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The A-infinity algebra of an Elliptic Curve and the j-invariant

We compute reduced Hochschild cohomology of B = Ext\star (O \oplus L, O \oplus L), where O is the structure sheaf of an elliptic curve and L is a line bundle of degree 1. The result suggests an A-infinity equivalence between the A-infinity structure computed for B in [1] and a structure in which m_3=m_4=m_5=0. We see that the j-invariant of the curve is explicitly related to this new structure.


Introduction
Let C = C/ Z ⊕ τ Z be an elliptic curve, D b (C) the bounded derived category of coherent sheaves on C. It is well-known that the category D b (C) is controlled by a natural A-infinity structure on the algebra B = Ext * (O ⊕ L, O ⊕ L), where O is the structure sheaf of C and L is a line bundle of degree 1.
The interest in this structure was motivated by questions of homological mirror symmetry. To determine this structure, one first chooses a resolution which calculates Ext. The total complex of the resolution has the structure of a dg-algebra A, which we consider as an A-infinity algebra with m n = 0 for n ≥ 3. According to the theorem of Kadeishvili ([2], [6]), B = H * (A) has an A-infinity structure with m 1 = 0, m 2 induced by m A 2 , such that A and B are equivalent as A-infinity algebras. This structure is unique up to strict A-infinity isomorphism.
Since G = O ⊕ L generates D b (C) as a triangulated category, D b (C) is equivalent to the derived category of perfect A-infinity modules over B = Ext * (G, G). For elliptic curves, it is also known that D b (C 1 ) ≃ D b (C 2 ) if and only if C 1 ≃ C 2 ; so this A-infinity structure on B determines C uniquely. In [1], Polishchuk uses the Dolbeault complex and a formula of Merkulov from [3] to compute an A-infinity structure on B in terms of the Eisenstein series of the curve.
By choosing a generator in the same way for all curves, the answer in [1] provides a family of A-infinity structures on the associative algebra B. Thus it is of interest to consider B as an associative algebra and examine its extensions to an A-infinity algebra with m 1 = 0 and m 2 = m B 2 . These extensions and their equivalences are governed by certain components of the Hochschild cohomology of B.
Written in part while hosted at the Institut Des HautesÉtudes Scientifiques.
If we let C be an elliptic curve over an algebraically closed field k, P a closed k-point, then the multiplicative structure of Ext * (O ⊕ O(P ), O ⊕ O(P )) is unchanged from the case where k = C. Thus it is valuable to compute Hochschild cohomology of this algebra for as general k as possible, despite the fact that the computations in [1] were done only for a complex curve.
The main goal of this paper is to compute the reduced Hochschild cohomology of B as an associative algebra over R = k id O , id L , where k is a field with char k = 2, 3. We use the reduced complex (as in [4] (20d)) as it is known to be quasi-isomorphic to the full complex for a unital algebra. The specific relationship of this cohomology to A ∞ -structures is well-known and recalled here in Section 2.
The fact that there are no nontrivial cocycles with internal degree -2 in HH 4 (B) implies the existence of a strict A-infinity isomorphism between the A-infinity algebra computed in [1] with a structure in which m 3 = m 4 = m 5 = 0. In this new structure, we see that the j-invariant of the curve is explicitly related to m 6 and m 8 , thus making the dependence of this A-infinity structure on the isomorphism class of the curve completely explicit.

