Moore Matrices and Ulrich Bundles on an Elliptic Curve

We give normal forms of determinantal representations of a smooth projective plane cubic in terms of Moore matrices. Building on this, we exhibit matrix factorizations for all indecomposable vector bundles of rank 2 and degree 0 without nonzero sections, also called Ulrich bundles, on such curves.


LINEAR DETERMINANTAL REPRESENTATIONS OF SMOOTH HESSE CUBICS AND STATEMENT OF THE RESULTS
1.1.Let K be a field and f ∈ S = K[x 0 , ..., x n ] a nonzero homogeneous polynomial of degree d 1.A matrix M = (m ij ) i,j=0,...,d−1 ∈ Mat d×d (S) is linear if its entries are linear homogeneous polynomials and it provides a linear determinantal representation of f if det M .= f , where .
= means that the two quantities are equal up to a nonzero scalar from K.More generally, if M is a linear matrix for which there exists a matrix M ′ with M • M ′ = f id d = M ′ • M, then M is an Ulrich matrix and F = Coker M, necessarily annihilated by f , is an Ulrich module over R = S/( f ).
If V( f ) = { f = 0} ⊂ P n , the projective hypersurface defined by f , is smooth, then the sheafification of F is a vector bundle, called an Ulrich bundle.
Our aim here is to give normal forms for all Ulrich bundles of rank 1 or 2 over a plane elliptic curve in Hesse form.
Linear determinantal representations of hypersurfaces have been studied since, at least, the middle of the 19 th century and for a very recent comprehensive treatment for curves and surfaces see Dolgachev's monograph [Do12].This reference contains as well a detailed study of the geometry of smooth cubic curves, especially of those in Hesse form and we refer to it for background material.
For smooth plane projective curves, the state-of-the-art result is due to Beauville.1.2.Theorem ([Be00, Prop.3.1]).Let C = V( f ) be a smooth plane projective curve of degree d defined by an equation f = 0 in P 2 over K.With g = 1 2 (d − 1)(d − 2) the genus of C, one has the following.In this result, the matrix M clearly determines L uniquely up to isomorphism, but L determines M only up to equivalence of matrices in that the cokernel of every matrix PMQ −1 for P, Q ∈ GL(d, K) yields a line bundle isomorphic to L. The so obtained action on these matrices of the group G d = GL(d, K) × GL(d, K)/G m (K), with G m (K) the diagonally embedded subgroup of nonzero multiples of the identity matrix, is free and proper; see [Be00,Prop.3.3].The geometric quotient by this group action identifies with the affine variety Jac g−1 (C) \ Θ, the Jacobian of C of line bundles of degree g − 1 minus the theta divisor Θ of those line bundles of that degree that have a nonzero section.
There is therefore the issue of finding useful representatives, or normal forms, for such linear representations in a given orbit.Realizing hyperelliptic curves as double covers of P 1 , Mumford [Mu84, IIIa, §2] exhibited canonical presentations for such line bundles, which motivated Beauville's work [Be90], and Laza-Pfister-Popescu in [LPP02] found such representative matrices for the Fermat cubic, while for general elliptic curves in Weierstraß form Galinat [Ga14] determines normal forms of those linear representations.
Our first result yields the following normal forms for plane elliptic curves in Hesse form.Here, and in the sequel, we denote [a 0 : • • • :a n ] ∈ P n (K) the K-rational point with homogeneous coordinates a i ∈ K, not all zero.1.3.Theorem A. Over an algebraically closed field 1 K of characteristic char K = 2, 3, each linear determinantal representation of the smooth plane projective curve E with equation Two such Moore matrices M (a 0 ,a 1 ,a 2 ),x and M (a 1.4.Remarks.(a) Choosing an inflection point as origin for the group law, the exceptional points a ∈ E with a 0 a 1 a 2 = 0 are precisely the 3-torsion points, equivalently, the inflection points of E. They form the subgroup isomorphic to the elementary abelian 3-group Z/3Z × Z/3Z of rank 2. (b) In geometric terms, the preceding result states that the map a → coker M a,x from of line bundles of degree 0 without nonzero sections is well defined and is (isomorphic to the restriction of) the isogeny 2 of degree 9 that is given by multiplication by 3 on E.
Building on the previous result, our second contribution is as follows.
