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The deformation theory of a two-dimensional singularity, which is isomorphic to an

arXiv:alg-geom/9303003v1 23 Mar 1993Deformations of conesover hyperelliptic curvesJan StevensThe deformation theory of a two-dimensional singularity, which is isomorphic to anaffine cone over a curve, is intimately linked with the (extrinsic) geometry of thiscurve. In recent times various authors have studied one-parameter deformations,partly under the guise of extensions of curves to surfaces (cf.

the survey [Wahl1989]). In this paper we consider the versal deformation of cones, in the simplestcase: cones over hyperelliptic curves of high degree.

In particular, we show thatfor degree 4g + 4, the highest degree for which interesting deformations exist, thenumber of smoothing components is 22g+1 (the case g = 3 is exceptional).Let X be the cone over a hyperelliptic curve C, embedded with a line bundleL.If d := deg L ≥2g + 3, then X has always infinitesimal deformations innegative degrees: dim T 1X(−1) = 2g + 2 [Drewes 1989]. On the other hand, oneknows that only conical deformations exist over reduced base spaces, if deg L >4g + 4 [Tendian 1992a].This implies that all deformations in negative degreemust be obstructed.If S is a surface with C as hyperplane section, then onecan degenerate S to the projective cone over C, or from another point of view,deform the projective cone over C to S; Pinkham calls this construction ‘sweepingout the cone’ [Pinkham 1970].Surfaces with hyperelliptic hyperplane sectionswere already classified by Castelnuovo, and the supernormal surfaces among themhave degree 4g + 4 [Castelnuovo 1890].They are rational ruled surfaces, andsuch surfaces come in two deformation types; therefore there are at least twosmoothing components.

This observation was the starting point of the presentpaper. A computer computation of the versal deformation in negative degree withMacaulay [Bayer–Stillman] gave for an example with g = 2 the number of 32smoothing components.As the versal deformation can be chosen C∗-equivariant, it makes sense torestrict to the part of negative degree.We want to show that the base spaceS−has 22g+1 one-dimensional components.

First of all we have to exhibit thisnumber of surfaces with C as hyperplane section, such that the normal bundleof C in the surface is L. The main point is that an elementary transformationon the ruled surface S in a Weierstraß point of C does not change the normalbundle of C. The composition of elementary transformations in all Weierstraßpoints gives an involution on S; we get 22g+2/2 surfaces. This construction worksfor every hyperelliptic curve C and every line bundle L on C. We obtain 22g+1smooth subspaces of dimension 3g of the base of the versal deformation; by a1

result of [Tendian 1992a] this dimension is exactly the dimension of smoothingcomponents. Therefore we have found 22g+1 irreducible components.The next thing to determine is T 2, the space where the obstructions lie.

ForT 2(ν), ν < −2, we have a general vanishing result [Wahl 1987], and it is not difficultto find the dimension in case ν > −2. We find the dimension of T 2 with the MainLemma of [Behnke–Christophersen 1991], which connects the number of generatorsof T 2 with the codimension of smoothing components in the base space of the versaldeformation of a general hyperplane section.

Therefore we compute the dimensionof T 1 for the cone over d points on a rational normal curve of degree d −g −1 inPd−g−1. We use explicit equations for the curve.

We also need the equations of Xto show that the OX-module T 2 is annihilated by the maximal ideal. Altogetherwe obtain that dim T 2(−2) = d −2g −3, dim T 2(−1) = (g −2)(d −g −3), and theother T 2(ν) vanish, if d > 2g + 3.Actually, the equations for the cone over a hyperelliptic curve, or for its hy-perplane sections, have a nice structure, which Miles Reid calls the rolling factorsformat [Reid 1989].

We give an interpretation of T 2(−2) in terms of the rollingfactors format, in (2.12). It is not difficult to compute the part of the versal de-formation in negative degrees, once one has represented T 1(−1) as perturbationsof the equations.

Unfortunately, this representation is given by complicated for-mula’s, which we only computed in the simplest case, that L is a multiple of theg12. The resulting equations for the base space S−are as complicated, and it isdifficult to see if they define a complete intersection; for d = 4g + 4 we have 2g + 1quadratic equations in 2g + 2 variables.

By determining explicitly the elements ofT 1, induced by our 22g+1 surfaces, we show that we have a complete intersection,and this allows us finally to conclude that there are no other components (exceptwhen g = 3).Powerful methods exist to compute T 1 for surface singularities, without usingexplicit equations. They are based on Schlessinger’s description of T 1 [Schlessinger1973]: suppose depth0 X ≥2 and write U = X \ 0, then T 1X = kerH1(U, ΘU) →H1(U, Θn|X).

In the special case that X is the cone over a projective variety Y ,all sheaves are graded, and the graded parts can be computed on Y , see [Schles-singer 1973] and Mumford’s ‘footnote’ to it [Mumford 1973].We describe thesituation in terms of the sheaf of differential operators of order ≤1 [EGA IV].In the case of cones over curves it is advantageous to dualise. Now the bundle ofprincipal parts comes in, and with it Wahl’s Gaussian map (cf.

[Wahl 1989]). Forcones over curves of high degree the computation of T 1(−1) is the most difficult;much of the work on the Gaussian map is connected with this case, and speciallywith vanishing results.

The most complete results on interesting deformations areobtained by Sonny Tendian [Tendian 1990].With a trick, which basically is contained in [Mumford 1973], one sees that fornon hyperelliptic curves, embedded with a non special line bundle L, the dimensionof T 1(−1) of the cone over C is equal to h0(C, NK ⊗L−1), where NK is the normalbundle of C in its canonical embedding. For low genus this gives quite preciseinformation, because then the normal bundle NK is easy to describe.

We takethe opportunity to remark that the computations in [Stevens 1989] imply thatfor a general curve of genus g ≥3 the Gaussian map ΦK,L is surjective for all2

line bundles L with deg L ≥2g + 11. In the hyperelliptic case a variant of theconstruction yields easily the dimension of T 1(−1).The paper [Looijenga–Wahl 1986] introduces a collection of smoothing datafor surface singularities, which in many cases distinguish between smoothing com-ponents.

In our case we determine a subset with 1 + 22g elements, onto whichthe set of smoothing components is mapped surjectively, so these smoothing datado not suffice to distinguish all components. The computations are similar to thecase of a simple elliptic singularity of degree 8 = 4g + 4, where the number ofsmoothing components is really 1 + 4.The organisation of this paper is as follows: in Section 1 we discuss thedescription of T 1 for cones, and give precise results in the hyperelliptic case.

Wealso give a formula for the graded parts of T 2.In Section 2 we describe theequations for hyperelliptic cones.We compute T 1 for the general hyperplanesection, and deduce the dimension of T 2 from it. In the last Section we prove theresults on smoothing components.1.

Cones over curves(1.1) The basic reference for the deformation theory of cones is a paper by Schles-singer [Schlessinger 1973] and Mumford’s ‘footnote’ to it [Mumford 1973]. Westart with the description of T 1X for a singularity (X, 0) ⊂(Cn, 0):0 −→ΘX −→Θn|X −→NX −→T 1X −→0.

(∗)Here NX = Hom(I/I2, OX) is the normal sheaf; the tangent sheaf of X is alsodefined as a dual: ΘX = Hom(Ω1X, OX).If Z ⊂X is a closed subset, containing the singular locus Sing X of X,and if depthZ X ≥2, then T 1X = cokerH0(U, Θn|X) →H0(U, NU), or T 1X =kerH1(U, ΘU) →H1(U, Θn|X), where U = X \ Z. In the special case that Xis the cone over a projective variety Y we want to interpret these groups on Y .From now on we work in the algebraic category.Let Y be a smooth projective variety, and L a very ample line bundle on Y .We set V = H0(Y, L).

Let φL: Y →P(V ∗) be the embedding of Y with L. LetX ⊂V ∗be the affine cone over φL(Y ). We will identify Y and φL(Y ); we haveL = OY (1).

Suppose that φL(Y ) is projectively normal, i.e. X is normal.

Thesmooth space U := X −0 is a C∗-bundle over Y . We denote with the same symbolπ the projections π: U →Y and π: V ∗−0 →P(V ∗).

If F is a sheaf on X witha natural C∗-action, then π∗F decomposes into direct sums of the eigenspaces forthe various characters of C∗. Let FY be the sheaf of C∗invariants.

Then:H0(X, F) = H0(U, F) =ν=∞Mν=−∞H0(Y, FY (ν)).The last equality does not hold in the analytic context; in that case one has afiltration on H0(X, F), whose associated graded space is the direct sum as above,which would suffice for our purposes.3

We describe the sheafs in the exact sequence (∗). Actually, when explicitlywriting down sections, it is more convenient to use homogeneous coordinates, i.e.

tocompute on X. We illustrate this with the tangent bundle of Pn.

One has theEuler sequence [Hartshorne 1977, II.8.20.1]:0 −→OPn −→OPn(1)n+1 −→ΘPn −→0.In homogeneous coordinates (z0, . .

. , zn) elements of a basis of H0(OPn(1)n+1)can be written as zi ∂∂zj , and the map from OPn is given by 1 7→P zi ∂∂zi .We recall the definition of the sheaf of principal parts [Kleiman 1977, IV.A,EGA IV.16.7].

Consider a scheme Y ; let J be the ideal sheaf of the diagonal∆in Y × Y , and let Y (n)∆be the n-th infinitesimal neighbourhood of ∆. Thecanonical projections p1 and p2 of the product induce maps p(n)1 : Y (n)∆→Y andp(n)2 : Y (n)∆→Y .

We define for a sheaf F [EGA IV. 16.7.2]:PnY (F) = (p(n)1 )∗((p(n)1 )∗(F)).We write PnY for PnY (OY ).

One has PnY (F) = PnY ⊗OY F, where the tensor productis taken with the OY -module structure, defined by p2. Because the diagonal is asection of Y × Y for both p1 and p2, both morphisms define a homomorphismOY →PnY , and therefore an OY -module structure.

Except when explicitly stated,we always consider the OY -module structure on PnY , induced by p1, and write itas left multiplication. One denotes by dn the morphism OY →PnY , induced by p2[EGA IV.16.3.6].

For every t ∈Γ(U, OY ), U ⊂Y open, dnt is the principal partof order n. In particular, dt = d1t −t ∈Γ(U, Ω1Y ) is the differential of t. We havethe exact sequence:0 −→Ω1Y −→P1Y −→OY −→0.On Pn this sequence is the dual of the Euler sequence.The sheaf PnY (F) has also two OY -module structures; it is convenient to writethem on the left and the right. For a ∈Γ(U, OY ), b ∈Γ(U, PnY ) and t ∈Γ(U, F)one has [EGA IV.16.7.4]:a(b ⊗t) = (ab) ⊗t,(b ⊗t)a = (b · dna) ⊗t = b ⊗(at) = (dna) · (b ⊗t).There is a map dnF: F →PnY (F) with dnF(t) = 1 ⊗t.

(1.2) Definition [EGA IV.16.8.1].Let F and G be two OY -modules.A ho-momorphism D: F →G is a differential operator of order ≤n if there exists ahomomorphism u: PnY (F) →G such that D = u ◦dnF.The differential operators form a group; by applying the construction on opensets we obtain a sheaf DiffnY (F, G), which is isomorphic to HomOY (PnY (F), G)[EGA IV.16.8.4]. We write simply DiffnY for DiffnY (OY , OY ), and DiffnY (F) forDiffnY (F, OY ).We return to our embedding φL: Y →P(V ∗).

Then φ∗L(P1P(V ∗)) = V ⊗C L−1.Here V is considered to be the vector space with dzi as basis. Let N ∗Y be theconormal bundle of Y in P(V ∗).

Because Y is smooth, one has the familiar exactsequence 0 −→N ∗Y −→φ∗L(Ω1P(V ∗)) −→Ω1Y −→0, with which we obtain thefollowing result [Kleiman 1977, (IV.19)]:4

(1.3) Proposition.In the situation as above the following sequence is exact:0 −→N ∗Y ⊗L −→V ⊗C OY −→P1Y (L) −→0,or dually:0 −→Diff1Y (L) −→V ∗⊗C OY −→NY ⊗L−1 −→0.In particular we can view the C∗-invariants of the exact sequence (∗) as ob-tained by taking global sections of the sequence 0 −→Diff1Y −→V ∗⊗C L −→NY −→0 on Y . We get the following formulation of a result of [Schlessinger 1973]:(1.4) Theorem.Let L be a very ample line bundle on a smooth projectivevariety Y , which embeds Y as projectively normal subvariety of P(V ∗), whereV = H0(Y, L).

