109 $$\mathfig{0.5}{smallblobs/tent}$$ |
109 $$\mathfig{0.5}{smallblobs/tent}$$ |
110 \caption{``Erecting a tent badly.'' We know where we want to send a simplex, and each of the iterated boundary components. However, these do not agree, and we need to stitch the pieces together. Note that these diagrams don't exactly match the situation in the text: a $k$-simplex has $k+1$ boundary components, while a $k$-blob has $k$ boundary terms.} |
110 \caption{``Erecting a tent badly.'' We know where we want to send a simplex, and each of the iterated boundary components. However, these do not agree, and we need to stitch the pieces together. Note that these diagrams don't exactly match the situation in the text: a $k$-simplex has $k+1$ boundary components, while a $k$-blob has $k$ boundary terms.} |
111 \label{fig:erectly-a-tent-badly} |
111 \label{fig:erectly-a-tent-badly} |
112 \end{figure} |
112 \end{figure} |
113 |
113 |
114 \nn{lots of signs are wrong ...} |
|
115 \begin{align*} |
114 \begin{align*} |
116 \bdy s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(\bdy b) + \\ |
115 \bdy s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(\bdy b) + \\ |
117 & \quad + (-1)^{\sigma(1)} \left( \restrict{\phi_{\beta}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) \right) + \\ |
116 & \quad + (-1)^{\sigma(1)} \left( \restrict{\phi_{\beta}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) \right) + \\ |
118 & \quad + (-1)^{\sigma(2)} \left( \restrict{\phi_{\beta}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) \right) + \\ |
117 & \quad + (-1)^{\sigma(2)} \left( \restrict{\phi_{\beta}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) \right) + \\ |
119 & \quad + (-1)^{\sigma(12)} \left( \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_2}}{x_0 = 0}(b_{12}) \right) + \\ |
118 & \quad + (-1)^{\sigma(12)} \left( \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_2}}{x_0 = 0}(b_{12}) \right) + \\ |
120 & \quad + (-1)^{\sigma(21)} \left( \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_1}}{x_0 = 0}(b_{12}) \right), \\ |
119 & \quad + (-1)^{\sigma(21)} \left( \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_1}}{x_0 = 0}(b_{12}) \right), \\ |
121 \intertext{while} |
120 \intertext{while} |
122 s(\bdy(b)) & = s(b_1) - s(b_2) \\ |
121 s(\bdy(b)) & = s(b_1) - s(b_2) \\ |
123 & = \restrict{\phi_{\beta_1}}{x_0=0}(b_1) - \restrict{\phi_{\eset \prec \beta_1}}{x_0=0}(b_{12}) - \restrict{\phi_{\beta_2}}{x_0=0}(b_2) + \restrict{\phi_{\eset \prec \beta_2}}{x_0=0}(b_{12}) . |
122 & = \restrict{\phi_{\beta_1}}{x_0=0}(b_1) - \restrict{\phi_{\eset \prec \beta_1}}{x_0=0}(b_{12}) - \restrict{\phi_{\beta_2}}{x_0=0}(b_2) + \restrict{\phi_{\eset \prec \beta_2}}{x_0=0}(b_{12}) . |
124 \end{align*} |
123 \end{align*} |
125 \nn{that does indeed work, modulo signs, with $\sigma() = 1,\sigma(1)=-1, \sigma(2)=1, \sigma(21)=-1, \sigma(12)=1$} |
124 \nn{that does indeed work, with $\sigma() = 1,\sigma(1)=-1, \sigma(2)=1, \sigma(21)=-1, \sigma(12)=1$} |
126 |
125 |
127 We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. |
126 We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. |
128 We first do some preliminary calculations, and introduce yet more notation. For $i \in \{1, \ldots, k\}^{m} \setminus \Delta$ and $1 \leq p \leq m$, we'll denote by $i \setminus i_p$ the sequence in $\{1, \ldots, k-1\}^{m-1} \setminus \Delta$ obtained by deleting the $p$-th entry of $i$, and reducing all entries which are greater than $i_p$ by one. Conversely, for $i \in \{1, \ldots, k-1\}^{m-1} \setminus \Delta$, $1 \leq p \leq m$ and $1 \leq q \leq k$, we'll denote by $i \ll_p q$ the sequence in $\{1, \ldots, k\}^{m} \setminus \Delta$ obtained by increasing any entries of $i$ which are at least $q$ by one, and inserting $q$ as the $p$-th entry, shifting later entries to the right. Note the natural bijection between the sets |
127 We first do some preliminary calculations, and introduce yet more notation. For $i \in \{1, \ldots, k\}^{m} \setminus \Delta$ and $1 \leq p \leq m$, we'll denote by $i \setminus i_p$ the sequence in $\{1, \ldots, k-1\}^{m-1} \setminus \Delta$ obtained by deleting the $p$-th entry of $i$, and reducing all entries which are greater than $i_p$ by one. Conversely, for $i \in \{1, \ldots, k-1\}^{m-1} \setminus \Delta$, $1 \leq p \leq m$ and $1 \leq q \leq k$, we'll denote by $i \ll_p q$ the sequence in $\{1, \ldots, k\}^{m} \setminus \Delta$ obtained by increasing any entries of $i$ which are at least $q$ by one, and inserting $q$ as the $p$-th entry, shifting later entries to the right. Note the natural bijection between the sets |
