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\begin{document}

\title{Lecture 4: Operads}
\author{Alexander A. Voronov}
%\address{Department of Mathematics\\ M.I.T.,
%2-246\\ 77 Massachusetts Ave.\\ Cambridge, MA 02139-4307}
%\curraddr{}
%\email{voronov@math.mit.edu}
%\urladdr{http://www-math.mit.edu/~voronov/}
%\thanks{Research supported in part by an AMS Centennial Fellowship.}

%\subjclass{Primary 14H10; Secondary 32G15, 55P62}
\date{October 1, 1997}

\maketitle

\section{Operads}

\subsection{Definitions and examples}

A commutative algebra is a Frobenius algebra without an invariant
inner product. Similarly, an operad is a PROP without things that can
possibly create something like an inner product in a
representation. More specifically, an operad is the part $\Mor(n,1)$,
$n \ge 0$, of a PROP. Of course, given only the collection of
morphisms $\Mor(n,1)$, it is not clear how to compose them. The idea
is to take the union of a $m$ elements from $\Mor(n,1)$ to be able to
compose them with an element of $\Mor(m,1)$. This leads to cumbersome
notation and ugly axioms, compared to those of PROP's. However operads
are in a sense more basic than the corresponding PROP's: as Victor Kac
put it on this lecture, the difference is similar to the difference
between Lie algebras and the universal enveloping algebras.

\begin{df}
Let $k$ be a ground field. An \emph{operad} $\OO$ is a collection of
sets (vector spaces, complexes, topological spaces, manifolds, \dots,
objects of a tensor category) $\OO (n)$, $n \ge 0$, with
\begin{enumerate}
\item A composition law:
\end{enumerate}
\noindent
\begin{equation*}
\gamma: \OO(m) \otimes \OO(n_1) \otimes \dots \otimes \OO(n_m) \to
\OO(n_1 + \dots + n_m).
\end{equation*}
\begin{enumerate}
\setcounter{enumi}{1}
\item A right action of the symmetric group $S_n$ on $\OO (n)$.
\item A unit $e \in \OO (1)$.
\end{enumerate}
such that the following properties are satisfied:
\begin{enumerate}
\item The composition is associative, i.e., the following diagram is
commutative:
\end{enumerate}
\noindent
\[
\begin{CD}
\left\{
\begin{aligned}
\OO(l) & \otimes \OO(m_1) \otimes \dots \otimes \OO(m_l) \\
& \otimes \OO(n_{11}) \otimes \dots \otimes \OO(n_{l,n_l})
\end{aligned}
\right\}
@>{\id \otimes \gamma^l}>>
\OO(l) \otimes \OO(n_1) \otimes \dots \otimes \OO(n_l)\\
@V{\gamma \otimes \id}VV        @VV{\gamma}V , \\
\OO(m) \otimes \OO(n_{11}) \otimes \dots \otimes \OO(n_{m,n_m})
@>\gamma>> \OO(n)
\end{CD}
\]
\noindent
\begin{quote}
where $m = \sum_i m_i$, $n_i = \sum_j n_{ij}$, and $n = \sum_i n_i$.
\end{quote}
\noindent
\begin{enumerate}
\setcounter{enumi}{1}
\item The composition is equivariant with respect to the symmetric group
actions: $S_m \times S_{n_1} \times \dots \times S_{n_m}$ acts on the
left-hand side and maps naturally to $S_{n_1 + \dots + n_m}$, acting
on the right-hand side.
\item The unit $e$ satisfies natural properties with respect to the
composition: $\gamma(e; f) \linebreak[0] = f$ and $\gamma(f; e, \dots,
e) = f$ for each $f \in \OO (k)$.
\end{enumerate}

The notion of a \emph{morphism of operads} is introduced naturally.
\end{df}

\begin{rem}
One can consider \emph{nonsymmetric operads}, not assuming the action
of the symmetric groups. Not requiring the existence of a unit $e$, we
arrive to \emph{nonunital operads}. Do not mix this up with operads
with no $\OO(0)$, algebras over which (see next section) have no
unit. There are also good examples of operads having only $n \ge 2$
components $\OO(n)$.