Let X ⊂V ∗be the affine cone over Y . Then the graded partsT 1X(ν) of T 1X are given by:T 1X(ν) = coker{V ∗⊗H0(Y, Lν+1) −→H0(Y, NY ⊗Lν)},or alternatively,T 1X(ν) = ker{H1(Y, Diff1Y ⊗Lν) −→V ∗⊗H1(Y, Lν+1)}.From now one we concentrate on the case that Y is one dimensional; Serreduality then transforms the second formula into one involving H0.

(1.5) Corollary.Let X be the cone over a curve C, embedded by L = OC(1).Write K for Ω1C. Then:(T 1X)∗(ν) = coker{V ⊗H0(C, K(−ν −1)) −→H0(C, P1C ⊗K(−ν))}.This Corollary makes it possible to determine the graded parts of T 1 in manycases.Vanishing results exist for line bundles of high degree.We recall someresults for the various degrees ν.

We assume that g(C) ≥2. (1.6) Case I: ν ≥1.Suppose first that Lν is non special, i.e.

H1(C, Lν) = 0. ThenT 1X(ν) = H1(C, Diff1C ⊗Lν) = H1(C, ΘC ⊗Lν).

In particular, if deg L > 4g −4,then T 1X(ν) = 0 [Mumford 1973]. If Lν is special, but H1(C, Lν+1) = 0, then stillT 1X(ν) = H1(C, Diff1C ⊗Lν).

One has the exact sequence:0 −→H1(C, Lν) −→H1(C, Diff1C ⊗Lν) −→H1(C, ΘC ⊗Lν) −→0.If Lν = K, this follows from the surjectivity of Γ(P1C) →Γ(OC). (1.7) Case II: ν = 0.One has T 1X(0) = ker{H1(C, Diff1C) −→V ∗⊗H1(C, L)}.The map here can also be thought of as a cup product H1(C, Diff1C)⊗H0(C, L) −→H1(C, L) [Arbarello–Cornalba 1981]; for a differential operator θ and a sections ∈Γ(L) the cup product θ · s = 0 ∈H1(L) if and only if the section s lifts tothe first order deformation of L →C, defined by θ.

In [loc.cit.] the vector spaceT 1X(0) is identified as tangent space to a space of grd’s on a variable curve.

Moreprecisely, let π: X →S be a miniversal family of smooth curves, and consider the5

relative Picard variety PicdX/S and the bundle Grd over S of grd’s on the fibres; thereis a map c: Wrd →PicdX/S with image Wrd. Because by assumption the linear seriesL is complete, the map c is injective.

Then H1(C, Diff1C) is the tangent space toPicdX/S in the point L →C, and T 1X(0) is the tangent space to Grd. The problemnow is to prove that Grd is smooth of expected dimension 3g −3 + ρ.

However, forρ < 0 not much is known.For non special L one has dim T 1X(0) = 4g + 3. In terms of deformations thismeans that the cone can be deformed by changing the moduli of the curve or bychanging the line bundle L in Picd.

For L = K the composed map H1(OC) →H1(Diff1C) →V ∗⊗H1(K) ∼= H1(OC) is an isomorphism, so dim T 1X(0) = 3g −3and the versal family is a family of curves in their canonical embedding. (1.8) Case III: ν ≤−2.In this case T 1X(ν) = H0(C, NC ⊗Lν) because Lν+1 is aline bundle of negative degree.

We have two vanishing results. (1.9) Lemma [Mumford 1973].If T 1X(−1) = 0, then T 1X(ν) = 0 for all ν ≤−2.Proof .

If Γ(NC ⊗Lν) ̸= 0, then NC ⊗L−2 has a non-zero section s, and for allt ∈V = Γ(L) the tensor product t⊗s is a non-zero section of NC ⊗L−1; thereforeh0(C, NC ⊗L−1) ≥dim V . Because T 1X(−1) = 0, the map V ∗→Γ(NC ⊗L−1) issurjective, so all sections are of the form t⊗s, therefore they are proportional, anddo not generate NC ⊗L−1.

But NC ⊗L−1 is generated by its sections, becauseV ∗⊗OC is, and V ∗⊗OC →NC ⊗L−1 is surjective.□(1.10) Remark.The argument in the proof shows that Γ(NC ⊗Lν) = 0, ν ≤−2,if dim T 1X(−1) < rank NC −1. (1.11) Proposition [Wahl 1987, 2.5].Let Y ⊂P = P(V ∗), with V ⊂Γ(L), bea projective variety, defined by a system of quadratic equations f. Suppose everynon-zero quadratic equation fi is involved in a linear relation r; this is true if therelations are generated by linear ones.

Then H0(Y, NY ⊗Lν) = 0 for ν ≤−2.Proof . Let I be the ideal sheaf of Y .

Consider the complex:OP(−3)⊕lr−→OP(−2)⊕kf−→I −→0,which is not necessarily exact at OP(−2)⊕k. Dualise, twist and restrict to Y toget:0 −→NY ⊗L−2 −→O⊕kYtr−→OY (1)⊕l.Let K be the kernel of the map tr: Ck →Γ(L)⊕l.

Here Ck can be identified withthe dual of the vector space Q of quadratic equations. The relations r involve onlyequations in K⊥.

Therefore K = 0.□By a theorem of Green the conditions are satisfied for an embedding of a curvewith a complete linear system of degree d ≥2g + 3 [Green 1984, Thm. 4.a.1].

(1.12) Example.Let C be hyperelliptic, with involution π: C →P1, and let L bevery ample of degree d. Then φL(C) lies on a scroll S, the image in P(Γ(C, L)∗)of S = P(π∗L), where π∗L ∼= O(a) ⊕O(b) with a + b = d −(g + 1), and a, b ≤d/2.Suppose b ≤a, write e = a −b, so 0 ≤e ≤g + 1, then S ∼= P(O ⊕O(−e)). The6

Picard group of S is generated by the section E0 with E20 = −e, and the class fof a fibre. We have C ∼2E0 + (g + 1 + e)f (the coefficient of f can be computedfrom the adjunction formula).

Therefore C2 = 4g + 4.Now suppose that d = deg L = 2g+2. Then S = S, except when b = 0, whichoccurs if L = (g + 1)g12; in that case X is the cone over a rational normal curve ofdegree g + 1, and C does not pass through the vertex (because E0 · C = 0).

Wehave the normal bundle exact sequence 0 →NC/S →NC →NS|C →0. Becausethe scroll S is defined by quadratic equations with linear relations, the argumentof the proposition gives that H0(C, NS ⊗L−2) = 0.Therefore Γ(NC(−2)) =Γ(NC/S(−2)).

Because L ∼E0+af, we have C·(C−2L) = C·(g+1+e−2a)f = 0,so NC/S(−2) ∼= OC and h0(C, NC/S(−2)) = 1. For L = (g + 1)g12 this result wasobtained by Drewes [Drewes 1989].We specialise to the case g = 3.

We proved in [Stevens 1991], that the coneX over φL(C) has 3 smoothing components, if L = 2K(= 4g12), and 2 componentsotherwise. The curve C is a complete intersection of the scroll and a quadric.

IfL = 2K, the scroll is the projective cone over a rational normal curve of degree4, which itself has two smoothing components; the deformation to the Veronesesurface occurs, if we deform C to a non hyperelliptic curve, with L = 2K.We compute T 1X(−2) for g(C) = 3, deg L = 8, C not hyperelliptic. If L = 2K,we find as above that Γ(NC(−2)) = Γ(NC/S(−2)) = Γ(OC), where S is theVeronese surface.

If L ̸= 2K, let D be a general divisor in the linear system; it iscut out on the canonical curve C4 in P2 by a cubic C3, and the linear system is thesystem of cubics through the residual intersection C4 ∩C3 −D. Therefore C ⊂P5lies on (non unique) Del Pezzo surface of degree 5.

One checks that all equations(which are quadratic) occur in linear relations, although the relations are notgenerated by linear ones. So T 1X(−2) = 0.

A smoothing is obtained by sweepingout the cone over the blow-up of P2 with the linear system of quartics with basepoints in D. By [Tendian 1992a, Prop. 6.1] the dimension of the correspondingsmoothing component is 15, which is also the dimension of T 1 (see the descriptionof T 1(−1) below), so the base space is smooth.For L = 2K this componenthas codimension 1, and there is a second component; if C = S ∩{Q = 0}, thenS ∩{Q = t2} is P2, branched in C4.

(1.13) Case IV: ν = −1.This is the most difficult case. A more specific knowl-edge of the maps in the theorem is necessary.

We review the relation with Wahl’sGaussian map [Wahl 1990], see also [Tendian 1990, Drewes 1989, Wahl 1989].We start with the following diagram:V ⊗OCy0→K ⊗L−→P1C(L)−→L→0The kernel of the composed map V ⊗OC →L is a vector bundle ML over C,and we get a map ML →K ⊗L. Let M be a second line bundle, and tensoreverything with M. Define:R(L, M) = ker{µL,M: Γ(L) ⊗Γ(M) →Γ(L ⊗M)}.7

Then R(L, M) = Γ(ML ⊗M) and we have the Gaussian map ΦL,M: R(L, M) →Γ(K ⊗L⊗M). This map is given explicitly by ΦL,M(α) = P d1Lli⊗mi, where α =P li⊗mi ∈R(L, M) with li ∈Γ(L) and mi ∈Γ(M).

A more symmetric definitioncan be given in local coordinates; represent sections on an open set U by functions,again denoted by li and mi. Then from P limi = 0 we get P(li dmi +mi dli) = 0,so ΦL,M(α) can be represented by the 1-form 1/2 P(li dmi −mi dli).For curves we have that H1(K ⊗L ⊗M) = 0, so if we denote the mapΓ(L) ⊗Γ(M) →Γ(P1C(L) ⊗M) by d1L ⊗1M, we have the exact sequence 0 →coker ΦL,M →coker d1L ⊗1M →coker µL,M →0.We can also start the construction above with M in stead of L. Up to apermutation of factors one has µL,M = µM,L and ΦL,M = −ΦM,L.

Therefore:coker d1L ⊗1M ∼= coker d1M ⊗1L.We apply this to the computation of T 1X(−1). From Corollary (1.5) we have(T 1X(ν))∗= coker d1L ⊗1K⊗L−(ν+1) and in particular (T 1X(−1))∗= coker d1L ⊗1K.Therefore (T 1X(−1))∗∼= coker d1K ⊗1L.If φL: C →P(H0(C, L)∗) is birational onto its image, then the multiplica-tion maps µL,K⊗L−(ν+1) are surjective [Arbarello–Sernesi 1978, Proof of Thm 1.6].This applies to our situation, so (T 1X(ν))∗= coker ΦL,K⊗L−(ν+1) and in particular(T 1X(−1))∗= coker ΦL,K = coker ΦK,L.Now suppose that C is not hyperelliptic.

Then K is very ample and we havethe exact sequence:0 −→N ∗K ⊗K −→Γ(K) ⊗C OC −→P1C(K) −→0,where N ∗K is the conormal bundle of the canonical embedding. This yields that(T 1X(−1))∗= ker{H1(N ∗K ⊗K ⊗L) →Γ(K) ⊗H1(L)}.

This map is surjective,because H1(P1C(K) ⊗L) = 0. Dually, T 1X(−1) = coker{Γ(K)∗⊗H0(K ⊗L−1) →H0(NK ⊗L−1)}.

In particular:(1.14) Lemma.For the cone X over a non hyperelliptic curve, embedded by anon special complete linear system, T 1X(−1) = H0(NK ⊗L−1).As Mumford remarks, there is an integer d0, depending only on C, such thatH0(NK ⊗L−1) = 0 for deg L ≥d0. The bundle E of [Mumford 1973] is NK ⊗K−1;in fact, Mumford’s construction can be understood as interchanching K and L.In general it is difficult to give sharp explicit bounds, but for low genus theresult is very effective.

Before we give some examples, we prove a result [Wahl1989, 2.2], which is based on a more general lemma of Lazarsfeld, giving sufficientconditions for surjectivity of Gaussian maps. (1.15) Lemma.Suppose C is not hyperelliptic, trigonal or a plane quintic.

