--- a/blob1.tex Fri Jul 04 00:26:46 2008 +0000
+++ b/blob1.tex Fri Jul 04 05:22:12 2008 +0000
@@ -1119,7 +1119,7 @@
We now define $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$, first giving an opaque formula, then explaining the combinatorics behind it.
\begin{align}
\notag \bdy(\tm_k(a_1 & \tensor \cdots \tensor a_k)) = \\
-\label{eq:bdy-tm-k-1} & \phantom{+} \sum_{i=1}^k (-1)^{\sum_{j=1}^{i-1} \deg(a_j)} \tm_k(a_1 \tensor \cdots \tensor \bdy a_i \tensor \cdots \tensor a_k) + \\
+\label{eq:bdy-tm-k-1} & \phantom{+} \sum_{\ell'=0}^{k-1} (-1)^{\sum_{j=1}^{\ell'} \deg(a_j)} \tm_k(a_1 \tensor \cdots \tensor \bdy a_{\ell'+1} \tensor \cdots \tensor a_k) + \\
\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) + \\
\label{eq:bdy-tm-k-3} & + \sum_{\ell=1}^{k-1} \sum_{\ell'=0}^{l-1} \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)
\end{align}
@@ -1161,6 +1161,39 @@
\label{fig:A4-terms}
\end{figure}
+\begin{lem}
+This definition actually results in a chain complex, that is $\bdy^2 = 0$.
+\end{lem}
+\begin{proof}
+\newcommand{\T}{\text{---}}
+\newcommand{\ssum}[1]{{\sum}^{(#1)}}
+For the duration of this proof, inside a summation over variables $l_1, \ldots, l_m$, an expression with $m$ dashes will be interpreted
+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
+takes the next $l_2$ factors, and so on. Further, we'll write $\ssum{m}$ for $\sum_{\sum_{i=1}^m l_i = k}$.
+In this notation, the formula for the differential becomes
+\begin{align}
+\notag
+\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} \\
+\intertext{and we calculate}
+\notag
+\bdy^2 \tm(\T) & = \ssum{2} (\bdy \tm(\T)) \tensor \tm(\T) \times \sigma_{0;l_1,l_2} \\
+\notag & \qquad + \ssum{2} \tm(\T) \tensor (\bdy \tm(\T)) \times \sigma_{0;l_1,l_2} \\
+\notag & \qquad + \ssum{3} \bdy \tm(\T \tensor m(\T) \tensor \T) \times \tau_{0;l_1,l_2,l_3} \\
+\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} \\
+\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} \\
+\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} \\
+\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} \\
+\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} ??? \\
+\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} \\
+\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} \\
+\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} ??? \\
+\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} ???
+\end{align}
+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
+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 $0$,
+by the usual relations between the $m_k$ in an $A_\infty$ algebra.
+\end{proof}
+
\nn{Need to let the input $n$-category $C$ be a graded thing (e.g. DG
$n$-category or $A_\infty$ $n$-category). DG $n$-category case is pretty
easy, I think, so maybe it should be done earlier??}