Equivariant Yamabe problem and Hebey-Vaugon conjecture

Universit¨ at Regensburg Mathematik

Equivariant Yamabe problem and Hebey-Vaugon conjecture

Farid Madani

Preprint Nr. 12/2011

FARIDMADANI

Abstrat. IntheirstudyoftheYamabeprobleminthepreseneofisometry

group,E.HebeyandM.Vaugonannounedaonjeture. Thisonjeturegen-

eralizesT.Aubin'sonjeture,whihhasalreadybeenprovenandissuient

to solvethe Yamabeproblem. Inthispaper,wegeneralizeAubin'stheorem

andweprovetheHebeyVaugononjetureinsomenewases.

1. Introdution

Let

(M, g)

^{be a ompat Riemannian manifold of dimension}

n ≥ 3

^{. Denote by}

I(M, g)

^,

C(M, g)

^and

R g

^{theisometrygroup,theonformal}transformationsgroup and thesalarurvature,respetively. Let

G

^{beasubgroupoftheisometrygroup}

I(M, g)

^{. E.HebeyandM.Vaugon[5℄onsideredthefollowingproblem:}

HebeyVaugonproblem.Istheresome

G −

^{invariantmetri}

g 0

^{whih minimizes}

the funtional

J (g ^′ ) = R

M R g ^′ dv(g ^′ ) ( R

M dv(g ^′ )) ^{n− n 2}

where

g ^′

^belongstothe

G −

^{invariant onformal lass ofmetris}

g

^denedby:

[g] ^G := { g ˜ = e ^f g/f ∈ C ^∞ (M ), σ ^∗ ˜ g = ˜ g ∀ σ ∈ G }

Thepositiveanswerwouldhavetwoonsequenes. Therstisthatthereexistsan

I(M, g) −

^{invariantmetri}

g 0

^onformalto

g

^{suh that thesalarurvature}

R g 0

^is

onstant. Theseondisthat theA.Lihnerowiz's onjeture[7℄, statedbelow,is

true. BytheworksofJ.Lelong-Ferrand[6℄andM.Obata[9℄,weknowthatif

(M, g)

is notonformalto

(S n , g can )

^{(the unit sphere endowed with its standardmetri}

g can

^{), then}

C(M, g)

^{is ompat and there exists a onformalmetri}

g ^′

^to

g

^suh

that

I(M, g ^′ ) = C(M, g)

^{. This impliesthat therst onsequeneis equivalentto}

the

A.Lihnerowizonjeture. Foreveryompat Riemannianmanifold

(M, g)

whihisnotonformaltotheunitsphere

S n

^{endowedwithitsstandardmetri,there}

exists ametri

˜ g

^{onformal to}

g

^{for whih}

I(M, ˜ g) = C(M, g)

^{,andthe salar ur-}

vature

R ˜ g

^isonstant.

TosuhmetrisorrespondfuntionswhihareneessarilysolutionsoftheYamabe

equation. Inotherwords,if

˜ g = ψ ^{n−2 4} g

^,

ψ

^isa

G −

^{invariantsmoothpositivefuntion}

then

ψ

^satises

4(n − 1)

n − 2 ∆ g ψ + R g ψ = R g ˜ ψ ^{n− n+2 2} .

1991MathematisSubjetClassiation. 53A30,53C21,35J20.

Key words and phrases. Conformal metri; Isometry group; Salar urvature; Yamabe

problem.

The lassial Yamabe problem, whih onsists to nd a onformal metri with

onstant salar urvature on a ompat Riemannian manifold, is the partiular

aseof theproblemabovewhen

G = { id }

^{. Denote by}

O G (P )

^theorbitof

P ∈ M

under

G

^,

W g

^{theWeyltensorassoiatedtothemanifold}

(M, g)

^and

ω n

^thevolume

oftheunit sphere

S n

^{. Wedenetheinteger}

ω(P )

^atthepoint

P

^as

ω(P ) = inf { i ∈ N / k∇ ⁱ W g (P ) k 6 = 0 } (ω(P ) = + ∞

^if

∀ i ∈ N , k∇ ⁱ W g (P ) k = 0)

HebeyVaugon onjeture. Let

(M, g)

^{bea ompat Riemannian manifold of}

dimension

n ≥ 3

^and

G

^{be a subgroup of}

I(M, g)

^{. If}

(M, g)

^{is not onformal to}

(S n , g can )

^{or if the ation of}

G

^{has no xed point, then the following inequality}

holds

(1)

inf

g ^′ ∈[g] ^G J (g ^′ ) < n(n − 1)ω ^2/n _n ( inf

Q∈M cardO G (Q)) ^2/n

Remarks1.1. (1) ThisonjetureisthegeneralizationoftheformerT.Aubin's

onjeture[1℄fortheYamabeproblemorrespondingto

G = { id }

^,wherethe

onstant in the right side of the inequality isequal to

inf g ^′ ∈[g can ] J(g ^′ )

^for

S n

^{. In thisase, the onjetureisompletely proved.}

(2) Theinequalityisobviousif

inf g ^′ ∈[g] ^G J (g ^′ )

^isnonpositive,itistheasewhen thereexistsaYamabemetri withnonpositive salar urvature.

