# A Finite Dimensional A∞ Algebra Example

Michael P. Allocca Department of Mathematics, North Carolina State University, Raleigh NC 27695, USA  and  Tom Lada Department of Mathematics, North Carolina State University, Raleigh NC 27695, USA Dedicated to Tornike Kadeishvili on the occasion of his 60th birthday
###### Abstract.

We construct an example of an algebra structure defined over a finite dimensional graded vector space.

## Introduction

algebras (or sha algebras) and (or sh Lie algebras) have been topics of current research. Construction of small examples of these algebras can play a role in gaining insight into deeper properties of these structures. These examples may prove useful in developing a deformation theory as well as a representation theory for these algebras.

In , an algebra structure on the graded vector space where is a 2 dimensional vector space, and is a 1 dimensional space, is discussed. This surprisingly rich structure on this small graded vector space was shown by Kadeishvili and Lada, , to be an example of an open-closed homotopy algebra (OCHA) defined by Kajiura and Stasheff . In an unpublished note  M. Daily constructs a variety of other algebra structures on this same vector space.

In this article we add to this collection of structures on the vector space by providing a detailed construction of non-trivial algebra data for .

## 1. A∞ Algebras

We first recall the definition of an algebra (Stasheff ).

###### Definition 1.1.

Let be a graded vector space. An structure on is a collection of linear maps of degree that satisfy the identity

 n−1∑λ=0n−λ∑k=1αmn−k+1(x1⊗⋯⊗xλ⊗mk(xλ+1⊗⋯⊗xλ+k)⊗xλ+k+1⊗⋯⊗xn)=0

where , for all .

This utilizes the cochain complex convention. One may alternatively utilize the chain complex convention by requiring each map to have degree .

We will define the desuspension of V (denoted ) as the graded vector space with indices given by , and the desuspension operator, (resp. suspension operator ) in the natural sense. We will also employ the usual Koszul sign convention in this setting: whenever two symbols (objects or maps) of degree and are commuted, a factor of is introduced. Subsequently, and .

Stasheff also showed that an structure on is equivalent to the existence of a degree 1 coderivation with the property . Here, is the tensor coalgebra on the graded vector space . Such a coderivation is constructed by defining by and then extending each to a coderivation on . By “abuse of notation”, can be described by

 m′k(↓x1⊗↓x2⊗⋯⊗↓xn)=n−1∑i=0(1⊗i⊗m′k⊗1⊗n−i−1)(↓x1⊗↓x2⊗⋯⊗↓xn) =n−1∑i=0(−1)(k−2)(|x1|+⋯+|xi|−i)(↓x1⊗⋯⊗↓xi⊗m′k(↓xi+1⊗⋯⊗↓xi+k)⊗↓xi+k+1⊗⋯⊗↓xn)

We then define .

## 2. A Finite Dimensional Example

Let denote the graded vector space given by where has basis , has basis , and for . Define a structure on by the following linear maps :

 m1(v1)=m1(v2) =w Forn≥2:mn(v1⊗w⊗k⊗v1⊗w⊗(n−2)−k) =(−1)ksnv1,0≤k≤n−2 mn(v1⊗w⊗(n−2)⊗v2) =sn+1v1 mn(v1⊗w⊗(n−1)) =sn+1w

where , and when evaluated on any element of that is not listed above. It is worth noting that this assumes the cochain convention regarding algebra structures. Hence and .

###### Theorem 2.1.

The maps defined above give the graded vector space an algebra structure.

The proof of this theorem relies on two lemmas:

###### Lemma 2.2.

Let

 m′1 =↓m1 Forn≥2:m′n(↓v1⊗(↓w)⊗k⊗↓v1⊗(↓w)⊗(n−2)−k) =↓v1,0≤k≤n−2 m′n(↓v1⊗(↓w)⊗(n−2)⊗↓v2) =↓v1 m′n(↓v1⊗(↓w)⊗(n−1)) =↓w
###### Remark 2.3.

Each is of degree 1.

###### Proof of Lemma 2.2.

for any .

