Maps onto Certain Fano Threefolds

We prove that if X is a smooth projective threefold with b 2 = 1 and Y is a Fano threefold with b 2 = 1, then for a non-constant map f : X ! Y , the degree of f is bounded in terms of the discrete invariants of X and Y . Also, we obtain some stronger restrictions on maps between certain Fano threefolds. 1991 Mathematics Subject Classi cation: 14E99, 14J45


Introduction
Let X, Y be smooth complex n-dimensional projective varieties with Pic(X) = Pic(Y ) = Z.Let f : X !Y be a non-constant morphism.It is a trivial consequence of Hurwitz's formula K X = f K Y + R that if Y is a variety of general type, then deg(f) is bounded in terms of the numerical invariants of X and Y , and in particular all the morphisms from X to Y t in a nite number of families.
If we drop the assumption that Y is of general type, then this assertion is no longer quite true.Indeed, if Y is a projective space P n , for any X we can construct in nitely many families of maps X !Y : take an embedding of X in P N by any very ample divisor on X and then project the image to P n .However, one might ask if P n is the only variety with this property (the following conjectures are suggested by A. Van de Ven) : Conjecture A: Let X, Y be as above and Y 6 = P n .Then there is only nitely many families of maps from X to Y .Moreover, the degree of a map f : X !Y can be bounded in terms of the discrete invariants of X and Y .
A weaker version is the following E. Amerik Conjecture B: Let X, Y be smooth n-dimensional projective varieties with b 2 (X) = b 2 (Y ) = 1.Suppose Y 6 = P n and, if n = 1, that Y is not an elliptic curve.Then the degree of a map f : X !Y can be bounded in terms of the discrete invariants of X and Y .
Remark: If n = 1, the Conjecture A is empty and the Conjecture B is trivial.If n = 2, one must check the Conjecture A with Y a K3-surface, and at the moment I do not know how to do this.This problem, of course, does not arise for Conjecture B, which again becomes a triviality in dimension two (note that if for a smooth complex projective variety V we have b 1 (V ) 6 = 0 and b 2 (V ) = 1, then V is a curve).The assumption in the Conjecture B that Y is not an elliptic curve is , of course, necessary: any torus has endomorphisms of arbitrarily high degree given by multiplication by an integer.
Evidence: It seems likely that \the more ample is the canonical sheaf on Y , the more di cult it becomes to produce maps from X to Y ".Of course, the projective space has the \least ample" canonical sheaf: K P n = (n + 1)H, where H is a hyperplane.The next case is that of a quadric: K Qn = nH with H a hyperplane section.For n = 3, it has been proved by C.Schuhmann ( S]) that the degree of a map from a smooth threefold X with Picard group Z to the three-dimensional quadric is bounded in terms of the invariants of X.In A], I have suggested a simpler method to prove results of this kind, which also generalizes to higher dimensions.
The main purpose of this paper is to show by a rather simple method that for Fano threefolds Y , at least for those with very ample generator of the Picard group, the above Conjecture B is true (we also show that for many of such threefolds Conjecture A holds).The boundedness results are proved in the next section.In Section 3, we obtain in a similar way a strong restriction on maps between \almost all" Fano threefolds with Picard group Z.This is related to the \index conjecture" of Peternell which states that if f : X !Y is a map between Fano varieties of the same dimension with cyclic Picard group, then the index of Y is not smaller than that of X.This conjecture is studied for Fano threefolds by C.Schuhmann in her thesis, and one of her main results is that there are no maps from such a Fano threefold of index two to a Fano threefold of index one with reduced Hilbert scheme of lines.An extension of this result appears also in Theorem 3.1 of this paper ; however, there is at least one Fano threefold of index one with non-reduced Hilbert scheme of lines, namely, Mukai and Umemura's V 22 .The last section of this paper deals with this variety: it is proved that a Fano threefold of index two with Picard group Z does not admit a map onto it.One would think that the Mukai-Umemura V 22 is the only Fano threefold of genus at least four with cyclic Picard group and non-reduced Hilbert scheme of lines.The proof of this would be a solution to the \index conjecture" in the three-dimensional case (recall that a Fano threefold of index one and genus at most three has the third Betti number which is bigger than the third Betti number of any Fano threefold of index two ( I1] ,table 3.5), so we do not have to consider the case of genus less than four to prove the index conjecture).In fact even a weaker statement would su ce (see Theorem 3.1).This paper can be viewed as a very extensive appendix to A], as a large part of the method is described there.
