Function theory on Kaehler manifolds. Note by Man-Chun LEE. g = g ij dx i dx j.

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1 Funtion theory on Kaehler manifolds Note by an-chun LEE Kaehler manifolds Let n be a smooth manifold. A riemannian metri g is a smooth setion of T T suh that g is symmetri and positive definite at any p. In loal oordinate (x,..., x n ), g = g ij dx i dx j. In addition, if is a omplex manifold with almost omplex struture J T T and g(x, Y ) = g(jx, JY ), X, Y T. Then is alled Hermitian manifold, (, g, J). One an define a 2-form ω g where ω g (X, Y ) = g(x, JY ). Let be the Levi-onnetion of a Hermitian manifold (, g, J). Definition.. A Kahler manifold (, g, J) is a Hermitian manifold suh that J = 0. In partiular, (, J) is a omplex manifold. Proposition.. J = 0 if and only if dω = 0. sympleti manifold. Thus, a Kahler manifold is also a Proof. For X, Y, Z Γ(T ). By invariant formula, we have dω(x, Y, Z) = Xg(Y, JZ) + Y g(x, JZ) Zg(X, JY ) + g([x, Y ], JZ) g([x, Z], JY ) + g([y, Z], JX) We hoose X = e, Y = e 2, Z = e 3 to be normal oordinate vetor field at p in order to simplify our alulation. By J 2 = Id, we have at p, dω(x, Y, Z) = g(y, ( X J)Z) g(x, ( Z J)Y ) + g(x, ( Y J)Z) = g(z, ( X J)Y ) + g(y, ( Z J)X) + g(x, ( Y J)Z). The above equation is independent of hoie of oordinate, so it is valid for arbitrary X, Y, Z. At the same time, sine J J + J J = 0, g(( X J)Y, Z) = g( X (JY ), Z) g(j X Y, Z) = Xg(JY, Z) g(jy, X Z) g(j X Y, Z) = Xω(Z, Y ) + ω( X Z, Y ) + ω(z, X Y ) Beause the onnetion is torsion free, using invariant formula 2g(( X J)Y, Z) = dω(x, Y, Z) dω(x, JY, JZ).

2 . Curvature on Kahler manifold Given a Kahler manifold (, g, J), extend g C-linearly to T R C = T,0 T 0,. Proposition.2. If u, v are both in T,0 or T 0,, then g(u, v) = 0. Thus, loally we have ω g = g 2 i j dz i d z j. i,j Denote h(u, v) = g(u, v) for u, v T,0, then h beomes a hermitian inner produt. Extend in a C way to Γ(T C ). z i z j = Γ k ij z k + Γk ij z k, z i Beause of the fat that J = 0, this will imply Γ k ij = Γ k i j = Γk i j = 0. So the only non-vanishing term will be Γ kī j and Γk ij. = Γ k i j + z j z k Γk i j z k. By definition, g j k z i = g( z i z j, ) + g( z k z j, z i z k ) = Γl i j g l k. So we have the following formula. Γ l ij = g l k i g j k. Proposition.3. is a Kahler manifold if and only if for all x, we an find a holomorphi hart (z,..., z n ) suh that g i j(x) = δ ij and dg i j(x) = 0. Reall Riemannian urvature tensor: R(u, v)w = u v w v u w [u,v] w, u, v T R. Extend Rm omplex linearly to T C and denote R(u, v, w, x) = g(r(u, v)w, x). Sine J is parallel, R(u, v)(jw) = J(R(u, v)w). Thus, R(u, v, Jw, Jx) = R(u, v, w, x). By symmetri of urvature operator, R(u, v, w, x) = 0 if w, x are of same type. The only non-vanishing term are R i jk l. By first Bianhi identity, we have a extra symmetri property for Rm on Kahler manifolds. R i jk l + R i l jk = 0. 2

