On the geography of Gorenstein minimal 3-folds of general type
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ON THE GEOGRAPHY OF GORENSTEIN MINIMAL 3-FOLDS OF GENERAL TYPE MENG CHEN AND CHRISTOPHER D.HACON Abstract.Let X be a minimal projective Gorenstein 3-fold of general type.We give two applications of an inequality between χ(ωX )and p g (X ):1)Assume that the canonical map Φ|K X |is of ?ber type.Let F be a smooth model of a generic irreducible component in the general ?ber of Φ|K X |.Then the birational invariants of F are bounded from above.2)If X is nonsingular,then c 31≤13.1.Introduction We work over the complex number ?eld C .The main purpose of this note is to study the geometry of Gorenstein minimal 3-folds X of general type.We improve the inequality χ(ωX )≤2p g (X )(see Proposition 2.2for a precise statement),and we show how this leads to several applications which we explain below:First,we improve the main theorem in [8]:Theorem 1.1.Let X be a minimal projective Gorenstein 3-fold of general type.Assume that the canonical map Φ|K X |is of ?ber type.Let F be a smooth model of a generic irreducible component in the general ?ber of Φ|K X |.Then the invariants of F are bounded from above as follows:(1)if F is a curve,then g (F )≤487;(2)if F is a surface,then p g (F )≤434.Remark 1.2.1)Theorem 1.1was veri?ed by the ?rst author in [8]under the assumption that p g (X )is su?ciently large.
2)When Φ|K X |is generically ?nite,the generic degree is bounded from above by the second author in [11].
3)In the surface case,the corresponding boundedness theorem was proved by Beauville in [1].
4)The numerical bounds in the above theorem might be far from sharp.
2Meng Chen and Christopher D.Hacon
Our second application is an inequality of Noether type between c1 and c2which improves the main theorem of[17].
Theorem1.3.Let X be a nonsingular projective minimal3-fold of general type.Then the following inequality holds:
K3X≥8
3
,or equivalently c31≤
1
3
.
Chen is grateful to De-Qi Zhang for pointing out an inequality(see the proof of Lemma2.1(3)in[19])similar to the one in Proposition2.2 and for an e?ective discussion.
2.Proof of Theorem1.1
Throughout this note,a minimal3-fold X is one with nef canonical divisor K X and with only Q-factorial terminal singularities.
2.1.Notations and the set up.Let X be a minimal projective 3-fold of general type.Since we are discussing the behavior of the canonical map,we may assume p g(X)≥2.Denote by?1the canonical map which is usually a rational map.Take the birational modi?cation π:X′?→X,which exists by Hironaka’s big theorem,such that (i)X′is smooth;
(ii)the movable part of|K X′|is base point free;
(iii)there exists a canonical divisor K X such thatπ?(K X)has sup-port with only normal crossings.
Denote by h the composition?1?π.So h:X′?→W′?P p g(X)?1is a morphism.Let h:X′f?→B s?→W′be the Stein factorization of h. We can write
K X′=π?(K X)+E=S+Z,
where S is the movable part of|K X′|,Z is the?xed part and E is an e?ective Q-divisor which is a sum of distinct exceptional divisors.
If dim?1(X)<3,f is a called an induced?bration of?1.If dim?1(X)=2,a general?ber F of f is a smooth curve C of genus g:=g(C)≥2.If dim?1(X)=1,a general?ber F of f is a smooth projective surface of general type.Denote by F0the smooth minimal model of F and byσ:F?→F0the smooth blow down map.Denote by b the genus of the base curve B.
Proposition2.2.Let V be a smooth projective3-fold of general type with p g(V)>0.Thenχ(ωV)≤p g(V)unless a generic irreducible component in the general?ber of the Albanese morphism is a surface V y with q(V y)=0,in which case one has the inequality
χ(ωV)≤(1+
1
On the geography of Gorenstein minimal3-folds3 Proof.Sinceχ(ωV)=p g(V)+q(V)?h2(O V)?1,the result is clear if q(V)≤1.So assume that q(V)≥2.Let a:V→Y be the Stein factorization of the Albanese morphism V→A(V).By the proof of Theorem1.1in[11],one sees that we may assume that dim Y=1and hence Y is a smooth curve.Recall also that by[11],p g(V)≥χ(a?ωV). Let y∈Y be a general point and V y the corresponding?ber.V y is
a smooth surface of general type.If q(V y)>0,then proceeding as in
[11],one sees thatχ(R1a?ωV)=χ(R1a?ωV/Y?ωY).Since the genus of Y is q(V),and deg R1a?ωV/Y≥0,one sees by an easy Riemann-Roch computation that
χ(R1a?ωV)≥(q(V)?1)q(V y).