Now let . The majority of the work here is centered around computing the signs associated with the graded setting. The elements and the maps , , and all contribute to an overall sign via their degrees. Observing these signs, we find

 m′n(↓x1⊗↓x2⊗⋯⊗↓xn)=(−1)n(n−1)2↓∘mn∘↑⊗n(↓x1⊗↓x2⊗⋯⊗↓xn)
 =⎧⎨⎩(−1)∑n/2i=1|x2i−1|↓mn(x1⊗x2⊗⋯⊗xn)if n is even.(−1)∑(n−1)/2i=1|x2i|↓mn(x1⊗x2⊗⋯⊗xn)if n is odd.

First consider

 m′n(↓v1⊗(↓w)⊗k⊗↓v1⊗(↓w)⊗(n−2)−k) =(−1)|v1|+(n2−1)|w|↓mn(v1⊗w⊗k⊗v1⊗w⊗(n−2)−k) =(−1)0+n2−1(−1)ksn↓v1 =(−1)n2−1(−1)(n+1)(n+2)2↓v1 =(−1)n2−1(−1)(n+1)(n2+1)↓v1(∗)

If is even, then where ‘odd’ denotes an odd number.
If is odd, then where ‘even’ denotes an even number.

A similar argument holds in the remaining 3 cases. Hence

 m′n(↓v1⊗(↓w)⊗k⊗↓v1⊗(↓w)⊗(n−2)−k)=↓v1,0≤k≤n−2

Now consider . This computation may be divided into 2 cases based on the parity of . If is even, then:

 m′n(↓v1⊗(↓w)⊗(n−2)⊗↓v2) =(−1)|v1|+(n2−1)|w|mn(↓v1⊗(↓w)⊗(n−2)⊗↓v2) =(−1)n2−1sn+1↓v1 =(−1)n2−1(−1)(n+2)(n+3)2↓v1 =(−1)n2−1(−1)(n2−1)(n+3)↓v1(∗)

If is even, then where ‘odd’ denotes an odd number.
If is odd, then where ‘even’ denotes an even number.

A similar argument holds in the case that is odd. Hence

 m′n(↓v1⊗(↓w)⊗(n−2)⊗↓v2)=↓v1

The preceding argument may also be repeated for . Hence

 m′n(↓v1⊗(↓w)⊗(n−1))=↓w

###### Lemma 2.4.

Let where is defined in Lemma 2.2. Let be a positive integer. Suppose

Then

###### Proof.

We first note that

 D2(↓x1⊗↓x2⊗⋯⊗↓xn)=∑i+j≤n+1m′im′j(↓x1⊗↓x2⊗⋯⊗↓xn)

since for . So

 D2(↓x1⊗↓x2⊗⋯⊗↓xn)= ∑i+j≤nm′im′j(↓x1⊗↓x2⊗⋯⊗↓xn) + ∑i+j=n+1m′im′j(↓x1⊗↓x2⊗⋯⊗↓xn)

Hence it suffices to show that

Consider : Since , we can break this sum up into 4 different types of of elements in based on whether the first and last terms in the tensor product contain or :

Type 1: Elements with first term and last term
(example:

Now each term of type 1 must be produced by with . Hence, by factorization of tensor products, all possible terms of type 1 are given by:

 (−1)2|x1|−2(↓x1⊗(∑i+j≤n−1m′im′j(↓x2⊗↓x3⊗⋯⊗↓xn−1)))⊗↓xn =(↓x1⊗(D2(↓x2⊗↓x3⊗⋯⊗↓xn−1)))⊗↓xn =(↓x1⊗0)⊗↓xn =0

since when evaluated on terms. A similar argument holds for the type 2 and type 3 summands.

We now consider type 4 terms. Consider an arbitrary element of type 4:

 m′i(↓x1⊗⋯⊗↓xi)⊗↓xi+1⊗⋯⊗↓xn−j⊗m′j(↓xn−j+1⊗⋯⊗↓xn)

Consider how this arbitrary element is generated: We begin with

 m′im′j(↓x1⊗⋯⊗↓xn)

We then apply to the last terms, which yields:

 (−1)|x1|+⋯+|xn−j|−(n−j)m′i(↓x1⊗⋯⊗↓xn−j⊗m′j(↓xn−j+1⊗⋯⊗↓xn))

Finally we apply to the first terms:

 (−1)|x1|+⋯+|xn−j|−(n−j)m′i(↓x1⊗⋯⊗↓xi)⊗↓xi+1⊗⋯⋯⊗↓xn−j⊗m′j(↓xn−j+1⊗⋯⊗↓xn)(∗)

