Hessian equations on closed Hermitian manifolds

0.16 MB·English

by  Dekai Zhang

HESSIAN EQUATIONS ON CLOSED HERMITIAN MANIFOLDS
DEKAIZHANG
5
1
0
2 Abstract. Inthispaper,usingthetechnicaltoolsin[14],wesolvethecomplexHessian
equationonclosedHermitianmanifolds,whichgeneralizesthetheKa¨hlercaseresultsin
n
a [4]and[3].
J

5 1. Introduction
1

Let(M,g)beacompactHermitianmanifoldofcomplexdimensionn 2,andωbethe
]
≥
G correspondingHermitianform. In local coordinates,wewriteωas
D n
ω = √ 1 g dzi dzj.
h. − i,j=1 ij ∧
P
t
a In thispaper, weconsiderthefollowingHessianequationon closedHermitianmanifolds
m
Ckωk ωn k = efωn, supu = 0
1 [ (1.1)  nωuu=∧ω+−√−1∂∂¯u ∈ ΓMk(M),
v
3 whereΓk(M)isaconvexconedefined in (2.2)in section2.
5 Whenk = n,theconditionω Γ (M)isequivalenttoω > 0. Equation(1.1)becomes
u k u
5 ∈
thefollowingMonge-Ampereequation
3
0 (1.2) ωn = efωn, supu = 0.
. u
1 M
0 In addition,when (M,ω)is aKa¨hler manifold,i.e., dω = 0, Yau [16]solvedtheequation
5
(1.2) now known as Calabi-Yau theorem. For general Hermitian manifolds, the equation
1
: (1.1) has been solved by Cherrier [1] in the case of dimensions 2 and Tosatti-Weinkove
v
i [11] for arbitrary dimension. For further background, we refer the reader to [10], [11],
X
[5], [17]and thereferences therein.
r
a When 2 k n 1, ω may not be positive,the analysis becomes more complicated.
u
≤ ≤ −
Suppose that (M,ω) is a Ka¨hle manifold and ω Γ (M) which is defined in section 2 ,
u k
∈
Hou-Ma-Wu [4]provedthefollowingsecondorderestimatesoftheequation(1.1)
(1.3) max ∂∂¯u C(1+max u2).
| |g ≤ |∇ |g
Theyalsopointedoutintheirpaperthat(1.3)maybeadaptedtotheblowingupanalysis.
Later on, Dinew-Kolodziej [3] obtained the gradient estimate by (1.3). Thus equation
(1.1)can besolvedonKa¨hlermanifoldsunderthecompatiblecondition
efωn = ωn.
M M
R 1 R
2 DEKAIZHANG
Tosatti-Weinkove[13]consideredanotherHessiantypedequationrelatedtotheGaudu-
chon conjecture
(1.4) det ω n 1 + √ 1∂∂¯u ωn 2 = eFdet ωn 1
0 − − −
− ∧
(cid:16) (cid:17) (cid:16) (cid:17)
ω n 1 + √ 1∂∂¯u ωn 2 > 0,supu = 0,
0 − −
− ∧
M
whereω andω are twoHermitianmetricson M.
0
In[13],Tosatti-Weinkovesolvedequation(1.4)ifωisKa¨hler. Oneofthemainpartsis
doing thesecond order estimate. They use the similarauxiliary function in [4]. Later on,
in [14], they can solve (1.4) if ω is Hermitian. The second order estimate becomes more
difficultintheHermitiancase,theauthorssucceededtoobtainthesecondorderestimates
by modifyingtheauxiliaryfunctionin [4].
In this paper, we solve equation (1.1) on closed Hermitian manifolds. More precisely,
ourmainresultis
Theorem 1.1. Let (M,g) be a closed Hermitian manifold of complex dimension n 2,
≥
f is a smooth real function on M. Then there is a unique real number b and a unique
smoothreal functionu on M solving
(1.5) Ckωk ωn k = ef+bωn
n u ∧ −
ω Γ (M), supu = 0.
u k
∈
M
We use the continuity method to solve the problem (1.5). The openness follows from
implicitfunction theory. The closeness argument can be reduced to a prioriestimates up
tothesecondbythestandardEvans-Krylovtheory. Actually,wecanderivethezeroorder
estimateand thesecond orderestimateofsolutionsofequation (1.1) and thus usea blow
up methodto obtainthegradient estimate.
