On the origin of higher braces and higher-order derivations

Abstract

The classical Koszul braces, sometimes also called the Koszul hierarchy, were introduced in 1985 by Koszul (Astérisque, (Numero Hors Serie):257–271, 1985). Their non-commutative counterparts came as a surprise much later, in 2013, in a preprint by Börjeson ($A_\infty$-algebras derived from associative algebras with a non-derivation differential, Preprint arXiv:1304.6231, 2013). In Part I we show that both braces are the twistings of the trivial $L_\infty$- (resp. $A_\infty$-) algebra by a specific automorphism of the underlying coalgebra. This gives an astonishingly simple proof of their properties. Using the twisting, we construct other surprising examples of $A_\infty$- and $L_\infty$-braces. We finish Part 1 by discussing $C_\infty$-braces related to Lie algebras. In Part 2 we prove that in fact all natural braces are the twistings by unique automorphisms. We also show that there is precisely one hierarchy of braces that leads to a sensible notion of higher-order derivations. Thus, the notion of higher-order derivations is independent of human choices. The results of the second part follow from the acyclicity of a certain space of natural operations.

1 Introduction

Advice for the reader. The article can be read in four possible ways:

• (1) as a tool for checking that the classical braces indeed form $L_\infty$- resp. $A_\infty$-algebras,

• (2) as a machine producing explicit examples of $L_\infty$-, $A_\infty$- and possibly also other types of strongly homotopy algebras,

• (3) as a justification that the higher-order derivations are God-given, not human, inventions existing since the beginning of time, or

• (4) as a vanilla version of [4].

The reader wanting only (1) and (2) may read Part 1 and skip the rest. Item (3) explains why higher-order derivations of commutative associative algebras appear e.g. in the interpretation of the algebraic structure of the combined conformal field theory of matter and ghosts given in [11]. It is natural to expect that higher-order derivations of associative algebras based on Börjeson’s braces would play a similar rôle for open strings.

Plan of the paper. Section 1 contains several examples of braces, including the classical Koszul $L_\infty$-hierarchy and Börjeson’s $A_\infty$-braces. We demonstrate various properties which the braces may posses, in particular those leading to a sensible definition of higher-order derivations.

In Sect. 2 we show how to generate braces by the twisting and interpret all examples in Sect. 1 as emerging this way. This offers a very simple verification that they indeed form $L_\infty$- resp. $A_\infty$-structures. We close this section by discussing possible generalizations to Lie and other types of algebras.

In Sect. 3 we analyze natural operations and prove that they form an acyclic space. The main results are Propositions 3.3 and 3.6. The material of this section is a baby version of the analysis of the Hochschild cochains in connection to Deligne’s conjecture as given in [4].

Section 4 formulates the consequences of Sect. 3. Theorems 4.1 and 4.9 state that all natural braces are the twistings by unique automorphisms, Corollaries 4.2 and 4.10 then describe the moduli space of all natural braces. By Theorems 4.3 and 4.11, Börjeson’s resp. Koszul braces are the unique ones leading to a meaningful notion of higher-order derivations of associative resp. commutative associative algebras.

Conventions. If not stated otherwise, all algebraic objects will be considered over a fixed field ${\mathbb {k}}$ of characteristic zero. The symbol $\otimes$ will denote the tensor product over ${\mathbb {k}}$ and $\mathrm{Span}(S)$ the ${\mathbb {k}}$-vector space spanned by a set $S$. We will denote by $1\!\!1_X$ or simply by $1\!\!1$ when $X$ is understood, the identity endomorphism of an object $X$ (set, vector space, algebra,&c.). We will usually write the product of elements $a$ and $b$ of an associative algebra as $a\cdot b$ or simply as $ab$.

A degree of a graded object will be denoted by $|w|$ though we will sometimes omit the vertical bars and write e.g. $(-1)^{a+b}$ instead of $(-1)^{|a|+|b|}$ to save the space. For a permutation $\sigma \in \Sigma _k$ and graded variables $w_{1},\dots ,w_{k}$, the Koszul sign $\varepsilon (\sigma ;w_{1},\dots ,w_{k}) \in \{-1,+1\}$ is defined by the equation

$$\begin{aligned} w_1\wedge \ldots \wedge w_k= \varepsilon (\sigma ;w_{1},\dots ,w_{k}) \cdot w_{\sigma (1)}\wedge \ldots \wedge w_{\sigma (k)} \end{aligned}$$

in the free graded commutative associative algebra ${\mathbb S}(w_{1},\dots ,w_{k})$ generated by $w_{1},\dots ,w_{k}$. We usually write $\varepsilon (\sigma )$ instead of $\varepsilon (\sigma ;w_{1},\dots ,w_{k})$ when the meaning of $w_{1},\dots ,w_{k}$ is clear from the context.

Given integers $a,b \ge 0$, an $(a,b)$-unshuffle is a permutation $\sigma \in \Sigma _{a+b}$ satisfying

$$\begin{aligned} \sigma (1)<\cdots <\sigma (a) \quad \hbox {and} \quad \sigma (a+1)< \cdots < \sigma (a+b). \end{aligned}$$

By braces we mean the structure operations of a strongly homotopy algebra.

Part 1. Examples and constructions

1.1 Examples in place of introduction

We start by recalling a construction attributed to Koszul [7] and sometimes referred to as the Koszul hierarchy, see also [1–3, 5, 17]. It is used to define higher order derivations of commutative associative algebras, see §1.2 below; they play a substantial rôle for instance in the BRST approach to closed string field theory [11, Section 4].

Example 1.1

(Classical $L_\infty$-braces) Let $A$ be a graded commutative associative algebra with a degree $+1$ differential $\Delta$ which is, very crucially, not necessarily a derivation. Koszul braces are linear degree $+1$ maps $\Phi _k^\Delta : {A}^{\otimes k} \rightarrow A$, $k \ge 1$, defined by the formulas

$$\begin{aligned} \Phi ^\Delta _1(a) \!&= \! \Delta (a),\\ \Phi ^\Delta _2(a_1,a_2) \!&= \! \Delta (a_1a_2) - \Delta (a_1)a_2 - (-1)^{a_1a_2}\Delta (a_2)a_1,\\ \Phi ^\Delta _3(a_1,a_2,a_3) \!&= \!\Delta (a_1a_2a_3)\\&\quad -\Delta (a_1a_2)a_3 \!-\! (-1)^{a_1(a_2+ a_3)} \Delta (a_2a_3)a_1 \!-\!(-1)^{a_3(a_1+ a_2)}\!\Delta (a_3a_1)a_2\\&\quad +\Delta (a_1)a_2a_3\!+ \! (-1)^{a_1(a_2\!+\! a_3)}\Delta (a_2)a_3a_1\!+\!(-1)^{a_3(a_1\! +\! a_2)} \Delta (a_3)a_1a_2,\\ \!&\vdots&\!\\ \Phi ^\Delta _k(a_{1},\dots ,a_{k}) \!&= \! \sum _{1 \le i \le k} (-1)^{k-i} \sum _\sigma \varepsilon (\sigma )\Delta (a_{\sigma (1)} \cdots a_{\sigma (i)})a_{\sigma (i+1)} \cdots a_{\sigma (k)}, \end{aligned}$$

for $a,a_1,a_2,a_3, \ldots \in A$. The summation in the last line runs over all $(i,k\!-\!i)$-unshuffles $\sigma$ and $\varepsilon (\sigma ) = \varepsilon (\sigma ;a_{1},\dots ,a_{k})$ is the Koszul sign. As proved for instance in [3], these braces form an $L_\infty$-algebra¹ and have moreover the property that

$$\begin{aligned} \hbox {if } \Phi ^\Delta _k = 0 \quad \hbox {identically on} \quad {A}^{\otimes k}\quad \hbox {then}\quad \Phi ^\Delta _{k+1} = 0\quad \hbox {identically on}\quad {A}^{\otimes k+1}. \end{aligned}$$

(1)

We call braces with this property hereditary.

1.2 Higher order derivations

Let $A$ be a graded commutative associative algebra with a differential $\Delta$ as in Example 1.1. One says [2] that $\Delta$ is an order $r$ derivation if $\Phi ^\Delta _{r+1} = 0$. Clearly, being an order $1$ derivation is the same as being a derivation in the usual sense. It is almost clear that an order $r$-derivation is determined by its values on the products $x_1 \ldots x_s$, $s \le r$, of generators of $A$; an explicit formula is given in [11, Proposition 3.4].

One may ask whether higher-order derivations are ‘God-given,’ i.e. whether the braces that define it are unique. Let us try to find out which properties the braces leading to a sensible notion of higher-order derivations should satisfy. First of all, they must be ‘natural’ in that they use only the data that are available for any graded associative commutative algebra with a differential. The exact meaning of naturality is analyzed in Sect. 3.

Given $L_\infty$-braces $(A,\Delta ,l^\Delta _2,l^\Delta _3,\ldots )$, we may call $\Delta$ an order $r$ $l$-derivation if and only if $l^\Delta _{r+1} = 0$. It is clear from the axioms for $L_\infty$-algebras recalled in §2.3 that, for arbitrary scalars $\alpha ,\beta \in {\mathbb {k}}$, the object

$$\begin{aligned} (A,\alpha \Delta ,\alpha \beta \, l^\Delta _2,\alpha \beta ^2\, l^\Delta _3, \alpha \beta ^3\, l^\Delta _4,\ldots ) \end{aligned}$$

(2)

is an $L_\infty$-algebra as well. The first property we want is that an order $1$ $l$-derivation is an ordinary derivation. This means that, after a suitable renormalization (2),

$$\begin{aligned} l^\Delta _2(a_1,a_2) = \Delta (a_1a_2) - \Delta (a_1)a_2 - (-1)^{|a_1|}a_1\Delta (a_2), \ \text{ for } \text{ each } a_1,a_2 \in A. \end{aligned}$$

(3)

We also certainly want that an order $r$ $l$-derivation is also an order $r+1$ $l$-derivation, that is:

$$\begin{aligned} \hbox {the braces}\, l^\Delta _2,l^\Delta _3,l^\Delta _4, \ldots \, \hbox {are hereditary}. \end{aligned}$$

(4)

The following example however shows that conditions (3)–(4) still do not determine the braces uniquely.

Example 1.2

(Hereditary exotic $L_\infty$-braces) Here $A$ is a graded commutative associative algebra with a differential $\Delta$ of degree $+1$ as in Example 1.1. For $a,a_1,a_2,a_3,\ldots \in A$ define

$$\begin{aligned} h^\Delta _1(a) \!&= \! \Delta (a),\\ h^\Delta _2(a_1,a_2) \!&= \! \Delta (a_1a_2) - \Delta (a_1)a_2 - (-1)^{a_1a_2}\Delta (a_2)a_1,\\ h^\Delta _3(a_1,a_2,a_3) \!&= \! 2\Delta (a_1)a_2a_3\! {+}\! (-1)^{a_1(a_2{+} a_3)} 2 \Delta (a_2)a_3a_1 \!+\! (-1)^{a_3(a_1\! +\! a_2)}2 \!\Delta (a_3)a_1a_2\\&{-} \Delta (a_1a_2)a_3 \!-\! (-1)^{a_1(a_2+ a_3)}\Delta (a_2a_3)a_1 \!-\! (-1)^{a_3(a_1+ a_2)}\Delta (a_3a_1)a_2 ,\\ \!&\vdots&\!\\ h^\Delta _k(a_{1},\dots ,a_{k}) \!&= \! {(-1)^{k+1}} (k\!-\!1)! \sum _\tau \varepsilon (\tau ) \Delta (a_{\tau (1)})a_{\tau (2)} \cdots a_{\tau (k)}\\&\quad + {(-1)^{k}} (k\!-\!2)! \sum _\sigma \varepsilon (\sigma ) \Delta (a_{\sigma (1)}a_{\sigma (2)})a_{\sigma (3)} \cdots a_{\sigma (k)} \end{aligned}$$

where $\tau$ runs over all $(1,k\!-\!1)$-unshuffles and $\sigma$ over all $(2,k\!-\!2)$-unshuffles. It is easy to verify that the above braces satisfy the induction

$$\begin{aligned} h^\Delta _{k+1}(a_{1},\dots ,a_{k+1}) = - \sum _{\sigma } \varepsilon (\sigma ) h^\Delta _k(a_{\sigma (1)},\dots ,a_{\sigma (k)}) a_{\sigma (k+1)} \end{aligned}$$

with $\sigma$ running over all $(k,1)$-unshuffles. This implies that they are hereditary.

Next, we want the recursivity of higher-order derivations, by which we mean that an order $r$ $l$-derivation is determined by its values on the products of $\le \!r$ generators. Moreover, the notion of higher-order derivations and therefore the braces as well must be defined over an arbitrary ring. The recursivity is thus equivalent to:

$$\begin{aligned} \hbox {The braces are defined over the ring}\,{\mathbb Z} \,\hbox {of integers and the coefficient}\, C_k\nonumber \\ \hbox {at the term }\, \Delta (a_1\cdots a_k)\, \hbox {in }\, l^\Delta _k(a_1,\ldots , a_k) \hbox { is}\, \hbox {either } +{1} \,\,or\, -{1} \hbox { for any}\, k \ge 1.\quad \end{aligned}$$

(5)

For, if $p := C_k \not \in \{-1,1\}$ for some $k$, then higher-order $l$-derivations will not be recursive over the ring ${\mathbb Z}/p{\mathbb Z}$ of integers modulo $p$. For the braces in Example 1.2, $C_k = 0$ for all $k \ge 3$, so they do not satisfy (5).