(3) Iffor any

Q ∈ M

^,

cardO G (Q) = + ∞

^{thenthis onjetureisalso obvious.}

Theonlyresultsknownaboutthisonjeturearegivenin thefollowingtheorem:

Theorem 1.1 (E.Hebeyand M.Vaugon). Let

(M, g)

^{be a smooth ompat Rie-}

mannian manifoldof dimension

n ≥ 3

^and

G

^{beasubgroup of}

I(M, g)

^{. Wealways}

have :

g ^′ inf ∈[g] ^G J(g ^′ ) ≤ n(n − 1)ω _n ^2/n ( inf

Q∈M cardO G (Q)) ^2/n

andinequality (1) holdsifone ofthe following itemsissatised.

(1) Theationof

G

^on

M

^isfree

(2)

3 ≤ dim M ≤ 11

(3) Thereexistsapoint

P

^{withminimalorbit(nite)under}

G

^suhthat

ω(P ) >

(n − 6)/2

^or

ω(P ) ∈ { 0, 1, 2 }

^.

Thease

ω = 3

^{wasstudiedbyA.Rauzy (private}ommuniations).

Inthisproveweprovethefollowingresults:

Maintheorem.TheHebeyVaugononjetureholdsifthereexistsapoint

P ∈ M

with minimal orbit(nite)for whih

ω(P ) ≤ 15

^{orif the degreeof theleading part}

of

R g

^{isgreateror equalto}

ω(P ) + 1

^{,in the}neighborhoodof thispoint

P

^.

Corollary 1.1. HebeyVaugon onjeture holds for every smooth ompat Rie-

mannian manifold

(M, g)

^{of dimension}

n ∈ [3, 37]

^.

To provethe main theorem, we need to onstrut a

G −

^{invariant test funtion}

φ

suhthat

I g (φ) < n(n − 1)ω ^2/n _n ( inf

Q∈M cardO G (Q)) ^2/n

Thus,allthediultiesareintheonstrutionofasuhfuntion. Forsomeases,

we anuse the test funtions onstruted by T. Aubin [1℄ and R. Shoen [10℄ in

M. Vaugon[5℄. Butthe item3,presentedin Theorem 1.1,usestest funtions dif-

ferentthanT.AubinandR.Shoenones.

WemultiplyT.Aubin'stestfuntion

u ε,P

^{byafuntionasfollows:}

(2)

ϕ ε (Q) = (1 − r ^ω+2 f (ξ))u ε,P (Q)

(3)

u ε,P (Q) =



 

  ε

r ² + ε ^{2 n−} 2 ²

− ε

δ ² + ε ^{2 n−} 2 ²

if

Q ∈ B P (δ)

0

^if

Q ∈ M − B P (δ)

for all

Q ∈ M

^{, where}

r = d(Q, P )

^{is thedistane between}

P

^and

Q

^.

(r, ξ ^j )

^{is a}

geodesioordinatessystemintheneighborhoodof

P

^and

B P (δ)

^{isthegeodesiball}

ofenter

P

^withradius

δ

^{xedsuientlysmall.}

f

^{isafuntiondependingonlyon}

ξ

^,

hosensuhthat

R

S n− 1 f dσ = 0

^{. Withoutlossof}generality,wesupposethatinthe oordinatessystem

(r, ξ ^j )

^wehave

det g = 1 + o(r ^m )

^for

m ≫ 1

^{. Infat,E.Hebey}

and M.Vaugonprovedthat thereexists

g ˜ ∈ [g] ^G

^{for whih}

det ˜ g = 1 + o(r ^m )

^and

inf g ^′ ∈[g] ^G J (g ^′ )

^{doesnotdependontheonformal}

G −

^{invariantmetri.}

2. Computationof

R

M R g ϕ ² _ε dv

Letbe

I _a ^b (ε) = Z δ/ε

0

t ^b

(1 + t ² ) ^a dt

^and

I _a ^b = lim

ε→0 I _a ^b (ε)

then

I _a ^2a−1 (ε) = log ε ⁻¹ + O(1)

^{. If}

2a − b > 1

^then

I _a ^b (ε) = I _a ^b + O(ε ^2a−b−1 )

^andby

integrationbyparts,weestablishthefollowingrelationships:

(4)

