FENGBO HANG AND PAUL C. YANG

Q CURVATURE ON A CLASS OF 3 ANIFOLDS FENGBO HANG AND PAUL C. YANG Abstract. otivated by the strong maximum principle for Paneitz operator in dimension 5 or higher found in [G] and the calculation of the second variation of the Green s function pole s value on S 3 in [HY], we study Riemannian metric on 3 manifolds with positive scalar and Q curvature. Among other things, we show it is always possible to nd a constant Q curvature metric in the conformal class. oreover the Green s function is always negative away from the pole and the pole s value vanishes if and only if the Riemannian manifold is conformal di eomorphic to the standard S 3. Compactness of constant Q curvature metrics in a conformal class and the associated Sobolev inequality are also discussed.. Introduction The study of Paneitz operator and Q curvature has improved our understanding of four dimensional conformal geometry ([CGY]). On three dimensional Riemannian manifold, much less is known. However the Paneitz operator and Q curvature may contain valuable information besides those related to conformal Laplacian which is associated with the scalar curvature. These additional information may help us distinguish some conformal classes from others. The aim of this article is to understand this fourth order operator for a class of metrics on three manifolds. Recall on three manifolds, the Q curvature is given by (.) Q = 4 R jrcj + 3 3 R ; and the fourth order Paneitz operator is de ned as (.) P ' = ' + 4 div [Rc (r'; e i ) e i ] 5 4 div (Rr') Q': Here e ; e ; e 3 is a local orthonormal frame with respect to the metric (see [B, P]). Under conformal transformation of the metric, the operator satis es (.3) P 4 g' = 7 P g (') : Note this is similar to the conformal Laplacian operator. As a consequence we know (.4) P 4 g' d 4 g = P g (') d g : Here g is the measure associated with metric g. oreover ker P g = 0, ker P 4 g = 0; and under this assumption, the Green s functions satisfy the transformation law (.5) G 4 g (p; q) = (p) (q) G g (p; q) :

FENGBO HANG AND PAUL C. YANG Assume (; g) is a smooth compact three dimensional Riemannian manifold, for u; v C (), we denote (.6) E (u; v) = P u vd = uv 4Rc (ru; rv) + 5 4 Rru rv Quv d and (.7) E (u) = E (u; u) : By the integration by parts formula in (.6) we know E (u; v) also makes sense for u; v H (). By (.3) the scaling invariant quantity g () 3 Q g d g satis es Hence 4 g () 3 As in [HY], we denote Q 4 gd 4 g = E g () L 6 (;g) : sup 4 g () 3 Q C ;>0 4 gd 4 g = inf C ;>0 E () L 6 = inf uh ();u>0 E (u) u L 6 : (.8) I (u) = E (u) u L 6 and (.9) Y 3 4 (; g) = inf uh ();u>0 E (u) u L 6 : From above discussion we see Y4 3 g = Y4 3 (g) for every positive smooth i.e. it is a conformal invariant quantity like the Yamabe invariant Y (g) ([LP]). An interesting question is whether Y4 3 (g) is nite and achieved by some particular metrics. This inequality is analytically di erent from the Y (g) case due to the negative power involved. On the other hand, the critical point for I satis es P u = const u 7 ; i.e. Q u 4 g = const. In [Y] it was shown that Y4 3 S 3 ; g S 3 is achieved by the standard metric gs 3 itself (see [H, HY] for di erent approaches). The closely related condition NN was introduced in [HY, De nition 5.]. We recall the de nition here for reference: De nition.. Given a smooth compact 3 dimensional Riemannian manifold (; g), if for any u H (), u (p) = 0 for some p implies E (u) 0, then we say (; g) satis es condition NN. If for any nonzero u H (), u (p) = 0 for some p implies E (u) > 0, then we say (; g) satis es condition P.

