Morphisms of $A_\infty$-algebras and modules

Introduction

$A_{\infty}$-algebras and modules are widely used tools in homological algebra and geometry. Although they behave nicely over a field and there are many strong results in this case, there are also important constructions over arbitrary rings such as the description of a particular objects in non-commutative geometry by Kontsevich and Soibelman in [3] or in Heegaard Floer theory in [5] and [8]. Therefore the question about generalization of the results over a field seems very natural. The two central results over a field are the following theorems:

These theorems combined imply functoriality of minimal models of $A_{\infty}$-module. The goal of this thesis is to present counterexamples to the naive generalizations of these theorems to arbitrary rings and propose some sufficient conditions for which minimal models are functorial. The main result is the following theorem:

$M \to L$ such that $f_1$ is an epimorphism. Then there exists a morphism $\psi: L \to M$, which is section of $f$.

A dual statement is given in Theorem 3.13 i Section 3 below.

We build on the results of [7] to show that without the assumption of projectivity, the result does not hold.

The structure of the paper is the following. Section 2 serves as a revision of basic notions of $A_\infty$-algebras, their morphisms and homotopy. We also present known results about them such as stated theorems of Keller and results over an arbitrary ring shown in [7]. Section 3 introduces $A_\infty$-modules in the same way as $A_\infty$-algebras and contains proofs of the main results of the thesis. Section 4 shows how presented theorems imply functoriality of minimal models under certain conditions. The paper is the expanded version of the author’s bachelor thesis.

AI declaration. The paper was written without using any AI tools.

Acknowledgments. The author is grateful to Karol Szumiło for the supervision of this thesis and Maciej Borodzik for many fruitful conversations

A-algebras

Definitions

We begin by giving the basic definitions of the theory of $A_\infty$-algebras.

$A_\infty$-algebras together with such morphisms form a category, in which identity morphisms are strict morphisms such that $f_1$ is an identity. Isomorphisms however don’t have to be strict.

The following definition introduces an algebraic analogue to topological homotopy and is a natural extension of chain homotopy to the $A_\infty$-algebras.

Known results over a field

In this section, let $R=K$ be a field. Over a field are known the following results:

These theorems allow to restrict the study only to minimal models of $A_\infty$-algebras and easily pass to a more convenient derived category, which, by Prouté’s theorem, is equal to a homotopy category of $A_\infty$-algebras.

General results over a ring

The results over a field directly use the structure of vector spaces e.g. the direct sum decompositions, which always exist. For modules over arbitrary rings such decomposition does not have to exist and there are known counterexamples for results listed in the previous section. The following example illustrates the failure of Kadeishvili’s theorem over a ring without the additional assumptions about the structure of the modules.

Proof. Step 1: There exists an $A_\infty$-algebra structure on $A$: We set

The maps $m_1$ and $m_2$ are homogenous of degree 1 and 0 accordingly. We also have $m_1^2=0$ and $m_2$ is associative, since it is just a matrix multiplication. Therefore we just have to check the Leibniz rule:

Thus $(A,m_1,m_2,0,\ldots)$ is indeed an $A_\infty$-algebra.

Step 2: Determining the homology $H^*A$:

Since $n$ is not a zero divisor, the set of cycles in $A$ is exactly

and the set of boundaries is

Hence we have $H^1A\cong H^0A\cong R/nR$, and the rest is $0$.

Step 3: There is no minimal $A_\infty$-structure on $H^*A$ together with a quasi-isomorphism $H^*A\longrightarrow A$:

By the choice of $n$ the module $R/nR$ is a nonzero torsion module annihilated by $n$. $A$, however, is torsion-free, therefore there is no nonzero $R$-linear map $f_1:H^*A\longrightarrow A$, in particular there is no quasi-isomorphism of chain complexes $f_1:(H^*A,0)\longrightarrow(A,m_1)$.

Step 4: There is no minimal $A_\infty$-structure on $H^*A$ together with a quasi-isomorphism $A\longrightarrow H^*A$:

Suppose that there is such structure $(H^*A,0,m'_2,\ldots)$ together with a quasi-isomorphism $(f_1,f_2,\ldots)$. Then the following equation must be satisfied

We have $m_1=n\cdot g$, where $g:A\longrightarrow A$, $g\left(\begin{pmatrix}a&b\\c&d\end{pmatrix}\right)=\begin{pmatrix}c&d-a\\0&c\end{pmatrix}$. Hence we obtain

since $H^*A$ is annihilated by $n$. Therefore the equation simplifies to

Let $\zeta=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\eta=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then $\zeta$ generates $Z^0A$, $\eta$—$Z^1A$. Since $f_1$ is a quasi-isomorphism, $f_1(\zeta)$ generates $H^0A$, and $f_1(\eta)$ generates $H^1A$, therefore there exists $r\in R$ such that

that is $f_1\left(\begin{pmatrix}1+r&0\\0&r\end{pmatrix}\right)=0$.