Notation, conventions, and recollections
When f : A → B is a map of graded objects, we say that f is homogeneous of internal degree n if deg f (x) = deg(x) + n for all x ∈ A. If C • is a cochain complex with a differential which preserves internal degree, then C • (n) will mean the subcomplex of maps of internal degree n, and similarly HH • (n) refers to the cohomology of such a subcomplex. When A is a graded algebra, we define the tensor algebra T (A) = ⊕ ∞ i=1 A ⊗i . T (A) inherits an internal grading from A by We let S be the functor on a category of graded objects that shifts grading by -1, i.e. (SA) i = A i+1 . When V is a graded space, f : T (V ) → T (V ) a homogeneous map with respect to the internal grading, we recall that f is a superderivation if for homogeneous v 1 , v 2 ∈ T (V ) we have We have a graded Lie superalgebra In particular if i, j, k are odd (as they will be in our case), then 1.1. Eisenstein series. For a lattice Λ ⊂ C with basis ω 1 , ω 2 , with m, n integers of the same parity, we set where a(Λ) = Im(ω 1 ω 2 ) is the area of C/Λ. Then for integers a, b ≥ 0 of different parity we set When m, n are of different parity or a, b the same parity, then f m,n (Λ) = g a,b (Λ) = 0. It is shown in [1] that g a,b is a polynomial in e * 2 , e 4 , . . . , e a+b+1 with rational coefficients, where e 2k are the Eisenstein series for Λ defined as for k ≥ 2, and A few of these polynomial relations we will use later. When the lattice is understood, we will write e 2k := e 2k (Λ), g a,b := g a,b (Λ), and so on.
When C = C/ Z ⊕ τ Z is a complex elliptic curve, we set Λ = Z + τ Z for the purpose of computing the Eisenstein series of the curve. We set t = ℑτ π , and for non-negative integers a, b, c, d define 1.2. Hochschild cohomology. We recall the appropriate definitions (see [7]). In particular if k is a field, B a k-algebra, and R ⊂ B a semisimple ring such that B = R ⊕ B + as an R-module, then we define here the reduced Hochschild cochain complex over R. If M is any B-bimodule, we define C n (B, M ) = Hom R (B ⊗n + , M ), where ⊗ is tensor over R unless otherwise specified. Similarly Hom will mean Hom R by default. The differential δ : C n → C n+1 is defined by the equation, δ(φ)(a 0 , a 1 , . . . , a n ) = a 0 φ(a 1 , . . . , a n ) + n i=1 (−1) i φ(a 0 , . . . , a i−1 a i , . . . , a n )+ (−1) n+1 φ(a 0 , . . . , a n−1 )a n .
The cohomology of this complex is the Hochschild cohomology of B with coefficients in M . When M = B, we call the result simply the Hochschild cohomology of B.
For a graded B-bimodule M , the Hochschild complex C • (B, M ) is bigraded, ). The Hochschild differential preserves m and increases n by 1. Therefore we may consider for a fixed m the complex C • (m) (B, M ) and compute its cohomology.
The lemma implies that the vanishing of HH k+1 (2−k) (A) guarantees that m 2 , . . . , m k−1 can be extended with m k (i.e., the equation ( * ) can be solved for m k ). We prove here that φ k (m 3 , . . . , m k−1 ) is a cocycle, as this was not located in the literature.
Proof. Each m i , i = 2, . . . , k − 1, defines a map m i : T (A) → A of internal degree 2 − n equal to m i on A ⊗i and zero otherwise. Following [9], let V = (SA) * , The m i correspond to maps d i : V → T (V ) of internal degree 1. Each d i can be uniquely extended to a superderivation d i : T (V ) → T (V ), also homoegeneous of internal degree 1.
The A ∞ -relations arise from certain relations among the brackets of the derivationsd i . The Lie bracket on derivations becomes the well-known Gerstenhaber bracket on Hochschild cochains in the dual picture. The Hochschild differential δ acts by δf = [m 2 , f ] and . . , m k−1 ) * is a derivation since its dual is defined by brackets of derivations.) We proceed by induction in k. For compactness we write φ * j := φ j (m 3 , . . . , m j−1 ) * . The base case k = 3 is obvious since φ * 3 = 0. Suppose d 2 , d 3 , . . . , d k−1 are defined such that [d 2 , d j ] = φ * j for all j < k with k even. Then When φ * k+2−i are expanded, the sum on the right will have terms: respectively) and their sum vanishes by the Jacobi identity. When k is odd we have with the same result as the k even case.