IfH0(K2 ⊗L−1) = 0 (in particular, if deg L > 4g −4), then ΦK,L is surjective.Proof . By Petri’s Theorem the ideal I of the canonical curve C is generated byquadrics, so there is a surjection O(−2)⊗a →I.

Dualising, restricting to C andtwisting gives an injection H0(NK ⊗L−1) →(H0(K2 ⊗L−1))⊗a.□We now look at low genera.8

(1.16) g = 3.The normal bundle is the line bundle K⊗4. Therefore d0 = 17.For deg L = 16 only L = K⊗4 has non zero T 1(−1), and the cone X(C, K⊗4) issmoothable, as hyperplane section of X(P2, O(4)).

For all L with deg L = 13 theGaussian map ΦK,L is not surjective, for the general L of degree 14 it is. (1.17) g = 4.The normal bundle is K⊗2 ⊕K⊗3, so d0 = 19.More generally, for C trigonal, the canonical curve sits on a scroll S as divisorof type 3H −(g −4)R [Schreyer 1986], and we have the normal bundle sequence0 →NC/S →NC →NS|C →0.

Here NC/S = K3 ⊗(g13)4−g, a bundle of degree3g +6. Sonny Tendian proves that H1(NS|C ⊗L−1) = 0, if H0(K2 ⊗L−1) = 0, orfor L = K2 [Tendian 1990].

In particular, d0 ≤max(3g +7, 4g −4). We conjecturethat H1(NS|C ⊗L−1) = 0 if deg L ≥d1 for some d1 < 3g + 3.

For deg L = 3g + 6we would then have that T 1(−1) ̸= 0 only if L is the characteristic linear system ofthe family of curves of type 3H −(g −4)R on S, which also gives a one parametersmoothing. (1.18) g = 5.The general curve is a complete intersection of three quadrics, withnormal bundle 3K2, and d0 = 17 (for a trigonal 5C we have d0 = 22).

(1.19) g = 6.Here the famous possibility of a plane quintic occurs. It givesd0 = 26.

The general curve lies on a (possibly singular) Del Pezzo surface S ofdegree 5 with normal bundle NC/S = K2.Tendian has studied the case L = K2 [Tendian 1990]. He proves that ΦK,K2is surjective, if the Clifford index of C is at least 3.

This leaves the hyperelliptic,trigonal and tetragonal curves, as well as smooth plane quintics and sextics. Wenow concentrate on the tetragonal case.

Then C lies on a three dimensional scrollX of type S(e1, e2, e3) of degree e1 + e −2 + e3 = g −3 [Schreyer 1986], but moreimportant are the surfaces on which C lies, cf. [L´evy-Bruhl-Mathieu 1953].

Weobtain C as complete intersection of divisors Y ∼2H −b1R and Z ∼2H −b2R onX with b1+b2 = g−5 and b2 ≤b1. Either Y is rational, or the given g14 is composedwith an elliptic or hyperelliptic involution C2:1−→E2:1−→P1, and Y is a ruled surfaceover E, with a rational curve of double points, which is the canonical image ofE; then deg Y = g −1 + b2, pa(E) = 1/2(b2 + 2), and C does not intersect thedouble curve.

The normal bundle of C in Y is NC/Y = K2 ⊗(g14)−b2. In the caseb2 = 0, and g > 5, c lies on a unique elliptic cone Y , and H1(C, NY ⊗K−2) = 0[Tendian 1990, 2.2.12], so dim T 1X(C,K2)(−1) = 1 and Y leads to a one parameterdeformation to a simple elliptic singularity of degree g −1.The examples above, where ΦK,L is not surjective, have in common that NKis unstable; the rank of NK is g −2, the degree is 2(g2 −1) = (g −2)(2g + 4) + 6,and we found surfaces Y with deg NC/Y > 2g + 4 + ⌊6/g −2⌋.

This motivates:(1.20) Question.Suppose that the canonical curve gC lies on a unique surface Ywith d0 = deg NC/Y > 2g + 4 + ⌊6/g −2⌋. Is then ΦK,L surjective for all L withdeg L > d0, and for L with deg L = d0 and L ̸= NC/Y ?For a general curve such surfaces do not exist; already for g = 7 the generalcurve has a plane representation as C7(8A2), so deg NC/Y = 17.

(1.21) Question [Wahl 1989, 2.5].Is ΦK,L surjective on a general curve of genusg ≥12 for deg L ≥2g −2? What is the best bound?9

We remark that for general C with g = 10 or g ≥12 the map ΦK,K is surjective[Ciliberto–Harris–Miranda 1988]. The problem is of course to find a general curve;the easiest to handle are singular curves.

In [Stevens 1989] we constructed for allg ≥5 a g-cuspidal canonically embedded curve Γ with a Weil divisor D of degreeg + 5 such that h0(NΓ(−D)) = 6. (1.22) Proposition.For a general curve of genus g ≥3 the map ΦK,L is sur-jective for all L with deg L ≥2g + 11.Proof .

From [Stevens 1989, Prop. 5] it follows that h0(NC ⊗L−1) ≤6 for ageneral curve C with g ≤5 and a general L with deg L = g + 5.Thereforeh0(NC ⊗L−1) = 0, if deg L ≥g + 11; presumably a general L of degree g + 6 willdo for large g. Fix a line bundle L0 of degree g + 11 with h0(NC ⊗L−10 ) = 0.Every line bundle L with deg L ≥2g + 11 can be written as L = L0(D) with Dan effective divisor of degree deg L −g −11 ≥q.

This implies that H0(NC ⊗L−1)is a subspace of H0(NC ⊗L−10 ). For g = 3, 4 see (1.16) and (1.17).□(1.23) The hyperelliptic case.The formula:(T 1X(−1))∗= coker d1L ⊗1K ∼= coker d1K ⊗1Lstill holds.

However the map d1K: Γ(K) ⊗OC →P1(K) is only generically surjec-tive, with cokernel equal to coker{φ∗KΩ1P ⊗K →K2} ∼= Ωφ ⊗K, where Ωφ are therelative differentials, so the cokernel is given by the Jacobian ideal [Piene 1978].Therefore, if B denotes the set of Weierstraß points, we have an exact sequence:Γ(K) ⊗OCd1K−→P1(K) −→Mp∈BCp →0.Let P = Imd1K; this is a rank 2 vector bundle. (1.24) Lemma.Let the rational normal curve R = φK(C) be the canonicalimage of C, and write φ: C →R; let H = OR(1) ∼= OP1(g −1).

Let NR be thenormal bundle of R in Pg−1. The following sequence is exact:0 →φ∗(N ∗R ⊗H−1) −→Γ(K) ⊗OCd1K−→P →0.Proof .

Consider the commutative diagram:0→φ∗(N ∗R ⊗H−1)−→Γ(K) ⊗φ∗(OR)φ∗◦d1H−→P1R(K)→0yy∼=y0→E−→Γ(K) ⊗OCd1K−→P1C(K)Because φ∗◦d1H = d1K ◦φ∗[EGA IV.16.4.3.4] the image of the right-hand verticalmap is P, and therefore E = φ∗(N ∗R ⊗H−1).□The normal bundle of R splits a direct sum of g−2 bundles of degree g+1, andtherefore E = ⊕g−2(g12)−2. Using the fact that H1(P1C(K) ⊗L) = 0 for deg L > 0,we obtain the following result; for deg L > 2g + 1 cf.

[Wahl 1988, 7.11.1, Drewes1989].10

(1.25) Proposition.Let C be a hyperelliptic curve of genus g, and L a linebundle with deg L > 0. Then dim coker d1K ⊗1L = 2g +2+(g −2)h1(L⊗(g12)−2)−gh1(L).

In particular, if deg L > 2g + 2, then dim T 1(−1) = 2g + 2.We remark that H1(L) = 0, if L defines a birational map of C. Then h1(L ⊗(g12)−2) ≤1, as one sees from the exact sequence 0 →L⊗(g12)−2 →L →C4 →0; ifh0(L ⊗(g12)−2) = 0, this follows because h0(L) ≥3; therefore we may assume thatH0(L ⊗(g12)−2) = {ψ, . .

. , ψzk} for some function ψ on C and a local coordinatez on P1.

Then H0(L) contains in addition the section ψzi+1, ψzi+2 and at leastone section which is not of this form. (1.26) T 2.We recall the definition of T 2X for a singularity (X, 0) ⊂(Cn, 0)[Schlessinger 1973]:let 0 →R →(On)⊕kj−→I →0 be a resolution of theideal I of X.Let R0 be the submodule of R, generated by the trivial rela-tions, i.e.

those of the form xj(y) −yj(x). Then R/R0 is a OX-module, andT 2X = coker{Hom((OX)⊕k, OX) →Hom(R/R0, OX)}.

We can also consider theexact sequence 0 →RX →(OX)⊕k →I/I2 →0 on X. There is a surjectionR/R0 →RX, whose kernel is a torsion module for reduced X.

Therefore we getthe alternative description:0 →NX −→Hom((OX)⊕k, OX) −→Hom(RX, OX) −→T 2X →0.Suppose the sheaf T 2 has support contained in Z with dpZ X ≥2, and let U =X \ Z.Then T 2X = cokerH0(U, O⊕kn ) →H0(U, R∗X)= kerH1(U, NX) →H1(U, (On)⊕k)).We specialise to the case of cones as before, so X is the cone over the smoothprojective variety Y , embedded with the line bundle L. Let the ideal of Y inP(V ∗), V = H0(Y, L), be generated by k equations of degree d1, . .

. , dk.

Thenthe graded parts of T 2X are given by the exact sequence:0 →T 2X(ν) −→H1(Y, NY (ν)) −→H1(OY (d1 + ν)) ⊕· · · ⊕H1(OY (dk + ν)).If dim Y = 1, then the group H1(Y, NY (ν)) occurs in the following exact sequence,where we write C for Y :0 →T 1X(ν) −→H1(Diff1C(ν)) −→V ⊗H1(C, Lν+1) −→H1(C, NC(ν)) →0.This gives a formula for the dimension of T 2X(ν):dim T 2X(ν) = dim ker{V ⊗H1(Lν+1) →H1(⊕iLdi+ν)}−h1(Diff1C(ν))+dim T 1X(ν).If H1(C, L) = 0, then H1(C, NC(ν)) = 0 for ν ≥0, and therefore also T 2X(ν) = 0.If C is defined by quadratic equations, and L is not special, then T 2X(−1) =H1(C, NC(−1)) and therefore dim T 2X(−1) = (g−2)h0(L)−6(g−1)+dimT 1X(−1),because h1(Diff1C(−1)) = 2 deg L + 4g −4.11

(1.27) Example.The above computation gives new examples of singularitiesfor which the obstruction map is not surjective, cf. [Tendian 1990, 2.4.1].Inparticular, if g(C) = 3 and L is general of degree d ≥14, then dim T 1X(−1) =0, and X has only conical deformations, so the base space is smooth, whereasdim T 2X(−1) = d −14.

For g = 4 and general L with 9 ≤d = deg L < 15 wehave dim T 1X(−1) = 15 −d, dim T 1X = 28 −d, which is also the dimension of asmoothing component [Tendian 1992a, 6.2]. Therefore the base space is smooth,but dim T 2X(−1) = d −9.

Finally, for a general X(5C, Ld) with d > 12 there areonly conical deformations, and dim T 2X(−1) = 3(d −12).For T 2X(ν) with ν < −2 we have the following general vanishing result:(1.28) Lemma [Wahl, Cor. 2.10].If Y ⊂P = Pn is a smooth projectively normalsubvariety, defined by quadratic equations, with a resolutionOP(−4)⊕m −→OP(−3)⊕l −→OP(−2)⊕k −→OP −→OX −→0,then the cone X over Y satisfies T 2X(ν) = 0 for ν < −2.By [Green 1984, Thm.

4.a.1] the conditions are satisfied for a curve, embeddedwith a complete linear system of degree d ≥2g + 4.2. The versal deformation of hyperelliptic cones(2.1) Our aim in this section is to compute equations of the versal deformation ofthe cone X = X(C, L) over a hyperelliptic curve gC, embedded with a completelinear system L of degree at least 2g+3.