Each of these arbitrary type 4 elements can be paired up with an element generated by as follows: Begin with

 m′jm′i(↓x1⊗⋯⊗↓xn)

Then apply to the first terms:

 m′j(m′i(↓x1⊗⋯⊗↓xi)⊗↓xi+1⊗⋯⊗↓xn)

Finally, apply to the last terms:

 (−1)|x1|+⋯+|xn−j|−(n−j)+1m′i(↓x1⊗⋯⊗↓xi)⊗↓xi+1⊗⋯⋯⊗↓xn−j⊗m′j(↓xn−j+1⊗⋯⊗↓xn)(∗∗)

Since these type 4 elements were arbitrary, and , all type 4 terms added together equal 0. Hence, all type 1, 2, 3, and 4 terms yield 0, and so

 ∑i+j≤nm′im′j(↓x1⊗↓x2⊗⋯⊗↓xn)=0

###### Proof of Theorem 2.1.

It is clear that each map is of degree . To prove that these maps yield an structure, one may verify that they satisfy the identity given in definition 1.1. However, this is a rather daunting task, due to the varying signs, , accompanying the maps. To utilize an alternative method of proof, we construct a degree 1 coderivation, , as described in section 1.

In the context of Theorem 2.1, we may use the definition for given by Lemma 2.2 to construct . It then suffices to show that .

We aim to prove by induction on the number of inputs for . It is worth first noting that , however since for .

For , we have

Now assume . We aim to show that :

###### Remark 2.5.

Since and are linear, it is sufficient to show that on only basis elements.

By Lemma 2.4, , hence it suffices to show that

It is advantageous to approach this problem from the bottom up, since implies calculating different combinations of elements. That is, we consider only nontrivial (nonzero) elements in the sum . Now since , we observe that

or

Therefore if , then is a sum of ’s or ’s.

We first consider the manner in which yields a :

By defintion of , must be produced by . To accomplish this, the original arrangement must satisfy and must contain exactly one more ‘’ ( or ).

Case 1: . Let us consider

We have

We have

Case 2: . Let us consider

For , the only nontrivial way to do this yields:

 m′n−k−1(m′k+2(↓v1⊗(↓w)⊗k⊗↓v2)⊗(↓w)⊗(n−2)−k)=↓w

and for , the only nontrivial way to do this yields:

 (−1)|v1|+k|w|−(k+1)m′n(↓v1⊗(↓w)⊗k⊗m′1(↓v2)⊗(↓w)⊗(n−2)−k)=−↓w

In either case, if produces ’s, then

 ∑i+j=n+1m′im′j(↓x1⊗⋯⊗↓xn)=0.

We now consider the manner in which yields a :

By defintion of , must be produced by either or .

Case 1: is produced by .

We examine the 4 different possibilities for which can yield this arrangement:

produces the first . produces a in .

produces the second . produces a in .

A key observation to make here is that and imply that the original arrangement must contain exactly 3 ’s, once again with . This yields 4 subcases:

Subcase 1: We have

Subcase 2: We have

Subcase 3: We have

Subcase 4: We have

Let us consider subcase 1:

must take the first two ’s. We have:

 m′i([m′j(↓v1⊗(↓w)⊗k⊗↓v1⊗(↓w)⊗j−k−2)⊗(↓w)⊗l−(j−k−2)]⊗↓v1⊗(↓w)⊗n−k−l−3)=↓v1

Now , so there are such terms.

must take only the second . We have:

 (−1)|v1|+k|w|−(k+1)m′i(↓v1⊗(↓w)⊗k⊗[m′j(↓v1⊗(↓w)⊗(j−1))⊗(↓w)⊗l−(j−1)]⊗↓v1⊗(↓w)⊗n−k−l−3)=−↓v1

Now , so there are such terms.

must take the second and third ’s. We have:

 (−1)|v1|+k|w|−(k+1)m′i(↓v1⊗(↓w)⊗k⊗[m′j(↓v1⊗(↓w)⊗l⊗↓v1⊗(↓w)⊗j−l−2)]⊗(↓w)⊗n−k−j+1)=−↓v1

Now , so there are such terms.

must take only the third . We have:

Now , so there are such terms.
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