Therefore if HH k . . , f k−1 (the start of a strict A ∞isomorphism from m to m ′ ) to f k are homotopic in this precise sense. (ii) Hom(O, L), a one-dimensional space, generated by a function θ; (iii) Ext 1 (L, O), a one-dimensional space generated by a function η; (iv) Ext 1 (L, L) and Ext 1 (O, O), both isomorphic to the one-dimensional space H 1 (O).
By Serre duality the products θη = ξ ∈ Ext 1 (O, O) and ηθ = ξ L ∈ Ext 1 (L, L) are nonzero, so we take ξ and ξ L as generators of those spaces. For degree reasons all other products (except those involving the identities) are zero.
Thus we consider the k-algebra B = B 0 ⊕ B 1 , graded with k-basis and nontrivial products We will also express B as B = R ⊕ B + where R = id L , id O and B + = θ, η, ξ, ξ L , and in this way treat B as an R-algebra.

Computing Hochschild cohomology
We start by computing Hochschild cohomology of B with coefficients in (ξ L , ξ), B 1 /(ξ L , ξ), B + /B 1 , and B/B + . We use the notation After making these calculations we will use the long exact sequences on cohomology associated to the short exact sequences (of R-bimodules) to finally compute HH n (1−n) (B), HH n (2−n) (B), and HH n (3−n) (B).

HH
we can consider each complex separately. Because the first and last terms of the Hochschild differential are zero, the cochain complex ( The chain complex C • (L) has an internal grading given by deg(a 1 ⊗ · · · ⊗ a n ) = deg a i . We let C (m) n (L) be the space of tensors of internal degree m in C n (L). The differential preserves m and decreases n by 1.
For m ≥ 0. We consider a decreasing filtration on the complex C (m) For fixed n, m, the space F i C (m) n (L) = 0 for i ≫ 0. The spectral sequence of this filtration therefore converges to the homology of the complex. The zero page has for each i ≥ 0 a complex and clearly d(ξ n−1 ) = 0. Since gr n−1 C where the first term on the right is 0 in the quotient. This proves the first claim.
(Note for future reference: this same trick will show that any tensor of the form such that all of a 2 , . . . , a n−1 = 0 and at least one b i = 0 is both in ker d and in im d. We will subsequently refer to this fact when needed the Lemma 4.1.3 trick.) We have This whole space is in ker d since d adds one to the sum of powers on ξ L . Since gr n−2 C (n−1) n+1 (L) = 0 for degree reasons, nothing in this space is in im d.
with the other terms vanishing in the quotient. Different choices of a, e therefore produce linearly independent boundaries, so no linear combination of such elements may be in ker d.
For i < n − 3 there are two cases.
Since either b = 0 or d = 0, this tensor is in im d by the Lemma 4.1.3 trick.
Thus for fixed a, e and different choices of b, d with b + d = n − 3 − i, we get an element of ker d by taking some combination of these elements. Tensors of the form More generally, the space of tensors with all interior powers of ξ L equal to 0, as here, form their own complex. Here d has the form . . , c m−1 } for all m gives an isomorphism with the reduced complex of the (n − 2 − i)-simplex. In this calculation we seekH 0 of the simplex, which is of course 0.
(Note for future reference: In subsequent sections we will encounter complexes which can be mapped to simplicial complexes in essentially this same way. When this is the case, we will refer to this procedure as the simplex trick.) Some combination of two tensors from (1) for fixed a, e, f and different choices of b, d are be in ker d. Similarly for pairs of tensors in (2). Both cases are covered by the simplex argument of Lemma 4.1.4 by tensoring the complex from that lemma with η(ξ) f θ(ξ L ) g on the right in the former case, or with (ξ L ) a η(ξ) b θ on the left in the latter.
Finally we must consider combinations of a tensor from (1) with powers b, d, f and a tensor from (2) Tensors of type (3) form a part of the complex considered in the simplex argument from 4.1.4. Here we must calculate H 1 of the simplex, which is 0.