Then, as we have seen, T 1X is concentratedin degrees 1, 0 and −1, and T 1X(1) = 0 if deg L > 4g −4. We restrict ourselves tothe part of the versal deformation in negative degree, because otherwise also nonhyperelliptic curves come in.For the computation of T 1 efficient methods exist, which avoid the explicituse of equations and relations.For the versal deformation there seems to beno alternative.

Actually, as Frank Schreyer repeatedly pointed out to me, theequations for φL(C) are rather simple in the hyperelliptic case: C is a divisoron a two-dimensional scroll, so besides the determinantal of the scroll we have‘essentially’ one equation. (2.2) Equations for X.Let L be any line bundle of degree d ≥2g + 3.

Denotethe involution by π: C →P1. Then φL(C) lies on the scroll S = PP1(π∗L), whereπ∗L ∼= O(a) ⊕O(b) with a + b = d −(g + 1), and a, b ≤d/2; in particular, ifL = kg12, then a = k and b = k −(g + 1).

Suppose b ≤a, write e = a −b, so0 ≤e ≤g + 1, and S ∼= P(O ⊕O(−e)). In Pic S = ZE0 ⊕Zf, where E0 is thesection with E20 = −e and f is the class of a fibre, we have C ∼2E0 +(g +1+e)f.Let L = kg12 with k > g +1.

Let C be the curve y2 −P2g+2i=0 aixi = 0. We canmake this equation quasi-homogeneous by introducing homogeneous coordinates(x, ¯x) on P1, so C = {y2 −P2g+2i=0 ai¯x2g+2−ixi = 0}.

A basis for H0(C, L) is givenby the functions:zi = ¯xk−ixi,i = 0, . .

. , k;wi = y¯xk−g−1−ixi,i = 0, .

. .

, k −g −1.12

The equations for X are those for the scroll, and 2k −2g −1 further equations φm,obtained by multiplying the equation of C by suitable powers of ¯x and x, suchthat the result can be expressed in the zi and wi. These equations are obtainedfrom one another by rolling factors, as Miles Reid puts it [Reid 1989].

For eachmonomial in an equation we get the corresponding monomial in the next equationby multiplying with x/¯x, that is, with the quotient of the entries in a suitablecolumn of the matrix defining the scroll. We make here a specific choice; we recallthe notation ⌈r⌉for the round up and ⌊r⌋for rounding offa real number r, andnote that m2= m+12for integers m. This gives the following equations:Rankz0z1.

. .zk−1w0.

. .wk−g−2z1z2.

. .zkw1.

. .wk−g−1≤1,φm = w⌊m2 ⌋w⌈m2 ⌉−2g+2Xi=0aiz⌊m+i2 ⌋z⌈m+i2 ⌉,m = 0, .

. .

, 2k −2g −2.If L is an arbitrary bundle of degree d ≥2g+3, then we can write L = kg12 +Dwith deg D = g +1−e, and k maximal such that D is effective. The divisor D canbe described by two polynomials U(x) and V (x), together with F(x) = P2g+2i=0 aixi[Mumford 1984, § 1].

Suppose D = P1 + · · · + Pg+1−e with all Pi distinct and let(xi, yi) be the coordinates of Pi. Define U(x) = Qi(x −xi), and take V (x) theunique polynomial of degree ≤g −e with V (xi) = yi.

The ideal (U, y −V ) definesD; this is indeed a subvariety of y2 −F, and one has F −V 2 = UW for somepolynomial W.The function (y + V )/U defines a section of H0(C, L); a basis of this vectorspace can be represented in inhomogeneous coordinates by the polynomial forms:zi = Uxi,i = 0, . .

., k;wi = (y + V )xi,i = 0, . .

. , k −e.Modulo the equation y2 −F we have the relation (y + V )2 = UW + 2(y + V )V .From it we obtain d −2g −1 further equations φm, by rolling factors.WriteU(x) = Pg+1−ei=0Uixi, V (x) = Pg−ei=0 Vixi and W(x) = Pg+1+ei=0Wixi; this definesF.

Then we have the following equations:Rankz0z1. .

.zk−1w0. .

.wk−e−1z1z2. .

.zkw1. .

.wk−e≤1,φm =g+1−eXi=0Uiw⌊m+i2 ⌋w⌈m+i2 ⌉−2g−eXi=0Viw⌊m+i2 ⌋z⌈m+i2 ⌉−g+1+eXi=0Wiz⌊m+i2 ⌋z⌈m+i2 ⌉,m = 0, . .

. , d −2g −2.

(2.3) Rolling factors format [Reid 1989].This format occurs often in connectionwith divisors on scrolls.We start with a k-dimensional rational normal scrollS ∈P = Pn; the classical construction is to take k complementary linear subspacesLi, spanning P, with a parametrised rational normal curve φi: P1 →Ci ⊂Li ofdegree di = dim Li in it, and to take for each p ∈P1 the span of the points φi(p).13

The degree of S is f = P di. The scroll S can be defined by the minors of thematrix: z(1)0. .

.z(1)d1−1. .

.z(k)0. . .z(k)dk−1z(1)1. .

.z(1)d1. .

.z(k)1. . .z(k)dk!.The Picard group of S is generated by the hyperplane class H and the ruling R[Schreyer 1986, Sect.

1]; let C be a divisor of type aH −bR. The resolution ofOC as OS-module is 0 →OS(−aH + bR) →OS →OC →0.

Schreyer describes,following Eisenbud, Eagon-Northcott type complexes Cb such that Cb(a) is theminimal resolution of OS(−aH + bR) as OP-module, if b ≥−1 [loc. cit., 1.2].The resolution of OC is then obtained by taking a mapping cone.

We obtain thefollowing first terms of the resolution:(OfP(−1) ⊗Symb−1 O2P)(−a) ⊕∧3OfP(−1) ⊗O2P −→−→(Symb O2P)(−a) ⊕∧2OfP(−1) −→OP −→OC →0.The b + 1 equations φm, describing C on S, are obtained by ‘rolling’ along P1: interms of an inhomogeneous coordinate x on P1 we have z(i)j= ψ(i)xj for some func-tion ψ(i), and in the transition from the equation φm to φm+1, which is obtainedby multiplying with x, we have to increase the lower index by one for exactly oneof the factors z(i)jin each monomial. (2.4)Infinitesimal deformations.For the cone X over a hyperelliptic curve ofdegree at least 2g + 3, the dimension of T 1X(−1) is 2g + 2, by Proposition (1.25).We want to describe the action on the equations.

For this we have to study thenormal bundle of C.Let d = deg L ≥2g +3. For the scroll S = P(π∗L), we have OS(1) ∼E0 +af,so the divisor of the line bundle NC/S(−1) is C · (L + (2g + 2 −d)f).

From theexact sequence:0 →H0(NC/S(−1)) −→H0(NC(−1)) −→H0(NS|C(−1)) −→H1(NC/S(−1))we obtain that h0(C, NS|C(−1)) ≤h0(C, NC(−1)) −χ(NC/S(−1)) = d + g + 3 −(3g+5−d) = 2d−2g−2, where h0(C, NC(−1)) can be computed with the sequence0 −→V ∗⊗H0(C, OC) −→H0(C, NC(−1)) −→T 1X(−1) −→0, by Theorem (1.4).On the other hand, h0(S, NS(−1)) = 2d−2g−2, and all deformations are obtainedby deforming the matrix of the scroll; because H0(C, OC(1)) = H0(S, OS(1)), therestriction map H0(S, NS(−1)) →H0(C, NS|C(−1)) is injective and so for dimen-sion reasons an isomorphism. This shows that every infinitesimal deformation ofnegative weight of X comes from a deformation of the scroll.We only compute the case L = kg12; then NC/S(−1) = (2g + 2 −k)g12.

(2.5) Proposition.Let L = kg12 with k > g+1. On T 1(−1) we take coordinates(s1, .

. .

, sk−1, t0, . .

. , t2g+2−k), subject to the relations P2g+2i=0 aisi+j = 0 for j = 1,. .

. , k −2g −3.

The first order deformations are the given by:Rankz0z1 + s1. .

.zk−1 + sk−1w0. .

.wk−g−2z1z2. .

.zkw1. .

.wk−g−1≤1,φm +2g+2−kXi=0tizm+i −2g+2Xj=02g+2Xi=m+jaisi−⌈j2⌉zm+⌈j2⌉+m−1Xj=1m−j−1Xi=0aisi+⌊j2⌋zm−⌊j2⌋,14

where φm is as above, m = 0, . .

. , 2k −2g −2.

Furthermore, aj = 0 for j > 2g +2,s0 = 0 and sj = 0 for j ≥k. Therefore in the last term the summation really runsto min(2g + 2, m −j −1), and in the other term to min(2g + 2, k −1 + j2).Proof .

Consider the following relation, involving φm:zj+1φm −zjφm+1 + w⌈m2 ⌉(zjw⌈m+12 ⌉−zj+1w⌊m2 ⌋)−2g+2Xi=0aiz⌈m+i2 ⌉(zjz⌈m+i+12⌉−zj+1z⌊m+i2 ⌋) = 0.Now deform φm to φm +φ′m, and the other equations as given by the matrix. Thecondition that this defines an infinitesimal deformation, is that in the local ring ofthe singularity the following equation holds:zj+1φ′m −zjφ′m+1 +2g+2Xi=0ais⌊m+i2 ⌋zjz⌈m+i2 ⌉= 0.In terms of the inhomogeneous coordinates (x, y) we can write:xφ′m −φ′m+1 + x2g+2Xi=0ais⌊m+i2 ⌋x⌈m+i2 ⌉= 0.This equation is independent of j.

The relations involving wj+1φm give the sameset of equations. It is easily checked that the given formula for φ′m satisfies thisset of equations.

Furthermore, it gives a well defined element of the local ring: wecheck that no zi with i > k occurs with non-zero coefficient. If in the last termm −j −1 < 2g + 2, then j2≥ m2−g −1 and m − j2≤ m2+ g + 1 ≤k,because m ≤2k −2g −2.

Likewise, if m + j ≤2g + 2, then m + j2≤k.□(2.6) T 2.To determine the dimension of T 2 we use the Main Lemma of [Behnke–Christophersen 1991, 1.3.2]. Let h: X →C define a hyperplane section Y = h−1(0)of X, then dim T 2X/hT 2X = dim T 1Y −eh, where eh is the dimension of the smoothingcomponent, on which the smoothing h of Y lies.

A general hyperplane section Yis the cone over d points in Pd−g−1, lying on a rational normal curve of degreed −g −1. Equations for this curve singularity can of course also be written inrolling factors format: let the polynomial F(¯x, x) = Pdi=0 ai¯xd−ixi determine thepoints on the rational curve, then we have:Rankz0z1.

. .zd−g−2z1z2.

. .zd−g−1≤1,φm =dXi=0aiz⌊m+i2 ⌋z⌈m+i2 ⌉,m = 0, .

. .

, d −2g −2. (2.7) Lemma.The OY -module T 2Y is annihilated by the maximal ideal mY .15

Proof . Let N =d−g−12+ d −2g −1 be the number of equations.

We want toshow that hψ ∈Im Hom((OX)⊕k, OX) for every ψ ∈Hom(R/R0, OX) and everyh ∈mX. In the determinantal case this is shown in [Behnke–Christophersen 1991,2.1.1]; we remark that the proof as written is not correct: some equalities do nothold modulo R0, but only modulo the larger submodule RI of relations with entriesin the ideal I of X; this is not a serious problem, because Hom(R/RI, OX) =Hom(R/R0, OX).

Using their computations we may assume that hψ vanishes ondeterminantal relations.To describe the additional relations, we introduce the notation fi,j = zizj+1 −zi+1zj. We get:Rj,m = φm+1zj −φmzj+1 −dXi=0aifj,⌊m+i2 ⌋z⌈m+i2 ⌉,where 0 ≤j < d −g −1 and 0 ≤m < d −2g −2.

The determinantal relations areRi,j,k = fi,jzk −fi,kzj + fj,kzi and Si,j,k = fi,jzk+1 −fi,kzj+1 + fj,kzi+1. We havethe following equality:Rj,mzk −Rk,mzj −dXi=0aiRj,k,⌊m+i2 ⌋z⌈m+i2 ⌉= φmj(fj,k) −fj,kj(φm).Now fix an element zj ∈mY .