In [11], Tosatti-Weinkove derived the key zero order estimate by proving a Cherrier-
type inequality which was originally proved in [1]. For equation (1.1), we can prove the
similarCherrier-typeinequalitybuttheanalysisbecomesabit complicatedsinceω may
u
not be positive. Some inequalities for k th elementary symmetric functions in [2] are
−
needed. For the second order estimate, the main difficulty is that there are new terms of
the form T D3u, where T is the torsion of ω and D3u represents the third derivatives
∗
of u. To control these terms, we use the auxiliary function due to Tosatti-Weinkove in
[14]. Themaindifferenceisthatforequation(1.1)weneed tousesomelemmasfork th
−
elementary symmetricfunctionsprovedby Hou-Ma-Wu in[4].
Therestofthepaperisorganizedasfollows. Insection2,wegivesomepreliminaries.
In section 3, the Cherrier-type inequality is derived , thus we obtain the C0 estimate. In
section4, wewillprovethesecond orderestimatebya similarauxiliaryfunctionin [14].
Acknowledgment: I would like to thank Professor Xinan Ma for his encouragement,
adviceand comments. I also thankProfessor Shengli Kongforcareful reading and many
suggestions.
HESSIANEQUATION 3
2. preliminaries
Let (M,g)be a compact Hermitian manifold and let denotethe Chern connection of
∇
g. In this section we will give some preliminaries about the k th elementary symmetric
−
functionand thecommutationformulaofcovariantderivatives.
2.1. Elementary symmetric function. The k th elementary symmetric function is de-
−
fined by
σ (λ) = λ λ ,
k i1 ··· ik
X
1 i1< <ik n
≤ ··· ≤
where λ = (λ , ,λ ) Rn. Let λ a denotetheeigenvalues ofHermitian matrix a ,
1 n i¯j i¯j
··· ∈
wedefine (cid:16) (cid:17) n o
σ a = σ λ a .
k i¯j k i¯j
(cid:16) (cid:17) (cid:16) n o(cid:17)
Thedefinitionofσ canbenaturallyextendedtoHermitianmanifold. Indeed,letA1,1(M,R)
k
bethespaceofsmoothreal (1,1)-formson M, forχ A1,1(M,R)wedefine
∈
n χk ωn k
σ (χ) = ∧ − .
k k ωn
(cid:18) (cid:19)
Definition 2.1.
(2.1) Γ := λ Rn : σ (λ) > 0, j = 1, ,k .
k j
{ ∈ ··· }
Similarly,wedefineΓ on M asfollows
k
(2.2) Γ (M) := χ A1,1(M,Rn) : σ (χ) > 0, j = 1, ,k .
k j
{ ∈ ··· }
Furthermore, σ (λi ...i), with i ,...,i being distinct, stands for the r–th symmetric
r 1 l 1 l
|
functionwithλ = = λ = 0. Formoredetailsaboutelementarysymmetricfunctions,
i1 ··· il
onecan seethelecturenotes[15].
To prove the C0 estimate, we need the following lemma of elementary symmetric
functions.
Lemma 2.2. Suppose that λ Γ ,3 k n and λ λ λ , then there exists a
k 1 2 n
∈ ≤ ≤ ≥ ≥ ··· ≥
positiveconstantC dependingonlyon k andn, suchthatfor0 i k 2.
≤ ≤ −
(2.3) λ λ λ Cσ (λ j),
j1 j2 ··· ji ≤ i |
1 ≤ j1 < j2 < ··· ji ≤ n,(cid:12)(cid:12)(cid:12)jl , j,1 ≤ l (cid:12)(cid:12)(cid:12)≤ i,1 ≤ j ≤ n.
