Cai, J X

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Available at: http://www.ictp.trieste.it/~pub o IC/2000/89United Nations Educational Scientic and Cultural OrganizationandInternational Atomic Energy AgencyTHE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICSTHE ALBANESE MAP OF A 3-FOLDOF GENERAL TYPE WHOSE CANONICAL MAPIS COMPOSED WITH A PENCILJin-Xing Cai1School of Mathematical Sciences, Peking University,Beijing 100871, People's Republic of ChinaandThe Abdus Salam International Centre for Theoretical Physics, Trieste, Italy.MIRAMARE { TRIESTEJuly 20001E-mail: cai@ictp.trieste.it; caijx@sxx0.math.pku.edu.cn1ext-2000-22401/11/2000IntroductionLet X be a complex projective manifold of general type. A natural question is, if X enjoyssuch and such analytic properties, then how is the topology of X? In this note we will studysome topological properities of X when the canonical map Xof X is composed with a pencil.It is known that, if X is a surface of general type whose canonical map is composed with apencil, then the irregularity q(X)  2 [X1]. (We refer the reader to [Be1] in which the canonicalmap of surface of general type was rst systematically studied.) However, when dimX  3,simple examples show that q(X) can be arbitrarily large whenever Xis composed with apencil.We recall that, the Albanese dimension a(X) of a manifoldX, is dened to be the dimensionof the image of the Albanese map alb:X ! Alb(X), and X is said of maximal Albanesedimension or of Albanese general type if a(X) = dimX. We note that a(X) is a topologicalinvariant of X (cf. [Ca, Proposition 1.4]). Our main result is the following.Theorem 1. Let X be a smooth projective 3-fold of general type over the complex numbereld. Assume that the canonical linear system is composed with a pencil, and the irregularityq(X)  6. Then X is not of maximal Albanese dimension.The condition that q(X)  6 is technical (see Remark below). Examples of 3-folds X withi) dim(ImX) = 2 and a(X) = 3, and ii) dim(ImX) = 1 and a(X) = 2 are easily constructed(see Example 1). Theorem 1 will be proved in section 1. In section 2 we will consider thebehaviour of the canonical map under smooth deformations.Theorem 2. Let X be a smooth projective 3-fold of general type whose canonical map iscomposed with a pencil. Let f :X ! C be the ber space associated to the canonical map. Ifg(C)  2, then the canonical map of any (global) smooth deformation of X is composed witha pencil.We use standard notations as in [BPV] or [Mo].Acknowledgments. I am grateful to the referee for helpful suggestions. This work is sup-ported by the Abdus Salam International Centre for Theoretical Physics, Trieste, Italy andpartially supported by the National Natural Science Foundation of China.1. Proof of Theorem 1Let X be a smooth complex projective 3-fold with pg(X)  2. The canonical mapX:X    ! Ppg(X) 1is dened by the moving part of the canonical linear system jKXj. Assume that Xis composedwith a pencil. Let :X0! X be a resolution of the base locus of the moving part of jKXj.2We have the following commutative diagramX0f ! C??y??yXX   !Ppg(X) 1;where f :X0! C is the Stein factorization of X . We call f :X0! C the canonicalbration associated with the canonical map. Clearly, such a bration is dened up to birationalequivalence. But this does not matter, since we only consider birational invariants (e.g., theAlbanese map, a(X), q(X), etc.). Using X0instead of X, for convenience, such a canonicalbration is denoted by f :X ! C.Proposition 1. (cf. [CC]) Let X be a 3-fold with pg(X)  2. Assume that the canonicalmap