I _a ^b = b − 1

2a − b − 1 I _a ^b−2 = b − 1

2a − 2 I _a−1 ^b−2 = 2a − b − 3

2a − 2 I _a−1 ^b , 4(n − 2)I _n ⁿ⁺¹ (I n ⁿ⁻² ) ^(n−2)/n = n

Using the inequality

(a − b) ^β ≥ a ^β − βa ^β−1 b

^for

0 < b < a

^{, wehave for}

β ≥ 2

^,

0 ≤ α < (n − 2)(β − 1) − n

(5)

Z

M

r ^α u ^β _ε,P dv = ω n−1 I _(n−2)β/2 ^α+n−1 ε α+n−β(n−2)/2 + O(ε ⁿ⁻² )

This integralappearsfrequentlyinthefollowingomputations,anditallowsusto

neglet the onstant term in the expression of

u ε

^{, when we hoose}

δ

^suiently

small and

ε

^smallerthan

δ

^.

Denote by

I g

^{theYamabefuntional denedforall}

ψ ∈ H ¹ (M )

^by

(6)

I g (ψ) = Z

M |∇ ^g ψ | ² dv + (n − 2) 4(n − 1)

Z

M

R g ψ ² dv

k ψ k ⁻² N

where

N = 2n/(n − 2)

^and

∇ g

^{isthegradientofthemetri}

g

^.

The seond integral of the funtional

I g

^{with the salar urvature term needs a}

speial onsideration. Let

µ(P)

^{beanintegerdened asfollows:}

|∇ ^β R g (P ) | = 0

forall

| β | < µ(P )

^{andthereexists}

γ ∈ N ^µ(P)

suhthat

|∇ ^γ R g (P ) | 6 = 0

^then

R g (Q) = ¯ R + O(r ^µ(P)+1 )

where

R ¯ = r ^µ(P) P

|β|=µ ∇ β R g (P )ξ ^β

^isahomogeneouspolynomialofdegree

µ(P)

^,

the

β

^aremulti-indies.

Forsimpliity,wedroptheletter

P

ⁱⁿ

ω(P)

^and

µ(P )

^.

Lemma 2.1.

µ ≥ ω

^,

g ij = δ ij + O(r ^ω+2 )

^and

R ¯

S(r) R g = O(r ^2ω+2 )

^{whih implies}

that

R

S(r) Rdσ ¯ = 0

^when

µ < 2ω + 2

¯ R

denotestheaverage. Then

Z

M

R g ϕ ² _ε dv = Z

M

R g u ² _ε,P dv − 2 Z

M

f u ² _ε,P R g r ^ω+2 dv + Z

M

f ² u ² _ε,P R g r ^2ω+4 dv

= ε ^2ω+4 ω n−1

¯ Z

S(r)

r ^−2ω−2 R g dσI _n−2 ^n+2ω+1 (ε) − 2ε ^ω+µ+4 I _n−2 ^ω+µ+n+1 (ε)ω n−1

¯ Z

S(r)

r ^−µ f (ξ) ¯ Rdσ(ξ) + O(ε ⁿ⁻² )

(7)

MoreoverT.Aubin[2℄provedthat:

Theorem2.1. If

µ ≥ ω + 1

^{thenthere exists}

C(n, ω) > 0

^suhthat

¯ Z

S n− 1 (r)

Rdσ = C(n, ω)( − ∆ g ) ^ω+1 R(P )r ^2ω+2 + o(r ^2ω+2 ) ( − ∆ g ) ^ω+1 R(P )

^{isnegative. Then}

I g (u ε,P ) < ⁿ⁽ⁿ⁻²⁾ ₄ ω ^2/n _n−1

^.

Fromnowuntiltheendofthissetion,wemaketheassumptionthat

µ = ω

^{. Now,}

wereallsomeresultsobtainedbyT.Aubin inhispapers[3,4℄:

R ¯

^ishomogeneouspolynomialofdegree

ω

^then

∆ E R ¯

^ishomogeneousofdegree

ω − 2

and

∆ E R ¯ = r ⁻² (∆ s R ¯ − ω(n + ω − 2) ¯ R)

where

∆ E

^{istheEulideanLaplaianand}

∆ s

^{istheLaplaianonthesphere}

S n−1

^.

∆ ^k−1 _E R ¯

^ishomogeneousofdegree

ω − 2k + 2

^and

∆ ^k _E R ¯ = r ⁻² (∆ s − ν k id)∆ ^k−1 _E R ¯ = r ^−2k

k

Y

p=1

(∆ S − ν p id) ¯ R

with

(8)

ν k = (ω − 2k + 2)(n + ω − 2k)

The sequeneof integers

(ν k ) {1≤k≤[ω/2]}

^{is dereasing. It willplaytherole of the}

eigenvaluesoftheLaplaianonthesphere

S n−1

^{. It isknownthat the}eigenvalues of the geometri Laplaian are non-negative and inreasing. Our

ν k

^{are in the}

oppositeorder.