We will often use the following notations: Generally, for X P n , H X denotes the hyperplane section divisor on X.Also, for X with cyclic Picard group, we will call H X the ample generator of Pic(X) (in this paper it will mostly be assumed that H X is very ample).By V k , following Iskovskih, we will often denote a Fano threefold with cyclic Picard group, which has index one and for which H 3 X = k (k will be called the degree of this Fano threefold).For Grassmann varieties, we use projective notation: G(k; n) denotes the variety of projective k-subspaces in the projective n-space.
Finally, throughout the paper we work over the eld of complex numbers.with i 3, then for any threefold X, Pic(X) = Z, the degree of a map f : X !Y is bounded in terms of the discrete invariants of X.
Proof: Let m be such that f H Y = mH X .Notice that by Hurwitz' formula, our conditions on U Y resp.S Y just mean that if deg(f) is big enough, then not the whole inverse image of U Y resp.S Y is contained in the rami cation.Indeed, if Y is, say, of index one, we have K Y = H Y .The Hurwitz formula reads K X = mH X + R: If the whole inverse image of S Y is in the rami cation, then R is at least 3 2 mH X , so m cannot get very big.Therefore one gets that the inverse image D of a general (-1,1)-line on Y (in the index-two case) or a general line on Y (in the index-one case) has a reduced irreducible component C. Let Y be a Fano threefold of index two satisfying U Y = iH Y with i 5.For C and D as above, there is a natural morphism and this map must be an isomorphism at a smooth point of D, i.e. at a su ciently general point of C, as C is reduced.Now, also due to the fact that C is reduced, the natural map is a generic surjection.Therefore if we nd an integer j such that T X (j) is globally generated, we must have m j.Such j depends only on the discrete invariants of X.Indeed, let A be a very ample multiple of H X .A linear subsystem of the sections of A gives an embedding of a threefold X into P 7 .We have 2 X is a quotient of 2 P 7 j X , and we deduce from this that 2 X (3A) is generated by the global sections.So T X (K X + 3A) is generated by the global sections, and j can be taken such that K X + 3A = jH X .So one only needs to know which multiple of H X is very ample, and this can be expressed in terms of the discrete invariants of X (see for example D] for many results in this direction).
The case of index one is completely analogous: a normal bundle of any line on a Fano threefold of index one has a negative summand.
Remark A: The assumption on the very ampleness of the generator of Pic(Y ) is not really necessary to prove Proposition 2.1.Otherwise, we call \lines" curves C satisfying C H Y =1.These curves are rational.One has then to count with the possibility that e. g. some of the \lines" on such a Fano 3-fold of index two can have normal bundle O P 1 ( 2) O P 1 (2), but this is not really essential for the argument: as soon as we can nd su ciently big 1-parameter family of smooth rational curves with a negative summand in the normal bundle, our method works.
Examples of Fano threefolds Y satisfying our assumptions on S Y , U Y : 1) Y a cubic in P 4 and 2) Y an intersection of two quadrics in P 5 .To check this is more or less standard and almost all details can be found in CG] for a cubic and in GH] (Chapter 6) for an intersection of two quadrics.For convenience of the reader, we give here the argument for Y an intersection of two quadrics in P 5 : Let F G(1; 5) be a surface which parametrizes lines on Y (Fano surface) , and let U ! F be the family of these lines.The rami cation locus of the natural nite map U ! Y consists exactly of (-1,1)-lines, that is, the surface M covered by (-1,1)-lines on Y is exactly the set of points of Y through which there pass less than four lines (of course there are four lines through a general point of Y ).F is the zero-scheme of a section of the bundle S 2 U S 2 U on G(1; 5).A standard computation with Chern classes yields then that K F = O F (in fact, F is an abelian variety ( GH])).