It will follow from Theorem 4.11 that assumptions (3)–(5) already imply that $l^\Delta _k = \Phi ^\Delta _k$ for each $k \ge 1$. Let us start our discussion of the non-commutative case by recalling one construction from a recent preprint [6] of Börjeson.

Example 1.3

(Börjeson’s $A_\infty$-braces) Given a graded associative (not necessarily commutative) algebra $A$ with a differential $\Delta$, define for $a,a_1,a_2,a_3,\ldots \in A$,

$$\begin{aligned} b^\Delta _1(a)&= \Delta (a),\\ b^\Delta _2(a_1,a_2)&= \Delta (a_1a_2) - \Delta (a_1)a_2 - {(-1)^{a_1}} a_1\Delta (a_2),\\ b^\Delta _3(a_1,a_2,a_3)&= \Delta (a_1a_2a_3) -\Delta (a_1a_2)a_3\! {-}\! {(-1)^{a_1}} a_1\Delta (a_2a_3) \!+\!{(-1)^{a_1}} a_1\Delta (a_2)a_3,\\ b^\Delta _4(a_1,a_2,a_3,a_4)&= \Delta (a_1a_2a_3a_4)\!\\&\ - \Delta (a_1a_2a_3)a_4 -{(-1)^{a_1}} a_1\Delta (a_2a_3a_4) +{(-1)^{a_1}} a_1\Delta (a_2a_3)a_4,\\&\vdots&\\ b^\Delta _k(a_{1},\dots ,a_{k})&= \Delta (a_1\cdots a_k) - \Delta (a_1\cdots a_{k-1})a_k\\&\quad - {(-1)^{a_1}}a_1\Delta (a_2\cdots a_{k}) + {(-1)^{a_1}}a_1 \Delta (a_2\cdots a_{k-1})a_k. \end{aligned}$$

As proved in [6], these braces form an $A_\infty$-algebra² and are hereditary.

It is obvious that Börjeson’s braces satisfy assumptions (3)–(5), so they lead to a sensible notion of higher-order derivations of graded associative (non-commutative) algebras. By Theorem 4.3, they are the only $A_\infty$-braces with these properties.

Example 1.4

(Non-recursive $A_\infty$-braces) The braces below lead to recursive higher-order derivations over ${\mathbb Z}$ but not over ${\mathbb Z}/5{\mathbb Z}$, the integers modulo $5$. They are, up to the obvious Koszul signs, given by

$$\begin{aligned} e^\Delta _1(a) \!&= \! \Delta (a),\\ e^\Delta _2(a_1,a_2) \!&= \! \Delta (a_1a_2) - \Delta (a_1)a_2 - a_1\Delta (a_2)\\ e^\Delta _3(a_1,a_2,a_3) \!&= \! 2\Delta (a_1a_2a_3)\! -\! \Delta (a_1)a_2a_3\! -\! a_1a_2\Delta (a_3)\! -\! \Delta (a_1a_2)a_3\! -\! a_1\Delta (a_2a_3),\\ e^\Delta _4(a_1,a_2,a_3,a_4) \!&= \! 5 \Delta (a_1a_2a_3a_4) {-} \Delta (a_1a_2)a_3a_4 {-} a_1a_2\Delta (a_3a_4)\\&\quad \! - 2\big (\Delta (a_1)a_2a_3a_4\! +\! a_1a_2a_3\Delta (a_4)\! +\! \Delta (a_1a_2a_3)a_4\! +\! a_1\Delta (a_2a_3a_4)\big )\\&\vdots&\\ e^\Delta _k(a_{1},\dots ,a_{k}) \!&= \! \alpha _k \Delta (a_1\cdots a_k)\\&\quad -\sum _{1\le u \le k-1} \alpha _u \alpha _{k-u} \big (\Delta (a_1\cdot \cdot \cdot a_u) a_{u+1} \cdot \cdot \cdot a_k \\&\quad +\,a_1 \cdot \cdot \cdot a_u \Delta (a_{u+1} \cdots a_k) \big ), \end{aligned}$$

where

$$\begin{aligned} \alpha _1 := 1 \ \text{ and } \ \alpha _k := \frac{1}{k-1} { {2k-2} \atopwithdelims (){k}} \ \text{ for } \ k \ge 2. \end{aligned}$$

(6)

Example 1.5

(Hereditary non-recursive $A_\infty$-braces) We define braces satisfying (3),(4) but not (5). Namely, for elements $a,a_1,a_2,a_3,\ldots$ of a graded associative algebra $A$ with a differential $\Delta$ we put

$$\begin{aligned} e^\Delta _1(a) \!&= \Delta (a),\\ e^\Delta _2(a_1,a_2) \!&= \! \Delta (a_1a_2) - \Delta (a_1)a_2 - a_1\Delta (a_2),\\ e^\Delta _3(a_1,a_2,a_3) \!&= \! -\Delta (a_1a_2)a_3 {-}a_1\Delta (a_2a_3) +\Delta (a_1)a_2a_3 {+} 2a_1\Delta (a_2)a_3\! +\! a_1a_2\Delta (a_3),\\ e^\Delta _4 \!(a_1,a_2,a_3,a_4) \!&= \! \Delta (a_1a_2)a_3a_4 +2a_1\Delta (a_2a_3)a_4 + a_1a_2\Delta (a_3a_4)\\&\quad - \Delta (a_1)a_2a_3a_4 \! -3a_1\Delta (a_2)a_3a_4 \! -3a_1a_2\Delta (a_3)a_4 \!-a_1a_2a_3\Delta (a_4), \\ \!&\vdots&\! \\ e^\Delta _k(a_{1},\dots ,a_{k}) \!&= \! (-1)^k \sum _{0 \le i \le k -2} {\textstyle {{k-2}\atopwithdelims ()i}} \ a_1 \cdots \Delta (a_{i+1}a_{i+2}) \cdots a_k \\&\quad - (-1)^k\sum _{0 \le i \le k -1} {\textstyle {{k-1}\atopwithdelims ()i}} \ a_1 \cdots \Delta (a_{i+1}) \cdots a_k. \end{aligned}$$

We omitted for clarity the obvious Koszul signs. It is easy to verify the inductive formula

$$\begin{aligned} e^\Delta _{k+1}(a_{1},\dots ,a_{k+1})=-{(-1)^{a_1}} a_1e^\Delta _k(a_{2},\dots ,a_{k+1}) - e^\Delta _k(a_{1},\dots ,a_{k})a_{k+1}, \ k \ge 1, \end{aligned}$$

which implies that they are hereditary. On the other hand, $e^\Delta _k(a_{1},\dots ,a_{k})$ does not contain the term $\Delta (a_1\cdots a_k)$, so the coefficients $C_k$ in (5) are $0$ for all $k \ge 2$.

Hereditarity is a very fine property; ‘randomly chosen’ braces will not be hereditary. A systematic method of producing non-hereditary braces, based surprisingly on a rather deep Proposition 4.4, is described in Example 4.7.

2 Constructions of higher braces

2.1 Non-commutative algebras and $A_\infty$-braces

Recall that an $A_\infty$-algebra consists of a graded vector space $V$ together with linear operations $\mu _k : {V}^{\otimes k} \rightarrow V$, $k \ge 1$, such that $\deg (\mu _k) = 2-k$, satisfying a system of axioms that say that $\mu _1$ is a differential, $\mu _2$ is associative up to the homotopy $\mu _3$,&c, see e.g. [15, 16].

It will be useful in the context of this paper to transfer the operations $\mu _k \! : \! {V}^{\otimes k} \rightarrow V$ to the desuspension $A :=\ \downarrow \! V$, i.e. to define new operations $m_k : {A}^{\otimes k} \rightarrow A$ by the commutativity of the diagram

where $\downarrow : V \rightarrow \ \downarrow \! V = A$ is the desuspension map. All $m_k$’s then are of degree $+1$ and they satisfy the axioms

$$\begin{aligned} \sum _{k+l = n+1}\sum _{1\le i \le k} m_k(1\!\!1_A^{\otimes {i-1}} \otimes m_l \otimes 1\!\!1_A^{\otimes {k-i}}) = 0, \ \hbox {for} \hbox { each }\, n \ge 1. \end{aligned}$$

(7)

We will use this version of $A_\infty$-algebras throughout the paper:

Definition 2.1

An $A_\infty$ -algebra is a structure ${\mathcal A}= (A,m_1,m_2,m_3,\ldots )$ consisting of a graded vector space $A$ and degree $+1$ linear maps $m_k : {A}^{\otimes k}\rightarrow A$, $k \ge 1$, satisfying (7).

Let ${\mathbb T}^cA$ be the coalgebra whose underlying space is the tensor algebra ${\mathbb T}A := \bigoplus _{n \ge 1} {A}^{\otimes n}$ and the diagonal (comultiplication) is the de-concatenation. It turns out that ${\mathbb T}^cA$ is a cofree conilpotent coassociative coalgebra cogenerated by $A$, see e.g. [13, §II.3.7].³ Its cofreeness implies that each coderivation $\vartheta$ of ${\mathbb T}^cA$ is given by its components $\vartheta _k : {A}^{\otimes k} \rightarrow A$, $k \ge 1$, defined by $\vartheta _k := \pi \circ \vartheta \circ \iota _k$, where $\pi : {\mathbb T}^cA \twoheadrightarrow A$ is the projection and $\iota _k : {A}^{\otimes k} \hookrightarrow {\mathbb T}^cA$ the inclusion. We write $\vartheta = (\vartheta _1,\vartheta _2,\vartheta _3,\ldots )$.

Let $m := (m_1,m_2,m_3,\ldots )$ be a degree $1$ coderivation of ${\mathbb T}^cA$ determined by the linear maps $m_k$ as in Definition 2.1. It is well-known that axiom (7) is equivalent to $m$ being a differential, i.e. to a single equation $m^2 = 0$. Therefore equivalently, an $A_\infty$-algebra is a pair $(A,m)$ consisting of a graded vector space $A$ and a degree $+1$ coderivation $m$ of ${\mathbb T}^cA$ which squares to zero.

In homological algebra one usually considers $A_\infty$-algebras $(A,m_1,m_2,m_3,\ldots )$ as objects living in the category of differential graded (dg) vector spaces, the linear operation (differential) $m_1$ being part of its underlying dg-vector space, not a structure operation. For this reason we call an $A_\infty$-algebra with $m_k = 0$ for $k\ge 2$ a trivial $A_\infty$-algebra.

Example 2.2

(Trivial $A_\infty$-algebra) Let $\Delta : A \rightarrow A$ be a degree $+1$ differential on a graded vector space $A$. It is clear that ${\mathcal A}_\Delta := (A,\Delta ,0,0,\ldots )$ is an $A_\infty$-algebra. The differential $\Delta$ extends to a linear coderivation $(\Delta ,0,0,\ldots )$ of ${\mathbb T}^cA$.

As coderivations, by the universal property of ${\mathbb T}^cA$ each endomorphism $\phi : {\mathbb T}^cA \rightarrow {\mathbb T}^cA$ is determined by its components $\phi _k : {A}^{\otimes k} \rightarrow A$, $k \ge 1$, defined by $\phi _k := \pi \circ \phi \circ \iota _n$, We will write $\phi = (\phi _1,\phi _2,\phi _3,\ldots )$. The sequence $(1\!\!1_A,0,0,\ldots )$ represents the identity automorphism. The composition $\psi \phi$ of $\phi$ with another endomorphism $\psi = (\psi _1,\psi _2,\psi _3,\ldots )$ has components

$$\begin{aligned} (\psi \phi )_k = \sum _{r \ge 1} \sum _{i_1 + \cdots + i_r = k} \psi _r(\phi _{i_1} \otimes \cdots \otimes \phi _{i_r}). \end{aligned}$$

(8)

It is well-known that $\phi : {\mathbb T}^cA \rightarrow {\mathbb T}^cA$ is an automorphism (i.e. invertible endomorphism) if and only if $\phi _1 : A \rightarrow A$ is invertible. We call $\phi$ linear if $\phi _k= 0$ for $k \ge 2$. Let us recall

Definition

Two $A_\infty$-algebras ${\mathcal A}' = (A',m')$ and ${\mathcal A}'' = (A',m'')$ are isomorphic if there exists an automorphism $\phi : {\mathbb T}^cA' \rightarrow {\mathbb T}^cA''$ such that $\phi m' = m'' \phi$.⁴ They are strictly isomorphic if there exist a linear $\phi$ as above.

Assume that we are given an $A_\infty$-algebra ${\mathcal A}= (A,m)$ and an automorphism $\phi : {\mathbb T}^cA \rightarrow {\mathbb T}^cA$. Then clearly $(A,\phi ^{-1} m \phi )$ is an $A_\infty$-algebra isomorphic to ${\mathcal A}$.

Definition 2.3

In the situation above, we denote $m^\phi :=\phi ^{-1} m \phi$ and call the $A_\infty$-algebra ${\mathcal A}^\phi := (A,m^\phi )$ the twisting of the $A_\infty$-algebra ${\mathcal A}= (A,m)$ by the automorphism $\phi$.