Xis composed with a pencil. Let f :X ! C be the ber space associated with X, andF a general ber of f . If g(C)  2, then we have that either pg(F ) = 1 or pg(F ) = pg(X) =g(C) = 2.Proof. Since pg(X) 6= 0, we have pg(F ) > 0. Now we assume that pg(F )  2. Let M andE be the moving part and the xed part of jKXj respectively. Since Xfactors through f ,OX(M) = fL0for some L02 PicC. Thus h0(L0) = pg(X) and !X= fL0 OX(E). Thereis an injective map of sheaves L0,! f!X. Let L be the saturated subsheaf of L0in f!X.Then L is invertible and the quotient Q := f!X=L is torsion free, hence locally free of rankpg(F )  1. We have an exact sequence of sheaves0! L! f!X! Q! 0:Consider the natural map H0(!C L 1)H0(L)! H0(!C), note that h0(L)  h0(L0)  2,by [Ha, Lemma IV.5.5]. We haveg(C)  1 h0(!CL 1)= h1(L) h0(Q) (note that h0(f!X) = h0(L))(pg(F )  1)(g(C)  1):The last inequality follows by the Riemann-Roch Theorem and the semipositivity of f!X=C=f!X ! 1C(cf. [Fu]). This implies pg(F ) = 2 and h1(L) = g(C)   1. Now by Cliord'stheorem, we have degL = 2 and L ' !C. So pg(X) = g(C) = 2. Lemma 1. Let X be a 3-fold of general type whose canonical map is composed with a pencil,and f :X ! C the ber space associated with X. If q(X)  5 and g(C)  2, then alb:X !Alb(X) is not of Albanese general type.Proof. Let F be a general ber of f . We have q(X)  g(C)  q(F ). Since F is (a surface) ofgeneral type, q(F )  pg(F ). If g(C)  2, by Proposition 1, we have that either q(X) g(C)  13or q(X)  g(C) + q(F )  2 + pg(F ) = 4. Thus fH0(1C) is of codimension  1 in H0(1X)by the assumption. So for any , ,  2 H0(1X), we have  ^  ^  = 0. Note that X is ofAlbanese general type if and only if Image(3^H0(1X)! H0(!X)) 6= 0. (cf. [Ca]) Notation. Let X be a complex projective manifold. For any 0 6=  2 H0(nX) (1  n dimX), we denote by Z() the zero-locus of the holomorphic n-form . When n = dimX, wehave Z() = div().Lemma 2. Let f :X ! C be as in Lemma 1. Assume that g(C)  1, and that there existlinearly independent 1-forms ,  2 H0(1X), such that  ^  = 0. Then alb:X ! Alb(X) isnot of Albanese general type.Proof. By a well-known theorem of Castelnuovo-de Franchis (see e.g. [B-P-V]), there exists asurjective morphism h:X ! B to a curve B (of genus  2) such that ,  2 hH0(1B). LetH be a general ber of h. Then H is not contained in the xed part of jKXj. Since g(C)  1by the assumption, f and h are dierent brations of X. So H is not contained in the movingpart of jKXj. Since ,  2 hH0(1B), we have that H  Z(t + t0) (t + t0 6= 0) for asuitable choice t; t02 C . If a(X) = 3, then there exist two 1-forms 1; 22 H0(1X) such that! := (t+ t0) ^ 1^ 26= 0. Then H  div(!) 2 jKXj. This is a contradiction. Proof of Theorem 1.Notations as in Lemma 1. By Lemma 1, we can assume that g(C)  1. We consider theAlbanese map alb:X ! alb(X)  Alb(X). Suppose that X is of Albanese general type.If (alb(X)) = 3, then alb(X)and hence Xis generically nite by a theorem of Grithsand Harris (cf. [Mo, Theorem 3.9]). This contradicts the assumption that Xis composedwith a pencil.Since alb(X) is a proper subvariety of Alb(X) generating Alb(X), one has that (alb(X)) >0. By a theorem of Ueno [Ue, Theorem 10.9], there exist an abelian subvariety A of Alb(X)and a projective variety S0which is a subvariety of an abelian variety such that alb(X) is aber bundle over S0(denoted by u0) whose ber is A, and (S0) = dim(S0) = (alb(X)).LetXalb    ! alb(X)u??yu0??yS     ! S0be the Stein factorization of u0 alb. Since the invariants we consider are birational, by astandard argument, we can assume that S is smooth.If (alb(X)) = 1, then S is