3 Proposition.4. In loal oordinate, R i jk l = k lg i j + g s t k g s j lg i t. We define Rii urvature on by R k l = g i j R i jk l = k l log det(g i j). Some notation for omplex manifold: For f C (), d = + where f = i f z i dzi, f = i f z i d zi. In general, : A p,q A p+,q, : A p,q A p,q+. And, satisfies 2 = 2 = 0, + = 0. The Rii form is then given by So loally Ri = whih is learly a losed, real (,) form. 2 R i jdz i dz j. Ri = 2 log det g Remark: One an hek that the definition is same as the Rii urvature in Riemannian ase. In partiular, if we hoose a orthonormal base {e,..., e n } suh that Je i = e n+i. And u i = (e i e n+i )/ 2. Then {u i } is a unitary frame on T,0. Diret omputation yield Ri(u i, ū i ) = Ri(e i, e i ). Definition.2. We say has a non-negative bisetional urvature, BK 0 if R(u, ū, v, v) 0 for all u, v T,0. We say has a non-negative holomorphi urvature, H 0, if R(u, ū, u, ū) 0 for all u T,0. Noted that BK 0 imply Ri 0. unknown. But the relationship between H and Ri remains.2 Result about struture of manifolds with urvature onstraint Theorem.3. (Uniformization theorem) Every simply onneted Riemann surfae is onformally equivalent to one of the three domains: the open unit disk, the omplex plane, or the Riemann sphere. 3

4 Uniformization onjeture by Yau: If is omplete non-ompat Kahler manifold with BK > 0, then is biholomorphi to C n. Theorem.4. (Soul theorem) If (, g) is a omplete onneted Riemannian manifold with setional urvature K 0, then there exists a ompat totally onvex, totally geodesi submanifold S suh that is diffeomorphi to the normal bundle of S. In partiular, if K > 0, then S is a singleton. That is to say being diffeomorphi to R n. Theorem.5. (Frankel Conjeture solved by Siu-Yau, ori) If is ompat Kahler manifold with BK > 0, then is biholomorphi to CP n. Definition.6. Given a Riemannian manifold n, let p. We say has maximal volume growth if V ol(b(p,r)) r n > 0 for all r > 0. Theorem.7. (ok-siu-yau,98) Let be a non-ompat Kahler manifold. If there exists C, ɛ > 0 suh that 0 BK(x) is biholomorphi and isometri to C n. C r(x) 2+ɛ and has maximal volume growth, then 2 ok-siu-yau s result Theorem 2.. (ok-siu-yau,98) Let be a non-ompat Kahler manifold. If there exists C, ɛ > 0 suh that 0 BK(x) is biholomorphi and isometri to C n. C r(x) 2+ɛ and has maximal volume growth, then Proposition 2.. n is Kahler manifold with BK 0. Let ρ be a real d-losed (,) form, f = ρ i jg i j. If f = 2 Ru for some funtion u, then u ρ 2 is sub-harmoni. Proof. Choose a normal oordinate at p. Then R i jk l = k lg i j at p. Let v = u ρ. 2 v 2 = i ī (g i lg k j v i jv l k) = R līp pv i jv l j + R j kp p v i jv i k + p v i j 2 + p v i j 2 + p p v i j v i j + v i j p p v i j By lemma, ρ = w for some w loally as ρ is d-losed (,) form. By assumption, tra(v) = 0. Taking derivative on tra(v) to yield R l ki jv k lv i j + k,l k i jv k k v i j = 0, i, j. By mean of transformation of U(n), we may further assume v i j = a i δ ij at p, where a i are real. Then, 2 v 2 R līp pv i jv l j + R j kp p v i jv i k + p p v i j v i j + v i j p p v i j 2R iīp pa 2 i 2R iīp pa i a p 0. 4