The components of the twisted coderivation $m^\phi := \phi ^{-1} m \phi$ can be expressed explicitly as

$$\begin{aligned} m^\phi _k = \sum _{r,u \ge 1}\sum _{1\le j \le r}\sum _{i_1 + \cdots + i_{r+u-1} = k} (\phi ^{-1})_r({1\!\!1}^{\otimes j-1} \otimes m_u \otimes {1\!\!1}^{\otimes r-j}) (\phi _{i_1} \otimes \cdots \otimes \phi _{i_{r+u-1}}). \end{aligned}$$

If $m$ is the linear coderivation $\Delta$ as in Example 2.2, the above formula simplifies to

$$\begin{aligned} \Delta ^\phi _k = \sum _{r \ge 1}\sum _{1\le j \le r}\sum _{i_1 + \cdots + i_{r} = k} (\phi ^{-1})_r({1\!\!1}^{\otimes j-1} \otimes \Delta \otimes {1\!\!1}^{\otimes r-j}) (\phi _{i_1} \otimes \cdots \otimes \phi _{i_{r}}). \end{aligned}$$

(9)

2.2 Explicit formulas

Let $A$ be graded associative algebra with the product $\mu : {A}^{\otimes 2} \rightarrow A$. Denote by ${\underline{\mathrm{Aut}}}(A)$ the group of automorphisms $\phi$ of ${\mathbb T}^c(A)$ of the form

$$\begin{aligned} \phi := (1\!\!1_A,f_2\, \mu _2,f_3\, \mu _3,f_4\,\mu _4,\ldots ), \end{aligned}$$

where $\mu _k: {A}^{\otimes k} \rightarrow A$ is the multiplication $\mu$ iterated $(k-1)$-times, and $f_k \in {\mathbb {k}}$ are scalars, $k \ge 2$.⁵ Such an automorphism is clearly determined by its generating series

$$\begin{aligned} \phi (t) := t + f_2t^2 + f_3 t^3 +\cdots \in {\mathbb {k}}[[t]]. \end{aligned}$$

(10)

It is easy to verify using (8) that the composition of automorphisms is translated into the composition of their generating series, i.e. $(\psi \phi )(t) = \psi \big (\phi (t)\big )$.

Let $\phi \in {\underline{\mathrm{Aut}}}(A)$ be the automorphism with the generating series (10) and $\psi (t) := \phi ^{-1}(t)$ its inverse with the generating series

$$\begin{aligned} \psi (t) = t + g_2t^2 + g_3 t^3 +\cdots \in {\mathbb {k}}[[t]]. \end{aligned}$$

Denote by

$$\begin{aligned} \psi '(t) := 1 + g_2t + g_3 t^2 + g_4t^3\cdots \in {\mathbb {k}}[[t]] \end{aligned}$$

the noncommutative derivative of $\psi (t)$. It is straightforward to verify that (9) gives

$$\begin{aligned} \Delta ^\phi _k = \sum _{r+s+p = k} c_{r,s} \cdot f_p \cdot \mu _{r+s+1}({1\!\!1}^{\otimes r} \otimes \Delta \mu _p \otimes {1\!\!1}^{\otimes s}), \end{aligned}$$

(11)

with the coefficients $c_{r,s} \in {\mathbb {k}}$ defined as

$$\begin{aligned} c_{r,s} := \left. \psi '\big (\phi (u) + \phi (v)\big ) \right| _{u^rv^s}\!, \end{aligned}$$

(12)

where $\big |_{u^rv^s}$ denotes the coefficient at $u^rv^s$ of the corresponding power series in the ring of noncommutative polynomials in $u$ and $v$, i.e. the noncommutative Taylor coefficient at $u^rv^s$. Explicitly,

$$\begin{aligned} c_{r,s} = \ \sum _{k,l \ge 0} g_{k+l+1} \sum _{a_1+\cdots +a_k = r} f_{a_1} \cdots f_{a_k} \sum _{b_1+\cdots +b_l = s} f_{b_1} \cdots f_{b_l}, \end{aligned}$$

(13)

where we put, by definition, $g_1=f_1 := 1$. Observe that the above sum makes sense even for $r$ or $s$ equaling $0$ provided we interpret the empty product as $1$.

Exercise If $\Delta$ is a derivation, then $\Delta ^\phi = \Delta$, i.e. $\Delta _k = 0$ in (11) for all $k \ge 2$.

Example 2.4

(Börjeson’s $A_\infty$-braces continued) We describe the braces constructed in [6] and recalled in Example 1.3 as a twisting of the trivial $A_\infty$-algebra ${\mathcal A}_\Delta = (A,\Delta ,0,0,\ldots )$. We take as $\phi$ the automorphism with the generating series

$$\begin{aligned} \phi (t) := t + t^2 + t^3 + \cdots = \frac{t}{1-t}, \end{aligned}$$

so that

$$\begin{aligned} \psi (t) := \phi ^{-1}(t) = t - t^2 + t^3 - \cdots = \frac{t}{1+t}. \end{aligned}$$

In this case, $g_{k+l+1}$ in (13) equals $(-1)^{k+l}$, therefore

$$\begin{aligned} c_{r,s}&= \sum _{a_1+\cdots +a_k = r} (-1)^k{f_1} \cdots f_{a_k} \sum _{b_1+\cdots +b_l = s} (-1)^l f_{b_1} \cdots f_{b_l}\\&= \left. \big (1 - \psi \phi (u)\big )\big (1-\psi \phi (v)\big ) \right| _{u^rv^s} = \left. (1-u)(1-v) \right| _{u^rv^s}. \end{aligned}$$

We conclude that

$$\begin{aligned} c_{r,s} = {\left\{ \begin{array}{ll} 1 &{} \hbox {if } (r,s) \in \big \{(0,0),(1,1)\big \},\\ -1 &{} \hbox {if } (r,s) \in \big \{(0,1),(1,0)\big \},\quad \hbox {and}\\ 0&{} \hbox {in the remaining cases}. \end{array}\right. } \end{aligned}$$

Since $f_p =1$, formula (11) obviously gives Börjeson’ braces, i.e. $b_k^\Delta = \Delta ^\phi _k$ for each $k \ge 1$.

Example 2.5

(Non-recursive $A_\infty$-braces continued) The braces described in Example 1.4 are the result of the twisting by the endomorphism $\phi$ with the generating function

$$\begin{aligned} \phi (t) {:=} \frac{1-\sqrt{1-4t}}{2} {=} t + \sum _{k \ge 2} \frac{t^k}{k-1} {{2k-2} \atopwithdelims ()k} {=} t+t^2+2t^3+5t^4+ 14t^5 +\cdots , \end{aligned}$$

whose inverse equals $\psi (t) = t-t^2$. Observe that $\psi '(t) = 1-t$, therefore, in (12)

$$\begin{aligned} c_{r,s}\! = \!\left. \big (1\! -\! \phi (u)\! -\! \phi (v)\big ) \right| _{u^rv^s}= {\left\{ \begin{array}{ll} -(\alpha _r+\alpha _s)&{} \hbox {if } r=0 \,\hbox {or}\, s=0 \,\hbox {but}\, (r,s) \not = (0,0),\\ 1&{} \hbox {if } (r,s) = (0,0), \hbox {and} \\ 0&{} \hbox {in the remaining cases},\\ \end{array}\right. } \end{aligned}$$

where $\alpha _i$’s are the Taylor coefficients of $\phi (t)$. It is simple to verify that formula (11) leads to the braces of Example 1.4.

Example

(Hereditary non-recursive $A_\infty$-braces continued)  The braces in Example 1.5 are generated by the automorphism with the generating series $\phi (t) = t+ t^2$ whose inverse is

$$\begin{aligned} \psi (t) \,:\,&= \, \frac{\sqrt{1+4t}-1}{2} \,=\, t - \sum _{k \ge 2} \frac{(-t)^k}{k-1} {{2k-2} \atopwithdelims ()k}\\&= t-t^2+2t^3-5t^4+14t^5 - \cdots \!. \end{aligned}$$

We leave as an exercise to perform the calculation. Notice that $\phi (t)$ and $\psi (t)$ are related with those from Example 1.5 via the transformation

$$\begin{aligned} \phi (t) \mapsto -\psi (-t),\ \psi (t) \mapsto -\phi (-t). \end{aligned}$$

2.3 Commutative algebras and $L_\infty$-braces

Lie counterparts of $A_\infty$-algebras are $L_\infty$-algebras. An $L_\infty$-algebra is a graded vector space $L$ with linear operations $\ell _k : {L}^{\otimes k} \rightarrow V$, $k \ge 1$, $\deg (\ell _k) = 2-k$, that are graded antisymmetric and satisfy axioms that say that $\ell _1$ is a differential, $\ell _2$ fulfills the Jacobi identity up to the homotopy $\ell _3$,&c, see e.g. [8, 9].

As for $A_\infty$-algebras, we will use the version transferred to the desuspension $A :=\ \downarrow \! L$. The transferred structure operations $l_k$’s have degree $+1$, are graded symmetric, and satisfy, for each $a_{1},\dots ,a_{n} \in A$, the ‘master identity’

$$\begin{aligned} \sum _{i+j=n+1}\sum _\sigma \varepsilon (\sigma )l_j\big (l_i(a_{\sigma (1)},\dots ,a_{\sigma (i)}),a_{\sigma (i+1)},\dots ,a_{\sigma (n)}\big )=0, \end{aligned}$$

(14)

where $\sigma$ runs over all $(i,n-i)$-unshuffles and $\varepsilon (\sigma )$ is the Koszul sign of $\sigma$. We thus use:

Definition 2.6

An $L_\infty$ -algebra is an object ${\mathcal L}= (A,l_1,l_2,l_3,\ldots )$ consisting of a graded vector space $A$ and degree $+1$ graded symmetric linear maps $l_k : {A}^{\otimes k}\rightarrow A$, $k \ge 1$, satisfying (14) for each $n \ge 1$.

Let ${\mathbb S}^cA = \bigoplus _{k \ge 1} {\mathbb S}^k A$ be the symmetric coalgebra with the diagonal given by the de-concatenation; it is the cofree conilpotent cocommutative coassociative coalgebra cogenerated by $A$. Each coderivation $\omega$ of ${\mathbb S}^cA$ is thus determined by its components $\omega _k := \pi \circ \omega \circ \iota _k$, $k \ge 1$, where $\pi : {\mathbb S}^cA \twoheadrightarrow A$ is the projection and $\iota _k : {\mathbb S}^k A \hookrightarrow {\mathbb S}^cA$ the inclusion of the $k$th symmetric power of $A$. We write $\omega = (\omega _1,\omega _2,\omega _3,\ldots )$.

Let $l := (l_1,l_2,l_3,\ldots )$ be a degree $1$ coderivation of ${\mathbb S}^cA$ determined by the linear maps $l_k$ of Definition 2.6. Axiom (14) is equivalent to a single equation $l^2 = 0$ [8, Theorem 2.3]. So an $L_\infty$-algebra is a pair $(A,l)$ a graded vector space and a degree $+1$ coderivation $l$ of ${\mathbb S}^cA$ which squares to zero.

Example 2.7

(Trivial $L_\infty$-algebra) The observations of Example 2.2 apply verbatim to the $L_\infty$-case – if $\Delta$ is a degree $+1$ differential on a graded vector space $A$, then ${\mathcal L}_\Delta := (A,\Delta ,0,0,\ldots )$ is an $L_\infty$-algebra.

As automorphisms of ${\mathbb T}^cA$ twist $A_\infty$-algebras, $L_\infty$-algebras can be twisted by automorphisms $\phi : {\mathbb S}^cA \rightarrow {\mathbb S}^cA$ determined by their components $\phi _k : {\mathbb S}^kA \rightarrow A$, $k \ge 1$. We leave as an exercise to derive formulas for the composition and for the twisting of ${\mathcal L}_\Delta$ analogous to (8) and (9).

2.4 Explicit formulas

For a graded associative commutative algebra with a multiplication $\mu : {A}^{\otimes 2} \rightarrow A$, denote by $\mathrm{Aut}(A)$ the group of automorphisms $\phi$ of ${\mathbb S}^c(A)$ of the form

$$\begin{aligned} \phi := (1\!\!1_A,f_2\, \mu _2,f_3\, \mu _3,f_4\,\mu _4,\ldots ), \end{aligned}$$

where $\mu _k: {A}^{\otimes k} \rightarrow A$ is the multiplication $\mu$ iterated $(k-1)$-times, and $f_k \in {\mathbb {k}}$ are scalars, $k \ge 2$. To such an automorphism we associate its generating series

$$\begin{aligned} \phi (t) = 1 + f_1t + \frac{f_2}{2!}t^2 + \frac{f_3}{3!} t^3 +\cdots \in {\mathbb {k}}[[t]]. \end{aligned}$$

(15)

It is simple to verify that the generating series of the composition of two automorphisms is the composition of their generating series, i.e. $(\psi \phi )(t) = \psi \big (\phi (t)\big )$. The situation is analogous to the non-commutative case, only the generating series involve factorials.