We look for ψm ∈OY with ψ(zjRk,m) = ψm+1zk −ψmzk+1 for all k and m.Because ψ(zjRk,m) = zkψ(Rj,m), we can determinethe ψm from the ψ(Rj,m), with the equations ¯xψ(Rj,m) −¯xψm+1 + xψm; takeψj = 0 and solve. For m > j we get that ¯xm−jψj,m = ¯xm−j−1xψ(Rj,m−1) +· · · + xm−jψ(Rj,j); because zj+kψ(Rj,l) = zjψ(Rj+k,l) we can divide by ¯xm−j.

Asimilar argument shows that we can solve for ψm with m < j.□Important for our application is that the corresponding result holds for T 2X;the computation is similar to the one for the hyperplane section, but the formula’sare more complicated. I expect that there is a general statement for the rollingfactors format, but it is not quite clear what the ‘generic’ rolling factors singularityis.We introduce more notation: gi,j = ziwj+1 −zi+1wj and hi,j = wiwj+1 −wi+1wj.

We have a relation Rj,m:Rj,m = φm+1zj −φmzj+1 −XUigj,⌊m+i2 ⌋w⌈m+i2 ⌉+ 2XVifj, m+i2 w m+i2+ 2XVigj, m+i−12z m+i+12+XWifj,⌊m+i2 ⌋z⌈m+i2 ⌉,where one term with Vi has to be chosen, depending on the parity of m + i;we have similar relations Sj,m, involving wj and wj+1. We find as above thatRj,mzk −Rk,mzj ≡0(mod RD), where RD is the submodule, generated bythe trivial and determinantal relations.

In the same way Sj,mzk −Rk,mwj ≡0(mod RD) and Sj,mwk −Sk,mwj ≡0(mod RD). This gives the desired result:(2.8) Lemma.The module T 2X is annihilated by the maximal ideal.16

(2.9) Proposition.Let Y be the cone over d points on the rational normal curveof degree d −g −1, with d < 2(d −g −1). Then dim T 1Y (−1) = d, dim T 1Y (0) =(g −1)(d −g −1), and dim T 1Y (ν) = 0 for ν ̸= 0, −1.Proof .

The d points are in generic position [Greuel 1982, 3.3]: a set of d points inPk is in general position, if for every n the images of the points under the n-tupleVeronese embedding of Pk span a linear space of maximal possible dimension, i.e.of dimension min{d, k+nn }−1. For points on a rational normal curve of degree k inPk we have to consider the composed kn-tuple embedding of P1; the dimension ofthe span is min{d, kn}−1.

In our case d < 2k, so every subset of the d points is ingeneral position, and by definition the d points are in generic position. ThereforeY is negatively graded in the sense of [Pinkham 1970], cf.

[Greuel 1982, 3.3], sodim T 1Y (ν) = 0 for ν > 0.The dimension of T 1Y (0) is equal to the number of moduli for d points inPd−g−1. Because every (quadratic) equations occurs in a linear relation, T 1Y van-ishes in degree < −1.

We finish the proof by computing T 1Y (−1) in three steps.Step 1.Every infinitesimal deformation of Y of degree −1 comes from a defor-mation of the cone over the rational normal curve of degree d −g −1.Proof . We may assume that (¯x, x) = (1, 0) or (0, 1) is not a root of the polynomialF, i.e.

a0 ̸= 0 and ad ̸= 0. Then ¯x and x are not zero divisors in OY .

Thereforewe get from the determinantal relations the equations:fj,j+k¯xk−1x = fj+1,j+k¯xk + fj,j+1xk.By induction we find:fj,j+k¯xk−1xk−1 =k−1Xi=0fj+i,j+i+1¯x2ix2(k−i−1). (∗j,k)Let n ∈Hom(I/I2, OY ) be a normal vector of degree −1 with n(fi,i+1) = gi.From equation (∗j,k) we get that n(fj,j+k)¯xk−1xk−1 = P gj+i¯x2ix2(k−i−1).

Weclaim that gj+i is divisible by ¯xk−i−1xi. Every gj+i occurs in (∗j+1,k) or (∗j,k−1),so by induction we obtain gj = ¯xk−2g′j, gj+k−1 = xk−2g′j+k−1 and the claim for0 < i < k −2.

Therefore we have x2g′j + ¯x2g′j+k−1 −¯xxψ ≡0(mod F) for somepolynomial ψ(¯x, x). The degree of x2g′j(¯x, x) is at most d −g + 1; because g > 1,we have in fact a polynomial equation in ¯x and x, so g′j is divisible by ¯x, andg′j+k−1 by x.

By taking k maximal we see that gj = λjzj + µjzj+1 + νjzj+2 forsome λj, µj, νj ∈C: this is the formula for the infinitesimal deformations of thecone over the rational normal curve.Step 2.The following formula’s define a d-dimensional subspace of T 1Y (−1);deformation parameters are (s1, . .

. , sd−g−2, t0, .

. .

, tg+1).Rankz0z1 + s1. .

.zd−g−2 + sd−g−2z1z2. .

.zd−g−1≤1,17

φm +g+1Xi=0tizm+i +d−g−2Xi=12i−2m−1Xj=0siai+1+⌈j2⌉zm+1+⌈j2⌉−d−g−2Xi=12m−2i−1Xj=0sia2i+1−m+⌈j2⌉zi+1+⌈j2⌉,where φm is as above, m = 0, . .

. , 2k −2g −2.

Furthermore, aj = 0 for j > d orj < 0.Proof . The deformation of the matrix gives the equation fj,k + sjzk+1 −skzj+1.Suppose φm is deformed to φm + φ′m.

The relations Rj,m give:¯xφ′m+1 −xφ′m +dXi=0ais⌊m+i2 ⌋x⌈m+i2 ⌉+1 = 0.The given formula satisfies these equations. The indices in it do not exceed d−g−1.Step 3.The deformations of the scroll, which deform fj,j+1 by gj = µjzj+1, donot extend to deformations of Y .Proof .

Suppose fi,j and φm deform with gi,j and φ′m. From equation (∗j,k) weget that gj,j+k = Pj+k−1i=jµix2j+k−i.

The relation Rj,m gives thatφ′m+1 −xφ′m +2j−m−1Xi=0aij−1Xk=⌊m+i2 ⌋µkxm+i−k −dXi=2j+2−mai⌊m+i2 ⌋−1Xk=jµkxm+i−k = 0.Because F(¯x, x) = 0, this expression is independent of j, so we take j = 0. Let αbe the smallest index such that µα ̸= 0.

We set m = 2d −2g −3:φ′2d−2g−2 −xφ′2d−2g−3 −dXi=0i−αXk=g+3−⌈d2⌉aiµi−kx2d−2g−3+k = 0.Because d −g −1 is the highest power of x, which can occur in φ′2d−2g−3, thecoefficient of 3d −2g −3 −α ≥2d −g −1 in the sum vanishes; so adµα = 0, whichcontradicts our assumption that ad ̸= 0. Therefore all µi are zero.□(2.10) Proposition.Let X = X(C, L) be the cone over a hyperelliptic curve ofgenus g, embedded with a complete linear system L of degree d, with d > 2g + 3.Then dim T 2X(−2) = d−2g−3, dim T 2X(−1) = (g−2)(d−g−3), and dim T 2X(ν) = 0for ν ̸= −1, −2.Proof .

For the hyperplane section Y the previous Proposition gives dim T 1Y =d + (g −1)(d −g −1); the dimension of a smoothing component is e = µ + t −1[Greuel 1982, 2.5. (3)], where t is the type of the singularity.

In our case t = g =δ(Y ) −d + 1, so e = 3g + d −2. Because T 2X is annihilated by the maximal ideal,the Main Lemma of [Behnke–Christophersen 1991] gives dim T 2X = dim T 1Y −e =(g −1)(d −g −4) −1.

By the formula of (1.26) we have T 2X(−1) = (g −2)h0(L) −6(g −1) + dim T 1X(−1). By Lemma (1.28), dim T 2X(ν) = 0 for ν < −1.□18

(2.11)Remark.One can also compute T 2X(−2) directly; we sketch this here forthe special case L = kg12. We have to show that dim ker{H0(L) ⊗H1(L−1) →H1(OC)⊕N} = 10k + 2g −7, where N is the number of equations.The el-ements y/(¯xg+1−ixi), i = 1, .

. ., g, form a basis of H1(OC); we abbreviate[i] = y/(¯xg+1−ixi), for all i ∈Z.

A basis of H0(L) ⊗H1(L−1) is:φi,j =1¯xk−ixi∂∂zji = 1, . .

. , k −1,j = 0, .

. .

, k,ψi,j =y¯xg+k+1−ixi∂∂zji = 1, . .

. , g + k,j = 0, .

. .

, k,χi,j =1¯xk−ixi∂∂wji = 1, . .

. , k −1,j = 0, .

. .

, k −g −1,θi,j =y¯xg+k+1−ixi∂∂wji = 1, . .

. , g + k,j = 0, .

. .

, k −g −1.On fl,m only the action of ψi,j is non trivial: ψi,l(fl,m) = [i−m−1], ψi,l+1(fl,m) =−[i−m], ψi,m+1(fl,m) = [i−l], ψi,m(fl,m) = [i−l−1] and ψi,j(fl,m) = 0 otherwise.Therefore:Xi,jαi,jψi,j(fl,m) =Xi(αi+m+1,l + αi+l,m+1 −αi+m,l+1 −αi+l+1,m)[i].The coefficient of [i] has to vanish for 1 ≤i ≤g. A set of solutions to theseequations is given by αi,j = αi+j for some constants αj, j = 1, .

. ., g + 2k.

Defineβi,j = αi+1,j −αi,j+1, where i = 1, . .

. , g+k−1 and j = 0, .

. ., k−1.

We have theequations βi+m,l = βi+l,m for i = 1, . .

. , g and 0 ≤l < m ≤k −1.

Consider pairs(i, j) and (i′, j′) with i + j = i′ + j′, and suppose that j = j′ + s(g −1) + (g −r).Write (i, j) ∼(i′, j′) if βi,j = βi′,j′. If i −j′ > g, we can use the equations toobtain (i, j) ∼(t + j, i −t) ∼(i + g −t, j −g + t), where t = r if s = 0, or t = 1for s > 0; for i −j′ < 0 we have a similar inductive procedure, whereas otherwisea suitable equation directly gives that (i, j) ∼(i′, j′) if i + j = i′ + j′.For the action of χi,j on hl,m we have analogous computations.

The next stepis to compute the action on φm:X((αi+j + jβi+j)ψi,j + (γi+j + jδi+j)χi,j)(φm) ==X(γi+⌊m2 ⌋+ m2δi+⌊m2 ⌋)[ m2−i] + (γi+⌈m2 ⌉+ m2δi+⌈m2 ⌉)[ m2−i]−al(αi+⌊m+l2 ⌋+ m+l2βi+⌊m+l2 ⌋)[ m+l2−i]−al(αi+⌈m+l2 ⌉+ m+l2βi+⌈m+l2 ⌉)[ m+l2−i]=X(2γm−c + mδm−c −2alαm+l−c −al(m + l)βm+l−c)[c]We get equations δj −P2g+2i=0 aiβj+i and 2γj −P2g+2i=0 ai(2αj+i −iβj+i), except forj = 2k −g −2, when there is only one equation; but there is no δ2k−g−2. So theparameters γj and δj are completely determined by αj and βj.

As β1 = β2k+g = 0,we have a (4k + 2g −2)-dimensional solution space.19

Finally, we compute the action on zlwm+1 −zl+1wm:X(αi,jφi,j + βi,jθi,j)(gl,m) =Xi(αi+m+1,l + βi+l,m+1 −αi+m,l+1 −βi+l+1,m)[i].The coefficient of [i] certainly vanishes, if αi,j and βi,j depend only on i+j. Defineγi,j = αi+1,j −αi,j+1 and δi,j = βi+1,j −βi,j+1.

Then γi+m,l = δi+l,m. One showsthat γi,j = δi,j = ǫi+j for all i, j.

The solution space has dimension 6k −5.□The dimension of T 2X(−2) is two less than the number of equations φm. Thisfact has an interpretation in terms of the rolling factors format, if all the φm arequadratic equations.

(2.12) Proposition.Let X be the cone over a divisor of type 2H −bR on ascroll S. Consider infinitesimal deformations of negative degree of X, which comefrom deformations of the cone over S; if the versal deformation of the cone over Sis: z(1)0z(1)1+ s(1)1. . .z(1)d1−1 + s(1)d1−1.

