SIGN OF GREEN S FUNCTION OF PANEITZ OPERATORS AND THE Q CURVATURE

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1 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND 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 su cient condition for the existence of metrics with positive Q curvature. The condition is conformally invariant. We also prove some inequality between the Green s functions of the conformal Laplacian operator and the Paneitz operator. 1. Introduction Since the fundamental work [CGY] in dimension, the Paneitz operator and associated Q curvature in dimension other than (see [B, P]) attracts much attention (see [DHL, HY1, HeR, HuR, QR] etc and the references therein). Let (; ) be a smooth compact n dimensional Riemannian manifold. For n 3, the Q curvature is iven by. Q = 1 (n 1) R () jrcj + n3 + 16n 16 8 (n 1) () R (1.1) Here R is the scalar curvature, Rc is the Ricci tensor. The Paneitz operator is iven by P ' (1.) = ' + div (Rc (r'; e i) e i ) n + 8 (n 1) () n div (Rr') + Q': Here e 1 ; ; e n is a local orthonormal frame with respect to. For n 6=, under conformal transformation of the metric, the operator satis es P ' = n+ n P (') : (1.3) n Note this is similar to the conformal Laplacian operator, which appears naturally when considerin transformation law of the scalar curvature under conformal chane of metric ([LP]). As a consequence we know P n ' d n = P (') d : (1.) Here is the measure associated with metric. oreover ker P = 0, ker P = 0; (1.5) n and under this assumption, the Green s functions G P satisfy the transformation law G (p; q) = (p) 1 (q) 1 G P; n P; (p; q) : (1.6) 1

2 FENGBO HANG AND PAUL C. YANG For u; v C 1 (), we denote the quadratic form associated with P as and = = E (u; v) (1.7) P u vd uv Rc (ru; rv) + n n + 8 Rru rv (n 1) () + n Quv d E (u) = E (u; u) : (1.8) By the interation by parts formula in (1.7) we know E (u; v) also makes sense for u; v H (). To continue we recall (see [LP]) for n 3, on a smooth compact Riemannian manifold ( n ; ), the conformal Laplacian operator is iven by L ' = The Yamabe invariant is de ned as (n 1) ' + R': (1.9) Y () (1.10) ( R Rde = inf e ) : e = for some positive smooth function (e ()) n 8 < R 9 = inf L ' 'd = : k'k : ' is a nonzero smooth function on ; : L n A basic but useful fact about the scalar curvature is Y () > 0, 1 (L ) > 0 (1.11), is conformal to a metric with scalar curvature > 0. Indeed more is true, namely the equivalence still holds if we replace all ">" by "= " or "<". Here 1 (L ) is the rst eienvalue of L. It is clear Y () is a conformal invariant, on the other hand the sin of 1 (L ) is also conformally invariant. The main reason that (1.11) holds is based on the fact the rst eienfunction of a second order symmetric di erential operator does not chane sin. Unfortunately such kind of property is known to be not valid for hiher order operators. The followin question keeps puzzlin people from the beinnin of research on Q curvature in dimension other than, namely: can we nd a conformal invariant condition which is equivalent to the existence of positive Q curvature in the conformal class (in the same spirit as (1.11))? Here we ive an answer to this question for conformal class with positive Yamabe invariant. Theorem 1.1. Let n > and ( n ; ) be a smooth compact Riemannian manifold with Yamabe invariant Y () > 0, then the followin statements are equivalent (1) there exists a positive smooth function with Q > 0. () ker P = 0 and the Green s function G P (p; q) > 0 for any p; q ; p 6= q. (3) ker P = 0 and there exists a p such that G P (p; q) > 0 for q n fp.

