A FORMULA FOR THE JACOBIAN OF A GENUS ONE CURVE OF ARBITRARY DEGREE

TOM FISHER

Abstract. We extend the formulae of classical invariant theory for the Jacobian of a genus one curve of degree n ≤ 4 to curves of arbitrary degree. To do this, we associate to each genus one normal curve of degree n, an n × n alternating matrix of quadratic forms in n variables, that represents the invariant differential. We then exhibit the invariants we need as homogeneous polynomials of degrees 4 and 6 in the coefficients of the entries of this matrix.

Introduction

Let C be a smooth curve of genus one defined over a field K. Its Jacobian is an elliptic curve E defined over the same field K. However it is only if C has a Krational point that C and E are isomorphic over K. Starting with equations for C we would like to compute a Weierstrass equation for E.

Let D be a K-rational divisor on C of degree n ≥ 1. It is natural to split into cases according to the value of n. If n = 1 then C has a K-rational point, and our task is that of writing an elliptic curve in Weierstrass form. If n ≥ 2 then the complete linear system |D| defines a morphism C → ℙ^{n−1}. Explicitly, the map is given by (f_{1} : …: f_{n}) where f_{1}, …, f_{n} are a basis for the Riemann-Roch space L(D). If n = 2 then C is a double cover of ℙ^{1} and is given by an equation of the form y^{2} = F (x_{1}, x_{2}) where F is a binary quartic. In this case Weil [W1], [W2] showed that the classical invariants of the binary quartic F give a formula for the Jacobian.

If n ≥ 3 then the morphism C → ℙ^{n−1} is an embedding. The image is a genus one normal curve of degree n. The word normal refers to the fact C is projectively normal (see for example [H, Proposition IV.1.2]), i.e. if H is the divisor of a hyperplane section then the natural map

S^{d}L(H) → L(dH)  (1)

is surjective for all d ≥ 1. If n = 3 then C ⊂ ℙ^{2} is a smooth plane cubic, say with equation F (x_{1}, x_{2}, x_{3}) = 0. The invariants of a ternary cubic F were computed by Aronhold [A], and again Weil (in the notes to [W1] in his collected papers) showed that these give a formula for the Jacobian. If n = 4 then C ⊂ ℙ^{3} is the complete intersection of two quadrics. If we represent these quadrics by 4 × 4 symmetric matrices A and B, then F (x_{1}, x_{2}) = det(Ax_{1} + Bx_{2}) is a binary quartic. The invariants of this binary quartic again give a formula for the Jacobian. For further details of these formulae in the cases n = 2, 3, 4 see [AKM^{3}P], [ARVT] or [F1].

If n = 5 then C ⊂ ℙ^{4} is no longer a complete intersection, and indeed the homogeneous ideal is generated by 5 quadrics. The Buchsbaum-Eisenbud structure theorem [BE1], [BE2] shows that these quadrics may be written as the 4×4 Pfaffians of a 5×5 alternating matrix of linear forms. The space of all such matrices is a 50-dimensional affine space, with a natural action of GL_{5} × GL_{5}. In [F1] we computed generators for the ring of invariants and showed that they again give a formula for the Jacobian. In fact the invariants are too large to write down as explicit polynomials, so instead we gave a practical algorithm for evaluating them (based in part on the case n = 5 of Proposition 9.3). More recently, B. Gross [G] gave a uniform description of the invariants in the cases n = 2, 3, 4, 5, using results of Vinberg, although this does not appear to give any way of evaluating the invariants in the case n = 5.

Date: 14th October 2015.

In this paper we extend these formulae for the Jacobian to genus one normal curves of arbitrary degree.

Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3. Since C has genus one, the space of regular differentials on C has dimension 1, say spanned by ω. We call ω an invariant differential, since geometrically it is invariant under all translation maps. There is a linear map

∧^{2}L(H) → L(2H) ; f ∧ g ↦ \frac{fdg − gdf}{ω} .  (2)

Since (1) is surjective for d = 2, we may represent this map by an n × n alternating matrix of quadratic forms in x_{1}, …, x_{n}. This matrix Ω represents ω in the sense that

ω = \frac{x_{j}^{2}d(x_{i}/x_{j})}{Ω_{ij}(x_{1}, …, x_{n})} for all i ≠ j.

However if n ≥ 4 then there are quadrics vanishing on C ⊂ ℙ^{n−1} and so this description does not determine Ω uniquely. Nonetheless we show, by proving [F3, Conjecture 7.4], that there is a canonical choice of Ω. We then define polynomials c_{4} and c_{6} of degrees 4 and 6 in the coefficients of the entries of Ω, and show that the Jacobian has Weierstrass equation

y^{2} = x^{3} − 27c_{4}(Ω)x − 54c_{6}(Ω).

These main results are stated in Section 1. In the next two sections we show that c_{4} and c_{6} are invariants for the appropriate action of GL_{n}, and that they reduce to the previously known formulae for n ≤ 5. At this point the proof of our results for any given value of n is a finite calculation. However finding a proof that works for all n is more challenging.

In Section 4 we show that if we can find a matrix Ω satisfying some apparently weaker hypotheses, then it will satisfy the properties claimed in Theorem 1.1. For the actual construction of Ω in Section 5 we reduce to the case C is an elliptic curve E embedded in ℙ^{n−1} via the complete linear system |n.0_{E}|. At first we specify Ω as a linear map ∧^{2}L(n.0_{E}) → S^{2}L(n.0_{E}), and use this in Section 6 to complete the proof of Theorem 1.1. Then in Section 7 we make a specific choice of basis for L(n.0_{E}), so that Ω becomes an alternating matrix of quadratic forms. We compute this matrix explicitly and, in Section 8, prove the formula for the Jacobian by computing c_{4}(Ω) and c_{6}(Ω). Much of the work here is in checking that the invariants c_{4} and c_{6} are scaled correctly for all n.

The description of Ω in Theorem 1.1 involves higher secant varieties. We quote any general results we need about these as required. Proofs, or references to the literature, are then given in Section 9.

In future work we plan to study the space of all matrices Ω. This appears to be defined by d_{1} + d_{2} quadrics in ℙ^{N−1} where N = (n^{2} − 1)(n^{2} − 4)/4 and

d_{1} = (n^{2} − 1)(n^{2} − 4)(n^{2} − 9)/36,
d_{2} = (n^{2} − 1)^{2}(n^{2} − 9)/9.

The numbers N, d_{1} and d_{2} are dimensions of irreducible representations for GL_{n}. Moreover, as suggested by Manjul Bhargava, we expect that d_{2} of the quadrics can be explained by an associative law, similar to that used in [B, Section 4].

We work throughout over a field K of characteristic 0, although it would in fact be sufficient that the characteristic is not too small compared to n. Except at the end of Section 1, where we give the application to computing Jacobians, we will assume that K is algebraically closed. For a projective variety X we write I(X) for its homogeneous ideal, and T_{P} X for the tangent space at P ∈ X.

1. Statement of results

Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3. For any integer r ≥ 1 the rth higher secant variety Sec^{r} C is the Zariski closure of the locus of all (r − 1)-planes through r points on C. For example, if r = 1 then Sec^{1} C = C. The codimension of Sec^{r} C in ℙ^{n−1} is max(n − 2r, 0). So according as n is odd or even there is a higher secant variety of codimension 1 or 2. If n = 2r + 1 then Sec^{r} C is a hypersurface of degree n, whereas if n = 2r + 2 then Sec^{r} C is the complete intersection of two forms of degree r + 1. In Section 9 we give references for these facts about higher secant varieties, and also explain how to compute equations for Sec^{r} C from equations for C.

We give the polynomial ring R = K[x_{1}, …, x_{n}] its usual grading by degree, say R = ⊕_{d}R_{d}, and write R(d) for the graded R-module with eth graded piece R_{d+e}. Maps between graded free R-modules are required to have relative degree 0, and are labelled by the matrices of forms that represent them. Our first main result is

Theorem 1.1. Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3.

(i) If n is odd, say n = 2r + 1, and Sec^{r} C = {F = 0} then there is a minimal free resolution

0⟶R(−2n) ⟶^{∇^{T}} R(−n − 1)^{n} ⟶^{Ω} R(−n + 1)^{n} ⟶^{∇} R

where Ω is an n × n alternating matrix of quadratic forms and

∇ = ∇(F) = (\frac{∂F}{∂x_{1}} ⋯ \frac{∂F}{∂x_{n}} ) .

(ii) If n is even, say n = 2r + 2, and Sec^{r} C = {F_{1} = F_{2} = 0} then there is a minimal free resolution

0⟶R(−n)^{2} ⟶^{∇^{T}} R(\frac{−n−2}{2})^{n} ⟶^{Ω} R(\frac{−n+2}{2})^{n} ⟶^{∇} R^{2}

where Ω is an n × n alternating matrix of quadratic forms and

∇ = ∇(F_{1}, F_{2}) = (\begin{matrix}\frac{∂F_{1}}{∂x_{1}} & ⋯ & \frac{∂F_{1}}{∂x_{n}} \\ \frac{∂F_{2}}{∂x_{1}} & ⋯ & \frac{∂F_{2}}{∂x_{n}} \end{matrix} ) .

We remarked in [F3, Section 7] that Theorem 1.1(i) follows from the Buchsbaum-Eisenbud structure theorem for Gorenstein ideals of codimension 3. In this paper we give a different proof, not only so that it runs in parallel with our proof of Theorem 1.1(ii), but also because this is needed for the proof of Theorem 1.2.

If the matrix Ω exists then, by the uniqueness of minimal free resolutions (see for example [E, Section 20.1] or [P, Section 7]), it is uniquely determined up to scalars. Moreover starting from equations for Sec^{r} C we can solve for Ω by linear algebra. The details are very similar to those in [F4, Section 4].

Let Ω = (Ω_{ij}) be as specified in Theorem 1.1. We put

M_{ij} = ∑_{r,s=1}^{n} \frac{∂Ω_{ir}}{∂x_{s}} \frac{∂Ω_{js}}{∂x_{r}} and N_{ijk} = ∑_{r=1}^{n} \frac{∂M_{ij}}{∂x_{r}} Ω_{rk}.  (3)

We then define

c_{4}(Ω) = \frac{3(n − 2)^{2}}{2^{4}n\binom{n+3}{5}} ∑_{i,j,r,s=1}^{n} \frac{∂^{2}M_{ij}}{∂x_{r}∂x_{s}} \frac{∂^{2}M_{rs}}{∂x_{i}∂x_{j}}  (4)

and

c_{6}(Ω) = \frac{−(n − 2)^{3}}{2^{6}n\binom{n+5}{7}} ∑_{i,j,k,r,s,t=1}^{n} \frac{∂^{3}N_{ijk}}{∂x_{r}∂x_{s}∂x_{t}} \frac{∂^{3}N_{rst}}{∂x_{i}∂x_{j}∂x_{k}} .  (5)

Let C_{1} and C_{2} be genus one curves with invariant differentials ω_{1} and ω_{2}. An isomorphism γ : (C_{1}, ω_{1}) → (C_{2}, ω_{2}) is an isomorphism of curves γ : C_{1} → C_{2} with γ^{∗}ω_{2} = ω_{1}.