a curve of genus  2. This is impossible by Lemma 2.In what follows, we treat the remaining case when (alb(X)) = 2. In this case, S is asurface of general type, and q(S) = q(X)  1. (Note that q(alb(X))  q(S0)  q(A) = 1, andif q(S) = q(X) then X is not of Albanese general type.)4We distinguish two cases according to whether f factors through u or not.Case 1. f does not factor through u.Let s 2 S be a general point such thati) u 1s is not contained in the xed part of jKXj, andii) f ju 1s:u 1s! C is surjective.Since q(S) > 3, there exists 0 6=  2 H0(1S) such that s 2 Z(). We have that u 1s Z(u). Let 0 6=  2 H0(1X) nuH0(1S), and  2 H0(1S) such that u, , u are linearlyindependent. By Lemma 2, we can assume that  ^  6= 0. Since  is closed, it is easy toverify that ! :=  ^u^u 6= 0. Now u 1s  div(!) 2 jKXj. This implies that there existsan irreducible component   of div(!) such that u 1s   . Clearly   is not a ber of f sincef ju 1s:u 1s! C is surjective by the choice of s. So   is an irreducible component of the xedpart of jKXj. This is a contradiction.Case 2. f factors through u.Let h:S ! C be the rational map such that f = hu. By a suitable choice of the birationalmodel of S and hence of X, we can assume that h is a morphism.We prove the case when g(C) = 1. The proof of the case when g(C) = 0 is similar and leftto the reader.Let s1; s22 S be general points, such thati) h(s1) 6= h(s2) and Fi= f(h(si)) (i = 1; 2) are general bers of f ;ii) u 1si(i = 1 and 2) is not contained in the xed part of jKXj.Since q(S) = q(X)  1  5, we have that there exists 0 6=  2 H0(1S) such that si2 Z()for i = 1 and 2. If Fi Z(u) for i = 1 and 2, then ft  u is a holomorphic 1-form onX, where t is a meromorphic function having simple poles exactly at h(s1) and h(s2). Clearlyft  u and u are linearly independent and (ft  u) ^ u = 0. By Lemma 2, we get acontradiction.Now we can assume that Fi6 Z(u) for some i (say i = 1). Let p 2 F1be a generalpoint. Then p 62 Z(u). Let 0 6=  2 H0(1X) n uH0(1S), and 1; 22 H0(1S) such that1; 2;  are linearly independent. By Lemma 2, we can assume that  ^ (t11+ t22) 6= 0 forany linear combination t11+ t226= 0. Then we have that !i:=  ^ u ^ ui6= 0, and!i(i = 1 and 2) are C -linearly independent. Thus div(!i) (i = 1 and 2) generate a linearpencil  of X. Since p is a general point of F1and F1is a general ber of f , we can assumethat p is not in the base locus of . So we have that p 62 div(!i) for some i. Fix such an i.Since u 1s1 div(!i) 2 jKXj, there exists an irreducible component   of div(!i) such thatu 1s   . By the above argument,   6= Fi. So   is a component of the xed part of jKXj.This contradicts the choice of s1. Corollary. Let X be as in Theorem 1. Assume that the canonical map Xis composed witha pencil, and q(X)  6. Let f :X ! C be the ber space associated with X. Assume thatg(C)  2. Then (alb(X)) = 1.5Proof. By Theorem 1 we have a(X)  2. The corollary is trivial when a(X) = 1.Now we assume that alb(X) is a surface. We claim that f factors through alb. Otherwise,there is a dominant rational mapX   ! Calb(X), which implies q(X)  q(alb(X))+g(C).We get necessarily g(C) = 0, contradicting the assumption. Let F be a general ber of f . SinceF is of general type, we have pg(F )  q(F ). By Proposition 1, we have that either pg(F ) = 1 orpg(F ) = g(C) = 2. The latter case does not occur since q(X)  g(C)+ q(F )  2+ pg(F ) = 4,contradicting the assumption. Now if (alb(X)) = 2, then the image albX(F ) of F under theAlbanese map of X is a curve of genus  2. So pg(F )  q(F )  g(albX(F ))  2. This is acontradiction. Example 1. 