86 \nn{revision marker ...} |
86 \nn{revision marker ...} |
87 |
87 |
88 \newcommand{\length}[1]{\operatorname{length}(#1)} |
88 \newcommand{\length}[1]{\operatorname{length}(#1)} |
89 |
89 |
90 We've finally reached the point where we can define a map $s: \bc_*(M) \to \bc^{\cU}_*(M)$, and then a homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ so that $dh+hd=i\circ s$. We have |
90 We've finally reached the point where we can define a map $s: \bc_*(M) \to \bc^{\cU}_*(M)$, and then a homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ so that $dh+hd=i\circ s$. We have |
91 $$s(b) = \sum_{i} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i)$$ |
91 $$s(b) = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i),$$ |
92 where the sum is over sequences $i=(i_1,\ldots,i_m)$ in $\{1,\ldots,k\}$, with $0\leq m \leq k$, $\sigma(i)$ is something to do with $i$, $i(b)$ denotes the increasing sequence of blob configurations |
92 where the sum is over sequences without repeats $i=(i_1,\ldots,i_m)$ in $\{1,\ldots,k\}$, with $0\leq m \leq k$ (we're using $\Delta$ here to indicate the generalized diagonal, where any two entries coincide), $\sigma(i)$ is something to do with $i$, $i(b)$ denotes the increasing sequence of blob configurations |
93 $$\beta_{(i_1,\ldots,i_m)} \prec \beta_{(i_2,\ldots,i_m)} \prec \cdots \prec \beta_{()},$$ |
93 $$\beta_{(i_1,\ldots,i_m)} \prec \beta_{(i_2,\ldots,i_m)} \prec \cdots \prec \beta_{()},$$ |
94 and, as usual, $b_i$ denotes $b$ with blobs $i_1, \ldots, i_m$ erased. We'll also write |
94 and, as usual, $b_i$ denotes $b$ with blobs $i_1, \ldots, i_m$ erased. |
95 $$s(b) = \sum_{m=0}^{k} \sum_{\length{i}=m} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i),$$ |
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96 where we arrange the sum according to the length of $i$. |
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97 The homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ is similarly given by |
95 The homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ is similarly given by |
98 $$h(b) = \sum_{i} (-1)^{\sigma(i)} \ev(\phi_{i(b)}, b_i).$$ |
96 $$h(b) = \sum_{m=0}^{k} \sum_{i} (-1)^{\sigma(i)} \ev(\phi_{i(b)}, b_i).$$ |
99 |
97 |
100 Before completing the proof, we unpack this definition for $b \in \bc_2(M)$, a $2$-blob. We'll write $\beta$ for the underlying balls (either nested or disjoint). |
98 Before completing the proof, we unpack this definition for $b \in \bc_2(M)$, a $2$-blob. We'll write $\beta$ for the underlying balls (either nested or disjoint). |
101 Now $s$ is the sum of $5$ terms, split into three groups depending on with the length of the sequence $i$ is $0, 1$ or $2$. Thus |
99 Now $s$ is the sum of $5$ terms, split into three groups depending on with the length of the sequence $i$ is $0, 1$ or $2$. Thus |
102 \begin{align*} |
100 \begin{align*} |
103 s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(b) + \\ |
101 s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(b) + \\ |
123 & \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 & \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), \\ |
124 \intertext{while} |
122 \intertext{while} |
125 s(\bdy(b)) & = s(b_1) - s(b_2) \\ |
123 s(\bdy(b)) & = s(b_1) - s(b_2) \\ |
126 & = \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 & = \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}) . |
127 \end{align*} |
125 \end{align*} |
128 \nn{that does indeed work, modulo signs} |
126 \nn{that does indeed work, modulo signs, with $\sigma() = 1,\sigma(1)=-1, \sigma(2)=1, \sigma(21)=-1, \sigma(12)=1$} |
129 |
127 |
130 We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. Calculate |
128 We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. |
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129 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 |
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130 \begin{align} |
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131 \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 \\ |
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132 \intertext{given by} |
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133 (i, p) & \mapsto (i \setminus i_p, p, i_p) \label{eq:reindexing-bijection} \\ |
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134 (i \ll_p q, p) & \mapsfrom (i,p,q) \notag |
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135 \end{align} |
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136 which we will use in a moment to re-index a summation. |
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137 |
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138 We then calculate |
131 \begin{align*} |
139 \begin{align*} |
132 \bdy(s(b)) & = \sum_{m=0}^{k} \sum_{\length{i}=m} (-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) \\ |
140 \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) \\ |
133 & = \sum_{m=0}^{k} \sum_{\length{i}=m}(-1)^{\sigma(i)} \ev\left(\sum_{p=1}^m \pm \restrict{\phi_{(i\setminus i_p)(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k \pm b_{i \cup \{q\}}\right) \\ |
141 & = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} \Bigg(\sum_{p=1}^m (-1)^{\sigma(i)+p+1} \ev\left(\restrict{\phi_{i(b)}}{x_0 = x_p = 0}\tensor b_i\right) \Bigg) + \\ |
134 \intertext{which we telescope as} |
142 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
135 & = \ev \left( \restrict{\phi_\beta}{x_0=0} \tensor \sum_{q=1}^k \pm b_{\{q\}}\right) + \\ |
143 \end{align*} |
136 & \qquad + \sum_{m=1}^{k-1} \sum_{\length{i}=m} \Bigg( \sum_{q=1}^{m+1} \sum_{\substack{i^+ \\ i = i^+ \setminus i^+_q}} (-1)^{\sigma(i^+)} \ev\left(\sum_{p=1}^m \pm \restrict{\phi_{i(b)}}{x_0 = 0})\tensor b_{i^+}\right) + \\ |
144 |
137 & \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 \pm b_{i \cup \{q\}}\right)\Bigg) \\ |
145 \nn{Crap follows:} |
138 & \qquad + (-1)^k \sum_{\length{i}=k}(-1)^{\sigma(i)} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k \pm b_{i \cup \{q\}}\right)\\ |
146 \begin{align*} |
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147 & = \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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148 & \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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149 \end{align*} |
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150 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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151 \begin{align*} |
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152 \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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153 & \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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154 \intertext{then reindex the first sum using the bijection from Equation \eqref{eq:reindexing-bijection}, giving} |
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155 & = \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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156 & \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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157 \end{align*} |
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158 |
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159 On the other hand, we have |
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160 \begin{align*} |
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161 s(\bdy b) & = \sum_{q=1}^k (-1)^{q+1} s(b_q) \\ |
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162 & = \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). |
139 \end{align*} |
163 \end{align*} |
140 \todo{to be continued...} |
164 \todo{to be continued...} |
141 |
165 |
142 Finally, we need to check that $dh+hd=i\circ s$. \todo{} |
166 Finally, we need to check that $dh+hd=i\circ s$. \todo{} |
143 \end{proof} |
167 \end{proof} |