Let $\omega=\begin{pmatrix}1+r&0\\0&r\end{pmatrix}\otimes\begin{pmatrix}0&1\\0&0\end{pmatrix}-\begin{pmatrix}0&1\\0&0\end{pmatrix}\otimes\begin{pmatrix}1+r&0\\0&r\end{pmatrix}$. Then we obtain

On the other hand

since $f_1(\eta)$ generates $H^1A\ne\{0\}$. This is a contradiction. □

An analogue of Kadeishvili’s theorem is true if we make additional assumptions about homology of our $A_{\infty}$-algebra.

A-modules

Definitions

Let $(A,m_1,m_2,\ldots)$ be an $A_{\infty}$-algebra.

Definition 3.1. An $A_{\infty}$-module over $A$ is a $\mathbb{Z}$-graded $R$-module $M$ together with maps

of degree $2-n$ for each $n\in\mathbb{N}$ such that

where we set $u=r+1+t$. The term $m^M_u(1^{\otimes r}\otimes m_s\otimes1^{\otimes t})$ is interpreted as $m^M_u(1^{\otimes r}\otimes m_s\otimes1^{\otimes t})$ for $r>0$ and as $m^M_u(m^M_s\otimes1^{\otimes t})$ for $r=0$.

Known results over a field

In this section, let $R=K$ be a field. We also set an $A_\infty$-algebra $A$.

The theorems about $A_\infty$-algebras in this situation are also true for $A_\infty$-modules.

The same as for $A_\infty$-algebras such structure is called a minimal model.

These theorems, analogously to their counterparts for $A_\infty$-algebras, allow us to easily pass to derived category, which has additional structure. An example of the application of this transition is described in detail in [2].

General results over a ring

The following examples show that the second theorem also fails over arbitrary ring and emphasize the importance of projectivity/injectivity assumptions for corresponding modules in later theorems.

Proof. Step 1: This structure is an $A_\infty$-module structure on $M$: The maps $m_1^M$ and $m_2^M$ are of degree 1 i 0 accordingly, $(m_1^M)^2=0$, and $m_2^M$ is associative. Thus it suffices to check the Leibniz rule.

Step 2: Homology $H^*M$: The homology $H^*M$ is a complex concentrated in degree 0 with trivial differential and we have $H^*M\cong\mathbb{Z}_2$. It is a $\mathbb{Z}$-module, therefore it is in particular an $A_\infty$-module.

Step 3: Quasi-isomorphism $M\longrightarrow H^*M$

We set $f=(f_1,0,0,\ldots)$, where $f_1$ is the projection $\mathbb{Z}\longrightarrow\mathbb{Z}_2$ in degree 0 and 0 everywhere else. It is both a morphism of complexes and a morphism of $\mathbb{Z}$-modules in each degree and since it is exactly the projection on homology, it is a quasi-isomorphism of $A_\infty$-modules.

Step 4: Lack of quasi-isomorphism $H^*M\longrightarrow M$

$\mathbb{Z}_2$ is a torsion module, and $\mathbb{Z}$ – torsion-free, therefore there is no nonzero morphism of complexes $H^*M\longrightarrow M$, therefore in particular there is no quasi-isomorphism of $A_\infty$-modules. □

Proof. Step 1: This structure is an $A_\infty$-module structure on $M$: The maps $m_1^M$ and $m_2^M$ are of degree 1 and 0 accordingly, $(m_1^M)^2=0$, and $m_2^M$ is associative. Thus it suffices to check the Leibniz rule.

Step 2: Homology $H^*M$: The homology $H^*M$ is a complex concentrated in degree 0 with trivial differential and we have $H^*M\cong\mathbb{Z}$. It is a $\mathbb{Z}$-module, therefore it is in particular an $A_\infty$-module.

Step 3: Quasi-isomorphism $H^*M\longrightarrow M$

We $f=(f_1,0,0,\ldots)$, where $f_1$ is an inclusion $\mathbb{Z}\longrightarrow2\mathbb{Z}\hookrightarrow\mathbb{Z}$ in degree 0 and 0 everywhere else. It is both a morphism of complexes and morphism of $\mathbb{Z}$-modules in each degree and since it is exactly an inclusion of homology, it is a quasi-isomorphism of $A_\infty$-modules.