Recall that R2a?ωV～=ωY and so
χ(ωV)=χ(a?ωV)?χ(R1a?ωV)+χ(R2a?ωV)≤χ(a?ωV)≤p g(V) whenever q(V y)>0.
We may therefore assume that q(V y)=0.Notice that by[14],the sheaf R1a?ωV is torsion free.Since its rank is given by h1(ωV
y
)= q(V y)=0,we have that R1a?ωV=0.Therefore,by a similar Riemann-Roch computation,one sees thatχ(a?ωV)≥(q(V)?1)p g(V y)and so
χ(ωV)=χ(a?ωV)+q(V)?1≤χ(a?ωV)(1+1
p g(V y)
).
Example2.3.Let S be a minimal surface of general type admitting a Z2action such that q(S)=0,p g(S)=1and p g(S/Z2)=0(cf.(2.6)of [10]).Let C be a curve admitting a?xed point free Z2action and let B=C/Z2.Assume that the genus of B is b≥2.Let V=S×C/Z2 be the quotient by the induced diagonal action.Then V is minimal, Gorenstein of general type such that p g(V)=b?1,q(V)=b and h2(O V)=0.It follows thatχ(ωV)=2b?2=(1+1/p g(V y))p g(V). This example shows that the above proposition is close to being optimal.
Lemma 2.4.Let X be a minimal3-fold of general type.Suppose dim?1(X)=1.Keep the same notations as in2.1.Replaceπ:X′?→X,if necessary,by a further birational modi?cation(we still denote it byπ).Then
π?(K X)|F?
p g(X)?1
4Meng Chen and Christopher D.Hacon
the?bers ofπare rationally connected1and b>0,it follows that f:X′→B factors through a morphism f1:X→B.But since X is minimal and terminal,it follows that a general?ber X b of f1is a smooth minimal surface of general type and hence it can be identi?ed with F0.It is now clear thatπ?(K X)|F～σ?(K F
).
Thus it su?ces to consider the case b=0.
Case2.If p g(X)=2,the lemma was veri?ed in section4(at page 526and page527)in[5].If p g(X)≥3,one may refer to Lemma3.4in [9]. Proposition2.5.Let X be a Gorenstein minimal projective3-fold of general type.Let d:=dim?1(X).The following inequalities hold: (1)If d=2,then K3X≥ 2
p g(X))2K2F
(p g(X)?1).
Proof.The inequality(1)is due to Theorem4.1(ii)in[6].
Suppose now that d=1.We may write
π?(K X)～S+Eπ
where S≡tF with t≥p g(X)?1and Eπis an e?ective divisor. Thus we have
K3X=π?(K X)3≥(π?(K X)2·F)(p g(X)?1)
≥(
p g(X)?1
1Shokurov([18])proved that if the pair(X,?)is klt and the MMP holds,then the?bres of the exceptional locus are always rationally chain connected.Further-more,the second author and M c Kernan(see[12])have recently extended Shokurov’s result to any dimension and without assuming MMP.
On the geography of Gorenstein minimal3-folds5 (1)Assume dim?1(X)= 2.The above argument implies that χ(ωX)≤3
2p g(X)by argument
(**)and Proposition2.2.The Miyaoka-Yau inequality yields K3X≤72χ(ωX)≤108p g(X).Again by Propositions2.2and2.5,we have
K2F
0≤108(
p g(X)
3
p g(X)?
10
2
p g(X)
unless a generic irreducible component in the general?ber of the Al-banese morphism is a surface V y with q(V y)=0and p g(V y)=1.
So in the general case by3.1one has the inequality
K3X≥8
3
or equivalently,
c31≤1
3
.
In the exceptional case with p g(X)>1,the argument(**)in2.6 says that|K X|is composed with a pencil of surfaces and?1generically factors through the Albanese map.Thus X is canonically?bred by surfaces with q(V y)=0and p g(V y)=1.According to Theorem4.1(iii) in[6],one has K3X≥2p g(X)?4.Since by Proposition2.2χ(ωX)≤2p g(X),one has the stronger inequality K3X≥χ(ωX)?4.