. .z(k)0. .

.z(k)dk−1 + s(k)dk−1z(1)1z(1)2. . .z(1)d1.

. .z(k)1. .

.z(k)dk!,then there are additional parameters ti, and linear equations on the ti and s(j)i ,such that the additional equations are given by φm + φ′m(t, s, z), m = 0, . .

. , b.Then this deformation can be extended over a base space given by b −1 equationsφ′m(t, s, s) −φm(s) = 0, m = 1, .

. .

, b −1.Proof . We have to lift the relations involving the φm.

The rolling factors assump-tion gives that we can write φm = Pα zαcα and φm+1 = Pα zα+1cα, where cαdepends (linearly) on z, and α runs through all possible indices (i)j , and cα depends(linearly) on z. This gives the relation φm+1z(i)j−φmz(i)j+1 −P f (i)jα cα, where f (i)jαis the determinantal equation z(i)j zα+1 −z(i)j+1zα.

We lift it to:(φm+1 + φ′m+1(z))(z(i)j+ s(i)j ) −(φm + φ′m(z + s))z(i)j+1 −X ˜f (i)j,αcα≡Xz(i)j+1sαcα(s)(mod I),where I is the ideal of the deformed scroll, and ˜f (i)jα is a deformed equation. Thelift up to first order is possible by assumption.

If 1 ≤m ≤b −1, then φm occursin a relation as first and as second term. Therefore φ′m(z) and φ′m(z + s) −φm(z)have to be equal.□(2.13) Example.Let X be the cone over a hyperelliptic curve, embedded with(2g + 2)g12.

With Proposition (2.5) we find as equations for the negative degreepart of the versal base:tsm −j=2g+2i=2g+2Xi=m+jj=0aisi−⌈j2⌉sm+⌈j2⌉+j=m−1i=m−j−1Xi=0j=1aisi+⌊j2⌋sm−⌊j2⌋+2g+2Xi=0ais⌈m+i2 ⌉s⌊m+i2 ⌋.20

In particular, if we take the curve y2 −1 + x2g+2, we have:tsm +2g+1−mXj=1s2g+2−⌈j2⌉sm+⌈j2⌉+mXj=2s⌊j2⌋sm−⌊j2⌋,m = 1, . .

. , 2g + 1.For g = 2 there are five equations: ts1 + 2s5s2 + 2s4s3, ts2 + 2s5s3 + s24 + s21,ts3 + 2s5s4 + 2s1s2, ts4 + s25 + 2s1s3 + s22, ts5 + 2s1s4 + 2s2s3.

This is a completeintersection of degree 25: it is the cone over 32 distinct points.We shall prove in the next Section that the base space is always a completeintersection. It is difficult to see this from the above equations.3.

Smoothing components(3.1) In this section we prove that the cone over a hyperelliptic curve of degree4g+4 has 22g+1 smoothing components. To each component corresponds a surfacewith the curve as hyperplane section; these are ruled surfaces, and we show how toobtain all by elementary transformations on a given one.

To this end we identifythe one dimensional subspace in T 1, determined by the surface.Let X ⊂PN+1 be the projective cone over a projectively normal algebraicvariety C ⊂PN of dimension n, and let S ⊂PN+1 be variety of dimension n + 1with C as hyperplane section. Then there exists a one parameter deformationof X with S as general fibre: let Y ⊂PN+2 be the projective cone over S, andconsider the pencil {Ht} of hyperplanes through PN ⊃C; let the hyperplane H0pass through the vertex, then Y ∩H0 = X, while for t ̸= 0 the projection fromthe vertex establishes an isomorphism between Y ∩Ht and S. Pinkham calls thisconstruction ‘sweeping out the cone’ [Pinkham 1970]; the degeneration to the coneoccurs already in the famous ‘Anhang F’ [Severi 1921].For the affine cone X over C we have a deformation with Milnor fibre S −C.The versal base S−X in negative degree is a fine moduli space for so called R-polarised schemes [Looijenga 1984], a notion defined in general for quasi-homo-geneous spaces; in our situation this are spaces S with C as hyperplane section.Basically one considers the coordinate t, which defines the hyperplane section, asdeformation parameter.Let S be given by equations Fi(x, t) = fi(x) + tf (1)i+ .

. .

+ tdif (di)i, in ho-mogeneous coordinates (x, t), where deg f (j)i= di −j. The base space of X has aC∗-action.

For simplicity we assume that the only occurring negative degree is −1,as in our application. The by S induced infinitesimal deformation of X is given byfi 7→∂∂tFi(x, t)|t=0 = f (1)i.

This can be interpreted as section of H0(C, NC(−1)).Let S be embedded by the line bundle L, and let P = P(H0(S, L)∗). LetC = H ∩S, with H a hyperplane in P, and let i: C →S be the inclusion.

ThenNC/S = L|C. In the exact sequence0 −→OC ∼= NC/S(−1) −→NC/P(−1) −→i∗NS/P(−1) −→0the infinitesimal deformation is the image of 1 ∈OC.

The curve C is minimallyembedded in the hyperplane H, so we really want a section of NC/H(−1); the21

exact sequence0 −→NC/H(−1) −→NC/P(−1) −→i∗NH/P(−1) −→0splits: NH/P(−1) is generated by the global section, which sends the equation t to1, and this section can be mapped to the section F 7→∂∂tF|t=0 of NC/P(−1).For curves we have a third description, which uses (T 1)∗. We recall fromProposition (1.3) the sequence 0 −→N ∗C ⊗L −→V ⊗C OC −→P1C(L) −→0.

Wehave seen that T 1X(−1)∗= coker{V ⊗H0(K) →H0(P1C(L) ⊗K)}. Consider thefollowing diagram of exact sequences:00yy0−→OC−→Ω1S ⊗OC(1)−→Ω1C(1)−→0yy0−→OC−→P1S(L) ⊗OC−→P1C(L)−→0yyL==Lyy00(3.2) Proposition.Let ξ ∈H0(C, NC(−1)) be the by S induced infinitesimaldeformation of the cone over C. The map ∪ξ: T 1(−1)∗→C is the connectinghomomorphism H0(P1C(L) ⊗K) →H1(K), obtained from the above sequence bytensoring with K = Ω1C; alternatively one may consider the map H0(K2L) →H1(K).Proof .

We first describe H0(P1(L) ⊗K). The sheaf P1(L) is generated by globalsections; if we take a local coordinate u on C and write a Newton dot for thederivative with respect to u, then the map V ⊗OC →P1(L) is given in accordancewith our earlier notations by dzi 7→zi(u) + ˙zi(u)du.

Let φ be a global section ofP1(L)⊗K. On Uj = {zj ̸= 0} the section φ can be represented by P ωijdzi, and therepresentations for different j are connected by the conditions P ωijzi = P ωikziand P ωij ˙zi = P ωik ˙zi.This allows us to compute φ ∪ξ: the cochain P(ωik −ωij)dzi represents anelement of H1(N ∗C(1) ⊗Ω1C) = H0(NC(−1))∗.

To express ξ in terms of the∂∂zi ,we take local coordinates (u, τ) on S with τ vanishing on C, and we denote dif-ferentation w.r.t. τ by ′.

For every equation F of S we have ∂F/∂τ = 0, so:t′(u, 0)∂F∂t +Xz′i(u, 0)∂F∂zi= 0,22

and therefore ξ = −P(z′i/t′) ∂∂zi . Now φ ∪ξ = P(ωij −ωik)(z′i/t′).To compute the connecting homomorphism δ in the exact sequence:0 →Ω1C →P1S(L) ⊗Ω1C →P1C(L) ⊗Ω1C →0,we lift φ on Uj to P1(L) ⊗K.

On Uj = {zj ̸= 0} we write φ as P ωijdzi + ωjdt,and the cochain conditions on S are: P ωijzi +ωjt = P ωikzi +ωkt, P ωij ˙zi +ωj ˙t =P ωik ˙zi+ωk ˙t and P ωijz′i+ωjt′ = P ωikz′i+ωkt′; for τ = 0 we have t = ˙t = 0, so thefirst two conditions are the same as before, and we have (ωk−ωj)t′ = P(ωij−ωik)z′i.The cochain ωk −ωj represents δ(φ) ∈H1(K), and therefore δ(φ) = φ ∪ξ.□We now concentrate on the hyperelliptic case. The cone over a hyperelipticcurve of degree d ≤4g + 4 is smoothable.

More precisely, we have:(3.3) Proposition [Tendian 1992b].Let gC be a hyperelliptic curve, embeddedwith a complete linear system L of degree 7g/3 + 1 ≤d ≤4g + 4.Then Cis a hyperplane section of a projectively normal surface S = φ|C|(S), where Scan be obtained from a rational ruled surface by blowing up 4g + 4 −d points.The dimension of the smoothing component is 7g + 4 −d + h0(C, K2L−1).Ifd > max{4g −4, 3g + 6} for g ̸= 6, and d > 25 for g = 6, then every smoothingcomponent is of this form, and has dimension 7g + 4 −d.In particular, for d = 4g + 4 the surface S is ruled, and the dimension ofthe smoothing component is 3g = (2g −1) + g + 1. The dimension of the hy-perelliptic locus is 2g −1, and g is the dimension of Picd.

So for general (C, L)each smoothing component determines a unique surface S with C as hyperplanesection. Furthermore dim T 1X(−1) = 2g + 2, and the base space in negative degreeis given by 2g + 1 equations.

From these facts we cannot yet conclude that theequations define a complete intersection.The existence of smoothings can be shown in the following way: given a linebundle L of degree 4g + 4 on C (which is not necessarily the linear system forwhich we want a smoothing — we denote that temporarily by N), we consider thescroll S of type (a, b), on which φL(C) lies: it is the image of the ruled surfaceS = P(π∗L), cf. (2.2).

On S the hyperplane class is L = E0 + af, and the curveC is a divisor of type 2E0 + (g + 1 + e)f = 2L −(2g + 2)f; here e = a −b. Sofor every L with L2 ∼= N ⊗(g12)2g+2 the normal bundle NC/S is isomorphic to ourfixed bundle N. The number of solutions to this equation is the order of the groupJ2(C) of 2-torsion points on Jac(C), which is 22g.For all surfaces S obtained by this construction we have e ≡g + 1(mod 2),because e = a −b, and a + b = 3g + 3.

This is not surprising, because for a fixedcurve and variable L the surfaces P(π∗L) form a continuous family, and the parityof e is conservated under deformations.Let C lie on on the ruled surface S ∼= Fe, with normal bundle NC/S = N, with0 ≤e ≤g + 1. Denote the elementary transformation [Hartshorne 1977, V.5.1.7]of S in Q ∈S by elmQ.Then we have the following simple, but importantobservation:(3.4) Lemma.Let C′ be the strict transform of C on S′ = elmP (S), where P isa Weierstraß point.

Then the normal bundle NC′/S′ is equal to the normal bundleNC/S.23

This lemma shows the existence of ruled surfaces C ⊂S ∼= Fe with NC/S = Nand g + 1 + e odd; again C is of type 2E0 + (g + 1 + e)f.(3.5) Proposition.Let B be the set of Weierstraß points on C. Let elmB bethe composition of the elementary transformations in all P ∈B. Then elmB(S) isisomorphic to S, under an isomorphism I, which leaves C pointwise fixed; on thegeneral fibre f of S →P1, I restricts to the unique involution with C ∩f as fixedpoints.

The isomorphims I maps the linear system |C| on S to the linear system|C′| on S′.Proof . The rational map elmB can be factorised as Sσ←−eSσ′−→S′, where the mapsσ and σ′ are the blowups in the points of B, with exceptional curves Ei and E′i.The involution on the general fibre f of eS →P1 with f ∩C as fixed points extendsto an involution eI on eS, which interchanges Ei and E′i.

This map descends tothe required isomorphism. The inverse image on eS of the linear system |C| is|C +P Ei|, where C is the strict transform of C. The involution eI transforms thissystem into |C + P E′i| = (σ′)∗|C′|.□(3.6) Proposition.Let C ⊂S ∼= Fe be a curve of type 2E0 + (g + 1 + e)f withnormal bundle NC/S = N. Denote for a subset T ⊂B by elmT the compositionof the elmPi, Pi ∈T.Two surfaces elmT1(S) and elmT2(S) induce the samedeformation of the cone X(C, N) if and only if T1 = T2 or T1 = B \ T2.Proof .

