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\begin{document}
\bibliographystyle{amsplain}
\relax \labelsONmargin
\title[Commutative and associative multiplication]{ On the ultra-commutative \\
homotopically associative multiplication \\
in cohomological complex }
\author{ Ilya Zakharevich }
\address{ Department of Mathematics, MIT, Cambridge, MA, 02139 }
\email {ilya@@math.mit.edu}
\date{ Interim!\quad January 1993\quad Printed: \today }
\setcounter{section}{-1}
\maketitle
\begin{abstract}
We construct a multiplication in the cohomological complex of a
simplicial set that is homotopically associative and, moreover, satisfies
some axioms that are strong amplifications of what is the commutativity
axioms in the associative case. This construction makes explicit some
usual constructions in the theory of simplicial categories.
\end{abstract}
\tableofcontents
\section{Introduction }
\subsection{Liftings of the product in cohomology to the cochain complex } Consider a
topological space $ M $ and the (some) cohomology $ H^{*}\left(M,K\right) $ of
this
space as a $ {\mathbb Z} $-graded (or $ {\mathbb Z}_{2} $-graded) $ K $-(super)module, $ K $ being a
fixed ring. A standard
construction determines a multiplication in this module that is both
associative and (super)commutative. However, by its definition $ H^{*}\left(M,K\right) $ is
the cohomology of some complex, and from a lot of viewpoints it would
be very
important to construct the corresponding multiplication in this complex.
If $ K={\mathbb R} $, $ M $ is smooth and we consider de Rham cohomology, then the
usual $ \wedge $-product of differential forms gives an example of such a
multiplication, since it is both associative and (super)commutative.
However, in most other cases it is very difficult to construct such a
product. Indeed, it was one of the technical problems in the
rational homotopy theory, where to construct a basic complex with
multiplication corresponding to a triangulated topological space one need
to consider a supernatural piecewise-de-Rham complex.
Let us consider for example the same case of triangulated
topological space (or {\em simplicial set\/}) and the usual cosimplicial complex of
this space. Then the $ \cap $-product determines an associative product on this
complex that is compatible with the multiplication in cohomology.
However, the $ \cap $-product is not (super)commutative.
One can consider a {\em symmetrization\/} of the $ \cap $-product, that is
commutative. However, this symmetrization ceases to be associative. It is
easy to understand that in the case $ K={\mathbb Z} $ or $ K={\mathbb F}_{q} $ it is impossible to
construct such a product at all. Indeed, the Steenrod powers are exactly
some obstacles for such an object to exist.
\subsection{A weaker condition on liftings } Though the symmetrization of the $ \cap $-product
is not
associative, it still induces an associative product in homology of the
complex. This means that the obstacle for the associativity is a
coboundary, therefore we can construct a {\em homotopy\/} that connects two
ternary products
\begin{equation}
\left(a,b,c\right)\mapsto\left(a\cap b\right)\cap c\quad \text{and\quad }\left(a,b,c\right)\mapsto a\cap\left(b\cap c\right).
\notag\end{equation}
(This vague description is expanded below.) This homotopy is itself a
ternary product.
Thus we should study in fact
pairs of products: a binary one that determines a multiplication, and a
ternary one that determines a homotopy for the non-associativity of this
multiplication. A crucial observation of Stasheff is the fact that we can
pose not only a question about
the {\em associativity\/} of the binary product but also about some
generalization of the {\em associativity\/} to the ternary products. Again,
the obstacle for the associativity of the {\em homotopy\/} is non-zero, but a
coboundary, and
the corresponding {\em homotopy\/} determines some $ 4 $-ary product and so on.
Moreover, one can consider the differential in the complex as a {\em unary
product}, and the resulting condition of homotopical associativy of this
system of products can be written as $ D^{2}=0 $ for some derivation $ D $ described
below.
A $ K $-module with such a system of products on it is called a
{\em(strictly) homotopically associative\/} $ K $-{\em algebra\/} by Stasheff, and in the
previous paragraph we
have ``shown'' that the differential together with the symmetrization
of the
$ \cap $-product on the cosimplicial complex can be extended to a structure of
such an algebra. However, now the situation is uncompleted in the same
sence as above: we have a lot of poly-ary products on a $ K $-module,
however, we pose the commutativity condition only on the binary one. It
seems that the first definition of what the commutativity of the higher
homotopies could mean was stated by V.~Drinfeld in his letter to
V.~Schechtman. We call such a product a {\em strictly commutative
homotopically associative\/} product.
One of the first topics of this paper is to show that
such a product on a cosimplicial complex does exist in the case $ K\supset{\mathbb Q} $, and
to study how unique this product is. We show that it is possible to make
this product compatible with mappings of triangulated topological spaces
(or simplicial sets), i.e., to define a {\em universal\/} product. Moreover,
in the process of studying the uniqueness we show that there exists
a universal product that satisfies much stronger conditions than ones
in the Drinfeld definition.
In the concluding part of the paper we show that any construction
of the universal product
leads to constructions of the known objects in the theory of simplicial
categories, such as a structure of a strictly commutative homotopically
associative algebra on a cosimplicial commutative algebra, and a
structure of a
homotopical Lie algebra on a cosimplicial Lie algebra (for example, on
\v Cech complexes of sheaves of commutative or Lie algebras). However, since
there is a universal product that satisfies much stronger conditions, the
above structures also satisfy the same conditions. This makes the
following question natural: What is the ``correct'' definition of the
commutativity of homotopically associative product.
Author expresses his profound gratitude to a lot of people for
fruitful discussions without which this text could not appear. Among them
are M.~Finkelberg, D.~Fucks, D.~Kaledin, D.~Kazhdan, H.~Miller,
V.~Serganova, V.~Schechtman. Espessial thanks go to
M.~Kontsevich. The idea to investigate this question
appeared when studying the preliminary variant of a currently deceased
text by him.
\section{Main definitions and statements }\label{h01}\myLabel{h01}\relax
Here we provide (without any justification) the definitions of algebraic
objects we need and the principal statements we are going to prove. We
postpone the discussion of motivations and the proofs until later.
\subsection{Homotopical algebras }\label{s01.1}\myLabel{s01.1}\relax The first object we discuss was first
introduced by Stasheff under the name of $ A_{\infty} $-algebras:
\begin{definition} \label{def1.10}\myLabel{def1.10}\relax A (strictly) homotopically associative $ K $-algebra $ A $
is a
$ K $-module and
an odd derivation $ D $ of
the free associative (super)algebra $ T^{\bullet}\left(\left(\Pi A\right)^{*}\right) $
generated by the $ K $-module $ \left(\Pi A\right)^{*} $
such that $ D^{2}=0 $. \end{definition}
Here we use standard notations for superalgebras, in particular, $ \Pi $ is the
functor of parity flipping. In this definition we can as well suppose
that $ A $ is a $ K $-supermodule, i.e., $ {\mathbb Z}_{2} $-graded $ K $-module.
\begin{definition} A (strictly) commutative (strictly) homotopically
associative $ K $-algebra $ A $ is a $ K $-module and an odd derivation $ D $ of the
free Lie (super)algebra $ {\mathcal L}^{\bullet}\left(\left(\Pi A\right)^{*}\right) $ generated by the $ K $-module $ \left(\Pi A\right)^{*} $ such
that $ D^{2}=0 $. \end{definition}
For the sake of completeness we provide also a definition of a
homotopical Lie algebra:
\begin{definition} \label{def01.5}\myLabel{def01.5}\relax A (strict) homotopical Lie $ K $-algebra $ {\mathfrak g} $ is a $ K $-module and an
odd derivation $ D $ of the free (super)commutative algebra $ S^{\geq1}\left(\left(\Pi{\mathfrak g}\right)^{*}\right) $
generated by the $ K $-module $ \left(\Pi{\mathfrak g}\right)^{*} $ such that $ D^{2}=0 $. \end{definition}
All these definitions allow a modification to the case of $ {\mathbb Z} $-graded
$ K $-modules (instead of $ {\mathbb Z}_{2} $-grading above). We provide a variant of the
Def.~\ref{def1.10}:
\begin{definition} \label{def01.6}\myLabel{def01.6}\relax A $ {\mathbb Z} $-graded (strictly) homotopically associative $ K $-algebra
$ A $ is
a $ {\mathbb Z} $-graded $ K $-module and an odd derivation $ D $ of grading 1 of the free
associative (super)algebra
$ T^{\bullet}\left(\left(A\left[-1\right]\right)^{*}\right) $ generated by the $ K $-module $ \left(A\left[-1\right]\right)^{*} $ such that $ D^{2}=0 $. \end{definition}
Here $ A\left[-1\right] $ denotes the same $ K $-module with the shifted $ {\mathbb Z} $-grading:
$ A\left[-1\right]_{k}=A_{k-1} $. In this definition we use the standard convension that the
parity grading on $ A $ is compatible with the $ {\mathbb Z} $-grading: $ A_{\bar{i}}\buildrel{\text{def}}\over{=}\bigoplus_{k}A_{i+2k} $.
Thus the supermodule structure on $ A\left[-1\right] $ coincides with the structure on
$ \Pi A $.
Below we denote the space $ \left(A\left[-1\right]\right)^{*} $ (or $ \left(\Pi A\right)^{*} $) by $ V $. In all the cases
we can consider the block of the linear mapping $ D $ that sends $ V $ to $ V $.
Denote by $ \partial $ the mapping $ A\to $A such that this block coincides with $ \left(\Pi\partial\right)^{*} $.
Then clearly $ \partial^{2}=0 $, $ \partial $ is odd, and in the $ {\mathbb Z} $-graded case $ \partial $ is of grading 1.
\subsection{Universal multiplication }\label{s01.2}\myLabel{s01.2}\relax For a triangulated topological
space $ X $ denote by $ \widehat{C}_{\bullet}\left(X\right) $ the {\em(full) simplicial complex\/} with coefficients
in $ K $ spanned by the all simplices in $ X $ (including the degenerate ones)
and by $ C_{\bullet}\left(X\right) $ the {\em(nondegenerate) simplicial complex\/} spanned by
nondegenerate simplices up to a change of ordering of vertices that
should multiply a simplex by the parity of substitution. We equip this
graded spaces with the standard differential $ \partial $.
\begin{definition} An {\em multiplication\/} for $ X $ (with coefficients in $ K $)
is a structure of strictly commutative strictly homotopically associative
algebra on the $ {\mathbb Z} $-graded space $ C_{\bullet}\left(X\right)^{*} $. We call a multiplication
{\em admissible\/} if
the corresponding differential in $ C_{\bullet}\left(X\right)^{*} $ coincides with $ \partial^{*} $. \end{definition}
\begin{remark} It is clear that any admissible multiplication induces a
(super)commutative associative multiplication in the cohomology
$ H\left(C\left(X\right)^{*}\right)\buildrel{\text{def}}\over{=}H^{*}\left(X,K\right) $. However, we {\em do not demand\/} that an admissible
multiplication induces {\em the standard\/} multiplication in cohomology. As we
will see, there is no much arbitrariness anyway, i.e., already the
condition of compatibility with the usual differential determines the
multiplication so uniquely as it is possible. \end{remark}
\begin{definition} A {\em universal multiplication\/} (with coefficients in $ K $) is a
structure of admissible
multiplication for any triangulated topological space that is compatible
with simplicial mappings of such spaces. \end{definition}
The structure of definitions of homotopical multiplication suggests
that the more natural object to consider is the comultiplication. Out
aim is to construct a universal (co)multiplication that satisfies some
additional restrictions, and to formulate this restrictions we need to
consider the mapping $ D $.
An intermediate notion between admissible multiplication and
universal multiplication is monotonic admissible multiplication, where
\begin{definition} A {\em monotonic multiplication\/} on $ X $ is a
multiplication on $ X $ that induces multiplications on all
tringulated subspaces in $ X $, i.e., is compatible with inclusion
\begin{equation}
{\mathcal L}\left(C_{\bullet}\left(Y\right)\left[1\right]\right)\hookrightarrow{\mathcal L}\left(C_{\bullet}\left(X\right)\left[1\right]\right)
\notag\end{equation}
for a triangulated subspace $ Y\subset X $.
\end{definition}
Analogous notions can be introduced in the complex $ \widehat{C}_{\bullet} $:
\begin{definition} An {\em order-preserving multiplication\/} on $ X $ is is a
structure of strictly commutative strictly homotopically associative
algebra on the $ {\mathbb Z} $-graded space $ \widehat{C}_{\bullet}\left(X\right)^{*} $ such that the
corresponding mapping $ D:
{\mathcal L}\left(C_{\bullet}\left(X\right)\left[1\right]\right) \to {\mathcal L}\left(C_{\bullet}\left(X\right)\left[1\right]\right) $ sends any ordered simplex $ S\subset X $ into the Lie
subalgebra generated by ordered subsimplices of $ S $. \end{definition}
The relation between monotonic multiplications and order-preserving
multiplication is summed up in
\begin{lemma} The natural mappings between complexes $ C_{\bullet} $ and $ \widehat{C}_{\bullet} $
identify the sets
of monotonic multiplications on $ X $ with a subset of order-preserving
multiplications on $ X $. \end{lemma}
\subsection{The standard simplex }\label{s01.3}\myLabel{s01.3}\relax Consider the standard $ k $-dimensional simplex
\begin{equation}
\sigma^{k}\buildrel{\text{def}}\over{=}\left\{\left(x_{0},\dots ,x_{k}\right) \mid x_{i}\geq0, \sum x_{i}=1\right\}.
\notag\end{equation}
This topological space is tautologically triangulated. There is a natural
action of the symmetric group $ {\mathfrak S}_{k+1} $ on $ \sigma^{k} $, and this
action is compatible with the natural embedding $ \sigma^{k}\hookrightarrow\sigma^{k+1} $ and the
corresponding embedding $ {\mathfrak S}_{k+1}\hookrightarrow{\mathfrak S}_{k+2} $. Thus we can consider the direct limit
\begin{equation}
\sigma^{\infty}=\bigcup_{k}\sigma^{k}
\notag\end{equation}
with the natural action of
\begin{equation}
{\mathfrak S}_{\infty}=\bigcup_{l}{\mathfrak S}_{l}.