Q CURVATURE ON A CLASS OF 3 ANIFOLDS 3 For detailed discussion of condition NN, we refer to [HY, section 5]. Here we simply note that both condition NN and P are conformally invariant properties. The main use of condition NN is (see [HY, Lemma 5.3]): if g satis es condition NN, u H (), u (p) = 0 and E (u) = 0, then P u = const p. In particular, when ker P = 0, u is simply a constant multiple of the Green function G p. If (; g) satis es condition P, then Y4 3 (g) is achieved and modulus scaling the set of minimizers are compact in C topology. S 3 ; g S 3 satis es condition NN but not condition P. Indeed, let x be the coordinate given by the stereographic projection with respect to north pole N, then the Green s function of P at N is given by G N = q : 4 jxj + In particular, E (G N ) = G N (N) = 0. In general condition NN is hard to check. In [HY] by explicit calculation of the eigenvalues and Green s function pole s value for Paneitz operator on Berger s spheres, condition NN was veri ed for all these special metrics. The main aim of this note is to prove the following Theorem.. Let be a smooth compact 3 manifold, consider g is a smooth metric such that there exists a positive = g : smooth function with R g > 0; Q g > 0 Endow with C topology. Then we have For every g, ker P = 0, the Greens function G p < 0 on n fpg. oreover if there exists a p with G p (p) = 0, then (; g) is conformal di eomorphic to the standard S 3. For every g, there exists C (), > 0 such that Q g = and R g > 0. oreover as long as (; g) is not conformal di eomorphic to the standard S 3, the set C () : > 0; Q g = is compact in C topology, and any such must satisfy R g > 0. Let N be a path connected component of. If there is a metric in N satisfying condition NN, then every metric in N satis es condition NN. Hence as long 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 : u L 6 () H () ; u > 0 > and is always achieved. Note that positive Q curvature Berger s sphere are contained in the path connected component of the standard sphere and condition NN follows. Theorem. is motivated by recent study of Paneitz operator in dimension 5 or higher in [G], where they found strong maximum principle and constant Q curvature metrics in conformal class of metrics with positive scalar and Q curvature, and the calculation of second variation of Green s function pole s value near the standard sphere in [HY]. To prove Theorem., the rst step is to show Green s function must be nonpositive. In another word, for any u C (), P u 0 implies u 0. To achieve this we use the method of continuity with the path used in [G] for dimension 5 :

4 FENGBO HANG AND PAUL C. YANG or higher. In dimension 3, the argument in [G] will not run through since certain function which is supharmonic in higher dimension becomes subharmonic and one can not apply strong maximum principle. Interestingly we found that use of weak Harnack principle for another function would resolve the obstruction. As a consequence we know ker P = 0 and the Green s function exists. Next note that the argument in [HR] can be adapted to dimension three to show the Greens function pole value is negative when the metric is not conformally equivalent to the standard sphere. In particular, in this case we know the Green s function are strictly negative everywhere. Based on this fact and apriori estimate for solutions of some integral equations we can show the existence of constant Q curvature metrics in the conformal class. It is not clear in general the minimizer for functional I (see (.8)) exists. But when the metric can be connected to another metric satisfying condition NN through a good path, we know that metric satis es condition NN and Y4 3 (g) is achieved hence nite. We would like to thank. Gursky for sending us their preprint [G].. Green function s sign and pole s value In this section we will show that for metrics with positive scalar curvature and positive Q curvature, the Green s function are in fact negative everywhere except at the pole. oreover, if the value at pole vanishes, then the metric is conformally equivalent to the standard sphere. The rst step is the following Proposition.. Let (; g) be a smooth compact Riemannian 3 manifold with R > 0, Q 0. If u C (), u 6= const and P u 0, then u > 0 and R u 4 g > 0. This proposition is in the same spirit as [G, Theorem.] and we will adapt their proof for dimension 5 or higher to dimension 3. The main new ingredient is when the strong maximum principle does not apply, we can replace it by the weak Harnack principle. Proof of Proposition.. For 0, let u = u +, then P u = P u (.) 0 = inf f 0 : u > 0g : For u > 0, let g = u 4 g, then Q 0. Let Q = P u 4 g = u7 P u 0: This implies 4 R jrc j + 3 3 R 0 and hence R + 3 8 R 0. In particular by strong maximum principle (or use the weak Harnack inequality [GT, Theorem 8.8, p94]) if R 0, then either R > 0 or R 0. The latter case is impossible since the Yamabe invariant Y (g) > 0 ([LP]). Hence R > 0. Now for, we have R = 4 R (+ u) 4 g > 0: Let = inf f > 0 : for all, R > 0g : Then = 0. If not, then R 0 and hence R > 0. This contradicts with the de nition of. Hence for any > 0, R > 0. Because R = u 5 8 u + Ru ;

