NOTES ON NEWLANDER-NIRENBERG THEOREM XU WANG

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1 NOTES ON NEWLANDER-NIRENBERG THEOREM XU WANG Abstract. In this short note we shall recall the classical Newler-Nirenberg theorem its vector bundle version. We shall also recall an L 2 -Hörmer-proof given by D ly. 1. Classical Newler-Nirenberg theorem Let us recall the definition of the complex number field C. Let us consider (1) 1 Z Q R R 2. Let us fix an inner product structure on R 2. Then we can talk about the rotation transform on R 2. Let e x, e y be an orthonormal base of R 2. Then a counterclockwise rotation of 90- degree on R 2 can be written as (2) J : ae x + be y be x + ae y. Let us write it as J(a, b) = ( b, a). Then we have (3) J 2 (a, b) = (a, b). So sometimes, we shall write J = 1 := i. By using the R-vector space structure on R, for every (x, y) R 2, we can define the following R-linear transform from R 2 to itself: (4) x + yi : (a, b) x(a, b) + y( b, a) = (xa yb, xb + ya). Let us denote the following set (5) {x + yi : (x, y) R 2 }, by C. Now let us define an addition operation, say, on C as follows: (6) x 1 + y 1 i x 2 + y 2 i := (x 1 + x 2 ) + (y 1 + y 2 )i. Then we know that (C, ) is an abelian group. Moreover, by using composition of mappings, one may define a product operation, say, on C. One may verify that (7) (x 1 + y 1 i) (x 2 + y 2 i) = (x 1 x 2 y 1 y 2 ) + (x 1 y 2 + y 1 x 2 )i. Then one may check that (C,, ) defines a number field. We call it complex number field. The fundamental theorem of algebra said that C can be seen as the algebraic closure of the real number field. Let 0 be the origin of R 2. Recall that the tangent space of R 2 at 0 is the space of equivalent classes of smooth curves passing through the origin. Let us denote it by T 0 R 2. In order to define the so called complex structure on T 0 R 2, let us fix a base, say / x, / y, of T 0 R 2. Then we call the R-linear isomorphism from T 0 R 2 to itself that Date: February 1,

2 2 XU WANG maps / x to / y maps / y to / x a complex structure on T 0 R 2. Let us denote it by J. Notice that, if we introduce an inner product on T 0 R 2 as follows: (8) dx dx + dy dy : (a / x + b / y, c / x + d / y) ac + bd. Then J is just the counterclockwise rotation of 90-degree on T 0 R 2. Since the eigenvalues of J are not real, we know that J has no real eigenvectors. Thus we have to complexify T 0 R 2 in order to find the complex eigenvectors of J. More precisely, we need to find eigenvectors of J in T 0 R 2 R C. It is easy to check that (9) J( / x ij / x) = i( / x ij / x), (10) J( / x + ij / x) = i( / x + ij / x), Let us denote by T 0 C the eigenspace of J spanned by / x ij / x, denote by T 0 C the eigenspace of J spanned by / x + ij / x. Now we have the following eigenspace decomposition (11) T 0 R 2 R C = T 0 C T 0 C. And according to this decomposition, we have (12) / x = / z + / z, where (13) / z := 1 2 ( / x ij / x) = 1 ( / x i / y), 2 (14) / z := 1 2 ( / x + ij / x) = 1 ( / x + i / y). 2 One may also define the notion of complex structure on the cotangent space T 0 R2. The complex structure on T 0 R2 that induced from the complex structure, J, on T 0 R 2 is defined as follows (15) (dx, dy) ( dy, dx). We shall also denote it by J. One may check that (16) Ju, V = u, JV, u T 0 R 2, V T 0 R 2, moreover (17) dz, / z = 1; dz, / z = 0. Now let (18) T 0 R 2 R C = T 0 C T 0 C. be the J-eigenspace decomposition of the complexified cotangent space, where T0 C T0 C are the eigenspace with eigenvalue i i respectively. Now let M be a smooth manifold. Let us recall the following definition: Definition 1. We call a smooth bundle mapping J from T M to itself an almost complex structure on M if J 2 = 1