3 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND THE Q CURVATURE 3 Alon the way we also nd the followin comparison inequality between Green s function of L and P. Proposition 1.1. Assume n >, ( n ; ) is a smooth compact Riemannian manifold with Y () > 0, Q 0 and not identically zero, then ker P = 0 and Here G n L c n G P : (1.1) c n = n 6 n n (n 1) () (n )! n ; (1.13)! n is the volume of unit ball in R n. oreover if G n L (p; q) = c n G P (p; q) for some p 6= q, then (; ) is conformal di eomorphic to the standard sphere. In dimension 3 we have Theorem 1.. Let (; ) be a smooth compact 3 dimensional Riemannian manifold with Yamabe invariant Y () > 0, then the followin statements are equivalent (1) there exists a positive smooth function with Q > 0. () ker P = 0 and G P (p; q) < 0 for any p; q ; p 6= q. (3) ker P = 0 and there exists a p such that G P (p; q) < 0 for q n fp. Similar to Proposition 1.1, we have Proposition 1.. Let (; ) be a smooth compact 3 dimensional Riemannian manifold with Y () > 0, Q 0 and not identically zero, then ker P = 0 and G 1 L 56 G P : (1.1) If for some p; q, G 1 L (p; q) = 6 G P (p; q), then (; ) is conformal di eomorphic to the standard S 3 (note here p can be equal to q). Theorem 1.1 and 1. are motivated by works on the Q curvature in dimension 5 or hiher ([G, HeR, HuR]) and in dimension 3 ([HY1, HY, HY3]). In [HeR, HuR], it was shown in some cases compactness property for solutions of the Q curvature equation can be derived under the assumption that the Green s function is positive. Recently [G] showed that the Green s function is indeed positive when both scalar curvature and Q curvature are positive. Theorem 1.1 says we could relax the assumption to Y () > 0; Q > 0. Whether these two kinds of assumptions are equivalent or not is still unknown. The main approach in [G] is rouhly speakin by applyin the maximum principle twice. In [HY3], by replacin maximum principle with the weak Harnack inequality it was shown that for metrics with R > 0 and Q > 0, P is invertible and G P (p; q) < 0 for p 6= q. Theorem 1. relax the assumption to Y () > 0 and Q > 0. The main new inredient in our proof of Theorem 1.1 and 1. is the formula (.1), which is closely related to the aruments in [HuR]. In a forthcomin article [HY], we will apply the results on Green s function to solution of Q curvature equations. In sectio we will prove the main formula (.1). Sections 3 and will prove Theorem 1.1 and 1. respectively. In section 5 we will derive the correspondin formula of (.1) in dimension. As a consequence we ive another elementary arument of Theorem B in [G] and more information on the ap between the two quantities considered there. In section 6, we will show that the positive mass theorem for Paneitz operator in [HuR, G] can also be deduced from (.1).

4 FENGBO HANG AND PAUL C. YANG. An identity connectin the Green s function of conformal Laplacian operator and Paneitz operator Here we will derive an interestin formula which illustrates the close relation between Green s function of conformal Laplacian operator and the Paneitz operator. This identity will play a crucial role in the future discussion. Propositio.1. Assume n 3, n 6=, (; ) is an n dimensional smooth com- pact Riemannian manifold with Y () > 0, p, then we have G n Rc G L 1 () and P G n in distribution sense. Here n = c n p () G n (.1) c n = n 6 n n (n 1) () (n )! n ; (.)! n is the volume of unit ball in R n, G is the Green s function of conformal Laplacian operator L = + R with pole at p. (n 1) It is worth pointin out that the metric G on n fp is exactly the stereoraphic projection of (; ) at p ([LP]). To prove the proposition, let us rst check what happens under a conformal chane of the metric. If C 1 () is a positive function, let e =, then usin we see G n e G e (q) = (p) 1 (q) 1 G (q) e e e Hence we only need to check G n de = (p) n n n G d: (.3) L 1 () for a conformal metric. By the existence of conformal normal coordinate ([LP]) we can assume exp p preserve the volume near p. Let x 1 ; ; x n be a normal coordinate at p, denote r = jxj, then (see [LP]) 1 G = r n 1 + O () (r) : (.) n (n 1)! n As usual, we say f = O (m) r to mean f is C m in the punctured neihborhood i1i k f = O r k for 0 k m. By (.) and the transformation law Rc G careful calculation shows = Rc D lo G + d lo G d lo G (.5) lo G + jr lo G j ; = O 1 : (.6) r