For a general line l Y consider a curve C l F which is the closure in F of lines intersecting l and di erent from l. C l contains l i l is (-1,1).C l is smooth for any l ( GH]).By adjunction, C l has genus 2. So the rami cation R of the natural 3:1 morphism h l : C l !l sending l 0 to l \ l 0 ( with l general, i.e. not a (-1,1)-line) has degree 8.The branch locus of h consists of intersection points of l and the surface M of (-1,1)-lines, and so we have that this surface is in jiH Y j with i 4 and i = 4 only if there are only 2 lines through a general point of M. This is again impossible: otherwise, for l a (-1,1)-line, C l would be birational to l.In fact, one gets that i = 8.
3) Y a quartic double solid.The computations are rather similar, and the best reference is W].Bitangent lines to the quartic surface give pairs of \lines" on Y as their inverse images under the covering map.hyperplane section of such a fourfold has reduced Hilbert scheme of lines.
5) Y any Fano threefold of index one and genus 10: Prokhorov shows in P] that the Hilbert scheme of lines on any such threefold is reduced.
6) Y any Fano threefold V 14 of index one and genus 8: such a threefold is a linear section of G(1; 5) in the Pl ucker embedding.Iskovskih shows in I], II, proof of th.6.1 (vi), that on such a threefold with reduced scheme of lines, lines will cover a surface which is linearly equivalent to 5H.So one sees that if the lines cover only H or 2H, the scheme of lines is non-reduced and the surface covered by lines consists of one or two components which are hyperplane sections of Y .Moreover, as a V 14 does not contain cones, all the lines in one of the components must be tangent to some curve A. One checks easily that this curve is a rational normal octic.A is then the Gauss image of a rational normal quintic B in P 5 ( A], proof of Proposition 3.1(ii)).This makes it possible to check that there is no smooth three-dimensional linear section of G(1; 5) containing the tangent surface to A. Indeed, one can assume that B is given as (x 5 0 : x 4 0 x 1 : ::: : x 5 1 ); (x 0 : x 1 ) 2 P 1 ; one computes then that the Gauss image of B in G(1; 5) P 14 (where G(1; 5) is embedded to P 14 by Pl ucker coordinates (z i ), the order of which we take as follows: for a line l through p = (p 0 : ::: : p 5 ) and q = (q 0 : ::: : q 5 ) we take z 0 = p 0 q 1 p 1 q 0 ; z 1 = p 0 q 2 p 2 q 0 ; ...; z 5 = p 1 q 2 p 2 q 1 ; ...; z 14 = p 4 q 5 p 5 q 4 ) generates the linear subspace L given by the following equations: z 2 = 3z 5 ; z 3 = 2z 6 ; z 4 = 5z 9 ; z 7 = 3z 9 ; z 8 = 2z 10 ; z 11 = 3z 12 : So we must consider all the projective 9-subspaces through L and prove that the intersection of every such space with G(1; 5) is singular.This can be done for example as follows: let L = P 5 be a parametrizing variety for these 9-subspaces.Notice that the points x = (1 : 0 : ::: : 0) and y = (0 : ::: : 0 : 1) belong to our curve A. Notice that if t is a point of A, then the set L t = fM 2 L : M \ G(1; 5) is singular at tg is a hyperplane in L. If we see that these sets are di erent at di erent points t, we are done.It is not di cult to check explicitly (writing down the matrix of partial derivatives) that for x = (1 : 0 : ::: : 0) 2 A and y = (0 : ::: : 0 : 1) 2 A, L x 6 = L y : if a 9-space M through L is given by the equations a 1i (z 2 3z 5 ) + a 2i (z 3 2z 6 ) + a 3i (z 7 3z 9 )+ +a 4i (z 8 2z 10 ) + a 5i (z 11 3z 12 ) + a 6i (z 4 5z 9 ) = 0 for i = 1; :::; 5, then M 2 L x if and only if det(a ki ) i=1;2;3;4;5 k=1;2;3;4;6 = 0 and M 2 L y if and only if det(a ki ) i=1;2;3;4;5 k=1;2;3;4;5 = 0: These conditions are clearly di erent.

Documenta Mathematica 2 (1997) 195{211
Examples of Fano threefolds not satisfying assumptions of Proposition 2.1: 1) Y is a linear section of G(1; 4) in the Pl ucker embedding: the surface U Y has degree 10.