We have to compute the connecting homomorphism from the exact se-quence:0 →Ω1C →Ω1S ⊗Ω1C(1) →(Ω1C)⊗2(1) →0,or, what amounts to the same, the hyperplane in H0(C, K2(1)), which is the imageof H0(C, Ω1S ⊗Ω1C(1)). We have on S the sequence:0 →π∗Ω1P1 →Ω1S →Ω1S/P1 →0,and on C:0 →π∗Ω1P1 →Ω1C →Ω1C/P1 →0,where Ω1C/P1 = T 1X(−1)∗∼= ⊕P ∈BCP .

The map Ω1S ⊗Ω1C(1) →(Ω1C)⊗2(1) is anisomorphism on the subspace π∗Ω1P1 ⊗Ω1C(1). Because KS = −2E0 −(e + 2)f, wehave Ω1S/P1 = OS(−2E0 −ef), and Ω1S/P1(C + (g −1)f) = OS(2gf).

We identifyT 1X(−1)∗with H0(C, (Ω1C/π∗ΩP1) ⊗Ω1C(1)) by taking in each Weierstraß point Pia generator of K2(1)/mPiK2(1); the map from H0(C, Ω1S/P1 ⊗Ω1C(1)) consists intaking the coefficients in the points Pi. The sections of (g12)⊗2g are 1, .

. .

, x2g,which come from sections on S, and y, . .

. , yxg−1, which vanish in the Weierstraßpoints.

Therefore, if we take coordinates si on T 1X(−1)∗, and if x-coordinate of Piis ai, then the image of H0(C, Ω1S/P1 ⊗Ω1C(1)) is given by the determinant:D(s1, . .

., s2g+2) =1a1a21. .

.a2g1s1..................1a2g+2a22g+2. .

.a2g2g+2s2g+2.24

Now we consider the surface S′ = elmT (S) for some T ⊂B. Again we can identifythe sections of H0(C, Ω1S′/P1 ⊗Ω1C(1)), coming from S′, with the polynomials 1,. .

. , x2g, but we have to express these in the same basis of T 1X(−1)∗, which we usedfor S. Consider local coordinates (x, y) in a neighbourhood of a point P ∈T, suchthat ruling is given by π(x, y) = x, and C is y2 = x.

We blow S up in P; the stricttransform of C passes through the origin of the (η, y) coordinate patch, where(x, y) = (ηy, y). Now blow down the y-axis: we have coordinates (ξ, η) = (ηy, η),so x = ξ, y = ξ/η.

Therefore the local generator dy of Ω1C,P is transformed intodη: we have dy = dξ/η −ξdη/η2; however, considered as section of Ω1S/π∗Ω1P1, theformula makes sense for η ̸= 0 and is on C the same as −ξdη/η2 = −dη. Thiscomputation shows that S′ yields the hyperplane:D((−1)χT (P1)s1, .

. .

, (−1)χT (P2g+2)s2g+2) = 0,where χT is the characteristic function of T, i.e χT (Pi) = 1 if and only if Pi ∈T.To finish the proof we remark that the coefficient of si in the equation D is aVandermonde determinant, and therefore non zero.□(3.7) Theorem.Let X be the cone over a hyperelliptic curve C of genus g,embedded with a complete linear system L of degree 4g + 4. Suppose L ̸= 4K, ifg = 3.

Then X has 22g+1 smoothing components.Proof . The number of subsets T of B modulo the equivalence relation T ∼B\T is22g+1.

So Proposition (3.6) gives this number of one parameter smoothings of X.After a rescaling these define in P(T 1X(−1)) the points (±1 : . .

. : ±1), and this isa complete intersection.

In coordinates si on P(T 1X(−1)) the ideal of these pointsis generated by s2i −s2j. The 2g + 1 quadratic equations for the base in negativedegree are contained in this ideal; they generate this ideal if and only if theyare linearly independent.

Because the dimension of each smoothing componentis 3g, the equations are independent for generic (C, L). Suppose that for somespecial (C, L) the equations are dependent.

Then the base space of X(C, L) isnot smooth along at least one of the parameter lines of the 22g+1 smoothing wejust constructed. But the only singularities in the fibres are cones over a rationalnormal curve of degree g + 1, which have a smooth reduced base space, if g ̸= 3.Therefore the equations are always independent.For g = 3 we still have the same description of S−, but in case L = 4K thecone over the rational normal curve of degree 4 appears as singularity over onecomponent.

Its Veronese smoothing leads to an additional smoothing component;it exists also for the non hyperelliptic curves.□(3.8)Remark.The set B of subsets T of B modulo the equivalence relationT ∼B \ T forms a group, isomorphic to Z2g+1; the subgroup B+ of T’s of evencardinality is isomorphic to J2(C): every η ∈J2(C) can be represented by a divisorD ∼kg12 −Pi∈T Pi, with |T| = 2k. (3.9) Lemma.Let η ∈J2(C) be represented by D ∼kg12 −Pi∈T Pi, withT ∈B+.

Let S = P(π∗L). Then Sη = P(π∗(L ⊗η)) is isomorphic to elmT (S).25

Proof . Factorise elmT as Sσ←−eSσ′−→S′ = elmT (S): as before, the exceptionaldivisors are Ei and E′i.

We have L = E0 + af; let T0 ⊂T be the index set ofpoints Pi on E0, with cardinality m. Let E0 be the strict transform of E0 on eS,and E′0 that on S′. Then:σ∗L = E0 +Xj∈T0Ej + af= (σ′)∗E′0 −Xi/∈T0E′i +Xj∈T0Ej + af∼((σ′)∗E′0 + (a + m −k)f) +XTEi −kf.On S′ the linear system |E′0 + (a + m −k)f| cuts out on C the series |L + D|.

Wehave E′0 · E′0 = −e + 2(k −m), so if 2(k −m) < e, the section E′0 is the uniquenegative section, and we see directly that E′0 +(a+m−k)f is the hyperplane class(2a = 3g +3+e). Otherwise there is a section E−with E−·E−= −e′, e′ ≥0, andE′0 ∼E−+(k −m−e/2+e′/2)f. The hyperplane class is E−+(3g +3+e′)/2f ∼E′0 + (a + m −k)f.□(3.10) Alternative description of the construction.We have given the surfaces Sas ruled surfaces, birationally embedded with the linear system |C| = |2E0 + af|.Castelnuovo decribes linear systems in the plane, of curves of degree g +e+1 withone (g + e −1)-ple point and e −1 infinitely near double points, if 1 ≤e ≤g + 1,or for e = 0 of curves of degree g +3 with a (g +1)-ple point and a double point ina different point [Castelnuovo 1890].

If there are at least two finite double points,a standard Cremona transformation will decrease the degree, for details see thebook [Conforto 1939].Now given a curve C in the plane of degree d with δ double points (finite ornot), and with multiplicity d −2 at the origin, and given a set T of Weierstraßpoints of cardinality k, we form the curve C ∪(∪i∈T Li) of degree d + k, whereLi is the line joining the origin with the Weierstraß point Pi, and we consider thelinear system of plane curves of degree d + k with multiplicity d + k −2 at theorigin, and δ + k double points in the double points of C and in the Pi, i ∈T.As the curves of this system do not intersect Li outside C, the rational mapφ|C|(P2) →φ|C+PLi|(P2) is elmT . (3.11)Example ((2.13) continued).Let X be the cone over the hyperellipticcurve y2 −1 + x6, embedded with 6g12.

The coordinates on T 1X(−1) used in (2.13)are different from the ones in the proof of Proposition (3.6); in principle one cancompute the coordinate transformation, but we will not do this here. We describetwo surfaces with C as hyperplane section, and identify the induced line in T 1X(−1).We remark that the Weierstraß points are the sixth roots of unity, and that thegroup µ6 operates on B.We start with the plane curve x2y2z2 −y6 + x6.

The curves of degree 6 withmultiplicity 4 in O, and A3 singularities, tangent to x = 0 and y = 0, are:zi = y6−ixi,i = 0, . .

. , 6;wi = zyxy3−ixi,i = 0, .

. .

, 3;u = z2y2x2.26

This gives equations:Rankz0. .

.z5w0w1w2z1. .

.z6w1w2w3≤1,φm = w⌊m2 ⌋w⌈m2 ⌉−uzm,m = 0, . .

. , 6.The image of P2 is isomorphic to the cone over the rational normal curve of degree3.

With t = u −z0 + z6 we get a deformation of C, with si = 0.Now consider the Weierstraß point P0 = (1 : 1 : 0), and the linear system|C + L0| of septic curves with additional double point at P0. A basis is given by:ζi = (x −y)2y5−ixi,i = 0, .

. .

, 5;u0 = z2y3x2,u1 = z2y2x3;wi = zyx(x −y)y3−ixi,i = 0, . .

. , 3.On C we have (y−x)−1ζi = zi−zi+1, and (y−x)−1u0 = z0+.

. .+z5, (y−x)−1u1 =z1 + .

. .

+ z6; therefore we find as equalities on C:6(y −x)z0 = u0 + 5ζ0 + 4ζ1 + 3ζ2 + 2ζ3 + ζ46(y −x)z1 = u0 −ζ0 + 4ζ1 + 3ζ2 + 2ζ3 + ζ46(y −x)z2 = u0 −ζ0 −2ζ1 + 3ζ2 + 2ζ3 + ζ46(y −x)z3 = u0 −ζ0 −2ζ1 −3ζ2 + 2ζ3 + ζ46(y −x)z4 = u0 −ζ0 −2ζ1 −3ζ2 −4ζ3 + ζ46(y −x)z5 = u0 −ζ0 −2ζ1 −3ζ2 −4ζ3 −5ζ46(y −x)z6 = u0 −ζ0 −2ζ1 −3ζ2 −4ζ3 −5ζ4 −6ζ5.Furthermore, C is given by u0 −u1 −P ζi = τ. The matrix for the scroll elmP (S)has as first entry 6(y−x)z0; i.e.

we make a coordinate change ρz0 := u0+5ζ0+4ζ1+3ζ2+2ζ3+ζ4 (we are allowed to multiply homogeneous coordinates with a commonfactor). Below 6(y−x)z0 we have to write 6(y−x)z1 = u1+5ζ1+4ζ2+3ζ3+2ζ4+ζ5;the second entry on the first row is u0 −ζ0 + 4ζ1 + 3ζ2 + 2ζ3 + ζ4 = 6(y −x)z1 + τ.Therefore s1 = τ; proceeding in this way we find si = τ for i = 1, .

. .

, 5. Forthe coordinates wi we take 6(y −x)zyxy3−ixi.

Finally we have to compute t interms of τ. This is now a simple question of making the explicitly given coordinatetransformation in the equations, and the best way to do this is by computer.

Onefinds t = −4τ. Indeed, si = τ and t = −4τ is a solution of the equations for thebase space.The other components are found with similar computations.

(3.12) The Milnor fibre.In their paper [Looijenga–Wahl 1986] Looijenga andWahl introduce a collection of smoothing data for singularities, which in manycases distinguish between components. We determine these data in our case.Because the cone over an hyperelliptic curve C of degree 4g + 4, is a homoge-neous singularity, the interior of the Milnor fibre of a one parameter smoothing,given by a surface S with C as hyperplane section, is isomorphic to the open sur-face S\C.

We have described all smoothing components by smooth ruled surfaces,27

except when S is isomorphic to the cone over a rational normal curve of degreeg+1; then NC/S = (2g+2)g12. In the last case there does not exist a one-parametersmoothing, which lies totally in S−, but by slightly changing the normal bundle wefind a smooth ruled surface S on the same smoothing component, and the Milnorfibre is again S \ C.(3.13) Lemma.Let C be a curve of type 2E0 + (g + 1 + e)f on S ∼= Fe.

Thenthe Milnor fibre F of the cone X over C has homology H2(F, Z) = Z2g+1 andH1(F) = Z/2. If g + e + 1 ∼= 0(mod 2), then H1(F) = Z/2 and H2(F, ∂F) =Z2g+1 ⊕Z/2; otherwise H1(F) = 0 and H2(F, ∂F) = Z2g+1.Proof .

We compute the homology of F = S \C with the exact homology sequenceof the pair (S, S \C). For the surface S we have H3(S) = H1(S) = 0 and H2(S) =Pic(S).