\notag\end{equation}
Call this topological space $ \sigma^{\infty} $ the {\em standard simplex}. It is triangulated,
since it is a union of its finite-dimensional faces. Denote by $ \left< l \right> $
the vertex $ \left\{x_{m}=0 \mid m\not=l\right\} $ and by $ \left< m_{0}\dots m_{k} \right> $ the face with vertices $ \left<
m_{0} \right>,\dots ,\left< m_{k} \right> $ and the orientation corresponding to the ordering $ \left(m_{l}\right) $.
A universal multiplication induces an admissible multiplication in
the standard simplex, and this multiplication is $ {\mathfrak S}_{\infty} $-invariant. This
construction can be
inverted:
\begin{proposition} Any $ {\mathfrak S}_{\infty} $-invariant admissible multiplication in the
standard simplex induces a unique universal multiplication. \end{proposition}
\begin{proof} For the sake of completeness we provide a proof of this
obvious
proposition. Consider the corresponding comultiplication and an arbitrary
face $ \left< m_{0}\dots m_{k} \right> $. Let $ W $ be a subspace of $ C\left(\sigma^{\infty}\right) $ spanned by the
subsimplices of $ \left< m_{0}\dots m_{k} \right> $. Consider a homogeneous component $ D_{i} $ of the
corresponding derivation of $ T^{\bullet}\left(C\left(\sigma^{\infty}\right)\left[1\right]\right) $. We claim that $ D_{i} $ sends
$ W\left[1\right] $
into $ W\left[1\right]^{\otimes i+1} $. Indeed, otherwise $ D_{i}\left(W\left[1\right]\right) $ would contain some simplex with
a vertex outside of $ \left< m_{0}\dots m_{k} \right> $. Consider now the stabilizer of $ \left<
m_{0}\dots m_{k} \right> $ in $ {\mathfrak S}_{\infty} $. The action of this stabilizer shows that {\em any\/} vertex
appears in $ D_{i}\left(W\left[1\right]\right) $, thus $ D_{i}\left(W\left[1\right]\right) $ contains a tensor of an infinite rank:
a contradiction.
Thus the comultiplication in $ \sigma^{\infty} $ induces a comultiplication in $ \sigma^{k} $,
and the latter comultiplication is $ {\mathfrak S}_{k+1} $-invariant. Now consider an
arbitrary triangulated topological space (or semisimplicial set) and a
$ k $-dimensional simplex in it. The comultiplication in $ \sigma^{k} $ induces a
comultiplication in this simplex, and this comultiplication is
independent on the identification due to $ {\mathfrak S}_{k+1} $-stability. The obtained
comultiplication is obviously admissible. \end{proof}
\begin{remark} Obviously a $ {\mathfrak S}_{\infty} $-invariant multiplication in $ \sigma^{\infty} $ of any particular
type (say, weakly homotopically associative) induces the corresponding
{\em universal\/} multiplication of the same kind. We study one such example in
the following section. \end{remark}
\subsection{The $ \cup $-product } Consider the following (cobinary) comultiplication in
$ \widehat{C}_{\bullet}\left(\sigma^{\infty}\right) $:
\begin{equation}
m^{*}\left(\left< m_{0}\dots m_{k} \right>\right)\buildrel{\text{def}}\over{=}\sum_{l=0}^{k}\left(-1\right)^{l}\left< m_{0}\dots m_{l} \right>\otimes\left< m_{l}\dots m_{k} \right>.
\notag\end{equation}
It obviously is coassociative and $ {\mathfrak S}_{\infty} $-invariant. However, it is not
cocommutative. Thus it induces an associative non-commutative universal
multiplication. This multiplication is called the $ \cup $-product by
topologists.
Moreover, this comultiplication is order-preserving, but is not
induced by the comultiplication in $ C_{\bullet} $.
Since this product satisfies the Leibniz relation, it induces
an associative product in cohomology. The key property of this product
that is not directly contained in the definition is
\begin{lemma} This product in cohomology is (super)commutative. \end{lemma}
In this paper we give some ``explanation'' why this is true.
\subsection{Simplest particular cases }\label{s01.4}\myLabel{s01.4}\relax We begin with the simplest case
of a triangulated space: a point.
\begin{proposition} Any multiplication on a point $ \sigma^{0} $ is
proportional to the $ \cup $-product:
\begin{equation}
\left< 0 \right>\mapsto\left< 0 \right>\otimes\left< 0 \right>=\frac{1}{2}\left[\left< 0 \right>,\left< 0 \right>\right].
\notag\end{equation}
\end{proposition}
This provides a monotonic multiplication on any $ 0 $-dimensional space
$ X $
(i.e., a union of points), and we can see that the space of monotonic
multiplications is of dimension card $ X $: we should provide one parameter
for any point.
The case of $ 1 $-dimensional space is much more complicated.
\begin{theorem} All monotonic multiplication on the interval $ \sigma^{1} $ have the
form:
\begin{equation}
D\left<0\right>=\alpha\left<0\right>^{2},\quad D\left<1\right>=\beta\left<1\right>^{2},
\notag\end{equation}
\begin{equation}
D\left<01\right>=f\left(\operatorname{ad}_{\left<01\right>}\right)\left<0\right>+g\left(\operatorname{ad}_{\left<01\right>}\right)\left<1\right>
\notag\end{equation}
where
\begin{enumerate}
\item
$ f\left(x\right)=-\alpha x,\quad g\left(x\right)=-\beta x $,
\item
or $ f=0,\quad g=\varphi_{\beta,\beta_{0}} $,
\item
or $ f=\varphi_{\alpha,\alpha_{0}},\quad g=0 $,
\item
or $ f=\varphi_{\alpha,\alpha_{0}},\quad g=\varphi_{\beta,\beta_{0}},\quad \alpha\beta_{0}+\alpha_{0}\beta=0 $.
\end{enumerate}
Here $ \alpha_{0} $, $ \beta_{0} $ are additional parameters such that $ \alpha_{0},\beta_{0}\not=0 $, a function
$ \varphi_{\alpha,\alpha_{0}}\left(x\right) $, $ \alpha_{0}\not=0 $, is defined by
\begin{equation}
\varphi_{\alpha,\alpha_{0}}\left(x\right)=\frac{\alpha x}{e^{\frac{\alpha}{\alpha_{0}}x}-1}\text{ if }\alpha,\alpha_{0}\not=0,
\notag\end{equation}
and as a limit $ \varphi_{0,\alpha_{0}}\left(x\right)=\alpha_{0} $.
\end{theorem}
\begin{corollary} All admissible monotonic multiplications on $ \sigma^{1} $ induce the
same
multiplications on the two ends of the interval. Moreover, any two of
them differ only by a simultaneous multiplication of all the chains by
the same constant (possibly infinite). \end{corollary}
\begin{proof} By definition in the admissible case $ f\left(0\right)=-1 $, $ g\left(0\right)=1 $, thus
only the fourth case is possible, $ \alpha_{0}=-1 $, $ \beta_{0}=1 $ and $ \alpha=\beta $. It turns
out that if $ \alpha=\beta\not=0 $, then the above $ 1 $-parametric family of products are all
isomophic under the multiplication of all the chains by a constant. The
case $ \alpha=\beta=0 $ is the limit case of these. \end{proof}
Generally speaking, under the condition of monotonicity alone we
have $ 3 $-parametric family of products. However, any two products in general
position in the same family are isomorphic under the multiplication of three
non-degenerate simplices in $ \sigma^{1} $ by different constants.
On the other hand, the 4 described above families of multiplication
are not limits of one another, thus the set of monotonic multiplications
on an interval has 4 irreducible components.
\begin{corollary} All admissible monotonic multiplications on a connected
graph induce the same
multiplications in the cohomology (up to a constant). \end{corollary}
\subsection{Symmetries of multiplications }\label{s01.5}\myLabel{s01.5}\relax We have seen above
that in
$ 1 $-dimensional case (i.e., on graphs) there is only one (up to
proportionality) multiplication that satisfies some natural conditions.
However, this is a specific feature of dimension 1, as shows the
following
\begin{lemma} Consider an even derivation $ D_{0} $ of a Lie superalgebra $ A $. Then
the vector field
\begin{equation}
D_{1}\mapsto\left[D_{0},D_{1}\right]
\notag\end{equation}
determines a vector field on the set of odd derivations, and this vector
field is tangent to the set of solutions of $ D_{1}^{2}=0 $.
\end{lemma}
Thus the Lie algebra of even derivations {\em acts\/} on the set of odd
solutions of $ D^{2}=0 $. Since the even derivations form a Lie algebra (as
opposoite to a Lie superalgebra), we can consider the corresponding Lie
group of (even) automorphisms of the initial Lie superalgebra. It clearly
acts on the set of odd solutions of $ D^{2}=0 $, thus we cannot hope that the
set of solutions is small if this Lie groop is huge.
\begin{remark} In the $ {\mathbb Z} $-graded case we should consider derivations of grading
0 and grading-preserving automorphisms instead. \end{remark}
\begin{proposition} The set of even derivations of grading 0 in $ {\mathcal L}\left(C_{\bullet}\left(X\right)\left[1\right]\right) $
acts on the set of multiplications in $ X $. The derivation preserves the
subset of admissible products iff the block $ C_{\bullet}\left(X\right)\left[1\right]\to C_{\bullet}\left(X\right)\left[1\right] $ of the
derivation is scalar. It preserves the set of monotonic multiplications
if it is monotonic itself. \end{proposition}
In the $ 0 $-dimensional and $ 1 $-dimensional cases we reproduce already
obtained results: on the set of multiplications in $ \sigma^{0} $ and of admissible
monotonic multiplications in $ \sigma^{1} $ we get an action of scalars, and on the
set of monotonic multiplications in $ \sigma^{1} $ we get an action of three copies
of scalars, one for any subsimplex in $ \sigma^{1} $. However, in the case of bigger
dimension we get an infinite-dimensional algebra, since any monotonic mapping
$ C_{\bullet}\left(X\right)\left[1\right]\to{\mathcal L}\left(C_{\bullet}\left(X\right)\left[1\right]\right) $ of grading 0 with a trivial block
$ C_{\bullet}\left(X\right)\left[1\right]\to C_{\bullet}\left(X\right)\left[1\right] $ induces such a derivation.
On the other hand, in the cases $ \dim =0 $ and $ \dim =1 $ we have seen that
the action of the corresponding group has a dense open orbit. This
situation is generalized by the following
\begin{theorem} There exist a universal multiplication. Moreover, for any
connected triangulated space $ X $ the action of the specified above group on
the (non-empty) set of admissible monotonic multiplications on $ X $ has an
open orbit. \end{theorem}
Our main task is to construct more or less explicit universal
multiplication. The best way to do it would be to find some natural
conditions and show that a universal multiplication satisfying these
conditions is unique. We could not do it completely, but there is one
nontrivial condition that we can pose on a universal multiplication.
However, before we can formulate this condition we need some
additional notations. If $ V $ is a graded vector space, then $ {\mathcal L}^{\bullet}\left(V\right) $ is graded
by (finite) sequences $ d=\left(d_{n}\right) $, where a Lie monomial is of grading $ d $ if it
contains exactly $ d_{n} $ occurrences of a generator from $ \operatorname{Gr}_{n}V $.
\begin{lemma} Consider the free Lie algebra $ {\mathcal L} $ generated by variables $ x_{0},\dots x_{k} $.
Let $ V $ be the vector space spanned by $ x_{1},\dots ,x_{k} $. Extend the natural
$ \operatorname{ad}_{V} $-action on $ {\mathcal L} $ to an action of a tensor power of $ V $ on $ {\mathcal L} $. Then the
mapping
\begin{equation}
AD\colon T^{\bullet}V\to{\mathcal L}\colon f\mapsto\operatorname{ad}_{f}\cdot x_{0}
\notag\end{equation}
is an isomorphism of a vector space $ T^{\bullet}V $ onto the subspace of $ {\mathcal L} $ consisting
of Lie polynomials of degree 1 in $ x_{0} $.
\end{lemma}
This shows that the components of grading
\begin{equation}
d=\left(\dots ,d_{i-1},d_{i},1,0,\dots ,0,\dots \right)
\notag\end{equation}
in $ {\mathcal L}^{\bullet}\left(V\right) $ can be identified with $ T^{\bullet}\left(\bigoplus_{k\leq i}V_{k}\right) $. Call a homogeneous
element of
such grading {\em a good element\/} if under this identification it goes to
\begin{equation}
{\mathcal L}^{\bullet}\left(\bigoplus_{k\leq i}V_{k}\right)\subset T^{\bullet}\left(\bigoplus_{k\leq i}V_{k}\right)\text{. Call a power series {\em good\/} if any homogeneous
component of it is good.}
\notag\end{equation}
\begin{definition} Call a derivation $ D $ in the Lie superalgebra $ {\mathcal L}^{\bullet}\left(V\right) $
(here $ V $ is a graded vector space) {\em acceptable in
degree\/} $ k $ if it sends any element of $ \operatorname{Gr}_{k}V $ to a Lie polynomial (or power
series) in $ \operatorname{Gr}_{\leq k}V $
that is linear in $ \operatorname{Gr}_{k}V $. Call it {\em good in degree\/} $ k $ if this Lie
polynomial is good. Call a multiplication in a triangulated space $ X $
{\em good\/} if the corresponding derivation is good in any degree $ >0 $. \end{definition}
\begin{remark} As we have seen even for $ 1 $-dimensional space $ X $ we cannot
construct a monotonic admissible product such that the corresponding
derivation is good in degree 0. In fact even the multiplication on a
point is not good in degree $ -1 $. Thus the following theorem is a kind of
surprise: \end{remark}
\begin{theorem} \label{th01.10}\myLabel{th01.10}\relax There exists a good universal multiplication. \end{theorem}
Unfortunately, the goodness condition does not specify this
multiplication uniquely, as the following proposition shows:
\begin{proposition} The set of good even derivations of grading 0 in $ {\mathcal L}\left(C_{\bullet}\left(X\right)\left[1\right]\right) $
acts on the set of good multiplications in $ X $. The derivation preserves the
subset of admissible products iff the block $ C_{\bullet}\left(X\right)\left[1\right]\to C_{\bullet}\left(X\right)\left[1\right] $ of the
derivation is scalar. It preserves the set of monotonic multiplications
if it is monotonic itself. \end{proposition}
In what follows we show that the set of good admissible
monotonic multiplications contains an open orbit for the action of
the corresponding group.