Theorem 1.2. Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3, and let Ω be an alternating matrix of quadratic forms as specified in Theorem 1.1. Then

(i) There is an invariant differential ω on C such that

ω = \frac{x_{j}^{2}d(x_{i}/x_{j})}{Ω_{ij}(x_{1}, …, x_{n})} for all i ≠ j.

(ii) The pair (C, ω) is isomorphic (over K = \overline{K}) to

(y^{2} = x^{3} − 27c_{4}(Ω)x − 54c_{6}(Ω), 3dx/y).

The following corollary gives the application of Theorem 1.2 to computing Jacobians. For this result only we drop our assumption that K is algebraically closed.

Corollary 1.3. Let C ⊂ ℙ^{n−1} be a genus one normal curve defined over a field K. Suppose we scale the matrix Ω in Theorem 1.1 so that the coefficients of its entries are in K. Then C has Jacobian elliptic curve y^{2} = x^{3} − 27c_{4}(Ω)x − 54c_{6}(Ω).

Proof: Let E be the elliptic curve y^{2} = x^{3} − 27c_{4}(Ω)x − 54c_{6}(Ω). By Theorem 1.2 there is an isomorphism γ : C → E with γ^{∗}(3dx/y) = ω. Let ξ_{σ} = σ(γ)γ^{−1} for σ ∈ Gal(\overline{K}/K). Since 3dx/y and ω are both K-rational it follows that ξ_{σ}^{∗}(3dx/y) = 3dx/y. This implies, as explained for example in [F1, Lemma 2.4], that ξ_{σ} : E → E is a translation map. Then C is the twist of E by the class of {ξ_{σ}} in H^{1}(K, E). It follows by Theorems 3.6 and 3.8 in [S, Chapter X] that C is a principal homogeneous space under E, and E is the Jacobian of C. □

Remark 1.4. Although we will not need it for the proofs of Theorems 1.1 and 1.2, it is natural to ask whether C ⊂ ℙ^{n−1} is uniquely determined by Ω. The answer is that it is. Indeed by the minimal free resolutions in Theorem 1.1 we can recover ∇ from Ω. Then by Euler’s identity we obtain equations for Sec^{r} C where n − 2r = 1 or 2. This then determines Sec^{1} C = C by Theorem 9.1(v).

2. Changes of co-ordinates

We show that the constructions in Section 1 behave well under all changes of coordinates. First we define an action of GL_{n} on the space of all n × n alternating matrices of quadratic forms in x_{1}, … x_{n}. For g ∈ GL_{n} we put

g ⋆ Ω = g^{−T} (Ω(∑_{i=1}^{n} g_{i1}x_{i}, …, ∑_{i=1}^{n} g_{in}x_{i})) g^{−1}

where g^{−T} is the inverse transpose of g. Since the scalar matrices act trivially, this could equally be viewed as an action of PGL_{n}.

Lemma 2.1. Let C ⊂ ℙ^{n−1} and C′ ⊂ ℙ^{n−1} be genus one normal curves. Let Ω and Ω′ be alternating matrices of quadratic forms that satisfy the conclusions of Theorem 1.1, and define invariant differentials ω and ω′ on C and C′. If γ : C′ → C is an isomorphism given by

(x_{1} : …: x_{n}) ↦ (∑_{i=1}^{n} g_{i1}x_{i} : …: ∑_{i=1}^{n} g_{in}x_{i})

for some g ∈ GL_{n} then there exists λ ∈ K^{×} such that g ⋆ Ω = λΩ′ and γ^{∗}ω = λ^{−1}ω′.

Proof: Suppose n is odd, say n = 2r + 1 and Sec^{r} C = {F = 0}. Then Sec^{r} C′ is defined by

F′(x_{1}, …, x_{n}) = F (y_{1}, …, y_{n})

where y_{j} = ∑_{i=1}^{n} g_{ij}x_{i}. By the chain rule

∇(F′)(x_{1}, …, x_{n}) = ∇(F)(y_{1}, …, y_{n}) g^{T} .

Then

∇(F)Ω = 0 ⟹ ∇(F′)(g ⋆ Ω) = 0.

It follows by the uniqueness of minimal free resolutions that g ⋆ Ω = λΩ′ for some λ ∈ K^{×}. The case n is even is similar.

We also have γ^{∗}ω = μω′ for some μ ∈ K^{×}. If y_{j} = ∑_{i=1}^{n} g_{ij}x_{i} then

y_{s}^{2}d(y_{r}/y_{s}) = ∑_{i,j=1}^{n} g_{ir}g_{js}x_{j}^{2}d(x_{i}/x_{j}).

Dividing by γ^{∗}ω = μω′ gives

Ω(y_{1}, …, y_{n}) = μ^{−1}g^{T} Ω′(x_{1}, …, x_{n})g.

Hence g ⋆ Ω = μ^{−1}Ω′ and so μ = λ^{−1}. □

Lemma 2.2. The polynomials c_{4} and c_{6} are invariants for the action of GL_{n}, i.e. c_{4}(g ⋆ Ω) = c_{4}(Ω) and c_{6}(g ⋆ Ω) = c_{6}(Ω) for all g ∈ GL_{n}.

Proof: Let Ω′ = g ⋆ Ω, i.e.

Ω′_{ij}(x_{1}, …, x_{n}) = ∑_{a,b=1}^{n}(g^{−1})_{ai}(g^{−1})_{bj}Ω_{ab}(y_{1}, …, y_{n})

where y_{j} = ∑_{i=1}^{n} g_{ij}x_{i}. Direct calculation using (3) shows that

M′_{ij}(x_{1}, …, x_{n}) = ∑_{a,b=1}^{n}(g^{−1})_{ai}(g^{−1})_{bj}M_{ab}(y_{1}, …, y_{n}),

N′_{ijk}(x_{1}, …, x_{n}) = ∑_{a,b,c=1}^{n}(g^{−1})_{ai}(g^{−1})_{bj}(g^{−1})_{ck}N_{abc}(y_{1}, …, y_{n}).

Then

\frac{∂^{2}M′_{ij}}{∂x_{r}∂x_{s}} = ∑_{a,b,c,d=1}^{n}(g^{−1})_{ai}(g^{−1})_{bj}g_{rc}g_{sd} \frac{∂^{2}M_{ab}}{∂x_{c}∂x_{d}} ,

\frac{∂^{2}M′_{rs}}{∂x_{i}∂x_{j}} = ∑_{A,B,C,D=1}^{n}(g^{−1})_{Cr}(g^{−1})_{Ds}g_{iA}g_{jB} \frac{∂^{2}M_{CD}}{∂x_{A}∂x_{B}} .

Multiplying these together and summing gives

∑_{i,j,r,s=1}^{n} \frac{∂^{2}M′_{ij}}{∂x_{r}∂x_{s}} \frac{∂^{2}M′_{rs}}{∂x_{i}∂x_{j}} = ∑_{a,b,c,d=1}^{n} \frac{∂^{2}M_{ab}}{∂x_{c}∂x_{d}} \frac{∂^{2}M_{cd}}{∂x_{a}∂x_{b}} .

Thus c_{4}(Ω′) = c_{4}(Ω). A similar argument shows that c_{6}(Ω′) = c_{6}(Ω). □

The following corollary shows that to prove Theorems 1.1 and 1.2 for a fixed value of n, it suffices to prove them for a family of curves covering the j-line.

Corollary 2.3. Let Ω_{1} and Ω_{2} correspond to pairs (C_{1}, ω_{1}) and (C_{2}, ω_{2}). If there is an isomorphism γ : C_{1} → C_{2} with γ^{∗}ω_{2} = λω_{1} then c_{4}(Ω_{1}) = λ^{4}c_{4}(Ω_{2}) and c_{6}(Ω_{1}) = λ^{6}c_{6}(Ω_{2}).

Proof: After composing the isomorphism γ with a translation map, we may suppose it is given by a change of co-ordinate on ℙ^{n−1}. The case λ = 1 is immediate from Lemmas 2.1 and 2.2. In general we use that c_{4} and c_{6} are homogeneous polynomials of degrees 4 and 6. □

3. Curves of small degree

We compare our general formula for the Jacobian with the formulae previously known for genus one normal curves of degrees 3, 4 and 5.

For curves of degrees 3 and 4 it is easy to write down a matrix Ω satisfying the conclusions of Theorems 1.1 and 1.2(i). Indeed for C = {F (x_{1}, x_{2}, x_{3}) = 0} ⊂ ℙ^{2} a plane cubic we put

Ω = (\begin{matrix}0 & \frac{∂F}{∂x_{3}} & −\frac{∂F}{∂x_{2}} \\ −\frac{∂F}{∂x_{3}} & 0 & \frac{∂F}{∂x_{1}} \\ \frac{∂F}{∂x_{2}} & −\frac{∂F}{∂x_{1}} & 0 \end{matrix} ) ,

and for C = {F_{1} = F_{2} = 0} ⊂ ℙ^{3} a quadric intersection we let Ω be the 4 × 4 alternating matrix with entries

Ω_{ij} = \frac{∂F_{1}}{∂x_{k}} \frac{∂F_{2}}{∂x_{l}} − \frac{∂F_{1}}{∂x_{l}} \frac{∂F_{2}}{∂x_{k}} ,

where (i, j, k, l) is an even permutation of (1, 2, 3, 4). To prove Theorem 1.2(ii) in these cases we may check by direct computation that c_{4}(Ω) and c_{6}(Ω) are the classical invariants of a ternary cubic or quadric intersection, as scaled in [F1, Section 7]. We note that these are polynomials of degrees 4 and 6 in the coefficients of F, respectively of degrees 8 and 12 in the coefficients of F_{1} and F_{2}.

As described for example in [F4, Section 4], a genus one normal curve of degree n = 5 is defined by the 4×4 Pfaffians p_{1}, …, p_{5} of a 5×5 alternating matrix of linear forms on ℙ^{4}. We call the matrix of linear forms Φ a genus one model of degree 5, and note that there is a natural action of GL_{5} × GL_{5} on the space of all such models. It is shown in [H, Proposition VIII.2.5] that the secant variety Sec^{2} C is a hypersurface of degree 5 with equation F = 0 where F is the determinant of the Jacobian matrix of p_{1}, …, p_{5}. In [F3, Section 7] we proved that there is a degree 5 covariant Ω satisfying the conclusions of Theorems 1.1 and 1.2(i). We gave an explicit formula for this covariant in [FS, Section 2].

We claim that c_{4}(Ω) and c_{6}(Ω) are invariants for the action of SL_{5} × SL_{5}. For the action of SL_{5} via changes of co-ordinates on ℙ^{4} this follows from Lemma 2.2. For the action of SL_{5} via Φ ↦ AΦA^{T} it turns out that the coefficients of the entries of Ω are already invariants. Since Ω is a covariant of degree 5, the invariants c_{4}(Ω) and c_{6}(Ω) have degrees 20 and 30 in the coefficients of the entries of Φ. Computing a single numerical example (to check the scaling) shows that c_{4}(Ω) and c_{6}(Ω) are the same as the invariants c_{4}(Φ) and c_{6}(Φ) constructed in [F1].