2) Y is a Fano variety of index one and genus 12 (V 22 ).The surface of lines belongs to j 2K Y j for all V 22 's but one ( P]), for which the scheme of lines is non-reduced and the surface covered by lines belongs to j K Y j.This threefold with non-reduced Hilbert scheme of lines (the Mukai-Umemura variety) will be denoted V s 22 .
Question: Are these the only examples?
Remark B: Though any V 22 violates the assumption of the Proposition 2.1, for a V 22 with the reduced Hilbert scheme of lines (therefore for all V 22 's but one) the boundedness of the degree of a map f : X !V 22 can be proved.The point is that a general line on such a V 22 has the normal bundle O P 1 O P 1 ( 1), so if U is the universal family of lines on V 22 and : U ! V 22 is the natural map, then is an immersion along a general line.Now if the preimage of a general line l is not contained in the rami cation R, one can proceed as before.If it is, then let C be the reduction of an irreducible component of f 1 (l), and let k be such that at a general point of the component of R containing C, the rami cation index is k 1 (i.e.\k points come together".)It turns out that using our observation about , we can then estimate the arithmetic genus of C (see A], section 5).Namely, let f H V22 = mH X and let K X = rH X .We get then 2p a (C) 2 (r m k )CH X : Suppose now that k 1 is a smallest rami cation index for R. Hurwitz' formula implies that if r < m 3 , then k = 2.So if m gets big, p a (C) becomes negative, and this is impossible.
Concerning the remaining Fano threefolds (in particular, V s 22 and G(1; 4) T P 6 ), we can prove a weaker result (as in Conjecture B): Proposition 2.2 Let Y be a Fano threefold with Pic(Y ) = Z and with H Y very ample, let X be a smooth threefold with b 2 (X) = 1 and let f : X !Y be a morphism.
If either Y is of index two, or Y is of index one with non-reduced Hilbert scheme of lines, then the degree of f is bounded in terms of the discrete invariants of X.
Proof: Consider for example the index one case.We have that Y has a one-dimensional family of ( 2; 1)-lines.If we take a smooth hyperplane section H through a line l of this family, the sequence of the normal bundles Therefore, if M is the inverse image of H and C is the inverse image of l (schemetheoretically), the sequence also splits.
It is not di cult to see that for a general choice of l and H, the surface M has only isolated singularities.As M is a Cartier divisor on a smooth variety X (say M 2 jO X (m)j), M is normal.
Now we are in the situation which is very similar to that of the following Theorem (R. Braun, B]): Let W be a Cartier divisor on a variety V of dimension n, 2 n < N, in P N such that W has an open neighborhood in V which is locally a complete intersection in P N .If the sequence of the normal bundles 0 !N W;V !N W;P N !N V;P N j W ! 0 ( ) splits, then W is numerically equivalent to a multiple of a hyperplane section of V .
It turns out that if we replace here W, V , P N by C, M, X as in our situation, the similar statement is true.The only additional assumption we must make is that M is su ciently ample, i.e. m is su ciently big: Claim: Let X be a smooth projective 3-fold with b 2 (X) = 1, and let M be a su ciently ample normal Cartier divisor on X.If C is a Cartier divisor on M and the sequence 0 !N C;M !N C;X !N M;X j C ! 0 splits, then C is numerically equivalent to a multiple of H X j M .
The proof of this claim is almost identical to that of Braun's theorem (which is itself a re nement of the argument of EGPS] where the theorem is proved for V a smooth surface).Recall that the main steps of this proof are: 1) The sequence ( ) splits i W is a restriction of a Cartier divisor from the second in nitesimal neighborhood V 2 of V in P N ; 2)The image of the natural map Pic(V 2 ) ! Num(V ) is one-dimensional.