Let $\phi \in \mathrm{Aut}(A)$ be the automorphism with the generating series (15) and $\psi (t) := \phi ^{-1}(t)$ the inverse of its generating series,

$$\begin{aligned} \psi (t) = 1 + g_1t + \frac{g_2}{2!}t^2 + \frac{g_3}{3!} t^3 +\cdots \in {\mathbb {k}}[[t]]. \end{aligned}$$

Denote by

$$\begin{aligned} \psi '(t) := g_1 + g_2t + \frac{g_3}{2!} t^2 + \frac{g_4}{3!}t^3\cdots \in {\mathbb {k}}[[t]] \end{aligned}$$

its (ordinary) derivative. It is simple to verify that the components of the twisting $\Delta ^\phi$ of $\Delta$ via $\phi$ are given by

$$\begin{aligned}&\Delta ^\phi _k(a_{1},\dots ,a_{k}) \nonumber \\&= \sum _{\sigma } \sum _{r+s = k} c_{r} \cdot f_s \cdot \varepsilon (\sigma ) \mu _{r+1} \big (\Delta \mu _s({a_{\sigma (1)}},\ldots ,{a_{\sigma (s)}}), {a_{\sigma (s+1)}},\ldots ,{a_{\sigma (k)}}\big )\qquad \end{aligned}$$

(16)

where $\sigma$ runs over all $(r,s)$-unshuffles, $\varepsilon (\sigma )$ is the Koszul sign of $\sigma$ and

$$\begin{aligned} c_{r} := \frac{d^r \psi '\big (\phi (t)\big )}{dt^r}\Big |_{t=0}. \end{aligned}$$

Example 2.8

(Classical $L_\infty$-braces) The classical Koszul braces recalled in Example 1.1 are the twisting of ${\mathcal L}_\Delta$ by the automorphism with the generating series

$$\begin{aligned} \phi (t) := e^t-1 = \sum _{k \ge 1} \frac{1}{k!}\ t^k = t + \frac{1}{2!}t^2 + \frac{1}{3!}t^3 + \cdots \end{aligned}$$

whose inverse $\psi (t)$ equals

$$\begin{aligned} \psi (t) = \ln (t+1) = \sum _{k \ge 1} \frac{(-1)^{k+1}}{k}\ t^k = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \cdots \!. \end{aligned}$$

Since $\psi '(t) = (1+t)^{-1}$, $\psi '\big (\phi (t)\big ) = e^{-t}$, therefore $c_{r} = (-1)^r$. As $f_s =1$ for each $s \ge 1$, formula (16) readily gives $\Delta ^\phi _k= \Phi ^\Delta _k$ for each $k \ge 1$.

Example

(Hereditary exotic $L_\infty$-braces continued) The braces in Example 1.2 are given by the automorphism with the generating series $\phi (t)= t + {t^2}/2$ whose inverse equals

$$\begin{aligned} \psi (t)&= -1 + \sqrt{1+2t} = t - \frac{1}{2}{t^2} + \frac{1}{2}t^3 - \frac{5 }{8}t^4 + \cdots \nonumber \\&= \,t - \sum _{k\ge 2}\frac{(-t)^k}{2^{k-1}(k-1)} {{2k-2}\atopwithdelims ()k}. \end{aligned}$$

Problem

It is clear that an automorphism $\phi (t)$ with the generating series (15) leads to recursive braces if and only if $f_k \in \{-1,+1\}$. Which property of the generating function guarantees the hereditarity?

2.5 The Lie case

One may ask how the previous material translates to the Lie algebra case. One could expect to have, for a graded Lie algebra $L$ with a differential $\Delta$, natural $C_\infty$-braces $(L,\Delta ,c_2,c_3,\ldots )$ emerging as the twistings of the trivial $C_\infty$-algebra by automorphisms of the Lie coalgebra ${\mathbb {L}}^{c}L$ and, among these structures, a particular one that leads to higher-order derivations of Lie algebras.

Recall that a $C_\infty$-algebra (also called, in [10, §1.4], a balanced $A_\infty$ -algebra) is an $A_\infty$-algebra as in Definition 2.1 whose structure operations vanish on decomposables of the shuffle product. As in the $A_\infty$- or $L_\infty$-cases, $C_\infty$-algebras can equivalently be described as square-zero coderivations of the cofree conilpotent Lie coalgebra ${\mathbb {L}}^{c}L$ cogenerated by $L$.

We may try to proceed as in the previous two cases. We have the trivial $C_\infty$-algebra ${\mathcal C}_\Delta = (L,\Delta ,0,0,\ldots )$, thus any natural automorphism $\phi : {\mathbb {L}}^{c}L \rightarrow {\mathbb {L}}^{c}L$ determines a coderivation $\Delta ^\phi := \phi ^{-1}\Delta \phi$ that squares to $0$, i.e. $C_\infty$-braces on $L$.

The sting lies in the notion of naturality. In constructing $A_\infty$-braces we very crucially relied on the fact that the cofree conilpotent coassociative coalgebra cogenerated by $A$ materialized as the tensor algebra ${\mathbb T}A$ equipped with the de-concatenation diagonal. Therefore natural operations ${\mathbb T}A \rightarrow A$ give rise to natural automorphisms of ${\mathbb T}^cA$ and thus also to natural $A_\infty$-braces. Similarly, the cofree conilpotent coalgebra cogenerated by $A$ can be realized as the symmetric algebra ${\mathbb S}V$ with the de-concatenation (unshuffle) diagonal.

We were however not able to find an explicit and natural (i.e. not depending e.g. on the choice of a basis) formula for a diagonal on the free Lie algebra $\mathbb LL$ that would make it a cofree conilpotent coalgebra; we were able to describe the diagonal for Lie words of length $\le$3 only. It is given, for $v,v_1,v_2,v_3 \in L$, by

$$\begin{aligned} \begin{aligned} D(v)&= 0\\ D\{v_1,v_2\}&= v_1\wedge v_2,\\ D\{v_1,\{v_2,v_3\}\}&= 2 v_1 \wedge \{v_2,v_3\} - v_2 \wedge \{v_3,v_1\} - v_3 \wedge \{v_1,v_2\}. \end{aligned} \end{aligned}$$

(17)

In the above display, we denoted the bracket in $\mathbb LL$ by $\{-,-\}$ to distinguish it from the bracket of $L$ which we will denote more traditionally by $[-,-]$.

Remark 2.9

The lack of an explicit diagonal for the free Lie algebra $\mathbb LL$ may be related to the problem of describing the Eulerian idempotents $e^{(1)}_k : {\mathbb T}^k X \rightarrow {\mathbb T}^k X$ [14, Corollary 1.6] in terms of iterated linearly-independent Lie braces. While, for $x,x_1,x_2,x_3 \in X$,

$$\begin{aligned} e^{(1)}_1(x)&= x, \\ e^{(1)}_2(x_1 \otimes x_2)&= \frac{1}{2!} [x_1,x_2] \quad \text{ and } \\ e^{(1)}_2(x_1 \otimes x_2 \otimes x_3)&= \frac{1}{3!}\big ([[x_1,x_2],x_3] + [x_1,[x_2,x_3]]\big ), \end{aligned}$$

a similar formula for $e^{(1)}_k$ with $k \ge 4$ is not known.

Let ${\mathbb L}^kL$ be the subspace of $\mathbb LL$ spanned by elements of the product length $k$. As in the previous cases, each coalgebra automorphism $\phi : {\mathbb {L}}^{c}L \rightarrow {\mathbb {L}}^{c}L$ is determined by its components $\phi _k : {\mathbb L}^kL \rightarrow L$, $k \ge 1$. One has also the canonical maps $\lambda _k : {\mathbb L}^kL \rightarrow L$ given by the multiplication in $L$. Assume we found a natural isomorphism between $\mathbb LL$ and ${\mathbb {L}}^{c}L$ such that the induced diagonal on $\mathbb LL$ agrees with (17) on elements of length $\le \! 3$.

Let $\phi = (1\!\!1_L, f_2\lambda _2,f_3\lambda _3,\ldots )$ be the automorphism whose $k$th component equals $f_k \lambda _k$ for some scalars $f_k \in {\mathbb {k}}$, and $\psi = (1\!\!1_L, g_2\lambda _2,g_3\lambda _3,\ldots )$ another one, with components $g_k \lambda _k$, $g_k \in {\mathbb {k}}$. Using (17), one derives the following formula for the first three components of the composition:

$$\begin{aligned} \psi \phi = \big (1\!\!1_L,(f_2+g_2)\lambda _2,(f_3 + 3 f_2g_2 + g_3) \lambda _3,\ldots \big ). \end{aligned}$$

With this formula, one easily verifies that the inverse of $\phi = (1\!\!1_L,\lambda _2,\lambda _3,\ldots )$ is of the form $\phi ^{-1} = (1\!\!1_L,-\lambda _2,+2\lambda _3,\ldots )$.

The twisting of the trivial $C_\infty$-algebra ${\mathcal C}_\Delta$ by $\phi = (1\!\!1_L,\lambda _2,\lambda _3,\ldots )$ leads to the following formulas for $c^\Delta _k = \Delta ^\phi _k$; we for clarity omit the Koszul signs:

$$\begin{aligned} c^\Delta _1(v)&= \Delta (v),\\ c^\Delta _2\{v_1,v_2\}&= \Delta [v_1,v_2] - [\Delta v_1,v_2]- [v_1,\Delta v_2], \\ c^\Delta _3\big \{v_1,\{v_2,v_3\}\big \}&= \Delta [v_1,[v_2,v_3]] -2[v_1,\Delta [v_2,v_3]]+ [\Delta v_2,[v_3,v_1]] \\&+ [v_2,\Delta [v_3,v_1]] + [\Delta v_3,[v_1,v_2]] + [v_3,\Delta [v_1,v_2]] \\&+2\big ([v_1,[\Delta (v_2),v_3]] + [v_1,[v_2,\Delta (v_3)]]\big ). \end{aligned}$$

It is easy to verify that

$$\begin{aligned} c^\Delta _3\big \{v_1,\{v_2,v_3\}\big \} = -2 \big [v_1,c_2^\Delta \{v_2,v_3\}\big ] - c_2^\Delta \big \{v_2,[v_3,v_1]\big \} - c_2^\Delta \big \{v_3,[v_1,v_2]\big \}. \end{aligned}$$

Therefore, if $\Delta$ is a derivation of the Lie algebra $L$, i.e. if $c^\Delta _2=0$, $c^\Delta _3$ vanishes as expected.

We saw in the $A_\infty$- resp. $L_\infty$-cases that the twisting by the automorphism whose components were the canonical maps $\mu _k : {\mathbb T}^k A \rightarrow A$ resp. $\mu _k: {\mathbb S}^k A \rightarrow A$, lead to the (unique) braces giving a sensible notion of higher-order derivations. This justifies:

Conjecture

The twisting of ${\mathcal C}_\Delta = (L,\Delta ,0,0,\ldots )$ by $\phi = (1\!\!1_L,\lambda _2,\lambda _3,\ldots )$ gives rise to $C_\infty$-braces satisfying the analogs of conditions (3)–(5).

Verifying this conjecture of course depends on describing the isomorphism

$$\begin{aligned} \mathbb LL \cong {\mathbb {L}}^{c}L. \end{aligned}$$

2.6 Other cases

Let us finish this part by formulating the most general context in which our approach may work. We will need the language of operads for which we refer for instance to [12, 13]. Let ${\mathcal P}$ be a quadratic Koszul operad and $A$ a ${\mathcal P}$-algebra. Denote by ${\mathcal P}^!$ the Koszul (quadratic) dual of ${\mathcal P}$ [13, Def. II.3.37] and by ${\mathbb F}_{{\mathcal P}}A$ (resp. ${\mathbb F}^c_{{\mathcal P}}A$) the free ${{\mathcal P}}$-algebra (resp. the cofree conilpotent ${{\mathcal P}}$-coalgebra) generated (resp. cogenerated) by $A$.

A strongly homotopy ${\mathcal P}^!$-algebra, also called a ${\mathcal P}^!_\infty$-algebra, is determined by a square-zero coderivation $p$ of ${\mathbb F}^c_{{\mathcal P}}A$. If $\Delta$ is a differential on $A$, one has as before the trivial ${\mathcal P}_\infty$-algebra ${\mathcal P}_\Delta = (A,\Delta ,0,0,\ldots )$ given by extending $\Delta$ to ${\mathbb F}^c_{{\mathcal P}}A$.

Under the presence of a natural identification ${\mathbb F}_{{\mathcal P}}A\cong {\mathbb F}^c_{{\mathcal P}}A$, one may speak about natural automorphisms that twist $P_\Delta$ to ${\mathcal P}^!_\infty$-braces. One has the automorphism $\phi : {\mathbb F}^c_{{\mathcal P}}A\rightarrow {\mathbb F}^c_{{\mathcal P}}A$ whose components are given by the structure map ${\mathbb F}_{{\mathcal P}}A\rightarrow A$. It is sensible to conjecture that the related ${\mathcal P}^!_\infty$-braces lead to a reasonable notion of higher-order derivations of ${\mathcal P}$-algebras.