3 Now similar as before, we have the following decompositon (19) k (T x M R C) = p+q=k ( p T x M C q T x M C ), x M, 1 k dim R M. Thus every complex valued k-form can be written as the sum of (p, q)-forms. Now let u be a smooth (p, q)-form on M, locally one may write (not unique) (20) u = u j 1 u jp ū k 1 ū kq, where each u jt is a section of T M C each ū ks is a section of T M C. By the Leibniz rule, du can be written as (21) du j 1 u jp ū k 1 ū kq + + ( 1) p+q 1 u j 1 u jp ū k 1 dū kq. Thus we have (22) du = du p+1,q + du p,q+1 + du p 1,q+2 + du p+2,q 1, where each du t,s denotes the (t, s)-part of du. Definition 2. u := du p,q+1, u := du p+1,q. We call J is integrable if d = +. Proposition 1. J is integrable if only if 2 = 0. Proof. It suffices to show that 2 = 0 implies that J is integrable. By definition, we have =. Thus 2 = 0 is equivalent to 2 = 0. By (21), it suffice to show that if u is an (1, 0) or (0, 1) form, then du = u + u. For example, if u is an (1, 0)-form, then one may write u = a j b j. Thus du = da j b j + a j d b j. Now it suffices to show that each d b j has no (0, 2)-part. Since each b j is a function, we have db j = b j + b j. Thus d b j = d(d )b j = d b j, the (0, 2)-part of d b j is just 2 b j, which vanishes by our assumption. Same argument works for (0, 1)-form. We shall show Hörmer s L 2 -proof the following theorem of Newler-Nirenberg: Theorem 1 (Newler-Nirenberg). If J is integrable then for every x M there exists n smooth complex valued functions, say z 1,, z n, near x such that z j = 0 near x for each j { z j (x)} 1 j n defines a base of T x M C. Let us show how to use this theorem first. Since { z j (x)} 1 j n defines a base of T x M C, we know that { z j (x)} 1 j n defines a base of T x M C. Since z j = 0 near x, we know that z j = 0 near x also. Thus {dimz j, drez j } 1 j n defines a R-base of T x M. Now we can use z := {Rez j, Imz j } 1 j n to define a coordinate covering of M. Let z : U R 2n w : V R 2n be two such coordinate charts. Assume that U V is a non-empty open subset of M. Then w z 1 is a smooth mapping from z(u V ) to w(u V ). If we look at z w as complex valued functions, then both z(u V ) w(u V ) can be seen as open subsets of C n. We shall show that w z 1 is holomorphic. Definition 3. Let U be an open subset in C n. Let f be a smooth mapping from U to C m. f is said to be holomorphic on U if (23) f k / z j 0, on U, 1 k m, 1 j n. For the definition of / z j see (14). 3

4 4 XU WANG Now let us prove f := w z 1 is holomorphic on z(u V ) in the sense of the above definition. Since w k = f k (z) {Rez j, Imz j } 1 j n defines a smooth local coordinate chart on M, we have (24) dw k = f k / Rez j drez j + f k / Imz j dimz j. Since w k = 0, we have (25) dw k = f k / Rez j drez j + f k / Imz j dimz j. Since w k = 0, thus by definition of w k above, we have (26) 0 = f k / Rez j Rez j + f k / Imz j Imz j. Since z j = 0 (27) Rez j = z j + z j 2 we have, Imz j = z j z j, 2i (28) Rez j = 1 2 z j; Imz j = 1 2i z j. Thus we have (29) 0 = ( f k / Rez j 1 ) f k / Imz j z j. i Since { z j } 1 j n are linearly independent, we have (30) f k / z j = 1 ( f k / Rez j + i ) f k / Imz j 0, 2 on z(u V ). Thus f = w z 1 is holomorphic on z(u V ). By definition, we know that z : U C n defines an complex structure on M. Thus the Newler-Nirenberg theorem told us that a smooth manifold with an integrable almost complex structure is in fact a complex manifold. Now let us recall the Hörmer proof [4] of the Newler-Nirenberg theorem. The basic idea of Hörmer is: by using the precise L 2 -estimate, the existence of good J- plurisubharmonic functions implies ampleness of J-holomorphic sections. Moreover, if J is integrable then locally there exist good J-plurisubharmonic functions. Step1: Construct good J-plurisubharmonic functions (see [3]). Let us fix a point x in M. Let us take 2n smooth functions, say x 1,, x 2n, near x such that x 1 (x) = = x 2n (x) = 0, {dx 1 (x),, dx 2n (x)} defines a base of T x M. Thus there exists a small open neighborhood, say U x, of x such that (x 1,, x 2n ) : U x R 2n, defines a smooth coordinate chart on U x. Now let us fix a base, say {σ 1,, σ n }, of T x M C. By definition, we have (31) σ j = a k j dx k (x), a k j C, 1 j n, 1 k 2n.