5 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND THE Q CURVATURE 5 It follows that G n = O r n hence it belons to L 1 (). To continue, we observe that equation (.1) is the same as G n P 'd = c n n' (p) () G n 'd (.7) for any ' C 1 (). A similar check as before shows (.7) is conformally invariant. Aain we assume exp p preserves the volume near p, then for > 0 small, it follows from (.) that P G n = c n + a L 1 function (.8) on B (p) in distribution sense. On the other hand, on n fp usin (1.) and (1.3) we have P G n = G n+ P 1 (.9) G = n G n+ n+ Here we have used the fact R = n+ G Q G n () G n : = 0. Combine (.8) and (.9) we et (.1). 3. The case dimension n > Throuhout this section we will assume (; ) is a smooth compact Riemannian manifold with dimension n >. Lemma 3.1. Assume n >, Y () > 0, u C 1 () such that u 0 and P u 0. If for some p, u (p) = 0, then u 0. Proof. By (.1) we have Hence P u = 0 and G n P ud = n () u = 0. G n ud: If u is not identically zero, then by unique continuation property we know fu 6= 0 is dense, hence Rc = 0. Since n fp ; G G is asymptotically at, it follows from relative volume comparison theorem that n fp ; G is isometric to the standard R n. In particular (; ) must be locally conformally at

6 6 FENGBO HANG AND PAUL C. YANG and 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 arument ives us the followin statement: If n >, Y () > 0, u L 1 () such that u 0 and P u 0 in distribution sense, for some p, u is smooth near p and u (p) = 0, then u 0. A straihtforward consequence of Lemma 3.1 is the followin useful fact. Proposition 3.1. Assume n >, Y () > 0, Q 0. If u C 1 () such that P u 0 and u is not identically constant, then u > 0. Proof. If the conclusion of the proposition is false, then u (p) = min u 0 for some p. Let = u (p) 0, then u + 0, u (p) + = 0 and P (u + ) = P u + n Q 0: It follows from the Lemma 3.1 that u + 0. This contradicts with the fact u is not a constant function. Proposition 3.1 helps us determine the null space of P without information on the rst eienvalue. Corollary 3.1. Assume n >, Y () > 0, Q 0, then ker P fconstant functions : If in addition, Q is not identically zero, then ker P = 0 i.e. 0 is not an eienvalue of P. Proof. Assume P u = 0. If u is not a constant function, then it follows from Proposition 3.1 that u > 0 and u > 0, a contradiction. Now we ready to prove half of Theorem 1.1. Lemma 3.. Assume n >, Y () > 0, Q 0 and not identically zero, then ker P = 0, moreover the Green s function G P;p (q) = G P (p; q) > 0 for p 6= q. Proof. By Corollary 3.1, we know ker P = 0. Hence for any f C 1 (), there exists a unique u C 1 () with P u = f, moreover u (p) = G P;p (q) f (q) d (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 = p 0 in distribution sense, we know from the Remark 3.1 that G P;p 0, a contradiction. Hence G P;p (q) > 0 for p 6= q. Next let us ive the full arument of Theorem 1.1. Proof of the theorem 1.1. (1))(): This follows from Lemma 3., (1.5) and (1.6). ())(1): This follows from the classical Krein-Rutman theorem ([L]). Since our case is relatively simple, we provide the arument here. De ne the interal operator T as T f (p) = G P (p; q) f (q) d (q) :

7 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND THE Q CURVATURE 7 T is the inverse operator of P. Let 1 = sup fl ()nf0 R T f fd kfk L > 0: 1 is an eienvalue of T. We note all eienfunctions of 1 does not chane sin. Indeed say T ' = 1 ', R ' d = 1, we have ' + + ' d = 1: Here ' + = max f'; 0, ' = max f '; 0. Without losin of enerality, we assume ' + is not identically zero. Then 1 = T ' 'd = T ' + ' + d + T ' ' d T ' + ' d 1 T ' + ' d: Hence R T ' + ' d = 0. Since T ' + > 0, we see ' = 0. Hence ' 0. Because T ' = 1 ' we see ' C 1 () and ' > 0. It follows that 1 must be a simple eienvalue and P ' = 1 1 ', hence Q = ' n n P ' n 1 = n ' n+ n P ' = n 1 1 ' 8 n > 0: ())(3): Assume p 0 such that G P;p0 > 0. For p, de ne (p) = min G P (p; q) (3.1) qnfp Then we have (p 0 ) > 0. We note that (p) 6= 0 for any p. Otherwise, say (p) = 0, then G P;p 0 and G P;p (q) = 0 for some q 6= p. It follows from Remark 3.1 that G P;p = const, a contradiction. Since is connected we see (p) > 0 for all p. In another word, G P (p; q) > 0 for p 6= q. Remark 3.. In the proof of ())(1), a similar arument tells us if is an eienvalue of T, 6= 1, then jj < 1. It follows that when G P is positive, the smallest positive eienvalue of P must be simple and its eienfunction must be either strictly positive or strictly neative. oreover if is a neative eienvalue of P, then jj is strictly bier than the smallest positive eienvalue. Proof of Proposition 1.1. By Lemma 3. we know ker P = 0 and G P > 0. From (.1) we know P G n n c n G P;p = () G n Rc 0: G Hence G n c ng P;p. If for some q 6= p, G n (q) = c ng P;p (q), then Rc = G 0, hence (; ) is conformal di eomorphic to the standard S n by the arument in the proof of Lemma 3.1.