5 Proposition 2.2. ( n, p), n 3 is a manifold with Ri 0 and maximal volume growth. Let f 0 be a smooth funtion on. Then u = f has a solution if. f(x) C +r(x) 2, and suh that log(r(x) + 2) u(x) log(r(x) + 2). 2. f(x) C +r(x) 2+ɛ, and C 2 > 0 suh that u(x) C 2. Proof. Here we will only demonstrate ase. (Some review for Green funtion [5]) For n with Ri 0, it is non-paraboli if and only if t V p (t) <. Due to the non-ollapsing assumption, Greeen funtion G(x, y) exists with the following properties.. G(x, y) > 0, 2. > 0 s.t. d n 2 (x,y) G(x, y) 3. G(x, y) = G(y, x), d n 2 (x,y), 4. G(x, y) f(y) dy = f(x) for f C 0 (), 5. x G(x, y) C d n (x,y). For R > 0, let f R (x) = ϕ(x)f(x) where ϕ is a non-inreasing smooth funtion in whih ϕ = on B(p, R + ), ϕ = 0 on B(p, R) and ϕ, ϕ are bounded. Let u R (x) = [G(p, y) G(x, y)]f R (y) dy. Then u R satisfy u R = f R and u R (p) = 0. For x, R >> r = r(x). By gradient estimate for G, G(p, y) G(x, y) f R (y) dy \B(p,2r) \B(p,2r) By volume omparsion, we further dedue that G(p, y) G(x, y) f R (y) dy B(p,R+)\B(p,2r) On the other hand, B(p,2r) G(p, y)f(y) dy = B(p,R+)\B(p,2r) B(p,2r) 2r 0 R 2r r f(y) dy (d(y, p) r) n Cr ( + r(y) 2 dy. )(r(y) r) n Ct n r ( + t 2 )(t r) n dt C n. r(y) n 2 ( + r(y) 2 ) dy t n 2 ( + t 2 ) da tdt C log(r + 2). 5

6 Also, for r > 0 B(p,2r)\B(x,r/2) B(x,r/2) G(x, y)f(y) dy G(x, y)f(y) dy C n r B(p,2r)\B(x,r/2) C 2r r/2 r/2 0 s + r s 2 + ds s ds C n. d(x, y) n 2 (d(p, y) 2 + ) dy = C 2 log 6r2 + r C r [artan(2r) + artan(r/2)] C n. Combine all this, we have u R (x) C n [log(r(x) + 2)]. We now laim that lim R u R (x) exists after passing to subsequene. First as has non-negative Rii urvature and is non-ollapsing, Sobolev inequality with ompat support is valid. Furthermore, the sobolev onstant is an absolute onstant. Thus we have Harnak inequality for positive harmoni funtions on geodesi ball and hene the Holder estimate for x, x 0 B(p, R). u(x) u(x 0 ) C R x x 0 α Let R < R 2 <... < R n <... orrespond the exhaustion, for eah j N, define v i = u Ri u Rj whih is harmoni on B(p, R j ). Thus for x, y B(p, R j /2), u Ri (x) u Ri (y) v i (x) v i (y) + u Rj (x) u Rj (y) C j x x 0 α + u Rj (x) u Rj (y). So, {u Ri } i=j is equiontinuous on B(p, R j/2). By ArzelAsoli theorem, we an extrat onvergent subsequene lim R u R (x) on B(p, R j /2). Using diagonal argument, we have limit funtion u(x) = lim R u R (x). By gradient estimate of G, we infer that u = O(/r). Lemma 2.2. Seond derivative estimate for both situations: Proof. Reall bohner s formula, 2 u 2 C for R >. vol(b(p, R)) R4 2 u 2 = 2 u 2 + u, u + Ri( u, u) 2 u 2 + u, f. Let φ be a ut-off funtion suh that φ = on B(p, R), φ = 0 outside B(p, 2r), and 6