In the situation of the lemma, 1) goes through without changes with W, V , P N replaced by C, M, X (M 2 will of course denote the second in nitesimal neighborhood of M in X).The second step is an obvious modi cation of that in B], EGPS]: as in these works, it is enough to prove that the image of the natural map Pic(M 2 ) ! H 1 (M; 1 M ) is contained in a one-dimensional complex subspace, and this follows from the commutative diagram restr: ? dlog ----* and the fact that for su ciently ample M, as follows from the restriction exact sequence 0 ! 1 X ( M) ! 1 X ! 1 X j M !0: Note that we can give an e ective estimate for \su cient ampleness" of M in terms of numerical invariants of X using the Gri ths vanishing theorem ( G]).Applying this to our situation of a map onto a Fano threefold Y of index one with non-reduced Hilbert scheme of lines, we get that C = f 1 (l) must be numerically equivalent to a multiple of the hyperplane section divisor on M = f 1 (H) if the number m (de ned by f H Y = mH X ) is large enough.As it is easy to show that C and the hyperplane section of M are independent in Num(M), it follows that m and therefore deg(f) must be bounded.The case of index two is exactly the same (use the existence of a divisor covered by (-1,1)-lines).So the Proposition is proved.
We summarize our results in the following Notice that in the rst three cases it is su cient that Pic(X) = Z.

Maps between Fano threefolds
It turns out that we obtain especially strong bound if X is also a Fano variety.In many cases,this even implies non-existence of maps: Before starting the proof, we formulate the following result from S]: Let f : X !Y be a non-trivial map between Fano threefolds with Picard group Z.
Then: A) If X,Y are of index two, then the inverse image of any line is a union of lines; B) If X,Y are of index one, then the inverse image of any conic is a union of conics; C) If X is of index one and Y is of index two, then the inverse image of any line is a union of conics; D) If X is of index two and Y is of index one, then the inverse image of any conic is a union of lines.
(here a conic is allowed to be reducible or non-reduced.Unions of lines and conics are understood in set-theoretical sense, i.e. a line or a conic from this union can, of course, have a multiple structure.) We will also need some facts on conics on a Fano threefold V of index one, with very ample K V and cyclic Picard group.Iskovskih proves ( I],II, Lemma 4.2) that if C is a smooth conic on such a threefold, then N C;V = O P 1 ( a) O P 1 (a) with a equal to 0,1,2 or 4. The following lemma is an almost obvious re nement of this: Lemma 3.2 a) Let C V be a smooth conic.Then N C;V = O P 1 ( 4) O P 1 (4) if and only if there is a plane tangent to V along C. In particular, such conics exist only if V is a quartic.b) Let C V be a reducible conic: C = l 1 S l 2 , l 1 6 = l 2 .Let N be the (locally free with trivial determinant) normal sheaf of C in V .Then Nj li = O P 1 ( a i ) O P 1 (a i ) with 0 a i 2, and if a i = 2 for both i, then there is a plane tangent to V along C (and V is a quartic ).
Proof: a) This is a simple consequence of the fact that for C V P n , N C;V N C;P n, and the only subbundle of degree 4 in N C;P n is N C;P with P the plane containing C. One concludes that V is a quartic as all the other Fano threefolds V considered here are intersections of quadrics and cubics which contain this V ( I], II, sections 1,2) and therefore must contain this P, which is impossible.
b) We have embeddings 0 !N li;V !Nj li ; this implies the rst statement: 0 a i 2. If a i = 2, then l i should be a (-2,1)-line; therefore there are planes P i tangent to V along l i , giving the degree 1 subbundle of N li;V and the exceptional section in P(N li;V ) = F 3 .In fact P 1 = P 2 .This is easy to see using so-called \ elementary modi cations" of Maruyama (of which I learned from AW] ,p.11): if we blow P(N l1;V ) up in the point p corresponding to the direction of l 2 and then contract the proper preimage of the ber, we will get P(Nj l1 ).Under our circumstances, p must lie on the exceptional section of P(N l1;V ), so l 2 P 1 .In the same way, l 1 P 2 , q.e.d..

Proof of the Theorem:
Let f : X !Y be a nite map between Fano threefolds as above.
Again, the condition on S Y , T Y means that not the whole inverse image of S Y , T Y can be contained in the rami cation.If Y is of index one resp.index two, we will denote by C be a reduced irreducible component of the inverse image of a general line C;D , supported on intersection points of C and other components of D. We see from our assumptions that it must have length one (so be supported at one point x).Suppose that C intersects non-reduced components of D at x. Let A be a local ring of D at x and p A a ber of I C;D .Of course p=p 2 6 = 0 by Nakayama.