Step 4: Lack of quasi-isomorphism $M\longrightarrow H^*M$

Such quasi-isomorphism has to be a projection on homology. If such projection were to exist $\mathbb{Z}\longrightarrow\mathbb{Z}$, which would be a retraction of an inclusion $\mathbb{Z}\longrightarrow2\mathbb{Z}\hookrightarrow\mathbb{Z}$, we would obtain $\mathbb{Z}\cong\mathbb{Z}\oplus\mathbb{Z}$, which is a contradiction. □

By making additional assumptions about the structure of modules we can prove the existence of sections or retractions of certain morphisms. The following theorems are dual to each other and each is preceded by a corresponding technical lemma which is analogous to Lemma 1.3.3.3 in [4].

Proof. Let $A'$ be the same grade module as $A$. We set $q_1=1_A$. Assume that we have $q_1,\ldots,q_n$ such that

We now take the $(n+1)$-th identity for compositions without a $p_1(q_{n+1})$ factor and compose it with the section of $p_1$. This is our $q_{n+1}$.

We now want to define the multiplications on $A'$. We set $m_1^{A'}=m_1^A$. Assume that we have $m_1^{A'},\ldots,m_n^{A'}$. The $(n+1)$-th identity for $q$ gives us

but $q_1 = 1_A$.

We now construct $q^{-1}$. Set $(q^{-1})_1 = 1_A$. Assume that we have $(q^{-1})_1,\ldots,(q^{-1})_n$. From the $(n+1)$-th identity of compositions we obtain

and in the sum there are no factors dependent on $(q^{-1})_{n+1}$, but $q_1 = 1_A$.

Proof. Let $\psi_1$ be inclusion $L\longrightarrow M$, which exists because of the projectivity of $L$. Then, by the previous lemma, we can without loss of generality assume that $f$ is a strict morphism. From this we get

for each $n \ge1$

and $M = \psi_1(L) \oplus M'$, where $M' = \ker f_1$.

Notice that in this situation $\psi= (\psi_1,0,0,\ldots)$ jest $A_\infty$-morphism. By composing this identity for $f_1$ with $\psi_1$ on the left, and with $\psi_1 \otimes1^{\otimes n-1}$ on the right we obtain

but $\psi_1 \circ f_1$ is an identity on $\psi_1(L)$, which completes the proof.

Proof. Let $D'$ be the same graded module as $D$. We set $k_1 = 1_D$. Assume that we have $k_1,\ldots,k_n$ such that

We take the $(n+1)$-th identity for compositions without a $k_{n+1}(j_1 \otimes1^{\otimes n})$ factor and compose it with the retraction of $j_1$. This is our $k_{n+1}$.

We now want to define the multiplications on $D'$. We set $m^{D'}_1 = m^D_1$. Assume that we have $m^{D'}_1,\ldots,m^{D'}_n$. From the $(n+1)$-th identity for $k$ we obtain

but $k_1 = 1_D$.

We now construct $k^{-1}$. We set $(k^{-1})_1 = 1_D$. Assume that we have $(k^{-1})_1,\ldots,(k^{-1})_n$. From the $(n+1)$-th identity of compositions we obtain

and in the sum there are no factors dependent on $(k^{-1})_{n+1}$, but $k_1=1_D$.

Proof. Let $\varphi_1$ be a projection $M\longrightarrow L$, which exists because of the injectivity of $L$. Therefore we can without loss of generality assume that $p$ is a strict morphism. From this we get:

and $M=p_1(L)\oplus M'$, where $M'=\ker\varphi_1$. Notice that in this situation $\varphi=(\varphi_1,0,0,\ldots)$ is an $A_\infty$-morphism. Composing this identity with $\varphi_1$ on the left, and with $\varphi_1\otimes1^{\otimes n-1}$ on the right we obtain, in the same way as in the previous theorem, the desired identity.

Functoriality of minimal models

A natural question is, when a morphism $A_\infty$-modules $M\longrightarrow L$ induces a morphism of their minimal models. First of all in this situation one must an make assumption that such models exist. The next natural assumption is the existence of morphisms between this modules and their minimal models. In this situation the theorems from the previous chapter give sufficient conditions for existence of sections or retractions of these morphisms and by taking corresponding compositions we obtain a morphism between the minimal models. This is precisely described in the following theorems:

Proof. From the Theorem 3.13 we obtain a retraction $L\longrightarrow H^*L$, by composing it with $H^*M\longrightarrow M\longrightarrow L$ we obtain the desired induced morphism.

The following theorem is dual to the previous and its proof proceeds exactly the same, only instead of the Theorem 3.13 we apply 3.11.

The following diagram illustrates these two theorems.

Vector (PDF)