In this general set-up, $A_\infty$-braces related to associative algebras correspond to the ${\mathcal P}= {\mathcal A{ ss}}$ case, $L_\infty$-braces related to commutative associative algebras to ${\mathcal P}= {\mathcal C{ om}}$, and $C_\infty$-braces related to Lie algebras to ${\mathcal P}= \mathcal L{ ie}$, where ${\mathcal A{ ss}}$, ${\mathcal C{ om}}$ and $\mathcal L{ ie}$ denote the operad for associative, commutative associative and Lie algebras, respectively.⁶

Part 2. Naturality and acyclicity

3 Naturality

This section is devoted to natural operations ${A}^{\otimes k} \rightarrow A$ (resp. ${\mathbb S}^k A \rightarrow A$), where $A$ is a graded associative (resp. graded commutative associative) algebra with a differential $\Delta$ of degree $+1$. Since in the commutative associative case the symmetric group action brings extra complications but nothing conceptually new, we analyze in detail only the associative case.

Associative case. We are going introduce the space ${\mathcal Nat}(k)$ of natural operations ${A}^{\otimes k} \rightarrow A$ and show that ${\mathcal Nat}(k)$, graded by the degrees of maps and equipped with the differential induced by $\Delta$, is acyclic for each $k \ge 2$. The content of this section is a kindergarten version of the analysis of Deligne’s conjecture given in [4].

3.1 Natural operations

Intuitively, natural operations ${A}^{\otimes k} \rightarrow A$ are linear maps composed from the data available for an arbitrary graded associative algebra $A$ with a differential. Equivalently, natural operations are linear combinations of compositions of ‘elementary’ operations, which are the multiplication, the differential, permutations of the inputs and projections to the homogeneous parts. Our categorial definition given below is chosen so that it excludes the projections; the reason is explained in Exercise 4.8. Our theory can, however, easily be extended to include the projections as well, cf. Exercise 3.7. Let us start with:

Example 3.1

The space ${\mathcal Nat}(1)$ of natural operations $A \rightarrow A$ is two-dimensional, spanned by the identity $1\!\!1:A \rightarrow A \in {\mathcal Nat}(1)^0$ in degree $0$ and $\Delta :A \rightarrow A \in {\mathcal Nat}(1)^1$ in degree $1$. The space ${\mathcal Nat}(2)^0$ is spanned by two operations,

$$\begin{aligned} a \otimes b \mapsto ab \ \text{ and } \ a \otimes b \mapsto {(-1)^{|a||b|}} ba, \end{aligned}$$

where $ab$ resp. $ba$ denotes the product in $A$. The space ${\mathcal Nat}(2)^1$ is $6$-dimensional, spanned by the operations

$$\begin{aligned} a \otimes b \mapsto \Delta (a)b,\ a \otimes b \mapsto {(-1)^{|a|}} a\Delta (b),\ a \otimes b \mapsto \Delta (ab), \end{aligned}$$

and compositions of these operations with the permutation $a\otimes b \mapsto {(-1)^{|a||b|}} b \otimes a$, i.e. by

$$\begin{aligned} a {\otimes } b \mapsto {(-1)^{|a||b|}}\Delta (b)a,\ a \otimes b \mapsto {(-1)^{|b|(|a| + 1)}} b\Delta (a),\ a \otimes b \mapsto {(-1)^{|a||b|}}\Delta (ba). \end{aligned}$$

Likewise, ${\mathcal Nat}(2)^2$ is spanned by

$$\begin{aligned} a \otimes b \mapsto \Delta (a)\Delta (b),\ a \otimes b \mapsto {(-1)^{|a|}} \Delta \big (a\Delta (b)\big ),\ a \otimes b \mapsto \Delta \big (\Delta (a)b\big ) \end{aligned}$$

and their permutations. Finally, ${\mathcal Nat}(2)^3$ is two-dimensional, spanned by

$$\begin{aligned} a\otimes b \mapsto \Delta \big (\Delta (a)\Delta (b)\big ) \ \text{ and } \ a\otimes b \mapsto {(-1)^{|b|(|a| + 1)}} \Delta \big (\Delta (b)\Delta (a)\big ). \end{aligned}$$

There are no natural operations ${A}^{\otimes 2} \rightarrow A$ of degrees $>4$. Observe that the Euler characteristic of the graded space ${\mathcal Nat}(2)^*$ is $2-6+6-2 = 0$. This indicates its acyclicity.

All operations $\beta : {A}^{\otimes 2} \rightarrow A \in {\mathcal Nat}(2)$ listed in Example 3.1 share the following property. Let $(A,\Delta _A)$ and $(B,\Delta _B)$ be graded associative algebras with differentials and $\varphi : A \rightarrow B$ a linear map such that

$$\begin{aligned} \varphi (a'a'') = \varphi (a')\varphi (a'')\ \text{ and } \ \varphi \Delta _A(a) = \Delta _B \varphi (a),\ \text{ for } \text{ each } \ a, a',a'' \in A. \end{aligned}$$

(18)

Then

$$\begin{aligned} \beta \big (\varphi (a')\otimes \varphi (a'')\big ) = \varphi \big (\beta (a'\otimes a'')\big )\ \text{ for } \text{ each } \ a',a'' \in A. \end{aligned}$$

(19)

Let us emphasize that we do not assume the map $\varphi$ to be homogeneous of degree $0$, it can be an arbitrary linear map $A \rightarrow B$ satisfying (18).

Example

This example explains why we did not require the homogeneity of the map $\varphi$ in (18). Consider the operation $\beta : {A}^{\otimes 2} \rightarrow A$ defined by

$$\begin{aligned} \beta (a \otimes b) := {\left\{ \begin{array}{ll} ab \in A&{} \hbox {if } |a| = 2\quad \hbox { and }\quad |b| = -13\quad \hbox {and} \\ 0,&{} \hbox {otherwise}. \end{array}\right. } \end{aligned}$$

It is certainly ‘natural’ in that it is defined using the data that are available for any graded associative algebra, but we do not want to consider this type of operation.⁷ What excludes $\beta$ from the family of well-behaved operations is precisely the lack of naturality (19) with respect to non-homogeneous maps.

To see why it is so, take $A := {\mathbb T}(a,b)$, the tensor algebra on two generators with $|a| := 2$ and $|b| := -13$, and $B := {\mathbb T}(u,v)$ generated by $u$, $v$ with $|u| = |v| := 0$. There clearly exists a unique ‘non-homogeneous’ homomorphism $\varphi : A \rightarrow B$ such that $\varphi (a) = u$ and $\varphi (b) = v$.⁸ But then

$$\begin{aligned} 0 = \beta (u\otimes v) = \beta \big (\varphi (a)\otimes \varphi (b)\big ) \not = \varphi \big (\beta (a\otimes b)\big ) = \varphi (ab) = \varphi (a)\varphi (b) = uv, \end{aligned}$$

so $\beta$ is not natural with respect to our extended notion of a homomorphism, though it is still natural with respect to conventional homomorphisms as can be easily checked.

Natural operations thus appear as natural transformations $\beta : \bigotimes ^k \rightarrow \Box$ from the tensor power functor $\bigotimes ^k \!\! : \mathtt{Algs}^\Delta \rightarrow \mathtt{Vect}$ to the forgetful functor $\Box \! : \mathtt{Algs}^\Delta \rightarrow \mathtt{Vect}$, where $\mathtt{Algs}^\Delta$ is the category of graded associative algebras with a differential, with morphisms as in (18), and $\mathtt{Vect}$ the category of vector spaces. We however prefer a more explicit:

Definition 3.2

For $k \ge 1$, let ${\mathcal Nat}(k)$ be the abelian group of families of linear maps

$$\begin{aligned} \beta _{A} : {A}^{\otimes k} \rightarrow A \end{aligned}$$

indexed by graded associative algebras $A = (A,\Delta )$ with a differential such that, for any linear map $\varphi :A \rightarrow B$ satisfying (18), the diagram

(20)

commutes.

We are going to prove a structure theorem for natural operations. Denote by $\mathcal Fr(x_1,\ldots ,x_k)$ the free graded associative algebra with a differential, generated by degree $0$ elements $x_1,\ldots ,x_k$ (an explicit description is given in §3.2). Denote also by $\mathcal Fr_{1,\ldots ,1}(x_1,\ldots ,x_k)$ the subspace of $\mathcal Fr(x_1,\ldots ,x_k)$ spanned by the words that contain each generator precisely once.

Proposition 3.3

For each $k \ge 1$ one has a natural isomorphism

$$\begin{aligned} \xi : {\mathcal Nat}(k) \cong \mathcal Fr_{1,\ldots ,1}(x_1,\ldots ,x_k). \end{aligned}$$

Proof

Denote, for brevity, $\mathcal Fr: = \mathcal Fr(x_1,\ldots ,x_k)$. Let $A = (A,\Delta )$ be an arbitrary algebra with a differential. Given elements $a_{1},\dots ,a_{k} \in A$, there exists a unique $\Phi ^A_{a_{1},\dots ,a_{n}} : \mathcal Fr\rightarrow A$ satisfying (18), specified by requiring $\Phi ^A_{a_{1},\dots ,a_{n}}(x_i) := a_i$ for $1 \le i \le k$.

Given a natural operation $\beta \in {\mathcal Nat}(k)$ and $a_{1},\dots ,a_{k} \in A$, one has the commutative diagram

(21)

by naturality (20). We will also need the particular case of (21) when $A = \mathcal Fr$ and $a_i = u_i x_i$, for some scalars $u_i \in {\mathbb {k}}$, $1 \le i \le k$:

(22)

Recall that $\mathcal Fr_{{1},\ldots ,{1}} := \mathcal Fr_{{1},\ldots ,{1}}(x_1,\ldots ,x_k)$ denotes the subspace of elements containing each generator $x_{1},\dots ,x_{k}$ precisely once. We begin the actual proof by observing that each natural operation $\beta \in {\mathcal Nat}(k)$ determines an element $\xi (\beta ) \in \mathcal Fr$ by

$$\begin{aligned} \xi (\beta ) := \beta _{\mathcal Fr}(x_1 \otimes \cdots \otimes x_n) \in \mathcal Fr. \end{aligned}$$

(23)

We will show that, quite miraculously, $\xi (\beta )$ belongs to $\mathcal Fr_{{1},\ldots ,{1}}$. Clearly, $\mathcal Fr$ decomposes as

$$\begin{aligned} \mathcal Fr= \textstyle \bigoplus _{j_{1},\dots ,j_{k} \ge 0} \mathcal Fr_{j_{1},\dots ,j_{k}}, \end{aligned}$$

where $\mathcal Fr_{j_{1},\dots ,j_{k}} \subset \mathcal Fr$ is the subspace of elements having precisely $j_i$ instances of $x_i$ for each $1 \le i \le k$. The endomorphism $\Phi ^A_{{u_1x_1},\ldots ,{u_kx_k}} : \mathcal Fr\rightarrow \mathcal Fr$ acts on $\mathcal Fr_{j_{1},\dots ,j_{k}}$ by the multiplication with $u_1^{j_1} \cdots u_k^{j_k}$; the subspace $\mathcal Fr_{j_{1},\dots ,j_{k}}\subset \mathcal Fr$ is, in fact, characterized by this property. The element $\xi (\beta )$ uniquely decomposes as $\xi (\beta ) = \sum _{j_{1},\dots ,j_{k} \ge 0} \xi (\beta )_{j_{1},\dots ,j_{k}}$, for some $\xi (\beta )_{j_{1},\dots ,j_{k}} \in \mathcal Fr_{j_{1},\dots ,j_{k}}$.

Let us turn our attention to (22). By the definition of the map $\Phi ^A_{{u_1x_1},\ldots ,{u_kx_k}}$, one has

$$\begin{aligned} \beta _{\mathcal Fr}\big ((\Phi ^A_{{u_1x_1},\ldots ,{u_kx_k}})^{\otimes k}\big ) (x_1 \otimes \cdots \otimes x_k) =\beta _{\mathcal Fr}(u_1x_1 \otimes \cdots \otimes u_kx_k), \end{aligned}$$

while the linearity of $\beta _{\mathcal Fr}$ implies

$$\begin{aligned} \beta _{\mathcal Fr}(u_1x_1 \otimes \cdots \otimes u_kx_k) = u_1\cdots u_k \cdot \beta _{\mathcal Fr} (x_1 \otimes \cdots \otimes x_k) = u_1\cdots u_k \cdot \xi (\beta ). \end{aligned}$$

On the other hand

$$\begin{aligned}&\Phi ^A_{{u_1x_1},\ldots ,{u_kx_k}}(\beta _{\mathcal Fr})(x_1 \otimes \cdots \otimes x_k) \\&\quad = \Phi ^A_{{u_1x_1},\ldots ,{u_kx_k}}\big (\xi (\beta )\big ) = \sum _{j_{1},\dots ,j_{k} \ge 0} u_1^{j_1}\cdots u_k^{j_k} \cdot \xi (\beta )_{j_{1},\dots ,j_{k}}, \end{aligned}$$

therefore the commutativity of (22) means that

$$\begin{aligned} u_1 \cdots u_k \cdot \xi (\beta ) = \sum _{j_{1},\dots ,j_{n} \ge 0} u_1^{j_1}\cdots u_k^{j_k} \cdot \xi (\beta )_{j_{1},\dots ,j_{k}} \end{aligned}$$

for each $u_{1},\dots ,u_{k}\! \in \! {\mathbb {k}}$. We conclude that $\xi (\beta )_{j_{1},\dots ,j_{k}} \!=\! 0$ if $(j_{1},\dots ,j_{k}) \!\not =\! ({1},\ldots ,{1})$, so $\xi (\beta ) = \xi (\beta )_{{1},\ldots ,{1}} \in \mathcal Fr_{{1},\ldots ,{1}}$ as claimed.