8 8 FENGBO HANG AND PAUL C. YANG. 3 dimensional case Throuhout this section we assume (; ) is a smooth compact Riemannian manifold of dimension 3. If Y () > 0, then for p, (.1) becomes P G 1 Note here G 1 H (). = 56 p + G 1 : (.1) Lemma.1. Assume Y () > 0, u H () such that u 0, P u 0 in distribution sense. If for some p, u (p) = 0, then either u 0 or (; ) is conformal di eomorphic to the standard S 3 and u is a constant multiple of G P;p. Proof. Usin the fact G 1 H (), it follows from (.1) that G 1 P ud G 1 ud = 0: Note here G 1 P ud = E G 1 ; u : Hence R G 1 P ud = 0 and R G 1 ud = 0. Hence RcG u = 0. Since P u must be a measure, we see P u = const p. In particular u is smooth on n fp. If u is not identically zero, it follows from unique continuation property that the set fu 6= 0 is dense, and hence = 0. Same arument as in the proof of Lemma 3.1 tells us (; ) must be conformal di eomorphic to the standard S 3, and hence u = const G P;p. Proposition.1. Assume Y () > 0, Q 0. If u C 1 () such that P u 0 and u is not identically constant, then u > 0. Proof. If the conclusion of the proposition is false, then u (p) = min u 0 for some p. Let = u (p) 0, then u + 0, u (p) + = 0 and P (u + ) = P u Q 0: It follows from the Lemma.1 that u + 0. This contradicts with the fact u is not a constant function. Corollary.1. Assume Y () > 0, Q 0, then ker P fconstant functions. If in addition, Q is not identically zero, then ker P = 0 i.e. 0 is not an eienvalue of P. Proof. Assume P u = 0. If u is not a constant function, then it follows from Proposition.1 that u > 0 and u > 0, a contradiction. Lemma.. Assume Y () > 0, Q 0 and not identically zero, then ker P = 0, and the Green s function G P;p (q) = G P (p; q) < 0 for p 6= q. oreover if for some p, G P;p (p) = 0, then (; ) is conformal di eomorphic to the standard S 3. Proof. By Corollary.1, we know ker P = 0. Hence for any f C 1 (), there exists a unique u C 1 () with P u = f, moreover u (p) = G P;p (q) f (q) d (q) :