WeknowbyT.Aubin'spaper[2℄that

∆ ^[ω/2] _E R ¯ = 0

^and

R

S(r) Rdσ ¯ = 0

^,then

q = min { k ∈ N /∆ ^k _E R ¯ = 0 }

is well dened and

r ^−ω R ¯ ∈ L q

k=1 E k

^{, with}

E k

^{the eigenspae assoiated to the}

positiveeigenvalues

ν k

^{oftheLaplaian}

∆ s

^onthesphere

S n−1

^{. If}

j 6 = k

^,then

E k

isorthogonalto

E j

^{,forthestandardsalarprodutin}

H ₁ ² (S n−1 )

^{. Moreover,sine}

R Rdσ ¯ = 0

^thereexist

ϕ k ∈ E k

(eigenfuntionsof

∆ s

^)suhthat

(9)

R ¯ = r ^ω ∆ s

q

X

k=1

ϕ k = r ^ω

q

X

k=1

ν k ϕ k

AordingtoLemma 2.1,weansplit themetri

g

^{inthefollowingway:}

(10)

g = E + h

where

E

^{is the Eulidean metri and}

h

^{is a symmetri 2-tensor dened in our}

geodesioordinatessystemby

(11)

h ij = r ^ω+2 ¯ g ij + r ^2(ω+2) g ˆ ij + ˜ h ij

^and

h ir = h rr = 0

where

¯ g

^,

g ˆ

^and

h ˜

^{are symmetri2-tensorsdened onthesphere}

S n−1

^{. Wedenote}

by

s

^{the standard metri on the sphere,}

∇

^,

∆

^{are the assoiated gradient and}

Laplaianon

S n−1

^{. By}straightforwardomputations, Aubin[3℄provedthat:

Lemma2.2.

R ¯ = ∇ ^ij ¯ g ij r ^ω

^and

¯ Z

S n− 1 (r)

Rdσ = [B/2 − C/4 − (1 + ω/2) ² Q]r ^2(ω+1) + o(r ^2(ω+1) )

where

B = ¯ R

S n− ∇ ⁱ 1 g ¯ ^jk ∇ j ¯ g ik dσ

^,

C = ¯ R

S n− ∇ ⁱ 1 g ¯ ^jk ∇ i g ¯ jk dσ

^and

Q = ¯ R

S n− 1 g ¯ ij ¯ g ^ij dσ

Forfurther detailsreferto [8℄.

Theintegrals

Q

^,

B

^and

C

^{aregivenintermsofthetensor}

g ¯

^{. Ourgoalistoompute}

them usingtheeigenfuntions

ϕ k

^{above. Letusdene}

b ij =

q

X

k=1

1

(n − 2)(ν k + 1 − n) [(n − 1) ∇ ^ij ϕ k + ν k ϕ k s ij ]

and

a ij

^{suh that}

¯ g ij = a ij + b ij

^{then,aordingto(9),wehekthat}

(12)

R ¯ = ¯ R b = ∇ ^ij b ij r ^ω

^and

R ¯ a = ∇ ^ij a ij r ^ω = 0

If

g ¯ ij = a ij

^then

R ¯ = ¯ R a = 0

^and

µ ≥ ω + 1

^{. ByTheorem2.1}

¯ Z

S n−1 (r)

Rdσ = Z ¯

S n−1 (r)

R a dσ < 0

If

g ¯ ij = b ij

^then

¯ Z

S n− 1 (r)

Rdσ = Z ¯

S n− 1 (r)

R b dσ = [B b /2 − C b /4 − (1 + ω/2) ² Q b ]r ^2(ω+1) + o(r ^2(ω+1) )

where

B b

^,

C b

^and

Q b

^{are thesameintegralsdened in Lemma 2.2when theon-}

sideredtensor

g ¯ ij = b ij

^{. Weomputethemintermsof}

ϕ k

Q b = Z ¯

S n− 1

¯ b ij ¯ b ^ij dσ = n − 1 n − 2

q

X

k=1

ν k

ν k − n + 1

¯ Z

S n− 1

ϕ ² _k dσ

B b = − (n − 1)Q b +

q

X

k=1

ν k

¯ Z

S n− 1

ϕ ² _k dσ

C b = − (n − 1)Q b + n − 1 n − 2

q

X

k=1

ν k

¯ Z

S n− 1

ϕ ² _k dσ

Tondtheseexpressions,weusedseveraltimestheidentity

∇ ⁱ b ij = − P q

k=1 ∇ ^j ϕ k

and Stokesformula (more detailsare given in [3,4℄ and [8℄). Inthe generalase,

wededuethat

Lemma2.3. If

µ = ω

^and

g ¯ ij = a ij + b ij

^,where

b ij

^{isdenedabove,}

(13)