Q CURVATURE ON A CLASS OF 3 ANIFOLDS 5 we know u R + 8 u > 0: It follows from weak Harnack principle [GT, Theorem 8.8, p94] that for any 0 < p < 3, inf u C (; g; p) u Lp () : Because u 6= const, we know inf u C, independent of > 0, and hence u L p () C: Let # 0, we see (.) u 0 L p () C < : Hence u 0 can not touch 0 i.e. u 0 > 0. Indeed if u 0 (p) = 0, then under the normal coordinate at p, by smoothness of u we have u 0 = O r, here r is the distance to p. In particular u = for 3 0 L p () p < 3, this contradicts with (.). Hence 0 = 0, u > 0. It follows that R u 4 g 0 and hence R u 4 g > 0. Corollary.. Assume R > 0, Q 0, then ker P fconstant functionsg. If in addition Q is not identically 0, then ker P = 0 i.e. 0 is not in the spectrum of P. Proof. Assume P u = 0, then u const. If not, by applying the previous proposition to u and u, we see u > 0 and u > 0, a contradiction. Next we will show the Green s function is always negative away from the pole. Proposition.. Assume R > 0, Q 0 and not identically zero. For any p, let G p be the Green s function of P at p i.e. P G p = p, then G p j nfpg < 0 and 8 Gp + RG p 0: If u H () such that P u 0 in distribution sense and u is not identically zero, then either u > 0 or u = cg p for some c > 0; p, moreover in distribution sense. 8 u + Ru 0 Proof. First we will show that if u H () such that P u 0 in distribution sense and u is not identically zero, then u 0 and for any 0 < p < 3, u <, Lp () moreover 8 u + Ru 0 in distribution sense. Indeed we can nd a sequence of smooth nonzero functions f i such that f i! P u in H () = H () 0 and f i 0. Because 0 is not in spectrum of P, we can nd u i C () such that P u i = f i. This implies u i! u in H (). It follows from the Proposition. that u i > 0 and R u 4 i g > 0 i.e. 8 u i + Ru i > 0. Because u is not identically zero and u i u we see inf u i C independent of i. Hence by weak Harnack inequality we have u i C for any xed Lp ()

6 FENGBO HANG AND PAUL C. YANG p (0; 3). By Fatou s lemma we have u C and u L p () i * u in L p, hence 8 u + Ru 0 in distribution sense. It follows that G p 0 and G p < for any 0 < q < 3. Because G Lq () p is smooth away from p, we see G p < 0 away from p. Now if P u 0, then P u = for some nonnegative measure. It follows that u (p) = G p (q) d (q) : It is not a constant multiple of Dirac mass, then clearly u (p) > 0 for every p. If is a constant multiple of Dirac mass, then u is simply a constant multiple of Green s function. For the Green s function pole s value, we adapt the argument in [HR] to prove the following Proposition.3. Assume the Yamabe invariant Y (g) > 0, ker P = 0. If p such that G p < 0 on n fpg, then G p (p) < 0 except when (; g) is conformal equivalent to the standard S 3. Proof. Under the conformal normal coordinate with respect to p, we have after multiplying 3, the Green s function of conformal Laplacian L = 8 + R can be written as (see [LP, section 6], indeed the remain term belongs to O () ()) = r + O(4) () : Here we write f = O (4) () to mean f is C 4 away from origin with @ ii k f = O r k for 0 k 4. We write G = G p, then (see [HY, section 4]) Let eg = 4 g on n fpg, then G = A + O (4) (r) : er = 0; e Q = f Rc : On n fpg, Denote u = 0 = P g G = P 4 egg = 7 P e ( G) : G < 0, then 0 = e u + 4div f h i frc eru; ei e i + Rc f u: Here e ; e ; e 3 is a local orthonormal base with respect to eg. For > 0 small, let B = B (p) with respect to the conformal normal coordinate, then integrate the above equation on nb we see 0 = @B @ e u @e d e S 4 @B We will let! 0 + and calculate all the limits. Claim.. Rc f eru; e ds e + @ u lim e!0 + @B @e d S e = 8A: nb Rc f ude:

