smooth compact dimensional Riemannian mani-fold with Yamabe invariant Y ( g ) > , then the following statements are equivalent (1) there exists a positive smooth function ρ with Q ρ g > . (2) ker P g = 0 and G P ( p, q ) < for any p, q ∈ M, p = q . (3) ker P g = 0 and there exists a p ∈ M such that G P ( p, q ) < for q ∈ M \ { p } . Similar to Proposition 1.1, we have
Proposition 1.2.
Let ( M, g ) be a smooth compact dimensional Riemannian man-ifold with Y ( g ) > , Q ≥ and not identically zero, then ker P = 0 and G − L ≤ − π G P . (1.14) If for some p, q ∈ M , G − L ( p, q ) = − π G P ( p, q ) , then ( M, g ) is conformaldiffeomorphic to the standard S (note here p can be equal to q ). In dimension 4 we have the following (see Corollary 5.1)
Proposition 1.3.
Assume ( M, g ) is a smooth compact dimensional Riemannianmanifold, Y ( g ) > , then for any p ∈ M , Z M Qdµ + 12 Z M (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) G L,p g dµ G L,p g = 16 π . (1.15) In particular, R M Qdµ ≤ π and equality holds if and only if ( M, g ) is conformaldiffeomorphic to the standard S . It is worthwhile to point out that the proof of Theorem B in [G], which givesthe inequality in Proposition 1.3, is elementary and does not use the positive masstheorem. Our argument is also elementary and identifies the difference between R M Qdµ and 16 π .Theorem 1.1 and 1.2 are motivated by works on the Q curvature in dimension5 or higher ([GM, HeR1, HeR2, HuR]) and in dimension 3 ([HY1, HY2, HY3]). In[HeR1, HeR2], it was shown in some cases compactness property for solutions ofthe Q curvature equation can be derived under the assumption that the Green’sfunction is positive. Recently [GM] showed that the Green’s function is indeedpositive when both scalar curvature and Q curvature are positive. Theorem 1.1says we could relax the assumption to Y ( g ) > , Q g >
0. Whether these twokinds of assumptions are equivalent or not is still unknown. The main approachin [GM] is roughly speaking by applying the maximum principle twice. In [HY3],
FENGBO HANG AND PAUL C. YANG by replacing maximum principle with the weak Harnack inequality it was shownthat for metrics with
R >
Q > P is invertible and G P ( p, q ) < p = q . Theorem 1.2 relax the assumption to Y ( g ) > Q >
0. The main newingredient in our proof of Theorem 1.1 and 1.2 is the formula (2.1), which is closelyrelated to the arguments in [HuR]. In [HY4], we will apply the results on Green’sfunction to solution of Q curvature equations. In section 2 we will prove the mainformula (2.1). In sections 3 and 4 we will prove Theorem 1.1 and 1.2 respectively.In section 5 we will derive the corresponding formula of (2.1) in dimension 4. Inparticular Proposition 1.3 follows from the formula. In section 6, we will show thatthe positive mass theorem for Paneitz operator in [GM, HuR] can be deduced from(2.1) too.2. An identity connecting the Green’s function of conformalLaplacian operator and Paneitz operator
Here we will derive an interesting formula which illustrates the close relationbetween Green’s function of conformal Laplacian operator and the Paneitz operator.This identity will play a crucial role later.To motivate the discussion, we note that positive Yamabe invariant implies wehave a positive Green’s function for the conformal Laplacian operator. Even thoughwe do not know whether P is invertible or not, we may still try to search for itsGreen’s function. Note that the possible Green’s function should have same highestorder singular term as G n − n − L,p (modulus dimension constant), we can use G n − n − L,p as afirst step approximation. Along this line we compute P (cid:18) G n − n − L,p (cid:19) and arrive at theinteresting formula (2.1).
Proposition 2.1.