An equivalent definition of an operad may be given in terms of
operations $f \circ_i g \linebreak[1] = \linebreak[0]
\gamma (f;, \id, \dots, \linebreak[1] \id, g, \id , \dots, \id)$, $i =
1, \dots, m$, for $f \in \OO(m), g \in \OO(n)$. Then the associativity
condition translates as $f \circ_i (g \circ_j h) = (f \circ_i g)
\circ_{i+j-1} h$.
\end{rem}

%\begin{sloppypar}
\begin{ex}[The Riemann surface and the endomorphism operad]
$\OO(n)$ is the space of all connected Riemann surfaces (with or
without complex structure) with $n$ inputs and 1 output.  Another
example is the \emph{endomorphism operad of a vector space $V$}:
$\End{V}(n) = \Hom(V^{\otimes n}, V)$, the space of $n$-linear
mappings from $V$ to $V$.
\end{ex}
%\end{sloppypar}

\subsection{Algebras over operads}

\begin{df}
An \emph{algebra over an operad $\OO$} (in other terminology, a
\emph{representation of an operad}) is a morphism of operads $\OO \to
\End{V}$, that is, a collection of maps
\[
\OO(n) \to \End{V}(n) \qquad \text{for $ n \ge 0$}
\]
compatible with the symmetric group action, the unit elements, and the
compositions. If the operad $\OO$ is an operad of vector spaces, then
we would usually require the morphism $\OO \to \End{V}$ to be a
morphism of operads of vector spaces. Otherwise, we would think of
this morphism as a morphism of operads of sets. Sometimes, we may also
need a morphism to be continuous or respect differentials, or have
other compatibility conditions.
\end{df}

\subsubsection{The commutative operad}
The \emph{commutative operad} is the operad of $k$-vector spaces with
the $n$th component $\Comm (n) = k$ for all $n \ge 0$. We assume that
the symmetric group acts trivially on $k$ and the compositions are
just the multiplication of elements in the ground field $k$. An
algebra over the commutative operad is nothing but a commutative
associative algebra with a unit, as we see from the following
exercise.

Another version of the commutative operad is $\Comm(n) = \text{point}$
for all $n \ge 0$. This is an operad of sets. An algebra over it is
also the same as a commutative associative unital algebra.

\begin{xca}
Show that the operad $\T(n) \linebreak[1] = \linebreak[0]$ \{the set
of diffeomorphism classes of connected Riemann surfaces of genus zero
with $n$ input holes and 1 output hole\} is isomorphic to the
commutative operad of sets.
\end{xca}

\begin{xca}
Prove that the structure of an algebra over the commutative operad
$\Comm$ on a vector space is equivalent to the structure of a
commutative associative algebra with a unit.
\end{xca}

\subsubsection{The associative operad}
The \emph{associative operad} $\Assoc$ can be considered as a 1d
analogue of the commutative operad $\T(n)$. $\Assoc(n)$ is the set of
equivalence classes of connected planar binary (each vertex being of
valence 3) trees that have a root edge and $n$ leaves labeled by
integers 1 through $n$:

\centerline{\epsfig{file=tree.eps,height=3cm}}

\noindent
If $n=1$, there is only one tree --- it has no vertices and only one
edge connecting a leaf and a root. If $n=0$, the only tree is the one
with no vertices and no leaves --- it only has a root. Unfortunately,
I have a problem sketching it: it probably exists only in the quantum
world.

Two trees are equivalent if they are related by a sequence of moves of
the kind
\medskip

\centerline{\epsfig{file=move.eps,height=2cm}}
\smallskip

\noindent
performed over pairs of two adjacent vertices of a tree. The symmetric
group acts by relabeling the leaves, as usual. The composition is
obtained by grafting the roots of $m$ trees to the leaves of an
$m$-tree; no new vertices being created at the grafting points. Note
that this is similar to sewing Riemann surfaces and erasing the seam,
just as we did to define operad composition in that case. By
definition, grafting a 0-tree to a leaf just removes the leaf and, if
this operation creates a vertex of valence 2, we should erase the
vertex.