¯ Z

S n − 1 (r)

Rdσ = Z ¯

S n − 1 (r)

R a + R b dσ ≤ [B b /2 − C b /4 − (1 + ω/2) ² Q b ]r ^2(ω+1) + o(r ^2(ω+1) )

and

(14)

B b /2 − C b /4 − (1 + ω/2) ² Q b =

q

X

k=1

u k

¯ Z

S n − 1

ϕ ² _k dσ

with

(15)

u k =

n − 3

4(n − 2) − (n − 1) ² + (n − 1)(ω + 2) ² 4(n − 2)(ν k − n + 1)

ν k

u k

^{isobtainedusingthe}expressionsof

Q b

^,

B b

^and

C b

^above.

3. Generalization of T.Aubin's theorem

Theorem3.1. Ifthereexists

P ∈ M

^suhthat

ω(P) ≤ (n − 6)/2

^{thenthereexists}

f ∈ C ^∞ (S n−1 )

^{withvanishingmeanintegral suhthat}

I g (ϕ ε ) < n(n − 2) 4 ω _n−1 ^2/n

The ase

ω = 0

^{of thethistheorem hasalreadybeenprovenby T.Aubin [1℄. He}

alsoprovedthetheoremwhen

µ ≥ ω + 1

^{(seeTheorem2.1).}

Fromnowuntiltheendofthispaper,wedroptheletter

P

ⁱⁿ

ω(P)

^and

µ(P)

^.

Proof. If

µ ≥ ω + 1

^{thentheinequalityholdsbyTheorem2.1. Sowesupposethat}

µ = ω

^{until theend of theproof. Westartby omputingthe rstintegralof the}

Yamabefuntional(6)with

ψ = ϕ ε

^{. Usingformula}

|∇ ^g ϕ ε | ² = (∂ r ϕ ε ) ² +r ⁻² |∇ ^s ϕ ε | ²

^,

weobtain:

Z

M |∇ g ϕ ε | ² dv = Z

M |∇ g u ε,P | ² dv + Z δ

0

[∂ r (r ^(ω+2) u ε,P )] ² r ⁿ⁻¹ dr Z

S n− 1

f ² dσ+

Z δ 0

u ² _ε,P r ^n+2ω+1 dr Z

S n− 1

|∇ f | ² dσ

Thesubstitution

t = r/ε

^gives

(16)

Z

M |∇ g ϕ ε | ² dv = (n − 2) ² ω n−1 I _n ⁿ⁺¹ (ε) + ε ^2ω+4 Z

S n−1

|∇ f | ² dσI _n−2 ^2ω+n+1 (ε)+

Z

S n− 1

f ² dσ[(ω − n+4) ² I _n ^2ω+n+5 (ε)+2(ω+2)(ω − n+4)I _n ^2ω+n+3 (ε)+(ω+2) ² I _n ^2ω+n+1 (ε)]

For

k ϕ ε k ⁻² N

^{, weneedto omputetheTaylorexpansionof:}

ϕ ^N _ε (Q) = [1 − N r ^ω+2 f (ξ) + N (N − 1)

2 r ^2ω+4 f ² (ξ) + o(r ^2ω+4 )]u ^N _ε,P

Usingthefat that

R

S n− 1 f dσ(ξ) = 0

^{andformula(5),weonludethat}

k ϕ ε k ^N N =

Z δ 0

Z

S n−1

[1 + N(N − 1)

2 r ^2(ω+2) f ² (ξ) + o(r ^2ω+4 )]r ⁿ⁻¹ u ^N _ε,P drdσ(ξ)

= ω n−1 I _n ⁿ⁻¹ + N (N − 1) 2 ε ^2(ω+2)

Z

S n− 1

f ² dσI _n ^2ω+n+3 + o(ε ^2ω+4 )

then

(17)

k ϕ ε k ⁻² N = (ω n−1 I _n ⁿ⁻¹ ) ^−2/N 1

− (N − 1)ε ^2(ω+2) Z

S n−1

f ² dσI _n ^2ω+n+3 /(ω n−1 I _n ⁿ⁻¹ ) + o(ε ^2ω+4 )

ByEqs(16),(17) ,(7)andtherelationship(4),if

n > 2ω + 6

^then:

I g (ϕ ε ) = n(n − 2)

4 ω ^2/n _n−1 + (ω n−1 I _n ⁿ⁻¹ ) ^−2/N I _n−2 ^n+2ω+1 ε ^2ω+4 × (n − 2)ω n−1

4(n − 1)

