247 The definitions for a topological $A_\infty$-$n$-category are very similar to the above

   248 $n=1$ case.

   249 One replaces intervals with manifolds diffeomorphic to the ball $B^n$.

   250 Marked points are replaced by copies of $B^{n-1}$ in $\bdy B^n$.

   251 

   252 \nn{give examples: $A(J^n) = \bc_*(Z\times J)$ and $A(J^n) = C_*(\Maps(J \to M))$.}

   253 

   254 \todo{the motivating example $C_*(\Maps(X, M))$}

   255 

   256 

   257 

   258 \newcommand{\skel}[1]{\operatorname{skeleton}(#1)}

   259 

   260 Given a topological $A_\infty$-category $\cC$, we can construct an `algebraic' $A_\infty$ category $\skel{\cC}$. First, pick your

   261 favorite diffeomorphism $\phi: I \cup I \to I$.

   262 \begin{defn}

   263 We'll write $\skel{\cC} = (A, m_k)$. Define $A = \cC(I)$, and $m_2 : A \tensor A \to A$ by

   264 \begin{equation*}

   265 m_2 \cC(I) \tensor \cC(I) \xrightarrow{\gl_{I,I}} \cC(I \cup I) \xrightarrow{\cC(\phi)} \cC(I).

   266 \end{equation*}

   267 Next, we define all the `higher associators' $m_k$ by

   268 \todo{}

   269 \end{defn}

   270 

   271 Give an `algebraic' $A_\infty$ category $(A, m_k)$, we can construct a topological $A_\infty$-category, which we call $\bc_*^A$. You should

   272 think of this as a generalisation of the blob complex, although the construction we give will \emph{not} specialise to exactly the usual definition

   273 in the case the $A$ is actually an associative category.

   274 

   275 We'll first define $\cT_{k,n}$ to be the set of planar forests consisting of $n-k$ trees, with a total of $n$ leaves. Thus

   276 \todo{$\cT_{0,n}$ has 1 element, with $n$ vertical lines, $\cT_{1,n}$ has $n-1$ elements, each with a single trivalent vertex, $\cT_{2,n}$ etc...}

   277 \begin{align*}

   278 \end{align*}

   279 

   280 \begin{defn}

   281 The topological $A_\infty$ category $\bc_*^A$ is doubly graded, by `blob degree' and `internal degree'. We'll write $\bc_k^A$ for the blob degree $k$ piece.

   282 The homological degree of an element $a \in \bc_*^A(J)$

   283 is the sum of the blob degree and the internal degree.

   284 

   285 We first define $\bc_0^A(J)$ as a vector space by

   286 \begin{equation*}

   287 \bc_0^A(J) = \DirectSum_{\substack{\{J_i\}_{i=1}^n \\ \mathclap{\bigcup_i J_i = J}}} \Tensor_{i=1}^n (\CD{J_i \to I} \tensor A).

   288 \end{equation*}

   289 (That is, for each division of $J$ into finitely many subintervals,

   290 we have the tensor product of chains of diffeomorphisms from each subinterval to the standard interval,

   291 and a copy of $A$ for each subinterval.)

   292 The internal degree of an element $(f_1 \tensor a_1, \ldots, f_n \tensor a_n)$ is the sum of the dimensions of the singular chains

   293 plus the sum of the homological degrees of the elements of $A$.

   294 The differential is defined just by the graded Leibniz rule and the differentials on $\CD{J_i \to I}$ and on $A$.

   295 

   296 Next,

   297 \begin{equation*}

   298 \bc_1^A(J) = \DirectSum_{\substack{\{J_i\}_{i=1}^n \\ \mathclap{\bigcup_i J_i = J}}} \DirectSum_{T \in \cT_{1,n}} \Tensor_{i=1}^n (\CD{J_i \to I} \tensor A).

   299 \end{equation*}

   300 \end{defn}

   301 

   302 \begin{figure}[!ht]

   303 \begin{equation*}

   304 \mathfig{0.7}{associahedron/A4-vertices}

   305 \end{equation*}

   306 \caption{The vertices of the $k$-dimensional associahedron are indexed by binary trees on $k+2$ leaves.}

   307 \label{fig:A4-vertices}

   308 \end{figure}

   309 

   310 \begin{figure}[!ht]

   311 \begin{equation*}

   312 \mathfig{0.7}{associahedron/A4-faces}

   313 \end{equation*}

   314 \caption{The faces of the $k$-dimensional associahedron are indexed by trees with $2$ vertices on $k+2$ leaves.}

   315 \label{fig:A4-vertices}

   316 \end{figure}

   317 

   318 \newcommand{\tm}{\widetilde{m}}

   319 

   320 Let $\tm_1(a) = a$.

   321 

   322 We now define $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$, first giving an opaque formula, then explaining the combinatorics behind it.