Proof. Since
n
λ =σ (λ 12 k 1) > 0,
p 1
| ··· −
Xp=k
and
λ λ λ
1 2 n
≥ ≥ ··· ≥
4 DEKAIZHANG
then
(2.4) λ (n k)λ ,k+1 p n.
p k
≤ − ≤ ≤
(cid:12) (cid:12)
We first prove the lemma fo(cid:12)r k(cid:12) = 3. In this case, it needs to prove that there exists a
(cid:12) (cid:12)
constantC suchthat
λ Cσ (λ j),
l 1
| | ≤ |
for 1 j,l n and l , j. Indeed, σ (λ j) = λ +σ (λ jl), thus λ σ (λ j). Now,we
1 l 1 l 1
≤ ≤ | | ≤ |
assumeλ < 0,then l 4. By (2.4), wehave
l
≥
λ (n 3)λ ,4 l n.
l 3
| | ≤ − ≤ ≤
Since λ j Γ , by the proof in [2] which used the result in [7], there exists a constant θ
2 1
| ∈
such that σ (λ j) θ λ if j = 1 and σ (λ j) θ λ if 2 j n. TakingC = n 3, we
1 | ≥ 1 2 1 | ≥ 1 1 ≤ ≤ θ−1
then provethelemmaforthecasek = 3.
Nextweprovethelemmaforthegeneral k,3 k n.
≤ ≤
If j > i, bytheresultin [15]
σ (λ j) θ(n,k)λ λ.
i 1 i
| ≥ ······
Thuswehave
λ λ λ =λ λ λ λ λ λ (n k)i qλi q
j1 j2 ··· ji j1 ··· jq jq+1··· ji ≤ 1··· q − − k−
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (n(cid:12)(cid:12) k)i
(cid:12) (cid:12) (n k)iλ(cid:12) λ (cid:12)− σ (λ j).
1 i i
≤ − ··· ≤ θ(n,k) |
If j i, thensimilarly
≤
σ (λ j) θ(n,k)λ λ λ λ .
i 1 j 1 j+1 i+1
| ≥ ··· − ···
Thuswehave
λ λ λ =λ λ λ λ λ λ λ λ (n k)i qλi q
j1 j2 ··· ji j1 ··· jq jq+1··· ji ≤ 1··· j−1 j+1··· q+1 − − k−
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (n k)i
(cid:12) (cid:12) (n k)iλ(cid:12) λ λ (cid:12) λ − σ (λ j).
1 j 1 j+1 i+1 i
≤ − ··· − ··· ≤ θ(n,k) |
(cid:3)
Usingthislemma,weimmediatelyobtainthefollowinglemmawhichisakeyingredi-
ent forprovinglemma3.2.
Lemma 2.3.
(2.5) k−2 √−1∂u∧∂¯u∧ωiu ∧Ti C k−2 √−1∂u∧∂¯u∧ωiu ∧ωn−i−1,
(cid:12)(cid:12) ωn (cid:12)(cid:12) ≤ ωn
Xi=0 (cid:12) (cid:12) Xi=0
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
HESSIANEQUATION 5
,where T isdefined asthecombinationsofω,∂ω,∂∂¯ω,moreprecisely
i
T = ωn i 3p 2q (√ 1)p(∂ω)p ∂¯ω p (√ 1)q ∂∂¯ω q
i −− −
∧ − ∧ ∧ −
0 3pX+2q n i (cid:16) (cid:17) (cid:16) (cid:17)
≤ ≤ −
Proof. For x M ,wechoosethecoordinatessuchthat
∈
n n
ω(x) = dzj dz¯j,ω (x) = λ dzj dz¯j,
u j
∧ ∧
Xj=1 Xj=1
and
λ λ λ .
1 2 n
≥ ≥ ··· ≥
Thuswehave
(2.6) k−2 √−1∂u∧∂¯u∧ωiu ∧Ti C k−2 u u λ λ λ
(cid:12)(cid:12) ωn (cid:12)(cid:12) ≤ j | l¯| j1 j2 ··· ji
Xi=0 (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) Xi=0 1≤j1<·X··<ji≤n,,j,l(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)
(cid:12) (cid:12) k 2 n
(cid:12) (cid:12) − 2
C u λ λ λ
≤ j j1 j2 ··· ji
Xi=0 Xj=1 1 ≤ j1 <X··· < ji ≤ n (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)
j , j
l
k 2 n
− 2
C σ (λ j) u
i j
≤ |
Xi=0 Xj=1 (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
=C k−2 √−1∂u∧∂¯u∧ωiu ∧ωn−i−1,
ωn
Xi=0
wherewehaveused thelemma2.1 inthelastinequality. (cid:3)
2.2. Commutationformulaofcovariantderivatives. Inlocalcomplexcoordinatesz , ,z ,
1 n
···
wehave
∂ ∂
(2.7) g = g( , ), gi¯j = g 1
i¯j ∂zi ∂z¯j { } { i¯j}−
FortheChern connection ,wedenotethecovariantderivativesas follows:
∇
(2.8) u = u,u = u,u = u
i ∂ i¯j ∂ ∂ i¯jk ∂ ∂ ∂
∇∂zi ∇∂z¯j∇∂zi ∇∂zk∇∂z¯j∇∂zi
we use the following commutation formula for covariant derivativeson Hermitian mani-
foldswhich can befoundedin [14]:
(2.9) u = u Tpu
i¯jl l¯ji − li p¯j
u = u +u R q
pi¯j p¯ji q i¯jp
u = u T¯q u .