7 φ /R. φ 2 2 u 2 B(p,2R) 2 B(p,2R) B(p,2R) Choose ɛ small enough to onlude the result. φ 2 u 2 φ 2 u, f φ φ u 2 φ 2 u, f [ ] C φ 2 f 2 + ɛφ 2 2 u 2 + ( + B(p,2R) ɛ ) φ 2 u 2 Theorem 2.3. Let n be a omplete Kahler manifold with BK 0 and maximal volume growth. In addition, if the salar urvature S r 2+ɛ, then is flat. Proof. By previous porposition, we an solve for u in whih u = s. In partiular, u C, u = o(/r). As Ri u is subharmoni, by Li s mean value inequality, for R >> r(x), Ri u 2 C (x) Ri V ol(b(p, R)) u 2 C 2 u 2 + Ri 2 = O( V ol(b(p, R)) R 4 ) 0. Thus, Ri = u. To finish our proof, it suffies to show that u is onstant. For n 2, B(p,r) ( u) n = B(p,r) u ( u) n = o(/r) ( r 2+ɛ )n O(r 2n ) 0. Thus, ( u) n = 0. Assuming is stein, let Φ : C N be the embedding, Φ(p) = (z (p), z 2 (p),..., z N (p)). Let φ be the restrition of N i= z i 2 on. If u is non-onstant, for eah < 0, = {u < } is preompat. We may further assume is smooth. Let x 0 be a point on suh that φ attains maximum. So loally is on one side of {φ = φ(x 0 )}. Therefore i u is bounded below by the omplex hessian of the distane funtion on the omplex tangent spae of at x 0. In partiular, i u is stritly positive on it. is then positive at x 0. However, on ( e u ) n = B(p,r) e u = e u u + e u u u = B(p,r) B(p,r) So, ( e u ) n 0. Contradition arised. e u ( e u ) n e nu u ( u + u u) n 0. Remark: It has been proved later by Ni-Tam that must be stein if it satisfies the assumption in this theorem. 7

8 Corollary 2.4. Suppose is a Kahler manifold with maximal volume growth and BK 0. Furthermore, if the salar urvature s C r 2, then Rin <. Proof. Solve for i u = Ri. Where C is independent of R. Ri n = i u Ri n C < Theorem 2.5. (Ni-Shi-Tam) The onlusion of Proposition 2.2 still hold if we only assume Ri 0 and 0 f(x) + r(x) 2 and f(x) C V p (r) B(p,r) + r 2. Corollary 2.6. If n has BK 0, and Ri > 0 at p. Furthermore, if the salar urvature satisfies s C r and 2 V p(r) B(p,r) s C r. Then has maximal volume growth 2 and = () > 0 suh that for r large s V p (r) B(p,r) r 2. Proof. Solve for u where i u = Ri by Theorem above. u /r. Then 0 < < Ri n = i u Ri n = whih implies maximal volume growth. Assume r >>, 0 < < B(p,r) i u Ri n C R Ri n = B(p,r) C r 2n 3 = C r 2n 3 i u Ri n B(p,r) B(p,r) vol(b(p, R)) R2n 2 Ri ω n s. Theorem 2.7. (Chen-Zhu) Let n be a Kahler manifold BK > 0. Let p. Then = (, p) > 0 suh that vol(b(p, r)) Cr n. Proof. Consider Busemann funtion b on, where b(x) = lim (r d(x, B(p, r))). r 8