We leave as an exercise to prove that, vice versa, each element $\xi \in \mathcal Fr_{{1},\ldots ,{1}}$ determines a natural operation $\beta (\xi ) \in {\mathcal Nat}(k)$ by the formula

$$\begin{aligned} \beta (\xi )_{A}(a_1 \otimes \cdots \otimes a_k) = \Phi ^A_{{a_1},\ldots ,{a_k}}(\xi ),\ \text{ for } \text{ each }\,\, a_{1},\dots ,a_{k}\in A, \end{aligned}$$

(24)

cf. the proof of [4, Proposition 2.9]. The above constructions define mutually inverse correspondences $\beta \mapsto \xi (\beta )$ and $\xi \mapsto \beta (\xi )$ that give the isomorphism of the proposition.

Remark

A crucial step of the previous proof was that $\xi (\beta )$ belonged to $\mathcal Fr_{{1},\ldots ,{1}}$. It was implied by the multilinearity, which is a particular feature of the monoidal structure given by $\otimes$. In the cartesian situation, $\xi (\beta )$ might have been an arbitrary element of the free algebra $\mathcal Fr(x_{1},\dots ,x_{k})$.

3.2 The algebra $\mathcal Fr(x_{1},\dots ,x_{k})$.

Elements of the free algebra $\mathcal Fr= \mathcal Fr(x_{1},\dots ,x_{k})$ are results of iterated applications of the associative multiplication and the differential on the generators $x_{1},\dots ,x_{k}$. The subspace $\mathcal Fr_{{1},\ldots ,{1}} = \mathcal Fr_{{1},\ldots ,{1}}(x_{1},\dots ,x_{k})$ is spanned by words containing each generator precisely once. A typical element of $\mathcal Fr_{{1},\ldots ,{1}}$ is thus an expression as

$$\begin{aligned} \Delta \big (\Delta (x_3)x_6\Delta (x_4)\big )x_5\Delta (x_2x_1). \end{aligned}$$

(25)

The algebra $\mathcal Fr$ and thus also $\mathcal Fr_{{1},\ldots ,{1}}$ is graded by the number of occurrences of $\Delta$; the element in (25) therefore belongs to $\mathcal Fr_{{1},\ldots ,{1}}^4$.

We can clearly encode elements of $\mathcal Fr_{{1},\ldots ,{1}}$ by ‘flow diagrams’ that record how the multiplication and the differential are applied. For instance, the diagram encoding (25) is

Its underlying graph is a rooted (=oriented) planar tree with the root pointing upwards. The labels of its leaves (=inputs) mark the position of the generators. The vertices symbolize iterated multiplication while the bullets the application of the differential.

We see that elements of $\mathcal Fr_{{1},\ldots ,{1}}(x_{1},\dots ,x_{k})$ can be represented by linear combinations of planar rooted trees $T$ such that

• each vertex of $T$ has at least two inputs,

• all internal edges and possibly some external edges are decorated by the bullet $\bullet$, and

• the leaves of $T$ are labelled by a permutation of $({1},\ldots ,{k})$.

Let us denote by ${\mathcal T}(k)$ the set of all trees as above. Proposition 3.3 together with our description of $\mathcal Fr_{{1},\ldots ,{1}}(x_{1},\dots ,x_{k})$ gives

Corollary 3.4

For each $k \ge 1$ one has a natural isomorphism ${\mathcal Nat}(k)\cong \mathrm{Span}\big ({\mathcal T}(k)\big )$.

It follows from general theory [13, Proposition II.1.27] that ${\mathcal Nat}(k)$ is the arity $k$-th piece of the operad ${\mathcal Nat}$ whose algebras are couples $(A,\Delta )$ consisting of an associative algebra and a differential. We will, however, not need this interpretation in the sequel.

Example

Corollary 3.4 offers the following description of ${\mathcal Nat}(2)$:

We leave as an exercise to relate the above description to the operations listed in Example 3.1.

Example 3.5

It easily follows from Corollary 3.4 that ${\mathcal Nat}(k)^0$ is spanned by operations

$$\begin{aligned} \beta (a_{1},\dots ,a_{k}) = \varepsilon (\sigma ) \cdot a_{\sigma (1)}\cdots a_{\sigma (k)}, \ \sigma \in \Sigma _n, \end{aligned}$$

i.e. ${\mathcal Nat}(k)^0 \cong {\mathbb {k}}[\Sigma _n]$ for $k \ge 1$.

3.3 Acyclicity

There is a differential on ${\mathcal Nat}(k)$ induced by $\Delta$. For $\beta : {A}^{\otimes k} \rightarrow A$ it is defined by

$$\begin{aligned} \delta (\beta ) := \Delta \beta - (-1)^{|\beta |} \sum _{1 \le i \le k} \beta \big ({1\!\!1}^{\otimes (i-1)} \otimes \Delta \otimes {1\!\!1}^{\otimes (k-i)}\big ). \end{aligned}$$

When evaluating the above formula, we shall of course take into account the Koszul sign convention. For instance, if $\beta \in {\mathcal Nat}(2)$ and $a,b,\in A$, then

$$\begin{aligned} \delta (\beta )(a\otimes b) = \Delta \beta (a\otimes b)- (-1)^{|\beta |} \beta \big (\Delta (a)\otimes b\big )- (-1)^{|a| + |\beta |} \beta \big (a \otimes \Delta (b)\big ). \end{aligned}$$

It is simple to verify that $\delta ^2 = 0$, so $\big ({\mathcal Nat}(k)^*,\delta \big )$ is a cochain complex. We leave as an exercise to describe $\delta$ in terms of trees.

Proposition 3.6

The cochain complex

$$\begin{aligned} {\mathbb {k}}= {\mathcal Nat}(k)^0 \mathop {\longrightarrow }\limits ^{\delta }{\mathcal Nat}(k)^{1} \mathop {\longrightarrow }\limits ^{\delta }{\mathcal Nat}(k)^{2} \mathop {\longrightarrow }\limits ^{\delta }\cdots \end{aligned}$$

(26)

is acyclic for each $k \ge 2$. In particular, the map $\delta : {\mathbb {k}}= {\mathcal Nat}(k)^0 \rightarrow {\mathcal Nat}(k)^{1}$ is monic.

The case of $k=1$ is a particular one, as the differential $\delta : {\mathcal Nat}(1)^0 \rightarrow {\mathcal Nat}(1)^1$ is the zero map $\mathrm{Span}(1\!\!1) \mathop {\rightarrow }\limits ^{0}\mathrm{Span}(\Delta )$. The explanation is that the identity $1\!\!1: A \rightarrow A$ is the only natural operation that is ‘generically’ a chain map.

Proof of Proposition 3.6

We describe a contracting homotopy. By Corollary 3.4, each natural operation $\beta \in {\mathcal Nat}(k)$ is represented by a unique linear combination of trees from ${\mathcal T}(k)$. It is therefore enough to specify how the homotopy acts on operations given by a single tree.

Let $\beta$ be represented by $T \in {\mathcal T}(k)$. If the root edge of $T$ is decorated by the bullet, we define $h(\beta )$ as the operation represented the tree $T'$ obtained from $T$ by removing the decoration of the root. We define $h(\beta ) := 0$ if the root edge of $T$ is not decorated by the bullet. We leave as an exercise to verify that $h \delta + \delta h = 1\!\!1$, so that $h$ is a contracting homotopy. $\square$

Notice that the complex ${\mathcal Nat}(k)$ is isomorphic to the direct sum of $k!$ copies of the contractible subcomplex $\underline{\mathcal Nat}(k) \subset {\mathcal Nat}(k)$ consisting of operations represented by trees with leaves indexed by the identity permutation $(1,\ldots ,k)$.

Exercise 3.7

Let $c{\mathcal Nat}(k)$ be the abelian group of families of linear maps $\beta _{A} : {A}^{\otimes k} \rightarrow A$ such that the diagram (20) commutes for all homogeneous $\varphi$’s.⁹Prove that then

$$\begin{aligned} c{\mathcal Nat}(k) \cong \bigoplus _{d_1,\ldots , d_k} c{\mathcal Nat}(d_1,\ldots , d_k), \end{aligned}$$

where $c{\mathcal Nat}(d_1,\ldots , d_k)$ consists of $\beta _A$’s such that

$$\begin{aligned} \beta _A(a_{1},\dots ,a_{k}) \not = 0\ \text{ only } \text{ if } \ (|a_1|,\ldots ,|a_k|) = (d_1,\ldots , d_k). \end{aligned}$$

Prove, moreover, that $c{\mathcal Nat}(d_1,\ldots , d_k) \cong {\mathcal Nat}(k)$ for each $d_1,\ldots , d_k$, and discuss the acyclicity of $c{\mathcal Nat}(k)$.

3.4 Natural $A_\infty$-algebras

Let $(A,\Delta )$ be a graded associative algebra with a differential. We call an $A_\infty$-algebra ${\mathcal A} =(A,m_1,m_2,\ldots )$ natural if $m_k : {A}^{\otimes k} \rightarrow A$ are natural operations from ${\mathcal Nat}(k)^1$ for each $k \ge 1$. Formally, a natural $A_\infty$-algebra should be considered as a family $\{{\mathcal A}_{(A,\Delta )}\}$ of $A_\infty$-algebras indexed by algebras with a differential, such that each $\varphi$ as in (18) induces a strict morphism ${\mathcal A}_{(A,\Delta _A)} \rightarrow {\mathcal A}_{(B,\Delta _B)}$. We however believe that our simplification will not lead to confusion.

An automorphism $\phi : {\mathbb T}^cA \rightarrow {\mathbb T}^cA$ is natural if all its components $\phi _k : {A}^{\otimes k} \rightarrow A$ are natural operations from ${\mathcal Nat}(k)^0$. Natural automorphisms with $\phi _1 = 1\!\!1_A$ form a monoid $\mathrm{Aut}(A)$. It is clear that the twisting of a natural $A_\infty$-algebra by a natural automorphism is a natural $A_\infty$-algebra.

Example 3.8

It follows from Example 3.5 that natural automorphisms $\phi \in \mathrm{Aut}(A)$ are encoded by sequences $(1\!\!1_A,\omega _2,\omega _3,\ldots )$ of elements $\omega _k \in {\mathbb {k}}[\Sigma _k]$. One has an important submonoid ${\underline{\mathrm{Aut}}}(A)\subset \mathrm{Aut}(A)$ of sequences such that all $\omega _k$’s are the identities $1\!\!1_{\Sigma _k}$.

Exercise 3.9

Assume that $\Delta$ is a derivation. The twisting of ${\mathcal A}_\Delta = (A,\Delta ,0,0,\ldots )$ by an arbitrary natural automorphism is then strictly isomorphic to ${\mathcal A}_\Delta$, i.e. $\Delta ^\phi = \Delta$.

Commutative associative case. We are going to formulate commutative versions of the main statements from the first part of this section. We omit the proofs which are analogous to the non-commutative case.

Let $A$ be a graded commutative associative algebra with a differential $\Delta$. Natural operations are, analogously to Definition 3.2, natural transformations $\beta _{A} : {\mathbb S}^k A \rightarrow A$ from the $k$th symmetric power of $A$ to $A$. Let us denote by ${\mathcal Nat}(k)$ the abelian group of all these natural operations. To see how ${\mathcal Nat}(k)$ differs from its non-commutative counterpart, we give commutative versions of Examples 3.1 and 3.5.

Example 3.10

(Commutative version of Example 3.1) The space ${\mathcal Nat}(1)$ is, as in the non-commutative case, spanned by the identity $1\!\!1:A \rightarrow A$ in degree $0$ and $\Delta :A \rightarrow A$ in degree $1$. The space ${\mathcal Nat}(2)^0$ is spanned by the multiplication $a \odot b \mapsto ab$, with $\odot$ denoting the symmetric product. The space ${\mathcal Nat}(2)^1$ is two-dimensional, spanned by the operations

$$\begin{aligned} a \odot b \mapsto \Delta (a)b + {(-1)^{|a||b|}} \Delta (b)a \ \text{ and } \ a \odot b \mapsto \Delta (ab). \end{aligned}$$

Likewise, ${\mathcal Nat}(2)^2$ is spanned by

$$\begin{aligned} a \odot b \mapsto \Delta (a)\Delta (b) \ \text{ and } a \odot b \mapsto \Delta \big (\Delta (a)b\big ) - {(-1)^{|a||b|}}\Delta \big (\Delta (b)a\big ). \end{aligned}$$

Finally, ${\mathcal Nat}(2)^3$ is spanned by

$$\begin{aligned} a\odot b \mapsto \Delta \big (\Delta (a)\Delta (b)\big ). \end{aligned}$$

The Euler characteristic of the graded space ${\mathcal Nat}(2)^*$ is $1-2+2-1 = 0$, so the acyclicity can be expected as in the non-commutative case.

Example 3.11

(Commutative version of Example 3.5) The space ${\mathcal Nat}(k)^0$ is one-dimensional, spanned by the iterated multiplication $\mu _k(a_{1},\dots ,a_{k}) = a_1\cdots a_k,~$so${\mathcal Nat}(k)^0 \cong {\mathbb {k}}$ for each $k \ge 1$.