Assume n ≥ , n = 4 , ( M, g ) is a n dimensional smooth compactRiemannian manifold with Y ( g ) > , p ∈ M , then we have G n − n − L,p (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 ∈ L ( M ) and P (cid:18) G n − n − L,p (cid:19) = c n δ p − n − n − G n − n − L,p (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 (2.1) in distribution sense. Here c n = 2 − n − n − n n − ( n − − n − n − ( n −
2) ( n − ω n − n , (2.2) ω n is the volume of unit ball in R n , G L,p is the Green’s function of conformalLaplacian operator L = − n − n − ∆ + R with pole at p . It is worth pointing out that the metric G n − L,p g on M \ { p } is exactly the stereo-graphic projection of ( M, g ) at p ([LP]). To prove the proposition, let us first checkwhat happens under a conformal change of the metric. If ρ ∈ C ∞ ( M ) is a positivefunction, let e g = ρ n − g , then using G e L,p ( q ) = ρ ( p ) − ρ ( q ) − G L,p ( q ) IGN OF GREEN’S FUNCTION AND THE Q CURVATURE 5 we see G n − n − e L,p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Rc G n − e L,p e g (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e g d e µ = ρ ( p ) − n − n − ρ n − n − G n − n − L,p (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 dµ. (2.3)Hence we only need to check G n − n − L,p (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 ∈ L ( M ) for a conformal metric.By the existence of conformal normal coordinate ([LP]) we can assume exp p preserve the volume near p . Let x , · · · , x n be a normal coordinate at p , denote r = | x | , then (see [LP]) G L,p = 14 n ( n − ω n r − n (cid:16) O (4) ( r ) (cid:17) . (2.4)As usual, we say f = O ( m ) (cid:0) r θ (cid:1) to mean f is C m in the punctured neighborhoodwith ∂ i ··· i k f = O (cid:0) r θ − k (cid:1) for 0 ≤ k ≤ m . By (2.4) and the transformation law Rc G n − L,p g = Rc − D log G L,p + 4 n − d log G L,p ⊗ d log G L,p (2.5) − (cid:18) n − G L,p + 4 n − |∇ log G L,p | (cid:19) g, careful calculation shows (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 = O (cid:18) r (cid:19) . (2.6)It follows that G n − n − L,p (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 = O (cid:0) r − n (cid:1) hence it belongs to L ( M ).To continue, we observe that equation (2.1) is the same as Z M G n − n − L,p
P ϕdµ = c n ϕ ( p ) − n − n − Z M G n − n − L,p (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 ϕdµ (2.7)for any ϕ ∈ C ∞ ( M ). A similar check as before shows (2.7) is conformally invariant.Again we assume exp p preserves the volume near p , then for δ > P G n − n − L,p = c n δ + a L function (2.8)on B δ ( p ) in distribution sense. On the other hand, on M \ { p } using (1.2) and (1.3)we have P g (cid:18) G n − n − L,p (cid:19) = G n +4 n − L,p P G n − L,p g n − G n +4 n − L,p Q G n − L,p g = − n − n − G n − n − L,p (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 . FENGBO HANG AND PAUL C. YANG
Here we have used the fact R G n − L,p g = 0. Combine (2.8) and (2.9) we get (2.1).3. The case dimension n >
M, g ) is a smooth compact Riemannianmanifold with dimension n > Lemma 3.1.
Assume n > , Y ( g ) > , u ∈ C ∞ ( M ) such that u ≥ and P u ≥ .If for some p ∈ M , u ( p ) = 0 , then u ≡ .Proof. By (2.1) we have Z M G n − n − L,p
P udµ = − n − n − Z M G n − n − L,p (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 udµ. Hence
P u = 0 and (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 u = 0.If u is not identically zero, then by unique continuation property we know { u = 0 } is dense, hence Rc G n − L,p g = 0. Since (cid:18) M \ { p } , G n − L,p g (cid:19) is asymptotically flat, itfollows from relative volume comparison theorem that (cid:18) M \ { p } , G n − L,p g (cid:19) is iso-metric to the standard R n . In particular ( M, g ) must be locally conformally flatand simply connected compact manifold, hence it is conformal to the standard S n by [K]. But in this case we have ker P = 0, hence u = 0, a contradiction. Remark 3.1.
Indeed the same argument gives us the following statement: If n > , Y ( g ) > , u ∈ L ( M ) such that u ≥ and P u ≥ in distribution sense, for some p ∈ M , u is smooth near p and u ( p ) = 0 , then u ≡ . A straightforward consequence of Lemma 3.1 is the following useful fact.
Proposition 3.1.