ip¯¯j i¯jp¯ − jp iq¯
6 DEKAIZHANG
(2.10) u = u +u R p u R p Tpu T¯p u TpT¯q u
i¯jlm¯ lm¯i¯j p¯j lm¯i − pm¯ i¯jl − li pm¯¯j − mj lp¯i − li mj pq¯
Forthedetailswerecommendthereaderto thereference [14].
3. zero order estimate
In this section we derive the zero order estimate by proving a Cherrier-type inequality
and thelemmasin [11]. Sincetheconstantb is inTheorem1.1 satisfies
b sup f +C,
| | ≤ | |
whereC isapositiveconstantdependingonlyon(M,ω). Thus,wewillassumeb = 0for
convenience.
Theorem 3.1. LetubeasolutionofTheorem1.1. ThenthereexistsaconstantC depend-
ingonlyon(M,ω)and sup f suchthat
| |
M
sup u C.
| | ≤
M
Due to Tosatti-Weinkove’s results, the zero order estimate can be reduced to derive a
Cherrier-typeinequalitywhichwasfirstlyprovedinCherrier’s paper[1]. FortheHessian
equation, the analysis becomes a bit complicated in the lack of the positivity of ω . Re-
u
cently1, Sun [8] also proved the following lemma for k = 2 and k 3 under some extra
≥
conditions.
Lemma 3.2. There exist constants p and C depending only on (M,ω) such that for any
0
p p
0
≥
Z |∂e−2pu|2gωn ≤ CpZ e−puωn
M M
Proof. By theequation,wehave
ωk ωn k ωn = (ef 1)ωn C ωn,
u ∧ − − − ≤ 0
whereC isa constantdependingonlyon f.
0
Ontheotherhand,
(3.1) ωk ωn k ωn = ωk ωk ωn k = √ 1∂∂¯u α,
u ∧ − − u − ∧ − − ∧
(cid:16) (cid:17)
k
whereα = ωi 1 ωn i.
u− ∧ −
i=1
P
1TheauthorindependentlyprovedtheC0estimatebefore[8]waspostedonarXiv.
HESSIANEQUATION 7
Nowmultiplyboth sidesin(3.1)bye pu and integrateby parts,
−
(3.2) C e puωn e pu√ 1∂∂¯u α
0 − −
Z ≥Z − ∧
M M
= ∂e pu√ 1∂¯u α+ e pu√ 1∂¯u ∂α
− −
−Z − ∧ Z − ∧
M M
1
=p e pu√ 1∂u ∂¯u α √ 1∂¯e pu ∂α
− −
Z − ∧ ∧ − p Z − ∧
M M
1
=p e pu√ 1∂u ∂¯u α+ e pu√ 1∂¯∂α
− −
Z − ∧ ∧ p Z −
M M
:=A+B,
wherewedenote
k
A =p e pu√ 1∂u ∂¯u ωi 1 ωn i
Z − − ∧ ∧ u− ∧ −
1 M Xi=1 
B = e pu√ 1∂¯∂α.