¯ Z

S(r)

r ^−2ω−2 R g dσ − n − 2 2(n − 1)

Z

S n− 1

f (ξ) ¯ Rdσ + Z

S n− 1

|∇ f | ² dσ+

− n(n − 2) ² − (ω + 2) ² (n ² + n + 2) (n − 1)(n − 2)

Z

S n− 1

f ² dσ

+ o(ε ^2ω+4 )

If

n = 2ω + 6

^then

I g (ϕ ε ) = n(n − 2)

4 ω ^2/n _n−1 + (ω n−1 I _n ⁿ⁻¹ ) ^−2/N ε ^2ω+4 log ε ⁻¹ × (n − 2)ω n−1

4(n − 1)

¯ Z

S(r)

r ^−2ω−2 R g dσ − n − 2 2(n − 1)

Z

S n− 1

f (ξ) ¯ Rdσ+

Z

S n− 1

|∇ f | ² dσ + (ω + 2) ² Z

S n− 1

f ² dσ

+ O(ε ^2ω+4 )

Forfurther detailsreferto [8℄.

Let

I S

^{be the funtional dened for a funtion}

f

^{on the sphere}

S n−1

^{, with zero}

meanintegral,by

I S (f ) = Z ¯

S n − 1

4(n − 1)(n − 2) |∇ f | ² − [4n(n − 2) ² − 4(ω + 2) ² (n ² + n + 2)]f ² +

− 2(n − 2) ² f Rdσ ¯

This impliesthat if

n > 2ω + 6

(18)

I g (ϕ ε ) = n(n − 2)

4 ω _n−1 ^2/n + ω _n−1 ^2/n I _n−2 ^n+2ω+1 ε ^2ω+4 4(n − 1)(n − 2)(I n ⁿ⁻¹ ) ^2/N × { (n − 2) ² Z ¯

S(r)

r ^−2ω−2 R g dσ + I S (f ) } + o(ε ^2ω+4 )

andif

n = 2ω + 6

(19)

I g (ϕ ε ) = n(n − 2)

4 ω _n−1 ^2/n + ω _n−1 ^2/n I _n−2 ^n+2ω+1 ε ^2ω+4 log ε ⁻¹ 4(n − 1)(n − 2)(I n ⁿ⁻¹ ) ^2/N × { (n − 2) ² Z ¯

S(r)

r ^−2ω−2 R g dσ + I S (f ) } + O(ε ^2ω+4 )

Notie that if

k 6 = j

^then

I S (ϕ k + ϕ j ) = I S (ϕ k ) + I S (ϕ j )

^{. Indeed,}

ϕ k

^and

ϕ j

^are

orthogonalforthestandardsalarprodutin

H ₁ ² (S n−1 )

^.

I S (c k ν k ϕ k ) =

d k c ² _k − 2(n − 2) ² c k ν _k ² Z ¯

S n−1

ϕ ² _k dσ

= − (n − 2) ⁴ d k

ν _k ² Z ¯

S n− 1

ϕ ² _k dσ

where

d k = 4[(n − 1)(n − 2)ν k − n(n − 2) ² + (ω + 2) ² (n ² + n + 2)]

^and

c k = (n − 2) ²

d k

Using (8) , weanhekeasily that

d k

^{is positiveforany}

1 ≤ k ≤ [ω/2]

^{. Now,let}

us onsider

f = P q

1 c k ν k ϕ k

^{. Then}

I S (f ) = −

q

X

1

(n − 2) ⁴ d k

ν ² _k Z ¯

S n− 1

ϕ ² _k dσ

andbyLemma 2.3

(n − 2) ² Z ¯

S(r)

r ^−2ω−2 R g dσ + I S (f ) ≤

q

X

1

(u k (n − 2) ² − (n − 2) ⁴ d k

ν _k ² ) Z ¯

S n−1

ϕ ² _k dσ + o(1)

Thefollowinglemmaimpliesthat

I g (ϕ ε ) < ⁿ⁽ⁿ⁻²⁾ ₄ ω _n−1 ^2/n

Lemma3.1. Forany

k ≤ q ≤ [ω/2]

^{the following inequality holds}

u k − (n − 2) ²

d k

ν _k ² < 0

Proof. Realltheexpressionof

ν k

^{givenin(8). Thesequene}

(U k )

^denedby

U k := (ν k − n + 1)d k { (n − 2) u k

ν k − (n − 2) ³ d k

ν k }

is polynomial dereasing in

ν k

^when

ν k ≥ 0

^{. In fat,}

U k = P (ν k )

^with

P

^the

dereasingpolynomialin

R ₊

,dened by

P (x) = [(n − 1)(n − 2)x − n(n − 2) ² + (ω + 2) ² (n ² + n + 2)] ×

[(n − 3)(x − n + 1) − (n − 1) ² − (n − 1)(ω + 2) ² ] − (n − 2) ³ (x ² − (n − 1)x)

Thederivativeof

P

^is

P ^′ (x) = − 2(n − 2)x − 2n(n − 2) ³ + 2(n ² − 3n − 2)(ω + 2) ²

Byassumption

ω + 2 ≤ (n − 2)/2

^then

P

^{isdereasingin}

R ₊

. Hene

U k = P (ν k ) ≤ P (ν ω/2 ) = U ω/2

forall

k ≤ ω/2

^{. Iteasytohekthat}

u ω/2

^isnegativeso

U k ≤ U ω/2 < 0

^.