1) Let S be a smooth surface of general type with q(S) = 2 such that i) alb:S !Alb(S) is surjective and ii) Sis composed with a pencil (cf. [X2, p. 600] for such examples).Let X = S C, where C is a smooth curve of genus  2. Then X is of Albanese general typeand dim(ImX) = 2.2) Let X = S  C, where S is a smooth surface of general type with pg(S) = q(S) = 1 andC is a smooth curve of genus  2. Then a(X) = 2 and Xis composed with a pencil.Remark. 1) The condition q(X)  6 in Theorem 1 is technically needed in the proof of Case 2when g(C) = 1. When g(C) = 0, a similar argument of Case 2 works only assuming q(X)  4.Thus from our proof we see that Theorem 1 holds if either i) q(X)  4 and g(C) = 0 org(C)  3, or ii) q(X)  6 and g(C) = 1, or iii) q(X)  5 and g(C) = 2.We conjecture that Theorem 1 is true under the assumption q(X)  4.Examples of 3-fold X of general type whose canonical map is composed with a pencil, withq(X) = 3 and a(X) = 3 are easily constructed. Let (A;) be a principal polarized Abeliann-fold (n  3), and D 2 j2j a smooth divisor. Let X ! A be the double cover determinedby the double cover data (D;). Then pg(X) = 2 and alb(X) ' A.2) The condition that X is of general type in Theorem 1 is indispensable. For example, letX = A  C, where A is an abelian variety and C is a smooth curve of genus  2. Then X isof Albanese general type and the map induced by jKXj is composed with a pencil.3) Let n  3 be an integer, and X be a smooth projective n-fold X of general type and ofmaximal Albanese dimension. Then it is easy to see that pg(X)  1 and by a theorem of Ueno[Ue], one has pg(X)  2 if q(X) > n.Question. Is there such an n-fold X with pg(X) = 1?Clearly, a statement similar to Theorem 1 for (n + 1)-folds of general type is false if theabove question has a positive answer.2. Deformations of the canonical mapLet f :X ! Y be a morphism of smooth projective manifolds with connected bers. It isknown that the morphism f :X ! Y (of the whole triple) has a semiuniversal deformation.6Its base space is denoted by Def(f :X ! Y ). There is a natural mapP : Def(f :X ! Y )! Def(X);where Def(X) is the base space of the semiuniversal deformation of X.Fact 1. If h0(R1fOX TY) = 0, where TYis the dual of Y, then P is surjective.We refer the reader to [Ca, Il, Fl, Ho, Pa, Ra ] for the basic results on deformations ofmorphisms and the proof of Fact 1.Proposition 2. Let f :X ! C be a bered 3-fold whose general ber F is a surface withpg(F ) = 1. Then Xis composed with a pencil if pg(X)  2.Proof. Since pg(F ) = 1 by assumption, L: = f!Xis an invertible sheaf. We have an exactsequence of sheaves0! fL ! !Xwith H0(fL) = H0(!X). This implies that Xfactors through f . We need the following result. For a proof, see [Be2] or [Si].Fact 2. (Siu-Beauville) Let g  2 be an integer, and 1(g) be the fundamental group of aRiemann surface of genus g. Let X be a compact Kaehler manifold of dimension  2. ThenX has a pencil of genus  g if and only if there exists a surjective homomorphism from 1(X)to 1(g).Theorem 2. Let X be a smooth projective 3-fold of general type whose canonical map iscomposed with a pencil. Let f :X ! C be the ber space associated with X. If g(C)  2, thenthe canonical map of any (global) smooth deformation of X is composed with a pencil.Proof. Since the geometric genus is a global deformation invariant, the theorem is triviallytrue when pg(X) = 2. So we can assume that pg(X)  3. By Proposition 1, we have thatpg(F ) = 1, where F is a general ber of f .Let X0be any (global) smooth deformation of X, that is, there is a relatively projectivesmooth family :X ! B such that Xb' X, Xb0' X0(b; b02 B) and B is connected. Lett 2 B be any given point. By Fact 2, there is a surjective morphism ft:Xt! Ctto a curveCtof genus  g(C). Note that q(X)  g(C) + q(F )  g(C) + 1 (q(F )  pg(F ) since F is ofgeneral type), we have that g(Ct) = g(C) again by Fact 2. Sinceh0(R1ftOXt TCt) = h1(R1ft!Xt !Ct) = 0by [Ko, Theorem 2.1], Def(ft:Xt! Ct)! Def(Xt) is surjective by Fact 1, there is a sucientlysmall neighborhood Utof t 2 B, such that for any t02 Ut, Xt0admits a holomorphic map~ft0:Xt0!