An obvious modification of Proposition 3.3 holds, with $\mathcal Fr(x_1,\ldots ,x_k)$ this time the free commutative associative algebra. Corollary 3.4 holds as well, with ${\mathcal T}(k)$ replaced by the space of all ‘abstract,’ i.e. non-planar, trees. As in the non-commutative case, $\Delta$ induces a differential $\delta$ so that $({\mathcal Nat}(k)^*,\delta )$ is acyclic for each $k \ge 2$.

The notions of a natural $L_\infty$-algebras and natural automorphisms $\phi : {\mathbb S}^c(A) \rightarrow {\mathbb S}^c(A)$ translate verbatim. The following example however shows that the space $\mathrm{Aut}(A)$ of natural automorphisms is much smaller than in the non-commutative case.

Example 3.12

The description of ${\mathcal Nat}(k)^0$ given in Example 3.11 implies that natural automorphisms $\phi \in \mathrm{Aut}(A)$ are encoded by sequences $(1\!\!1_A,f_2,f_3,\ldots )$ of scalars $f_k \in {\mathbb {k}}$.

4 Main results

We are going to formulate and prove the main theorems. As in Sect. 3, we treat in detail only the associative non-commutative case.

Associative case. Let $A$ be a graded associative algebra with a differential $\Delta$, and ${\mathcal A}_\Delta = (A,\Delta ,0,0,\ldots )$ the trivial $A_\infty$-algebra of Example 2.2. According to the following result, each natural $A_\infty$-algebra whose linear operation $m_1$ equals $\Delta$, is uniquely given by a twisting of ${\mathcal A}_\Delta$, see §3.4 and Definition 2.3 for the meaning of naturality and twisting. In particular, each such an $A_\infty$-algebra is (weakly) isomorphic to ${\mathcal A}_\Delta$.

Theorem 4.1

For each natural $A_\infty$-algebra ${\mathcal A}= (A,m_1,m_2,m_3,\ldots )$ such that

$$\begin{aligned} m_1 = \Delta \end{aligned}$$

(27)

there exist a unique natural automorphism $\phi = (1\!\!1_A,\phi _2,\phi _3,\ldots )$ of the tensor coalgebra ${\mathbb T}^cA$ such that ${\mathcal A}$ equals the twisting of ${\mathcal A}_\Delta$ via $\phi$. Explicitly

$$\begin{aligned} m_k(a_1,\ldots ,a_k) = \pi \phi ^{-1}\Delta \phi (a_1,\ldots ,a_k), \ \text{ for } a_{n},\dots ,a_{k} \in A, \end{aligned}$$

where $\pi : {\mathbb T}^cA \rightarrow A$ is the canonical projection.

The $A_\infty$-algebra ${\mathcal A}= (A,m_1,m_2,m_3,\ldots )$ satisfies also

$$\begin{aligned} m_2(a_1,a_2) = \Delta (a_1a_2) - {(-1)^{|a_1|}} a_1\Delta (a_2) - \Delta (a_1)a_2,\ a_1,a_2 \in A, \end{aligned}$$

(28)

if and only if the $\phi _2$-part of the automorphism $\phi$ equals the product of $A$, that is if $\phi _2(a,b) = ab$ for each $a,b \in A$.

Proof

Let $m$ denote the coderivation of ${\mathbb T}^cA$ determined by $(m_1,m_2,m_3,\ldots )$.   Assume that we have already constructed an automorphism $\vartheta = (1\!\!1_A,\vartheta _2,\vartheta _3,\ldots )$ of  ${\mathbb T}^cA$ such that

$$\begin{aligned} (\vartheta m \vartheta ^{-1})_1= \Delta \ \text{ and } \ (\vartheta m \vartheta ^{-1})_k= 0 ,\ \text{ for } 1 < k \le n, \end{aligned}$$

with some $n \ge 1$. To simplify the notation, denote $\mu :=\vartheta m \vartheta ^{-1}$. The coderivation $\mu$ determines an $A_\infty$-structure of the form $(A,\Delta ,0,\ldots ,0,\mu _{n+1},\ldots )$. Axiom (7) for $n+1$ implies that $\delta (\mu _{n+1}) = 0$. Consider any automorphism $\alpha$ of ${\mathbb T}^cA$ of the form $\alpha = (1\!\!1_A,0,\ldots ,0,\alpha _{n+1},\ldots )$. Clearly, $(\alpha \mu \alpha ^{-1})_1 = \Delta$, $(\alpha \mu \alpha ^{-1})_k = 0$ for $1< k \le n$ and

$$\begin{aligned} (\alpha \mu \alpha ^{-1})_{n+1}&= \mu _{n+1} + \sum _{1 \le j \le n+1} \alpha _{n+1}\left( 1\!\!1_A^{\otimes (j-1)}\otimes \Delta \otimes 1\!\!1_A^{\otimes (n-j+1)}\right) -\Delta \alpha _{n+1} \\&= \mu _{n+1} - \delta (\alpha _{n+1}). \end{aligned}$$

Since $\delta (\mu _{n+1}) = 0$, by Proposition 3.6 one finds $\alpha _{n+1}$ such that $\mu _{n+1} = \delta (\alpha _{n+1})$. With this choice, $(\alpha \mu \alpha ^{-1})_{n+1} = 0$, thus $\phi ' := \alpha \phi$ satisfies

$$\begin{aligned} (\phi ' m \phi '^{-1})_1= \Delta \ \text{ and } \ (\phi ' m \phi '^{-1})_k= 0 ,\ \text{ for } 1 < k \le n+1. \end{aligned}$$

This shows that we can inductively construct an automorphism $\phi$ of ${\mathbb T}^cA$ such that $\phi m \phi ^{-1} = \Delta$ or, equivalently, $m = \phi ^{-1} \Delta \phi$. The first part of the theorem is proven.

To demonstrate that the twisting automorphism is unique, assume that $m = \phi ^{-1} \Delta \phi = \psi ^{-1} \Delta \psi$. Then $\omega := \phi \psi ^{-1}$ satisfies $\omega ^{-1} \Delta \omega = \Delta$. For the bilinear part $\omega _2$ of $\omega$ this gives

$$\begin{aligned} \omega _2 (\Delta \otimes 1\!\!1_A) +\omega _2 (1\!\!1_A \otimes \Delta ) - \Delta \omega _2 = 0, \end{aligned}$$

i.e. $\delta (\omega _2)=0$. Since $\omega \in {\mathcal Nat}(k)^0$, $\omega _2 = 0$ by Proposition 3.6. In the same vein we prove inductively that $\omega _k = 0$ for all $k \ge 2$, therefore $\omega = 1\!\!1: {\mathbb T}^cA \rightarrow {\mathbb T}^cA$ so $\phi = \psi$.

Theorem 4.1 combined with Example 3.8 gives:

Corollary 4.2

Natural $A_\infty$-algebras satisfying (27) are parametrized by power series

$$\begin{aligned} \phi (t) := t + \omega _2t^2 + \omega _3t^3 + \omega _4 t^4 +\cdots \end{aligned}$$

(29)

with $\omega _k \in {\mathbb {k}}[\Sigma _k]$, $k \ge 2$. Natural $A_\infty$-algebras satisfying also (28) are parametrized by expressions (29) with $\omega _2$ the identity permutation $1\!\!1_{\Sigma _2}$.

A weak isomorphism of $A_\infty$-algebras induces a strict isomorphism of their cohomology algebras. We therefore get another

Corollary

The cohomology $H^*(A,\Delta )$ of any natural $A_\infty$-algebra $(A,\Delta ,m_2,m_3,...)$ is the trivial associative algebra with the underlying space $H^*(A,\Delta )$.

The observation made in Example 3.9 combined with Theorem 4.1 lead to another

Corollary

Assume that $\Delta$ is a derivation of $A$. Then any natural $A_\infty$-braces vanish on $A$.

We finally formulate our characterization of Börjeson braces [6] recalled in Example 1.3.

Theorem 4.3

Börjeson’s braces $(A,\Delta ,b^\Delta _2,b^\Delta _3,b^\Delta _4,\ldots )$ are the unique, up to a strict isomorphism, natural recursive $A_\infty$-braces defined over ${\mathbb Z}$ such that

• (i) $b^\Delta _2$ measures the deviation of $\Delta$ from being a derivation, i.e.

  $$\begin{aligned} b^\Delta _2(a_1,a_2) = \Delta (a_1a_2) - \Delta (a_1)a_2 - {(-1)^{|a_1|}}a_1\Delta (a_2), \quad a_1,a_2 \in A, \end{aligned}$$

• (ii) the coefficient at $\Delta (a_1a_2a_3)$ in $b^\Delta _3(a_1,a_2,a_3)$ is either $+1$ or $-1$ and,

• (iii) the hereditarity is satisfied, that is for all $k \ge 1$,

  $$\begin{aligned} b^\Delta _k = 0 \ \text{ implies } \ b^\Delta _{k+1} = 0. \end{aligned}$$

It is obvious that the Börjeson braces are recursive and satisfy (i) and (ii). Their hereditarity established in [6] follows from an inductive formula mentioned in Remark 4.6. It remains to prove that (i)–(iii) characterize Börjeson’s braces up to a strict isomorphism. This will follow from Propositions 4.4 and 4.5 below.

Proposition 4.4

Suppose that $A = (A,\Delta ,m_2,m_3,m_4,\ldots )$ are natural hereditary $A_\infty$-braces such that $m_2 = b^\Delta _2$ and $m_3 = b^\Delta _3$. Then $m_k = b^\Delta _k$ for any $k \ge 2$.

Proof

Assume we have already proved that

$$\begin{aligned} b^\Delta _k = m_k \ \text{ for } 2 \le k \le n, \end{aligned}$$

with some $n \ge 3$. The $A_\infty$-axiom (7) taken with $n = k+1$ implies that $\delta (m_{n+1}) = \delta (b^\Delta _{n+1})$, i.e.  $\delta (m_{n+1} -b^\Delta _{n+1}) = 0$. By Proposition 3.6 there exists $z_{n+1} \in {\mathcal Nat}(n+1)^0$ such that

$$\begin{aligned} m_{n+1} -b^\Delta _{n+1} = \delta z_{n+1}. \end{aligned}$$

(30)

Consider the free associative algebra $\mathcal Fr(x_{1},\dots ,x_{n+1})$ on degree $0$ variables $x_{1},\dots ,x_{n+1}$. Let $A_{n+1}$ be $\mathcal Fr(x_{1},\dots ,x_{n+1})$ quotiented by the ideal $I_{n+1}$ generated by

$$\begin{aligned} m_n(a_{1},\dots ,a_{n}) \ \text{ for } a_{1},\dots ,a_{n} \in \mathcal Fr(x_{1},\dots ,x_{n+1}). \end{aligned}$$

(31)

Then both $b^\Delta _{n+1}$ and $m_{n+1}$ vanish on $A_{n+1}$, since both braces are hereditary. By (30), $\delta z_{n+1}$ must vanish on $A_{n+1}$, too. In particular, $\delta z_{n+1}(x_{1},\dots ,x_{n+1}) = 0$.¹⁰ We are going to prove that this implies that $z_{n+1} = 0$, so $m_{n+1} = b^\Delta _{n+1}$ again by (30).

It follows from the description of ${\mathcal Nat}(n+1)^0$ given in Example 3.11 that

$$\begin{aligned} z_{n+1}(a_1,\ldots ,a_{n+1}) = \sum _{\sigma \in \Sigma _{n+1}} \xi _\sigma a_{\sigma (1)} a_{\sigma (1)} \cdots a_{\sigma (n+1)}, \end{aligned}$$

with some $\xi _\sigma \in {\mathbb {k}}$, therefore

$$\begin{aligned}&\delta z_{n+1}(x_1,\ldots ,x_{n+1}) \nonumber \\&\quad =\sum _{\sigma \in \Sigma _{n+1}} \xi _\sigma \big (\Delta (x_{\sigma (1)} \cdots x_{\sigma (n+1)})\nonumber \\&\qquad - \sum _{1 \le i \le n+1} x_{\sigma (1)} \cdots \Delta (x_{\sigma (i)}) \cdots x_{\sigma (n+1)}\big ).\qquad \quad \end{aligned}$$

(32)

The crucial observation is that modding out by the ideal $I_{n+1}$ generated by (31) does not introduce any relations involving the monomials

$$\begin{aligned} \Delta (x_{\sigma (1)}) x_{\sigma (2)}\cdots x_{\sigma (n+1)}, \ \sigma \in \Sigma _{n+1}. \end{aligned}$$

(33)

Let us show for instance that the degree $1$ part $I^1_{n+1}$ of the ideal $I_{n+1}$ does not involve the element $\Delta (x_1)x_2 \cdots x_{n+1}$. It follows from the definition of an ideal that the subspace of $I^1_{n+1}$ spanned by words containing $x_{1},\dots ,x_{n+1}$ in this order consists of linear combinations of the monomials

$$\begin{aligned} x_1m_n(x_2,{...}, x_{n+1}),\, m_n(x_1x_2,{...}, x_{n+1}),{...}, m_n(x_1,{...}, x_nx_{n+1}),\, m_n(x_1,{...},x_{n})x_{n+1}. \end{aligned}$$

Looking at the explicit form of $m_n = b^\Delta _n$ we immediately realize that none of the above terms contains the monomial $\Delta (x_1)x_2 \cdots x_{n+1}$. The argument for other permutations is similar.