4. Proof of the maintheorem

ByRemarks 1.1, we onsideronly thepositivease(i.e.,

inf _g ′ ∈[g] ^G J(g ^′ ) > 0

^)and

theasewhenthereexists

P ∈ M

^suhthat

O G (P) = { P i } 1≤i≤m , m = cardO G (P ) = inf

Q∈M cardO G (Q), ω ≤ n − 6

2

^and

P 1 = P

Let

ϕ ˜ ε,i

^{beafuntion denedasfollows:}

(20)

ϕ ˜ ε,i (Q) = (1 − r ^ω+2 _i f i (ξ))u ε,P i (Q)

where

r i = d(Q, P i )

^,thefuntion

u ε,P i

^{isdenedasin(3)and}

f i

^isdenedby:

(21)

f i (Q) = cr _i ^−ω ∇ ^ω g R (P i ) (exp ⁻¹ _{P i} Q, · · · , exp ⁻¹ _{P i} Q)

exp P i

^istheexponentialmap. Inageodesioordinatessystem

{ r, ξ ^j }

^withorigin

P

^,induedbytheexponentialmap

f 1 = cr ^−ω R ¯ = c

q

X

k=1

ν k ϕ k

where

R ¯

^,

ϕ k

^and

ν k

^{are denedin Setion2. Thus thefuntions}

f i

^{aredened on}

thesphere

S n−1

^{. Thehoieoftheonstant}

c

^isimportant.

Lemma 4.1. Suppose that

ω ≤ (n − 6)/2

^{. If}

ω ∈ [3, 15]

^orif

deg ¯ R ≥ ω + 1

^then

thereexists

c ∈ R

suhthat the orresponding funtions

ϕ ˜ ε,i

^{satisfy :}

(22)

I g ( ˜ ϕ ε,i ) < 1

4 n(n − 2)ω _n ^2/n

Remarks4.1. (1) Weprovedinequalityof thislemmafor any

ω ≤ (n − 6)/2

^,

using test funtion

ϕ ε

^{(see Theorem 3.1 ). We notie that the dierene}

between

ϕ ε

^and

ϕ ˜ ε,i

^{isontheonstrutionofthe}orrespondingfuntions

f

and

f i

respetively. From

ϕ ˜ ε,i

^{wedene a}

G −

^{invariantfuntion (seeproof}

ofthe maintheorem below), this property is notpossible with the funtion

ϕ ε

^.

(2) For

ω = 16

^and

n

^{suientlybig,weanhekthatforany}

c ∈ R

,inequality (22)isfalse.

Proof. 1. If

deg ¯ R ≥ ω + 1

^{,thenbyTheorem2.1}

I g (u ε,P i ) < n(n − 2)

4 ω _n ^2/n

It issuientto take

c = 0

^,hene

ϕ ˜ ε,i = u ε,P i

^.

2. If

deg ¯ R = ω

^{. UsingestimatesgivenintheproofofTheorem3.1(see(18),(19)),}

itissuienttoshowthat thereexists

c ∈ R

suh that (23)

I S (f 1 ) + (n − 2) ² Z ¯

S(r)

r ^−2ω−2 R g dσ r < 0

WekeepthenotationsusedintheproofofTheorem3.1. Thus

I S (f 1 ) =

q

X

k=1

I S (cν k ϕ k ) =

d k c ² − 2(n − 2) ² c ν _k ² Z ¯

S n− 1

ϕ ² _k dσ

and

¯ Z

S(r)

r ^−2ω−2 R g dσ r =

q

X

k=1

u k

¯ Z

S n−1

ϕ ² _k dσ

Toproveinequality(23) ,itissuienttoprovethat

(24)

∀ k ≤ q d k

2(n − 2) c ² − (n − 2)c + (n − 2) u k

2ν _k ² < 0

Theleftsideoftheinequalityaboveisaseonddegreepolynomialwithvariable

c

^,

his disriminantis:

(25)

