IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME OF SMOOTH CURVES

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1 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME OF SMOOTH CURVES CHANGHO KEEM*, YUN-HWAN KIM AND ANGELO FELICE LOPEZ** Abstract. Denote by H d,g,r the Hilbert scheme of smooth curves, that is the union of components whose general point corresponds to a smooth irreducible and non-degenerate curve of degree d and genus g in P r. A component of H d,g,r is rigid in moduli if its image under the natural map π : H d,g,r M g is a one point set. In this note, we provide a proof of the fact that H d,g,r has no components rigid in moduli for g > 0 and r =. In case r 4, we also prove the non-existence of a component of H d,g,r rigid in moduli in a certain restricted range of d, g > 0 and r. In the course of the proofs, we establish the irreducibility of H d,g, beyond the range which has been known before. 1. Basic set up, terminologies and preliminary results Given non-negative integers d, g and r, let H d,g,r be the Hilbert scheme parametrizing curves of degree d and genus g in P r and let H d,g,r be the Hilbert scheme of smooth curves, that is the union of components of H d,g,r whose general point corresponds to a smooth irreducible and non-degenerate curve of degree d and genus g in P r. Let M g be the moduli space of smooth curves of genus g and consider the natural rational map π : H d,g,r M g which sends each point c H d,g,r representing a smooth irreducible non-degenerate curve C in P r to the corresponding isomorphism class [C] M g. In this article, we concern ourselves with the question regarding the existence of an irreducible component Z of H d,g,r whose image under the map π is just a one point set in M g, which we call a component rigid in moduli. It is a folklore conjecture that such components should not exist, except when g = 0. It is also expected [9, 1.47] that there are no rigid curves in P r, that is curves that admit no deformations other than those given by projectivities of P r, except for rational normal curves. In the next two sections, we provide a proof of the fact that H g+1,g, is irreducible and H d,g, does not have a component rigid in moduli if g > 0. In the subsequent section we also prove that, for r 4, H d,g,r does not carry any component rigid in moduli in a certain restricted range with respect to d, g > 0 and r. In proving the results, we utilize several classical theorems including the so-called Accola- Griffiths-Harris bound on the dimension of a component consisting of birationally Key words and phrases. Hilbert scheme, algebraic curves, linear series, gonality. * Supported in part by NRF Grant # ** Supported in part by the MIUR national project Geometria delle varietà algebriche PRIN Mathematics Subject Classification: Primary 14C0, 14C0. 1

2 C. KEEM, Y.H. KIM AND A.F. LOPEZ very ample linear series in the variety of special linear series on a smooth algebraic curve. We work over the field of complex numbers. For notations and conventions, we usually follow those in []; e.g. π(d, r) is the maximal possible arithmetic genus of an irreducible and non-degenerate curve of degree d in P r. Before proceeding, we recall several related results that are rather well known; cf. [1]. For any given isomorphism class [C] M g corresponding to a smooth irreducible curve C, there exist a neighborhood U M g of [C] and a smooth connected variety M which is a finite ramified covering h : M U, together with varieties C, Wd r and Gd r which are proper over M with the following properties: (1) ξ : C M is a universal curve, that is for every p M, ξ 1 (p) is a smooth curve of genus g whose isomorphism class is h(p), () Wd r parametrizes pairs (p, L), where L is a line bundle of degree d with h 0 (L) r + 1, () Gd r parametrizes couples (p, D), where D is possibly an incomplete linear series of degree d and of dimension r, which is denoted by gd r, on ξ 1 (p). Let G be the union of components of Gd r whose general element (p, D) corresponds to a very ample linear series D on the curve C = ξ 1 (p). Note that the open subset of H d,g,r consisting of points corresponding to smooth curves is a PGL(r+1)-bundle over an open subset of G. We also make a note of the following fact which is basic in the theory; cf. [1] or [8, Chapter ]. Proposition 1.1. There exists a unique component G 0 of G which dominates M (or M g ) if the Brill-Noether number ρ(d, g, r) := g (r + 1)(g d + r) is nonnegative. Furthermore in this case, for any possible component G of G other than G 0, a general element (p, D) of G is such that D is a special linear system on C = ξ 1 (p). Remark 1.. In the Brill-Noether range, that is ρ(d, g, r) 0, the unique component G 0 of G (and the corresponding component H 0 of H d,g,r as well) which dominates M or M g is called the principal component. We call the other possible components exceptional components. We recall the following well-known fact on the dimension of a component of the Hilbert scheme H d,g,r ; cf. [8, Chapter.a] or [9, 1.E]. Theorem 1.. Let c H d,g,r be a point representing a curve C in P r. The tangent space of H d,g,r at c can be identified as T c H d,g,r = H 0 ( C, N C/P r), where N C/P r is the normal sheaf of C in P r. Moreover, if C is a locally complete intersection, in particular if C is smooth, then χ(n C/P r) dim c H d,g,r h 0 ( C, N C/P r), where χ(n C/P r) = h 0 ( C, N C/P r) h 1 ( C, N C/P r). For a locally complete intersection c H d,g,r, we have χ(n C/P r) = (r + 1)d (r )(g 1)