−
p Z −
M
Wewillusetheterm A tocontrol theterms B. Direct calculationgives
k 1
−
∂α = n ωi 1 ωn i 1 ∂ω+(n k)ωk 1 ωn k 1 ∂ω
u− ∧ −− ∧ − u− ∧ − − ∧
Xi=1
∂¯∂α =(n k)(n k 1)ωk 1 ωn k 2 ∂¯ω ∂ω+(n k)ωk 1 ωn k 1 ∂¯∂ω
− − − u− ∧ − − ∧ ∧ − u− ∧ − − ∧
+(n k)(n+k 1)ωk 2 ωn k 1 ∂¯ω ∂ω
− − u− ∧ − − ∧ ∧
k 3 k 2
+n(n 1) − ωi ωn i 3 ∂¯ω ∂ω+n − ωi ωn i 2 ∂¯∂ω
− u ∧ −− ∧ ∧ u ∧ −− ∧
Xi=0 Xi=1
Therefore, wehave
(n k)(n k 1)
B = − − − √ 1e puωk 1 ωn k 2 ∂¯ω ∂ω
p Z − − u− ∧ − − ∧ ∧
M
(n k)
+ − √ 1e puωk 1 ωn k 1 ∂¯∂ω
p Z − − u− ∧ − − ∧
M
(n+k 1)(n k)
+ − − √ 1e puωk 2 ωn k 1 ∂¯ω ∂ω
p Z − − u− ∧ − − ∧ ∧
M
k 3 k 2
+ n(n−1) − √ 1e puωi ωn i 3 ∂¯ω ∂ω+ n − √ 1e puωi ωn i 2 ∂¯∂ω
p Z − − u ∧ −− ∧ ∧ p Z − − u ∧ −− ∧
Xi=0 M Xi=1 M
8 DEKAIZHANG
When k = 2,theterm B justbecomes
(3.3)
(n 2)(n 3) (n 2)
B = − − √ 1e puω ωn 4 ∂¯ω ∂ω+ − √ 1e puω ωn 3 ∂¯∂ω
− u − − u −
p Z − ∧ ∧ ∧ p Z − ∧ ∧
M M
(n+1)(n 2)
+ − √ 1e puωn 3 ∂¯ω ∂ω
− −
p Z − ∧ ∧
M
(n 2)(n 3) (n 2)
= − − √ 1e pu√ 1∂∂¯u ωn 4 ∂¯ω ∂ω+ − √ 1e pu√ 1∂∂¯u ωn 3 ∂¯∂ω
− − − −
p Z − − ∧ ∧ ∧ p Z − − ∧ ∧
M M
2(n 1)(n 2) (n 2)
+ − − √ 1e puωn 3 ∂¯ω ∂ω+ − √ 1e puωn 2 ∂¯∂ω
− − − −
p Z − ∧ ∧ p Z − ∧
M M
(n 2)(n 3)
− − √ 1e pu√ 1∂∂¯u ωn 4 ∂¯ω ∂ω
− −
≥ p Z − − ∧ ∧ ∧
M
(n 2) C
+ − √ 1e pu√ 1∂∂¯u ωn 3 ∂¯∂ω 1 e puωn
− − −
p Z − − ∧ ∧ − p Z
M M
We next use integration by parts again to deal with the first term and second term on the
righthand sideoftheaboveequality. Indeed,
√ 1e pu√ 1∂∂¯u ωn 4 ∂¯ω ∂ω
− −
Z − − ∧ ∧ ∧
M
(3.4)
=p √ 1e pu√ 1∂u ∂¯u ωn 4 ∂¯ω ∂ω+ √ 1e pu√ 1∂¯u ∂(ωn 4 ∂¯ω ∂ω)
− − − −
Z − − ∧ ∧ ∧ ∧ Z − − ∧ ∧ ∧
M M
1
=p √ 1e pu√ 1∂u ∂¯u ωn 4 ∂¯ω ∂ω+ √ 1e pu√ 1∂¯∂(ωn 4 ∂¯ω ∂ω)
− − − −
Z − − ∧ ∧ ∧ ∧ p Z − − ∧ ∧
M M
C
pC e pu√ 1∂u ∂¯u ωn 1 1 e puωn
1 − − −
≥− Z − ∧ ∧ − p Z
M M
C
C A 1 e puωn
1 −
≥− − p Z
M
Thesimilarcalculationgives
C
(3.5) √ 1e pu√ 1∂∂¯u ωn 3 ∂¯∂ω C A 1 e puωn
− − 1 −
Z − − ∧ ∧ ≥ − − p Z
M M
Inserting(3.4)and (3.5) into(3.3), wehave
C C
B 1A 1 e puωn
−
≥ − p − p Z
M
HESSIANEQUATION 9
By (3.2)andchoosing p = 2C +1,weobtainfor p p
0 1 0
≥
A C C
(1 1)A ( 1 +C ) e puωn (C +1) e puωn
0 − 0 −
2 ≤ − p ≤ p Z ≤ Z
M M
By (3.7)inthenextpage, wethusprovethelemma.