\subsection{Step of induction } The Theo.~\ref{th01.10} is an immediate corollary of
\begin{lemma} Suppose that there is a good admissible monotonic
multiplication $ D $
in the boundary $ \partial\sigma^{n} $ of $ n $-dimensional simplex, $ n\geq2 $. Then it can be extended to
the simplex $ \sigma^{n} $ preserving the above properties. All possible
extensions form an orbit of the specified above group. \end{lemma}
If we denote
\begin{equation}
D\left(\sigma^{n}\right)=\operatorname{ad}_{P_{1}}\cdot\sigma^{n}+P_{0},
\notag\end{equation}
we can reformulate the above lemma as
\begin{lemma} \label{lm01.20}\myLabel{lm01.20}\relax Suppose that there is a good admissible
monotonic multiplication $ D $
in the boundary $ \partial\sigma^{n} $ of $ n $-dimensional simplex, $ n\geq2 $. Then there exist two
Lie polynomials $ P_{0,1}\in{\mathcal L}\left(C_{\bullet}\left(\partial\sigma^{n}\right)\right) $ such that
\begin{equation}
D\left(P_{1}\right)-P_{1}^{2}=0
\label{equ4.20}\end{equation}\myLabel{equ4.20,}\relax
and
\begin{equation}
-\operatorname{ad}_{P_{1}}\cdot P_{0}+D\left(P_{0}\right)=0.
\label{equ4.25}\end{equation}\myLabel{equ4.25,}\relax
All solutions form an orbit of the specified above group. \end{lemma}
\subsection{Construction of the linear part $ P_{1} $ and the zero curvature equation
}\label{s4.5}\myLabel{s4.5}\relax Here we solve the Equ.~\eqref{equ4.20}. Note that the Lie algebra of
derivations of grading 0 acts on the set of solutions of Equ.~\eqref{equ4.20} as
\begin{equation}
V_{P_{0}}\left(P\right)=DP_{0}-\left[P,P_{0}\right],
\notag\end{equation}
thus the corresponding gauge group acts on this set.
\begin{proposition} Let $ V $ be a $ {\mathbb Z} $-graded vector space, $ D $ be a derivation in
$ {\mathcal L}^{\bullet}\left(V\right) $ of grading $ -1 $ such that $ D^{2}=0 $. Let $ \partial $ be the component of $ D $ of degree 0.
Suppose
that comultiplication in $ H_{\partial}\left({\mathcal L}^{\bullet}\left(V\right)\right) $ has a counity $ \varepsilon $ (necessarily of grading 1)
and that {\em reduced\/} $ \partial $-homology of $ V $ vanishes in negative
gradings. Then the zero curvature equation
\begin{equation}
DP-P^{2}=0,\qquad D\in{\mathcal L}^{\bullet}\left(V\right)
\notag\end{equation}
has a nontrivial solution of negative grading such that $ \varepsilon\left(P\right)=1 $, and all such
solutions form an orbit with respect to the natural gauge group action.
\end{proposition}
\begin{remark} Consider the case of an interval $ \sigma^{1} $. In this case $ V=C_{\bullet}\left(\partial\sigma^{1}\right)\left[1\right] $,
and the reduced homology of $ \partial\sigma^{1} $ are one-dimensional and concentrated in
grading 0. Thus the reduced $ \partial $-homology of $ V $ are concentrated in
non-positive
degree, which explains why the $ D $-image of $ \sigma^{1} $ {\em is not\/} linear in $ \sigma^{1} $.
However, beginning from a triangle, reduced homology of $ \partial\sigma^{n} $ are
concentrated in positive degrees, thus the analysis of this section is
applicable, what will lead us to the inductive construction. \end{remark}
Consider now the Equ.~\eqref{equ4.25}. Suppose that $ P $ is a solution of
Equ.~\eqref{equ4.20} in a differential Lie algebra $ {\mathcal L} $. Then the mapping
\begin{equation}
{\mathcal D}\colon {\mathcal L}\to{\mathcal L}\colon l\mapsto Dl-\left[P,l\right]
\notag\end{equation}
satisfies the condition $ {\mathcal D}^{2}=0 $, thus we can consider homology $ H_{{\mathcal D}}\left({\mathcal L}\right) $.
\begin{proposition} Suppose that $ {\mathcal L} $ has a decreasing filtration $ {\mathcal F}^{k}{\mathcal L} $
such that $ {\mathcal F}^{1}{\mathcal L}={\mathcal L} $, $ \left[{\mathcal F}^{k}{\mathcal L},{\mathcal F}^{l}{\mathcal L}\right]\subset{\mathcal F}^{k+l}{\mathcal L} $, and $ {\mathcal D}\left({\mathcal F}^{k}{\mathcal L}\right)\subset{\mathcal F}^{k}{\mathcal L} $. Let $ V={\mathcal L}/{\mathcal F}^{2}{\mathcal L} $, $ \partial $ be the
image of $ {\mathcal D} $ in $ V $. Then in the conditions of the previous
proposition the natural mapping
\begin{equation}
H_{{\mathcal D}}\left({\mathcal L}\right)\to H_{\partial}\left(V\right)
\notag\end{equation}
is an isomorphism.
\end{proposition}
Below we show that the last two propositons imply the Lemma
~\ref{lm01.20}.
\subsection{$ 2 $-dimensional case } Here we are going to construct a some
multiplication on a triangle $ \sigma^{2} $. This multiplication is not $ {\mathfrak S}_{3} $-invariant,
thus it would not give us a universal multiplication on $ 2 $-dimensional
triangulated topological spaces. However, we still get
an analogous operation: a universal multiplication on $ 2 $-dimensional
triangulated topological spaces with an ordering of $ 0 $-dimensional faces.
As we have seen above, it is sufficient to construct two elements
$ P_{1} $ and $ P_{0} $ of $ {\mathcal L}^{\bullet}\left(C\left(\sigma^{2}\right)\right) $ such that
\begin{equation}
DP_{1}=P_{1}^{2},\qquad DP_{0}=\left[P_{1},P_{0}\right].
\notag\end{equation}
However, $ D \left<0\right> = \left<0\right>\otimes\left<0\right> $, as we have seen in the section~\ref{s4.4},
thus we can put $ P_{1}=\left<0\right> $. To find the corresponding $ P_{0} $ recall the
fundamental property of multiplication on $ 1 $-dimensional spaces:
\begin{equation}
D e^{\left} = -\lefte^{\left}+e^{\left}\left.
\notag\end{equation}
Thus
\begin{equation}
D e^{\left<01\right>} e^{\left<12\right>} e^{\left<20\right>} = \left[e^{\left<01\right>} e^{\left<12\right>} e^{\left<20\right>},\left<0\right>\right],
\notag\end{equation}
what would give the solution of the equation on $ P_{0} $ if
$ e^{\left<01\right>} e^{\left<12\right>} e^{\left<20\right>} $
were a Lie power series. However,
$ e^{\left<01\right>} e^{\left<12\right>} e^{\left<20\right>} $ is a group element, thus
\begin{equation}
\log e^{\left<01\right>} e^{\left<12\right>} e^{\left<20\right>}
\notag\end{equation}
is a Lie power series, and it clearly satisfies the same equation.
So we have proved the following
\begin{theorem} Consider a $ 2 $-dimensional triangulated topological space
$ X $ with the set $ I $ of $ 1 $-dimensional faces. Suppose that for any face $ \left* $ of $ X $ a vertex of it $ I_{ij k}\in\left\{i,j,k\right\}\subset I $ is fixed. Then the following
differential in $ {\mathcal L}^{\bullet}\left(C\left(X\right)\left[1\right]\right) $ defines a structure of commutative
homotopically associative algebra in $ C^{*}\left(X\right) $:
\begin{align} D\left< i \right> & =\frac{1}{2}\left[\left< i \right>,\left< i \right>\right],
\notag\\
D\left< ij \right> & = \varphi\left(\operatorname{ad}_{\left}\right)\left< i \right>-\varphi\left(-\operatorname{ad}_{\left}\right)\left< j\right>,
\notag\\
D\left & = \left[i,\left< ij k \right>\right]+\log \left(e^{\left} e^{\left} e^{\left}\right).
\notag\end{align}
Here $ i,j,k\in I $, and in the third formula we suppose that $ I_{ij k}=i $, and the
orientation of $ \left< ij k \right> $ fixes the cyclic ordering $ i,j,k $ of vertices of
this triangle.
\end{theorem}
\subsection{An explicit construction } Here we enhance the above proof of existance
of a good universal multiplication and provide an explicit construction
of such a multiplication. However, in doing this we get a double
recurrence relation for the coefficients of this multiplication, thus the
result is not so explicit as in the cases of dimension 0, 1, and (in
non-symmetrical case) 2.
Since we do have already an explicit construction in the
$ 1 $-dimensional case, it is sufficient to provide an explicit formula for a
solution $ \delta_{k} $ of equation~\eqref{equ4.30}. Thus we need some inverse of the
operator $ \partial $ of boundary.
\begin{definition} Define an operator $ s^{\left(n\right)} $ of grading 1 in $ C\left(\partial\sigma^{n}\right) $ by the rule
\begin{equation}
s^{\left(n\right)}\left=\sum\Sb m\in\left\{0,\dots ,n\right\} \\ m\notin\left\{i_{0},\dots ,i_{l}\right\}\endSb \frac{1}{n-l} .
\notag\end{equation}
Define a $ 2 $-dimensional subspace $ V_{0} $ of $ C\left(\partial\sigma^{n}\right) $ as the space spanned by
\begin{equation}
\sum_{m=0}^{n}\left< m \right>\quad \text{and\quad }\sum_{m=0}^{n}\left(-1\right)^{m}\left<\text{ 01\dots \^m\dots }n \right>,
\notag\end{equation}
and the subspace $ V_{1} $ as the orthogonal complement to $ V_{0} $ in the metric on
$ C\left(\partial\sigma^{n}\right) $ in which all the faces are mutually orthogonal. Expand $ s^{\left(n\right)} $ to
$ T^{\bullet}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right) $ as an odd derivation, and denote it by the same symbol $ s^{\left(n\right)} $.
Let $ \bar{s}^{\left(n\right)} $ be
equal to $ \frac{s^{\left(n\right)}}{l} $ on monomials from $ T^{m}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right) $ with $ l $ occurences of
variables from $ V_{1}\left[1\right] $ and $ m-l $ occurences of variables from $ V_{0}\left[1\right] $ if $ l\not=0 $,
and be equal to 0 if $ l=0 $.
\end{definition}
\begin{theorem} Let the elements $ \delta_{k}^{\left(n\right)},\varepsilon_{k}^{\left(n\right)} \in T^{k}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right) $, $ k>0 $, and
the odd derivations $ D_{k}^{\left(n\right)} $, $ k\geq0 $, of $ T^{\bullet}\left(C\left(\sigma^{n}\right)\left[1\right]\right) $ be defined by the
following recurrence relations:
\begin{enumerate}
\item
All $ D_{k}^{\left(0\right)}\left< 0 \right> $ vanish but $ D_{2}^{\left(0\right)}\left< 0 \right>=\left< 0 \right> \otimes \left< 0 \right> $;
\item
Let
\begin{equation}
D_{k}^{\left(1\right)}\left< 01 \right>=B_{k}\left(\operatorname{ad}_{\left<01\right>}\right)^{k}\left< 0 \right>-B_{k}\left(-\operatorname{ad}_{\left<01\right>}\right)^{k}\left< 0 \right>,
\notag\end{equation}
Here $ \operatorname{ad}_{X}Y=X\otimes Y-Y\otimes X $, $ B_{k} $ are Bernouilli numbers;
\item
The value of $ D_{k}^{\left(n\right)} $ on a face $ \left< i_{0}i_{1}\dots i_{m} \right> $ of $ \sigma^{n} $, $ m $. Here we consider $ I=\left(i_{0}\dots i_{n}\right) $ as an element
of the symmetric group $ {\mathfrak S}_{n+1} $ with the corresponding action on
$ T^{\bullet}\left(C\left(\sigma^{n}\right)\left[1\right]\right) $.
\item
The value of $ D_{k}^{\left(n\right)} $ on the face $ \left< 01\dots n \right> $ of $ \sigma^{n} $ is
defined as
\begin{equation}
D_{k}^{\left(n\right)}\left<0\dots n\right>= \delta_{k}^{\left(n\right)}\left<0\dots n\right>+\left(-1\right)^{n}\left<0\dots n\right>\delta_{k}^{\left(n\right)}+\varepsilon_{k+1}^{\left(n\right)}.
\notag\end{equation}
Here we put $ \delta_{0}^{\left(n\right)}=0 $;
\item
Let for $ n\geq2 $
\begin{equation}
\delta_{1}^{\left(n\right)} = \frac{1}{n+1}\sum_{m=0}^{n}\left< m \right>,\quad \varepsilon_{1}^{\left(n\right)} = \frac{1}{n+1}\sum_{m=0}^{n}\left(-1\right)^{m}\left<\text{ 0\dots \^m\dots }n \right>.
\notag\end{equation}
\item
Let for $ n,k\geq2 $
\begin{align}\delta_{k}^{\left(n\right)} & =\bar{s}^{\left(n\right)}\left(\sum_{l=1}^{k-1}\delta_{k-l}^{\left(n\right)}\otimes\delta_{l}^{\left(n\right)}-\sum_{l=1}^{k-1}D_{k-l}^{\left(n\right)}\delta_{l}^{\left(n\right)}\right),
\notag\\
\varepsilon_{k}^{\left(n\right)} & =\bar{s}^{\left(n\right)}\left(\sum_{l=1}^{k-1}\varepsilon_{k-l}^{\left(n\right)}\otimes\delta_{l}^{\left(n\right)}-\left(-1\right)^{n}\sum_{l=1}^{k-1}\delta_{k-l}^{\left(n\right)}\otimes\varepsilon_{l}^{\left(n\right)}-\sum_{l=1}^{k-1}
D_{k-l}^{\left(n\right)} \varepsilon_{l}^{\left(n\right)}\right).
\notag\end{align}
\end{enumerate}
Then these relations determine these data uniquely, and the derivation $ D $
in $ T^{\bullet}\left(C\left(\sigma^{\infty}\right)\left[1\right]\right) $ defined as
\begin{equation}
D|_{T^{\bullet}\left(C\left(\sigma^{n}\right)\left[1\right]\right)}=\sum_{k}D_{k}^{\left(n\right)}
\notag\end{equation}
defines a good universal multiplication in $ C^{*}\left(\sigma^{n}\right) $.