4. Minimal free resolutions

Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3. Let Ω be an n × n alternating matrix of quadratic forms in x_{1}, …, x_{n}. In Sections 5 and 6 we exhibit Ω satisfying the following three hypotheses.

(H1) If n − 2r ≥ 1 and f ∈ I(Sec^{r} C) then ∑_{i=1}^{n} \frac{∂f}{∂x_{i}} Ω_{ij} ∈ I(Sec^{r} C) for all j.

(H2) If n − 2r = 2 and Sec^{r} C = \{F_{1} = F_{2} = 0\} then ∑_{i,j=1}^{n} \frac{∂F_{1}}{∂x_{i}} Ω_{ij} \frac{∂F_{2}}{∂x_{j}} = 0.

(H3) If n − 2r ≥ 1 then there exists P ∈ Sec^{r} C with rank Ω(P) = 2r.

In this section we prove:

Theorem 4.1. Let Ω be an n × n alternating matrix of quadratic forms, satisfying the hypotheses (H1), (H2) and (H3). Then there is a minimal free resolution as described in Theorem 1.1.

The next two propositions are proved in Section 9. By abuse of notation we write P both for a point in ℙ^{n−1} and for a vector of length n representing this point.

Proposition 4.2. If n − 2r ≥ 1 and P = ∑_{i=1}^{r} ξ_{i}P_{i} for some P_{1}, …, P_{r} ∈ C distinct and ξ_{1}, …, ξ_{r} ≠ 0 then the tangent space T_{P} Sec^{r} C is the linear span of the tangent lines T_{P_{1}}C, …, T_{P_{r}}C.

Proposition 4.3. Let ∇(F) and ∇(F_{1}, F_{2}) be as defined in Theorem 1.1.

(i) If n − 2r = 1 and Sec^{r} C = {F = 0} then the entries of ∇(F) define a variety in ℙ^{n−1} of codimension 3.

(ii) If n−2r = 2 and Sec^{r} C = {F_{1} = F_{2} = 0} then the 2×2 minors of ∇(F_{1}, F_{2}) define a variety in ℙ^{n−1} of codimension 3.

Proof: (i) Theorem 9.1 tells us that Sec^{r} C has singular locus Sec^{r−1} C, and that this has codimension 3. (ii) This is proved in Section 9.3. □

We start the proof of Theorem 4.1 with the following lemma.

Lemma 4.4. Let C ⊂ ℙ^{n−1} be a genus one normal curve. Suppose that n − 2r ≥ 1 and ℓ_{1}, …, ℓ_{n} are linear forms in x_{1}, …, x_{n} such that

∑_{i=1}^{n} ℓ_{i} \frac{∂f}{∂x_{i}} ∈ I(Sec^{r} C)  for all f ∈ I(Sec^{r} C).  (6)

Then there exists λ ∈ K such that ℓ_{i} = λx_{i} for all 1 ≤ i ≤ n.

Proof: The coefficients of ℓ_{1}, …, ℓ_{n} form an n × n matrix. Let V ⊂ Mat_{n}(K) be the subspace of all solutions to (6). We must show that V consists only of scalar matrices. Let E be the Jacobian of C. Translation by T ∈ E[n] is an automorphism of C that extends to an automorphism of ℙ^{n−1}, say given by a matrix M_{T}. Now V is stable under conjugation by each M_{T}. By considering the standard representation of the Heisenberg group (see for example [F2, Section 3]) it follows that V has a basis {M_{T} : T ∈ X} for some subset X ⊂ E[n].

We suppose for a contradiction that M_{T} ∈ V for some 0_{E} ≠ T ∈ E[n]. Then translation by T on C extends to an automorphism of ℙ^{n−1} that sends each point P ∈ Sec^{r} C to a point in the tangent space T_{P} Sec^{r} C. Let H be the divisor of a hyperplane section on C. For D an effective divisor on C we write \overline{D} ⊂ ℙ^{n−1} for the linear subspace cut out by L(H − D) ⊂ L(H). For example, if D is a sum of distinct points on C then \overline{D} is the linear span of these points. We also write D_{T} for D translated by T. We choose D = P_{1} + … + P_{r} an effective divisor of degree r such that

(i) P_{1}, …, P_{r} ∈ C are distinct,

(ii) D and D_{T} have disjoint support,

(iii) 2D + D_{T} ≁ H. Proposition 4.2 shows that for generic P ∈ \overline{D} we have T_{P} Sec^{r} C = \overline{2D}. It follows from our assumption M_{T} ∈ V that \overline{D_{T}} ⊂ \overline{2D}, equivalently L(H−2D) ⊂ L(H−D_{T}). Then by (ii) we have

L(H − 2D) = L(H − 2D) ∩ L(H − D_{T}) = L(H − 2D − D_{T}).

However by (iii) and the Riemann-Roch theorem these spaces do not have the same dimension. Indeed, since r ≥ 1 and n − 2r ≥ 1 we have

dim L(H − 2D) = n − 2r ≠ max(n − 3r, 0) = dim L(H − 2D − D_{T}).

This is the required contradiction. □

We show that the resolution in Theorem 1.1 is a complex.

Lemma 4.5. Let C ⊂ ℙ^{n−1} be a genus one normal curve, and let Ω be an alternating matrix of quadratic forms satisfying the hypotheses (H1) and (H2).

(i) If n = 2r + 1 and Sec^{r} C = {F = 0} then

∑_{i=1}^{n} \frac{∂F}{∂x_{i}} Ω_{ij} = 0 for all 1 ≤ j ≤ n.

(ii) If n = 2r + 2 and Sec^{r} C = {F_{1} = F_{2} = 0} then

∑_{i=1}^{n} \frac{∂F_{1}}{∂x_{i}} Ω_{ij} = ∑_{i=1}^{n} \frac{∂F_{2}}{∂x_{i}} Ω_{ij} = 0 for all 1 ≤ j ≤ n.

Proof: (i) By the hypothesis (H1) we have

∑_{i=1}^{n} \frac{∂F}{∂x_{i}} Ω_{ij} = ℓ_{j}F for all 1 ≤ j ≤ n,

for some linear forms ℓ_{1}, …, ℓ_{n}. We multiply by \frac{∂F}{∂x_{j}} and sum over j. Since Ω is alternating the left hand side is zero. Therefore

∑_{j=1}^{n} ℓ_{j} \frac{∂F}{∂x_{j}} = 0.

By Lemma 4.4 and Euler’s identity it follows that ℓ_{1} = … = ℓ_{n} = 0 as required. (ii) By the hypothesis (H1) we have

∑_{i=1}^{n} \frac{∂F_{1}}{∂x_{i}} Ω_{ij} = ℓ_{j}F_{1} + m_{j}F_{2}  for all 1 ≤ j ≤ n,  (7)

for some linear forms ℓ_{1}, …, ℓ_{n} and m_{1}, …, m_{n}. We multiply by \frac{∂F_{1}}{∂x_{j}} and sum over j. Since Ω is alternating the left hand side is zero. Since F_{1} and F_{2} are forms defining a variety of codimension 2 they must be coprime. Therefore

∑_{j=1}^{n} ℓ_{j} \frac{∂F_{1}}{∂x_{j}} = ξF_{2} and ∑_{j=1}^{n} m_{j} \frac{∂F_{1}}{∂x_{j}} = −ξF_{1}

for some ξ ∈ K. If instead we multiply (7) by \frac{∂F_{2}}{∂x_{j}} and sum over j then using the hypothesis (H2) we find that

∑_{j=1}^{n} ℓ_{j} \frac{∂F_{2}}{∂x_{j}} = ηF_{2} and ∑_{j=1}^{n} m_{j} \frac{∂F_{2}}{∂x_{j}} = −ηF_{1}

for some η ∈ K.

By Lemma 4.4 there exist λ, μ ∈ K such that ℓ_{i} = λx_{i} and m_{i} = μx_{i} for all 1 ≤ i ≤ n. By Euler’s identity and the linear independence of F_{1} and F_{2} it follows that λ = μ = 0. Therefore

∑_{i=1}^{n} \frac{∂F_{1}}{∂x_{i}} Ω_{ij} = 0 for all 1 ≤ j ≤ n.

The corresponding result for F_{2} follows by symmetry. □

To complete the proof of Theorem 4.1 we must show that the complex is exact. First we need some linear algebra. If B is an n×n matrix and S ⊂ {1, …, n} then we write B^{S} for the (n − |S|) × (n − |S|) matrix obtained by deleting the rows and columns indexed by S. The Pfaffian pf(M) of an alternating matrix M is a polynomial in the matrix entries with the property that det(M) = pf(M)^{2}.

Lemma 4.6. (i) Let A be a 1 × n matrix and B an n × n alternating matrix over a field K. Suppose that rank A = 1, rank B = n − 1 and AB = 0. Then there exists λ ∈ K^{×} such that

(−1)^{i} pf(B^{{i}}) = λa_{i}

for all 1 ≤ i ≤ n. (ii) Let A be a 2 × n matrix and B an n × n alternating matrix over a field K.

Suppose that rank A = 2, rank B = n − 2 and AB = 0. Then there exists λ ∈ K^{×} such that

(−1)^{i+j} pf(B^{{i,j}}) = λ(a_{1i}a_{2j} − a_{1j}a_{2i})

for all 1 ≤ i < j ≤ n.

Proof: (i) It is well known that the vector with ith entry (−1)^{i} pf(B^{{i}}) belongs to the kernel of B. See for example [BH, Section 3.4]. Since rank B = n − 1 this vector is non-zero and the kernel is 1-dimensional. The result follows. (ii) We first claim there exist λ_{1}, …, λ_{n} ∈ K such that

(−1)^{i+j} pf(B^{\{i,j\}}) = \{ \begin{matrix}λ_{i}(a_{1i}a_{2j} − a_{1j}a_{2i}) & if i < j, \\ −λ_{i}(a_{1i}a_{2j} − a_{1j}a_{2i}) & if i > j. \end{matrix}

Indeed taking a_{2i} times the first row of A minus a_{1i} times the second row of A, gives a non-zero vector in the kernel of B^{{i}}. If rank B^{{i}} = n − 2 then we argue as in (i). Otherwise we can simply take λ_{i} = 0. This proves the claim.

Now let C = (a_{1i}a_{2j} − a_{1j}a_{2i})_{i,j=1,...,n} and let D be the diagonal matrix with entries λ_{1}, …, λ_{n}. We must show that if CD = DC then CD is a scalar multiple of C. More generally this is true for any rank 2 alternating matrix C and diagonal matrix D. Indeed we may re-order the rows and columns so that the diagonal entries of D which are equal are grouped together. Then C is in block diagonal form. Since C is alternating of rank 2, exactly one of these blocks is non-zero. The result is then clear.

□

Lemma 4.7. Let C ⊂ ℙ^{n−1} be a genus one normal curve, and let Ω be an alternating matrix of quadratic forms satisfying the hypotheses (H1), (H2) and (H3).

(i) If n = 2r + 1 and Sec^{r} C = {F = 0} then the (n − 1) × (n − 1) Pfaffians of Ω are (scalar multiples of) the partial derivatives of F.

(ii) If n = 2r + 2 and Sec^{r} C = {F_{1} = F_{2} = 0} then the (n − 2) × (n − 2) Pfaffians of Ω are (scalar multiples of) the 2 × 2 minors of ∇(F_{1}, F_{2}).

Proof: We apply Lemma 4.6 over the function field K(x_{1}, …, x_{n}).

(i) By Lemma 4.5 we have ∑_{i=1}^{n} \frac{∂F}{∂x_{i}} Ω_{ij} = 0. By the hypothesis (H3) the generic rank of Ω is n − 1. So by Lemma 4.6(i) there exists λ ∈ K(x_{1}, …, x_{n}) such that

(−1)^{i} pf(Ω^{\{i\}}) = λ\frac{∂F}{∂x_{i}} for all 1 ≤ i ≤ n.

Since pf(Ω^{\{i\}}) and \frac{∂F}{∂x_{i}} are forms of degree n − 1, we can write λ = u/v where u and v are coprime forms of the same degree. Then v divides \frac{∂F}{∂x_{i}} for all i, and so must be a constant by Proposition 4.3(i). Therefore λ is a constant. (ii) By Lemma 4.5 we have ∑_{i=1}^{n} \frac{∂F_{1}}{∂x_{i}} Ω_{ij} = ∑_{i=1}^{n} \frac{∂F_{2}}{∂x_{i}} Ω_{ij} = 0. By the hypothesis (H3) the generic rank of Ω is n − 2. So by Lemma 4.6(ii) there exists λ ∈ K(x_{1}, …, x_{n}) such that

(−1)^{i+j} pf(Ω^{\{i,j\}}) = λ\frac{∂(F_{1}, F_{2})}{∂(x_{i}, x_{j})} for all 1 ≤ i < j ≤ n.

Since pf(Ω^{\{i,j\}}) and \frac{∂(F_{1},F_{2})}{∂(x_{i},x_{j})} are forms of degree n − 2, we can write λ = u/v where

u and v are coprime forms of the same degree. Then v divides \frac{∂(F_{1},F_{2})}{∂(x_{i},x_{j})} for all i, j, and so must be a constant by Proposition 4.3(ii). Therefore λ is a constant. □

Let R = K[x_{1}, …, x_{n}]. Consider a complex of graded free R-modules

0 → F_{m} ⟶^{φ_{m}} F_{m−1}⟶ … ⟶F_{1} ⟶^{φ_{1}} F_{0}.  (8)

We write V_{k} ⊂ ℙ^{n−1} for the subvariety defined by the r_{k} ×r_{k} minors of φ_{k} where r_{k} = rank(φ_{k}). The Buchsbaum-Eisenbud acyclicity criterion (see [BH, Theorem 1.4.13] or [E, Theorem 20.9]) states that (8) is exact if and only if rank F_{k} = rank φ_{k} + rank φ_{k+1} and codim V_{k} ≥ k for all 1 ≤ k ≤ m.

Proof of Theorem 4.1: We already saw in Lemma 4.5 that the resolution in Theorem 1.1 is a complex. We must prove it is exact. If n is odd then the free R-modules have ranks 1, n, n, 1 and the maps have ranks 1, n − 1, 1. If n is even then the free R-modules have ranks 2, n, n, 2 and the maps have ranks 2, n − 2, 2. By Lemma 4.7 we have V_{1} = V_{2} = V_{3} and Proposition 4.3 shows that this variety has codimension 3. We now apply the Buchsbaum-Eisenbud acyclicity criterion. □

5. A basis-free construction

The results of Section 2 show that for the proof of Theorems 1.1 and 1.2 we are free to make changes of co-ordinates on ℙ^{n−1}. Since we are working over an algebraically closed field we can therefore reduce to the following situation. Let E be the elliptic curve

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}

with point at infinity 0_{E} and invariant differential

ω = dx/(2y + a_{1}x + a_{3}) = dy/(3x^{2} + 2a_{2}x + a_{4} − a_{1}y).

Let C ⊂ ℙ^{n−1} be the image of E embedded via the complete linear system |n.0_{E}|. The embedding depends on a choice of basis for the Riemann-Roch space L(n.0_{E}), but the only effect of changing this is to make a change of co-ordinates on ℙ^{n−1}. In this section we define a linear map Ω : ∧^{2}L(n.0_{E}) → S^{2}L(n.0_{E}). In the next section we show that the corresponding alternating matrix of quadratic forms satisfies the hypotheses (H1), (H2) and (H3).

For f ∈ L(n.0_{E}) we put ḟ = df/ω ∈ L((n + 1).0_{E}). Motivated by (2) we define a linear map

A : ∧^{2}L(n.0_{E}) → S^{2}L((n + 1).0_{E})
f ∧ g ↦ f ⊗ ġ − g ⊗ ḟ.

Lemma 5.1. Let f, g ∈ L(n.0_{E}). Then the rational function on E × E given by

(P, Q) ↦ \frac{y_{P} + y_{Q} + a_{1}x_{Q} + a_{3}}{x_{P} − x_{Q}} (f(Q)g(P) − f(P)g(Q))

belongs to L((n + 1).0_{E}) ⊗ L((n + 1).0_{E}).

Proof: (i) If we fix Q = (x_{Q}, y_{Q}) then as rational functions of P = (x, y),

\frac{y + y_{Q} + a_{1}x_{Q} + a_{3}}{x − x_{Q}} ∈ L(0_{E} + Q) and f(Q)g − g(Q)f ∈ L(n.0_{E} − Q).

Therefore the product belongs to L((n + 1).0_{E}). (ii) If we fix P = (x_{P} , y_{P}) then as rational functions of Q = (x, y),

\frac{y_{P} + y + a_{1}x + a_{3}}{x_{P} − x} ∈ L(0_{E} + P) and g(P)f − f(P)g ∈ L(n.0_{E} − P).

Therefore the product belongs to L((n + 1).0_{E}). □

We define a second linear map

B : ∧^{2}L(n.0_{E}) → S^{2}L((n + 1).0_{E})
f ∧ g ↦ \frac{y_{P} + y_{Q} + a_{1}x_{Q} + a_{3}}{x_{P} − x_{Q}} (f(Q)g(P) − f(P)g(Q))|_{P =Q}

where |_{P =Q} is our notation for the natural map

L((n + 1).0_{E}) ⊗ L((n + 1).0_{E}) → S^{2}L((n + 1).0_{E}).

We show that A and B both represent the invariant differential ω, in the sense of Theorem 1.2(i).

Lemma 5.2. As rational functions on E we have

A(f ∧ g) = B(f ∧ g) = fġ − gḟ = \frac{fdg − gdf}{ω} .

Proof: This is clear for A. For B we apply l’Hôpital’s rule to get

\frac{f(Q)g − g(Q)f}{x − x_{Q}} |_{P =Q} = \frac{f(Q)ġ − g(Q)ḟ}{ẋ}|_{P =Q},

and then use that ẋ = 2y + a_{1}x + a_{3}. □

If we pick bases for L(n.0_{E}) and L((n + 1).0_{E}) then A and B are (represented by) n × n alternating matrices of quadratic forms in n + 1 variables. However the matrix Ω we seek is an n × n alternating matrix of quadratic forms in n variables. It turns out that the correct choice of Ω is a linear combination of A and B.

We may expand rational functions on E as Laurent power series in the local parameter t = x/y at 0_{E}. Let ϕ be the linear map that reads off the coefficient of t^{−n−1}. There are exact sequences

0 → L(n.0_{E}) → L((n + 1).0_{E}) →^{ϕ} K → 0

and

0 → S^{2}L(n.0_{E}) → S^{2}L((n + 1).0_{E}) →^{ϕ_{2}} L((n + 1).0_{E}) → 0  (9)

where ϕ_{2}(f ⊗ g) = ϕ(f)g + ϕ(g)f.

Lemma 5.3. Let f, g ∈ L(n.0_{E}) be rational functions whose coefficients of t^{−n} (when expanded as Laurent power series in t) are 0, 1 respectively. Then

ϕ_{2}(A(f ∧ g)) = nf and ϕ_{2}(B(f ∧ g)) = 2f.

Proof: (i) We have x = t^{−2} +… and y = t^{−3} +. . .. Then ẋ = 2y +a_{1}x+a_{3} = 2t^{−3} + … and ẏ = 3x^{2}+2a_{2}x+a_{4}−a_{1}y = 3t^{−4}+. . .. Writing g as a polynomial in x and y it follows that g = t^{−n} + … and ġ = nt^{−n−1} + . . .. Therefore ϕ(f) = ϕ(g) = ϕ(ḟ) = 0 and ϕ(ġ) = n. We compute

ϕ_{2}(A(f ∧ g)) = ϕ_{2}(f ⊗ ġ − g ⊗ ḟ) = nf.

(ii) If we fix Q = (x_{Q}, y_{Q}) then as rational functions of P = (x, y),

\frac{y + y_{Q} + a_{1}x_{Q} + a_{3}}{x − x_{Q}} = t^{−1} + … and f(Q)g − g(Q)f = f(Q)t^{−n} + …

with product f(Q)t^{−n−1} + . . ..

If we fix P = (x_{P} , y_{P}) then as rational functions of Q = (x, y),

\frac{y_{P} + y + a_{1}x + a_{3}}{x_{P} − x} = −t^{−1} + … and g(P)f − f(P)g = −f(P)t^{−n} + …

with product f(P)t^{−n−1} + . . .. In both cases the leading coefficient is f. Adding these together gives ϕ_{2}(B(f ∧ g)) = 2f. □

Corollary 5.4. Let Ω = nB − 2A. Then Ω is a linear map ∧^{2}L(n.0_{E}) → S^{2}L(n.0_{E}).

Proof: This follows from Lemma 5.3 and the exact sequence (9). □

6. Proof of Theorem 1.1

If we pick a basis for L(n.0_{E}) then the linear map defined in Corollary 5.4 is represented by an n × n alternating matrix of quadratic forms in n variables. In this section we complete the proof of Theorem 1.1 by showing that this matrix Ω satisfies the hypotheses (H1), (H2) and (H3), as stated at the start of Section 4.

For 0_{E} ≠ P ∈ E we write P and dP for the linear maps f ↦ f(P) and f ↦ ḟ(P) in the dual space L(n.0_{E})^{∗}. For example, if L(n.0_{E}) has basis 1, x, y, x^{2}, xy, … then

P = (1, x_{P} , y_{P} , x_{P}^{2} , x_{P} y_{P} , …),
dP = (0, 2y_{P} + a_{1}x_{P} + a_{3}, 3x_{P}^{2} + 2a_{2}x_{P} + a_{4} − a_{1}y_{P} , …).

We note that [P] is a point on C ⊂ ℙ^{n−1} = ℙ(L(n.0_{E})^{∗}), with tangent line passing through [dP]. The square brackets indicate that we are taking the 1-dimensional subspaces spanned by these vectors, i.e. the corresponding points in projective space. For 0_{E} ≠ Q ∈ E we likewise define Q and dQ.

For P, Q ∈ E let λ_{P,Q} be the slope of the chord (or tangent line if P = Q) joining P and Q. In the following lemma the vectors P, Q, dP, dQ in L(n.0_{E})^{∗} are extended to L((n + 1).0_{E})^{∗} using exactly the same definition. Evaluating A or B at a linear combination ξP + ηQ gives an element of (∧^{2}L(n.0_{E}))^{∗} = ∧^{2}(L(n.0_{E})^{∗}).

Lemma 6.1. Let 0_{E} ≠ P, Q ∈ E and ξ, η ∈ K. Then

A(ξP + ηQ) = ξ^{2}(P ∧ dP) + ξη(P ∧ dQ + Q ∧ dP) + η^{2}(Q ∧ dQ),
B(ξP + ηQ) = ξ^{2}(P ∧ dP) + ξη(λ_{Q,−P} − λ_{P,−Q})(P ∧ Q) + η^{2}(Q ∧ dQ).

Proof: (i) For f, g ∈ L(n.0_{E}) we compute

A(P)(f ∧ g) = (fġ − gḟ)(P) = (P ∧ dP)(f ∧ g).

The formula for A(ξP + ηQ) follows by bilinearity. (ii) For f, g ∈ L(n.0_{E}) we write

B(ξP + ηQ)(f ∧ g) = ξ^{2}B_{0} + ξηB_{1} + η^{2}B_{2}

By Lemma 5.2 we have

B_{0} = (fġ − gḟ)(P) = (P ∧ dP)(f ∧ g),
B_{2} = (fġ − gḟ)(Q) = (Q ∧ dQ)(f ∧ g).

Since for s, t ∈ L((n + 1).0_{E}) we have

(s ⊗ t)(ξP + ηQ) = s(ξP + ηQ)t(ξP + ηQ)

= ξ^{2}s(P)t(P) + ξη(s(P)t(Q) + s(Q)t(P)) + η^{2}s(Q)t(Q),

it follows from the definition of B that

B_{1} = λ_{P,−Q}(f(Q)g(P) − f(P)g(Q)) + λ_{Q,−P} (f(P)g(Q) − f(Q)g(P))
= (λ_{Q,−P} − λ_{P,−Q})(P ∧ Q)(f ∧ g).
□

We pick a basis for L(n.0_{E}), so that now Ω(P) is an n × n alternating matrix, and P, Q, dP, dQ are column vectors.

Lemma 6.2. Let 0_{E} ≠ P_{1}, …, P_{r} ∈ E distinct and ξ_{1}, …, ξ_{r} ∈ K. Then

Ω(∑_{i=1}^{r} ξ_{i}P_{i}) = Π (\begin{matrix}∗ & Ξ \\ −Ξ & 0 \end{matrix}) Π^{T}

where

Ξ = (\begin{matrix}(n − 2)ξ_{1}^{2} & −2ξ_{1}ξ_{2} & … & −2ξ_{1}ξ_{r} \\ −2ξ_{1}ξ_{2} & (n − 2)ξ_{2}^{2} & … & −2ξ_{2}ξ_{r} \\ ⋮ & ⋮ & ... & ⋮ \\ −2ξ_{1}ξ_{r} & −2ξ_{2}ξ_{r} & … & (n − 2)ξ_{r}^{2} \end{matrix})  (10)

and Π is the n × 2r matrix with columns P_{1}, …, P_{r}, dP_{1}, …, dP_{r}.

Proof: Recall that Ω = nB − 2A. The case r = 2 is immediate from Lemma 6.1. Since the entries of Ω are quadratic forms the general case follows. □

We now check the hypotheses (H1), (H2) and (H3).

Proof of (H1) and (H3): Suppose n − 2r ≥ 1. A generic point P ∈ Sec^{r} C may be written P = [∑_{i=1}^{r} ξ_{i}P_{i}] for some 0_{E} ≠ P_{1}, …, P_{r} ∈ E distinct and ξ_{1}, …, ξ_{r} ≠ 0. By Proposition 4.2 the tangent space T_{P} Sec^{r} C ⊂ ℙ^{n−1} is spanned by P_{1}, …, P_{r}, dP_{1}, … dP_{r}. In particular these 2r vectors are linearly independent.

For f ∈ I(Sec^{r} C) we have ∑_{i=1}^{n} \frac{∂f}{∂x_{i}} (P)v_{i} = 0 for any v in the linear span of P_{1}, …, P_{r}, dP_{1}, … dP_{r}. By Lemma 6.2 the columns of Ω are linear combinations of these vectors. So for each 1 ≤ j ≤ n the form ∑_{i=1}^{n} \frac{∂f}{∂x_{i}} Ω_{ij} vanishes at P. Since P ∈ Sec^{r} C is generic, this proves (H1). Since n ∉ {0, 2r} and ξ_{1}, …, ξ_{r} ≠ 0, the matrix (10) is non-singular. Therefore rank Ω(P) = 2r and this proves (H3). □

Proof of (H2): We write n = 2r and Sec^{r−1} C = {F_{1} = F_{2} = 0} where F_{1} and F_{2} are forms of degree r. We must show that the form

∑_{i,j=1}^{n} \frac{∂F_{1}}{∂x_{i}} Ω_{ij} \frac{∂F_{2}}{∂x_{j}}  (11)

is identically zero. A generic point P ∈ Sec^{r} C = ℙ^{n−1} may be written P = [∑_{i=1}^{r} ξ_{i}P_{i}] for some 0_{E} ≠ P_{1}, …, P_{r} ∈ E distinct and ξ_{1}, …, ξ_{r} ≠ 0. In addition we may assume that 2(P_{1} + … + P_{r}) ≁ H where H is the hyperplane section. This ensures that the vectors P_{1}, …, P_{r}, dP_{1}, … dP_{r} are linearly independent. We choose co-ordinates on ℙ^{n−1} so that [P_{1}] = (1 : 0 : …: 0), [P_{2}] = (0 : 1 : 0 : …: 0), …, dP_{r} = (0 : …: 0 : 1). Since F_{1} and F_{2} vanish on Sec^{r−1} C they vanish on the linear span of any r − 1 of the [P_{i}]. Replacing F_{1} and F_{2} by suitable linear combinations we may assume

F_{1}(x_{1}, …, x_{r}, 0, …, 0) = 0,
F_{2}(x_{1}, …, x_{r}, 0, …, 0) = x_{1}x_{2} … x_{r}.

Therefore at P = (ξ_{1} : …: ξ_{r} : 0 : …: 0) we have

(\frac{∂F_{1}}{∂x_{1}} (P), …, \frac{∂F_{1}}{∂x_{n}} (P)) = (0, …, 0, ∗, …, ∗),
(\frac{∂F_{2}}{∂x_{1}} (P), …, \frac{∂F_{2}}{∂x_{n}} (P)) = (∏_{i≠1} ξ_{i}, …, ∏_{i≠r} ξ_{i}, ∗, …, ∗).

By Lemma 6.2 we have

Ω(P) = (\begin{matrix}∗ & Ξ \\ −Ξ & 0 \end{matrix})

where Ξ is the matrix (10). Since n = 2r the coefficients in each row and column of Ξ sum to zero. Therefore the form (11) vanishes at P. Since P ∈ ℙ^{n−1} is generic, this shows that the form is identically zero. □

This completes the proof of Theorem 1.1.

7. Explicit formulae

In this section we give an explicit formula for the matrix Ω defined in Section 5. As before E is the elliptic curve

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.

with invariant differential ω = dx/(2y + a_{1}x + a_{3}). We embed E in ℙ^{n−1} via

(x_{0} : x_{2} : x_{3} : …: x_{n}) = (1, x, y, x^{2}, xy, x^{3}, x^{2}y, x^{4}, …)  (12)

Notice there is no x_{1}. The indicator function of a set X is denoted 1_{X}. We define linear forms in indeterminates {x_{m} : m ∈ ℤ} as follows

ẋ_{m} = \frac{m}{2}(2x_{m+1} + a_{1}x_{m} + a_{3}x_{m−2}) + 1_{m odd} ∑_{i=1}^{6}(−1)^{i}(m − \frac{i}{2})a_{i}x_{m+1−i}

\overline{x}_{m} = \frac{1}{2}(2x_{m+1} + a_{1}x_{m} + a_{3}x_{m−2}) + 1_{m odd} ∑_{i=1}^{6}(−1)^{i}a_{i}x_{m+1−i}

where by convention a_{5} = 0. For x ∈ ℝ we let sign(x) = −1, 0, 1 according as x is negative, zero or positive. For r, s ∈ ℤ we define

A_{rs} = x_{r}ẋ_{s} − x_{s}ẋ_{r},
B_{rs} = ∑_{k=−∞}^{∞} sign(k + \frac{1}{2})(x_{r+2k}\overline{x}_{s−2k} − x_{s+2k}\overline{x}_{r−2k}).

Theorem 7.1. Let C ⊂ ℙ^{n−1} be the image of E under the embedding (12).

(i) A = (A_{rs})_{r,s=0,2,...,n} and B = (B_{rs})_{r,s=0,2,...,n} are n × n alternating matrices of quadratic forms in x_{0}, x_{2}, …, x_{n+1}.

(ii) Ω = nB − 2A is an n × n alternating matrix of quadratic forms in x_{0}, x_{2}, …, x_{n}. It satisfies the conclusions of Theorem 1.1 and

(n − 2)ω = \frac{x_{j}^{2}d(x_{i}/x_{j})}{Ω_{ij}(x_{1}, …, x_{n})}  for all i ≠ j.  (13)

Proof: It is part of the theorem that the indeterminates x_{m} for m ∉ {0, 2, 3, …, n} cancel from the formula for Ω. So when applying the theorem we simply set them to be zero. However we will not do this in the proof. Since \overline{x}_{m} is a linear combination of x_{m+1}, x_{m}, …, x_{m−5} each B_{rs} is of the form ∑_{ij} c_{ij}x_{i}x_{j} where each c_{ij} is a finite sum. But it is not immediately clear that the B_{rs} are polynomials, i.e. that c_{ij} = 0 for all but finitely many pairs (i, j). We check this first.

If r ≡ s (mod 2) and r < s then

B_{rs} = 2(x_{r}\overline{x}_{s} + x_{r+2}\overline{x}_{s−2} + … + x_{s−2}\overline{x}_{r+2})  (14)

whereas if r is even and s is odd then

B_{rs} = −a_{1}x_{r}x_{s} + Q_{r,s+1} + a_{2}Q_{r,s−1} + a_{4}Q_{r,s−3} + a_{6}Q_{r,s−5} − Q_{s,r+1}  (15)

where

Q_{ij} = \{ \begin{matrix}x_{i}x_{j} + x_{i+2}x_{j−2} + … + x_{j}x_{i} & if i < j + 2, \\ 0 & if i = j + 2, \\ −(x_{i−2}x_{j+2} + x_{i−4}x_{j+4} + … + x_{j+2}x_{i−2}) & if i > j + 2. \end{matrix}

Since B_{sr} = −B_{rs} this proves that the B_{rs} are polynomials.

We show that the matrices A and B defined in the statement of the theorem represent the linear maps A and B defined in Section 5. The theorem then follows from the results of Sections 4, 5 and 6. In particular (13) follows from Lemma 5.2.

In the statement of the theorem the {x_{m} : m ∈ ℤ} are indeterminates. However for the proof they will be the following rational functions on E,

x_{m} = \{ \begin{matrix}x^{m/2} & if m is even, \\ x^{(m−3)/2}y & if m is odd. \end{matrix}

As rational functions on E, we claim that ẋ_{m} = dx_{m}/ω (in agreement with the notation in Section 5) and \overline{x}_{m} = \frac{1}{2}x_{m−2}(2y + a_{1}x + a_{3}). In checking these claims, we start with the right hand sides, since this also serves to motivate the definitions of ẋ_{m} and \overline{x}_{m}. For m even we have

dx_{m}/ω = \frac{m}{2}x^{(m−2)/2}(dx/ω)
= \frac{m}{2}x^{(m−2)/2}(2y + a_{1}x + a_{3})
= \frac{m}{2}(2x_{m+1} + a_{1}x_{m} + a_{3}x_{m−2}),

and

\frac{1}{2}x_{m−2}(2y + a_{1}x + a_{3}) = \frac{1}{2}(2x_{m+1} + a_{1}x_{m} + a_{3}x_{m−2}).

For m odd we have

dx_{m}/ω = \frac{m−3}{2}x^{(m−5)/2}y(dx/ω) + x^{(m−3)/2}(dy/ω)
= \frac{m−3}{2}x^{(m−5)/2}(2y^{2} + a_{1}xy + a_{3}y) + x^{(m−3)/2}(3x^{2} + 2a_{2}x + a_{4} − a_{1}y)
= \frac{m−3}{2}x^{(m−5)/2}(−a_{1}xy − a_{3}y + 2x^{3} + 2a_{2}x^{2} + 2a_{4}x + 2a_{6})
+ x^{(m−5)/2}(3x^{3} + 2a_{2}x^{2} + a_{4}x − a_{1}xy)
= x^{(m−5)/2}(mx^{3} − \frac{m−1}{2}a_{1}xy − \frac{m−3}{2}a_{3}y + ∑_{i=1}^{3}(m − i)a_{2i}x^{3−i})
= \frac{m}{2}(2x_{m+1} + a_{1}x_{m} + a_{3}x_{m−2}) + ∑_{i=1}^{6}(−1)^{i}(m − \frac{i}{2})a_{i}x_{m+1−i},

and

\frac{1}{2}x_{m−2}(2y + a_{1}x + a_{3}) = \frac{1}{2}x^{(m−5)/2}(2y^{2} + a_{1}xy + a_{3}y)
= \frac{1}{2}x^{(m−5)/2}(−a_{1}xy − a_{3}y + 2x^{3} + 2a_{2}x^{2} + 2a_{4}x + 2a_{6})
= \frac{1}{2}(2x_{m+1} + a_{1}x_{m} + a_{3}x_{m−2}) + ∑_{i=1}^{6}(−1)^{i}a_{i}x_{m+1−i}.

It is now clear that A(x_{r} ∧ x_{s}) = A_{rs} for all r, s ∈ ℤ. It remains to prove the same for B. By definition of B we have

B(x_{r} ∧ x_{s}) = \frac{y_{P} + y_{Q} + a_{1}x_{Q} + a_{3}}{x_{P} − x_{Q}} (x_{r}(Q)x_{s}(P) − x_{r}(P)x_{s}(Q))|_{P =Q}

where P, Q are points on E. Since \overline{x}_{m} = \frac{1}{2}x_{m−2}(2y + a_{1}x + a_{3}) we have

2x_{r}(P)\overline{x}_{s}(Q) = (2y_{Q} + a_{1}x_{Q} + a_{3})x_{r}(P)x_{s−2}(Q)

= \frac{2y_{Q} + a_{1}x_{Q} + a_{3}}{x_{P} − x_{Q}} (x_{r+2}(P)x_{s−2}(Q) − x_{r}(P)x_{s}(Q)).

Adding this to the same expression with (r, s) replaced by (s − 2, r + 2) and then setting P = Q gives

B_{rs} − B_{r+2,s−2} = 2(x_{r}\overline{x}_{s} + x_{s−2}\overline{x}_{r+2}) = B(x_{r} ∧ x_{s}) − B(x_{r+2} ∧ x_{s−2}). (16)

Rather more obviously, replacing (r, s) by (r+2, s+2) changes B_{rs} and B(x_{r} ∧x_{s}) in the same way, that is, by shifting the subscripts up by 2. So to prove B(x_{r} ∧x_{s}) = B_{rs} for all r, s ∈ ℤ it suffices to prove it for all r ∈ {0, 1} and s ∈ {0, 1, 2, 3}. This is a finite calculation. We give two examples:

B(x_{0} ∧ x_{3}) = \frac{y_{P} + y_{Q} + a_{1}x_{Q} + a_{3}}{x_{P} − x_{Q}} (y_{P} − y_{Q})|_{P =Q}

= \frac{(y_{P}^{2} + a_{1}x_{P} y_{P} + a_{3}y_{P}) − (y_{Q}^{2} + a_{1}x_{Q}y_{Q} + a_{3}y_{Q})}{x_{P} − x_{Q}} − a_{1}y_{P} |_{P =Q}

= (x_{P}^{2} + x_{P} x_{Q} + x_{Q}^{2} − a_{1}y_{P} + a_{2}(x_{P} + x_{Q}) + a_{4})|_{P =Q}

= 2x_{0}x_{4} + x_{2}^{2} − a_{1}x_{0}x_{3} + 2a_{2}x_{0}x_{2} + a_{4}x_{0}^{2},

and

B(x_{2} ∧ x_{3}) = \frac{y_{P} + y_{Q} + a_{1}x_{Q} + a_{3}}{x_{P} − x_{Q}} (y_{P} (x_{Q} − x_{P}) + x_{P} (y_{P} − y_{Q}))|_{P =Q}
= (−y_{P} (y_{P} + y_{Q} + a_{1}x_{Q} + a_{3}) + x_{P} (x_{P}^{2} + x_{P} x_{Q} + … + a_{4}))|_{P =Q}
= (x_{P}^{2} x_{Q} + x_{P} x_{Q}^{2} − y_{P} y_{Q} − a_{1}x_{Q}y_{P} + a_{2}x_{P} x_{Q} − a_{6})|_{P =Q}
= 2x_{2}x_{4} − x_{3}^{2} − a_{1}x_{2}x_{3} + a_{2}x_{2}^{2} − a_{6}x_{0}^{2}.

It is easy to check using (15) that these are equal to B_{03} and B_{23}. The other cases we need can then be checked using (16) and the fact that B is alternating. □

8. Proof of Theorem 1.2

Let Ω = nB − 2A be as in Theorem 7.1. Then c_{4}(Ω) = f_{n}(a_{1}, …, a_{6}) and c_{6}(Ω) = g_{n}(a_{1}, …, a_{6}) for some polynomials f_{n} and g_{n}. We consider the effect of a change of Weierstrass equation, with notation as in [S, Chapter III].

Lemma 8.1. Let a_{1}, …, a_{6} and a′_{1}, …, a′_{6} be the coefficients of two Weierstrass equations related by x = u^{2}x′ + r and y = u^{3}y′ + u^{2}sx′ + t. Then

f_{n}(a_{1}, …, a_{6}) = u^{4}f_{n}(a′_{1}, …, a′_{6})
g_{n}(a_{1}, …, a_{6}) = u^{6}g_{n}(a′_{1}, …, a′_{6})

Proof: This follows from Corollary 2.3 and u^{−1}ω′ = ω. □

It follows by Lemma 8.1, and the standard procedure for converting a Weierstrass equation to the shorter form y^{2} = x^{3} + ax + b, that f_{n} and g_{n} are scalar multiples of the usual polynomials c_{4} and c_{6} in a_{1}, …, a_{6}. Explicitly,

\begin{aligned}f_{n}(a_{1}, …, a_{6}) & = ξ_{n}(b_{2}^{2} − 24b_{4}) = ξ_{n}(a_{1}^{4} + …), \\ g_{n}(a_{1}, …, a_{6}) & = η_{n}(−b_{2}^{3} + 36b_{2}b_{4} − 216b_{6}) = η_{n}(−a_{1}^{6} + …), \end{aligned}  (17)

where b_{2} = a_{1}^{2} + 4a_{2}, b_{4} = 2a_{4} + a_{1}a_{3} and b_{6} = a_{3}^{2} + 4a_{6}.

To complete the proof of Theorem 1.2 we must compute the constants ξ_{n} and η_{n}. For any given value of n these can be read off from a single numerical example. However we need to compute these constants for all n. We write

Ω = Ω^{(0)} + a_{1}Ω^{(1)} + a_{2}Ω^{(2)} + a_{3}Ω^{(3)} + a_{4}Ω^{(4)} + a_{6}Ω^{(6)}.

Since c_{4}(Ω) and c_{6}(Ω) have degrees 4 and 6 in the coefficients of the entries of Ω, we see by (17) that it suffices to compute the invariants of Ω^{(1)}.

We put

γ_{rs} = (−1)^{max(r,s)} sign(s − r)n − 2 ((−1)^{s}⌊\frac{s}{2}⌋ − (−1)^{r}⌊\frac{r}{2}⌋) .

Lemma 8.2. The alternating matrix Ω^{(1)} has entries above the diagonal

γ_{rs}x_{r}x_{s} + (−1)^{s}n1_{r+s even} ∑_{k=1}^{(s−r)/2−1} x_{r+2k}x_{s−2k}.  (18)

Proof: Since Ω = nB − 2A we have Ω^{(1)} = nB^{(1)} − 2A^{(1)} where the superscripts indicate that we are taking the coefficient of a_{1}. Then A^{(1)} has (r, s) entry

((−1)^{s}⌊\frac{s}{2}⌋ − (−1)^{r}⌊\frac{r}{2}⌋) x_{r}x_{s}

whereas (14) and (15) show that if r < s then B^{(1)} has (r, s) entry

\{ \begin{matrix}(−1)^{s}(x_{r}x_{s} + x_{r+2}x_{s−2} + … x_{s−2}x_{r+2}) & if r ≡ s (mod 2) \\ (−1)^{s}x_{r}x_{s} & if r ≢ s (mod 2). \end{matrix}
□

Lemma 8.3. The matrices Ω^{(1)}, Ω′ = (γ_{rs}x_{r}x_{s})_{r,s=0,2,3,...,n} and

Λ = ((sign(j − i)n − 2(j − i))x_{i}x_{j})_{i,j=0,1,...,n−1}

all have the same invariants c_{4} and c_{6}.

Proof: We first explain why Ω^{(1)} and Ω′ have the same invariants, despite the “extra terms” in (18). We start with Ω^{(1)}. The only entries involving x_{0} are in the first row and column. We replace x_{0} by λ^{−1}x_{0} and multiply the first row and column by λ. By Lemma 2.2 this does not change the invariants, but setting λ = 0 removes the extra terms from the first row and column. Now the only entries involving x_{2} are in the second row and column. We replace x_{2} by λ^{−1}x_{2} and multiply the second row and column by λ. This does not change the invariants, but setting λ = 0 removes the extra terms from the second row and column. We repeat this procedure for all subsequent rows and columns. In the end we remove all the extra terms, and are left with the matrix Ω′.

We define a bijection π : {0, 1, …, n − 1} → {0, 2, 3, …, n} by

π(i) = \{ \begin{matrix}2i & if i ≤ n/2, \\ 2(n − i) + 1 & if i > n/2. \end{matrix}

We then compute

γ_{π(i),π(j)} = \{ \begin{matrix}sign(j − i)n − 2(j − i) & if i ≤ n/2 and j ≤ n/2, \\ −n − 2(−(n − j) − i) & if i ≤ n/2 and j > n/2, \\ n − 2(j + (n − i)) & if i > n/2 and j ≤ n/2, \\ sign(j − i)n − 2(−(n − j) + (n − i)) & if i > n/2 and j > n/2. \end{matrix}

In all cases we have γ_{π(i),π(j)} = sign(j −i)n−2(j −i). Therefore Ω′ and Λ are related by a permutation matrix. It follows by Lemma 2.2 that they have the same invariants.

□ Lemma 8.4. The alternating matrix of quadratic forms

Λ = (\begin{matrix}0 & (n − 2)x_{1}x_{2} & (n − 4)x_{1}x_{3} & (n − 6)x_{1}x_{4} & ⋯ & (2 − n)x_{1}x_{n} \\ & 0 & (n − 2)x_{2}x_{3} & (n − 4)x_{2}x_{4} & ⋯ & (4 − n)x_{2}x_{n} \\ & & 0 & (n − 2)x_{3}x_{4} & ⋯ & (6 − n)x_{3}x_{n} \\ & − & & & & ⋮ \\ & & & ... & & (n − 2)x_{n−1}x_{n} \\ & & & & & 0 \end{matrix})

has invariants c_{4}(Λ) = (n − 2)^{4} and c_{6}(Λ) = −(n − 2)^{6}.

Proof: We have Λ = (λ_{rs}x_{r}x_{s})_{r,s=1,...,n} where λ_{rs} = sign(s−r)n−2(s−r). Following the definitions of c_{4} and c_{6} in Section 1 we put

M_{ij} = ∑_{r,s=1}^{n} \frac{∂Λ_{ir}}{∂x_{s}} \frac{∂Λ_{js}}{∂x_{r}} = μ_{ij}x_{i}x_{j}

N_{ijk} = ∑_{r=1}^{n} \frac{∂M_{ij}}{∂x_{r}} Λ_{rk} = ν_{ijk}x_{i}x_{j}x_{k}

where μ_{ij} = (∑_{r=1}^{n} λ_{ir}λ_{jr}) − λ_{ij}^{2} and ν_{ijk} = μ_{ij}(λ_{ik} + λ_{jk}). It is not hard to show that

∑_{r=1}^{n} sign(i − r) sign(j − r) = n − 2|i − j| − δ_{ij},

∑_{r=1}^{n}(i − r) sign(j − r) = 2ij − j^{2} − (n + 1)i + n(n + 1)/2,

∑_{r=1}^{n}(i − r)(j − r) = nij − (i + j)n(n + 1)/2 + n(n + 1)(2n + 1)/6.

We use these to compute

∑_{r=1}^{n} λ_{ir}λ_{jr} = 2n|i − j|^{2} − 2n^{2}|i − j| − δ_{ij}n^{2} + (n^{3} + 2n)/3

and then subtract off

λ_{ij}^{2} = 4|i − j|^{2} − 4n|i − j| + (1 − δ_{ij})n^{2}

to get

μ_{ij} = 2(n − 2)(|i − j|^{2} − n|i − j|) + n(n − 1)(n − 2)/3.

Noting the symmetries μ_{ij} = μ_{ji} and ν_{ijk} = ν_{jik}, and using computer algebra to check our calculations, we find

∑_{i,j,r,s=1}^{n} \frac{∂^{2}M_{ij}}{∂x_{r}∂x_{s}} \frac{∂^{2}M_{rs}}{∂x_{i}∂x_{j}} = 4 ∑_{i≤j} μ_{ij}^{2} = (16/3)n(n − 2)^{2}\binom{n + 3}{5}

and

∑_{i,j,k,r,s,t=1}^{n} \frac{∂^{3}N_{ijk}}{∂x_{r}∂x_{s}∂x_{t}} \frac{∂^{3}N_{rst}}{∂x_{i}∂x_{j}∂x_{k}}

= 4 ∑_{i≤j≤k}(ν_{ijk} + ν_{jki} + ν_{kij})^{2}

= 4 ∑_{i≤j≤k}(λ_{ij}(μ_{ik} − μ_{jk}) + λ_{jk}(μ_{ij} − μ_{ik}) + λ_{ik}(μ_{ij} − μ_{jk}))^{2}

= 64(n − 2)^{2} ∑_{i≤j≤k}(i − 2j + k)^{2}(n + i + j − 2k)^{2}(n + 2i − j − k)^{2}

= 64n(n − 2)^{3}\binom{n + 5}{7}.

The final sums are evaluated using the standard formulae for ∑_{i=1}^{n} i, ∑_{i=1}^{n} i^{2}, etc. In practice it is simpler to observe that the answer is a polynomial in n, say of degree at most d, and then check the result for d + 1 distinct values of n.

Finally scaling by the constants included in the definitions (4) and (5) it follows that c_{4}(Λ) = (n − 2)^{4} and c_{6}(Λ) = −(n − 2)^{6}. □

The last two lemmas show that ξ_{n} = (n − 2)^{4} and η_{n} = (n − 2)^{6}. Therefore c_{4}(Ω) = (n − 2)^{4}c_{4}(E) and c_{6}(Ω) = (n − 2)^{6}c_{6}(E). Let ω = dx/(2y + a_{1}x + a_{3}). By the formulae in [S, Chapter III] we have

(E, ω) ≅ (y^{2} = x^{3} − 27c_{4}(E)x − 54c_{6}(E), 3dx/y).

Therefore

(E, (n − 2)ω) ≅ (y^{2} = x^{3} − 27c_{4}(Ω)x − 54c_{6}(Ω), 3dx/y).

Recalling from Theorem 7.1 that Ω = nB − 2A represents the invariant differential (n − 2)ω, this completes the proof of Theorem 1.2.

9. Higher secant varieties

In this final section we give references and proofs for the facts about higher secant varieties we used earlier in the paper.

Theorem 9.1. Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3. (i) Sec^{r} C ⊂ ℙ^{n−1} is an irreducible variety of codimension max(n − 2r, 0). (ii) The vector space of forms of degree r + 1 vanishing on Sec^{r} C has dimension

β(r + 1, n), where

β(r, n) = \binom{n − r}{r} + \binom{n − r − 1}{r − 1}

is the number of ways of choosing r elements from ℤ/nℤ such that no two elements are adjacent.

(iii) If n − 2r ≥ 2 then the homogeneous ideal I(Sec^{r} C) is generated by forms of

degree r + 1. (iv) if n − 2r = 1 then Sec^{r} C is a hypersurface of degree n.

(v) If n − 2r ≥ 1 then Sec^{r} C has singular locus Sec^{r−1} C.

Proof: (i) This is a general fact about curves. See for example [L, Section 1]. (ii), (iii), (iv). More generally the minimal free resolution for I(Sec^{r} C) was computed in [vBH, Section 8]. See [GP, Section 5] for the cases r = 1, 2, and [F2, Section 4] for further discussion.

(v) This is [vBH, Proposition 8.15]. □

9.1. Computing equations for higher secant varieties. The following two propositions may be used to compute equations for Sec^{r} C from equations for C. We say that a form f vanishes on C with multiplicity r if (passing to affine co-ordinates) the Taylor expansion of f at each point P ∈ C begins with terms of order greater than or equal to r.

Proposition 9.2. Let C ⊂ ℙ^{n−1} be a variety contained in no hyperplane. Let f be a form of degree r + 1.

(i) If r ≥ 1 then

f ∈ I(Sec^{r} C) ⟺ f vanishes on C with multiplicity r.

(ii) If r ≥ 2 then

f ∈ I(Sec^{r} C) ⟺ \frac{∂f}{∂x_{i}} ∈ I(Sec^{r−1} C) for all i = 1, …, n.

Proof: (i) We choose P_{1}, …, P_{n} ∈ C spanning ℙ^{n−1}. By a change of co-ordinates we may assume P_{1} = (1 : 0 : …: 0), P_{2} = (0 : 1 : 0 : …: 0), …, P_{n} = (0 : 0 : …: 1). If f ∈ I(Sec^{r} C) then it vanishes on the linear span of any r of the P_{i}. Therefore the monomials appearing in f involve at least r + 1 of the x_{i}, and since f has degree r + 1 must be squarefree. But then f vanishes at P_{1} with multiplicity r. Since P_{1} ∈ C was arbitrary it follows that f vanishes on C with multiplicity r.

Conversely, suppose f vanishes on C with multiplicity r. Let Π be an (r − 1)- plane spanned by points P_{1}, …, P_{r} ∈ C. By a change of co-ordinates we may assume P_{1} = (1 : 0 : …: 0), P_{2} = (0 : 1 : 0 : …: 0), . . .. Then f(x_{1}, …, x_{r}, 0, …, 0) has total degree r + 1, but has degree at most 1 in each of the variables. It follows that f vanishes on Π. By definition Sec^{r} C is the Zariski closure of the union of all such (r − 1)-planes. Therefore f ∈ I(Sec^{r} C) as required. (ii) Since char(K) = 0 this follows from (i). □

Now let C ⊂ ℙ^{n−1} be a genus one normal curve. Taking r = 1 in Theorem 9.1 shows that the homogeneous ideal I(C) is generated by a vector space of quadrics of dimension n(n − 3)/2. Suppose we know a basis for this space. Then by repeatedly applying Proposition 9.2(ii) we can find a basis for the space of forms of degree r + 1 vanishing on Sec^{r} C. Theorem 9.1(iii) tells us that if n − 2r ≥ 2 then these forms define Sec^{r} C. The following proposition covers the remaining case.

Proposition 9.3. Suppose n − 2r = 1. Let f be a form of degree n. If r ≥ 2 then

f ∈ I(Sec^{r} C) ⟺ \frac{∂f}{∂x_{i}} ∈ I(Sec^{r−1} C)^{2} for all i = 1, …, n.

Proof: “⇒” Let H be the divisor of a hyperplane section, and let P ∈ C be any point. Let C_{+} ⊂ ℙ^{n} and C_{−} ⊂ ℙ^{n−2} be the images of C embedded via the linear systems |H ± P |. We choose co-ordinates so that the isomorphisms C_{+} → C → C_{−} are given by

(x_{1} : …: x_{n+1}) ↦ (x_{1} : …: x_{n}) ↦ (x_{1} : …: x_{n−1}).

In particular P is the point (x_{1} : …: x_{n}) = (0 : …: 0 : 1). By Theorem 9.1 we know that I(Sec^{r−1} C_{−}) is generated by forms g_{1}, g_{2} ∈ K[x_{1}, …, x_{n−1}] of degree r. By [F2, Corollary 2.3] there exist forms h_{1}, h_{2} ∈ K[x_{1}, …, x_{n}] of degree r + 1 such that f_{i} = x_{n+1}g_{i} + h_{i} ∈ I(Sec^{r} C_{+}) for i = 1, 2. Then F = g_{1}h_{2} − g_{2}h_{1} belongs to

I(Sec^{r} C_{+}) ∩ K[x_{1}, …, x_{n}] = I(Sec^{r} C).

Since g_{1}, g_{2} are coprime and f_{1}, f_{2} are irreducible it is clear that F is non-zero. By Theorem 9.1(iv) we have I(Sec^{r} C) = (F). We compute

\frac{∂F}{∂x_{n}} = \frac{∂f_{1}}{∂x_{n+1}} \frac{∂f_{2}}{∂x_{n}} − \frac{∂f_{1}}{∂x_{n}} \frac{∂f_{2}}{∂x_{n+1}} .

On the other hand, for i = 1, 2 and j = n, n + 1 we have

\frac{∂f_{i}}{∂x_{j}} ∈ I(Sec^{r−1} C_{+}) ∩ K[x_{1}, …, x_{n}] = I(Sec^{r−1} C).

Therefore \frac{∂F}{∂x_{n}} ∈ I(Sec^{r−1} C)^{2}. Since P ∈ C was arbitrary, and C spans ℙ^{n−1}, the result follows. “⇐” Let P_{1}, …, P_{r} be r distinct points on C. By a change of co-ordinates we may assume P_{1} = (1 : 0 : …: 0), P_{2} = (0 : 1 : 0 : …: 0), . . .. By Proposition 9.2 we know that f vanishes on C with multiplicity 2(r − 1) + 1 = n − 2. Therefore f(x_{1}, …, x_{r}, 0, …, 0) has total degree n, but has degree at most 2 in each of the variables. Since 2r < n it follows that f vanishes on the linear span of P_{1}, …, P_{r}. By definition Sec^{r} C is the Zariski closure of the union of all such (r − 1)-planes. Therefore f ∈ I(Sec^{r} C) as required. □

9.2. Proof of Proposition 4.2. Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n. Let H be the divisor of a hyperplane section. We identify L(H) with the space of linear forms on ℙ^{n−1}. For D an effective divisor on C we write \overline{D} ⊂ ℙ^{n−1} for the linear subspace cut out by L(H − D) ⊂ L(H). We have

Sec^{r} C = ⋃_{deg D=r} \overline{D}.

We also put D^{∘} = \overline{D} \ ∪_{D′<D}\overline{D′}. The gcd and lcm of divisors ∑ m_{P} P and ∑ m′_{P} P are ∑ min(m_{P} , m′_{P})P and ∑ max(m_{P} , m′_{P})P.

Lemma 9.4. Let D, D_{1}, D_{2} be effective divisors on C.

(i) If deg D < n then dim \overline{D} = deg D − 1. (ii) The linear span of \overline{D_{1}} and \overline{D_{2}} is \overline{lcm(D_{1}, D_{2})}. (iii) If deg(lcm(D_{1}, D_{2})) < n then \overline{D_{1}} ∩ \overline{D_{2}} = \overline{gcd(D_{1}, D_{2})}.

Proof: (i) By Riemann-Roch we have dim L(H − D) = n − deg D. (ii) We have L(H − D_{1}) ∩ L(H − D_{2}) = L(H − lcm(D_{1}, D_{2})). (iii) The inclusion “⊃” is clear. Equality follows by counting dimensions using (i) and (ii). □

With the above notation, Proposition 4.2 becomes Proposition 9.5. Suppose n − 2r ≥ 1. Let D = P_{1} + … + P_{r} be an effective divisor of degree r with P_{1}, …, P_{r} ∈ C distinct. Then for any P ∈ D^{∘} we have T_{P} Sec^{r} C = \overline{2D}.

Proof: If P ∈ \overline{D′} for D′ an effective divisor of degree at most r, then by Lemma 9.4(iii) we have D = D′. In particular P ∉ Sec^{r−1} C. It follows by Theorem 9.1(v) that P is a smooth point on Sec^{r} C. The next lemma shows that \overline{2D} ⊂ T_{P} Sec^{r} C, and equality follows by comparing dimensions, using Lemma 9.4(i) and Theorem 9.1(i). □

Lemma 9.6. Let X be an affine variety and P_{1}, …, P_{r} ∈ X. Let P = ∑ ξ_{i}P_{i} where ∑ ξ_{i} = 1. If ξ_{i} ≠ 0 then T_{P_{i}}X ⊂ T_{P} (Sec^{r} X).

Proof: There is a morphism X × … × X → Sec^{r} X; (a_{1}, …, a_{r}) ↦ ∑ ξ_{i}a_{i} with derivative T_{P_{1}}X × … × T_{P_{r}}X → T_{P} (Sec^{r} X) ; (b_{1}, …, b_{r}) ↦ ∑ ξ_{i}b_{i}. □

In fact Proposition 9.5 is true without the hypothesis that P_{1}, …, P_{r} are distinct. However, since we do not need this, we omit the details.

9.3. Proof of Proposition 4.3. We must prove the following.

Proposition 9.7. Suppose n − 2r = 2 and write Sec^{r} C = {F_{1} = F_{2} = 0}. Then the variety X ⊂ ℙ^{n−1} defined by

rank (\begin{matrix}\frac{∂F_{1}}{∂x_{1}} & ⋯ & \frac{∂F_{1}}{∂x_{n}} \\ \frac{∂F_{2}}{∂x_{1}} & ⋯ & \frac{∂F_{2}}{∂x_{n}} \end{matrix} ) ≤ 1

has codimension 3.

If n = 4 then C = {F_{1} = F_{2} = 0} ⊂ ℙ^{3} is the intersection of two quadrics. There are 4 singular quadrics in the pencil spanned by F_{1} and F_{2}, and each is singular at just one point. Then X is the union of these 4 singular points, and so has codimension 3.

We now generalise this argument. Let H be the divisor of a hyperplane section. We identify L(H) with the space of linear forms on ℙ^{n−1}. Let D_{1} and D_{2} be divisors on C of degree r + 1 with D_{1} + D_{2} = H. Let Φ(D_{1}, D_{2}) be the (r + 1) × (r + 1) matrix of linear forms representing the multiplication map

L(D_{1}) × L(D_{2}) → L(H).

Since Φ(D_{1}, D_{2}) has rank at most 1 on C, it has rank at most r on Sec^{r} C. Therefore det Φ(D_{1}, D_{2}) is a form of degree r + 1 vanishing on Sec^{r} C. In particular it belongs to the pencil spanned by F_{1} and F_{2}.

Lemma 9.8. Every linear combination of F_{1} and F_{2} arises in this way. Moreover there are exactly 4 forms in the pencil arising as det Φ(D_{1}, D_{2}) with D_{1} ∼ D_{2}.

Proof: We say that divisor pairs (D_{1}, D_{2}) and (D′_{1}, D′_{2}) are equivalent if D_{1} ∼ D′_{1} or D_{1} ∼ D′_{2}. It is shown in [F2, Lemma 2.9] that if (D_{1}, D_{2}) and (D′_{1}, D′_{2}) are inequivalent then Sec^{r} C = {det Φ(D_{1}, D_{2}) = det Φ(D′_{1}, D′_{2}) = 0} ⊂ ℙ^{n−1}. In particular these two forms are linearly independent.

We claim that the map (D_{1}, D_{2}) ↦ Φ(D_{1}, D_{2}) is a bijection between the equivalence classes of divisor pairs and the pencil of forms spanned by F_{1} and F_{2}. To prove this let C be the image of an elliptic curve E embedded in ℙ^{n−1} by |n.0_{E}|. Then writing

det Φ(r.0_{E} + P, (r + 2).0_{E} − P) = s(P)F_{1} + t(P)F_{2},

for P ∈ E, we can see that s/t is a rational function on E. It therefore defines a morphism (s : t) : E → ℙ^{1}. By the previous paragraph, this morphism is nonconstant, and indeed has fibres of the form {P, −P }. It must therefore be surjective. This proves the claim.

For the final statement we note that r.0_{E} + P ∼ (r + 2).0_{E} − P if and only if P ∈ E[2]. □

Lemma 9.9. Let S be the singular locus of V = {det Φ(D_{1}, D_{2}) = 0} ⊂ ℙ^{n−1}. Then S contains Sec^{r−1} C. Moreover

(i) If D_{1} ≁ D_{2} then S = Sec^{r−1} C.

(ii) If D_{1} ∼ D_{2} then S has codimension 3.

Proof: Since C spans ℙ^{n−1} it is clear that for each P ∈ Sec^{r−1} C we have T_{P} Sec^{r} C = ℙ^{n−1}. Therefore S contains Sec^{r−1} C.

(i) Let P ∈ V \ Sec^{r−1} C be any point. According to [F2, Theorem 1.3] the r × r minors of Φ(D_{1}, D_{2}) generate I(Sec^{r−1} C). Therefore evaluating Φ(D_{1}, D_{2}) at P gives a matrix of rank r. Moving P to (1 : 0 : …: 0) and picking suitable bases for L(D_{1}) and L(D_{2}) we have

Φ(D_{1}, D_{2}) = x_{1} (\begin{matrix}0 & 0 & ⋯ & 0 \\ 0 & 1 & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 1 \end{matrix}) + Φ′

where Φ′ is an (r + 1) × (r + 1) matrix of linear forms in x_{2}, …, x_{n}. Now the top left entry of Φ(D_{1}, D_{2}) is an equation for T_{P} V. Since the product of non-zero rational functions on C is again non-zero, the entries of Φ(D_{1}, D_{2}) are non-zero. Therefore P ∈ V is a smooth point. (ii) Picking suitable bases for L(D_{1}) and L(D_{2}) we may suppose that Φ(D_{1}, D_{2}) is symmetric. Since {rank Φ(D_{1}, D_{2}) ≤ r − 1} ⊂ S, and the quadratic forms of rank at most m − 2 have codimension 3 in the space of all quadratic forms in m variables, it follows that S has codimension at most 3. Suppose for a contradiction that S has codimension at most 2. Then its intersection with Sec^{r} C = {F_{1} = F_{2} = 0} has codimension at most 3. But this intersection is contained in the singular locus of Sec^{r} C, which by Theorem 9.1 has codimension 4. This is the required contradiction.

□

To complete the proof of Proposition 9.7, we note that X is the union of the singular loci of the hypersurfaces defined by linear combinations of F_{1} and F_{2}. It follows by Lemmas 9.8 and 9.9 that X has codimension 3.

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University of Cambridge, DPMMS, Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK

  E-mail address: T.A.Fisher@dpmms.cam.ac.uk