\begin{xca}
Prove that the structure of an algebra over the associative operad
$\Assoc$ on a vector space is equivalent to the structure of an
associative algebra.
\end{xca}

\subsubsection{The Lie operad}

The \emph{Lie operad} $\Lie$ is another variation on the theme of a
tree operad. Consider the same binary trees as for the associative
operad, except that we do not include the zero-tree, \emph{i.e}., the
operad has only positive components $\Lie(n)$, $n \ge 1$, and there
are now two kinds of equivalence relations:
\medskip

\centerline{\epsfig{file=skew.eps,height=2.5cm}}
\smallskip

\noindent
and
\medskip

\centerline{\epsfig{file=jacobi.eps,height=2.5cm}}
\smallskip

\noindent
Now that we have arithmetic operations in the equivalence relations,
we consider the Lie operad as an operad of vector spaces. We also
assume that the ground field is of characteristic zero, because
otherwise we will arrive to the wrong definition of a Lie algebra.

\begin{xca}
Prove that the structure of an algebra over the Lie operad $\Lie$ on a
vector space over a field of characteristic zero is equivalent to the
structure of a Lie algebra.
\end{xca}

\begin{xca}
Describe algebraically an algebra over the operad $\Lie$, if we modify
it by including the zero-tree, whose composition with any other tree
is defined as (a) zero, (b) for the associative operad.
\end{xca}

\subsubsection{The Poisson operad}
Recall that a \emph{Poisson algebra} is a vector space $V$ (over a
field of characteristic zero) with a unit element $e$, a dot product
$ab$, and a bracket $[a,b]$ defined, so that the dot product defines
the structure of commutative associative unital algebra, the bracket
defines the structure of a Lie algebra, and the bracket is a
derivation of the dot product:
\[
[a,bc] = [a,b] c + b [a,c] \qquad \text{for all $a$, $b$, and $c \in
V$}.
\]

\begin{xca}
Define the \emph{Poisson operad}, using a tree model similar to the
previous examples. Show that an algebra over it is nothing but a
Poisson algebra. [\emph{Hint}: Use two kinds of vertices, one for the
dot product and the other one for the bracket.]
\end{xca}

\subsubsection{The Riemann surface operad and vertex operator algebras}

Just for a change, let us return to the operad of Riemann surfaces
$\PP$ --- this time, of isomorphism classes of genus zero Riemann
surfaces with holomorphic holes. What is an algebra over it? Since
there are infinitely many nonisomorphic pairs of pants, there are
infinitely many (at least) binary operations. In fact, we have an
infinite dimensional family of binary operations parameterized by
classes of pairs of pants. However modulo the unary operations, those
which correspond to cylinders, we have only one fundamental binary
operation corresponding to a fixed pair of pants. An algebra over this
operad $\PP$ is part of a CFT data at the tree level, $c=0$. If we
consider a holomorphic algebra over this operad, that is, require that
the defining mappings $\PP(n) \to \End{V}(n)$, where $V$ is a complex
vector space, be holomorphic, then we get part of a chiral CFT, or an
object which may be called a \emph{vertex operator algebra
$($VOA$)$}. This kind of object is not equivalent to what people used
to call a VOA, but according to Huang's Theorem, a true VOA is a
holomorphic algebra over a ``partial pseudo-operad of Riemann surfaces
with rescaling'', which is a version of $\PP$, where the disks are
allowed to overlap. The fundamental binary operation $Y(a,z)b$ for
$a, b \in V$ of a VOA is commonly chosen to be the one corresponding to
a pair of pants which is the Riemann sphere with a standard
holomorphic coordinate and three unit disks around the points 0, $z$,
and $\infty$ (No doubt, these disks overlap badly, but we shrink them
on the figure to look better):
\medskip

\centerline{\epsfig{file=voapants.eps,height=1.8cm}}
\smallskip

The famous associativity identity 
\[
Y(a,z-w)Y(b,-w) c = Y(Y(a,z)b,-w) c
\]
for vertex operator algebras comes from the following natural
isomorphism of the Riemann surfaces:
\medskip

\centerline{\epsfig{file=voass.eps,width=2.5cm,angle=-90}}
\smallskip

\end{document}