Assume n > , Y ( g ) > , Q ≥ . If u ∈ C ∞ ( M ) such that P u ≥ and u is not identically constant, then u > .Proof. If the conclusion of the proposition is false, then u ( p ) = min M u ≤ p . Let λ = − u ( p ) ≥
0, then u + λ ≥ u ( p ) + λ = 0 and P ( u + λ ) = P u + n − λQ ≥ . It follows from the Lemma 3.1 that u + λ ≡
0. This contradicts with the fact u isnot a constant function.Proposition 3.1 helps us determine the null space of P without information onthe first eigenvalue. Corollary 3.1.
Assume n > , Y ( g ) > , Q ≥ , then ker P ⊂ { constant functions } . If in addition, Q is not identically zero, then ker P = 0 i.e. is not an eigenvalueof P .Proof. Assume
P u = 0. If u is not a constant function, then it follows fromProposition 3.1 that u > − u >
0, a contradiction.
IGN OF GREEN’S FUNCTION AND THE Q CURVATURE 7
Now we ready to prove half of Theorem 1.1.
Lemma 3.2.
Assume n > , Y ( g ) > , Q ≥ and not identically zero, then ker P = 0 , moreover the Green’s function G P,p ( q ) = G P ( p, q ) > for p = q .Proof. By Corollary 3.1, we know ker P = 0. Hence for any f ∈ C ∞ ( M ), thereexists a unique u ∈ C ∞ ( M ) with P u = f , moreover u ( p ) = Z M G P,p ( q ) f ( q ) dµ ( q ) . If f ≥
0, it follows from the Proposition 3.1 that u ≥
0. Hence G P,p ≥
0. If G P,p ( q ) = 0 for some q , since P G
P,p = δ p ≥ G P,p ≡
0, a contradiction. Hence G P,p ( q ) > p = q .Next let us give the full argument of Theorem 1.1. Proof of Theorem 1.1. (1) ⇒ (2): This follows from Lemma 3.2, (1.5) and (1.6).(2) ⇒ (1): This follows from the classical Krein-Rutman theorem ([L]). Since ourcase is relatively simple, we provide the argument here. Define the integral operator T as T f ( p ) = Z M G P ( p, q ) f ( q ) dµ ( q ) .T is the inverse operator of P . Let α = sup f ∈ L ( M ) \{ } R M T f · f dµ k f k L > .α is an eigenvalue of T . We note all eigenfunctions of α does not change sign.Indeed say T ϕ = α ϕ , R M ϕ dµ = 1, we have Z M (cid:0) ϕ + ϕ − (cid:1) dµ = 1 . Here ϕ + = max { ϕ, } , ϕ − = max {− ϕ, } . Without losing of generality, we assume ϕ + is not identically zero. Then α = Z M T ϕ · ϕdµ = Z M T ϕ + · ϕ + dµ + Z M T ϕ − · ϕ − dµ − Z M T ϕ + · ϕ − dµ ≤ α − Z M T ϕ + · ϕ − dµ. Hence R M T ϕ + · ϕ − dµ = 0. Since T ϕ + >
0, we see ϕ − = 0. Hence ϕ ≥
0. Because
T ϕ = α ϕ we see ϕ ∈ C ∞ ( M ) and ϕ >
0. It follows that α must be a simpleeigenvalue and P ϕ = α − ϕ , hence Q ϕ n − g = 2 n − P ϕ n − g n − ϕ − n +4 n − P g ϕ = 2 n − α − ϕ − n − > . (2) ⇒ (3): Assume p ∈ M such that G P,p >
0. For p ∈ M , defineΘ ( p ) = min q ∈ M \{ p } G P ( p, q ) (3.1)Then we have Θ ( p ) >
0. We note that Θ ( p ) = 0 for any p ∈ M . Otherwise, sayΘ ( p ) = 0, then G P,p ≥ G P,p ( q ) = 0 for some q = p . It follows from Remark FENGBO HANG AND PAUL C. YANG G P,p = const , a contradiction. Since M is connected we see Θ ( p ) > p . In another word, G P ( p, q ) > p = q . Remark 3.2.