5 5 Since each σ j has no (0, 1)-part, we know that (32) a k j x k (x) = 0. (33) f j = a k j x k, 1 j n, then f j are smooth complex valued functions on U x (34) f j (x) = 0. By choosing sufficiently small U x, one may assume that {Ref j, Imf j } 1 j n defines a coordinate chart on U x. (35) ψ = f 2 := f j fj, then ψ is a smooth function on U x. Thus choose δ > 0 sufficienely small, we know that (36) := {ψ < δ} is a relatively compact open subset of U x the gradient of ψ has no zero point on the boundary, {ψ = 1}, of. (37) ω = i ψ, ˆω = i ( log(δ ψ)). If J is integrable then both ω ˆω are real d-closed (1, 1)-forms on. Since (38) f j (x) = f j (x) = 0, we have (39) i ψ(x) = i f j (x) f j (x) = i σ j σ j > 0. Thus if δ is small enough then i ψ > 0 on. We call ω a J-Kähler form on. In order to constuct a singular J-plurisubharmonic function, we have to use the following lemma of D ly [3]: Lemma 1 (D ly). If J is integrable then there exist n smooth complex valued functions, say g 1,, g n, on a neighborhood of x such that for every 1 j n. g j (x) = 0, g j (x) = σ j, g j = O(ψ), Proof. Since { f k } 1 k n defines a base of (0, 1)-form on U x, we have (40) f j = p kl j f l f k + qj kl f l f k, where p kl j, qkl j are smooth function on U x. Since J is integrable, we know that (41) 0 = 2 f j (x) = q kl j (x) f l (x) f k (x), which implies that (42) q kl j (x) = q lk j (x). Let us consider (43) g j = f j + a kl fk f l + b kl fk fl, where a kl, b kl are complex constants b kl = b lk. Thus (44) g j = f j + a kl f l f k + a kl fk f l + 2 b kl fl f k.

6 6 XU WANG It suffices to choose a kl, b kl with b kl = b lk such that (45) 2b kl + q kl j (x) = 0, the existence of a kl, b kl follows from (42). a kl + p kl j (x) = 0. By D ly s lemma, one may assume that (consider g j instead of f j ) (46) f j = O(ψ) Then we have (47) ψ = fj f j + O(ψ 3/2 ), thus (48) i ψ ψ ψ i f j f j + O(ψ 2 ). Moreover, f j = O(ψ) implies that (49) i f j = O(ψ 1/2 ), thus (50) i ψ = i f j f j + O(ψ). Hence (51) i ψ ψ ψi ψ + O(ψ 2 ). Thus (choose a smaller δ if necessary) there exists a sufficiently big positive number N such that (52) i ψ ψ (ψ Nψ 2 )i ψ, on, which implies that (53) i log(ψ + ε) Ni ψ, for every ε > 0. Step2: -Bochner formula for integrable J. Let γ be an arbitrary smooth (n q, 0)-form on let φ be a real valued smooth function on. ω q = ω q /q!, u = γ ω q, T = i (n q)2 γ γ e φ ω q 1. By definition, we have = φ, where φ := φ. Since J is integrable, we have 2 φ = 2 = 0. Thus i T can be written as (54) 2Re u, u e φ ω n + u 2 e φ ω n + i φ T S, where S = i (n q+1)2 γ γ e φ ω q 1.