3 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME which is denoted by λ(d, g, r). The following bound on the dimension of the variety of special linear series on a fixed smooth algebraic curve shall become useful in subsequent sections. Theorem 1.4 (Accola-Griffiths-Harris Theorem; [8, p.7] ). Let C be a curve of genus g, D a birationally very ample special gd r, that is a special linear system of dimension r and degree d inducing a birational morphism from C onto a curve of degree d in P r. Then in a neighborhood of D on J(C), either dim Wd r (C) = 0 or { dim Wd r (C) h 0 d r + 1 if d g (O C (D)) r d r g + 1 if d g. where J(C) denotes the Jacobian variety of C. We will also use the following lemmas that are a simple application of the dimension estimate of multiples of the hyperplane linear system on a curve of degree d in P r ; cf. [, p.11] or [8, Chapter.a]. Lemma 1.. Let r and let C be a smooth irreducible non-degenerate curve of degree d and genus g in P r. Then r { d+1 1 if d g. (d g + 1) if d g Proof. Set m = d 1 d+1 r 1. Suppose d g and assume that r >, so that 1 m. If m = 1 we have g π(d, r) = d r g r, a contradiction. If m = we get g π(d, r) = d r + 1 < d g, a contradiction. Then m = and g π(d, r) = d 6r + d 1 < g, again a contradiction. Therefore r d+1 when d g. Suppose now d g, so that d > g and H 1 (O C ()) = 0. By [, p.11] we have h 0 (O C ()) r, whence, by Riemann-Roch, d g + 1 r, that is r 1 (d g + 1). In the next result we will use the second and third Castelnuovo bounds π 1 (d, r) and π (d, r). Lemma 1.6. Let r 4 and let C be a smooth irreducible non-degenerate curve of degree d and genus g in P r. Assume that either (i) d r + 1 and g > π 1 (d, r) or (ii) C is linearly normal, r 8, d r + and either g > π (d, r) or g = π 1 (d, r). Then C admits a degeneration {C t P r } t P 1 to a singular stable curve. Proof. Under hypothesis (i) it follows by [8, Theorem.1] that C lies on a surface of degree r 1 in P r. Under hypothesis (ii) it follows by [17, Theorem.10] or [8, Theorem.1] respectively, that C lies on a surface of degree r 1 or r in P r. To do the case of the surface of degree r 1, we recall the following notation (see [10,.]). Let e 0 be an integer and let E = O P 1 O P 1( e). On the ruled surface X e = PE, let C 0 be a section in O PE (1) and let f be a fiber. Then any curve D ac 0 +bf has arithmetic genus 1 (a 1)(b ae ). Write r 1 = n e for some n e and let S n,e be the image of X e under the linear system C 0 + nf.

4 4 C. KEEM, Y.H. KIM AND A.F. LOPEZ This linear system embeds X e when n > e, while, when n = e, it contracts C 0 to a point and is an isomorphism elsewhere, thus S e,e is a cone. As is well-known (see for example [8, Proposition.10]), every irreducible surface of degree r 1 in P r is either the Veronese surface v (P ) P or an S n,e. If C v (P ) then C aq where Q is a conic and d = a 11, so that a 6. Let C 1 be general in (a 1)Q and let C be a general in Q. Now C 1 and C are smooth irreducible, C 1 C = a 1 and they intersect transversally. Therefore C specializes to the singular stable curve C 1 C. Now suppose that C S n,e. We deal first with the case n > e. We have C ac 0 + bf for some integers a, b such that a (because g ) and, as C is smooth irreducible, we get by [10, Corollary V..18(b)] that either e = 0, b (if b = 1 then g = 0) or e > 0, b ae. Consider first the case a. If e = 0 or if e > 0 and b > ae, let C 1 be general in C f and C be a general fiber. As above C 1 and C are smooth irreducible, intersect transversally and C 1 C = a. Therefore C 1 C is a singular stable curve and C specializes to it. If e > 0 and b = ae let C 1 be general in C 0 + ef and C be general in (a 1)(C 0 + ef). As above both C 1 and C are smooth irreducible, intersect transversally and C 1 C = e(a 1) unless a =, e = 1, b =, which is not possible since then g = 1. Therefore C 1 C is a singular stable curve and C specializes to it. Now suppose that a =. Then g = b e 1 and therefore b e +. Let C 1 = C 0 and let C be general in C 0 + bf. As above C 1 and C are smooth irreducible, intersect transversally and C 1 C = b e. Therefore C 1 C is a singular stable curve and C specializes to it. This concludes the case n > e. If n = e let C be the strict transform of C on X e. Then C = C ac 0 + bf for some integers a, b. We get d = C (C 0 +ef) = b and, as C is smooth, d ae = C C 0 = 0, 1. Since e = r 1, setting η = 0, 1 we get d = a(r 1) + η. Note that if a we have d r 1, a contradiction. Hence a. If η = 1 let C 1 be general in C f and C be a general fiber. As above C 1 and C are smooth irreducible, intersect transversally and C 1 C = a. Therefore C 1 C is a singular stable curve and C specializes to it. On the other hand C, C 1 and C get mapped isomorphically in P r, therefore also C specializes to a singular stable curve. If η = 0 let C 1 be general in C 0 + ef and C be general in (a 1)(C 0 + ef). Again C 1 and C are smooth irreducible, intersect transversally, C1 C = (a 1)(r 1) 6 and they get mapped isomorphically in P r, therefore C specializes to a singular stable curve image of C 1 C. This concludes the case of the surface of degree r 1. We now consider the case r 8, d r +, C is linearly normal and lies on a surface of degree r in P r. By a classical theorem of del Pezzo and Nagata (see [16, Thorem 8]) we have that such a surface is either a cone over an elliptic normal curve in P r 1 or the -Veronese surface v (P ) P 9 or the image of X e, e = 0, 1, with the linear system C 0 + f, C 0 + f or C 0 + 4f respectively. The case v (P ) is done exactly as the case v (P ) above, while the cases e = 0, 1 are done exactly as the case S n,e, n > e above, since the linear systems C 0 + f, C 0 + f are very ample and therefore a degeneration on X e gives a degeneration in P 8. In the case e = let C be the strict transform of C on X. Then C = C ac 0 + bf for some integers a, b. We get d = C (C 0 + 4f) = b and, as C is smooth,