\end{theorem}
\begin{proof} Evidently $ \bar{s}^{\left(n\right)} $ preserves the subspace
$ {\mathcal L}^{\bullet}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right)\subset T^{\bullet}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right) $, thus $ \delta_{k}^{\left(n\right)},\varepsilon_{k}^{\left(n\right)}\in{\mathcal L}^{k}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right) $ and $ D_{k}^{\left(n\right)} $
preserves the subspace $ {\mathcal L}^{\bullet}\left(C\left(\sigma^{n}\right)\left[1\right]\right)\subset T^{\bullet}\left(C\left(\sigma^{n}\right)\left[1\right]\right) $. The goodness and
admissibility, as well as $ {\mathfrak S}_{\infty} $-invariance
are evident by construction. Thus the only thing to prove is the identity
$ D^{2}=0 $.
This is evidently equivalent to
\begin{align}\partial\delta_{k}^{\left(n\right)} & =\sum_{l=1}^{k-1}\delta_{k-l}^{\left(n\right)}\otimes\delta_{l}^{\left(n\right)}-\sum_{l=1}^{k-1}D_{k-l}^{\left(n\right)}\delta_{l}^{\left(n\right)},
\notag\\
\partial\varepsilon_{k}^{\left(n\right)} & =\sum_{l=1}^{k-1}\varepsilon_{k-l}^{\left(n\right)}\otimes\delta_{l}^{\left(n\right)}-\left(-1\right)^{n}\sum_{l=1}^{k-1}\delta_{k-l}^{\left(n\right)}\otimes\varepsilon_{l}^{\left(n\right)}-\sum_{l=1}^{k-1}
D_{k-l}^{\left(n\right)} \varepsilon_{l}^{\left(n\right)}.
\notag\end{align}
However the commutator $ \left[\partial,s^{\left(n\right)}\right] $ in $ C\left(\partial\sigma^{n}\right) $ evidently coincides with the
projector on $ V_{1} $ along $ V_{0} $, hence the commutator $ \left[\partial,\bar{s}^{\left(n\right)}\right] $ in $ T^{\bullet}\left(C\left(\partial\sigma^{n}\right)\left[1\right]\right) $
coincides with the projector on the two-sided ideal generated by $ V_{1} $ along
$ T^{\bullet}V_{0} $. Thus it is sufficient to show that
\begin{equation}
\sum_{l=1}^{k-1}\delta_{k-l}^{\left(n\right)}\otimes\delta_{l}^{\left(n\right)}-\sum_{l=1}^{k-1}D_{k-l}^{\left(n\right)}\delta_{l}^{\left(n\right)}
\label{equ4.50}\end{equation}\myLabel{equ4.50,}\relax
and
\begin{equation}
\sum_{l=1}^{k-1}\varepsilon_{k-l}^{\left(n\right)}\otimes\delta_{l}^{\left(n\right)}-\left(-1\right)^{n}\sum_{l=1}^{k-1}\delta_{k-l}^{\left(n\right)}\otimes\varepsilon_{l}^{\left(n\right)}-\sum_{l=1}^{k-1} D_{k-l}^{\left(n\right)} \varepsilon_{l}^{\left(n\right)}
\label{equ4.55}\end{equation}\myLabel{equ4.55,}\relax
are closed elements in this ideal for $ k\geq2 $. On the other hand, we do
already
proved inductively that these elements are closed. Thus it is sufficient
to show that these elements are in the ideal. However, closed elements of
this ideal are exactly the exact elements. In the case $ k=2 $ we can
either easily check that
the above elements have trivial class in cohomology, or apply the
arguments from the sections~\ref{s4.5} and~\ref{s4.6}. This shows that $ \delta_{2}^{\left(n\right)} $ and
$ \varepsilon_{2}^{\left(n\right)} $ are in the ideal, since $ \bar{s} $ preserves this ideal. Thus to finish the
proof it is sufficient to show that $ D_{l}^{\left(n\right)} $, $ l\geq1 $, preserve the ideal.
Indeed, $ D_{l}^{\left(n\right)}\delta_{1}^{\left(n\right)}=0 $ for $ l>2 $, hence we can show by simultaneous
induction that~\eqref{equ4.50} and $ \delta_{k}^{\left(n\right)} $ are in the ideal if the above fact is
true. On the other hand, the only terms in~\eqref{equ4.55} not in the ideal are
\begin{equation}
\varepsilon_{k-1}^{\left(n\right)}\otimes\delta_{1}^{\left(n\right)}-\left(-1\right)^{n}\delta_{1}^{\left(n\right)}\otimes\varepsilon_{k-1}^{\left(n\right)}-\sum_{l=1}^{k-1}D_{k-l}^{\left(n\right)}\varepsilon_{l}^{\left(n\right)}.
\notag\end{equation}
We can prove by simultaneous induction that $ \varepsilon_{k}^{\left(n\right)} $, $ k\geq2 $, and~\eqref{equ4.55}
are in the ideal if we know that $ D_{k-1}^{\left(n\right)}\varepsilon_{1}^{\left(n\right)} $ is in the ideal and $ D_{l}^{\left(n\right)} $
preserves the ideal. However, any simplex of dimension between 0 and $ n-1 $ is
in the ideal, thus it is sufficient to consider components of
$ D_{k-1}^{\left(n\right)}\varepsilon_{1}^{\left(n\right)} $ that are monomials in simplices of dimension 1 and $ n-1 $.
However, $ \varepsilon_{1}^{\left(n\right)} $ is a sum of simplices of dimension $ n-1 $ and $ D_{k-1}^{\left(n\right)} $ is of
grading $ -1 $, thus for $ n>2 $ such monomials are products of $ 0 $-dimensional
simplex and
$ n-1 $-dimensional simplex, since $ D_{k-1}^{\left(n\right)}\varepsilon_{1}^{\left(n\right)} $ is linear in simplices of
dimension $ n-1 $. Hence for $ n>2 $ there is no such monomials if $ k>2 $, i.e., in
the case we consider!
The case $ n=2 $ demands a special consideration. In this case $ D_{k}^{\left(2\right)}\left(\left<
01 \right>\right) $ vanishes for odd $ k>1 $, since the Bernouilli number vanishes, and is
of a form
\begin{equation}
\left(\operatorname{ad}_{\left<01\right>}\right)^{k}\left(\left<0\right>-\left<1\right>\right)
\notag\end{equation}
in the case of even $ k $, thus it always is in the ideal for $ k>1 $, since
$ \left(\left<0\right>-\left<1\right>\right) $ is.
Let us prove that $ D_{l}^{\left(n\right)} $ preserves the ideal for any $ l $. It is
obvious in action on $ 0 $-dimensional simplices. We have already shown it
in the action on $ 1 $-dimensional simplices in the case $ l\not=1 $. It is obvious
for $ l=1 $ in the action on $ 1 $-dimensional simplices if $ n>2 $, and can be
easily checked in the case $ l=1 $, $ n=2 $. In the general case again, the
only interesting monomial are products of simplices of dimension 0 and
$ n-1 $, and there is no such monomials for the case $ l>1 $ and the action on
simplices of dimension bigger than 1. Thus the only two cases to check
are the action of $ D_{0}^{\left(n\right)} $ and $ D_{1}^{\left(n\right)} $ on simplices of dimension $ n-1 $. However,
we know the explicit formulae for these cases up to the term $ \varepsilon_{2}^{\left(n-1\right)} $ (which
we know is in the ideal), hence it is simple to check this fact in those
cases.\end{proof}
\section{Cosimplicial commutative and Lie algebras }
Here we want to show how a universal multiplication allows one to
construct a structure of commutative homotopically associative algebra on
a \v Cech complex of a cosimplicial commutative algebra over $ {\mathbb Q} $, and a
structure of
homotopical Lie algebra on a \v Cech complex of a cosimplicial Lie algebra
over $ {\mathbb Q} $.
However, we want to consider first the opposite direction: we want
to show that the construction of universal multiplication is a particular
case of the above construction in the case of cosimplicial commutative
algebra. Indeed, the crucial part of the above construction was a
construction of a structure of a commutative homotopically associative
algebra on the simplicial complex of a simplex. However, the simplicial
complex of a simplex is evidently a particular case of a \v Cech complex:
it is sufficient to consider a constant sheaf $ \underline{{\mathbb Q}} $ on this simplex
and the covering by neighborhoods of faces of codimension 1. Evidently
the sections of this sheaf on different intersections of these
neighborhoods form a cosimplicial commutative algebra.
In this section we are going to show this simplest case is already
sufficient to consider the general case, as well as the case of
cosimplicial Lie algebras. So let
\begin{equation}
D\left< 01\dots n \right>=F_{n}\left( \left<0\right>,\left<1\right>,\dots ,\left,\dots \right)
\notag\end{equation}
be representation of the image of the simplex $ \left< 01\dots n \right> $ under the
differential $ D $ that corresponds to universal multiplication. This means
that $ F_{n} $ is a Lie polynomial in faces of the simplex $ \left< 01\dots n \right> $. This
polynomial is $ {\mathfrak S}_{n+1} $-skewsymmetric, if we consider the natural action of
$ {\mathfrak S}_{n+1} $ on its arguments.
\subsection{Cosimplicial commutative algebras } Consider now cosimplicial
commutative (super)algebra $ {\mathfrak A} $, i.e., a
set of
commutative algebras $ A_{\left} $ parametrized by faces of the
simplex $ \sigma^{\infty} $ with
fixed mappings from the algebra that corresponds to a subface to the
algebra that corresponds to a bigger face. This mappings satisfy some
natural compatibility conditions. A natural example of this
situation is the case of a countable covering $ \left(U_{i}\right) $ of a topological space $ X $
with a sheaf of commutative algebras $ {\mathcal F} $. Associate to the face $ \left<
i_{0}i_{1}\dots i_{l} \right> $ of $ \sigma^{\infty} $ the algebra $ \Gamma\left(U_{i_{0}}\cap U_{i_{1}}\cap\dots \cap U_{i_{l}},{\mathcal F}\right) $, and let the
corresponding mappings are mappings of restriction with appropriate signs.
By definition the \v C{\em ech complex\/} $ C\left({\mathfrak A}\right) $ is the direct sum of these algebras
with an appropriate differential. We want to construct a structure of
commutative homotopically associative algebra on this direct sum.
Consider the dual space to the flipped-parity space
\begin{equation}
V=\left(\Pi C\left({\mathfrak A}\right)\right)^{*}= \bigoplus_{\left}\left(\Pi A_{\left}\right)^{*}.
\notag\end{equation}
We want
to construct a derivation $ {\mathfrak D} $ in $ {\mathcal L}^{\bullet}\left(V\right) $ that satisfies the equation $ {\mathfrak D}^{2}=0 $. In
particular, we need to know a mapping
\begin{equation}
{\mathfrak D}\colon \left(\Pi A_{\left<01\dots n\right>}\right)^{*} \to {\mathcal L}^{\bullet}\left(\bigoplus_{\left}\left(\Pi A_{\left}\right)^{*}\right).
\notag\end{equation}
Moreover, it is sufficient to construct the specified mappings.
In fact the mapping we will construct sends
$ \left(\Pi A_{\left<01\dots n\right>}\right)^{*} $ into a smaller space $ {\mathcal L}^{\bullet}\left(\bigoplus_{i_{j}\leq n^{\left}}\left(\Pi A_{\left}\right)^{*}\right) $.
To construct this mapping we use the Lie polynomial $ F_{n} $. Consider one
tensor monomial in this polynomial $ f_{1}\otimes\dots \otimes f_{M} $, where $ f_{j} $ are some faces of
$ \left< 01\dots n \right> $, $ f_{j}=\left< i_{0}^{j}\dots i_{l_{j}}^{j} \right> $. Associate to this monomial a mapping
\begin{equation}
\left(\Pi A_{\left<01\dots n\right>}\right)^{*} \to \bigotimes_{j=1}^{M}\left(\Pi A_{\left}\right)^{*}
\notag\end{equation}
constructed using the following rule: first, let $ m^{\left(M\right)} $ be the mapping of
iterated multiplication (correctly defined due to commutativity of
multiplication)
\begin{equation}
m^{\left(M\right)}\colon A_{\left<01\dots n\right>}^{\otimes M}\to A_{\left<01\dots n\right>},
\notag\end{equation}
and let $ \left(\Pi m^{\left(M\right)}\right)^{*} $ be the flipped-parity dual mapping to $ m^{\left(M\right)} $
\begin{equation}
\left(\Pi m^{\left(M\right)}\right)^{*}\colon \left(\Pi A_{\left<01\dots n\right>}\right)^{*} \to \left(\Pi A_{\left<01\dots n\right>}\right)^{*\otimes M}.
\notag\end{equation}
Now the natural mappings between algebras corresponding to included faces
of $ \sigma^{\infty} $ induce a mapping
\begin{equation}
\left(\Pi A_{\left<01\dots n\right>}\right)^{*\otimes M} \to \bigotimes_{j=1}^{M}\left(\Pi A_{\left}\right)^{*}.
\notag\end{equation}
Associate to monomial $ f_{1}\otimes\dots \otimes f_{M} $, the composition of these two mappings.
Now extend this mapping to polynomials linearly.
\begin{lemma} If a polynomial in faces of $ \left< 01\dots n \right> $ is in fact a Lie
polynomial, then the image of the corresponding mapping
\begin{equation}
\left(\Pi A_{\left<01\dots n\right>}\right)^{*} \to T^{\bullet}\left(\bigoplus_{\left}\left(\Pi A_{\left}\right)^{*}\right)
\notag\end{equation}
is contained in $ {\mathcal L}^{\bullet}\left(\bigoplus_{\left}\left(\Pi A_{\left}\right)^{*}\right) $.
\end{lemma}
Now define $ {\mathfrak D}|_{\left(\Pi A_{\left<01\dots n\right>}\right)^{*}} $ as the mapping that corresponds to the
Lie polynomial $ F_{n} $.
\begin{theorem} The defined above derivation of the \v Cech complex satisfies
the relation $ {\mathfrak D}^{2}=0 $. \end{theorem}
\begin{proof} Let us consider the case of a constant simplicial algebra
first. This is a simplicial algebra were all the faces morphisms are
identical. If the corresponding algebras are $ K $, then we can consider
them as $ 1 $-dimensional $ K $-algebras. However, in this case we have already
constructed the structure of commutative homotopically associative
$ K $-algebra on the \v Cech complex, and it is obviously compatible with the
structure of $ {\mathbb Q} $-algebra constructed
above. Hence the latter structure is also commutative homotopically
associative, thus $ {\mathfrak D}^{2}=0 $.
