RIEMANNIAN MANIFOLDS WITH POSITIVE YAMABE INVARIANT

RIEMANNIAN MANIFOLDS WITH POSITIVE YAMABE INVARIANT AND PANEITZ OPERATOR MATTHEW J. GURSKY, FENGBO HANG, AND YUEH-JU LIN Abstract. For a Riemannian manifold with dimension at least six, we prove that the existence of a conformal metric with positive scalar and Q curvature is equivalent to the positivity of both the Yamabe invariant and the Paneitz operator.

1. Introduction Let (M; g) be a smooth compact Riemannian manifold with dimension n 3. Denote [g] as the conformal class of metrics associated with g. The Yamabe invariant is given by (see [LP]) R e Rde M Y (M; g) = inf (1.1) n 2 ; g e2[g] e (M ) n e is the scalar curvature of ge and e is the measure associated with ge. In terms here R of the conformal Laplacian operator L=

we have Y (M; g)

=

=

inf 1

u2C (M ) u>0

inf

4 (n 1) n 2 R

M

+ R;

(1.2)

Lu ud

(1.3)

2

kuk n2n2 L R 4(n M

u2H 1 (M )nf0g

1) n 2

2

jruj + Ru2 d 2

kuk

:

2n 2

Ln

In particular Y (M; g) > 0 if and only if the …rst eigenvalue 1 (L) > 0. Moreover based on the fact that the …rst eigenfunction of L must be strictly positive or negative, we know e > 0: e 2 [g] with R (1.4) 1 (L) > 0 , there exists a g

Here we are looking for similar characterization for Paneitz operator and Q curvature (see [B, P]). More precisely we are interested in the solution to

Problem 1.1. For a Riemannian manifold with dimension at least …ve, can we …nd a conformal invariant condition which is equivalent to the existence of a conformal metric with positive scalar and Q curvature? In view of the results on positivity of Paneitz operator in [GM, XY], many people suspect the conformal invariant condition wanted in Problem 1.1 should be the positivity of Yamabe invariant and Paneitz operator. As a consequence of the 1

2

M ATTHEW J. GURSKY, FENGBO HANG, AND YUEH-JU LIN

main result below, this is veri…ed for dimension at least six (see Corollary 1.1). It is very likely the statement is still true for dimension …ve. However due to technical constrains in our approach, dimension …ve case remains an open problem. To write down the formula of Q curvature and Paneitz operator, following [B], let 1 R ; A= (Rc Jg) ; (1.5) J= 2 (n 1) n 2 here Rc is the Ricci curvature tensor. The Q curvature is given by n 2 Q= J 2 jAj + J 2 : (1.6) 2 The Paneitz operator is given by 2

n

4

Q': (1.7) 2 Here e1 ; ; en is a local orthonormal frame with respect to g. For n 5, under a conformal change of metric, the Paneitz operator satis…es P' =

' + div (4A (r'; ei ) ei

P

n

4

4

g

(n

n+4 n 4

'=

2) Jr') +

Pg ( ') ;

(1.8)

(see [B]); compare to the conformal covariant property of the conformal Laplacian (1.2). In addition, the Q curvature is transformed by the formula Q

4

4

g

=

2

n+4 n 4

Pg : (1.9) n 4 We now de…ne two conformal invariants related to the Q-curvature. First, in analogy with the Yamabe invariant, we de…ne R R e Qde P u ud n 4 + M M Y4 (M; g) = inf inf : (1.10) n 4 = 2 1 2 ge2[g] (e (M )) n u2C (M ) kuk 2n n

u>0

Ln

4

We use Y4+ (M; g) to emphasize the in…mum is taken over positive functions (i.e. conformal factors). To de…ne the second invariant, denote E (')

= =

Z

ZM M

P ' 'd 2

( ')

(1.11) 4A (r'; r') + (n

2

2) J jr'j +

n

4 2

Q'2 d :

It is clear that E (') is well de…ned for ' 2 H 2 (M ). De…ne Y4 (M; g) =

inf

u2H 2 (M )nf0g

E (u) kuk

2

:

(1.12)

2n 4

Ln

Clearly Y4 (M; g)

Y4+ (M; g) ;

but in general (and in contrast to the usual Yamabe invariant) we have an inequality instead of equality, due to the fact Paneitz operator is fourth order. Note that from standard elliptic theory Y4 (M; g) > 0 if and only if the …rst eigenvalue 1 (P ) > 0 i.e. P is positive de…nite. There are several criterions for the positivity of P (see [CHY, Theorem 1.6] and [GM, XY]).