5 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME b a = C C 0 = η. Also if a we have d 10, a contradiction. Hence a. Now exactly as in the case n = e above we conclude that C specializes to a singular stable curve. It remains to do the case when C is contained in the cone over an elliptic normal curve in P r 1. Let E P r 1 be a linearly normal smooth irreducible elliptic curve of degree r, set E = O E O E ( 1) and let π : PE E be the standard map. By [10, Example V..11.4] the cone is the image of PE under the linear system C 0 + π O E (1), which contracts C 0 to the vertex and is an isomorphism elsewhere. In particular it follows that C 0 + π O E (1) is big and base-point free. Let C be the strict transform of C on PE, so that C = C ac 0 + π M for some integer a and some line bundle M on E of degree b. As before we have d = C (C 0 + rf) = b and, as C is smooth, d ar = C C 0 = η, so that a. Assume that η = 0. We claim that M = O E (a). In fact if M = OE (a) we compute h 0 (PE, ac 0 + π M) = h 0 (E, π (ac 0 + π M)) = while = h 0 ((Sym a E) M) = h 0 ( a a 1 M( i)) = (a i)r i=0 a 1 h 0 (PE, (a 1)C 0 + π M) = h 0 ( M( i)) = i=0 i=0 a 1 (a i)r and therefore the linear system ac 0 + π M has C 0 as base component. But C is irreducible and C C 0, whence a contradiction. Hence M = O E (a) and C a(c 0 + π O E (1)). Let C 1 be general in C 0 + π O E (1) and let C be general in (a 1)(C 0 + π O E (1)). Now C 1 and C are smooth irreducible by Bertini s theorem, intersect transversally and C 1 C = (a 1)r 16. Therefore C 1 C is a singular stable curve and C specializes to it. On the other hand C, C 1 and C get mapped isomorphically in P r, therefore also C specializes to a singular stable curve. Finally let is do the case η = 1. Then M( a) has degree 1 and thefore there is a point P E such that M = O E (a)(p ). Let F = π (P ) be a fiber and let C 1 be general in C F. Note that C F a(c 0 + π O E (1)) and therefore C 1 is smooth irreducible. Again C 1 and C intersect transversally and C 1 C = a. Hence C 1 C is a singular stable curve and C specializes to it. Also C, C 1 and C get mapped isomorphically in P r, therefore also C specializes to a singular stable curve. i=0. Irreducibility of H g+1,g, for small genus g The irreducibility of H g+1,g, has been known for g 9; cf. [11, Theorem.6 and Theorem.7 ]. In this section we prove that any non-empty H g+1,g, is irreducible of expected dimension for g 8, whence for all g without any restriction on the genus g. Proposition.1. H g+1,g, is irreducible of expected dimension 4(g+1) if g 6 and is empty if g. Moreover, dim π(h g+1,g, ) = g if g 8, dim π(h 8,7, ) = 17 and dim π(h 7,6, ) = 1.