In general case consider $ {\mathfrak D}^{2}\left(\left< 01\dots n \right>\right) $. This is an element of the
free Lie algebra generated by a direct sum of faces coalgebras
$ \left(\Pi A_{\left}\right)^{*} $. However, the above construction of $ {\mathfrak D} $ shows that in
fact this element is an image of an element in the free Lie algebra
generated by the direct sum of several copies of $ \left(\Pi A_{\left<01\dots n\right>}\right)^{*} $. On the
other hand the latter element vanishes since it is equal to $ {\mathfrak D}^{2}\left(\left< 01\dots n
\right>\right) $ {\em in the constant cosimplicial algebra}.
Slightly rephrasing, consider $ {\mathfrak A}|_{\left<01\dots n\right>} $ and the constant
complicial algebra
\underline{A}$ _{\left<01\dots n\right>} $ with the stalk $ A_{\left<01\dots n\right>} $. There is a natural mapping
\begin{equation}
{\mathfrak A}|_{\left<01\dots n\right>} \to\text{ \underline{A}}_{\left<01\dots n\right>}
\notag\end{equation}
and a mapping of corresponding simplicial coalgebras
\begin{equation}
\Pi\underline{A}_{\left<01\dots n\right>}^{*} \to \Pi{\mathfrak A}|_{\left<01\dots n\right>}^{*}.
\notag\end{equation}
The structure of homotopical algebra in $ {\mathfrak A} $ is ultimately some copolylinear
operations for algebras in $ {\mathfrak A} $. Thus the knowledge of some identities on
compositions of these operations restricted on $ \Pi A_{\left<01\dots n\right>}^{*} $ can be
transferred from $ \Pi\underline{A}_{\left<01\dots n\right>}^{*} $ onto $ \Pi{\mathfrak A}|_{\left<01\dots n\right>}^{*} $.\end{proof}
\subsection{Cosimplicial Lie algebras } Let $ {\mathfrak A} $ be a cosimplicial Lie algebra, and
let $ C\left({\mathfrak A}\right) $ be the corresponding \v Cech complex. We want to construct a
structure of homotopical Lie algebra on $ C\left({\mathfrak A}\right) $. Thus we need to provide the
mappings
\begin{equation}
{\mathfrak D}\colon \left(\Pi A_{\left<01\dots n\right>}\right)^{*} \to S^{\bullet}\left(\bigoplus_{\left}\left(\Pi A_{\left}\right)^{*}\right).
\notag\end{equation}
The construction is somewhat simpler than in the commutative case.
The right-hand side consists of functions on
$ \bigoplus_{\left}\left(\Pi A_{\left}\right) $. Thus to a linear functional on
$ \Pi A_{\left<01\dots n\right>} $ we should associate a (non-linear function) on
$ \bigoplus_{\left}\left(\Pi A_{\left}\right) $. It is sufficient to consider a
corresponding {\em(nonlinear) mapping\/}
\begin{equation}
\bigoplus_{\left}\left(\Pi A_{\left}\right) \to \Pi A_{\left<01\dots n\right>}.
\notag\end{equation}
We will construct this mapping basing on Lie polynomials $ F_{n} $ again.
Again consider a case of a constant cosimplicial Lie algebra first. Given
an element of this Lie algebra for any face of $ \sigma^{\infty} $ we can {\em plug\/} these
elements into the corresponding slots in the Lie polynomial $ F_{n} $. Now
we can consider the result as an element of the same Lie algebra.
This gives us some mapping between the above spaces in the constant case.
To define the corresponding mapping in non-constant case we instead of
plugging the element into the polynomial should plug {\em its image in\/}
$ \Pi A_{\left<01\dots n\right>} $ {\em under the faces mapping}.
However, the non-intuitive nature of plugging in the odd case makes
us specify explicitely what it is. Consider one variable $ \left< i_{1}\dots i_{m} \right> $ in
the Lie monomial. Suppose that it appears $ l $ times. Now take $ l $ different
variables in $ \left< i_{1}\dots i_{m} \right> $, plug them into the corresponding places in
this monomial, and take a supersymmetrization. The result coincides
with usual plugging in
the case of even variables, but works generally. Thus is the operation of
plugging.
\begin{theorem} The resulting derivation $ {\mathfrak D} $ of $ S^{\bullet}\left(C\left({\mathfrak A}\right)^{*}\left[1\right]\right) $ satisfies the equation
$ {\mathfrak D}^{2}=0 $. \end{theorem}
\begin{proof} We have constructed a mapping that associates to any
derivation in $ {\mathcal L}^{\bullet}\left(C\left(\sigma^{\infty}\right)\left[1\right]\right) $ a derivation in $ S^{\bullet}\left(C\left({\mathfrak A}\right)^{*}\left[1\right]\right) $. It is sufficient
to show that this mapping is a homomorphism of Lie superalgebras
of (super)derivations.
However, in both cases derivations are uniquely determined by theirs
values on generators. Fix two derivations $ D_{1} $, $ D_{2} $ of
$ {\mathcal L}^{\bullet}\left(C\left(\sigma^{\infty}\right)\left[1\right]\right) $. Consider the corresponding derivations $ \widetilde{D}_{1} $, $ \widetilde{D}_{2} $ of $ S^{\bullet}\left(C\left({\mathfrak A}\right)^{*}\left[1\right]\right) $.
A straightforward calculation shows that the image of a generator of
$ S^{\bullet}\left(C\left({\mathfrak A}\right)^{*}\left[1\right]\right) $
under the action of $ \widetilde{D}_{1}\circ\widetilde{D}_{2} $ is the result of application of above procedure
to the mapping $ D_{1}\circ D_{2} $. Hence the commutator $ \left[\widetilde{D}_{1},\widetilde{D}_{2}\right] $ is related to the
commutator $ \left[D_{1},D_{2}\right] $ as above. However, both of them are derivations, thus
they are uniquely determined by their values on generators, hence $ \left[\widetilde{D}_{1},\widetilde{D}_{2}\right] $
is the image of $ \left[D_{1},D_{2}\right] $. \end{proof}
\section{Appendix on associativity and the condition $ D^{2}=0 $ }\label{h1}\myLabel{h1}\relax
We are going to provide here some motivations for the Def.~\ref{def1.10} and
the following definitions. To do this we consider several special cases
of these definitions.
\subsection{Associative algebras } Suppose that in conditions of the Def.
~\ref{def1.10} the derivation $ D $ send $ V=\left(\Pi A\right)^{*} $ to $ V\otimes V\subset T^{\bullet}V $. Any such mapping can
be extended to a derivation of a free algebra, thus the condition $ D^{2}=0 $
imposes some quadratic condition on this mapping. On the other hand, the
associativity condition is also quadratic.
If we begin from the other end, consider a product in a $ K $-algebra $ A $:
\begin{equation}
\left(a,b\right)\mapsto\operatorname{ab}.
\notag\end{equation}
This product determines the multiplication mapping and the
comultiplication mapping
\begin{equation}
m\colon A\otimes A\to\text{A and }m^{*}\colon A^{*}\to A^{*}\otimes A^{*}.
\notag\end{equation}
The usual condition of associativity can be written in terms of $ m $ as
\begin{equation}
m\circ\left(m\otimes\operatorname{id}-\operatorname{id}\otimes m\right)=0\in\operatorname{Mappings}\left(A\otimes A\otimes A\to A\right),
\notag\end{equation}
and in terms of $ m^{*} $ as
\begin{equation}
\left(m^{*}\otimes\operatorname{id}-\operatorname{id}\otimes m^{*}\right)\circ m^{*}=0\in\operatorname{Mappings}\left(A^{*}\to A^{*}\otimes A^{*}\otimes A^{*}\right).
\notag\end{equation}
The crucial observation now is that
\begin{equation}
m^{*}\otimes\operatorname{id}-\operatorname{id}\otimes m^{*}\colon a\otimes b\mapsto m^{*}\left(a\right)\otimes b-a\otimes m^{*}\left(b\right)\left(1.5\right)
\notag\end{equation}
looks very alike the right-hand side of the Leibniz condition on a
derivation:
\begin{equation}
D\left(a\otimes b\right)=D\left(a\right)\otimes b+a\otimes D\left(b\right)=\left(D\otimes\operatorname{id}+\operatorname{id}\otimes D\right)\left(a\otimes b\right)
\notag\end{equation}
in an algebra with a product $ \otimes $ (note the difference with the case of the
mapping $ m $, when $ m\otimes\operatorname{id}-\operatorname{id}\otimes m $ had three arguments!). To close the gap we
can switch on the most common kind of black magic in mathematics: the
superization. In fact the latter formula
becomes {\em exactly\/} like the formula~\eqref{equ1.5} if we consider the Leibniz
condition
\begin{equation}
D\left(a\otimes b\right)=D\left(a\right)\otimes b+\left(-1\right)^{\bar{D}\cdot\bar{a}}a\otimes D\left(b\right)
\notag\end{equation}
in the case when both $ D $ and $ a $ are odd (the
Leibniz condition is compatible with the convention
$ \left(\operatorname{id}\otimes D\right)\left(a\otimes b\right)\buildrel{\text{def}}\over{=}\left(-1\right)^{\bar{D}\cdot\bar{a}}\operatorname{id}\left(a\right)\otimes D\left(b\right) $). These convention and many others can
be found in any book on superalgebras, say \cite{Man88Gau}.
Thus we are forced to consider a $ K $-supermodule (i.e., a $ {\mathbb Z}_{2} $-graded
$ K $-module) $ V=V_{\bar{0}}\oplus V_{\bar{1}} $ such that $ V_{\bar{0}}=\left\{0\right\} $, $ V_{\bar{1}}=A^{*} $, and the algebra with a
multiplication $ \otimes $ that contains this space $ V $, i.e., the tensor algebra
$ T^{*}\left(V\right) $ of the module $ V $. Throughout this paper we consider the tensor
power without constants:
\begin{definition} For a $ K $-module $ W $ define the tensor power as
\begin{equation}
T^{\bullet}\left(W\right)=\bigoplus_{k=1}^{\infty}W^{\otimes k}.
\notag\end{equation}
\end{definition}
Now the comultiplication defines an odd mapping
\begin{equation}
m^{*}\colon V\to V\otimes V,
\notag\end{equation}
and $ m^{*}\otimes\operatorname{id}-\operatorname{id}\otimes m^{*} $ is the extension of this mapping on $ V\otimes V $ by the Leibniz
rule. By the last claim we mean the following construction: we can
consider $ m^{*} $ as a mapping
\begin{equation}
T^{\bullet}\left(V\right)\supset V\to T^{\bullet}\left(V\right)
\notag\end{equation}
defined on the set of generators of the algebra $ T^{\bullet}\left(V\right) $. However, by
the universality property of $ T^{\bullet}\left(V\right) $ we can extend any $ K $-linear mapping from $ V $
to an algebra $ B $ to a homomophism from $ T^{\bullet}\left(V\right) $ to $ B $. If we consider the case
$ B=T^{\bullet}\left(V\right) $ and the mapping $ \operatorname{id}+\varepsilon m^{*} $ (here $ \varepsilon $ is an infinitesimal parameter), we
can conclude that any mapping $ m^{*}\colon V\to T^{\bullet}\left(V\right) $ can be extended to a derivation
$ T^{\bullet}\left(V\right)\to T^{\bullet}\left(V\right) $.
Consider an odd derivation $ D $ of $ T^{\bullet}\left(V\right) $ that is the only extension
of the odd mapping $ m^{*}\colon V\to V\otimes V $. We conclude that the condition of the
associativity of $ m $ can be written as
\begin{equation}
D^{2}=0.
\notag\end{equation}
Indeed, it is clear that $ D^{2} $ maps $ V $ into $ V^{\otimes3} $, and that the vanishing of
the restriction $ D^{2}|_{V} $ is equivalent to the associativity of $ V $. However,
the derivations form a Lie superalgebra, therefore $ D^{2}=\frac{1}{2}\left[D,D\right] $ is also
a derivation. Thus $ D^{2} $ is uniquely determined by the restriction on $ V $, and
the condition $ D^{2}|_{V}=0 $ is equivalent to $ D^{2}=0 $.
\begin{remark} If the reader prefers to work with algebras instead of
coalgebras, he can recognize in the above space with a differential a
known object, a Hochshild complex. For a $ K $-algebra $ A $ it is the complex
\begin{equation}
\dots \to A^{\otimes k+2}\xrightarrow[]{m_{k+2}}A^{\otimes k+1}\xrightarrow[]{m_{k+1}}A^{\otimes k}\xrightarrow[]{m_{k}}A^{\otimes k-1}\to\dots \to A^{2}\xrightarrow[]{m_{2}}A
\notag\end{equation}
with $ m_{2}=m $ and
\begin{equation}
m_{k}=m\otimes\operatorname{id}^{\otimes k-2}-\operatorname{id}\otimes m\otimes\operatorname{id}^{\otimes k-3}+\dots \pm\operatorname{id}^{\otimes k-2}\otimes m.
\notag\end{equation}
Indeed, we can identify the dual space to $ T^{\bullet}\left(V\right) $ with $ T^{\bullet}\left(A\right) $ with
components of the even degree considered as even part of a superspace,
and visa versa. Thus the space $ T^{\bullet}\left(A\right) $ has
a natural grading and is equipped with a differential $ D^{*} $ of degree $ -1 $. The
changes of
signs in the above formula correspond to a fact that instead of the
module $ A^{*} $ we consider an odd supermodule $ V $.
\end{remark}
\subsection{Associative superalgebras } Return back to a case of a derivative $ D $ in
$ T^{\bullet}V $ that sends $ V $ to $ V\otimes V $. Define the (super)space $ A $ via $ \left(\Pi A\right)^{*}=V $ and
define a mapping $ m\colon A\otimes A\to $A via $ \left(\Pi m\right)^{*}=D|_{V} $. We have seen that if $ V $ is
purely odd, then $ m $ is an associative multiplication in $ A $. However, the
same is true in the general case also:
\begin{proposition} \label{prop1.01}\myLabel{prop1.01}\relax The even multiplication $ m $ in the $ K $-module $ A $ is
associative then and only then when $ D^{2}=0 $. \end{proposition}
Recall the sign convensions for mappings:
\begin{definition} The dual mapping to $ l $ is defined as
\begin{equation}
l^{*}\left(\alpha\right)\buildrel{\text{def}}\over{=}\left(-1\right)^{\bar{\alpha}\cdot\bar{l}}\alpha\circ l,\qquad l\colon W\to W',\quad l^{*}\colon W'{}^{*}\to W^{*},\quad \alpha\in W'{}^{*}.