7 The following lemma follows from the primitive decomposition of γ (see Lemma 4.2 in Berndtsson s lecture notes [1]). Lemma 2. S = ( u 2 φ u 2 )e φ ω n. Thus we have the following -Bochner formula for (n, q)-forms: ( (55) i T = 2Re u, u + u 2 u 2 + φ u 2) e φ ω n + i φ T. 7 Step3: Hörmer s L 2 -estimate. In order to construct -closed functions that we need, we have to solve the -equation for (0, 1)-form. But we only have -Bochner formula for (n, q)-forms, thus it is necessary to solve -equation for (n, 1)-form construst a good -closed holomorphic (n, 0)-form, then we can use it to solve the -equation for (0, 1)-form. We shall use Chen s method [2] to solve the -equation. More precisely, put (56) v = ( f 1 f n ), (57) ψ ε = nnψ + n log(ψ + ε) log(δ ψ), By (49), (37) (53), we have (58) i ψ ε ˆω, (59) v 2ˆω e ψε ˆω n (60) Q(α, β) := By (55), we have (61) Q(α, α) v 2ˆω e ψ0 ˆω n := I(v) <, ε > 0. α, β ˆω e ψε ˆω n + α, β ˆω e ψε ˆω n. α 2ˆω e ψε ˆω n, for every smooth (n, 1)-form α with compact support in. Denote by H the completion under Hermitian form Q of the space of smooth (n, 1)-forms with compact support in. Thus (62) α α, v ˆω e ψε ˆω n, extends to a Q-bounded linear functional on H. By the Riesz representation theorem (61), we know that there exists a H such that (63) Q(α, a) = α, v ˆω e ψε ˆω n, α H, (64) Q(a, a) Notice that (63) implies that (65) v 2ˆω e ψε ˆω n I(v). ( ) + a = v,

8 8 XU WANG in the sense of current. Since + is elliptic v is smooth, we know that there exists a smooth representative, say a, of the current a such that ( ) (66) + a = v, on D (in fact, by using the Fourier transform, we get a Gårding inequality for elliptic operator with constant coefficients, then one can get the Gårding inequality for general elliptic operator by comparing with the case of constant coefficients. In our case, by (64) convolution of a with a smooth function, say the Gaussian kernel, using the Arzelà- Ascoli theorem, we get that the current solution a locally has a smooth representative. By using the partition of unity, we finally get a global smooth representative, say a, of a) (67) Q(a, a) I(v). Since J is integrabel, (66) implies that (68) a 0 on. Let χ be a smooth function on R such that χ 1 on (, 1/2), χ 0 on (1, ) χ 3 on R. Then (69) χ t := χ(t log Moreover, 1 δ ψ ) C 0 (, R), t > 0. (70) (χ t ) ˆω 3t, on. Since (71) (χ 2 t a, a) = χ t a 2 (2χ t χ t a, a), by (68) (70), we have (72) χ t a 2 6t χ t a. Let t goes to zero, we know that (73) a 0, on. Thus we have (74) a = v, on. We know that u ε = v (75) u ε = a. u ε 2ˆω e ψε ˆω n Q(a, a) I(v). Let ε goes to zero, by taking the weak limit, we get an (n, 0)-current u such that u = v in the sense of current (76) u 2ˆω e ψ0 ˆω n I(v). Again since is elliptic on (n, 0)-forms, we know that the current u has a smooth representative u such that u = v (77) u 2ˆω e ψ0 ˆω n I(v).

9 Since e ψ 0 is not integrable near x, we know that (78) u(x) = 0. (79) σ = f 1 f n u. we know that (80) σ = 0, σ(x) 0. (81) 1 = {ψ < δ 1 }. Choose 0 < δ 1 < δ small enough, one may assume that σ has no zero point in the closure of 1. This time, put (82) v j = (f j σ), 1 j n, (83) φ ε = (n + 1)Nψ + (n + 1) log(ψ + ε) log(δ 1 ψ). Consider the complete J-Kähler form (84) ω := i ( log(δ 1 ψ)). Similar as before, for each j, we get a smooth (n, 0)-form u j on 1 such that (85) u j = v j, u j (x) = 0, du j (x) = 0, 1 j n. (86) u j = h j σ, 1 j n. Then (87) (f j h j ) = 0, h j (x) = 0, dh j (x) = 0, 1 j n. We know that (88) z j := f j h j, 1 j n. fit our needs. Thus the Newler-Nirenberg theorem is proved. 2. Vector bundle version of the Newler-Nirenberg theorem In this section, we shall prove the following theorem: Theorem 2. Let E be a complex smooth vector bundle, say complex rank r, over a complex manifold X. Let D be a smooth connection on E. Denote by D 0,1 the (0, 1)-part of D. If (D 0,1 ) 2 S 0 on X for every smooth section S of E over X then for every x X there exist r smooth sections, say S 1,, S r, near x such that D 0,1 S j = 0 near x for each j {S j (x)} 1 j r defines a base of E x. Idea of the proof: Fix a Hermitian metric, say h, of E define the almost Chern connection, say D h, on E with respect to D 0,1 h. Then use D h to prove that the -Bochner formula is true for smooth section of E. Finally, solve the D 0,1 -equation with singular weight to get the sections that we need. Step1: Construct the almost Chern connection. 9