1.5.Theorem B. Let E be the smooth plane cubic curve from above.
(a) Let F be an indecomposable vectorbundle of rank 2 and degree 0 on E. If H 0 (E, F) = 0, then there exists a = (a 0 , a 1 , a 2 ) ∈ K 3 representing a point a ∈ E with a 0 a 1 a 2 = 0 such that the sequence of O P 2 -modules is an indecomposable vectorbundle F of rank 2 and degree 0 on E that has no nonzero sections, H 0 (E, F) = 0.
1 Likely, it suffices that K contains six distinct sixth roots of unity, which forces char K = 2, 3.However, one key ingredient in the proof, see (2.22) below, is stated in the literature only over algebraically closed fields, thus, we are compelled to make that assumption too. 2 We thank Steve Kudla for suggesting this interpretation.
(c) Replacing a by a ′ results in a vector bundle isomorphic to F if, and only if, 3 In the next section we will review some known facts about elliptic curves in Hesse form and will prove Theorem A. In section 3 we will establish Theorem B.

PROOF OF THEOREM A
To lead up to the proof of Theorem A, we first review some ingredients.To begin with, we review how Moore matrices encode conveniently the group law on a smooth Hesse cubic E.
Moore matrices and their rank.
= f := x 3 0 + x 3 1 + x 3 2 − λx 0 x 1 x 2 , whence M a,x indeed yields a determinantal presentation of the cubic curve C = V( f ) and the point a = [a 0 :a 1 :a 2 ] in P 2 (K) underlying a lies on C.
Note that C will be smooth if, and only if, λ 3 = 27 in K.In the smooth case we write E = V( f ) to remind the reader that this curve is elliptic over K in that it is smooth of genus 1 and contains at least one point, for example [0: − 1:1], defined over K. [Pa15], Moore matrices have already appeared in the literature in a variety of contexts: to describe equations of projective embeddings of elliptic curves; see [GP98]; to give an explicit formula for the group operation on a cubic in Hesse form; see [Fr02] and [Ra97]; as differential in a projective resolution of the field over elliptic algebras; see [ATvdB90].

Remark. As pointed out in
Although the matrices above were known, and are easily established to form a matrix factorization of a Hesse cubic cf.[EG10, example 3.6.5],their relation to line bundles of degree 0 via representations of the Heisenberg group, as we establish below, seems first to have been observed in [Pa15].
The next result is well-known and easily established through, say, explicit calculation as in [Fr02, Lemma 3].
2.3.Lemma.For E = V( f ) a smooth cubic in Hesse form and every pair a, b with a, b ∈ E, the (specialized) Moore matrix M a,b is of rank 2.
In the situation of the preceding Lemma, basic Linear Algebra tells us that the one- the point underlying the one-dimensional space M ⊥ a,b in P 2 .A less obvious result, part of [Fr02,Thm.4], is that c will again be a point of E along with a, b.
2.4.Before we turn to the group structure, let us note that the transpose of a Moore matrix is again a Moore matrix, , where ι is the involution ι(x 0 , x 1 , x 2 ) = (x 0 , x 2 , x 1 ), or, counting indices modulo 3, ι(x j ) = x −j .For use below, we follow again Dolgachev and set Before continuing towards the proof of Theorem A, we take the opportunity to interpret Moore matrices geometrically in two ways, following Artin-Tate-van den Bergh [ATvdB90] in the first and Dolgachev [Do12] in the second.
2.6.In the introduction to [ATvdB90] the authors consider 3 the trilinear forms that can as well be interpreted as a system of three linear equations in, at least, two ways: or, shorter, in terms of Moore matrices, 3 Their indexing of the ingredients is different, but that is just due to the fact that the authors of [ATvdB90] chose the point at infinity [1: − 1:0] as the origin for the group law on E.
2.7.Viewing, for a fixed a ∈ E, the f i as sections of O(1, 1) on P 2 x × P 2 y , these equations imply, by Lemma (2.3), that the subscheme X = V( f 0 , f 1 , f 2 ) ⊆ P 2 x × P 2 y is mapped isomorphically by each of the projections p x , p y : P 2 x × P 2 y −→P 2 onto E ⊆ P 2 , and, in light of the preceding Theorem, the subscheme X constitutes the graph of the translation by −a on the elliptic curve in that when going from P 2 x to P 2 y , while it represents the graph of the translation by a, when going in the opposite direction.In other words, Geometric Interpretation à la Dolgachev.