9 It is known that b = a.e. and b is stritly plurisubharmoni in the support sense. By smoothing argument, for R > 2, > 0 suh that 0 < R 2n (i b) n (R r) 2n = 2n (R r) 2n b (i b) n r C n whih implies V ol(b(p, r)) r n, for all r > 2. C n (R r) 2n (i b) n ω (R r) n ω n C nr n V p (R), 3 Appliation of Hörmander L 2 estimate. Theorem 3.. (Hörmander L 2 estimate): Let (, ω) be a omplete Kaehler manifold. L be holomorphi line bundle with Hermitian metri h (loally given by e 2φ, s 2 h = e 2φ. Let Θ be the urvature (,)-form, Θ = 2i φ. Let g be a smooth setion in Λ n, L (i.e. L valued (n,) form.), with g = 0. Assume Θ ɛω where ɛ is positive funtion on. If g 2 hɛ <, then there exists f Γ(Λ (n,0) L) suh that f = g and f 2 h g 2 h ɛ. 3. Some Appliations of L 2 -estimate Theorem 3.2. (ok) Let n be a omplete Kahler manifold with BK > 0 and maximal volume growth. Furthermore suppose the salar urvature s satisfies s r 2, r(x) = d(x, x 0 ) for some fixed point x 0, then there exists a non onstant holomorphi funtion with polynomial growth on. Proof. Due to the growth ondition of salar urvature s, one an solve u in whih i u = Ri. Furthermore, u satisfies u(x) C log(r(x) + 2). Let p, there exists a holomorphi hart (z,..., z n ) on B(p, δ) with z i (p) = 0. Let ϕ be ut-off funtion with ϕ = on B(p, δ/5) and ϕ = 0 outside B(p, δ/3). Now we look for a good weight funtion. Lemma 3.3. i (λu + 2nϕ log( z 2 )) ω > 0 on B(p, δ/3) for some λ >>, > 0. Proof. Noted that i log z 2 0 in the urrent sense. Thus, i [ϕ log z 2 ] = iϕ log z 2 + i ϕ log z 2 + iϕ log z 2 + i log z 2 ϕ, 9

10 where depends on the C 2 bounds of ϕ and δ. So for suffiiently λ, we have on B(p, δ/3) i ( λu + 2nϕ log z 2) ω. Take L = T n,0 (), hoose metri h = e ψ = e (λu+2nϕ log z 2). Thus, the urvature form Θ = i log h ω > 0. Let φ be a smooth ut-off funtion suh that φ = on B(p, δ/5) and φ = 0 outside B(p, δ/4). Apply L 2 estimate to solve g = (φz ) with g 2 e ψ 2 (φz ) 2 e ψ <. Noted that on B(p, δ/5), e ψ = e λu z 4n. Thus, it is not loally integrable whih implies g(p) = 0 and dg(p) = 0. Now f = g φz is holomorphi on with f(p) = 0, df(p) = d(φz ) 0. f is thus non-onstant and f = g outside B(p, δ). It suffies to show that g is of polynomial growth. Let y B(p, R) in whih R > 00, C g 2 e ψ g 2 e λu B(y,R/2) B(y,R/2) g 2 (r + 2) C R B(y,R/2) By Li s mean value inequality together with maximal volume growth ondition, we onlude that C > 0 independent of y suh that g(y) 2 C n V y (R/2) B(y,R/2) g 2 CR 2n. g 2. Theorem 3.4. (Chen-Zhu) Let n be a Kahler manifold with BK > 0, then the salar urvature s satisfies V p(r) B(p,r) s r, where = (, p). Proof. Let p. Let (z,...z n ) be a hart on B(p, δ). Consider the Busemann funtion b, it satisfies i b > ɛω for some ɛ > 0 on B(p, δ). Define ψ = λb + 2nφ log z 2, where φ is a ut-off funtion same as the proof above. λ is large enough suh that i ψ ω on B(p, δ). Consider a smooth setion on the anonial line bundle u = ϕdz dz 2... dz n, where ϕ is a smooth funtion with ompat support on B(p, δ/4) and equals to on B(p, δ/5). By L 2 estimate, we obtain a nontrival holomorphi setion θ on K suh that θ(p) = and θ 2 e λb <. Sine the Busemann funtion is distane like funtion and θ 2 0, by Li s mean value inequality [4], we have θ(x) C n e r(x). 0