To prove the claim for i = n − 4, note that all of the tensors described are obviously in ker d. Other possible tensors and their boundaries are, For different choices of a, c, e, g such that the appropriate sums equal n − 4, the set of tensors on the right are linearly independent and so produce no class in homology.
Proof. These elements are all clearly in ker d and cannot be in im d. The boundaries of tensors with either k 2 = 0, k 3 = 0, k 4 = 0 are all linearly independent, similar to the previous lemma.
Proof of Proposition 4.1.1. On the first page of the spectral sequence we have from this collection of lemmas one nontrivial complex, The differential takes the form We use the simplex trick as in Lemma 4.1.4, to map to a simplicial complex. In particular, we map isomorphically to the dimension 3, 2, 1, 0, −1 part of the simplicial complex ∆[n − 1] considered in Appendix A. The condition that the middle k i are nonzero is equivalent to the middle k ′ i ≥ 2, thus the difference of at least two between adject vertices. Since From Proposition A.0.5, the resulting simplicial complex has (reduced) homology in dimension 0 for n − 1 = 2, 3 (so the chain complex in dimension n for n = 3, 4); in dimension 1 for n − 1 = 5, 6 (so the chain complex in dimension n + 1 for n + 1 = 7, 8); and in dimension 2 for n − 1 = 8, 9 (so the chain complex in dimension n + 2 for n + 2 = 11, 12). This is the result.
Furthermore the correspondence with the simplicial complex allows us to find explicit representatives of all classes.
location representative 12 (L) σ (9) 12 := σ (8) 11 ⊗ ξ L We would like to find representatives that are a little more manageable, and they are available.

So an updated version of representatives is
So we get representatives,   We consider the decreasing filtration with Then the zero page of the spectral sequence of the filtration has complexes such that the sum of powers of ξ L is exactly i. The proof of Proposition 4.2.1 will follow from the following lemmas. n−1 (η) is one-dimensional for i ≤ n − 2 spanned by (ξ L ) i η(ξ) n−2−i . If i < n − 2 then the power on ξ is nonzero and thus this tensor is a boundary, and otherwise it is not.
Proof. Consider i < n − 3 first. In this case there is at least one ξ present and there are two cases: c = 0, c = 0. If c = 0 then we get an element of ker d and im d as usual.
If c = 0 then For fixed a the complex of such tensors is isomorphic to that from Lemma 4.1.4, via tensoring this present complex with θ(ξ L ) e on the right. Thus this complex has no homology.
The expressions on the right are linearly independent for different choices of a, c, e. If both c, e = 0 then we get an element of ker d and not im d as usual. Proof. Indecomposable tensors in ker d (that is, where c, e, g are all nonzero) are in im d for i < n − 5 since they have some nonzero power of ξ.
With i < n − 5 and at least one of c, g, e = 0 we have, Combinations of pairs of tensors with fixed a, e, g, f, h in line (1) are in im d, following from the calculation in Lemma 4.1.5, by tensoring those with η(ξ) h on the right, and similarly for pairs from (2) and pairs from (3).
Combinations of cross-terms from (1) and (2) in ker d are in im d following also from tensoring calculations from Lemma 4.1.5 on the right with η(ξ) h ; similarly cross-terms from (2) and (3), dividing the calculation in Lemma 4.1.5 by θ(ξ L ) g (in an appropriate sense) on the right and tensoring on the left with (ξ L ) a 0 η(ξ) b 0 θ. It follows that all combinations of cross-terms from (1) and (3) whose boundaries vanish also appear in im d.
All tensors with i = n − 5 and c, e, g = 0 are in ker d and not im d. Then note that if any of c, e, g = 0, the resulting boundaries are all linearly independent. This fact is immediate after writing them down. H n+3 (gr n−6 C (n−1) • (η)) = (ξ L ) a ηθ(ξ L ) c ηθ(ξ L ) e ηθ(ξ L ) g ηθ(ξ L ) i η| a + c + e + g + i = n − 6 and c, e, g, i = 0 .