M ANIFOLDS W ITH POSITIVE YAM ABE INVARIANTS AND PANEITZ OPERATOR

3

When the Yamabe invariant Y (M; g) > 0, there is another closely related quantity Y4 (M; g) de…ned by R e Qde n 4 M Y4 (M; g) = : (1.13) inf 2 ge2[g] (e (M )) nn 4 e R>0

Obviously

Y4 (M; g) Y4+ (M; g) Y4 (M; g) : (1.14) One of the main goals of this paper is to understand the relationship between these quantities, and their connection to the existence of a metric with positive scalar and Q-curvature. Our motivation for studying these problems comes from recent progress in the understanding of Q curvature equations for dimension at least …ve in [GM, HY1, HY2]. Paneitz operator and Q curvature were brought to attention in [CGY]. For dimension at least …ve, in various work [HeR, HuR, R], people had realized the important role of positivity of Green’s function of Paneitz operator in understanding the Q curvature equations. Such kind of positivity is hard to get due to the lack of maximum principle for fourth order equations. A breakthrough was achieved in [GM], namely for n 5, if R > 0 and Q > 0, then P > 0 and the Green’s function of Paneitz operator GP is strictly positive. Subsequently in [HY1], for positive Yamabe invariant case it was found the positivity of Green’s function of Paneitz operator is equivalent to the existence of a conformal metric with positive Q curvature. Indeed it was shown that if n 5, Y (M; g) > 0, then there exists a conformal metric with positive Q curvature if and only if ker P = 0 and GP > 0, it is also equivalent to ker P = 0 and GP;p > 0 for a …xed p. Note the positivity of Green’s function is a conformal invariant condition. In [GM], it was shown that R > 0 and Q > 0 implies Y4 (M; g) is achieved at a positive smooth function. The assumption was relaxed to Y (M; g) > 0; Q > 0 and P > 0 in [HY2]. Trying to understand relations between various assumptions motivates problems considered here. Theorem 1.1. Let (M; g) be a smooth compact Riemannian manifold with dimension n 6. If Y (M; g) > 0 and Y4 (M; g) > 0, then there exists a metric ge 2 [g] e > 0 and Q e > 0. satisfying R

Combine Theorem 1.1 with existence and positivity results in [GM, HY2, XY] we have the following corollaries. Corollary 1.1. Let (M; g) be a smooth compact Riemannian manifold with dimension n 6. Then the following statements are equivalent (1) Y (M; g) > 0; P > 0. (2) Y (M; g) > 0; Y4 (M; g) > 0. e > 0 and Q e > 0. (3) there exists a metric ge 2 [g] satisfying R Corollary 1.1 answers Problem 1.1 for dimension at least six.

Corollary 1.2. Let (M; g) be a smooth compact Riemannian manifold with dimension n 6. If Y (M; g) > 0 and Y4 (M; g) > 0, then P > 0, the Green’s function GP > 0, and Y4 (M; g) is achieved at a positive smooth function u with R n 4 4 > 0 u

and Q

un

4

4

g

= const. In particular, Y4 (M; g) = Y4+ (M; g) = Y4 (M; g) :

g

4


As a consequence of Theorem 1.1 and [CHY, Theorem 1.6], we have Corollary 1.3. Let (M; g) be a smooth compact Riemannian manifold with dimension n 6. If J 0, 2 (A) 0 and (M; g) is not Ricci ‡at, then P > 0, the Green’s function GP > 0, and Y4 (M; g) is achieved at a positive smooth function u with R n 4 4 > 0 and Q n 4 4 = const. In particular, u

g

u

g

Y4 (M; g) = Y4+ (M; g) = Y4 (M; g) : The dimensional restriction n 6 is an unfortunate by-product of our technique and it is very likely the result holds in dimension …ve as well. To explain our approach, we …rst point out that the Q curvature equation is variational: a metric has constant R Q curvature if and only if it is a critical point of the total Q curvature functional M Qg d g with g running through the set of conformal metrics with unit volume. A closely related quantity is 1 2 J 2 jAj : (1.15) 2 (A) = 2 Indeed 2 (A) equation is also variational in similar sense. Since Q= we have