∆ k = (n − 2) ² − d k u k

ν _k ²

Using Lemma 3.1, wededuethat forany

k ≤ q

^,

∆ k > 0

^{. Hene,thepolynomial}

aboveadmitstwodierentrootsdenoted

x k < y k

^andgivenby

x k = (n − 2) ² − (n − 2) √

∆ k

d k

, y k = (n − 2) ² + (n − 2) √

∆ k

d k

Inequality(24)holdsifandonlyif

(26)

q

\

k=1

(x k , y k ) 6 = ∅

Thesequene

(d k ) k≤[ω/2]

^{dereases. Itiseasytohekthat}

(27)

∀ k < j ≤ [ ω

2 ] x k < y j

Hene intersetion(26)isnotemptyif

(28)

∀ k < j ≤ [ ω

2 ] x j < y k

Wealsohekthat if

ω

^iseven,

u ω/2 < 0

^,whihimplies

x ω/2 < 0

^.

i.

^If

ω = 3

^then

q = 1

^,intersetion aboveisnotempty. Itissuienttotake

c = (x 1 + y 2 )/2

^.

ii.

^If

ω = 4

^then

k ∈ { 1, 2 }

^,

x 2 < 0

^(beause

u 2 < 0

^)and

0 < x 1 < y 2

^{. Hene}

intersetion

]x 1 , y 1 [ ∩ ]x 2 , y 2 [

^isnotempty.

iii.

^If

5 ≤ ω ≤ 15

^{,itissuienttoprove(28)whihisequivalenttoprovethat}

(29)

∀ k < j ≤ [ ω

2 ] (n − 2)(d j − d k ) + d k p

∆ j + d j

p ∆ k > 0

Notiethat

∆ k

^{givenby(25)isarationalfrationin}

n

^{. By}straightforward omputations,wehekthat there existsreel numbers

a k , b k , e k , h k

^and

s k

^{whih dependon}

k

^and

ω

^suhthat

∆ k = a k n ² + b k n + e k + h k

n − 2 + s k

ν k + 1 − n

(30)

p ∆ k > √ a k (n + b k

2a k

)

(31)

Inequality(29)holdsifweuse(31).

Theexpressions of thereel numbers aboveare known expliitly (weused

thesoftwareMaple to ompute them, see [8℄). Forsimpliity, we omitto

givetheseexpressions.

Proof of themain theorem. Theorbitof

P

^{under theationof}

G

^issupposed

to beminimal(i.e.

cardO G (P) = inf Q∈M cardO G (Q)

^{). Without lossof} generality, wesupposethat

3 ≤ ω ≤ (n − 6)/2

^,beauseif

ω > (n − 6)/2

^or

ω ≤ 2

^,weonlude

using Theorem1.1. Fromfuntions

ϕ ˜ ε,i

^{dened by(20),wedenethefuntion}

φ ε

asfollows:

φ ε =

m

X

k=1

˜ ϕ ε,i

φ ε

^is

G −

^{invariant. Infat,forany}

σ ∈ G

^,suhthat

σ(P i ) = P j

u ε,P i = u ε,P j ◦ σ

^and

f i = f j ◦ σ f i

^{aredenedby(21),wededuethat}

˜

ϕ ε,i = ˜ ϕ ε,j ◦ σ

The support of

ϕ ˜ ε,i

^{is inludedin the ball}

B P i (δ)

^{. Wehoose}

δ

^{suientlysmall}

suhthat forallintegers

i 6 = j

ⁱⁿ

[1, m]

^,intersetion

B P j (δ) ∩ B P i (δ) = ∅

. Thus

I g (φ ε ) = (cardO G (P)) ^2/n I g (ϕ ε )

ByLemma4.1,weonludethat

I g (φ ε ) < n(n − 2)

4 ω ^2/n _n−1 (cardO G (P )) ^2/n

It remainsto notiethatif

˜ g = φ ^4/(n−2) ε g

^then

J(˜ g) = 4 n − 1

n − 2 I g (φ ε ) < n(n − 1)ω _n−1 ^2/n (cardO G (P )) ^2/n

where

ε

^{issuientlysmallerthan}

δ

^.

Proof of theCorollary 1.1. Supposethat theorbitof

P

^{undertheationof}

G

isminimal (otherwisetheonjetureis obvious).

If

ω = ω(P) > [(n − 6)/2]

^{,weonludeusingTheorem 1.1.}

If

ω ≤ [(n − 6)/2] ≤ 15

^{,weonludeusingmain theorem.}

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C ⁰

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241286.

6. J.Lelong-Ferrand,Mém.Aad.RoyaleBelgique,ClassedesSienes39(1971).

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Institut de Mathématiques de Jussieu, Université Pierre et Marie Curie, Équipe:

d'AnalyseComplexeetGéométrie,175,rue Chevaleret,75013Paris,Frane.

E-mail address: madanimath.jussieu.fr