10 10 XU WANG Let us fix a smooth Hermitian metric, say h, on E. They there is a natural sesquilinear product (see [3]), say {, }, on, T X E with respect to h. A connection D is said to be h-hermitian if (89) d{s, T } = {DS, T } + ( 1) deg S {S, DT }, for every E-valued differantial forms S, T. We shall prove that: Lemma 3. Let D be the connection in Theorem 2. Then there exists a unique h-hermitian connection, say D h, on E such that D h is h-hermitian D 0,1 h = D 0,1. Moreover, (D 0,1 ) 2 = 0 implies that (D 1,0 h )2 = 0. Proof. Let us define D 1,0 h by requiring (90) {S, T } = {D 1,0 h S, T } + ( 1)deg S {S, D 0,1 T }, for every smooth E-valued differantial forms S, T. D h = D 0,1 + D 1,0 h. Then one may verify that for every smooth differential form f on X, we have D h (f S) = df S + ( 1) deg f f D h S, (91) d{s, T } = {D h S, T } + ( 1) deg S {S, D h T }. Thus D h fits our needs. Now it suffices to show that (D 1,0 h )2 = 0. Since 2 = 0, by (92), if (D 0,1 ) 2 = 0 then (92) 0 = {S, T } = {(D 1,0 h )2 S, T }, for every smooth E-valued differantial forms S, T. Thus (D 1,0 h )2 = 0. Definition 4. The curvature of the almost Chern connection D h in the above lemma is defined to be Θ(E, h) := D 2 h. Definition 5. Let us write D 1,0 h = E D 0,1 =. Step2: -Bochner formula for smooth sections of E. Since a complex manifold is locall Kähler. Let ω be a Kähler on a pesudoconvex open neighborhood, say, of x. Let γ be an arbitrary E-valued smooth (n q, 0)-form on. (93) T = i (n q)2 {γ, γ} ω q 1, u = γ ω q. By tha above lemma Step2 in the last section, still we have ( (94) i T = 2Re u, u + u 2 u 2 + ( E ) u 2) ω n + iθ(e, h) T, Step3: Solve D 0,1 -equation with singular weight. Let us choose a smooth base, say {σ 1,, σ r } of E over a pesudoconvex open neighborhood, say 1, of x such that (95) D h σ j (x) = 0, 1 j r.

11 Assume that 1 is relatively compact in has global holomorphic coordinates, say z 1,, z n, such that z j (x) = 0 for each j. (96) τ j := D 0,1 ((dz 1 dz n ) σ j ), 1 j r. Similar as before, one may solve the D 0,1 -equation with singular weight whose singular part is n log z 2 get r smooth E-valued (n, 0)-form a j, 1 j r, on 1 such that (97) a j (x) = 0, D 0,1 a j = τ j, 1 j r. a j = (dz 1 dz n ) f j, then (98) S j := σ j f j, 1 j r. are sections that we need. Remark. By Theorem 2, each {S j } 1 j r will give a smooth local trivialization of E, since each S j is in the kernel of D 0,1, we know that transition maps between these local trivializations are in fact holomorphic. Thus E has a holomorphic vector bundle structure. References [1] B. Berndtsson, An introduction to things, Analytic algebraic geometry, IAS/Park City Math. Ser., vol. 17, Amer. Math. Soc., Providence, RI, 2010, [2] B. Y. Chen, J. J. Wu X. Wang, Ohsawa-Takegoshi type theorem extension of plurisubharmonic functions, Mathematische Annalen, 362 (2015), [3] J.-P. D ly, Complex analytic differential geometry. Book available from the author s homepage. [4] L. Hörmer, An introduction to complex analysis in several variables, Van Nostr, Princeton, Chalmers Maths Department, Gothenburg, Sweden address: wangxu1113@gmail.com 11