\notag\end{equation}
\end{definition}
\begin{definition} For a supervector space $ W $ define the flipped-parity space $ \Pi W $
with an odd mapping $ \Pi\colon W\to\Pi W $ as
\begin{equation}
\left(\Pi W\right)_{\bar{0}}\buildrel{\text{def}}\over{=}W_{\bar{1}},\qquad \left(\Pi W\right)_{\bar{1}}\buildrel{\text{def}}\over{=}W_{\bar{0}},
\notag\end{equation}
the mapping $ \Pi $ is tautological. For a mapping $ l\colon W\to W' $ we define the
mapping $ \Pi l\colon \Pi W\to\Pi W' $ as $ \Pi l=\Pi\circ l\circ\Pi^{-1} $, i.e., $ \Pi l\left(\Pi w\right)\buildrel{\text{def}}\over{=}\Pi\left(l\left(w\right)\right) $.
Define a parity operator $ p $ on $ W $ via
\begin{equation}
p|_{W_{\bar{0}}}=\operatorname{id}_{W_{\bar{0}}},\qquad p|_{W_{\bar{1}}}=-\operatorname{id}_{W_{\bar{1}}}.
\notag\end{equation}
Define flipping on a tensor product $ W_{1}\otimes W_{2} $ as the action of $ \Pi_{W_{1}}\otimes\Pi_{W_{2}} $, and
flipping on a mapping, say $ f\colon W_{1}\otimes W_{2}\to W_{3} $, as
\begin{equation}
\Pi f\buildrel{\text{def}}\over{=}\Pi_{W_{3}}\circ f\circ\left(\Pi_{W_{1}}\otimes\Pi_{W_{2}}\right)^{-1},\qquad \Pi f\colon \Pi W_{1}\otimes\Pi W_{2}\to\Pi W_{3}.
\notag\end{equation}
\end{definition}
\begin{remark} It is clear that $ \Pi^{2}=0 $, but $ \left(\Pi\otimes\Pi\right)^{2}=-1 $. The action of $ \Pi f $ can be
written as
\begin{equation}
\Pi f\left(\Pi w_{1}\otimes\Pi w_{2}\right)=-\left(-1\right)^{\overline{\Pi w_{1}}}\Pi\left(f\left(w_{1}\otimes w_{2}\right)\right).
\notag\end{equation}
\end{remark}
The Prop.~\ref{prop1.01} follows immediately from the following lemma:
\begin{lemma} Consider a $ K $-supermodule $ A $ with an odd multiplication $ m:
A\otimes A\to A $. The multiplication $ m $ is associative iff the odd derivation $ D $ of
$ T^{\bullet}\left(A^{*}\right) $ uniquely defined by
\begin{equation}
D|_{A^{*}}=-\left(p\otimes\operatorname{id}\right)\circ m^{*}
\notag\end{equation}
satisfies the condition $ D^{2}|_{A^{*}}=0 $.
\end{lemma}
\begin{proof} Consider $ \alpha\in A^{*} $. Let $ D\alpha=\sum_{i}\beta_{i}\otimes\gamma_{i} $,
$ \beta,\gamma\in A^{*} $. Now for $ a,b\in A $
\begin{equation}
\begin{split}
\sum_{i}\left< \beta_{i},a \right>\left< \gamma_{i},b \right> & =\sum_{i}\left(-1\right)^{\bar{a}\cdot\bar{\gamma}_{i}}\left< \beta_{i}\otimes\gamma_{i},a\otimes b \right>=\left(-1\right)^{\bar{a}\cdot\bar{b}}\left< D\alpha,a\otimes b \right>
\\
& =-\left(-1\right)^{\bar{a}\cdot\bar{b}+\bar{\alpha}+\bar{a}}\left< \alpha,ab \right>=\left(-1\right)^{\bar{a}\cdot\bar{b}+\bar{b}}\left< \alpha,ab \right>
\end{split}
,
\notag\end{equation}
the term $ \bar{a} $ corresponds to $ p\otimes\operatorname{id} $, the term $ \bar{\alpha} $ to the definition of the dual
mapping, we used the fact that the parities of $ a $ and $ \alpha $, $ b $ and $ \beta $, and $ c $
and $ \gamma $ should coincide for the expression to be non-zero. Let $ D\beta_{i}=\sum_{j}\delta_{ij}\otimes\varepsilon_{ij} $,
$ D\gamma_{i}=\sum_{j}\zeta_{ij}\otimes\xi_{ij} $. Now
\begin{equation}
\begin{split}
\left< \left(\left(D\otimes\operatorname{id}\right)\circ D\right)\left(\alpha\right),a\otimes b\otimes c \right> & = \sum_{ij}\left< \delta_{ij}\otimes\varepsilon_{ij}\otimes\gamma_{i},a\otimes b\otimes c \right>
\\
& = \sum_{ij}\left(-1\right)^{\bar{a}\left(\bar{b}+\bar{c}\right)+\bar{b}\cdot\bar{c}}\left< \delta_{ij},a\right>\left<\varepsilon_{ij},b\right>\left<\gamma_{i},c \right>
\\
& = \left(-1\right)^{\bar{a}\left(\bar{b}+\bar{c}\right)+\bar{b}\cdot\bar{c}+\left(\bar{a}+1\right)\bar{b}+\left(\bar{a}+\bar{b}\right)\bar{c}}\left<\alpha,\left(ab\right)c \right>
\\
& = \left(-1\right)^{\bar{b}}\left<\alpha,\left(ab\right)c \right>
\end{split}
\notag\end{equation}
On the other hand,
\begin{equation}
\begin{split}
\left< \left(\left(\operatorname{id}\otimes D\right)\circ D\right)\left(\alpha\right),a\otimes b\otimes c \right> & = \sum_{ij}\left(-1\right)^{\bar{a}}\left< \beta_{i}\otimes\zeta_{ij}\otimes\xi_{ij},a\otimes b\otimes c \right>
\\
& = \sum_{ij}\left(-1\right)^{\bar{a}+\bar{a}\left(\bar{b}+\bar{c}\right)+\bar{b}\cdot\bar{c}}\left< \beta_{ij},a\right>\left<\zeta_{ij},b\right>\left<\xi_{ij},c \right>
\\
& = \left(-1\right)^{\bar{a}+\bar{a}\left(\bar{b}+\bar{c}\right)+\bar{b}\cdot\bar{c}+\left(\bar{b}+1\right)\bar{c}+\left(\bar{a}+1\right)\left(\bar{b}+\bar{c}+1\right)}\left<\alpha,a\left(bc\right) \right>
\\
& = -\left(-1\right)^{\bar{b}}\left<\alpha,a\left(bc\right) \right>
\end{split}
\notag\end{equation}
Thus $ \left(D\otimes\operatorname{id}+\operatorname{id}\otimes D\right)\circ D=0 $ is equivalent to $ \left(ab\right)c-a\left(bc\right)=0 $.\end{proof}
\subsection{Differential associative algebras } Consider now the case when the
derivation $ D $ of $ T^{\bullet}V $ send $ V $ to $ V\oplus V\otimes V $. It has now two blocks $ V\to V $ and
$ V\to V\otimes V $. Again let $ V=\left(\Pi A\right)^{*} $, define the mapping $ d\colon A\to $A and $ m\colon A\otimes A\to $A such
that the above blocks are $ \left(\Pi d\right)^{*} $ and $ \left(\Pi m\right)^{*} $ correspondingly. Now the
condition $ D^{2}=0 $ determines some quadratic relations between $ d $ and $ m $. Let
us show that they are the standard relations for an associative
differential algebra.
Indeed, $ D^{2}|_{V} $ has three components, sending $ V $ correspondingly to $ V $,
$ V\otimes V $ and $ V\otimes V\otimes V $. From the discussion in the previous section it is clear
that the vanishing of the third component is equivalent to associativity
of $ m $. Vanishing of the first is obviously equivalent to $ d^{2}=0 $. A simple
check shows that the second component gives the Leibniz relation between
$ d $ and $ m $.
Thus we have proved the following
\begin{proposition} Consider a $ K $-supermodule $ V $ and an odd derivation $ D $ of $ T^{\bullet}\left(V\right) $
such that
\begin{equation}
D\left(V\right)\subset V\oplus\left(V\otimes V\right).
\notag\end{equation}
Let $ D|_{V}=D_{0}+D_{1} $, where $ D_{0}\colon V\to V $ and $ D_{1}\colon V\to V\otimes V $. Let $ V=\left(\Pi A\right)^{*} $. Define $ d\colon A\to $A
and $ m\colon A\otimes A\to $A via $ D_{0}=\left(\Pi d\right)^{*} $ and $ D_{1}=\left(\Pi m\right)^{*} $. Then the condition
$ D^{2}=0 $ is
equivalent to the condition $ d^{2}=0 $, and to the Leibniz and the
associativity conditions on the multiplication $ m $ in the complex $ \left(A,d\right) $. \end{proposition}
\subsection{(Weakly) homotopically associative algebras } In the previous
section we considered the case of a multiplication in a differential
module that is compatible with the differential {\em and\/} associative. However,
if the multiplication satisfies the first property only, it induces some
multiplication in the homologies. In many important cases the latter
multiplication is associative, but the former is not.
Formally speaking, the multiplication mapping in the differential
module
\begin{equation}
m\colon A\otimes A\to A
\notag\end{equation}
induces a multiplication mapping
\begin{equation}
\bar{m}\colon H\left(A\right)\otimes H\left(A\right)=H\left(A\otimes A\right)\to H\left(A\right)
\notag\end{equation}
in homology and the not-associativity mapping in the differential module
\begin{equation}
m\circ\left(m\otimes\operatorname{id}-\operatorname{id}\otimes m\right)\colon A\otimes A\otimes A\to A
\notag\end{equation}
induces the corresponding not-associativity mapping
\begin{equation}
\bar{m}\circ\left(\bar{m}\otimes\operatorname{id}-\operatorname{id}\otimes\bar{m} \right)\colon H\left(A\right)\otimes H\left(A\right)\otimes H\left(A\right)\to H\left(A\right)
\notag\end{equation}
in homology. The above case corresponds to the condition
\begin{equation}
\bar{m}\circ\left(\bar{m}\otimes\operatorname{id}-\operatorname{id}\otimes\bar{m} \right)=0,
\notag\end{equation}
and we want to consider a condition on the multiplication in the
differential module $ A $ that insures that the multiplication in homology
is associative.
In the simplest and most important case the differential module $ A $
is in fact a complex, i.e., it is equipped with a $ {\mathbb Z}_{\pm} $-grading that is
compatible with the $ {\mathbb Z}_{2} $-grading, the differential is of degree $ -1 $, and the
graded components are free $ K $-modules. In this case $ A\otimes A $ and $ A\otimes A\otimes A $ also
have the same properties, thus any mapping
\begin{equation}
f\colon A\otimes A\otimes A \to A
\notag\end{equation}
that induces a trivial mapping in homologies allows a {\em homotopy\/} $ s $, i.e., a
mapping $ s\colon A\otimes A\otimes A\to A $ such that
\begin{equation}
f=s\circ d_{A\otimes A\otimes A}+d_{A}\circ s.
\notag\end{equation}
It is easy to see that a mapping $ s\circ d_{A\otimes A\otimes A}+d_{A}\circ s $ is compatible with a
differentials for any mapping $ s\colon A\otimes A\otimes A\to $A, and $ s\circ d_{A\otimes A\otimes A}+d_{A}\circ s $ always
induces a trivial mapping in homology. Thus we can introduce the
following definition
(that is supported not only by the vague discussion above, but also by
ideas from topology):
\begin{definition} Consider a $ K $-supermodule $ A $ and a mapping of multiplication
\begin{equation}
m\colon A\otimes A\to A
\notag\end{equation}
that is compatible with differentials (i.e., is a mapping of complexes).
We say that the multiplication $ m $ is {\em(weakly) homotopically associative\/} if
there exists a mapping $ s\colon A\otimes A\otimes A\to A $ such that
\begin{equation}
m\circ\left(m\otimes\operatorname{id}-\operatorname{id}\otimes m\right) = s\circ d_{A\otimes A\otimes A}+d_{A}\circ s.
\label{equ1.10}\end{equation}\myLabel{equ1.10,}\relax
We call such mapping $ s $ a {\em homotopy for the associativity\/} of $ m $.
\end{definition}
Let us consider the translation of this definition to the language
of coalgebras. First of all, it is easy to see that the parity of $ s $ is
always odd independently of the parity of $ m $. As in the previous sections,
if the multiplication $ m $ is even, we can consider the flipped parity
module $ \Pi A $ with the odd multiplication $ \Pi m $, i.e., it is sufficient to
consider the case of odd $ m $. However, in this case all three mappings $ d $, $ m $
and $ s $ are odd, therefore it is reasonable to consider the {\em sum\/} of these
mappings, or, better, the sum of dual mappings.
Thus introduce a $ K $-supermodule $ V $ that is $ \left(\Pi A\right)^{*} $ in the case of even $ m $
and $ A^{*} $ in the case of odd $ m $ and the odd mapping
\begin{equation}
\widetilde{D}=\Pi d^{*}+\Pi m^{*}+\Pi s^{*}\quad \text{or\quad }\widetilde{D}=d^{*}+m^{*}+s^{*}\colon V\to V\oplus\left(V\otimes V\right)\oplus\left(V\otimes V\otimes V\right).
\notag\end{equation}
If $ m $ is strictly associative, we can take $ s=0 $, and we know that (after
some correction of signs) $ D $ extends to $ T^{\bullet}\left(V\right) $ as a derivation with
condition $ D^{2}=0 $. Not very surprising now, we can introduce the analogous
sign corrections for $ s $ in the case $ s\not=0 $ such that the condition~\eqref{equ1.10}
becomes some vanishing condition on the corresponding derivation $ D^{2} $.