Z

n

J+

Qd =

n

4

4 2

Z

J2 + 4

J 2d + 4

2

(A) ;

(1.16)

Z

(1.17) 2 (A) d : 2 M M Obviously M J 2 d is always nonnegative. For t 1 consider the functional Z Z n 4 t Jg2 d g + 4 (1.18) 2 (Ag ) d g : 2 M M M

R

A critical metric of this functional restricted to the space of conformal metrics of unit volume satis…es n 4 2 t J+ J + 4 2 (A) = const: (1.19) 2

In the appendix, we will use elementary methods to show that if Y (M; g) > 0, 4

then there exists g0 2 [g], g0 = u0n t0

0 J0

4

+

g and t0

n

4 2

J02

1 such that +4

2

(A0 ) > 0

(1.20)

and J0 > 0. De…ne f as t0 Then for 1

t t

Let

0 J0

+

n

4 2

J02

+4

2

(A0 ) = f u0

n+4 n 4

:

(1.21)

t0 we consider the following 1-parameter family of equations: eJ + n

2

4 e2 J

+4

2

e = fu A

n+4 n 4

;

ge = u n

4 4

g:

(1.22)

S (1.23) n o e>0 : = t 2 [1; t0 ] : there exists positive smooth u solving (1.22), with R


5

Then t0 2 S. We will show S is both open and closed by the implicit function theorem and apriori estimates. Hence S = [1; t0 ]. It follows that there exists a e > 0 and Q e > 0. ge 2 [g] with R We conclude the introduction with some remarks. The dimensional restriction n 6 appears in both the open and closed part of the argument. The power of the conformal factor u on the right hand side of (1.22) is chosen to be negative to give better estimate of solutions. Moreover this choice also leads to a good sign of the zeroth order term in the linearized operator. We also observe that our path of equations is variational, it has a divergence structure which we will exploit in apriori estimates. At last we note that in dimension four a path of equations which is analogous to (1.22) is considered in [CGY] to produce a conformal metric with positive scalar curvature and 2 (A) R curvature assuming the positivity of Yamabe invariant and conformal invariant M 2 (A) d . Acknowledgements. The …rst author is supported in part by the NSF grant DMS–1206661. We would like to thank Professor Alice Chang and Paul Yang for valuable discussions.

2. The method of continuity and openness In this section we set up the continuity method. It will be more convenient if we …rst rewrite equation (1.22) in terms of Q and 2 (A). Using (1.16), equation (1.22) can be expressed as e t Q

hence

4

2

e A

+4

e + 4 (1 tQ

Dividing by t (recall t

t)

2

1) and denoting =

4 (t

1) t

2

n+4 n 4

e = fu A

e = fu A

;

=

n+4 n 4

;

(2.1)

:

(2.2)

f ; t

(2.3)

equation (1.22) is equivalent to e Q

4

2

e = u A

n+4 n 4

;

(2.4)

here ge = u n 4 g and 0 < 4. To write this equation in terms of the conformal factor u, observe that the Schouten tensor of the conformal metric ge is given by eij = Aij A

2

n

4

u

1

uij

2

(n

2u

4)

2

2

jruj gij +

2 (n (n

2) 2

4)

u

2

ui uj ;

(2.5)

6


hence 2

e A

8

= u

n

+

4

(2.6) "

4 (n

2

2) 3

(n

4) 2

(n

(A) + 3

u

4)

(n

( u)

4)

2 n

2) 2

(n

4)