   323 \begin{align}

   324 \notag \bdy(\tm_k(a_1 & \tensor \cdots \tensor a_k)) = \\

   325 \label{eq:bdy-tm-k-1}   & \phantom{+} \sum_{\ell'=0}^{k-1} (-1)^{\abs{\tm_k}+\sum_{j=1}^{\ell'} \abs{a_j}} \tm_k(a_1 \tensor \cdots \tensor \bdy a_{\ell'+1} \tensor \cdots \tensor a_k) + \\

   326 \label{eq:bdy-tm-k-2}   &          +  \sum_{\ell=1}^{k-1} \tm_{\ell}(a_1 \tensor \cdots \tensor a_{\ell}) \tensor \tm_{k-\ell}(a_{\ell+1} \tensor \cdots \tensor a_k) + \\

   327 \label{eq:bdy-tm-k-3}   &          +  \sum_{\ell=1}^{k-1} \sum_{\ell'=0}^{l-1} (-1)^{\abs{\tm_k}+\sum_{j=1}^{\ell'} \abs{a_j}} \tm_{\ell}(a_1 \tensor \cdots \tensor m_{k-\ell + 1}(a_{\ell' + 1} \tensor \cdots \tensor a_{\ell' + k - \ell + 1}) \tensor \cdots \tensor a_k)

   328 \end{align}

   329 The first set of terms in $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$ just have $\bdy$ acting on each argument $a_i$.

   330 The terms appearing in \eqref{eq:bdy-tm-k-2} and \eqref{eq:bdy-tm-k-3} are indexed by trees with $2$ vertices on $k+1$ leaves.

   331 Note here that we have one more leaf than there arguments of $\tm_k$.

   332 (See Figure \ref{fig:A4-vertices}, in which the rightmost branches are helpfully drawn in red.)

   333 We will treat the vertices which involve a rightmost (red) branch differently from the vertices which only involve the first $k$ leaves.

   334 The terms in \eqref{eq:bdy-tm-k-2} arise in the cases in which both

   335 vertices are rightmost, and the corresponding term in $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$ is a tensor product of the form

   336 $$\tm_{\ell}(a_1 \tensor \cdots \tensor a_{\ell}) \tensor \tm_{k-\ell}(a_{\ell+1} \tensor \cdots \tensor a_k)$$

   337 where $\ell + 1$ and $k - \ell + 1$ are the number of branches entering the vertices.

   338 If only one vertex is rightmost, we get the term $$\tm_{\ell}(a_1 \tensor \cdots \tensor m_{k-\ell+1}(a_{\ell' + 1} \tensor \cdots \tensor a_{\ell' + k - \ell}) \tensor \cdots \tensor a_k)$$

   339 in \eqref{eq:bdy-tm-k-3},

   340 where again $\ell + 1$ is the number of branches entering the rightmost vertex, $k-\ell+1$ is the number of branches entering the other vertex, and $\ell'$ is the number of edges meeting the rightmost vertex which start to the left of the other vertex.

   341 For example, we have

   342 \begin{align*}

   343 \bdy(\tm_2(a \tensor b)) & = \left(\tm_2(\bdy a \tensor b) + (-1)^{\abs{a}} \tm_2(a \tensor \bdy b)\right) + \\

   344                          & \qquad - a \tensor b + m_2(a \tensor b) \\

   345 \bdy(\tm_3(a \tensor b \tensor c)) & = \left(- \tm_3(\bdy a \tensor b \tensor c) + (-1)^{\abs{a} + 1} \tm_3(a \tensor \bdy b \tensor c) + (-1)^{\abs{a} + \abs{b} + 1} \tm_3(a \tensor b \tensor \bdy c)\right) + \\

   346                                    & \qquad + \left(- \tm_2(a \tensor b) \tensor c + a \tensor \tm_2(b \tensor c)\right) + \\

   347                                    & \qquad + \left(- \tm_2(m_2(a \tensor b) \tensor c) + \tm_2(a, m_2(b \tensor c)) + m_3(a \tensor b \tensor c)\right)

   348 \end{align*}

   349 \begin{align*}

   350 \bdy(& \tm_4(a \tensor b \tensor c \tensor d)) = \left(\tm_4(\bdy a \tensor b \tensor c \tensor d) + \cdots + \tm_4(a \tensor b \tensor c \tensor \bdy d)\right) + \\

   351                                              & + \left(\tm_3(a \tensor b \tensor c) \tensor d + \tm_2(a \tensor b) \tensor \tm_2(c \tensor d) + a \tensor \tm_3(b \tensor c \tensor d)\right) + \\

   352                                              & + \left(\tm_3(m_2(a \tensor b) \tensor c \tensor d) + \tm_3(a \tensor m_2(b \tensor c) \tensor d) + \tm_3(a \tensor b \tensor m_2(c \tensor d))\right. + \\

   353                                              & + \left.\tm_2(m_3(a \tensor b \tensor c) \tensor d) + \tm_2(a \tensor m_3(b \tensor c \tensor d)) + m_4(a \tensor b \tensor c \tensor d)\right) \\

   354 \end{align*}

   355 See Figure \ref{fig:A4-terms}, comparing it against Figure \ref{fig:A4-faces}, to see this illustrated in the case $k=4$. There the $3$ faces closest

   356 to the top of the diagram have two rightmost vertices, while the other $6$ faces have only one.