6 6 C. KEEM, Y.H. KIM AND A.F. LOPEZ Proof. By the Castelnuovo genus bound, one can easily see that there is no smooth non-degenerate curve in P of degree g + 1 and genus g if g. Hence H g+1,g, is empty for g. We now treat separately the other cases. (i) g = 6: A smooth curve C of genus 6 with a very ample g7 is trigonal; K g7 = g. 1 Furthermore, C has a unique trigonal pencil by Castelnuovo-Severi inequality and the g7 is unique as well. Conversely a trigonal curve of genus 6 has a unique trigonal pencil and the residual series g7 = K g 1 is very ample which is the unique g7. Hence G G7 is birationally equivalent to the irreducible locus of trigonal curves M 1 g,. Therefore it follows that H 7,6, is irreducible which is a PGL(4)- bundle over the irreducible locus M 1 g, and dim H 7,6, = dim M 1 g, +dim PGL(4) = (g + 1) + 1 = 8. (ii) g = 7: First we note that a smooth curve C of degree 8 in P of genus 7 does not lie on a quadric surface; there is no integer solution to the equation a + b = 8, (a 1)(b 1) = 7 assuming C is of type (a, b) on a quadric surface. We then claim that C is residual to a line in a complete intersection of two cubic surfaces; from the exact sequence 0 I C () O P () O C () 0, one sees that h 0 (P, I C ()) and hence C lies on two irreducible cubics. Note that deg C = g + 1 = 8 = 1 and therefore C is a curve residual to a line in a complete intersection of two cubics, that is C L = X where L is a line and X is a complete intersection of two cubics. Upon fixing a line L P, we consider the linear system D = P(H 0 (P, I L ())) consisting of cubics containing the line L. Note that any 4 given points on L imposes independent conditions on cubics and hence dim D = dim P(H 0 (P, O())) 4 = 19 4 = 1. Since our curve C is completely determined by a pencil of cubics containing a line L P, we see that H 8,7, is a G(1, 1) bundle over G(1, ), the space of lines in P. Hence H 8,7, is irreducible of dimension dim G(1, 1) + dim G(1, ) = = = 4 8. By taking the residual series K C g8 = g4 1 of a very ample g8, we see that H 8,7, maps into the irreducible closed locus M 1 g,4 consisting 4-gonal curves, which is of dimension g +. We also note that dim W8 (C) = dim W4 1 (C) = 0. For if dim W4 1 (C) 1, then C is either trigonal, bielliptic or a smooth plane quintic by Mumford s theorem; cf. [, p.19]. Because g = 7, C cannot be a smooth plane quintic. If C is trigonal with the trigonal pencil g, 1 one may deduce that g 1 + g 1 is our very ample g8 by the base-point-free pencil trick [, p.16]; ker ν = H 0 (C, F L 1 ) where ν : H 0 (C, F) H 0 (C, L) H 0 (C, F L) is the natural cup-product map with F = g8, L = g 1 and F L 1 turns out to be a g. 1 Therefore it follows that C is a smooth curve of type (, ) on a smooth quadric in P. However, we have already ruled out the possibility for C lying on a quadric. If C is bi-elliptic with a two sheeted map φ : C E onto an elliptic curve E, one sees that K g8 = g4 1 = φ (p + q) by Castelnuovo-Severi inequality and hence g8 = K φ (p + q) where p, q E. Therefore for any r E, we have g8 φ (r) = K φ (p + q + r) = K g6 = g6 whereas g8 is very ample, a contradiction. Furthermore we see that H 8,7, dominates the locus M 1 g,4, for otherwise the inequality dim π(h 8,7, ) < dim M 1 4,g = g + which would lead to the inequality dim H 8,7, = dim PGL(4) + dim W 8 (C) + dim π(h 8,7, ) < , which is an absurdity. (iii) g = 8: Since we have the non-negative Brill-Noether number ρ(d, g, ) = ρ(9, 8, ) = 0, there exists the principal component of H 9,8, dominating M 8 by

7 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME 7 Proposition 1.1. Because almost the same argument as in the proof of [11, Theorem.6 ] works for this case, we provide only the essential ingredient and important issue adopted for our case g = 8. Indeed, the crucial step in the proof of [11, Theorem.6 ] was [11, Lemma.4 ] (for a given g 9) in which the author used a rather strong result [11, Lemma. ]; e.g. if dim W 1 (C) = 1 on a fixed curve C of genus g = 9 then dim W4 1 (C) = 0. However a similar statement for g = 8 was not known at that time. In other words, it was not clear at all that the condition dim W 1 (C) = 1 would imply dim W4 1 (C) = 0 for a curve C of genus g = 8. However by the results of Mukai [1] and Ballico et al. [, Theorem 1] it has been shown that the above statement holds for a curve of genus 8. Therefore the same proof as in [11, Lemma.4, Theorem.6] works (even without changing any paragraphs or notations therein). The authors apologize for not being kind enough to provide a full proof; otherwise this article may become unnecessarily lengthy and tedious.. Non-existence of components of H d,g, rigid in moduli In this section, we give a strictly positive lower bound for the dimension of the image π(z) of an irreducible component Z of the Hilbert scheme H d,g, under the natural map π : H d,g,r M g, which will in turn imply that H d,g, has no components rigid in moduli. The non-existence of a such component of H d,g, has certainly been known to some people (e.g. cf. [14, p.487]). However the authors could not find an adequate source of a proof in any literature. We start with the following fact about the irreducibility of H d,g, which has been proved by Ein [7, Theorem 4] and Keem-Kim [1, Theorems 1. and.6]. Theorem.1. H d,g, is irreducible for d g + and for d = g +, g. Using Proposition 1.1 and Theorem.1, one can prove the following rather elementary facts, well known to experts and included for self-containedness, when the genus or the degree of the curves under consideration is relatively low. Proposition.. For 1 g 4, every non-empty H d,g, is irreducible of dimension λ(d, g, ) = 4d. Moreover, H d,g, dominates M g. Proof. For d g +, we have ρ(d, g, ) = g 4(g d+) 0 and hence there exists a principal component H 0 which dominates M g by Proposition 1.1. Since H d,g, is irreducible for d g + by Theorem.1, it follows that H d,g, = H 0 dominates M g. Therefore it suffices to prove the statement when d g +. If 1 g, one has π(d, ) < g for d g + and hence H d,g, =. If g = 4 we have d 6. Since π(d, ) for d, one has H d,4, = for d and hence we just need to consider H 6,4,. We note that a smooth curve in P of degree 6 and genus 4 is a canonical curve, that is a curve embedded by the canonical linear series and vice versa. Hence G G6 is birationally equivalent to the irreducible variety M 4 and it follows that H 6,4, is irreducible which is a PGL(4)-bundle over an open subset of M 4 or G. Proposition.. The Hilbert schemes H 8,8,, H 8,9, and H 9,g, for g = 9, 1 are irreducible, while H 9,11, is empty and H 9,10, has two irreducible components. Moreover, under the natural map π : H d,g, M g, we have (i) dim π(h 8,8, ) = 17; (ii) dim π(h 8,9, ) = 18;