(n

1

Ju

4

2 (n

2

jruj

2

2

2

2u

uij ui uj 2

2 Ju

2

u

u+ 2

2u

2

u

1

D2 u

4) 2 n

4 #

2

4

+

(n 2 (n

Aij uij

3u

1) 3

(n

4)

u

2

3

jruj

4

jruj

4)

4

Aij ui uj :

Using the formula (1.9) we also have 2

e= Q

n

4

u

n+4 n 4

Pu

(2.7)

Substituting (2.6) and (2.7) into (2.4), then multiplying through by have

=

Pu " u

+

n+4 n 4 n 4 2 u

we

(2.8) n

2 (n

4 2

2 2) 2

(n 4) 1 Ju n 4

1

(A) +

u 2

3

n

4

uij ui uj 2

jruj

n n

( u) 1

Ju

1

2

2

u

2 u 4

u+u 2

2

u

n

4

1

Aij uij

Aij ui uj +

n

D2 u

2

+

n

1

(n

4)

4

2 (n

2u

4

2u

4)

2

n

4

u

1

2

jruj +

4

:

2

n

4

2

2

jruj

jruj

Also, the condition that Je > 0 is equivalent to the inequality u
0, then there exists a metric ge 2 [g] with e Je + n

2

4 e2 J > 0;

Je > 0:

(2.10)

Je > 0:

(2.11)

Note this proposition is clearly a consequence of the solution to the Yamabe problem ([LP]). In the Appendix we will provide an elementary proof. In view of 4

Proposition 2.1 we can …nd ge = u0n

e Je + n

Since e Q

2

e = A

4

g such that

2

4 e2 J > 0;

e Je + n

2

4 e2 J + (4

)

2

e ; A

(2.12)

u

u


we can …nd 0
0. In particular A

n+4 4

e A

0 2

7

u0n

(2.13)

o : 9u 2 C 1 (M ) ; u > 0; satisfying (2.4) and Je > 0 :

(2.14)

It is clear that 0 2 . Indeed in this case u0 is a solution. On the other hand if we can show 0 2 , then it follows there exists a metric ge 2 [g] with Je > 0 and e > 0. To achieve this we will show is both open and closed. Q To prove that is open, we consider the linearized operator. To this end, de…ne the map n+4 4 e e ge = u n 4 g 7! N [u] = Q u n 4: (2.15) 2 A

Then N [u] = 0 if and only if ge is a solution of (2.4). Let Se denote the linearization of N at u: e = d N [u + t'] : (2.16) S' dt t=0 To compute Se we use the standard formulas for the variation of the Q curvature and the Schouten tensor d dt

d dt

Q(1+t

)g

=

t=0 2

A(1+t

)g

=

t=0

n+4 Qg ; 4 1 2 g + g Ag ; Dg 2

1 Pg 2 1 Jg 2

(2.17)

2

2

(Ag ) : (2.18)

In our setting, using (u + t') n

4 4

1

g = 1 + tu

4

'

n

4

and d dt

1 + tu

1

4

4

'

n

4

=

t=0

n

4

u

ge

1

';

(2.19)

(2.20)

we see e = S'

where

2 n

4

e u H

1

' ;

(2.21)

e H'

(2.22) n+4 n + 4 n + 4 2 e + e D e ' +4 2 A e ' + = Pe' Q' Jee ' ge A; u n 4 ': 2 2 The openness of follows from implicit function theorem and standard elliptic theory, together with the following lemma: Lemma 2.1. Let (M; g) be a smooth compact Riemannian manifold with dimension n 6. If 0 4; Q J > 0; (2.23) 2 (A) > 0;

8


then the operator H'

(2.24)

n+4 = P' Q' + 2 is positive de…nite.

n+4 Aij 'ij + 4 2 (A) ' + (Q 2

J '

Proof. For any smooth function ', Z H' 'd M Z 2 2 = ( ') + (n 2 ) J jr'j

M

we see

M

(4

n

) A (r'; r') +

2

M

(A)) '

(2.25)

M

Using the Bochner formula Z Z Z 2 2 2 ( ') d = D ' d +

2

J jr'j d +(n

2)

Z

M

4 2

(Q

2

(A)) '2 d :