We may take a new basis, consisting of E = E0 + g+1+e2f and f. ThenC ∼2E+ǫf with ǫ = 0 if g+1+e is even, and 1 otherwise, and E·E = g+1−ǫ. Wehave already encountered this division in cases.

Let N be a tubular neighbourhoodof C; by excision and Poincar´e Duality Hi(S, S \ C) = Hi(N, ∂N) ∼= H4−i(C).Therefore we have:H3(S, S \ C)֒→H2(S \ C)→H2(S)→H2(S, S \ C)։H1(S \ C)∥≀Z2g∥≀Z2ցyP DH2(C)The composed map H2(S) →H2(C) is given by intersecting with the curve C; sof 7→2 and E 7→2g + 2 −ǫ. If ǫ = 0, then H1(F; Z) = Z/2, and H1(F; Z) = 0, ifǫ = 1.

In the odd case ker{H2(S) →H2(S, S\C)} is generated by 2E−(2g+2−ǫ)f,and (2E −(2g +2−ǫ)f)2 = −4(g +1); in the even case a generator is E −(g +1)f,with self-intersection −(g + 1).We compute H2(F, ∂F) ∼= H2(F) with the cohomology sequence of the pair(S, S \ C). We obtain:0 −→H2(S, S \ C) −→H2(S) −→H2(S \ C) −→H3(S, S \ C) −→0.By Poincare duality Hi(S, S \ C; Z) ∼= H4−i(C), and the map H2(C) →H2(S) isthe transpose of the map H2(S) →H2(C) above.□The boundary ∂F of the Milnor fibre is diffeomorphic to the link M of thesingularity.

As M is a circle bundle over C, we have that H1(M) = Z2g⊕Z/(4g+4).The linking form on H1(M)t, the torsion part, lifts to a quadratic function, whichcan be computed from the resolution [Looijenga–Wahl 1986, 4.2, 4.6]. We have,with ¯e a generator of H1(M)t:q: H1(M)t →Q/Z,q(m¯e) = m(m −6g −2)8g + 8.By [loc.

cit., Thm. 4.5] the group H2(F), with intersection form I, carries thestructure of a quadratic lattice with associated non-degenerate lattice H2(F) =28

H2(F)/radI, whose discriminant quadratic function is canonically isomorphic to(I⊥/I, qI), where I is the q-isotropic subgroup Im{∂t: H1(F, M)t →H1(M)t}.In our case the radical of the intersection form has rank 2g: µ0 = 2g, µ−= 1and µ+ = 0. If g + 1 + e is odd, then H1(F, M)t = 0, so I = 0 and I⊥= H1(M)t.For g + 1 + e even H2(F) is generated by E −(g + 1)f, with self-intersection−(g + 1).

In this case the isotropic subgroup I has order two.We recall the definition of smoothing data [loc. cit., 4.16] in the case µ+ = 0.Let A be a finitely generated abelian group, and q: At →Q/Z a nonsingularquadratic function on its torsion part.

Then S(A, q) is the set of equivalence classesof 4-tuples (V, Q, I, i), where (V, Q) is a negative semi definite ordinary lattice, Iis a q-isotropic subspace of At, and i: V ∗/B′(V ) →A/I is an injective homomor-phism with finite cokernel, which induces an isomorphism V#/V →i⊥/I (hereB′: V →V ∗is the adjoint of the bilinear part of Q); two 4-tuples (V1, Q1, I1, i1)and (V2, Q2, I2, i2) are equivalent, if I1 = I2 and there exists an isomorphismΦ: (V1, Q1) →(V2, Q2) such that i2 = φ ◦i1, where φ is induced by Φ.Let J be the preimage of Im(i) in A. Then J ∩At = I⊥(so J determinesI).

By [loc. cit., 4.17] the elements in S(A, q), which give the same triple (V, Q, J)form O(qI)-orbit, where O(qI) is the group of linear automorphims preserving thequadratic function qI on I⊥/I.

Each orbit is equivalent to a coset space in O(qI)of Im{O(Q) →O(qI)}.For all our smoothings we have H1(M)t/I⊥∼= H1(F). In terms of J thismeans that At + J = A.In [loc.

cit.] this condition is shown to hold for allminimally elliptic singularities, with the concept of permissible quotients.

Denotefor our singularity X by S(X) the subset of S(H1(M), q) of lattices with µ−= 1,µ0 = 2g, for which H1(M)t+J = H1(M). The set of smoothing components mapsto S(X).

(3.14) Proposition.Suppose g is even. Then the set S(X) has 1+22g elements.On the set of even smoothing components (i.e.

with g + 1 + e even) the map toS(X) is injective, whereas the odd components all map to the same element.Proof . We first do not impose restrictions on g. By [Looijenga–Wahl 1986, Ex.4.6] the group H1(M)t has an isotropic subgroup of order r, if 4g + 4 = r2s,2g −2 = ru with s(1 + r) ≡u(mod 2).

So 8 ≡0(mod r). Possible valuesare r = 2, u = g −1, s = g + 1, for arbitrary g.If g = 8k + 3, then also(r, s, u) = (4, 2k + 1, 4k + 1) is a solution.Next we determine O(qI).

Write g +1 = h. First consider the case I = 0. Welook for a map ¯e 7→d¯e such that for all m:8h(q(md¯e) −q(m¯e) = (d −1)(m2(d + 1) −6hm + 4m) ≡0(mod 8h).Let G = gcd(d −1, 8h), and 8h = aG, then m2(d + 1) −6hm + 4m ≡0(mod a),and therefore d + 1 ≡6h −4, 2(d + 1) ≡0, so 16 ≡0(mod a).

Because d < 4h,possible values for a are 4, 8 and 16. Only for a = 8 we find solutions, given byd = h + 1, h = 8k −2 or d = 3h + 1, h = 8k + 2.

In these cases O(qI) is a groupof order two. If I is a group of order 2, with 2h¯e as generator, then I⊥/I ∼= Z/h,and we find that for h = 8k, d = 4k + 1, the group O(qI) has order two.29

So if h is odd, then the group O(qI) is trivial, and the number of elements S(X)is the number of triples (V, Q, J). If I = 0, then J = H1(M)t, and this is the onlypossibility.

Now suppose that I is a group of order two. In [loc.

cit., 6.2] it is shownthat Hom(H1(M)/H1(M)t, H1(M)t/I⊥) acts simply transitive on the collectionof subgroups J of H1(M) with H1(M)t ∩J = I⊥and H1(M)t + J = H1(M). Inour case this group is isomorphic to (Z/2)2g.To show injectivity, we have to identify this action with the action of the B+.The group H2(F, ∂F) is isomorphic to H2(S, C); we descibe this group with theexact sequence of the pair (S, C).

For H2(S) we have two generators, E and f,and C ∼2E. Take a basis of H1(C), in the traditional way (cf.

[Mumford 1984, p.3.75]): choose a certain system of 2g paths in P1 −B, B the set of branch points ofπ: C →P1, which each encircle an even number of branch points, and lift them toC; in P1 such a path αi bounds a disc Di. Now we consider the surface Sπ−→P1.In S a lift of αi to C is the boundary of a lift ∆i of Di to S. Then E, f and the∆i generate H2(F, ∂F).

The map to H1(M) is given by intersecting on S with theboundary M of a tubular neighbourhood of C (with as orientation the one inducedby F). The manifold M is a circle bundle over C, with Euler number −(4g + 4);let γ be the homology class of a fibre, then γ generates H1(M)t. A general fibreof Sπ−→P1 intersects M in two circles; let furthermore Ai be a lift of αi to M.Then the map ∂: H2(F, ∂F) →H1(M) is given by ∂(f) = 2γ, ∂(E) = 2(g + 1)γand ∂(∆i) = Ai.Consider now S′ = elmT (S), giving the Milnor fibre F ′.

Let αi encircle a pointP ∈T; we may suppose that ∆i does not pass through P, but intersects the fibreπ−1(P) once, outside the tubular neighbourhood of C.. The strict transform ∆′i onS′ is the element of H2(F ′, ∂F ′), which lifts αi.

But on S′ the disc ∆i intersects C′transversally, and therefore intersects M in a circle. So ∂′(∆′i) = Ai + tiγ, whereti is the number of branch points lying in Di.

If ti is odd, then ∂′(∆′i) does notlie in ∂(H2F, ∂F). This construction establishes an isomorphism between B+ andHom(H1(M)/H1(M)t, H1(M)t/I⊥).□(3.15) Remark.In the case g = 3 an isotropic subgroup I ⊂H1(M)t of order4 occurs for a smoothing of X(C, K4): a non hyperelliptic curve is hyperplanesection of 4-fold Veronese embedding of P2, i.e.

P2 embedded with O(4). It is wellknown that H1(P2 \ C) = π1(P2 \ C) = Z/4.

(3.16) Remark.For a simple elliptic singularity of multiplicity 8 = 4g + 4 thetopological computation still work; in this case for I = 0 the group O(qI) is trivial.Then the permissible smoothing data form a set of 1+22 = 5 elements [Looijenga–Wahl 1986, 6.4]. The number of smoothing components is also five; there are four‘even’ components and one ‘odd’ component.Let X be the cone over an elliptic curve E of degree 8, embedded with thelinear system L; so φL: E →P7.

We identify Pic8(E) with Jac(E) ∼= E in such away that L represents the origin of the group law: we write L = 8P. There are 4line bundles M of degree 4 with M ⊗2 = L. Each of these embeds E in P3; thenφL is the composition of φM with the the Veronese embedding V : P3 →P9.

Thecurve φM(E) lies on a pencil of quadrics, which is transformed by V in a pencilof hyperplane sections of V (P3); the common intersection is E. This construction30

gives a 2-dimensional linear subspace (a cone over P1) in the negative part ofthe base space of the versal deformation of X. The general quadric in the pencilis non-singular and has two rulings, but there are four singular quadrics, whichare cones with one ruling; the resulting double cover of P1 is isomorphic to E.The fifth component is isomorphic to the cone over E; given a point Q ∈E, weembed E in the plane with the linear system |2P + Q|, then embed P2 with thetriple Veronese embedding V3 in P9, and project the surface V3(P2) from the pointV3 ◦φ|2P +Q|(3Q) to P8.

Each of the resulting surfaces has a unique ruling.Each ruling determines a g12 ∈Pic2 ∼= E; on the even components a given g12occurs exactly once, and on the odd component four times. The operations elmT ,with T a subset of the set of branch points B of the g12, permute the 23 points inthe projectivised base space.

(3.17)Other degrees.The structure of the equations (see (2.2) and (2.12))and the dimension of T 2X(−2), which is d −2g −3, two less than the number ofequations, which cut out the curve on the scroll, lead us to expect that the basespace S−in negative degree is a complete intersection of dimension 4g + 5 −d, if2g + 4 ≤d ≤4g + 5. The number of equations will probably increase linearly ford > 4g +5, until it is (g +1)(2g +3), the number of quadratic monomials in 2g +2variables.For d = 4g + 5 the cone X has only conical deformations [Tendian 1992a],and therefore all infinitesimal deformations of negative degree are obstructed.

Inthis case we have indeed a zero-dimensional complete intersection.For d = 4g + 3 the space S−should be the cone over a complete intersectioncurve of degree 22g, which is a 22g-fold unbranched covering of C. One finds thisdegree from the genus formula’s. From the point of view of the surfaces, whichcorrespond to one parameter deformations, we get the following picture.

A surfaceS of degree 4g + 3 is a scroll, blown up in one point P of the hyperelliptic curveC. The fibration π: S →P1 has one singular fibre, which intersects C in P andin P, the image of P under the hyperelliptic involution.

By blowing down one ofthe two (−1)-curves in the exceptional fibre we get a ruled surface with a minimalsection E0 with E20 = −e; if we blow down the other curve, then the parity of eis just the opposite. Suppose that blowing down the curve through P gives aneven e. As we have seen, there are 22g different surfaces with e even, such thatthe characteristic linear system on C is a given linear system |L + P|.

Altogetherwe get a covering of P1 of degree 22g+1, with simple branching at the Weierstraßpoints, so indeed a 22g-fold unbranched covering of C. In particular, there is onlyone smoothing component, except of course when g = 3.We leave it to the reader to continue this description for still lower values ofthe degree d.31

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