8 8 C. KEEM, Y.H. KIM AND A.F. LOPEZ (iii) dim π(z) = 1 both when Z = H 9,9, and when Z is one of the two irreducible components of H 9,10, ; (iv) dim π(h 9,1, ) =. Proof. To see that H 9,11, = we use [8, Corollary.14]. In fact note that there is no pair of integers a b 0 such that a+b = 9, (a 1)(b 1) = 11. Since the second Castelnuovo bound π 1 (9, ) = 10 and π(9, ) = 1, it follows that H 9,11, =. As for the other cases, we start with a few general remarks. We will first prove that the Hilbert schemes H d,g, or the components Z H d,g, to be considered are irreducible, generically smooth and that their general point represents a smooth irreducible non-degenerate linearly normal curve C P. Moreover we will show that the standard multiplication map µ 0 : H 0 (O C (1)) H 0 (ω C ( 1)) H 0 (ω C ) is surjective. From the above it will then follow, by well-known facts about the Kodaira-Spencer map (see e.g. [18, Proof of Proposition.], [1, Proof of Theorem 1.], that if N C is the normal bundle of C, then dim π(z) = g + ρ + h 1 (N C ) = 4d 1 + h 1 (N C ) (.1) and this will give the results in (i)-(iv). In general, given a smooth surface S P containing C, we have the commutative diagram H 0 (O S (1)) H 0 (ω S (C)( 1)) ν H 0 (ω S (C)) (.) H 0 (O C (1)) H 0 (ω C ( 1)) µ 0 H 0 (ω C ) H 1 (ω S ) = 0 so that µ 0 is surjective when ν is. Now consider a smooth irreducible curve C of type (a, b), with a b, on a smooth quadric surface Q P. In the exact sequence 0 N C/Q N C N Q C 0 we have that H 1 (N C/Q ) = 0 since C = g + d > g and N Q C = OC (), so that h 1 (N C ) = h 1 (O C ()). (.) Moreover we claim that In fact from the exact sequence C is linearly normal and µ 0 is surjective. (.4) 0 O Q (1 a, 1 b) O Q (1) O C (1) 0 and the fact that H i (O Q (1 a, 1 b)) = 0 for i = 0, 1, we see that h 0 (O C (1)) = h 0 (O Q (1)) = 4. Setting S = Q in (.) we find that ν is the surjective multiplication map of bihomogeneous polynomials H 0 (O Q (1, 1)) H 0 (O Q (a, b )) H 0 (O Q (a, b )) on P 1 P 1. This proves (.4).

9 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME 9 To see (i) note that, since π 1 (8, ) = 7 and π(8, ) = 9, it follows by [8, Corollary.14] that H 8,8, is irreducible of dimension and its general point represents a curve of type (, ) on a smooth quadric. Moreover H 1 (O C ()) = 0 since d = 16 > g = 14 and therefore H 1 (N C ) = 0 by (.) and H 8,8, is smooth at the point representing C. Now (.4) and (.1) give (i). To see (ii) observe that, by [, Example (10.4)], H 8,9, is smooth irreducible of dimension and its general point represents a curve of type (4, 4) on a smooth quadric. Moreover h 1 (O C ()) = h 1 (ω C ) = 1 and therefore h 1 (N C ) = 1 by (.). Hence (.4) and (.1) give (ii). Finally, to prove (iii) and (iv), consider H 9,g, for g = 9, 10 or 1. By [6, Theorem..1] we know that H 9,9, is irreducible of dimension 6 and its general point represents a curve C residual of a twisted cubic D in the complete intersection of a smooth cubic S and a quartic T. In the exact sequence 0 N C/S N C N S C 0 we have that H 1 (N C/S ) = 0 since C = > g = 16 and N S C = OC () that has degree 7, so that again H 1 (N S C ) = 0 and therefore also H 1 (N C ) = 0. Hence H 9,9, is smooth at the point representing C. Moreover, as D is projectively normal, C is also projectively normal. It remains to prove that µ 0 is surjective, whence, by (.), that ν is surjective. To this end observe that, if H is the hyperplane divisor of S, then K S + C H H D. A general element D H D is again a twisted cubic and we get the commutative diagram 0 H 0 (O S (H)) H 0 (O S ) ν 1 0 H 0 (O S (H)) H 0 (O S (H)) H 0 (O S (D )) ν H 0 (O S (H + D )) H 0 (O D (H)) H 0 (O D (D )) τ H 0 (O D (H + D )) H 0 (O S (H)) H 1 (O S ) = 0 H 1 (O S (H)) = 0 from which we see that ν is surjective since ν 1 is and so is τ, being the standard multiplication map H 0 (O P 1()) H 0 (O P 1(1)) H 0 (O P 1(4)). Now (.1) gives (iii) for g = 9. In the case g = 10 it follows by [, Example (10.4)], that H 9,10, has two generically smooth irreducible components Z 1 and Z, both of dimension 6, and their general point represents a curve C of type (6, ) on a smooth quadric for Z 1 and a complete intersection of two cubics for Z. In the first case, from the exact sequence 0 O Q ( 4, 1) O Q () O C () 0