If we define $ D $ (we write down only the even
case) as a derivation $ D\colon T^{\bullet}\left(V\right)\to T^{\bullet}\left(V\right) $,
\begin{equation}
D|_{V}=\Pi d^{*}+\Pi m^{*}+\Pi s^{*}\colon V\to V\oplus\left(V\otimes V\right)\oplus\left(V\otimes V\otimes V\right),
\notag\end{equation}
then $ D^{2}=\frac{1}{2}\left[D,D\right] $ is also a derivation of $ T^{\bullet}\left(V\right) $, and
\begin{equation}
D^{2}|_{V}\colon V\to V\oplus\left(V\otimes V\right)\oplus\left(V\otimes V\otimes V\right)\oplus\left(V\otimes V\otimes V\otimes V\right)\oplus\left(V\otimes V\otimes V\otimes V\otimes V\right).
\notag\end{equation}
It turns out that the condition~\eqref{equ1.10} together with the condition
$ d^{2}=0 $ and the condition of compatibility between $ m $ and $ d $ can be written
down as vanishing of the first three components of $ D^{2}|_{V} $. We can write
down this condition as the vanishing of
\begin{equation}
D^{2}|_{V}/_{T^{\geq4}\left(V\right)}\colon V\to T^{\bullet}\left(V\right)/T^{\geq4}\left(V\right).
\notag\end{equation}
Moreover, if we add to $ D $ some additional components ``of higher
degree''
\begin{equation}
\delta D|_{V}\colon V\to T^{\geq4}\left(V\right),
\notag\end{equation}
then $ D'=D+\delta D $ satisfies the same condition $ D'{}^{2}|_{V}/_{T^{\geq4}\left(V\right)}=0 $, thus we can
drop the condition that $ D $ maps $ V $ into the first three components of $ T^{\bullet}\left(V\right) $
and consider $ \Pi d^{*} $, $ \Pi m^{*} $ and $ \Pi s^{*} $ as the first three components
of an arbitrary odd derivation $ D $ of $ T^{\bullet}\left(V\right) $:
\begin{theorem} Consider a $ K $-supermodule $ V $ and an odd derivation $ D $ of
$ T^{\bullet}\left(V\right) $. Let
\begin{equation}
D|_{V}= \bigoplus_{i=0}^{\infty}D_{i}\colon V\to \bigoplus_{i=0}^{\infty}V^{\otimes i+1}
\notag\end{equation}
be the decomposition of $ D|_{V} $ into a sum of homogeneous components. Let
$ V=\left(\Pi A\right)^{*} $, define $ d\colon A\to $A, $ m\colon A\otimes A\to $A and $ s\colon A\otimes A\otimes A\to $A via $ D_{0}=\left(\Pi d\right)^{*} $,
$ D_{1}=\left(\Pi m\right)^{*} $
and $ D_{2}=\left(\Pi s\right)^{*} $. Consider the derivation $ D^{2} $ of $ T^{\bullet}\left(V\right) $. Then the
condition
that $ D^{2}|_{V} $ sends $ V $ into $ \bigoplus_{i=4}^{\infty}V^{\otimes i} $ (instead of $ \bigoplus_{i=1}^{\infty}V^{\otimes i} $) is equivalent to
the condition $ d^{2}=0 $, the condition of compatibility of $ m $ and $ d $, and the
fact that $ s $ is a homotopy for associativity of the multiplication $ m $.
\end{theorem}
\begin{proof} The calculations in the previous sections show that the
condition $ d^{2}=0 $ is equivalent to the vanishing of the $ \left(D^{2}\right)_{0} $ (we use
the same notation as with $ D $, $ D^{2}|_{V}=\bigoplus \left(D^{2}\right)_{i} $, $ \left(D^{2}\right)_{i} $ sends $ V $ into $ V^{\otimes i+1} $),
and the
condition of compatibility between $ d $ and $ m $ is equivalent to the vanishing
of $ \left(D^{2}\right)_{1} $. Now we need to show that the condition $ \left(D^{2}\right)_{2}=0 $ is equivalent to
the equation~\eqref{equ1.10}. This condition can be written as
\begin{equation}
D_{0}\circ D_{2}+D_{1}\circ D_{1}+D_{2}\circ D_{0}=0.
\notag\end{equation}
We know already that $ D_{1}\circ D_{1} $ is the left-hand side of the~\eqref{equ1.10}. The
other two terms correspond evidently to two terms in the right-hand side.\end{proof}
\begin{remark} This theorem suggests that we can consider the tensor algebra
as a direct sum with summands of decreasing weights. Indeed, since the
terms in $ T^{\geq4}\left(V\right) $ are considered negligible, we can consider as well the
completion $ \widehat{T}^{\bullet}\left(V\right) $ of $ T^{\bullet}\left(V\right) $ with respect to the system $ \left\{T^{\geq k}\left(V\right)\right\} $, $ k\to\infty $, of
neighborhoods of zero:
\begin{equation}
\widehat{T}^{\bullet}\left(V\right) =\text{ proj }\lim _{k\to\infty} T^{\bullet}\left(V\right)/T^{\geq k}\left(V\right).
\notag\end{equation}
In this approach we can denote $ T^{\geq k}\left(V\right) $ as $ O\left(V^{\otimes k}\right) $, and denote the set
of mapping sending $ T^{\geq m}\left(V\right) $ into $ T^{\geq m+k}\left(V\right) $ for any $ m $ by the same symbol. Now
we can write the condition
\begin{equation}
D^{2}|_{V}\colon V \to\bigoplus_{i=4}^{\infty}V^{\otimes i}
\notag\end{equation}
as $ D^{2}=O\left(V^{3}\right) $.
On the other hand we can consider the natural grading on $ T^{\bullet}\left(V\right) $, $ \deg
V^{\otimes k}=k $. This grading induces the grading on the set of derivations as
well. Since any derivation is uniquely determined by its restriction on
the set of generators $ V $, the latter grading is necessary non-negative.
Thus the restriction in the above theorem is the vanishing of the first three
graded components of $ D^{2} $.
Yet another point of view lies around the topological approach
disseminated by M.~Kontsevich: it is much better to work with {\em discrete\/}
vector spaces (i.e., with inductive limits of finite-dimensional spaces)
than with {\em compact\/} vector spaces (i.e., with projective limits
of finite-dimensional spaces). Thus if we are forced to work with, say, a
space of power series (as $ \widehat{T}^{\bullet}\left(V\right) $ is), then it is better to consider the
dual vector space of polynomials (as $ \left(\widehat{T}^{\bullet}\left(V\right)\right)^{*}= T^{\bullet}\left(V^{*}\right) $ is). Since $ \widehat{T}^{\bullet}\left(V\right) $
has a natural structure of a free associative algebra, the dual
space is a (co)free coassociative coalgebra, and the mapping $ D^{*} $ (and
$ D^{*2} $) is a coderivation.
On the latter language a structure of a weakly homotopically associative
algebra on a (super)vector space $ A $ is the same as a coderivation $ D^{*} $ of the
(co)free coassociative coalgebra $ T^{\bullet}\left(\Pi A\right) $ (co)generated by $ \Pi A $ such that
the coderivation $ D^{*2} $ lowers the degree at least by three.
\end{remark}
\subsection{(Strictly) homotopically associative algebras } We have seen that that
a strictly homotopically associative
$ K $-algebra
is automatically a weakly homotopically associative $ K $-algebra. To
construct the corresponding differential, multiplication, and the
homotopy for associativity of the multiplication we should take the first
three graded components of $ D $, flip the parity and take the duals. Thus
the condition of the definition~\ref{def1.10} is an amplification of the
condition of weak homotopical associativity.
To understand what is this amplification let us consider the first
graded component of $ D^{2} $ that is not used in the definition of weak
associativity. It is
\begin{equation}
D_{0}\circ D_{3}+D_{1}\circ D_{2}+D_{2}\circ D_{1}+D_{3}\circ D_{0}.
\notag\end{equation}
Thus the vanishing condition for this component is equivalent to the fact
that the commutator of $ D_{1} $ and $ D_{2} $ is homotopic to 0, and $ D_{3} $ supplies the
corresponding homotopy. Thus the cocommutator of $ m $ and $ s $ is also
homotopically zero. The condition that the cocommutator of $ m $ with itself
is homotopically zero is an analogue of the associativity condition for
the corresponding multiplication in homology. Hence we can consider the
condition $ \left(D^{2}\right)_{3}=0 $ as some analogue of homotopical associativity for the
{\em ternary product\/} $ s $.
Denote the homotopy for this analogue by $ t $. Then $ \left(D^{2}\right)_{4}=0 $ is an
analogue of condition of homotopical associativity for $ t $ and so on. Thus
we can interpret the condition of strong homotopical associativity as a
condition of homotopical associativity for all $ n $-ary products with the
$ \left(n+1\right) $-ary product being the homotopy for associativity of the $ n $-ary
product.
\subsection{Commutative algebras }\label{s2.1}\myLabel{s2.1}\relax Our principal definitions of different kinds of
associativity deal with coalgebra structures and not with algebra
structures. Thus the question how to formulate the commutativity
condition in the same terms is not absolutely obvious.
The commutativity condition means that
\begin{equation}
\left[x,y\right]\buildrel{\text{def}}\over{=}xy-\left(-1\right)^{\bar{x}\cdot\bar{y}}yx=0\qquad \forall x,y\in A.
\notag\end{equation}
Thus the image of comultiplication $ m^{*}\colon A^{*}\to A^{*}\otimes A^{*} $ is orthogonal to
tensors of the form $ x\otimes y-\left(-1\right)^{\bar{x}\cdot\bar{y}}y\otimes x $, $ x,y\in A $, and
the image of the flipped-parity comultiplication $ \left(\Pi m\right)^{*}:
\left(\Pi A\right)^{*}\to\left(\Pi A\right)^{*}\otimes\left(\Pi A\right)^{*} $ is orthogonal to tensors of the form
\begin{equation}
-\left(-1\right)^{\bar{x}}x\otimes y+\left(-1\right)^{\bar{y}}\left(-1\right)^{\left(\bar{x}+1\right)\cdot\left(\bar{y}+1\right)}y\otimes x,\qquad x,y\in\Pi A,
\notag\end{equation}
or to
\begin{equation}
x\otimes y+\left(-1\right)^{\bar{x}\cdot\bar{y}}y\otimes x\text{, }x,y\in\left(\Pi A\right)^{*}.
\notag\end{equation}
Let us try to write down a general
form for an element $ \omega $ of $ \left(\Pi A\right)^{*}\otimes\left(\Pi A\right)^{*} $ that satisfies this condition. Since
\begin{equation}
\begin{split}
\left< \alpha\otimes\beta,x\otimes y+\left(-1\right)^{\bar{x}\cdot\bar{y}}y\otimes x \right> & = \left(-1\right)^{\bar{\beta}\cdot\bar{x}}\left< \alpha,x \right>\left< \beta,y \right>+\left(-1\right)^{\bar{\beta}\cdot\bar{y}+\bar{x}\cdot\bar{y}}\left<
\alpha,y \right>\left< \beta,x \right>
\\
& =\left< \alpha\otimes\beta+\left(-1\right)^{\bar{\beta}\cdot\bar{y}+\bar{x}\cdot\bar{y}+\bar{\alpha}\cdot\bar{x}}\beta\otimes\alpha,x\otimes y \right>,
\end{split}
\label{equ2.3}\end{equation}\myLabel{equ2.3,}\relax
we conclude that such an element satisfies the condition
\begin{equation}
\omega+\left(\Pi P\right)\omega=0.
\notag\end{equation}
Here $ P $ is the mapping
\begin{equation}
P\colon A^{*}\otimes A^{*}\to A^{*}\otimes A^{*}\colon \alpha\otimes\beta\to\left(-1\right)^{\bar{\alpha}\cdot\bar{\beta}}\beta\otimes\alpha.
\notag\end{equation}
Indeed, in the calculation of
$ \left(-1\right)^{\bar{\beta}\cdot\bar{y}+\bar{x}\cdot\bar{y}+\bar{\alpha}\cdot\bar{x}}
$ in the formula~\eqref{equ2.3} we can suppose that $ \bar{\alpha}=\bar{y} $, $ \bar{\beta}=\bar{x} $, in the other case
the paring $ \left<\beta\otimes\alpha,x\otimes y \right> $ would be zero automatically. Thus we pair $ x\otimes y $
with
\begin{equation}
\alpha\otimes\beta+\left(-1\right)^{\bar{\alpha}\cdot\bar{\beta}}\beta\otimes\alpha.
\notag\end{equation}
We conclude that $ \omega\in\operatorname{Im}\left(1-P\right) $, or $ \omega $ is in the image of the
tensor commutator mapping
\begin{equation}
\left[\alpha,\beta\right]\buildrel{\text{def}}\over{=}\alpha\otimes\beta-\left(-1\right)^{\bar{\alpha}\cdot\bar{\beta}}\beta\otimes\alpha,\qquad \alpha,\beta\in\left(\Pi A\right)^{*}.
\notag\end{equation}
Thus we have proved the
\begin{proposition} Consider a multiplication $ m $ in a $ K $-module $ A $. Let
$ V=\left(\Pi A\right)^{*} $. This multiplication is commutative iff the flipped-parity
comultiplication mapping $ \left(\Pi m\right)^{*}\colon V\to V\otimes V $ sends $ V $ into the
linear span $ \left[V,V\right] $ of commutators of elements of $ V $ in the tensor algebra
\begin{equation}
\left[v_{1},v_{2}\right]\buildrel{\text{def}}\over{=}v_{1}\otimes v_{2}-\left(-1\right)^{\bar{v}_{1}\cdot\bar{v}_{2}}v_{2}\otimes v_{1}.
\notag\end{equation}
\end{proposition}
\subsection{Commutative differential algebras } Let $ A $ be a $ K $-(super)module with
an odd differential $ d $ and multiplication $ m $. In the previous section we
found the condition on $ \left(\Pi m\right)^{*} $ that corresponds to commutativity of $ m $.