129 \begin{align} |
128 \begin{align} |
130 \setc{(i,p)}{i \in \{1, \ldots, k\}^{m} \setminus \Delta, 1 \leq p \leq m} & \iso \setc{(i,p,q)}{i \in \{1, \ldots, k-1\}^{m-1} \setminus \Delta, 1 \leq p \leq m, 1 \leq q \leq k} \notag \\ |
129 \setc{(i,p)}{i \in \{1, \ldots, k\}^{m} \setminus \Delta, 1 \leq p \leq m} & \iso \setc{(i,p,q)}{i \in \{1, \ldots, k-1\}^{m-1} \setminus \Delta, 1 \leq p \leq m, 1 \leq q \leq k} \notag \\ |
135 which we will use in a moment to re-index a summation. |
134 which we will use in a moment to re-index a summation. |
136 |
135 |
137 We then calculate |
136 We then calculate |
138 \begin{align*} |
137 \begin{align*} |
139 \bdy(s(b)) & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev\left(\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) \\ |
138 \bdy(s(b)) & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev\left(\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) \\ |
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139 \intertext{and begin by expanding out $\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})$,} |
140 & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} \Bigg(\sum_{p=1}^{m+1} (-1)^{\sigma(i)+p+1} \ev\left(\restrict{\phi_{i(b)}}{x_0 = x_p = 0}\tensor b_i\right) \Bigg) + \\ |
140 & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} \Bigg(\sum_{p=1}^{m+1} (-1)^{\sigma(i)+p+1} \ev\left(\restrict{\phi_{i(b)}}{x_0 = x_p = 0}\tensor b_i\right) \Bigg) + \\ |
141 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
141 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
142 \end{align*} |
142 \end{align*} |
143 First, write $s_{p_1,p}(i)$ to indicate the sequence obtained from $i$ by transposing its $p-1$-th and $p$-th entries. Now note that for $2 \leq p \leq m$, |
143 Now, write $s_{p_1,p}(i)$ to indicate the sequence obtained from $i$ by transposing its $p-1$-th and $p$-th entries and note that for $2 \leq p \leq m$, |
144 \begin{align*} |
144 \begin{align*} |
145 \restrict{\phi_{i(b)}}{x_0=x_p=0} & = \restrict{\phi_{\beta_{i_1\cdots i_m} \prec \beta_{i_2 \cdots i_m} \prec \cdots \prec \beta}}{x_0=x_p=0} \\ |
145 \restrict{\phi_{i(b)}}{x_0=x_p=0} & = \restrict{\phi_{\beta_{i_1\cdots i_m} \prec \beta_{i_2 \cdots i_m} \prec \cdots \prec \beta}}{x_0=x_p=0} \\ |
146 & = \restrict{\phi_{\beta_{i_1\cdots i_m} \prec \beta_{i_2 \cdots i_m} \prec \cdots \prec \beta_{i_{p-1} i_p \cdots i_m} \prec \beta_{i_{p+1} \cdots i_m} \prec \cdots \prec \beta}}{x_0=0} \\ |
146 & = \restrict{\phi_{\beta_{i_1\cdots i_m} \prec \beta_{i_2 \cdots i_m} \prec \cdots \prec \beta_{i_{p-1} i_p \cdots i_m} \prec \beta_{i_{p+1} \cdots i_m} \prec \cdots \prec \beta}}{x_0=0} \\ |
147 & = \restrict{\phi_{s_{p-1,p}(i)(b)}}{x_0=x_p=0}. |
147 & = \restrict{\phi_{s_{p-1,p}(i)(b)}}{x_0=x_p=0}. |
148 \end{align*} |
148 \end{align*} |
149 |
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150 Since $\sigma(i) = - \sigma(s_{p_1,p}(i))$, we can cancel out in pairs all the terms above except those with $p=1$ or $p=m+1$. Thus |
149 Since $\sigma(i) = - \sigma(s_{p_1,p}(i))$, we can cancel out in pairs all the terms above except those with $p=1$ or $p=m+1$. Thus |
151 \begin{align*} |
150 \begin{align*} |
152 \bdy(s(b)) & = ... |
151 \bdy(s(b)) & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} \Bigg((-1)^{\sigma(i)} \ev\left(\restrict{\phi_{\rest(i)(b)}}{x_0 = 0}\tensor b_i\right) + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{\most(i)(b_{i_m})}}{x_0 = 0}\tensor b_i\right)\Bigg) + \\ |
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152 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
153 \end{align*} |
153 \end{align*} |
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154 where we write $\rest(i)$ for the `tail' of $i$ and $\most(i)$ for the `head' of $i$, so $\rest(i_1 i_2 \cdots i_m) = (i_2 \cdots i_m)$ and $\most(i_1 \cdots i_{m-1} i_m) = (i_1 \cdots i_{m-1})$. Next, we note that $b_i = (b_{i_1})_{\rest(i)} = (b_{i_m})_{\most(i)}$, and then rewrite the sum of $i$ as a double sum over $i_1$ and $\rest(i)$, with $i = \rest(i) \ll_1 i_1$, for the first term, and as a double sum over $\most(i)$ and $i_m$, with $i = \most(i) \ll_{m} i_m$, for the second term. |
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155 \begin{align*} |
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156 \bdy(s(b)) & = \sum_{m=0}^{k} \Bigg( \sum_{\rest(i) \in \{1, \ldots, k\}^{m-1} \setminus \Delta} \sum_{\substack{i_1 = 1 \\ i_1 \not\in \rest(i)}}^{k} (-1)^{\sigma(i_1\rest(i))} \ev\left(\restrict{\phi_{\rest(i)(b)}}{x_0 = 0}\tensor b_{i_1\rest(i)}\right) \Bigg)+ \\ |
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157 & \qquad \Bigg( \sum_{\most(i) \in \{1, \ldots, k\}^{m-1} \setminus \Delta} \sum_{\substack{i_m = 1 \\ i_1 \not\in \most(i)}}^{k} (-1)^{\sigma(\most(i) i_m) + m} \ev\left(\restrict{\phi_{\most(i)(b_{i_m})}}{x_0 = 0}\tensor b_{\most(i) i_m}\right)\Bigg) + \\ |
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158 & \qquad \Bigg( \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right)\Bigg) |
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159 \end{align*} |
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160 Now $$\sum_{\substack{i_1 = 1 \\ i_1 \not\in \rest(i)}}^{k} (-1)^{\sigma(i_1\rest(i))} b_{i_1\rest(i)} = (-1)^{\sigma(\rest(i))} \bdy (b_{\rest(i)})$$ |
154 |
161 |
155 |
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156 \nn{Crap follows:} |
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157 \begin{align*} |
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158 & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta}(-1)^{\sigma(i)} \ev\left(\sum_{p=1}^m (-1)^{p+1} \restrict{\phi_{(i\setminus i_p)(b_{i_p})}}{x_0 = 0})\tensor (b_{i_p})_{(i \setminus i_p)}\right) + \\ |
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159 & \qquad \qquad \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k (-1)^{q+1+\card{\setc{r}{i_r < q}}} b_{i \cup \{q\}}\right). |
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160 \end{align*} |
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161 Notice the first term vanishes when $m=0$, and the second term vanishes when $m=k$, so it is convenient to rearrange the terms according to the degree of the family of diffeomorphisms. We obtain |
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162 \begin{align*} |
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163 \bdy(s(b)) & = \sum_{m=0}^{k-1} \sum_{i \in \{1, \ldots, k\}^{m+1} \setminus \Delta}(-1)^{\sigma(i) + p + 1} \ev\left(\sum_{p=1}^{m+1} \restrict{\phi_{(i\setminus i_p)(b_{i_p})}}{x_0 = 0})\tensor (b_{i_p})_{(i \setminus i_p)}\right) + \\ |
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164 & \qquad \sum_{m=0}^{k-1} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k (-1)^{q+1+\card{\setc{r}{i_r < q}}} b_{i \cup \{q\}}\right) \\ |
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165 \intertext{then reindex the first sum using the bijection from Equation \eqref{eq:reindexing-bijection}, giving} |
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166 & = \sum_{m=0}^{k-1} \sum_{i \in \{1, \ldots, k-1\}^{m} \setminus \Delta} \sum_{p=1}^{m+1} \sum_{q=1}^k (-1)^{\sigma(i \ll_p q) + p + 1} \ev\left( \restrict{\phi_{i(b_{q})}}{x_0 = 0})\tensor (b_{q})_{i}\right) + \\ |
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167 & \qquad \sum_{m=0}^{k-1} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k (-1)^{q+1+\card{\setc{r}{i_r < q}}} b_{i \cup \{q\}}\right) \\ |
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168 \end{align*} |
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169 |
162 |
170 On the other hand, we have |
163 On the other hand, we have |
171 \begin{align*} |
164 \begin{align*} |
172 s(\bdy b) & = \sum_{q=1}^k (-1)^{q+1} s(b_q) \\ |
165 s(\bdy b) & = \sum_{q=1}^k (-1)^{q+1} s(b_q) \\ |
173 & = \sum_{q=1}^k (-1)^{q+1} \sum_{m=0}^{k-1} \sum_{i \in \{1, \ldots, k-1\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b_q)}}{x_0 = 0} \tensor (b_q)_i). |
166 & = \sum_{q=1}^k (-1)^{q+1} \sum_{m=0}^{k-1} \sum_{i \in \{1, \ldots, k-1\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b_q)}}{x_0 = 0} \tensor (b_q)_i). |