Inspecting the coefficients at the terms (33) in (32), we see that $\delta z_{n+1}=0$ in $A_{n+1}$ only if all $\xi _\sigma$, $\sigma \in \Sigma _{n+1}$, are trivial. So $z_{n+1} = 0$ as required, and the induction goes on.

In the proof of Proposition 4.4, the requirement that $n \ge 3$ in (30) was crucial. Indeed, the monomial $\Delta (x_1)x_2x_3$ does occur in $m_2(x_1,x_2)x_3$, so the argument following formula (33) does not work for $n=2$. We must therefore prove also the following

Proposition 4.5

Let ${\mathcal A}= (A,\Delta ,m_2,m_3,m_4,\ldots )$ be a natural recursive $A_\infty$-algebra such that $m_2=b^\Delta _2$, the coefficient $C_3$ at $\Delta (a_1a_2a_3)$ in $m_3(a_1,a_2,a_3)$ is either $1$ or $-1$, and $m_3 = 0$ implies $m_4=0$. Then in fact $C_3 = 1$ and $m_3 = b^\Delta _3$.

Proof

Boring tour de force. It follows from the assumptions that the automorphism $\phi$ inducing ${\mathcal A}$ must be of the form $\phi = (1\!\!1_A,\mu ,C_3\mu _3,\ldots )$. This means that $a_1,a_2,a_3$ appear in $m_3$ only in this order. It is simple to prove that $m_3$’s with this property form the family:

$$\begin{aligned} m_3(a_1,a_2,a_3) \!&= \! (1\!+\!\alpha ) \Delta (a_1a_2a_3) - \alpha \big ( (\Delta (a_1)a_2a_3\! +\! (-1)^\varepsilon a_1a_2\Delta (a_3)\big )\nonumber \\&+(1\!-\!\alpha ) (-1)^\varepsilon a_1\Delta (a_2)a_3\! -\!(\Delta (a_1a_2)a_3 + (-1)^\varepsilon a_1\Delta (a_2a_3)\big ),\qquad \quad \end{aligned}$$

(34)

depending on a parameter $\alpha \in {\mathbb {k}}$. We did not write the signs explicitly, because they equal the Koszul sign of the permutation of $(\Delta ,a_1,a_2,a_3)$ in the corresponding term. For instance, $\varepsilon$ at the last term in the first line means $|a_1| + |a_2|$.

The operation $m_4$ decomposes into the sum $m'_4 + m''_4$, where $m'_4$ is the part that contains $a_1,a_2,a_3,a_4$ in this order. The $m'_4$-part is parametrized by $\alpha \in {\mathbb {k}}$ as above and another parameter $\beta \in {\mathbb {k}}$ by

$$\begin{aligned}&m'_4(a_1,a_2,a_3,a_4)\\&\quad =(1+\beta )\Delta (a_1a_2a_3a_4) +(2\alpha -\beta )\big (\Delta (a_1)a_2a_3a_4 +(-1)^\varepsilon a_1a_2a_3\Delta (a_4)\big )\\&\qquad +\,\,(4\alpha -\beta )\big ((-1)^\varepsilon a_1\Delta (a_2)a_3a_4 + (-1)^\varepsilon a_1a_2\Delta (a_3)a_4\big )\\&\qquad -\,\, (1+\alpha )\big (\Delta (a_1a_2a_3)a_4 + (-1)^\varepsilon a_1\Delta (a_2a_3a_4)\big )\\&\qquad -\,\, \alpha \big (\Delta (a_1a_2)a_3a_4 + (-1)^\varepsilon a_1a_2 \Delta (a_3a_4)\big ) +(1-\alpha ) (-1)^\varepsilon a_1 \Delta (a_2a_3)a_4. \end{aligned}$$

If the vanishing of $m_3$ implies the vanishing of $m_4$, then $m'_4$ must equal the linear combination

$$\begin{aligned} A m_3(a_1a_2,&a_3,a_4) + B m_3(a_1,a_2a_3,a_4) +C m_3(a_1,a_2,a_3a_4)\nonumber \\&+ Da_1m_3(a_2,a_3,a_4) + E m_3(a_1,a_2,a_3)a_4 \end{aligned}$$

(35)

with some $A,\ldots ,E \in {\mathbb {k}}$. Expanding $m_3$’s in the above display gives the system:

$$\begin{aligned} \begin{aligned} (1+\beta )&= (1+\alpha )(A+B+C),\\ (2\alpha -\beta )&= -\alpha (B+C+E)= -\alpha (A+B+D),\\ (4\alpha - \beta )&= (1 -\alpha )C - \alpha D - (\alpha -1) E = (1 -\alpha )A - \alpha E - (\alpha -1) D,\\ \alpha&= \alpha A + C + E= \alpha C + A + D,\\ (1-\alpha )&= (1-\alpha ) B - D -E,\ \text{ and } \\ (\alpha +1)&= A+B - (\alpha +1)E = B+C - (\alpha +1)D. \end{aligned} \end{aligned}$$

(36)

The case $C_3=1$ corresponds to $\alpha =0$ by (34). The 2nd equation of (36) immediately gives that $\beta =0$, so the corresponding $m_2$ and $m_3$ equal the Börjeson braces $b^\Delta _2$ and $b^\Delta _3$ as claimed.

The $C_3=-1$ case happens when $\alpha = -2$. One may verify that then the system (36) has the unique solution, namely

$$\begin{aligned} \alpha = \beta = -2,\ A = C = 1/2, B=0 \text{ and } D=E = -3/2. \end{aligned}$$

Therefore, the only way how to express $m'_4$ as a linear combination (35) is

$$\begin{aligned}&m'_4(a_1,a_2,a_3,a_4) \\&\quad =\frac{1}{2}\big ( m_3(a_1a_2,a_3,a_4) +m_3(a_1,a_2,a_3a_4)\big )\\\quad&\qquad -\frac{3}{2} \big (a_1m_3(a_2,a_3,a_4) + m_3(a_1,a_2,a_3)a_4\big ), \end{aligned}$$

so $m_3 = 0$ does not imply $m_4=0$ over ${\mathbb Z}$. This excludes this case.

Remark 4.6

In the $C_3=1$ case of the above proof, the solutions of the system (36) form a 2-parametric family depending on $A,B \in {\mathbb {k}}$:

$$\begin{aligned} \alpha = \beta = 0,\ C = 1-(A+B),\ D = -A \text{ and } E = (A+B)-1. \end{aligned}$$

The particular solution with $B=1$ and all other parameters trivial gives rise to the ‘canonical’ recursion

$$\begin{aligned} b^\Delta _4(a_1,a_2,a_3,a_4) = b^\Delta _3(a_1,a_2a_3,a_4) \end{aligned}$$

that generalizes to all higher Börjeson’s braces in the obvious way, proving their hereditarity.

Example 4.7

(Non-hereditary braces) Proposition 4.4 can be used to produce examples of non-hereditary braces. Recall from Example 2.4 that the Börjeson braces are generated by the automorphism $\phi$ with the generating series $\phi (t) := t + t^2 + t^3 + t^4 + \cdots$. Assume that ${\mathcal A}= (A,\Delta ,m_2,m_3,\ldots )$ is the $A_\infty$-algebra given by the automorphism $\tilde{\phi }$ with the generating series of the form

$$\begin{aligned} \tilde{\phi }(t) := t + t^2 + t^3 + \hbox { higher order terms.} \end{aligned}$$

Clearly $m_2 = b^\Delta _2,\ m_3 = b^\Delta _3$. By Proposition 4.4, ${\mathcal A}$ is hereditary if and only if $\phi (t)= \tilde{\phi }(t)$. So choosing e.g. $\tilde{\phi }(t) := t + t^2 + t^3$ produces non-hereditary braces.

Example 4.8

(Super-exotic $A_\infty$-braces) We present $A_\infty$-braces that are natural in the extended sense of Example 3.7. They are given, for elements $a,a_1,a_2,\ldots$ of a graded associative algebra $A$ with a differential $\Delta$, by

$$\begin{aligned} s^\Delta _1(a)&= \Delta (a),\\ s^\Delta _2(a_1,a_2)&= {\left\{ \begin{array}{ll} {\Delta (a_1a_2)},&{}\mathrm{if }\quad (|a_1|,|a_2|) = (0,0),\\ {-\Delta (a_1)a_2},&{}\mathrm{if }\quad (|a_1|,|a_2|) = (-1,0),\\ {-a_1\Delta (a_2)},&{}\mathrm{if }\quad (|a_1|,|a_2|) = (0,-1),\quad \hbox {and}\\ 0,&{}\hbox { in the remaining cases,}\\ \end{array}\right. }\\ s^\Delta _3(a_1,a_2,a_3)&= {\left\{ \begin{array}{ll} \Delta (a_1a_2a_3),&{} \hbox {if }\quad (|a_1|,|a_2|,|a_3|) = (0,0,0),\\ {a_1\Delta (a_2)a_3}&{} \hbox {if }\quad (|a_1|,|a_2|,|a_3|) = (0,-1,0),\quad \hbox {and}\\ 0&{} \hbox {in the remaining cases},\\ \end{array}\right. }\\&\vdots&\\ s^\Delta _k(a_1,\ldots ,a_k)&= {\left\{ \begin{array}{ll} {\Delta (a_1\cdots a_k)},&{} {\hbox {if}}\quad (|a_1|,\ldots ,|a_k|) = (0,\ldots ,0),\quad \hbox {and}\\ 0&{} \hbox {in the remaining cases},\ k \ge 4. \end{array}\right. } \end{aligned}$$

The above braces are the Börjeson braces of the couple $(A',\Delta )$, where $A'$ is the associative algebra with the same underlying space as $A$ but the multiplication $\cdot '$ defined as

$$\begin{aligned} a \cdot ' b := {\left\{ \begin{array}{ll} {ab},\quad \hbox {if } (|a|,|b|) = (0,0), \hbox {and}\\ {0},\quad \,\, \hbox { in the remaining cases.} \end{array}\right. } \end{aligned}$$

Notice that the above braces are nontrivial even when $\Delta$ is a derivation! For this reason we did not consider this kind of extended naturality.

Commutative associative case. The first part of this section translates to the commutative case in a straightforward manner, so we formulate only the commutative versions of the main theorems. We start with the commutative variant of Theorem 4.1. Recall that ${\mathcal L}_\Delta$ denotes the trivial $L_\infty$-algebra from Example 2.7.

Theorem 4.9

For each natural $L_\infty$-algebra ${\mathcal L}= (A,l_1,l_2,l_3,\ldots )$ such that

$$\begin{aligned} l_1 = \Delta \end{aligned}$$

(37)

there exist a unique natural automorphism $\phi = (1\!\!1,\phi _2,\phi _3,\ldots )$ of the symmetric coalgebra ${\mathbb S}^cA$ such that ${\mathcal L}$ equals the twisting of ${\mathcal L}_\Delta$ via $\phi$. The $L_\infty$-algebra ${\mathcal A}= (A,l_1,l_2,l_3,\ldots )$ satisfies

$$\begin{aligned} l_2(a_1,a_2) = \Delta (a_1a_2) - {(-1)^{|a_1|}} a_1\Delta (a_2) - \Delta (a_1)a_2,\ a_1,a_2 \in A, \end{aligned}$$

(38)

if and only if $\phi _2$ equals the product of $A$.

Theorem 4.9 combined with the description of natural automorphisms given in Example 3.12 leads to:

Corollary 4.10

Natural $L_\infty$-algebras satisfying (37) are parametrized by power series

$$\begin{aligned} \phi (t) := t + f_2t^2 + f_3t^3 + f_4 t^4 +\cdots \in {\mathbb {k}}[[t]]. \end{aligned}$$

(39)

Natural $L_\infty$-algebras satisfying also (38) are parametrized by series (39) with $f_2 = 1$.

The following theorem offers a characterization of Koszul $L_\infty$-braces analogous to that of Börjeson $A_\infty$-braces given in Theorem 4.3.

Theorem 4.11

The Koszul braces $(A,\Delta ,\Phi ^\Delta _2,\Phi ^\Delta _3,\Phi ^\Delta _4,\ldots )$ are the unique natural $L_\infty$-braces defined over ${\mathbb Z}$ such that

• (i) $\Phi ^\Delta _2$ measures the deviation of $\Delta$ from being a derivation, i.e.

  $$\begin{aligned} \Phi ^\Delta _2(a_1,a_2) = \Delta (a_1a_2) - \Delta (a_1)a_2 - {(-1)^{|a_1|}}a_1\Delta (a_2), \ a_1,a_2 \in A, \end{aligned}$$

• (ii) the coefficient at $\Delta (a_1a_2a_3)$ in $\Phi ^\Delta _3(a_1,a_2,a_3)$ is either $+1$ or $-1$ and,

• (iii) the hereditarity is satisfied, that is for all $k \ge 1$, that is

  $$\begin{aligned} \Phi ^\Delta _k = 0 \ \text{ implies } \ \Phi ^\Delta _{k+1} = 0. \end{aligned}$$

Exercise

Explain how to construct non-hereditary $L_\infty$-braces.