   357 

   358 \begin{figure}[!ht]

   359 \begin{equation*}

   360 \mathfig{1.0}{associahedron/A4-terms}

   361 \end{equation*}

   362 \caption{The terms of $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$ correspond to the faces of the $k-1$ dimensional associahedron.}

   363 \label{fig:A4-terms}

   364 \end{figure}

   365 

   366 \begin{lem}

   367 This definition actually results in a chain complex, that is $\bdy^2 = 0$.

   368 \end{lem}

   369 \begin{proof}

   370 \newcommand{\T}{\text{---}}

   371 \newcommand{\ssum}[1]{{\sum}^{(#1)}}

   372 For the duration of this proof, inside a summation over variables $l_1, \ldots, l_m$, an expression with $m$ dashes will be interpreted

   373 by replacing each dash with contiguous factors from $a_1 \tensor \cdots \tensor a_k$, so the first dash takes the first $l_1$ factors, the second

   374 takes the next $l_2$ factors, and so on. Further, we'll write $\ssum{m}$ for $\sum_{\sum_{i=1}^m l_i = k}$.

   375 In this notation, the formula for the differential becomes

   376 \begin{align}

   377 \notag

   378 \bdy \tm(\T) & = \ssum{2} \tm(\T) \tensor \tm(\T) \times \sigma_{0;l_1,l_2} + \ssum{3} \tm(\T \tensor m(\T) \tensor \T) \times \tau_{0;l_1,l_2,l_3} \\

   379 \intertext{and we calculate}

   380 \notag

   381 \bdy^2 \tm(\T) & = \ssum{2} \bdy \tm(\T) \tensor \tm(\T) \times \sigma_{0;l_1,l_2} \\

   382 \notag         & \qquad + \ssum{2} \tm(\T) \tensor \bdy \tm(\T) \times \sigma_{0;l_1,l_2} \\

   383 \notag         & \qquad + \ssum{3} \bdy \tm(\T \tensor m(\T) \tensor \T) \times \tau_{0;l_1,l_2,l_3} \\

   384 \label{eq:d21} & = \ssum{3} \tm(\T) \tensor \tm(\T) \tensor \tm(\T) \times \sigma_{0;l_1+l_2,l_3} \sigma_{0;l_1,l_2} \\

   385 \label{eq:d22} & \qquad + \ssum{4} \tm(\T \tensor m(\T) \tensor \T) \tensor \tm(\T) \times \sigma_{0;l_1+l_2+l_3,l_4} \tau_{0;l_1,l_2,l_3} \\

   386 \label{eq:d23} & \qquad + \ssum{3} \tm(\T) \tensor \tm(\T) \tensor \tm(\T) \times \sigma_{0;l_1,l_2+l_3} \sigma_{l_1;l_2,l_3} \\

   387 \label{eq:d24} & \qquad + \ssum{4} \tm(\T) \tensor \tm(\T \tensor m(\T) \tensor \T) \times \sigma_{0;l_1,l_2+l_3+l_4} \tau_{l_1;l_2,l_3,l_4} \\

   388 \label{eq:d25} & \qquad + \ssum{4} \tm(\T \tensor m(\T) \tensor \T) \tensor \tm(\T) \times \tau_{0;l_1,l_2,l_3+l_4} ??? \\

   389 \label{eq:d26} & \qquad + \ssum{4} \tm(\T) \tensor \tm(\T \tensor m(\T) \tensor \T) \times \tau_{0;l_1+l_2,l_3,l_4} \sigma_{0;l_1,l_2} \\

   390 \label{eq:d27} & \qquad + \ssum{5} \tm(\T \tensor m(\T) \tensor \T \tensor m(\T) \tensor \T) \times \tau_{0;l_1+l_2+l_3,l_4,l_5} \tau_{0;l_1,l_2,l_3}  \\

   391 \label{eq:d28} & \qquad + \ssum{5} \tm(\T \tensor m(\T \tensor m(\T) \tensor \T) \tensor \T) \times \tau_{0;l_1,l_2+l_3+l_4,l_5} ??? \\

   392 \label{eq:d29} & \qquad + \ssum{5} \tm(\T \tensor m(\T) \tensor \T \tensor m(\T) \tensor \T) \times \tau_{0;l_1,l_2,l_3+l_4+l_5} ???

   393 \end{align}

   394 Now, we see the the expressions on the right hand side of line \eqref{eq:d21} and those on \eqref{eq:d23} cancel. Similarly, line \eqref{eq:d22} cancels

   395 with \eqref{eq:d25}, \eqref{eq:d24} with \eqref{eq:d26}, and \eqref{eq:d27} with \eqref{eq:d29}. Finally, we need to see that \eqref{eq:d28} gives $0$,

   396 by the usual relations between the $m_k$ in an $A_\infty$ algebra.

   397 \end{proof}