To see that dimp=p 2 2, we nd an ideal a p, not contained in p 2 .For example, we can take an ideal de ning the union of C and the reduction of an irreducible but non-reduced component of D intersecting C. We have a surjection p=p 2 !(p=a)=(p 2 =(p 2 \ a)) !0; which has non-trivial (again by Nakayama) image and non-trivial kernel, q. e. d.. Corollary 3.3 Let X, Y be Fano threefolds of index one as in Theorem 3.1 i).Then any map between X and Y is an isomorphism.
Proof: Iskovskih computed the third Betti numbers of all Fano threefolds ( see also M]), and in fact as soon as deg(X) > deg(Y ), then b 3 (X) < b 3 (Y ), so a morphism f : X !Y cannot exist.
Remark C: Some part of the argument of Theorem 3.1 goes through without assumptions on the very ampleness of the generator H of the Picard group.For example, when X is a quartic double solid, which is a Fano threefold of index two, all the \lines" C on X except possibly a nite number, have either trivial normal bundle, or the normal bundle O C (H) O C ( H) (in other words, the surface which parametrizes lines on X, has only isolated singularities).One can then replace the words \T X (H) is globally generated", which are not true in general, by some \normal bundle arguments" as in the above proof.The same should hold for the Veronese double cone.See W], T] for details.As for maps to the quartic double solid, the argument goes through without changes.
Examples: Any cubic in P 4 satis es the assumption we made on Y .By our Theorem 3.1 , we get that if a Fano threefold X of index one with cyclic Picard group is mapped onto a cubic, then the degree of this map can only be degX 3 .So if X admits such a map, then deg(X) is divisible by 3. Of course there are Fano threefolds of index one which admit a map onto a cubic: we can take an intersection of a cubic cone and a quadric in P 5 .Theorem 3.1 shows that if a smooth complete intersection of type (2,3) in P 5 maps to a cubic, then it is contained in a cubic cone and the map is the projection from the vertex of this cone.The same applies of course to maps from a complete intersection of three quadrics in P 6 to a complete intersection of two quadrics in P 5 .Notice that any smooth complete intersection of two quadrics in P 5 admits a map g onto a quadric in P 4 such that the inverse image of the hyperplane section is the hyperplane section (any pencil of quadrics with non-singular base locus contains a quadratic cone).Therefore if a smooth intersection of three quadrics in P 6 can be mapped onto a smooth complete Documenta Mathematica 2 (1997) 195{211 intersection of two quadrics in P 5 , it must lie in a quadric of corank 2 in P 6 .Of course a general intersection of three quadrics in P 6 does not have this property, as the space of quadrics of corank 2 is of codimension four in the space of all quadrics.
Additional examples of varieties satisfying the assumption of Theorem 3.1: 1) any complete intersection of a cubic and a quadric in P 5 or 2) any complete intersection of three quadrics in P 6 .Indeed, if lines on these varieties cover only a hyperplane section divisor, then the scheme of lines must be non-reduced, i.e. each line must have normal bundle O P 1 ( 2) O P 1 (1).So the surface of lines is either a cone or the tangent surface to a curve.But one can check that these varieties do not contain cones; neither do they contain a tangent surface to a curve as a hyperplane section, because by a version of Zak's theorem on tangencies (see for example FL]), a hyperplane section of a complete intersection has only isolated singularities.
3) Any V 22 with reduced Hilbert scheme of lines.By ( P]), there is exactly one V 22 such that its Hilbert scheme of lines is non-reduced.4) any Fano threefold V 16 of index one and genus 9.This can be shown by the method of Prokhorov ( P]) : First, notice that if the lines on V 16 cover only a hyperplane section, the scheme of lines is non-reduced.So all the lines are tangent to a curve.This is actually a rational normal curve, so the lines never intersect.For convenience of the reader, we recall from I2] the notion of double projection from a line and its application to V 16 : Let X be a Fano threefold of index one, g(X) 7, and let l be a line on X.On X, the blow-up of X, we consider the linear system j H 2Ej, where is the blow-up, H = K Y and E is the exceptional divisor.This is not base-point-free, namely, its base locus consists of proper preimages of lines intersecting l, and, if l is (-2,1), from the exceptional section of the ruled surface E = F 3 .However, after a op (i.e. a birational transformation which is an isomorphism outside this locus) we can make it into a base-point-free system j( H) + 2E + j on the variety X+ .