10 10 C. KEEM, Y.H. KIM AND A.F. LOPEZ and the fact that H 1 (O Q ()) = H (O Q ( 4, 1)) = 0, we get h 1 (O C ()) = 0 and therefore h 1 (N C ) = 0 by (.). Then (.4) and (.1) give (iii) for Z 1. As for Z, we have that N C = OC () and ω C = OC (), whence H 1 (N C ) = 0. Moreover, if S is one of the two cubics containing C, we have the diagram H 0 (O P (1)) H 0 (O P (1)) ν 1 H 0 (O P ()) H 0 (O S (1)) H 0 (O S (1)) α ν H 0 (O S ()). As is well known both α and ν 1 are surjective, whence so is ν and then µ 0 by (.). Therefore (.1) gives (iii) for Z. Finally, since π(9, ) = 1, it follows by [8, Corollary.14] that H 9,1, is irreducible of dimension 8 and its general point represents a curve of type (, 4) on a smooth quadric. Moreover, from the exact sequence 0 O Q (, ) O Q () O C () 0 and the fact that H i (O Q ()) = 0 for i = 1,, we get h 1 (O C ()) = h (O Q (, )) = h 0 (O Q (1, 0)) = and therefore h 1 (N C ) = by (.). Hence h 0 (N C ) = 8 and H 9,1, is smooth at the point representing C. Now (.4) and (.1) give (iv). We can now prove our result for r =. Theorem.4. Let Z be an irreducible component of H d,g, and g. under the natural map π : H d,g, M g, the following possibilities occur: (i) Z dominates M g ; (ii) Z = H 7,6, and dim π(z) = 1; (iii) Z = H 8,7, or H 8,8, and dim π(z) = 17; (iv) Z = H 8,9, and dim π(z) = 18; (v) Z = H 9,9, or Z H 9,10, and dim π(z) = 1; (vi) dim π(z). Then, Proof. We first make the following general remark, which will also be used in the proof of Theorem 4.1. Let r, let Z be an irreducible component of H d,g,r not dominating M g and let C be a smooth irreducible non-degenerate curve of degree d and genus g in P r corresponding to a general point c Z. We claim that O C (1) is special. In fact Z is a PGL(r + 1)-bundle over an open subset of a component G 1 of G (see 1). If O C (1) is non-special then, by Riemann-Roch, d g + r and G 1 must coincide with G 0 of Proposition 1.1. But G 0 dominates M g, so that also Z does, a contradiction. Therefore O C (1) is special. Set α = dim O C (1), so that α r. We now specialize to the case r =. First we notice that, in cases (ii)-(v), using Propositions.1 and., H d,g, is irreducible, except for H 9,10,, and the dimension of the image under π of each component is as listed. Also we have d 7, for if d 6 then g π(6, ) = 4. Assume that Z is not as in (i), (ii) or the first case of (iii) and that dim π(z). By Theorem.1, Proposition.1 and Remark 1., we can assume that d g. For any component G 1 G Gd, there exists a component W of Wα d and a closed subset W 1 W Wd α such that G 1 is a Grassmannian G(, α)-bundle over

11 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME 11 a non-empty open subset of W 1. Thus we have λ(d, g, ) = 4d dim Z By Lemma 1. we have α d+1 If dim Wd α (C) = 0, (.) gives dim π(z) + dim W α d (C) + dim G(, α) + dim PGL(4) dim W α d (C) + 4α +. (.). 4d 4 (d + 1) + therefore d 9. If dim Wd α(c) 1 then Theorem 1.4 implies α d. By (.) and Theorem 1.4 again, we find 4d d d + α + 6 that is again d 9. If d = 7 we find the contradiction 7 g π(7, ) = 6. If d = 8 it follows that 8 g π(8, ) = 9 and we get that Z is as in the second case of (iii) or as in case (iv). If d = 9 we find that 9 g π(9, ) = 1. Since, by Proposition., H 9,11, is empty and dim π(h 9,1, ) =, we get case (v). Remark.. (i) There are no reasons to believe that our estimate on the lower bound of dim π(z) is sharp. On the other hand, it would be interesting to have a better estimate (hopefully sharp) on the lower bound of dim π(z) and come up with (irreducible or reducible) examples of Hilbert scheme H d,g, with a component Z achieving the bound. (ii)if d g there is a better lower bound for the dimension of components Z of H d,g, in [4, Theorem 1.]. This leads, in this case, to a better lower bound of dim π(z). Theorem.4 and Proposition. yield the following immediate corollary. Corollary.6. H d,g, has no component that is rigid in moduli if g > Non-existence of components of H d,g,r rigid in moduli with r 4 In this section, we prove the non-existence of a component of H d,g,r rigid in moduli in a certain restricted range of d, g > 0 and r 4. Theorem 4.1. H d,g,r has no components rigid in moduli if g > 0 and (i) d > min{ 17g+7 64, 4g+1 1, max{ g+18 4, 17g }} if r = 4; (ii) d > min{ 9g+0 0, 10g+17, max{ g+, 9g+10 0 }} and d > g+ for 101 d 11, if r = ; (iii) d > min{ 1g+0, g+, max{ g+10, 1g+10 }, max{ g+10, g 1 }} if r = 6; (iv) d > min{ 19g+4 7, max{ 4g+9 7, 76g }} if r = 7; (v) d > min{ 4g+1, g 4 6 } if r = 8; (vi) d > min{ 9g 10, 9g+ } and (d, g) (0, 4) if r = 9; (vii) d > min{ 1g 4, 17g+1 18 } if r = 10; (viii) d > g if r = 11;