2.8.Applying the treatment from [Do12, 4.1.2]to the special case of plane elliptic curves gives a geometric interpretation of the adjugate of a Moore matrix as follows.
Fixing again a ∈ E, consider the closed embedding and follow it with the Segre embedding s 2 : P 2 × P 2 ֒→P 8 that sends (x, y) to the class of the 3 thus, E gets embedded into the Segre variety s 2 (P 2 × P 2 ) ⊂ P 8 through the adjugate of the Moore matrix.As to the image of E in P 2 × P 2 , if we set y = x − E a, then x = y + E a and − E x − E a = − E y − E 2 • E a so that (l, r) a (E) is the graph of the involution 4 y → − E y − E 2 • E a on E that one may view as the "reflection" in − E a.

Doubling and Tripling Points on E.
2.9.As an immediate application of Theorem 2.5 one can easily determine 5 2 • E a and 3 , where b = 2 • E a, and explicit coordinates are obtained from the columns of the corresponding adjugate matrices.Now 4 That Moore matrices define an involution on E in this way we learned from Kristian Ranestad who kindly shared his notes [Ra97] with us. 5 The formulas for 2 • E a are already contained in [Fr02].We recall them here for completeness und later use.
whence 2 and evaluate through a straightforward though somewhat lengthy expansion: For an additional check, note that [a 0 :a 2 : as it has to be.
When a 0 a 1 a 2 = 0, the case we are interested in, these results can be simplified a bit.2.10.Corollary.For a = (a 0 , a 1 , a 2 ) as above representing a point a ∈ E with a 0 a 1 a 2 = 0, doubling, respectively tripling a on E results in 2.11.Example.As an immediate application, one obtains the set of 6-torsion points on E in that 2 • E a is a 3-torsion point if, and only if, a is a 6-torsion point.Now x 1 x 2 ) as was noted above.The formulae for doubling a point thus show that ) is the intersection of E with the indicated 12 lines.As the four lines V(x 0 x 1 x 2 (x 2 − x 1 )) cut out the 3-and 2-torsion points, the remaining 8 lines cut out the 24 primitive 6-torsion points as stated in [Fr02].
It follows that E[6] ∼ = Z/6Z × Z/6Z and that all 36 points of E[6] are defined over K, as soon as char K = 2, 3 and K contains three distinct third roots of unity.

The Algebraic Heisenberg Group.
It is a classical result in the theory of elliptic curves that translation by a 3-torsion point on a smooth cubic is afforded by a projective linear transformation; see [Mu66, §5 Case (b)].We first recall the precise result and then show that the action of the relevant algebraic Heisenberg group lifts to a free action on the Moore matrices.
2.12.Definition.Let K be a field that contains three distinct third roots of unity, µ 3 (K) = {1, ω, ω 2 } K * with ω 3 = 1.In terms of the matrices of order 3, the algebraic Heisenberg group is That this is indeed a subgroup is due to the equality ΣT = ωTΣ.This same equality also shows that H 3 , the subset of Heis 3 , where µ is restricted to powers of ω, is a finite subgroup of SL(3, K) of order 27.
The crucial property of the algebraic Heiseisenberg group is then the following.2.13.Proposition.Let a = (a 0 , a 1 , a 2 ) as before represent a point a ∈ E with a 0 a 1 a 2 = 0.For a (2) a ′ is in the Heis 3 -orbit of a, thus, a ′ ∈ Heis 3 • a.
(3) The Moore matrices M a,x and M a ′ ,x are equivalent.
Moreover the action of Heis 3 on the Moore matrices is free.