9 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND THE Q CURVATURE 9 If f 0, it follows from the Proposition.1 that u 0. Hence G P;p 0. If G P;p (q) = 0 for some q, since P G P;p = p 0, it follows from Lemma.1 that (; ) must be conformal di eomorphic to the standard S 3 and G P;p is a constant multiple of G P;q, this implies p = q. Hence G P;p < 0 on n fp. Now we are ready to prove Theorem 1.. Proof of Theorem 1.. (1))(): This follows from Lemma. and (1.5), (1.6). ())(1): This follows from Krein-Rutman theorem, or one may use the arument in the proof of Theorem 1.1. We also remark it follows that the larest neative eienvalue of P must be simple and its eienfunction must be strictly positive or strictly neative. oreover if is a positive eienvalue of P, then is strictly bier than the absolute value of the larest neative eienvalue. (3))(): We can assume (; ) is not conformal di eomorphic to the standard S 3. For any p, we let (p) = max G P;p: Then it follows from Lemma.1 that (p) 6= 0 for any p. Since (p 0 ) < 0 for some p 0, we see (p) < 0 for all p. In another word, G P < 0. With all the above analysis, we can easily deduce Proposition 1.. Proof of Proposition 1.. Under the assumption of Proposition 1., it follows from Lemma. that ker P = 0 and G P (p; q) < 0 for p 6= q. From (.1) we see P G G P;p = G 1 0: Hence G G P;p 0. If it achieves 0 somewhere, then = 0 and hence (; ) is conformal di eomorphic to the standard S 3. At last we want to point out based on Proposition 1., usin the aruments in the proof of Theorem 1.1 in [HY3] we have the followin statement: Let is a smooth metric with Y () > 0 and there exists = : a positive smooth function such that Q > 0 be endowed with C 1 topoloy. Then (1) For every, there exists C 1 (), > 0 such that Q = 1. oreover as lon as (; ) is not conformal di eomorphic to the standard S 3, the set C 1 () : > 0; Q = 1 is compact in C 1 topoloy. () Let N be a path connected component of. If there is a metric in N satisfyin condition NN, then every metric in N satis es condition NN. Hence as lon as the metric is not conformal to the standard S 3, it satis es condition P. As a consequence, for any metric in N, n inf E (u) o u 1 : u L 6 () H () ; u > 0 > 1 and is always achieved. We omit the details here.

10 10 FENGBO HANG AND PAUL C. YANG 5. dimension case revisited Throuhout this section we will assume (; ) is a smooth compact Riemannian manifold of dimension. In this dimension the Q curvature is written as The Paneitz operator can be written as Q = 1 6 R 1 jrcj R : (5.1) P ' = ' + div (Rc (r'; e i ) e i ) div (Rr') : (5.) 3 Here e 1 ; e ; e 3 ; e is a local orthonormal frame with respect to. P satis es P e w ' = e w P ' (5.3) for any smooth function w. The Q curvature transforms as In the spirit of Propositio.1, we have Q e w = e w (P w + Q ) : (5.) Proposition 5.1. Assume (; ) is a dimensional smooth compact Riemannian manifold with Y () > 0, p, then we have L1 () and P (lo G ) = 16 1 p Rc G Q (5.5) in distribution sense. Here G is the Green s function of conformal Laplacian operator L = 6 + R with pole at p. Proof. If is a positive smooth function on, e =, then e e de = e d: (5.6) Hence to show L1 (), in view of the existence of conformal normal coordinate, we can assume exp p preserves volume near p. Let x 1 ; x ; x 3 ; x be normal coordinate at p, r = jxj, then (see [LP]) G = 1 1 r 1 + O () r : (5.7) Usin = Rc D lo G + d lo G d lo G (5.8) lo G + jr lo G j ; we see = O (1), hence L1 (). On the other hand, (5.5) means lo G P 'd = 16 1 ' (p) Rc G 'd Q'd: (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 (p) for > 0 small, P (lo G ) = 16 p + a L 1 function (5.10)

11 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND THE Q CURVATURE 11 in distribution sense. On n fp, we have P (lo G ) = G Q G Q = 1 Rc G Q: (5.11) (5.5) follows. By interatin (5.5) on, we immediately et Corollary 5.1. Assume Y () > 0, then for any p, Qd + 1 Rc G d = 16 : (5.1) In particular, R Qd 16 and equality holds if and only if (; ) is conformal di eomorphic to the standard S. This ives another elementary arument for Theorem B in [G] (elementary in the sense that we do not use the solution to Yamabe problem). oreover it tells us the ap between R Qd and 16 is exactly the interal of square of Ricci curvature of the stereoraphic projection. 6. Positive mass theorem for Paneitz operator revisited Throuhout this section we will assume (; ) is a smooth compact Riemannian manifold with dimension n >. In [HuR], for locally conformally at manifold with Y () > 0 and positive Green s function G P, a positive mass theorem for Paneitz operator was proved by a nice calculation. Note that this result plays similar role for Q curvature equation as the classical positive mass theorem for the Yamabe problem ([LP]). It was observed that similar calculation works for n = 5; 6; 7 in [G] and for n = 3 in [HY3]. Since the case n = 3 can be covered by Lemma.1, we concentrate on the case n >. The main aim of this section is to show the positive mass theorem for Paneitz operator follows from the formula (.1). Lemma 6.1. Assume n >, Y () > 0, ker P = 0. Let x 1 ; ; x n be a coordinate near p with x i (p) = 0, r = jxj. If either is locally conformally at or n = 5; 6; 7, then c n G P;p G n = const + O() (r) : (6.1) Here c n is the constant iven by (1.13). Proof. First we observe that if is a positive smooth function on, e = n, then c n G ep ;p G n e = (p) 1 1 c n G P;p G n : (6.) Hence we only need to verify (6.1) for a conformal metric. For locally conformally at manifold, by conformal chane of metric, we can assume is Euclidean near p. Then under the normal coordinate at p we have 1 G P;p = r n + A + O () (r) : (6.3) n () (n )! n Here! 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 1 G = r n + O () r 1 : (6.) n (n 1)! n

12 1 FENGBO HANG AND PAUL C. YANG It is worth pointin out one has better estimate for the Green s function then the one in (6.3) and (6.), but the formula we wrote above also works for n = 5; 6; 7 without locally conformally at assumption. ore precisely, for n = 5; 6; 7, under the conformal normal coordinate, (6.3) and (6.) remain true (see [LP, G]). It follows that c n G P;p G n = (n (n 1)! n) n A + O () (r) : (6.5) To continue, we note that under the assumption of Lemma 6.1, by (.1) we have P c n G P;p G n = n () G n Rc ; (6.6) G hence and G n (n (n 1)! n ) n A = n () = O r 3 (6.7) G P;p G n d: (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 (; ) is conformal equivalent to the standard S n. This proves the positive mass theorem for Paneitz operator. References [B] T. Branson. Di erential operators canonically associated to a conformal structure. ath. Scand. 57 (1985), no., [CGY] S.-Y. A. Chan,. J. Gursky and P. C. Yan. An equation of one-ampere type in conformal eometry, and four-manifolds of positive Ricci curvature. Ann. of ath. () 155 (00), [DHL]. Djadli, E. Hebey and. Ledoux. Paneitz-type operators and applications. Duke ath. Jour. 10 (000), [G]. J. Gursky. The principal eienvalue of a conformally invariant di erential operator. Comm. ath. Phys. 07 (1999), no. 1, [G]. J. Gursky and A. alchiodi. A stron maximum principle for the Paneitz operator and a nonlocal ow for the Q curvature. Preprint (01). [HY1] F. B. Han and P. Yan. The Sobolev inequality for Paneitz operator on three manifolds. Calculus of Variations anf PDE. 1 (00), [HY] F. B. Han and P. Yan. Paneitz operator for metrics near S 3. Preprint (01). [HY3] F. B. Han and P. Yan. Q curvature on a class of 3 manifolds. Preprint (01). [HY] F. B. Han and P. Yan. Q curvature equation on a class of manifolds with dimension 5. In preparation. [HeR] E. Hebey and F. Robert. Compactness and lobal estimates for the eometric Paneitz equation in hih dimensions. Electron Res Ann Amer ath Soc. 10 (00), [HuR] E. Humbert and S. Raulot. Positive mass theorem for the Paneitz-Branson operator. Calculus of Variations and PDE. 36 (009), [K] N. H. Kuiper. On conformally- at spaces in the lare. Ann of ath. 50 (199), [L] P. Lax. Functional analysis. John Wiley & Sons, Inc. 00. [LP] J.. Lee and T. H. Parker. The Yamabe problem. Bull AS. 17 (1987), no. 1, [P] S. Paneitz. A quartic conformally covariant di erential operator for arbitrary pseudo- Riemannian manifolds. Preprint (1983).

13 SIGN OF GREEN S FUNCTION OF PANEIT OPERATORS AND THE Q CURVATURE 13 [QR] J. Qin and D. Raske. On positive solutions to semilinear conformally invariant equations on locally conformally at manifolds. Int ath Res Not. Art. id 917 (006). 51 ercer Street, New York NY address: fenbo@cims.nyu.edu Department of athematics, Princeton University, Fine Hall, Washinton Road, Princeton NJ 085 address: yan@math.princeton.edu