6Meng Chen and Christopher D.Hacon
In the exceptional case with p g(X)=1,by Proposition2.2,one has χ(ωX)≤2and so the inequality in Theorem2.5holds.
Case2.p g(X)=0.
We can not rely on3.1in this case.Sinceχ(ωX)>0,one has q(X)>1.Thus we can study the Albanese map.Let a:X?→Y be the Stein factorization of the Albanese morphism.We claim that dim(Y)=1.In fact,if dim(Y)≥2,then the Proof of Theorem1.1of [11]shows p g(X)≥χ(a?ωX)≥χ(ωX)>0,a contradiction.
So we have a?bration a:X?→Y onto a smooth curve Y with g(Y)=q(X)>1.Denote by F a general?ber of a.If p g(F)>0,then the Proof of Theorem1.1of[11]also shows0=2p g(X)≥χ(ωX)>0, which is also a contradiction.Thus one must have p g(F)=0.Because F is of general type,one has q(F)=0.Therefore,the sheaves a?ωX and R1a?ωX have rank h0(ωF)=p g(F)=0and h1(ωF)=q(F)=0. Since,by[14],they are torsion free,it follows that they are both zero. So
χ(ωX)=χ(R2a?ωX)=χ(ωY)=q(X)?1.
Still looking at the?bration a:X?→Y,one sees that a is relatively minimal since X is minimal.Therefore K X/Y is nef by Theorem1.4of [16].Thus one has K3X≥(2q(X)?2)K2F≥2χ(ωX),which is stronger than the required inequality.
4.Examples
In Example2(e)of[4],one may?nd a smooth projective3-fold of general type which is composed with a pencil of surfaces of p g(F)=5, the biggest value among known examples.Here we present another example which is composed with curves of genus g=5.
Example4.1.We follow the Example in§4of[3].We consider bi-double covers f i:C i→E i of curves where,g(E i)=0,0,2.We assume that
(d i)?O C
i =O E
i
⊕L∨i⊕P∨i⊕L∨i?P∨i
where for we have deg(L1)=d1,deg(L2)=d2,deg(P1)=deg(P2)=1 and L3,P3are distinct2-torsion elements in Pic0(E3).In particular g(C i)=2d i?1for i∈{1,2}and f3is′e tale.It follows that
δ:D1×D2×D3→E1×E2×E3
is a Z62cover.We denote by l i,p i,l i p i the elements of Z22whose eigen-sheaves with eigenvalues1are L∨i,P∨i and(L i?P i)∨.Let X:= D1×D2×D3/G where G～=Z42is the group generated by
{(1,p2,l3),(p1,l2,1),(l1,1,p3),(p1,p2,p3)}.
Then one sees that X is Gorenstein and for the induced morphism f:X→E1×E2×E3,one has
f?O X=(δ?O D
1×D2×D3)G～=O E
1
×O E
2
×O E
3
⊕
On the geography of Gorenstein minimal3-folds7 (L∨1?L∨2?P∨2?P∨3)⊕(P∨1?L∨2?L∨3?P∨3)⊕(L∨1?P∨1?P∨2?L∨3). Since f?ωX=ωE
1×E2×E3
?f?O X,it follows easily that
H0(ωX)～=H0(ωE
1?L1)?H0(ωE
2
?L2?P2)?H0(ωE
3
?P3).
In particular p g(X)=(d1?1)(d2?1)and?1factors through the map X→C1/Z2×C2/Z2.The?bers of?1are then isomorphic to C3and hence have genus5.
Question4.2.A very natural open problem is to?nd sharp upper bounds of the invariants of F in Theorem1.1.It is very interesting to ?nd a new example X which is a Gorenstein minimal3-fold of general
type such thatΦ|K
X|is of?ber type and that the generic irreducible
component in a general?ber has larger birational invariants.
We remark that the above question is still open in the surface case. So Question4.2is probably quite di?cult.A?rst step should be to construct new examples with bigger?ber invariants.
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Institute of Mathematics,School of Mathematical Sciences,Fudan University, Shanghai,200433,People’s Republic of China
E-mail address:mchen@https://www.sodocs.net/doc/1915834473.html,
Department of Mathematics,University of Utah,155South1400East,Room 233,Salt Lake City,Utah84112-0090,USA
E-mail address:hacon@https://www.sodocs.net/doc/1915834473.html,