Proof. All of the described terms are obviously in ker d and not im d. Once again the boundaries of all terms with any of c, e, g, i = 0 are linearly independent.
Proof of Proposition 4.2.1. From the lemmas we find that on the first page of the spectral sequence we have one non-trivial complex, We use the simplex trick as in the proof of We also need consider n = 1, 2 for which ∆[n − 3] does not make sense. In both of these cases H n+1 (gr n−2 C (n−1) • (η)) = 0 and H n (gr n−1 C (n−1) • (η)) is one-dimensional, generated by η and (ξ L )η respectively.
The simplicial correspondence allows us to find representatives of classes. We see these are the representatives for the corresponding classes of H * (L) tensored with (ξ L )η on the right. location representative    Proof. There is an isomorphism of complexes C (m) We have representatives, location representative give rise to long exact sequences in Hochschild cohomology.  Proof.
We check, So δ is injective.
b. We consider the sequence So the image of ψ under δ is the negative of the image of φ from Proposition 4.3.1(b), but this was made 0 already there. Thus δ is zero here.
d. We have a sequence The class ψ generating HH 4 Thus again as in the proof of Proposition 4.3.1(d) the kernel is φ 1 + φ 2 , meaning that in this case δ = 0.  So we check So δ is an isomorphism.
i. We have a sequence For precisely the reasons as in (c,f) above δ is an isomorphism.
j. We have a sequence For precisely the reasons as in (d,g) above δ = 0.
All products m n with odd n vanish.
We set about amending this structure based on the result in Theorem 1 and the discussion in Section 2.
The associative algebra B obtained by restricting the A ∞ -structure from Theorem 2 is independent of τ . The calculation that m 3 = 0 also does not depend on τ . Theorem 2 has a nonzero m 4 , but the calculation in Theorem 1 that HH 4 (−2) (B) = 0 implies that all possible m 4 extending m 1 = 0, m 2 , m 3 = 0 are equivalent to m 4 = 0 (since m 3 = m 4 = m 5 = · · · = 0 is one possible A ∞ -structure on B). The discussion in Section 2 therefore implies that there is a strict equivalence f : B → B such that f * m 4 = 0.
Since f 1 = id B and f 2 = 0 it follows that m ′ 3 = m 3 = 0, and in order for m ′ 4 = 0 we must have δf 3 = m 4 . That this is true is verified by direct calculation.
(2) We show by induction that m ′ 2k+1 = 0 for k ∈ Z ≥1 . The base case is above, and for k ≥ 2 the left side of the relation vanishes. This is clear since f i = 0 except for i = 1, 3, and the terms involving f 1 , f 3 have m 2k+1 and m 2k−1 , respectively, which are both zero. The right side vanishes except for the term m ′ 2k+1 , since in order for i 1 + · · · + i r = 2k + 1 where all i j are 1 or 3, we must have r odd. The induction hypothesis guarantees that in these cases m ′ r = 0.
(3) Finally, the A ∞ -relations for (m ′ n ) give where Φ k is a quadratic expression. The right side of the first must be zero since The right side of the second must be zero since only m ′ 6 = 0 and in this quadratic expression, m ′ 6 is paired with m ′ 4 .
We determine from the relation in the proof of (1) that 1 ⊗r ⊗ f 3 ⊗ 1 ⊗t .
The vanishing of HH 3 (−2) (B) implies that all choices of f 3 for the equivalence in Proposition 5.1.1 are homotopic, so that the cohomology classes of m ′ 6 , m ′ 8 are well-defined in the following sense.   We also use this class in ∆ [6]. (The choice between + and − when offered is not of importance for our application of this calculation, so we do not make it.) The class in ∆ [8] we realize similarly. The intersection of the two contractible parts is ∆ [6], which we consider as the loop above. We make a cone over this loop to the points {7} and {8} to get the class representative This class will also work for ∆ [9].