Proof.The equivalence of (1) and ( 2) is, of course, classical.For (2) =⇒ (3), it suffices to show that the Moore matrices for a, T(a), Σ(a) and µa, µ ∈ K * , are equivalent.This is obvious for µa as M µa,x = (µ id 3 )M a,x .For T(a) = (a 0 , ωa 1 , ω 2 a 2 ), the Moore matrix is It remains to prove (3) =⇒ (1).If M a,x is equivalent to M a ′ ,x then these two matrices have the same determinant up to a nonzero scalar.This shows that a ′ represents a point on E along with a and that a ′ 0 a ′ 1 a ′ 2 = 0. Now write where M i = ∂M a,x /∂x i ∈ Mat 3×3 (K).The matrix M 0 = diag(a 0 , a 2 , a 1 ) is invertible by assumption and we set analogously with a ′ replacing a. Then the matrices M a,x , M a ′ ,x are equivalent under the action of G 3 if, and only if N and N ′ are equivalent under that action.As N 0 = N ′ 0 = 1 3 , the identity matrix, the matrices N, N ′ are equivalent with respect to G 3 if, and only if, N and N ′ are equivalent under conjugation by a matrix P ∈ GL(3, K), that is, In other words, the pairs of 3 × 3 matrices (N 1 , N 2 ) and (N ′ 1 , N ′ 2 ) are related by simultaneous conjugation.Clearly the trace functions tr(A 1 • • • A n ), for any n-tuple A i ∈ {U, V}, i = 1, ..., n, are constant on the class of a pair (U, V) ∈ Mat 3×3 (K) 2 under simultaneous conjugation.Moreover, Teranishi [Te86] showed that 11 of these traces suffice to generate the ring of invariants.See [Fo87] for a survey of these results, especially the list of the generating traces on the bottom of page 25.
We will not need any details of that invariant theory, but we easily extract from those classical results the traces that are relevant here.
2.14.Lemma.With notation as just introduced, set further a ij = a i /a j for i, j = 0, 1, 2.
(i) The matrices N 1 , N 2 have the form (ii) Taking traces yields Proof.Straightforward verification.
Combining item (ii) in this Lemma with Corollary 2.10 shows that equivalence of M a,x and M a ′ ,x forces 3 As for the final claim, this follows from Beauville's result that the action of G 3 on linear matrices is free.
2.15.Corollary.The subgroup of G 3 that transforms Moore matrices into such is isomorphic to the algebraic Heisenberg group Heis 3 .
2.16.Remark.In light of the preceding result, we sometimes write M a,x to denote any representative of the equivalence class of M a,x under the action of Heis 3 , with a ∈ E as before representing the point underlying a ∈ K 3 .
It is indeed the representation theory of the Heisenberg groups that allows us to finish the proof of Theorem A. Instead of working with the algebraic Heisenberg groups, it suffices to restrict to the finite Heisenberg groups and their representations.
The Schr ödinger Representations of the Finite Heisenberg Groups.
2.17.The general Heisenberg group H(R) over a commutative ring R is usually understood to be the subgroup of unipotent upper triangular 3 × 3 matrices in GL(3, R).For R = Z/nZ, n 1 an integer, we call these the finite Heisenberg groups and abbreviate H n = H(Z/nZ).The group H n is of order n 3 and admits the presentation Each element of H n has a unique representation as [σ, τ] r σ s τ t with r, s, t ∈ Z/nZ.
Note that H 3 as defined here is indeed isomorphic to the group H 3 that we exhibited as a subgroup of Heis 3 above.2.18.Over a field K that contains a primitive n th root of unity ζ ∈ K * , the group H n carries the K-linear Schrödinger representations ρ j : H n → GL(n, K), parametrized by j ∈ Z/nZ, that in a suitable Schrödinger basis v i , i ∈ Z/nZ, of a vector space V of dimension n over K are given by and thus, for a general element, In particular, the character χ j of the representation ρ j satisfies 2.19.Remark.For a complete and detailed discussion of the irreducible representations of the finite Heisenberg groups see [GH01].
2.20.If d 2 is a divisor of n, say n = dm, then the subgroup of H n generated by σ m , τ m is a homomorphic image of H d , in that surely σ m and τ m are of order d, and these elements commute with [σ m , τ m ] = [σ, τ] m 2 .If we restrict the Schr ödinger representation ρ j of H n along the resulting homomorphism H d → H n , then it decomposes in that the actions of σ m and τ m , given by For i = αm + k, in the basis w α = v αm+k , for α = 0, ..., d − 1, of W jk the action is given by ρ j (σ m )(w α ) = w α−1 , ρ j (τ m )(w α ) = (ζ m ) j(αm+k) w α , and, for a general element The corresponding character is thus 2.22.Returning to elliptic curves, let, more generally, L be an ample line bundle on an abelian variety defined over an algebraically closed field K whose characteristic does not divide the degree n > 0 of L. It is a deep result from the theory of abelian varieties; see [Mu91, Prop.3.6] for the general case or [Hu86] for an explicit treatment over the complex numbers; that then the vector space of sections of L comes naturally equipped with the Schr ödinger representation ρ 1 of H n -in fact this is the restriction of the Schr ödinger representation of the larger algebraic Heisenberg group Heis n that is defined in analogous fashion to Heis 3 .