11 For δ > 0, log( θ 2 + δ) is smooth, and by diretly omputation, log( θ 2 + δ) s θ 2 θ 2 + δ. Consider C 2 instead of, positive green funtion exists and the volume growth is at least of order 4. For α, β >, ɛ > 0, 0 < δ <, for (z, t) C 2 s θ 2 θ 2 + δ (G(z, p) α)+ɛ log( θ 2 + δ)(g α) +ɛ β>g(z)>α Noted that on {β > G > α}, So we get β>g(z)>α Letting ɛ 0 it follows Sine G and G n and = β>g(z)>α β>g(z)>α + G=β G=β log( θ 2 + δ) (G α) +ɛ n [log( θ 2 + δ)](g α) +ɛ (G α) +ɛ = ɛ( + ɛ)(g α) ɛ G 2 0. log( θ 2 + δ) (G α) +ɛ G>α β>g>α s θ 2 θ 2 (G(z, p) α) + δ sup are asymptoti to r 2n (G α) n G=β G=β ( + ɛ) log( θ 2 + δ)(g α) ɛ G n. sup log( θ 2 + δ) (G α) +ɛ β>g>α β>g>α sup log( θ 2 + δ) β>g>α G=β log( θ 2 G + δ) β>g>α G=β n + G=β G=β n r 2n 2 and n r 2n G n n [log( θ 2 + δ)](g α) log( θ 2 + δ) G n. as β, thus = (n)r 0, as β. r2n 2 (n) as β. ( + ɛ)(g α) ɛ G n. Sine θ(p) =, letting β yield s θ 2 [ ] θ 2 + δ (G(x, p) α) n sup log( θ 2 + δ) log( θ 2 ( p) + δ) G>α n sup log( θ 2 + δ). G>α

12 Letting δ 0 implies G>2α s(z)g(z, p) n sup log θ 2. G>α For α > 0, let r α be the maximum positive number suh that B(p, r α ) {G > α}. For equiped with Ri 0, the minimal postive Green s funtion satisfies By volume omparsion, On the other hand, G(z, p) C C r(z) r(z) G(z, p) C r(z) t dt V p (t) G(x, p) C t dt V p (t) C t dt r(z) V p (t) Cr(z)4 r(z) t dt V p (t). tr 2n V p (r)t 2n dt C r(z) 2 n V p (r(z)). r(z) t dt V p (r)t 4 = Cr(z)2 V p (r(z)). As a result, there exists C = C(n) suh that for all α > 0, r α satisfies C r 2 α V p (r α ) α Cr2 α V p (r α ). Thus, if G(z, p) > α, C r(z) 2 V p (r(z)) α G(z, p) > α C r2 V p (r α ). by the above inequality, V p (r(z)) V p (r α ) C n r 2 (z) rα 2. Sine B(p, r) B(0, r) B( p, r) B(p, r) B(0, r) for eah r > 0, V p (r(z)/2) V p (r α ) ( ) 4 rα n Vp(r(z)/2) (r(z)/2) 4 r(z) 2 4 V p (r α ) rα 4 n ( rα r(z) ) 4 V p (r(z)/2) V p (r α ) nr 2 α r 2 (z). That is r(z) n rα 2 Vp(r(z)/2). V p (r α ) In both of the ases r(z)/2 r α or r(z)/2 r α, r(z) n r α. Thus {G > α} B( p, n r α ). Combine everythings, one an get Crα 2 s(z) α s(z) α s(x) G s(x) V p (r α ) B( p,r α) B( p,r α) G>α G>α n sup G>α/2 log θ 2 n sup B( p, nr α/2 ) log θ 2. On the other hand, r α and r α/2 an be related in the following relation. r 2 α/2 C nα V p (r α/2 ) C n V p (r α/2 ) V p (r α ) r2 α C n r 2 α ( rα/2 r α ) 2n. 2