~Ct0which is a smooth deformation of ft:Xt! Ct. On the other hand, Xt0admits at7most one holomorphic map to a curve of genus g(C) up to isomorphism since q(Xt0)  g(C)+1.So~ft0and ft0are the same brations up to isomorphism. This implies that the geometricgenus of the general ber of ft0:Xt0! Ct0(t02 Ut) is constant on Ut. Consequently, since Bis connected, we have that pg(F0) = pg(F ) = 1, where F0is a general ber of fb0:Xb0! Cb0.By Proposition 2, we get the result. Remark. Let X be a smooth complex projective manifold of general type with pg(X)  3.Assume that Xis composed with a pencil. In general, the canonical map of the (global)deformation of X is not necessarily composed with a pencil even when dimX = 2.Question. Is there a maximal family (i.e., moduli space component) of smooth projectivesurfaces of general type with pg 0 whose canonical maps are composed with a pencil?References[Be1] Beauville, A., L'application canonique pour les surfaces de type general, Invent. Math. 55 (1979), 121{140.[Be2] Beauville, A., Appendix to [Ca].[BPV] Barth, W., Peters, C., Van de Ven, A., Compact complex surfaces, Ergeb. Math. Grenzgeb., 1984.[Ca] Catanese, F., Moduli and classication of irregular Kaehler manifolds (and algebraic varieties) withAlbanese general type brations, Invent. Math. 104 (1991), 263{289.[CC] Chen, M., Chen, Z. J., Irregularity of canonical pencils for a threefold of general type, Math. Proc.Camb. Phil. Soc. 125 (1999), 83{87.[Fl] Flenner, H.,Uber deformationen holomorpher abbildungen, Osnabrucker Schriften zur mathematik I-VIIand 1-142, 1979.[Fu] Fujita, T., On Kaehler bre spaces over curves, J. Math. Soc. Japan 30 (1978), 779{794.[Ha] Hartshorne, R., Algebraic Geometry, GTM 52, Springer-Verlag, 1977.[Il] Illusie, L., Complexe cotangent et deformations I, II, LNM 239, LNM 283 Springer-Verlag, (1971).[Ho] Horikawa, E., On deformation of holomorphic maps, I, II, J. Math. Soc. Japan vol 25 (1973), 372-397;vol 26(1974) 647-667; III, Math. Ann. 222, 275-282 (1976).[Ko] Kollar, J., Higher direct images of dualizing sheaves I, Ann. Math. 123 (1986), 11{42.[Pa] Palamodov, V.P., Deformations of complex spaces, Usp. Mat. Nauk. 31 (1976), 129{194.[Ra] Ran, Z., Deformations of maps, LNM 439, Springer-Verlag, (1989), 246{253.[Mo] Mori, S., Classication of higher-dimensional varieties, Proceedings of Symposia in Pure Math. 46(1987), 269{331.[Si] Siu, Y. T., Strong rigidity for Kaehler manifolds and the construction of bounded holomorphic functions,in Discrete groups and analysis (Howe, R., eds.), Birkh..auser, 1987, pp. 124{151.[Ue] Ueno, K., Classication theory of algebraic varieties and compact complex spaces, LNM 439, Springer-Verlag, (1975).[X1] Xiao, G., L'irregularite des surfaces de type general dont le systeme canonique est compose d'un pinceau,Comp. Math. 56 (1985), 251{257.[X2] Xiao, G., Irregularity of surfaces with a linear pencil, Duke Math. J. 55 (1987), 597{602.8

The albanese map of a 3-fold of general type whose canonical map is composed with a pencil

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