In case L is a line bundle on an elliptic curve E over K this representation lifts the translation by n-torsion points on E, thus, the action of E[n] ∼ = Z/nZ × Z/nZ on P(H 0 (E, L)) to an action by linear automorphisms on V = H 0 (E, L).For an elliptic curve E, embedded as a smooth projective plane cubic curve, the special case L = O E (1) with n = deg L = 3 was discussed in detail above.
Using the preceding Lemma, the following result is an easy consequence of the fundamental fact just recalled.
2.23.Proposition.Let L, L ′ be locally free sheaves of degree 3 and L ′′ a locally free sheaf of degree 6 on an elliptic curve E over an algebraically closed field K whose characteristic does not divide 6.
In particular, the kernel of that homomorphism is a Schrödinger representation ρ 2 of H 3 .
For Part (b), let (x 0 , x 1 , x 2 ) be a Schr ödinger basis of V = H 0 (E, L) and (y 0 , y 1 , y 2 ) be a Schr ödinger basis of V ′ = H 0 (E, L ′ ).With x i y j = x i ⊗ y j and a 0 , a 1 , a 2 ∈ K, the tensor f 0 = a 0 x 0 y 0 + a 1 x 2 y 1 + a 2 x 1 y 2 is a fixed vector for the action of Therefore, f 0 , f 1 , f 2 form indeed a Schr ödinger basis for a representation of H 3 that is equivalent to ρ 2 as soon as (a 0 , a 1 , a 2 ) = (0, 0, 0) ∈ K 3 .Choosing in turn (a 0 , a 1 , a 2 ) = e i , for i ∈ Z/3Z and (e i ) i=0,1,2 the standard basis of K 3 , it follows that indeed 2 as H 3 -representations -which fact could have been established as well by just looking at the corresponding group characters.The reader will note that viewed as trilinear forms, the f i are precisely the forms from (2.6) above.
In Part (c), surjectivity of the multiplication map is well known and the H 3 -equivariance follows as translation is compatible with tensor products, t a (L) for any point a ∈ E. Applied to 3-torsion or 6-torsion points and using that translations by those points manifest themselves through the Schr ödinger representation ρ 1 of H 3 , respectively H 6 , the proof of the Proposition is complete.
End of the Proof of Theorem A.

2.24.
Let L be a line bundle of degree 0 on the smooth cubic curve E ⊂ P 2 with defining equation f = 0.According to Beauville's result stated above in Theorem 1.2, if H 0 (E, L) = 0 then there exists a 3 × 3 linear matrix M such that f = det M and an exact sequence of Twisting this sequence by O E (1), respectively O E (2), and taking sections, one can identify this exact sequence as where W is the kernel of the H 3 -equivariant multiplication map as in Proposition 2.23(c) above for L = O E (1), L ′ = L(1).
Choosing Schr ödinger bases f 0 , f 1 , f 2 for W, x 0 , x 1 , x 2 for H 0 (E, O E (1)), and y 0 , y 1 , y 2 for H 0 (E, L(1)) as in the proof of Proposition 2.23(b), M becomes identified with a Moore ma- trix M a,x = (a i+j x i−j ) i,j∈Z/3Z for some a = (a 0 , a 1 , a 2 ) ∈ K 3 .As det M .= f by Beauville's result, it follows that a ∈ E with a 0 a 1 a 2 = 0.This completes the proof of Theorem A from the introduction.

PROOF OF THEOREM B
The starting point is the following result from Atiyah's seminal paper [At57].

A Result of Atiyah and Ulrich Bundles.
3.1.Theorem (cf.Atiyah [At57, Thm. 5 (ii)]).Let F be an indecomposable vector bundle of rank 2 on an elliptic curve E over a field K.If deg F = 0, then there exists a unique line bundle L of degree 0 that fits into an exact sequence of vector bundles Conversely, if L is a line bundle of degree 0 then there exists an indecomposable vector bundle F, unique up to isomorphism and necessarily of rank 2 and degree 0, that fits into such an exact sequence.