13 So for any α > 0, rα 2 s dṽ V p (r α ) B( p,r α) C(n, p) sup log θ 2 C(n, p)(r α + ). B( p, n rα) Projeting everythings bak to implies our desired result. 4 Heat flow tehnique on Kaehler manifolds Theorem 4.. (Ni-Tam) Let n be a omplete nonompat Kahler manifold with nonnegative holomorphi bisetional urvature and let u be a ontinuous plurisubharmoni funtion on satisfying u (x) Ce ar(x)2 for some onstants a, C > 0 where r(x) is the distane of x from a fixed point. Let v be the solution of the heat equation with initial data u. There exists T 0 > 0 depending only on a and there exists T (0, T 0 ) suh that the following are true.. For 0 < t < T 0, v(, t) is a smooth plurisubharmoni funtion. 2. Let K(x, t) = {w Tx,0 () : v α β(x, t)w α = 0, β.} be the null spae of v α β(x, t). Then for any 0 < t < T, K(x, t) is distribution on. oreover the distribution is invariant under parallel translations. 3. If the holomorphi bisetional urvature is positive at some point, then v(x, t) is stritly plurisubharmoni for all 0 < t < T. Before we proeed to the proof, we first state some of its appliations. Corollary 4.2. Let n be a omplete Kahler manifold with nonnegative holomorphi bisetional urvature. Suppose is of maximal volume growth, then is stein. Proof. Assume first that is simply onneted. Let b(x) be the Busemann funtion on. By a result of Shen, if n has nonnegative Rii urvature and maximal volume growth. Then the Busemann funtion is a exhaustion funtion. Consider the heat flow v(t) with v(0) = b(x). Let u = v(t 0 ) where t 0 < T, then we obtain a smooth plurisubharmoni funtion on. By heat kernel estimate, u is also a exhaustion funtion. By the main theorem, K(x, t) is parallel invariant, By de Rham deomposition, = N isometrially and holomorphially so that u α β 0 when restrited on N and u α β > 0 on. Due to the non-ollapsing ondition, N must be nonompat. On N, define h = log( + 3

14 u 2 ), h is plurisubharmoni. To see this, it suffies to show that h γ γ 0 in normal oordinate. Let F = u 2, h γ γ = ( + F ) 2 [( + F )F γ γ F γ F γ ] [ ( = ( + F ) 2 ( + F ) u γα u γᾱ + ) R γ γα s u s uᾱ α α,s α ( ( + F ) 2 u αγ uᾱ γ + ) R γ γα s u s uᾱ 0. α α,s (u αγ uᾱ) α (u α uᾱ γ ) ] The Busemann funtion b is distane like, and b = almost everywhere on. Thus, u = O(r(x)) by Heat Kernel estimate. By gradient estimate, u = O(). And hene h = o(log(r(x)). By three irle theorem [7], h is onstant funtion on N and hene u is onstant on N. By Bohner s formula, on N, 0 = 2 u 2 = 2 u 2 + u, u + Ri( u, u) 2 u 2. u is parallel. Hene J u is parallel. So N = N C as N is simply onneted. Repeating the proess, we onlude that N = N C k for some k and N is ompat. But Due to the non-ollapsing assumption, N does not exist. So = C k where u is a stritly plurisubharmoni exhaustion funtion on. Thus, is stein and so is. If is not simply onneted, let be its universal over, is of maximal volume growth and nonnegative Rii urvature. By a result in [6], π () is finite. Let f be a smooth stritly plurisubharmoni exhaustion funtion on. Then h(x) = f(p) π (x) is a stritly plurisubharmoni exhaustion funtion on. Thus is stein. Referenes [] N. ok, Y. T. Siu and S. T. Yau, The Poinare-Lelong equation on omplete Kahler manifolds, Compositio. ath(44), 98, [2] Hrmander, Lars An introdution to omplex analysis in several variables. Third edition. North-Holland athematial Library, 7. North-Holland Publishing Co., Amsterdam, 990. xii+254 pp. ISBN: [3] B.-L. Chen and X.-P. Zhu, Volume growth and urvature deay of positively urved Kahler manifolds, preprint. f. R [4] Li, Peter Geometri analysis. Cambridge Studies in Advaned athematis, 34. Cambridge University Press, Cambridge, 202. x+406 pp. ISBN:

15 [5] Li, Peter Curvature and funtion theory on Riemannian manifolds. Surveys in differential geometry, , Surv. Differ. Geom., VII, Int. Press, Somerville, A, [6] P. Li, Large time behavior of the heat equation on omplete Riemannian manifolds with nonnegative Rii urvature, Ann. of ath. 24 (986), 2.. [7] G. Liu, Three irle theorems on Kahler manifolds and appliations, arxiv: