1084 \bc_0^A(J) = \DirectSum_{\substack{\{J_i\}_{i=1}^n \\ \mathclap{\bigcup_i J_i = J}}} \left(\{J_i\}, \Tensor_{i=1}^n (\CD{J_i \to I} \tensor A) \right).

  1084 \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).

  1095 \bc_1^A(J) = \DirectSum_{\substack{\{J_i\}_{i=1}^n \\ \mathclap{\bigcup_i J_i = J}}} \DirectSum_{T \in \cT_{1,n}} \left(\{J_i\}, T, \Tensor_{i=1}^n (\CD{J_i \to I} \tensor A) \right).

  1095 \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).

  1100 \newcommand{\ttm}{\widetilde{\widetilde{m}}}

  1116 

  1101 

  1117 Let $\tm_1(a) = a$.

  1102 Define $\ttm_k$ by

  1118 

  1119 We now define $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$, first giving an opaque formula, then explaining the combinatorics behind it.

  1120 \begin{align}

  1121 \notag \bdy(\tm_k(a_1 & \tensor \cdots \tensor a_k)) = \\

  1122 \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) + \\

  1123 \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) + \\

  1124 \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)

  1125 \end{align}

  1126 The first set of terms in $\bdy(\tm_k(a_1 \tensor \cdots \tensor a_k))$ just have $\bdy$ acting on each argument $a_i$.

  1127 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.

  1128 Note here that we have one more leaf than there arguments of $\tm_k$.

  1129 (See Figure \ref{fig:A4-vertices}, in which the rightmost branches are helpfully drawn in red.)

  1130 We will treat the vertices which involve a rightmost (red) branch differently from the vertices which only involve the first $k$ leaves.

  1131 The terms in \eqref{eq:bdy-tm-k-2} arise in the cases in which both

  1132 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

  1133 $$\tm_{\ell}(a_1 \tensor \cdots \tensor a_{\ell}) \tensor \tm_{k-\ell}(a_{\ell+1} \tensor \cdots \tensor a_k)$$

  1134 where $\ell + 1$ and $k - \ell + 1$ are the number of branches entering the vertices.

  1135 If only one vertex is rightmost, we get the term $$\tm_{\ell}(a_1 \tensor \cdots \tensor m_{k-\ell+1}(a_{\ell' + 1} \tensor \cdots \tensor a_{\ell' + k - \ell}) \tensor \cdots \tensor a_k)$$

  1136 in \eqref{eq:bdy-tm-k-3},

  1137 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.

  1138 For example, we have

  1104 \ttm_k(a_1 \tensor \cdots \tensor a_k) & = m_k(a_1 \tensor \cdots \tensor a_k) \\

  1140 \bdy(\tm_2(a \tensor b)) & = \left(\tm_2(\bdy a \tensor b) + \tm_2(a \tensor \bdy b)\right) + \\

  1105 \ttm_k(a_1 \tensor \cdots \tensor a_{k-1} \tensor z) & = z \tensor \tm_{k-1}(a_1 \tensor \cdots \tensor a_{k-1}) \\

  1106 \intertext{and}

  1107 \ttm_k(a_1 \tensor \cdots \tensor a_{k-2} \tensor z \tensor a_k) & = z \tensor \tm_{k-2}(a_1 \tensor \cdots \tensor a_{k-2}) \tensor a_k.

  1108 \end{align*}

  1109 

  1110 Let $\tm_1(a) = a$.

  1111 

  1112 Then define

  1113 \begin{align*}

  1114 \bdy(\tm_k(a_1 \tensor \cdots \tensor a_k)) & = \sum_{j=1}^{k} \tm_k(a_1 \tensor \cdots \tensor \bdy a_j \tensor \cdots \tensor a_k) + \\

  1115    & z \perp \sum_{q=2}^{k-1} \sum_{p=1}^{k-q+2} \ttm_{k-q+1}(a_1 \tensor \cdots a_{p-1} \tensor \ttm_q(a_p \tensor \cdots \tensor a_{p+q-1}) \tensor a_{p+q} \tensor \cdots \tensor a_{k+1}).

  1116 \end{align*}

  1117 where here $a_{k+1}$ is just notation for $z$.

  1118 \todo{err... here I mean $z \perp z \tensor x = x$...}

  1119 \todo{actually, if you let $q$ start from 1 you don't need the first term}

  1120 

  1121 \begin{align*}

  1122 \bdy(\tm_2(a \tensor b)) & = (\tm_2(\bdy a \tensor b) + \tm_2(a \tensor \bdy b)) + \\

  1125 \bdy(\tm_3(a \tensor b \tensor c)) & = (\tm_3(\bdy a \tensor b \tensor c) + \tm_3(a \tensor \bdy b \tensor c) + \tm_3(a \tensor b \tensor \bdy c)) + \\

  1143 \bdy(\tm_3(a \tensor b \tensor c)) & = \left(\tm_3(\bdy a \tensor b \tensor c) + \tm_3(a \tensor \bdy b \tensor c) + \tm_3(a \tensor b \tensor \bdy c)\right) + \\

  1126                                    & + (\tm_2(a \tensor b) \tensor c + a \tensor \tm_2(b \tensor c)) + \\

  1144                                    & \qquad + \left(\tm_2(a \tensor b) \tensor c + a \tensor \tm_2(b \tensor c)\right) + \\

  1127                                    & + (\tm_2(m_2(a \tensor b) \tensor c) + \tm_2(a, m_2(b \tensor c)) + m_3(a \tensor b \tensor c)) \\

  1145                                    & \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)

  1128 \bdy(\tm_4(a \tensor b \tensor c \tensor d)) & = (\tm_4(\bdy a \tensor b \tensor c \tensor d) + \cdots + \tm_4(a \tensor b \tensor c \tensor \bdy d)) + \\

  1129                                              & + (\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)) + \\

  1130                                              & + (\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)) + \\

  1131                                              & + \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)) \\

  1132 %d(\tm_k(x_1 \tensor \cdots \tensor x_k)) & = \sum_{i=1}^k (-1)^{\sum_{j=1}^{i-1} \deg(x_j)} \tm_k(x_1 \tensor \cdots \tensor d x_i \tensor \cdots \tensor x_k) + \\

  1133 %                                         & \qquad + + \\

  1134 %                                         & \qquad +