By our Theorem
A we know that a line bundle of degree 0 with no nonzero sections is obtained as L = coker M a,x , where M a,x is a Moore matrix, a ∈ K 3 representing a point a ∈ E on the elliptic curve E ⊂ P 2 with a 0 a 1 a 2 = 0. We fix these data in the following.
Let S = K[x 0 , x 1 , x 2 ] be the homogeneous coordinate ring of P 2 (K), with its homogeneous components S m = H 0 (P 2 , O P 2 (m)) the vector space over K of homogeneous polynomials of degree m ∈ Z.
Applying the functor Γ * = ⊕ i∈Z H 0 (P 2 , ( )(i)) to the exact sequence of coherent O P 2 - modules 0 The module L = Γ * (L), cokernel of the map between graded free S-modules represented by M a,x , is an Ulrich module of rank one over the homogeneous coordinate ring R = S/( f ) of E, and each Ulrich module over R of rank one (and generated in degree 1) can be so realized by Theorem A.

Matrix Factorizations and Extensions.
In view of Atiyah's result cited above, our aim here is to find a similar description for Ulrich modules over R of rank two, namely the one stated in Theorem B. To simplify notation a bit, we fix for now the point a and set A = M a,x , B = M adj a,x , viewed as matrices over S. The pair (A, B) represents a matrix factorization of f = det A ∈ S and so, by [Ei80], L admits the graded R-free resolution that is 2-periodic up to the shift in degrees by − deg f = −3.
3.3.Now consider an element6 of Extgr 1 R (L, L(m)) for some m ∈ Z.It can be represented by a homotopy class of morphisms between graded free resolutions and, invoking again [Ei80], such morphisms and their homotopies can again be chosen to be 2-periodic so that one has a diagram as follows ) are 3 × 3 matrices whose entries are homogeneous poynomials of the indicated degrees.• The pair of matrices (C, D) satisfies AD + CB = 0 = DA + BC over S, with 0 the zero matrix, and so defines a morphism of complexes over R. • U, V ∈ Mat 3×3 (S m ) represent the possible homotopies, in that the morphisms of complexes L → L(m) induced by run through the homotopy class of (C, D) for the various choices of U, V.If A, as in our case of interest, is a determinantal representation of a reduced polynomial, one can reduce the description of extensions further.Proof.As f is reduced, it is generically regular.For a regular point x ∈ V( f ) this implies that rank A(x) = n − 1, thus, rank B(x) = 1, as the cokernel of A is locally free of rank 1 at such point.Accordingly there are vectors u, v ∈ k(x) n such that B(x) = u T • v. Therefore, • u T is an element of the residue field k(x) at x and so, considering it as a 1 × 1 matrix, Embedding this observation into the right-hand side of the previous equality, it follows that Therefore, BCB − tr(BC)B vanishes at each regular point of { f = 0}, thus, it vanishes everywhere on that hypersurface.Moreover, B(x) = 0 at each regular point, whence at such points BCB(x) = 0 if, and only if, tr(B(x)C(x)) = 0.The claim follows.
3.7.Putting these facts together, we get the following description of the extension groups we are interested in here: This description shows immediately that Extgr 1 R (L, L(m)) = 0 when m < −1, because there is then no nonzero choice for C. It also shows that there are no nonzero homotopies for m = −1, a fact we will exploit below.Concerning shifts by m −1, we determine the size of the extension group directly in terms of possible Yoneda extensions, that is, short exact sequences, as follows.Given a short exact sequence Conversely, applying Γ * to such a short exact sequence of O E -modules with m −1 returns a short exact sequence of graded R-modules as above in that the connecting homomorphism in cohomology H It follows that Γ * and sheafification yield inverse isomorphisms between Ext 1 E (L, L(m)) and Extgr 1 R (L, L(m)) for m −1.
We thus have the following result.