A (r'; r') d ; (2.26)

Z

H' 'd (2.27) Z Z 4 n 6+ 2 2 = ( ') d + D2 ' d n 2 n 2 M Z M Z n 4 4 + (n 2) (n 2 ) 2 2 J jr'j d + (Q + 2 (A)) ' d : n 2 2 M M Note for n 6 and 0 4, all coe¢ cients before the integral sign are nonnegative. As a consequence H is positive de…nite. R 2 For n = 5, note that the coe¢ cient of M J jr'j d is equal to 13 3 4 and it is negative when is close to 4. M

3. Apriori estimate In this section we prove apriori estimate for smooth positive solutions to (2.4) with positive scalar curvature for 0 0 . An immediate consequence is that the set (see (2.14)) is closed. Lemma 3.1. Assume (M; g) is a smooth compact Riemannian manifold with dimension n 6. If Y (M; g) > 0, Y4 (M; g) > 0 and 0 0 < 4, then any e smooth positive solution u to (2.4) with J > 0 satis…es kuk

2n 4

Ln

c

(3.1)

and u

c > 0:

(3.2)

Here c is independent of u and . Proof. Let ge = u n

4 4

Z

g. Using

M

e =4 Qde

Z

M

2

e de + n A

4 2

and (2.4) we get Z Z Z n 4 2 e e 1 Qde + J de = u 4 8 M M M

Z

M

n+4 n 4

Je2 de;

de =

(3.3)

Z

M

ud :

(3.4)


9

By the de…nition of Y4 (M; g) (see (1.13)) we have 2

kuk

=

2n 4

Ln

n

4

e (M )

1 Y4 (M; g)

c

ud

n

4

2 Z

(3.5)

n

Z

M

e Qde

M

c kuk

2n 4

Ln

:

Hence kuk

c:

2n 4

Ln

(3.6)

Multiplying both sides of equation (2.8) by u and doing integration by parts we get Z n 4 u d (3.7) 2 M Z Z 3 1 2 2 = u 1 ( u) d + ( 1) + u 2 jruj ud 2 (n 4) M M Z Z (n 4) 2 (n 8) 2 4 2 3 u jruj d + (n 2 ) Ju 1 jruj d + 2 2 (n 4) M M Z Z n 4 1 +1 (4 ) u A (ru; ru) d + (Q d : 2 (A)) u 2 M M

If

3 1 > 0; 2 (n 4) (this happens when j j is large enough), then using (2.9) we get Z n 4 u d 2 M #Z " Z 4 16 2 n (4 ) 2 1 2 u u ( u) d + 2 2 (n 4) 2 (n 4) M M Z Z n 4 2 2n 2 1 + + Ju jruj d (4 ) u 2 4 4 M M Z n 4 +1 (Q d : + 2 (A)) u 2 M (

Now let

=

(p + 1) with p u

we get

Z

1) +

u

p 1

d

p 2

1, using 2

jruj c

M

+c

Z

ZM

u

p 2

p

4

3

jruj d

1

A (ru; ru) d

< 4 and the fact for all " > 0, 1 4 4 jruj + u p ; 4" 2

( u) d

c

Z

u

p 4

M

u

(3.9)

0

p

"u

(3.8)

4

jruj d

(3.10)

d :

M

Hence u

1 p+1 Lp+1

c u

1 p Lp

c u

1 p Lp+1

:

(3.11)

10


It follows that 1

u To continue let u

1

c:

Lp+1

(3.12)

= U , then the inequality (2.9) implies U

0:

c, in another (3.14)

To derive further estimates on u, we denote v=u here q

q 1

u+

n

4 2

Ju ;

(3.15)

0 is a number to be determined. It follows from (2.9) that v>

2 n

4

u

q 2

2

jruj :

(3.16)

uq+1 v;

(3.17)

Since u=

n

4 2

Ju

we see that under local orthonormal frame with respect to g, 2

u

=

uq+1 v

2 (q + 1) uq ui vi + (q + 1) u2q+1 v 2 n 4 2 (q + 2) Juq+1 v q (q + 1) uq 1 jruj v 2 2 (n 4) 2 n 4 + (n 4) Ji ui + J u+ J u: 4 2