In the proof of (2) ⇒ (1), a similar argument tells us if β is aneigenvalue of T , β = α , then | β | < α . It follows that when G P is positive, thesmallest positive eigenvalue of P must be simple and its eigenfunction must beeither strictly positive or strictly negative. Moreover if λ is a negative eigenvalueof P , then | λ | is strictly bigger than the smallest positive eigenvalue.Proof of Proposition 1.1. By Lemma 3.2 we know ker P = 0 and G P >
0. From(2.1) we know P (cid:18) G n − n − L,p − c n G P,p (cid:19) = − n − n − G n − n − L,p (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 ≤ . Hence G n − n − L,p ≤ c n G P,p . If for some q = p , G n − n − L,p ( q ) = c n G P,p ( q ), then Rc G n − L,p g =0, hence ( M, g ) is conformal diffeomorphic to the standard S n by the argument inthe proof of Lemma 3.1. 4. 3 dimensional case Throughout this section we assume (
M, g ) is a smooth compact Riemannianmanifold of dimension 3.If Y ( g ) >
0, then for p ∈ M , (2.1) becomes P (cid:16) G − L,p (cid:17) = − π δ p + G − L,p (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g . (4.1)Note here G − L,p ∈ H ( M ). Lemma 4.1.
Assume Y ( g ) > , u ∈ H ( M ) such that u ≥ , P u ≤ in distribu-tion sense. If for some p ∈ M , u ( p ) = 0 , then either u ≡ or ( M, g ) is conformaldiffeomorphic to the standard S and u is a constant multiple of G P,p .Proof.
Using the fact G − L,p ∈ H ( M ), it follows from (4.1) that Z M G − L,p
P udµ − Z M G − L,p (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g udµ = 0 . Note here Z M G − L,p
P udµ = E (cid:16) G − L,p , u (cid:17) . Hence R M G − L,p
P udµ = 0 and R M G − L,p (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g udµ = 0. Hence (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g u = 0.Since P u must be a measure, we see
P u = const · δ p . In particular u is smooth on M \ { p } . If u is not identically zero, it follows from unique continuation propertythat the set { u = 0 } is dense, and hence Rc G L,p g = 0. Same argument as in theproof of Lemma 3.1 tells us ( M, g ) must be conformal diffeomorphic to the standard S , and hence u = const · G P,p . Proposition 4.1.
Assume Y ( g ) > , Q ≥ . If u ∈ C ∞ ( M ) such that P u ≤ and u is not identically constant, then u > . IGN OF GREEN’S FUNCTION AND THE Q CURVATURE 9
Proof.
If the conclusion of the proposition is false, then u ( p ) = min M u ≤ p . Let λ = − u ( p ) ≥
0, then u + λ ≥ u ( p ) + λ = 0 and P ( u + λ ) = P u − λQ ≤ . It follows from the Lemma 4.1 that u + λ ≡
0. This contradicts with the fact u isnot a constant function. Corollary 4.1.
Assume Y ( g ) > , Q ≥ , then ker P ⊂ { constant functions } . Ifin addition, Q is not identically zero, then ker P = 0 i.e. is not an eigenvalue of P .Proof. Assume
P u = 0. If u is not a constant function, then it follows fromProposition 4.1 that u > − u >
0, a contradiction.
Lemma 4.2.
Assume Y ( g ) > , Q ≥ and not identically zero, then ker P = 0 ,and the Green’s function G P,p ( q ) = G P ( p, q ) < for p = q . Moreover if for some p ∈ M , G P,p ( p ) = 0 , then ( M, g ) is conformal diffeomorphic to the standard S .Proof. By Corollary 4.1, we know ker P = 0. Hence for any f ∈ C ∞ ( M ), thereexists a unique u ∈ C ∞ ( M ) with P u = f , moreover u ( p ) = Z M G P,p ( q ) f ( q ) dµ ( q ) . If f ≤
0, it follows from the Proposition 4.1 that u ≥
0. Hence G P,p ≤
0. If G P,p ( q ) = 0 for some q , since P G
P,p = δ p ≥
0, it follows from Lemma 4.1 that(
M, g ) must be conformal diffeomorphic to the standard S and G P,p is a constantmultiple of G P,q , this implies p = q . Hence G P,p < M \ { p } .Now we are ready to prove Theorem 1.2. Proof of Theorem 1.2. (1) ⇒ (2): This follows from Lemma 4.2 and (1.5), (1.6).(2) ⇒ (1): This follows from Krein-Rutman theorem, or one may use the argumentin the proof of Theorem 1.1. We also remark it follows that the largest negative eigenvalue of P must be simple and its eigenfunction must be strictly positive orstrictly negative. Moreover if λ is a positive eigenvalue of P , then λ is strictlybigger than the absolute value of the largest negative eigenvalue.(3) ⇒ (2): We can assume ( M, g ) is not conformal diffeomorphic to the standard S . For any p ∈ M , we let Θ ( p ) = max M G P,p . Then it follows from Lemma 4.1 that Θ ( p ) = 0 for any p ∈ M . Since Θ ( p ) < p ∈ M , we see Θ ( p ) < p ∈ M . In another word, G P < Proof of Proposition 1.2.
Under the assumption of Proposition 1.2, it follows fromLemma 4.2 that ker P = 0 and G P ( p, q ) < p = q . From (4.1) we see P (cid:16) G − L,p + 256 π G P,p (cid:17) = G − L,p (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g ≥ . Hence G − L,p + 256 π G P,p ≤
0. If it achieves 0 somewhere, then Rc G L,p g = 0 andhence ( M, g ) is conformal diffeomorphic to the standard S . At last we want to point out based on Proposition 1.2, using the arguments in[HY3] we have the following statement: Let M = (cid:26) g : g is a smooth metric with Y ( g ) > ρ such that Q ρ g > (cid:27) be endowed with C ∞ topology. Then(1) For every g ∈ M , there exists ρ ∈ C ∞ ( M ), ρ > Q ρ g = 1.Moreover as long as ( M, g ) is not conformal diffeomorphic to the standard S , the set (cid:8) ρ ∈ C ∞ ( M ) : ρ > , Q ρ g = 1 (cid:9) is compact in C ∞ topology.(2) Let N be a path connected component of M . If there is a metric in N satisfying condition NN, then every metric in N satisfies condition NN.Hence as long as the metric is not conformal to the standard S , it satisfiescondition P . As a consequence, for any metric in N ,inf n E ( u ) (cid:13)(cid:13) u − (cid:13)(cid:13) L ( M ) : u ∈ H ( M ) , u > o > −∞ and is always achieved.We omit the details here.5. 4 dimension case revisited Throughout this section we will assume (
M, g ) is a smooth compact Riemannianmanifold of dimension 4. In this dimension the Q curvature is written as Q = −
16 ∆ R − | Rc | + 16 R . (5.1)The Paneitz operator can be written as P ϕ = ∆ ϕ + 2 div ( Rc ( ∇ ϕ, e i ) e i ) −
23 div ( R ∇ ϕ ) . (5.2)Here e , e , e , e is a local orthonormal frame with respect to g . P satisfies P e w g ϕ = e − w P g ϕ (5.3)for any smooth function w . The Q curvature transforms as Q e w g = e − w ( P g w + Q g ) . (5.4)In the spirit of Proposition 2.1, we have Proposition 5.1.
Assume ( M, g ) is a dimensional smooth compact Riemannianmanifold with Y ( g ) > , p ∈ M , then we have (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g ∈ L ( M ) and P (log G L,p ) = 16 π δ p − (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g − Q (5.5) in distribution sense. Here G L,p is the Green’s function of conformal Laplacianoperator L = −
6∆ + R with pole at p . IGN OF GREEN’S FUNCTION AND THE Q CURVATURE 11
Proof. If ρ is a positive smooth function on M , e g = ρ g , then (cid:12)(cid:12)(cid:12) Rc G e L,p e g (cid:12)(cid:12)(cid:12) e g d e µ = (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g dµ. (5.6)Hence to show (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g ∈ L ( M ), in view of the existence of conformal normalcoordinate, we can assume exp p preserves volume near p . Let x , x , x , x benormal coordinate at p , r = | x | , then (see [LP]) G L,p = 124 π r (cid:16) O (4) (cid:0) r (cid:1)(cid:17) . (5.7)Using Rc G L,p g = Rc − D log G L,p + 2 d log G L,p ⊗ d log G L,p (5.8) − (cid:16) ∆ log G L,p + 2 |∇ log G L,p | (cid:17) g, we see (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g = O (1), hence (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g ∈ L ( M ).On the other hand, (5.5) means Z M log G L,p · P ϕdµ = 16 π ϕ ( p ) − Z M (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g ϕdµ − Z M Qϕdµ. (5.9)Careful check shows (5.9) is conformally invariant. Hence we can assume exp p preserves volume near p . It follows from (5.7) that on B δ ( p ) for δ > P (log G L,p ) = 16 π δ p + a L function (5.10)in distribution sense. On M \ { p } , we have P (log G L,p ) = G L,p Q G L,p g − Q = − (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g − Q. (5.11)(5.5) follows.By integrating (5.5) on M and observing that (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) g dµ g = (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) G L,p g dµ G L,p g we immediately get Corollary 5.1.
Assume Y ( g ) > , then for any p ∈ M , Z M Qdµ + 12 Z M (cid:12)(cid:12)(cid:12) Rc G L,p g (cid:12)(cid:12)(cid:12) G L,p g dµ G L,p g = 16 π . (5.12) In particular, R M Qdµ ≤ π and equality holds if and only if ( M, g ) is conformaldiffeomorphic to the standard S . Positive mass theorem for Paneitz operator revisited
Throughout this section we will assume (
M, g ) is a smooth compact Riemannianmanifold with dimension n > Y ( g ) > G P , a positive mass theorem for Paneitz operator was provedby a nice calculation. Note that this result plays similar role for Q curvatureequation as the classical positive mass theorem for the Yamabe problem ([LP]). Itwas observed that similar calculation works for n = 5 , , n = 3 in[HY3]. Since the case n = 3 can be covered by Lemma 4.1, we concentrate on thecase n >
4. The main aim of this section is to show the positive mass theorem forPaneitz operator follows from the formula (2.1).
Lemma 6.1.
Assume n > , Y ( g ) > , ker P = 0 . Let x , · · · , x n be a coordinatenear p with x i ( p ) = 0 , r = | x | . If either M is conformally flat near p or n = 5 , , ,then c n G P,p − G n − n − L,p = const + O (4) ( r ) . (6.1) Here c n is the constant given by (1.13).Proof. First we observe that if ρ is a positive smooth function on M , e g = ρ n − g ,then c n G e P ,p − G n − n − e L,p = ρ ( p ) − ρ − (cid:18) c n G P,p − G n − n − L,p (cid:19) . (6.2)Hence we only need to verify (6.1) for a conformal metric.For the case M is conformally flat near p , by a conformal change of metric, wecan assume g is Euclidean near p . Then under the normal coordinate at p we have G P,p = 12 n ( n −
2) ( n − ω n (cid:16) r − n + A + O (4) ( r ) (cid:17) . (6.3)Here ω n is the volume of unit ball in R n and A is a constant. People usually call A as the mass of Paneitz operator. The Green’s function of conformal Laplacian G L,p = 14 n ( n − ω n (cid:16) r − n + O (4) (cid:0) r − (cid:1)(cid:17) . (6.4)It is worth pointing out one has better estimate for the Green’s function than theone in (6.3) and (6.4), but the formula we wrote above also works for n = 5 , , n = 5 , ,
7, underthe conformal normal coordinate, (6.3) and (6.4) remain true (see [GM, LP]). Itfollows that c n G P,p − G n − n − L,p = (4 n ( n − ω n ) − n − n − A + O (4) ( r ) . (6.5)To continue, we note that under the assumption of Lemma 6.1, by (2.1) we have P (cid:18) c n G P,p − G n − n − L,p (cid:19) = n − n − G n − n − L,p (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 , (6.6)hence G n − n − L,p (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 = O (cid:0) r − (cid:1) (6.7) IGN OF GREEN’S FUNCTION AND THE Q CURVATURE 13 and (4 n ( n − ω n ) − n − n − A = n − n − Z M G P,p G n − n − L,p (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 dµ. (6.8)If in addition we know the Green’s function G P,p >
0, then it follows from (6.8)that A ≥
0, moreover A = 0 if and only if ( M, g ) is conformal equivalent to thestandard S n . This proves the positive mass theorem for Paneitz operator. References [B] T. Branson.
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