However, in the section~\ref{h1} we have seen that it is very important to
consider the sum $ \left(\Pi d\right)^{*}+\left(\Pi m\right)^{*} $.
\begin{proposition} Consider a $ K $-supermodule $ V $ and an odd derivation $ D $ in
$ T^{\bullet}\left(V\right) $. Let $ V=\left(\Pi A\right)^{*} $. Suppose that $ D $ sends $ V $ into $ V\oplus\left(V\otimes V\right) $, and $ D|_{V}=D_{0}+D_{1} $
is the corresponding decomposition of $ D|_{V} $. Define
$ d\colon A\to $A and $ m\colon A\otimes A\to $A via $ D_{0}=\left(\Pi d\right)^{*} $, $ D_{1}=\left(\Pi m\right)^{*} $. Then the
conditions of
commutativity and associativity of the product $ m $, the
compatibility
condition for $ m $ and $ d $, and the condition $ d^{2}=0 $ are equivalent to $ D^{2}=0 $
together with condition that $ D $ preserves the set of Lie polynomials in
$ T^{\bullet}\left(V\right) $. \end{proposition}
Recall that Lie polynomials are expressions involving the generators
and the operation of commutation.
\begin{proof} We know that all the conditions on $ m $ and $ d $ but commutativity
are equivalent to $ D^{2}=0 $. On the other side, the commutativity is
equivalent to the fact that $ D_{1} $ sends $ V $ into $ \left[V,V\right] $. However, since $ D_{0} $
sends $ V $ to $ V $, i.e., to the space of generators, $ D $ sends $ V $ to Lie
polynomials iff $ D_{1} $ does so. Moreover, since $ D $ is a derivation, if $ D $
sends $ V $ to the set of Lie polynomials, then it sends any Lie polynomial
to a Lie polynomial. \end{proof}
\begin{remark} Thus we can express the conditions of this proposition in
several different ways. First of all we can change the additional
condition to the condition that $ D $ sends $ V $ to Lie polynomials. Secondly,
the subset of Lie polynomials in $ T^{\bullet}\left(V\right) $ forms a free Lie algebra $ {\mathcal L}^{\bullet}\left(V\right) $ with
generators in $ V $. Thus $ D $ is a derivative in $ {\mathcal L}^{\bullet}\left(V\right) $ as well. Thirdly, since
one can write $ {\mathcal L}^{\bullet}\left(V\right) $ as a set of elements in $ T^{\bullet}\left(V\right) $ that satisfy equation
$ \Delta x=1\otimes x+x\otimes1 $, here $ \Delta $ is the standard comultiplication. Thus we $ D $ is a
derivative in $ T^{\bullet}\left(V\right) $ that preserves $ \Delta $ (i.e., is a coderivative as well).
Moreover, the inclusion $ {\mathcal L}^{\bullet}\left(V\right)\hookrightarrow T^{\bullet}\left(V\right) $ induces a mapping of the
universal enveloping algebra $ U\left({\mathcal L}^{\bullet}\left(V\right)\right)\to T^{\bullet}\left(V\right) $, and this mapping is an
isomorphism. Hence any derivation of $ {\mathcal L}^{\bullet}\left(V\right) $ induces a unique derivation
of $ T^{\bullet}\left(V\right) $. Hence we can reformulate the conditions of the proposition as \end{remark}
\begin{corollary} Consider a $ K $-supermodule $ V $ and an odd derivation $ D $ in
$ {\mathcal L}^{\bullet}\left(V\right) $. Let $ V=\left(\Pi A\right)^{*} $. Suppose that $ D $ sends $ V $ into $ V\oplus\left[V,V\right] $, $ D=D_{0}+D_{1} $, and
$ D|_{V}=D_{0}+D_{1} $ is the corresponding decomposition of $ D|_{V} $. Define $ d\colon A\to $A and
$ m\colon A\otimes A\to $A via $ D_{0}=\left(\Pi d\right)^{*} $, $ D_{1}=\left(\Pi m\right)^{*} $. Then the conditions of
commutativity and associativity of the product $ m $, the compatibility
condition for $ m $ and $ d $, and the condition $ d^{2}=0 $ are equivalent to $ D^{2}=0 $. \end{corollary}
\subsection{(Strictly) commutative (strictly) homotopically associative algebras }
We can see now that the definition of stricly commutative homotopically
associative algebra is the amplification of the condition that the binary
multiplication in the homotopically associative algebra is commutative.
This condition poses additional restrictions not only on the binary
product, but also on all the other $ n $-ary products. However, it
looks much harder to write these additional conditions in terms of the
multiplication itself.
\begin{remark} It seems that the relevant definition was introduced first by
V.~Drinfeld in his letter to V.~Schechtman \cite{DriLetter}. \end{remark}
\subsection{$ {\mathbb Z} $-graded variant } Since we will work with the tensor powers of
graded spaces, we
should avoid a natural confusion with two possible gradings on the tensor
power, and we need to work with both of them. So in the following we
use the word ``degree'' to denote the number of variables in a monomial,
and the word ``grading'' to denote the sum of gradings of these
variables. Contrary to the usual conventions, we are
forced to use
the same two words ``degree'' and ``grading'' not only for components of vector
spaces, but
also for linear mappings between such spaces. Thus we can, say, consider
a component of degree 1 of a mapping of grading $ -1 $.
It is easy to see that the only modification in the definitions
is in the
definition of the flipping $ \Pi $ of the parity. One of the reasons to flip
the parity was the fact that for the formulae to work we should have the
differential and the multiplication of the same parity in the sum $ d^{*}+m^{*} $.
Note that shifting of the grading does not change the grading of $ d $,
but changes the grading of $ m $. Since the comultiplication in chains is of
grading 0, and the differential of grading $ -1 $, to equate their gradings
we need to shift the grading by 1, what lead to Def.~\ref{def01.6}.
\subsection{Lie algebras } The case of Lie algebras is much simpler than the
commutative case since the corresponding complex is well known. Consider a
vector space $ {\mathfrak g} $. If it has a structure of a
Lie algebra, then we can consider the cohomological complex
$ C^{*}\left({\mathfrak g}\right) $ that is usually defined as $ \Lambda^{\geq1}{\mathfrak g}^{*} $.
This complex has a natural algebra structure, but it is easy to see that
the multiplication in this complex (and division of it into even and odd
parts) is compatible not with this definition, but with
$ S^{\geq1}\left(\Pi{\mathfrak g}\right)^{*} $, i.e., we consider the elements of $ {\mathfrak g}^{*} $ as odd (that automatically
makes the exterior power into symmetrical one). It turns out that the
differential
is a derivation with respect to this algebra structure. On the other
hand, we can define a similar mapping $ D $ in $ C^{*}\left({\mathfrak g}\right) $ for {\em any
skewsymmetric\/}
multiplication in $ {\mathfrak g} $. A simple check shows that this mapping satisfies the
differential condition $ D^{2}=0 $ then and only then when the product in $ {\mathfrak g} $
satisfies the Jacobi condition:
\begin{lemma} Consider a vector space $ {\mathfrak g} $ and a differential $ D $ of degree 1 in
$ S^{\geq1}\left(\Pi{\mathfrak g}\right)^{*} $ that is a derivation of this algebra. Then $ {\mathfrak g} $ has a natural
structure of a Lie algebra defined by the component
\begin{equation}
\Pi{\mathfrak g}^{*}\to S^{2}\Pi{\mathfrak g}^{*}
\notag\end{equation}
of the differential $ D $ (that induces a even mapping $ {\mathfrak g}^{*}\to{\mathfrak g}^{*}\wedge{\mathfrak g}^{*} $). \end{lemma}
In the same way we can consider a Lie superalgebra case:
\begin{lemma} Consider a supervector space $ {\mathfrak g} $ and a differential $ D $ of degree 1 in
$ S^{\geq1}\left(\Pi{\mathfrak g}\right)^{*} $ that is a derivation of this algebra. Then $ {\mathfrak g} $ has a natural
structure of a Lie superalgebra defined by the component
\begin{equation}
\Pi{\mathfrak g}^{*}\to S^{2}\Pi{\mathfrak g}^{*}
\notag\end{equation}
of the differential $ D $. \end{lemma}
Let us check for completeness the symmetry condition for the above
Lie superalgebra structure. The superbracket satisfies the supersymmetry
condition
\begin{equation}
\left[x,y\right]+\left(-1\right)^{\bar{x}\cdot\bar{y}}\left[y,x\right]=0,
\notag\end{equation}
thus the image of comultiplication is orthogonal to elements of
the form $ x\otimes y+\left(-1\right)^{\bar{x}\cdot\bar{y}}y\otimes x $, $ x,y\in{\mathfrak g} $, the flipped-parity
comultiplication is orthogonal to
\begin{equation}
-\left(-1\right)^{\bar{x}}x\otimes y-\left(-1\right)^{\bar{y}}\left(-1\right)^{\left(\bar{x}+1\right)\cdot\left(\bar{y}+1\right)}y\otimes x,\qquad x,y\in\Pi{\mathfrak g},
\notag\end{equation}
or to $ x\otimes y-\left(-1\right)^{\bar{x}\cdot\bar{y}}y\otimes x $, $ x,y\in\Pi{\mathfrak g} $. Now
the same discussion as in the section~\ref{s2.1} shows that the image of
comultiplication is in the symmetric algebra.
In the same way the case of differential Lie algebra can be
described as a derivation in $ S^{\bullet}\left(\left(\Pi{\mathfrak g}\right)^{*}\right) $ that sends $ \left(\Pi{\mathfrak g}\right)^{*} $ to
$ \left(\Pi{\mathfrak g}\right)^{*}\oplus S^{2}\left(\left(\Pi{\mathfrak g}\right)^{*}\right) $. Now the Def.~\ref{def01.5} is a natural merging of Def.
~\ref{def1.10} and the description of a differential Lie algebra.
\subsection{Caveats } Some natural relations between associative, commutative and
Lie algebras preserve in the homotopical situation, but some other cease
to be true.
\begin{example} Consider a (strongly) homotopically associative algebra $ A $.
The derivation $ D $ in $ T^{\bullet}\left(\Pi A\right)^{*} $ induces a derivation $ D_{S} $ in $ S^{\bullet}\left(\Pi A\right)^{*} $ making the
diagram
\begin{equation}
\begin{CD}
T^{\bullet}\left(\Pi A\right)^{*} @>{D}>{}> T^{\bullet}\left(\Pi A\right)^{*}
\\
@VVV @VVV
\\
S^{\bullet}\left(\Pi A\right)^{*} @>{D_{\text{Lie}}}>{}> S^{\bullet}\left(\Pi A\right)^{*}
\end{CD}
\notag\end{equation}
commutative (vertical arrows are natural projection). There are two ways
to see it: first, $ D|_{\left(\Pi A\right)^{*}}\to T^{\bullet}\left(\Pi A\right)^{*} $ induces $ D_{\text{Lie}}|_{\left(\Pi A\right)^{*}}\to S^{\bullet}\left(\Pi A\right)^{*} $ that
determines $ D_{\text{Lie}} $ by the Leibniz rule. Secondly, the kernel of natural
projection $ T^{\bullet}\left(\Pi A\right)^{*} \to T^{\bullet}\left(\Pi A\right)^{*} $ is generated by commutators, thus $ D $ sends it
into itself. Thus any homotopically assocative algebra $ A $
carries simultaneously a homotopical Lie (super)algebra structure that we
denote $ \operatorname{Lie}\left(A\right) $. \end{example}
\begin{example} Consider a homotopically associative algebra $ A $ again. Under
which conditions the corresponding homotopical Lie algebra $ \operatorname{Lie}\left(A\right) $ has
trivial
multiplications? It is clear that the differential in $ \operatorname{Lie}\left(A\right) $ is the same
as in $ A $, so the only interesting question is the question about the
components of $ D_{\text{Lie}} $ beginning from the second. It is clear also that the
condition on the binary multiplication is the supercommutativity
condition. However, already the condition on the ternary product is
weaker than commutativity: the image of differential should be in the
ideal of $ T^{\bullet}\left(\Pi A\right)^{*} $ generated by Lie brackets, not in the set of Lie
polynomials. That (together with the calculations for simplicial Lie
and commutative algebras) shows that the definitions of
homotopical Lie and commutative algebra may be in need of some correction. \end{example}
\begin{example} Consider a homotopically associative algebra over $ K\left[\left[t\right]\right] $
such that as a $ K\left[\left[t\right]\right] $-module it is isomorphic to $ A\left[\left[t\right]\right] $ for some $ K $-module
$ A $. Suppose that it becomes commutative when we specify $ t=0 $, so $ A $
carries a structure of commutative homotopically associative algebra.
There is a family with parameter $ t $ of associative products on $ A $. Denote
the corresponding family of derivations of $ T^{\bullet}\left(\Pi A\right)^{*} $ by $ D_{t} $. In a situation
without homotopy such a structure induces a Lie algebra structure on $ A $
induced by $ D^{\left(1\right)}=\frac{\partial D}{\partial t}|_{t=0} $. Let us look what we get in the homotopical
situation.
Since the differential $ D_{0} $ in $ S^{\bullet}\left(\Pi A\right)^{*} $ is not necessary 0, but
coincides with $ \partial_{0} $, we get a different compatibility condition for $ D^{\left(1\right)} $.
It turns out that first of all, we get
\begin{equation}
\left[\partial, D_{\text{Lie}}^{\left(1\right)}\right]=0,
\notag\end{equation}
(here the Lie subscript means the corresponding derivation of $ S^{\bullet}\left(\Pi A\right)^{*} $),
and second,
\begin{equation}
\left(D_{\text{Lie}}^{\left(1\right)}\right)^{2}+\left[\partial,D_{\text{Lie}}^{\left(2\right)}\right]=0.
\notag\end{equation}
Here $ D^{\left(2\right)}=\frac{1}{2}\frac{\partial^{2}D}{\partial t^{2}}|_{t=0} $. In both these formulae we consider the
extension of $ \partial $ on $ S^{\bullet}\left(\Pi A\right)^{*} $.
We see that these two conditions ensure that $ D_{\text{Lie}}^{\left(1\right)} $ and $ \partial+D_{\text{Lie}}^{\left(1\right)} $
define structures of homotopical Lie algebra on $ A $. However, these
conditions become trivial if $ \partial=0 $, so consideration of differential
algebras makes the usual correspondence between Lie algebras and
deformations weaker.
\end{example}
\begin{remark} We see that there is a kind of duality between the definition
of a commutative homotopically associative algebras and homotopical Lie
algebras. Any one is defined as a derivation in a free algebra of the
second type. This duality is an important component of the flavor of
noncommutative geometry invented by M.~Kontsevich. The part of this
duality that concerns non-homotopical algebras was known for a long
time. \end{remark}
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