(3.18)

Plug (3.17) and (3.18) into (2.8) we get 2

2 (q + 1) u 1 ui vi + (q + 1) uq v 2 q (q + 1) u 2 jruj v (3.19) n 4 2 + q + 2 Jv + 4u q 1 Aij uij + 2u q 1 Ji ui (n 4) jAj u q 2 2 2 (n 2) 2 2 q 3 uq v 2 u 2 jruj v u q 2 D2 u + uij ui uj = 2 2 u n 4 n 4 (n 4) (n 4) (n 1) (n 2) (n 4) 4 2 q 4 + u q 1 Aij uij jruj u q 2 Aij ui uj jAj u q 2 u n 4 4 (n 4) n 4 + u q 1: 2 v


11

Multiplying v on both sides and doing integration by parts we have

1

v

2

jrvj + q 1

4 u

q 1

2u n

4

uq v

(n

1)

(n

4)

u

q 1

2

q 4

u

q

2

2

2u

4) 4

(n

4)

to mean

1

v

2

jrvj d + 2 +1

2 (n

2)

(n Z

4)

2

v

M

Z 1 2

v

R

M

(

(n

+1

n

u

q 1

2

q

jAj u

4

(q + 1) q +

q

jruj v

jruj v

J i ui v

(q + 1) q +

+c

2

4) jAj u

d =

4)

n

v +

R

4

1

4

q 1

2 (n

2)

(n

4)

2

q

n

u

4

2

q 2

n u

2

M

Z

4

uq v

2

+1

Z

q 3

M

uij ui uj

2

= j j + =

uij ui uj v

Aij ui uj v

+2

d

(3.21)

d

n

q 4

4

Z

u

q 2

M

4

jruj v d

M

D2 u =

+

u g: n

1 2q+2 2 u v n

n

4

u n g

and trace-free component

(3.22)

Consequently by (3.17),

2

q 3

v :

M

jruj v

To continue we split D2 u into its trace component ,

D2 u

u

d . In view of (3.2) and (3.16) we

M

(n 1) uij ui uj v d u 2 (n 4) M M Z Z jrvj d + c v +1 d + c v d : u

D2 u v +

+

u

2

2

q 2

u

Aij ui vj v

1) (q + 1) +1 #Z 2 2

2 2 u 2 jruj v +1 (3.20) +1 (n 4) ( 1) 2 +n 2 2 Aij ui uj v + q+ Jv +1 2 ( + 1) +1

+2

+ 4 (q + 1) u

(n 2

+2

M

+

1

Aij ui vj v

Here we write see

"

1) (q + 1) q u v +1

Ji ui v

(n

Z

(

n

Juq+2 v +

1 q+1 n 4 2 2 u jruj v + Ju jruj + n 2n

(n

2

4) 2 2 J u ; 4n

ij ui uj :

(3.23) (3.24)

2

D2 u v d

12


Plug these equalities into (3.21), we get Z Z ( 1) (q + 1) (n 1) 2 v 1 jrvj d + uq v +2 d (3.25) + 1 n (n 4) M M " #Z 4 (n 1) 2 2 (q + 1) q + u 2 jruj v +1 d 2 +1 n (n 4) M Z 2 (n 2) 1 n 1 2 4 2 u q 2v j j u u jruj d ij ui uj + n 4 M n 4 n 4 Z Z (n 2) (n 4) 2 q 2 + Ju J 2u q v d jruj v d n (n 4) M 4n M Z Z Z 1 v +1 d + c +c v d : v 2 jrvj d + c M

M

M

By (3.16) we have Z

(n 2) n (n 4) This and c

Z

v

M

1 2

Ju

q 2

M

jrvj d

2

2

jruj v d Z

v

1

M

c

Z

v

+1

d :

(3.26)

M

c2 jrvj d + 2 2

Z

v

together with (3.25) implies for 1, Z Z ( 1) (q + 1) (n 1) 2 v 1 jrvj d + uq v 2 M +1 n (n 4) M #Z " 4 (n 1) 2 2 u 2 jruj v +1 d (q + 1) q + 2 +1 n (n 4) M Z 2 (n 2) 1 n 1 2 u q 2v j j u u ij ui uj + n 4 M n 4 n 4 Z Z +c v +1 d + c v d : M

Because

(3.27)

M

+2

2

d

(3.28)

4

jruj

d

M

is trace-free, j

r

ij ui uj j

n

1 n

2

j j jruj ;

(3.29)

hence 2 (n 2) 1 n 1 2 4 u u jruj ij ui uj + n 4 n 4 r 2 (n 2) n 1 1 n 1 2 2 2 4 u j j jruj + u jruj j j n 4 n n 4 !2 r n 2 n 1 1 4 (n 1) 2 4 2 j j u jruj jruj 2u n 4 n n (n 4) 2

j j

=

4 (n

1)

n (n

4)

2u

2

4

jruj :

(3.30)


Plug this inequality into (3.28) we get Z 2 v 1 jrvj d 2 M ( 1) (q + 1) (n 1) + +1 n (n 4) " 4 (n 2 (q + 1) q + +1 n (n Z 4 (n 1) 4 + u q 4 jruj v 3 n (n 4) M By (3.16) we have Z 4 (n 1) 3

n (n

4)

q 4

u

M

hence

Z

2

4

jruj v d

1

v

M

(

+ "

+c

v

Z

u

2

Z

v

1

M

2

4)

+1

d +c

2

2

jruj v

2

u

Z

v

Z

u

2

+1

jruj v

+1

d +c

M

2 (n

1)

n (n

4)

2

d

Z

v d :

M

2

M

Z

(n 1) n (n 4)

uq v M #Z 1)

2 (n

2

n (n

Z

2

jruj v

+1

d ;

(3.32)

4)

+2

u

d 2

M

2

jruj v

+1

d

v d :

+1

d

n

4 2

Z

uq v

+2

d ;

(3.34)

M

2

jrvj d

(3.35)

(

M

Z

uq v

+2

d

M

(n 1) ;0 n (n 4)

Z

uq v

+2

d

M

M

In another word Z 2 v 1 jrvj d 2 M ( 1) (q + 1) (n 1) + min ; (q + 1) +1 n (n 4) Z Z c v +1 d + c v d : M

d

M

d +c

1) (q + 1) (n 1) +1 n (n 4) n 4 2 max (q + 1) q + 2 +1 Z Z +c v +1 d + c v d : +

+2

M

M

we get

1)

uq v #Z

(3.33)

2 +1

M

By (3.16) again,

M

2

1) (q + 1) +1

Z

(3.31) Z

jrvj d

(q + 1) q +

13

M

(3.36) n

4 2

q+

n+3 +1

Z

M

uq v

+2

d

14


Because n

6, we see n n

Fix a q

2 4 (n > 4 n (n

1) : 4)

(3.37)

0 such that n n

then for Z

large enough, v

M

1

2

jrvj d +

By (3.2) and (3.39) we get

Z

uq v

+2

d

c

M

+2

kvkL

2 4 (n >q+1> 4 n (n

+2

Z

1) ; 4) +1

v

(3.38)

d +c

M

+1

c ( ) kvkL

+2

Z

v d :

(3.39)

M

+ kvkL

:

+2

(3.40)

Hence for

kvkL +2 c ( ) large enough. To continue, we observe that for large enough, Z Z Z +1 2 d c v +1 d + c 2 v d : rv 2 M

M

Hence

v

Replacing

+1 2

+ 1 by

Z

2 2n Ln 2

we see

c

M

kvkL

Here

+1 2

2

d +

Z

v

+1

d

(3.43)

M

+1 +1

2

+c

c kvkL + c =

(3.42)

M

rv

c kvkL

(3.41)

n n

2

2

kvkL

kvkL

+1

1

:

:

:

(3.44) (3.45)

Let b = max fkvkL ; 1g ; large enough, we have

then for

b

c

It follows from iteration that for a …xed b

k

0

0

cb

2

b :

(3.46) (3.47)

large, c:

0

(3.48)

Hence Letting k ! 1 we see

kvkL

k

0

c:

kvkL1 c: Now we go back to the equation of u, n 4 Ju = vuq+1 : u+ 2 Since kuk n2n4 c and q < n 4 4 , standard bootstrap method tells us

(3.49) (3.50)

(3.51)

L

Elliptic estimate tells us

kukL1 kukW 2;p

c:

(3.52)

c (p)

(3.53)


15

for any 1 < p < 1. This together with (2.8) and (3.2) imply kukC k

c (k)

(3.54)

for all k 2 N. Proposition 3.1. Under the assumption of Theorem 1.1, the set Proof. Assume i 2 , satisfying Ji > 0, then

i

!

1

as i ! 1, ui is a solution to (2.4) for

kui kC k

for all k 2 N and

ui

4

g;

n 4 g1 = u1 g:

(Ai ) = ui

n+4 n 4

Let i ! 1, we get Q1

1 2

c > 0. Denote

4

Then i 2

i

c > 0:

4

gi = uin

=

c (k) ;

After passing to a subsequence we have ui ! u1 in C 1 and u1

Qi

is closed.

;

n+4 4

(A1 ) = u1n

(3.55)

Ji > 0:

;

J1

0:

(3.56)

In another way, 1 J1

4 2

1

2

jA1 j1 +

n 2

1

n+4 4

2 J1 = u1n

> 0:

(3.57)

Hence n

1

2 J1 > 0: (3.58) 2 Since J1 0, it follows from strong maximum principle that either J1 > 0 or J1 0. The latter case contradicts with (3.58). Hence J1 > 0 and 1 2 . It follows that is closed. 1 J1

+

This …nishes the proof of Theorem 1.1. Proof of Corollary 1.1. We only need to show (3))(1). However this follows from [GM, Proposition B]. Corollary 1.1 answers Problems 1.1 for dimension greater then …ve. Proof of Corollary 1.2. By Corollary 1.1 we know P > 0 and there exists a metric e and Q. e The fact GP > 0 follows from [GM, Theorem A] ge 2 [g] with positive R and [HY1, Theorem 1.1]. The remaining conclusion follows from [GM, remark (1) after Theorem D] and [HY2, Theorem 1.2]. Using the second order criterion for the positivity of Paneitz operator in [CHY, Theorem 1.6], we can deduce Corollary 1.3. Proof of Corollary 1.3. Because g is not Ricci ‡at, A does not vanish somewhere. 2 Using J 0 and 2 (A) = 12 J 2 jAj 0, we see J > 0 somewhere. Hence Y (M; g) > 0. The fact P > 0 follows from [CHY, Theorem 1.6]. With these facts at hand, Corollary 1.3 follows from Corollary 1.2.

16


4. Appendix Here we give an elementary proof of Proposition 2.1. Assume Y (M; g) > 0, we can assume R > 0. For 1 < p < nn+22 , based on the compact embedding H 1 (M ) Lp+1 (M ) we know there exists a positive smooth function u satisfying Lu = up

(4.1)

(see [LP]). Here L is the conformal Laplacian operator (1.2). Let then hence Je > 0. Note that e Je + n

Since

2

4 e2 J =

using (4.1) and (4.3) we have eR e+ n 4 R e2 4 (n 1) u

4

n

2

n+2 n 2

e=u R

e' = u

=

ge = u n

e + 2u R

Lu = up

(4.2) n+2 n 2

> 0;

eR e+

n+2 n 2

(4.3)

n 4 e2 R : 4 (n 1)

g (ru; r') ;

(4.4) (4.5)

(4.6) n+2 n 2

n 4 e2 R 4 (n 1) n+2 n 2 + 4 (n 1) n 2

e g ru; rR

max 1; this is possible since n

g;

' + 2u

2

n 2 6 p u2p 4 (n 1) n 2 n+2 4 + p p n 2 n 2 We choose p such that =

2

1 2 (n 1)

4

n

4

2(n+2) n 2

up

6 n

2

3n+2 n 2

+

p Rup

n+6 n 2

2

jruj :