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