3.8.Lemma.Let L be the Ulrich module that is the cokernel of the Moore matrix M a,x as in Theorem A. The vector spaces of graded self-extensions of L over R satisfy With these preparations we now determine Extgr 1 R (L, L(−1)).3.9.Proposition.Let L be the cokernel of a Moore matrix M a,x : S(−2) 3 −→S(−1) 3 for a ∈ K 3 with a 0 a 1 a 2 = 0 representing a point a ∈ E. The three-dimensional vector space Extgr 1 R (L, L(−1)) over K is isomorphic to the space of specialized Moore matrices Because b = (0, 0, 0), the matrices M 0 = M b,(1,0,0) , M 1 = M b,(0,1,0) , M 2 = M b,(0,0,1) are clearly linearly independent in the vector space Mat 3×3 (K) in that their nonzero entries are located at different positions in these matrices.Further, as tr(B • −) is an S-linear function on Mat 3×3 (S), and Extgr 1 R (L, L(−1)) is known to be of dimension 3 over K, it suffices to show that for each of the three matrices M i one has tr(BM i ) ≡ 0 mod ( f ), where B = M adj a,x .In fact, as the entries of BM i are quadratic polynomials, but f is of degree 3, the congruence is equivalent to tr(BM i ) = 0.
One now verifies this easily directly for the three matrices in question.
For a more conceptual explanation of the identities tr(BM i ) = 0, note that, say, M 0 = M b,(1,0,0) = diag(b) is a diagonal matrix with the coordinates b on the diagonal representing 2 • E a. On the other hand, the diagonal elements in B = M adj a,x involve only the quadratic monomials x 2 0 and x 1 x 2 , and the coefficients of x 2 0 along the diagonal are the entries from the third row, those of x 1 x 2 the entries from the first row of M ι(a),a ; see (2.9).The column vector b T spans the kernel of that matrix by construction.As the trace tr(BM 0 ) is the scalar product of the two diagonals, the vanishing of the trace becomes obvious.The case of the remaining two matrices yields to analogous arguments.
Note in particular that div(M b,x ) = 3 ∈ K, thus, is not zero when char(K) = 3.For a characteristic-free statement, note that div(M b,ι(x) ) = 1 ∈ K.
The one-dimensional vector space Extgr 1 R (L, L) over K can be realized as and the divergence M b,y → div(M b,y ) ∈ K induces an isomorphism Proof.As mentioned before, with B = M adj a,x , the function tr(B • −) : Mat 3×3 (S) → S is S-linear, whence each matrix M b,y satisfies tr(B • M b,y ) = 0 as we know this for the matrices M 0 = M b,(1,0,0) , M 1 = M b,(0,1,0) , M 2 = M b,(0,0,1) from above.Thus, the vector space M b,y ∈ Mat 3×3 (S 1 ) is contained in the numerator of the description of Extgr 1 R (L, L) in (3.7)( * ).As we know from Lemma 3.8 that this extension group is one-dimensional, and, say, divM b,ι(x) = 1 ∈ K as noted above, it remains only to show that the denominator in the description here lies in the kernel of the divergence.
To this end, assume M b,y = U M a,x − M a,x V for some linear forms y i and some U, V ∈ Mat 3×3 (K).Differentiating both sides with respect to x 0 and comparing the diagonal entries yields the system of equations a 2i (a 3 2i−1 − a 3 2i+1 ) Dividing by a 2i , which is not zero by assumption, and then adding up shows that necessarily ∑ i∈Z/3Z u ii = ∑ i∈Z/3Z v ii .Differentiating as well with respect to x 1 , x 2 , comparing entries on both sides of the matrix equation and eliminating common factors of the form a i leads to the system of equations (a 3 2 − a 3 1 ) Now at least one of the terms (a 3 i − a 3 i−1 ), i ∈ Z/3Z, that occur as coefficients on the lefthand sides is nonzero, as not all entries of b are zero.Picking one such nonzero term and using it to solve for ∂y i ∂x i , i = 0, 1, 2, shows immediately that div(M b,y ) = ∂y 0 ∂x 0 + ∂y 1 ∂x 1 + ∂y 0 ∂x 1 = 0 is a necessary condition on M b,y to be representable as U M a,x − M a,x V.In fact, we also know that this condition is sufficient.