Forthegeneralk,3 k n,weclaimthatthereexistconstantsC dependingonlyon
1i
≤ ≤
n,k,(M,ω)suchthat thefollowingholdsfor0 i k 1,
≤ ≤ −
k 2
(3.6) e puωi T pC − e pu√ 1∂u ∂¯u ωj ωn j 1 C e puωn
Z − u ∧ i ≥ − 1i Z − − ∧ ∧ u ∧ − − − 1iZ −
M Xj=0 M M
,whereT is defined as thecombinationsofω,∂ω,∂∂¯ω,moreprecisely
i
T = ωn i 3p 2q (√ 1)p(∂ω)p ∂¯ω p (√ 1)q ∂∂¯ω q
i −− −
∧ − ∧ ∧ −
0 3pX+2q n i (cid:16) (cid:17) (cid:16) (cid:17)
≤ ≤ −
Weusetheclaim(3.6)toprovethelemma
k
C
B C e pu√ 1∂u ∂¯u ωk i ωn+i k 1 1 e puωn
≥− 1 Z − − ∧ ∧ u− ∧ − − − p Z −
Xi=2 M M
C C
1A 1 e puωn
−
≥− p − p Z
M
Thuswehave
C C
(1 1)A ( 1 +C ) e puωn
0 −
− p ≤ p Z
M
Nowwechoose p = 2C +1, thenforany p p ,
0 1 0
≥
p2 e pu√ 1∂u ∂¯u ωn 1 2p(C +1) e puωn
− − 0 −
Z − ∧ ∧ ≤ Z
M M
Thereforewehave
np2
(3.7) Z |∂e−2pu|2gωn = 4 Z e−pu√−1∂u∧∂¯u∧ωn−1
M M
np(C +1)
0 e puωn = pC e puωn
− −
≤ 2 Z Z
M M
Now,weprovetheclaim(3.6)byinductiveargument.
10 DEKAIZHANG
Wheni = 1,we have
e puω T = e puω T + e pu√ 1∂∂¯u T
− u 1 − 1 − 1
Z ∧ Z ∧ Z − ∧
M M M
= e puω T ∂e pu √ 1∂¯u T + e pu√ 1∂¯u ∂T
− 1 − 1 − 1
Z ∧ −Z ∧ − ∧ Z − ∧
M M M
1
= e puω T + p e pu√ 1∂u ∂¯u T √ 1∂¯e pu ∂T
− 1 − 1 − 1
Z ∧ Z − ∧ ∧ − p Z − ∧
M M M
1
=p e pu√ 1∂u ∂¯u T + e puω T e pu √ 1∂∂¯T
− 1 − 1 − 1
Z − ∧ ∧ Z ∧ − p Z ∧ −
M M M
C p e pu√ 1∂u ∂¯u T C e puωn
1 − 1 1 −
≥ − Z − ∧ ∧ − Z
M M
Supposethat theclaimis true forl i 1, we willprovethatthe claimis also truefor
≤ −
l = i. Indeed,
e puωi T = e puωi 1 ω T + e pu√ 1∂∂¯u ωi 1 T
Z − u ∧ i Z − u− ∧ ∧ i Z − − ∧ u− ∧ i
M M M
= e puωi 1 ω T + p e pu√ 1∂u ∂¯u ωi 1 T
Z − u− ∧ ∧ i Z − − ∧ ∧ u− ∧ i
M M
+ e pu∂¯u √ 1∂ ωi 1 T
Z − ∧ − u− ∧ i
M (cid:16) (cid:17)
:=A +A +A
i,1 i,2 i,3
By theinduction,
A = e puωi 1 ω T
i,1 Z − u− ∧ ∧ i
M
k 2
pC (n,k,ω) − e pu√ 1∂u ∂¯u ωj ωn j 1 C (n,k,ω) e puωn
≥− 1i Z − − ∧ ∧ u ∧ − − − 1i Z −
Xj=0 M M
By theinequality(2.5)inlemma2.3, wehave
(3.8)
A = p e pu√ 1∂u ∂¯u ωi 1 T pC e pu√ 1∂u ∂¯u ωi 1 ωn i
i,2 Z − − ∧ ∧ u− ∧ i ≥ − 2iZ − − ∧ ∧ u− ∧ −
M M