Indeed since eg = On the other hand, Hence This implies Claim.. Indeed we have Q CURVATURE ON A CLASS OF 3 ANIFOLDS 7 4 g, we see eu = e = d e S = = @ r ; 4 ds: 4 u + = Ar + O () r : @ e u @e d e S = @B = @ r e u ds A r 4 g (r log ; ru) + O ds: r @ e u @e d e S = 8A + O () : lim Rc f eru; e ds e = 0:!0 + @B frc = Rc D log + 4d log d log log = log r + O (4) (r) ; eru = Ar @ r + O r 3 ; log + 4 jr log j g; e = @ r ; hence f Rc = O r and The claim follows. By letting! 0 + we get frc eru; e ds e = O ds: r nfpg Rc f ude = 8A: Note A = G p (p). If A = 0, then Rc f = 0. Since (n fpg ; eg) is asymptotically at, it follows from relative volume comparison theorem that (n fpg ; eg) is isometric to the standard R 3. In particular, (; g) must be locally conformally at and simply connected compact manifold, hence it is conformal to the standard S 3 by [K]. 3. Proof of Theorem. Based on preparations in section, we will complete the proof for Theorem.. Assume (; g) is not conformal to the standard S 3, then the Green s function G (p; q) < 0 for all p; q. For u C (), u > 0, Q u 4 g =

8 FENGBO HANG AND PAUL C. YANG if and only if and this is equivalent to (3.) u (p) = P u = u 7 G (p; q) u (q) 7 d (q) : For convenience we write K (p; q) = G (p; q). We will derive the existence of solution by degree theory. Indeed for 0 t let (3.) K t (p; q) = t + tk (p; q) : Then for some 0 ; > 0, we have 0 K t (p; q). Consider the equation (3.3) u (p) = K t (p; q) u (q) 7 d (q) for u C () and u > 0. Claim 3.. here c 0 u (p) c : () 8 c 0 = 0 7 8 ; c = Indeed by the equation we see u 7 d u (p) 0 Hence for p ; p, u (p ) 0 u (p ) : () 8 7 8 0 : u 7 d: It follows that u (p) u 7 d () 7 0 8 u (p) 7 ; hence the upper bound follows. Lower bound can be proven similarly. Let n = u C () : c o 0 u c C () with uniform convergence topology. Let (T t u) (p) = K t (p; q) u (q) 7 d (q) : Claim 3. tells us deg (I T ; ; 0) = deg (I T 0 ; ; 0) = : The existence of solution follows. Let 0 = fg : (; g) satis es condition NNg. It is clear that 0 is closed. We only need to show 0 is open. Indeed assume g 0, if (; g) is not conformal di eomorphic to the standard S 3, then G p (p) < 0 for all p. It follows that (; g) satis es condition P, and hence nearby metric also satis es condition P (see [HY, Lemma 5.]). Assume (; g) is conformal di eomorphic to the standard S 3, we may assume = S 3 and g = g S 3, the

Q CURVATURE ON A CLASS OF 3 ANIFOLDS 9 standard metric. Recall we have the so called Berger s metric g t for t > 0, with g = g S 3 (see [HY, section 8]). Claim 3.. If eg is close to g S 3 in the C topology and S 3 ; eg is not conformally equivalent to S 3 ; g S 3, then there exists a t 6=, jt j small, and a continuous path (with respect to C topology) of metrics h = h (s), with h (0) = eg, h () = g t, S 3 ; h (s) is not conformally equivalent to S 3 ; g S 3 and h (s) is close to gs 3 in C topology. Assume the claim has been proven, then we will show eg satis es condition P. Indeed the set fs [0; ] : h (s) satis es condition P g is clearly open, but it is also closed since for any limit point s, h (s) satis es condition NN, and the Green s function pole value does not vanish (by Proposition.3) imply it satis es condition P. To prove Claim 3., we recall in three dimension, the metric is locally conformally at if and only if the Cotton tensor equals to zero. Note jc gt j = 9t t 6= 0 for t 6= (see [HY, section 8]). Since S 3 ; eg is not conformally equivalent to S 3 ; g S 3, by [K] we know it is not locally conformally at, hence it s Cotton tensor ec 6= 0 somewhere. By rotation we can assume e C (N) 6= 0, here N is the north pole. To continue we x a C S 3 with 0, = near north pole N and 0 near the south pole S. Then the path h (s) = ( s) eg + s [eg + ( ) gt ] ; 0 s ( s) [eg + ( ) g t ] + (s ) g t ; s : satis es all the requirement. This nishes proof of the claim. Note Added in Proof. In [HY3], it is shown that on a smooth compact 3 dimensional Riemannian manifold (; g) with positive Yamabe invariant Y (g), there exists a positive smooth function with Q g > 0 if and only if ker P g = 0 and the Green s function of P satis es G P (p; q) < 0 for all p 6= q. Similar statements are also proven when dimension is at least 5 and they are applied in [HY4] to solve the Q curvature equation on a class of manifolds with dimension larger than or equal to 5. Acknowledgement. The research of Yang is supported by NSF grant 04536. References [B] T. Branson. Di erential operators canonically associated to a conformal structure. ath. Scand. 57 (985), no., 93 345. [CGY] S.-Y. A. Chang,. J. Gursky and P. C. Yang. An equation of onge-ampere type in conformal geometry, and four-manifolds of positive Ricci curvature. Ann. of ath. () 55 (00), 709 787. [GT] D. Gilbarg and N. S. Trudinger. Elliptic partial di erential equations of second order. nd edition, 3rd printing, Berlin, Springer-Verlag, 998. [G]. J. Gursky and A. alchiodi. A strong maximum principle for the Paneitz operator and a nonlocal ow for the Q curvature. Preprint (04). [H] F. B. Hang. On the higher order conformal covariant operators on the sphere. Commun Contemp ath. 9 (007), no. 3, 79 99. [HY] F. B. Hang and P. Yang. The Sobolev inequality for Paneitz operator on three manifolds. Calculus of Variations and PDE. (004), 57 83. [HY] F. B. Hang and P. Yang. Paneitz operator for metrics near S 3. Preprint (04). [HY3] F. B. Hang and P. Yang. Sign of Green s function of Paneitz operators and the Q curvature. International athematics Research Notices 04; doi: 0.093/imrn/rnu47.

0 FENGBO HANG AND PAUL C. YANG [HY4] F. B. Hang and P. Yang. Q curvature on a class of manifolds with dimension at least 5. Preprint (04). [HR] E. Humbert and S. Raulot. Positive mass theorem for the Paneitz-Branson operator. Calculus of Variations and PDE. 36 (009), 55 53. [K] N. H. Kuiper. On conformally- at spaces in the large. Ann of ath. 50 (949), 96 94. [LP] J.. Lee and T. H. Parker. The Yamabe problem. Bull AS. 7 (987), no., 37 9. [P] S. Paneitz. A quartic conformally covariant di erential operator for arbitrary pseudo- Riemannian manifolds. Preprint (983). [Y] P. Yang and. J. hu. On the Paneitz energy on standard three sphere. ESAI: Control, Optimization and Calculus of Variations. 0 (004), 3. Courant Institute, New York University, 5 ercer Street, New York NY 00 E-mail address: fengbo@cims.nyu.edu Department of athematics, Princeton University, Fine Hall, Washington Road, Princeton NJ 08544 E-mail address: yang@math.princeton.edu