( a )
Let L be a line bundle of degree g − 1 on C with H 0 (X, L) = 0. Then there exists a d × d linear matrix M such that f = det M and an exact sequence of O P 2 -modules 0 / / O P 2 (−2) d M / / O P 2 (−1) d / / L / / 0 .(b) Conversely, let M be a d × d linear matrix such that f .= det M. Then the cokernel of M : O P 2 (−2) d → O P 2 (−1) d is a line bundle L on C of degree g − 1 with H 0 (X, L) = 0.
thus represents the space ⊥ M a,x of solutions to the linear system of equations d • M a,b = 0, spanned by the row vectors of M adj a,b .2.5.Theorem (see [Fr02, Thm.4]).The assignment (a, b) → c = l(M a,b ) for a, b ∈ E defines the group law on E by setting b − E a = c.The identity element is given by o = [0: − 1:1] and the inverse of a is − E a = ι(a) = [a 0 :a 2 :a 1 ].

( a )
Restricting the translations by 6-torsion points to the 3-torsion points restricts the representation ρ 1 of H 6 on H 0 (E, L ′′ ) to the direct sum of two copies of the Schrödinger representationρ 2 of H 3 .(b) The tensor product H 0 (E, L) ⊗ K H 0 (E, L ′ ) ofthe Schrödinger representations ρ 1 of H 3 on each of the two factors decomposes as the direct sum of three copies of the Schrödinger representation ρ 2 of H 3 .(c) With L ′′ = L ⊗ O E L ′ , the natural multiplication map on global sections represents a surjective H 3 -equivariant homomorphism

3. 4 .
Given a pair of matrices (C, D) with AD + CB = 0 = DA + BC as above, the block matrices factorization of f and give rise to the commutative diagram of graded S-modules with exact rows and columns the rightmost column representing the extension defined by (C, D) over R.The following observation cuts down considerably on the work of finding solutions to the equations AD + CB = 0 = DA + BC, whenever (A, B) is a matrix factorization of a non-zero-divisor f in a commutative ring S.3.5.Lemma.Assume A, B ∈ Mat n×n (S) is a matrix factorization of a non-zero-divisor f ∈ S, in that AB = f id n = BA.For a matrix C ∈ Mat n×n (S) the following are equivalent.(a) There exists a matrix D ∈ Mat n×n (S) such that AD + CB = 0. (b) There exists a matrix D ′ ∈ Mat n×n (S) such that D ′ A + BC = 0. (c) There exists a matrix D ′′ ∈ Mat n×n (S) such that f D ′′ + BCB = 0. Equivalently, each entry of BCB ∈ Mat n×n (S) is divisible by f .If either equivalent condition holds then D = D ′ = D ′′ and that matrix is the unique one satisfying f D = −BCB.Moreover, one can recover C from D in that C is the unique matrix such that f C = −ADA.Proof.If AD + CB = 0 then multiplying from the left with B yields 0 = BAD + BCB = f D + BCB , whence BCB ≡ 0 mod ( f ) and one can take D ′′ = D. Conversely, if that congruence holds then there exists a matrix D ′′ with f D ′′ = −BCB.Multiplying this equation with A from the left results inA f D ′′ + ABCB = f (AD ′′ + CB) = 0 .As f is a non-zero-divisor, this implies AD ′′ + CB = 0, whence one can take D = D ′′ .Thus (a) ⇐⇒ (c).The equivalence (b) ⇐⇒ (c) is completely analogous.Uniqueness of D follows as AD 1 + CB = AD 2 + CB implies A(D 1 − D 2 ) = 0. Now the linear map represented by A is injective because f id n = BA and multiplication with f is injective by assumption.Thus, D 1 = D 2 as claimed and, in particular, D = D ′′ must hold.Analogously one must have D ′ = D ′′ .Concerning the final assertion, multiply the equation f D = −BCB on both sides with A to obtain ADA f = − f C f , thus, f C = −ADA, as f is a non-zero-divisor.Uniqueness of C follows as above.
and only if jt ≡ 0 mod d and one recognizes the Schr ödinger representation ρ jm of H d .Therefore, we have the following result.2.21.Lemma.For d a positive divisor of n with gcd(d, n/d) = 1, under the group homomorphism H d → H n described above the Schrödinger representation ρ j of H n restricts to the direct sum of n/d copies of the Schrödinger representation ρ (jn/d) mod d of H d .
3.6.Lemma.If the determinant f = det A ∈ S, for a matrix A ∈ Mat n×n (S), is reduced, then with B = A adj one has for any matrix C ∈ Mat n×n (S) that BCB ≡ tr(BC)B mod ( f ) , and BCB ≡ 0 mod ( f ) if, and only if, tr(BC) ≡ 0 mod ( f ).