12 1 C. KEEM, Y.H. KIM AND A.F. LOPEZ (ix) d > (r )g r+14 r+1 if r 1. Proof. Suppose that there is a component Z of H d,g,r rigid in moduli and let C be a smooth irreducible non-degenerate curve of degree d and genus g in P r corresponding to a general point c Z. Let α = dim O C (1), so that α r and note that, as in the proof of Theorem.4, using Proposition 1.1, we have that O C (1) is special. In particular d g and g. Moreover we claim that C does not admit a degeneration {C t P α } t P 1 to a singular stable curve. In fact such a degeneration gives a rational map P 1 M g whose image contains two distinct points, namely the points representing C and the singular stable curve. Hence the image must be a curve and therefore the curves in the pencil {C t P α } t P 1 cannot be all isomorphic. Now we have a projection p : P α P r that sends C P α isomorphically to p(c) = C P r. Thus the pencil gets projected and gives rise to a deformation p(c t ) P r of C in Z (recall that C represents a general point of Z). For general t we have that p(c t ) is therefore smooth, whence p(c t ) = C t. Since Z is rigid in moduli we get the contradiction C = C t for general t. This proves the claim and now, by Lemma 1.6, we can and will assume that g π 1 (d, α) when d α + 1 and that g π (d, α), g < π 1 (d, α) when α 8 and d α +. Recall again that for any component G 1 G Gd r, there exists a component W of Wd α and a closed subset W 1 W Wd α such that G 1 is a Grassmannian G(r, α)-bundle over a non-empty open subset of W 1. By noting that W 1 is a sublocus inside Wd α (C) in our current situation, we come up with an inequality similar to (.): λ(d, g, r) = (r + 1)d (r )(g 1) dim Z dim W 1 + dim G(r, α) + dim PGL(r + 1) = dim W 1 + (r + 1)(α r) + r + r (4.1) dim W α d (C) + (r + 1)α + r. This leads to the following four cases. CASE 1: d < g and dim Wd α (C) = 0. We have α (d + 1)/ by Lemma 1. and (4.1) gives d < g, α (d + 1)/ and (r + 1)(d α) (r )g. (4.) CASE : d < g and dim Wd α (C) 1. By Theorem 1.4 we get α d/. By (4.1) and Theorem 1.4 again, we find d < g, α d/ and rd (r )α 4 (r )g. (4.) CASE : d g and dim Wd α (C) = 0. We have α (d g+1)/ by Lemma 1. whence, in particular, d (g+r 1)/. Now (4.1) gives d g, α (d g + 1)/ and (r + 1)(d α) (r )g. (4.4) CASE 4: d g and dim Wd α (C) 1. By Theorem 1.4 we have α (d g)/ whence, in particular, d (g + r)/. By (4.1) and Theorem 1.4 again, we find d g, α (d g)/ and (r 1)d (r )α 4 (r 4)g. (4.)