If g(X) = 9, i.e.X is a V 16 , the variety X+ is birationally mapped by this linear system onto P 3 .This map, say g, is a blow-down of the surface of conics intersecting l to a curve Y P 3 , which is smooth of degree 7 and genus three (smoothness of Y is obtained from Mori's extremal contraction theory).Y lies on a cubic surface which is the image of E + .Moreover, the inverse rational map from P 3 to X is given by the linear system j7H 2Y j.
One has therefore that the lines from X, di erent from l, must be mapped by g to trisecants of Y .Note that if lines on X form only a hyperplane section, the desingularization of the surface of lines on X is rational ruled, and it remains so after the blow-up and the op.So, as in P], we must have a morphism F e !P 3 , which is given by some linear system jC + kFj with C the canonical section and f a ber, such that the inverse image of Y belongs to the system j3C + lFj.deg(Y ) = 7 implies (3C + lF)(C + kF) = 3e + 3k + l = 7; Documenta Mathematica 2 (1997) 195{211 such that L 4 1 + ::: + L 4 6 = F where F = 0 de nes C; \the variety of polar hexagons" means here the closure of the set of 6-tuples l 1 ; :::l 6 with L 4 1 + ::: + L 4 6 = F in the Hilbert scheme Hilb 6 (P 2 ); a general V 22 is isomorphic to such a variety for a certain curve C; V s 22 is the variety of polar hexagons of a double conic).Then there is a birational isomorphism between (P 2 ) and T given as follows: for a general l P 2 the curve of polar hexagons to C containing l is a conic on V 22 .This description and the fact that through any point on a V 22 there is only a nite number of conics ( I], II, Theorem 4.4) gives that there are six conics through a general point of V 22 .
In M], Mukai claims that the Fano surface of conics on a V 22 is even isomorphic to P 2 .Unfortunately, this paper does not contain a proof of this fact.The proof appears in the paper of A. Kuznetsov ( K]): he uses another description of a general V 22 as a subvariety of G(2; 6).Namely, if V and N are 7-and 3-dimensional vector spaces respectfully and f : N ! 2 V is a general net of skew-symmetric forms on V , then a general V 22 (including V s 22 , Sch]) appears as a set of all 3-subspaces of V which are isotropic with respect to this net (i.e. to all forms of the net simultaneously).Let U (resp.Q) denote restriction on a V 22 of the universal (resp.universal quotient) bundle on G(2; 6).Kuznetsov proves that every (possibly singular) conic on a V 22 is a degeneracy locus of a homomorphism U ! Q ; the Fano surface of conics is thus P(Hom(U; Q )) = P 2 .Now if there is a nite map f : X !V 22 as above, then X must be a cubic: indeed, a Fano threefold with cyclic Picard group and with 6 lines through a general point is a cubic.Let f H V22 = mH X , then one easily computes that the inverse image of a general conic consists of deg(g) = s = 3 11 m 2 lines.For simplicity, we will use the same notation for points of T (resp.S) and corresponding conics on V 22 (resp.lines on X).We have T = P 2 .Let a be such that conics on V 22 intersecting a given (general) conic A, form a divisor D A from jO P 2 (a)j On S, denote as E L the divisor of lines intersecting a given line L. It is well-known and easy to compute that E L E M = 5 for any L; M.
Documenta Mathematica 2 (1997) 195{211 Theorem 2.3 Let X be a smooth projective threefold with b 2 (X) = 1, let Y be a Fano threefold with b 2 (Y ) = 1 and very ample H Y and let f : X !Y be a morphism.If Y P 3 , then the degree of f is bounded in terms of the discrete invariants of X; Y .Proof: Indeed, there are only four possibilities: a) Y is a quadric: this is proved in S], A]. b) Proposition 2.1 applies; c) Y is V 22 with reduced scheme of lines: the boundedness for deg(f) is obtained in Remark B; d) Y is either G(1; 4) \ P 6 , or a Fano threefold with non-reduced Hilbert scheme of lines: then Proposition 2.2 applies.
Welters proves the following results:the Fano surface F Y has only isolated singularities (and is smooth for a general Y ); the curve C l for a general l is smooth except for one double point; there are 12 \lines" through a general point of Y ; p a (C l ) = 71.We use these results to conclude that Y