13 IRREDUCIBILITY AND COMPONENTS RIGID IN MODULI OF THE HILBERT SCHEME 1 The plan is to show that, given the hypotheses, the inequalities (4.) (4.) contradict g π 1 (d, α) when d α + 1 or g π (d, α), g < π 1 (d, α) when α 8 and d α +. To this end let us observe that d α + in cases (4.) (4.): In fact this is obvious in cases (4.) and (4.), while in cases (4.4) and (4.), using g d α+1 and g d α respectively, if d α+, we get 4α r 14 and 4α r 10, both contradicting α r. Therefore in the sequel we will always have that g π 1 (d, α) and that either α 7 or α 8 and g π (d, α), g < π 1 (d, α). We now recall the notation. Set m 1 = d 1 α, m = d 1 α + 1, ε 1 = d m 1 α 1, ε = d m (α + 1) 1 and so that π 1 (d, α) = µ 1 = ( m1 { 1 if ε 1 = α 1 0 if 0 ε 1 α, µ = ) α + m 1 (ε 1 + 1) + µ 1, π (d, α) = if ε = α 1 if α ε α 1 0 if 0 ε α ( m ) (α + 1) + m (ε + ) + µ. We now deal with the case r 11 (and hence α 11). We start with (4.). If α d/ then either α = (d + 1)/ or α = d/. Then m =, µ = 0 and π (d, α) d < g. Therefore α (d 1)/ and (4.) gives d (r )g r+8 (r+1), contradicting (viii)-(ix). Similarly, in (4.), if α = d/ then m =, µ = 0 and π (d, α) = d 1 < g. Therefore α (d 1)/ and (4.) gives d (r )g r+14 (r+1), contradicting (viii)-(ix). Now in (4.4), if α (d g)/ then either α = (d g)/ or α = (d g + 1)/. We find m =, µ = 0 and π (d, α) g 1. Therefore α (d g 1)/ and (4.4) gives d (r )g r+8 r+1, contradicting (viii)-(ix). Instead in (4.), if α = (d g)/ we find m =, µ = 0 and π (d, α) = g 1. Therefore α (d g 1)/ and (4.) gives d (r )g r+14 r+1, contradicting (viii)-(ix). This concludes the case r 11. Assume now that 4 r 10. We first claim that (4.4) and (4.) do not occur. In fact note that we have d max{r +, g, (g + r 1)/} in (4.4) and d max{r +, g, (g + r)/} in (4.). (4.6) Plugging in α (d g + 1)/ in (4.4) and α (d g)/ in (4.) we get d (r )g + r + 10 r + 1 in case (4.4) and d (r )g + 1 r + 1 and it is easily seen that these contradict (4.6). Therefore, in the sequel, we consider only (4.) and (4.). If α 7 (whence r 7), we see that (4.) gives and (4.) gives d d (r )g + 7r + 10 r + 1 in case (4.) (4.7) (r )g + 7r 10. (4.8) r

14 14 C. KEEM, Y.H. KIM AND A.F. LOPEZ Now assume α 8, so that g π (d, α), g < π 1 (d, α). Set i = d + 1 α and j = d α. Then (4.) implies d < g, i 0, α(m 1 1)[ r m 1 r 1] + (ε 1 + 1)[(r )m 1 r 1] + + µ 1 (r ) > 0 (4.9) and (α+1)(m 1)[ r m r 1]+(ε +1)[(r )m r 1] r++(m +µ )(r ) 0. (4.10) On the other hand (4.) implies d < g, j 0, ( ) m1 α[(r ) m 1 r + r ] + (ε 1 + 1)[(r )m 1 r] µ 1 (r ) > 0 (4.11) and ( ) m (α+1)[(r ) m r+r ]+(ε +1)[(r )m r] r+6+(m +µ )(r ) 0. (4.1) Suppose now r = 4. It is easily seen that (4.9) implies m 1 9 and i 7α + 1, so that α d 10. Plugging in (4.) we contradict (i). On the other hand (4.11) implies m 1 8, so that α d 1 11α 8, and also j, so that α d+ 17. Moreover (4.1) implies m 8, so that α d 9 11α+1 8, and also j, so that α d Plugging in (4.) and using (4.7) and (4.8) we contradict (i) and the case r = 4 is concluded. If r = it is easily seen that (4.9) implies m 1 and i > α + 1, so that α < d 6. Also (4.10) implies m, so that i α+, hence α d 4 6. Plugging in (4.) we contradict (ii). On the other hand (4.11) implies m 1, so that α d 1, and also j 1α 4, so that α d+4 7. Moreover (4.1) implies m and also j 1α+16, so that α d Plugging in (4.) and using (4.7) and (4.8) we contradict (ii) and the case r = is done. When r = 6 we see that (4.9) implies m 1 4 and α d 1 8α+ 4, but also i >, so that α < d+. Also (4.10) implies m 4, so that i 8α+0, hence α d 1. Plugging in (4.) we contradict (iii). On the other hand (4.11) implies m 1 4, so that α < d 1 4α 4, and also j >, so that α < d+ 1. Moreover (4.1) implies m 4 so that α d 4α+7 4, and also j, so that α d 7 1. Plugging in (4.) and using (4.7) and (4.8) we contradict (iii) and we have finished the case r = 6. Now assume that r = 7. We have that (4.9) implies m 1 and i α + 1, so that α d 4. Plugging in (4.) we contradict (iv). On the other hand (4.11) implies m 1 and j > 4α 4, so that α < d Moreover (4.1) implies m and j 4α+1, so that α d Plugging in (4.) and using (4.7) and (4.8) we contradict (iv). This concludes the case r = 7. If r = 8 we find from (4.10) that m and i α+6 d 4, so that α 7. Plugging in (4.) we contradict (v). On the other hand (4.1) implies m so that α d 4 and also j α+11 7, so that α 7d Plugging in (4.) we contradict (v) and the case r = 8 is proved. Now let r = 9. Then (4.10) gives m and i α+ 8, so that α 8d 1 6. Plugging in (4.) we contradict (vi). On the other hand (4.1) gives m and j, so that α d. We also get m = if and only if (d, g) = (0, ), (0, 4). Then, if (d, g) = (0, ), (0, 4), we see that (4.1) implies m and j α+14 9,

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