96 |
96 |
97 \newcommand{\length}[1]{\operatorname{length}(#1)} |
97 \newcommand{\length}[1]{\operatorname{length}(#1)} |
98 |
98 |
99 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 |
99 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 |
100 $$s(b) = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i),$$ |
100 $$s(b) = \sum_{m=0}^{k} \sum_{i \in \{1, \ldots, k\}^{m} \setminus \Delta} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i),$$ |
101 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 |
101 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 defined blow, $i(b)$ denotes the increasing sequence of blob configurations |
102 $$\beta_{(i_1,\ldots,i_m)} \prec \beta_{(i_2,\ldots,i_m)} \prec \cdots \prec \beta_{()},$$ |
102 $$\beta_{(i_1,\ldots,i_m)} \prec \beta_{(i_2,\ldots,i_m)} \prec \cdots \prec \beta_{()},$$ |
103 and, as usual, $b_i$ denotes $b$ with blobs $i_1, \ldots, i_m$ erased. |
103 and, as usual, $b_i$ denotes $b$ with blobs $i_1, \ldots, i_m$ erased. |
104 The homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ is similarly given by |
104 The homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ is similarly given by |
105 $$h(b) = \sum_{m=0}^{k} \sum_{i} (-1)^{\sigma(i)} \ev(\phi_{i(b)}, b_i).$$ |
105 $$h(b) = \sum_{m=0}^{k} \sum_{i} (-1)^{\sigma(i)} \ev(\phi_{i(b)}, b_i).$$ |
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106 |
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107 Given a sequence $a \in \{1, \ldots, k\}^{m-1} \setminus \Delta$ and $1 \leq b \leq k$ with $b \not \in a$, denote by $a \!\downarrow_b\in \{1, \ldots, k-1\}^{m-1} \setminus \Delta$ the sequence obtained by reducing by 1 each entry of $a$ which is greater than $b$. Then $\sigma(i) = 0$ or $1$ is defined recursively by $\sigma() = 0$, and |
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108 $$\sigma(ab) = \sigma(a \!\downarrow_b) + m + b + 1 \mod{2}.$$ |
106 |
109 |
107 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). |
110 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). |
108 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 |
111 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 |
109 \begin{align*} |
112 \begin{align*} |
110 s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(b) + \\ |
113 s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(b) + \\ |
132 & = \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}) . |
135 & = \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}) . |
133 \end{align*} |
136 \end{align*} |
134 \nn{that does indeed work, with $\sigma() = 1,\sigma(1)=-1, \sigma(2)=1, \sigma(21)=-1, \sigma(12)=1$} |
137 \nn{that does indeed work, with $\sigma() = 1,\sigma(1)=-1, \sigma(2)=1, \sigma(21)=-1, \sigma(12)=1$} |
135 |
138 |
136 We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. |
139 We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. |
137 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 |
140 %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 |
138 \begin{align} |
141 %\begin{align} |
139 \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 \\ |
142 %\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 \\ |
140 \intertext{given by} |
143 %\intertext{given by} |
141 (i, p) & \mapsto (i \setminus i_p, p, i_p) \label{eq:reindexing-bijection} \\ |
144 %(i, p) & \mapsto (i \setminus i_p, p, i_p) \label{eq:reindexing-bijection} \\ |
142 (i \ll_p q, p) & \mapsfrom (i,p,q) \notag |
145 %(i \ll_p q, p) & \mapsfrom (i,p,q) \notag |
143 \end{align} |
146 %\end{align} |
144 which we will use in a moment to re-index a summation. |
147 %which we will use in a moment to re-index a summation. |
145 |
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146 We then calculate |
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147 \begin{align*} |
148 \begin{align*} |
148 \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) \\ |
149 \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) \\ |
149 \intertext{and begin by expanding out $\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})$,} |
150 \intertext{and begin by expanding out $\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})$,} |
150 & = \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) + \\ |
151 & = \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) + \\ |
151 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
152 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
152 \end{align*} |
153 \end{align*} |
153 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 $2 \leq p \leq m$, |
154 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 $2 \leq p \leq m$, |
154 \begin{align*} |
155 \begin{align*} |
155 \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} \\ |
156 \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} \\ |
156 & = \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} \\ |
157 & = \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} \\ |
157 & = \restrict{\phi_{s_{p-1,p}(i)(b)}}{x_0=x_p=0}. |
158 & = \restrict{\phi_{s_{p-1,p}(i)(b)}}{x_0=x_p=0}. |
158 \end{align*} |
159 \end{align*} |
159 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 |
160 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 |
160 \begin{align*} |
161 \begin{align*} |
161 \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) + \\ |
162 \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) + \\ |
162 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
163 & \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) |
163 \end{align*} |
164 \end{align*} |
164 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. |
165 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 $i = i_1\rest(i)$, for the first term, and as a double sum over $\most(i)$ and $i_m$, with $i = \most(i)i_m$, for the second term. |
165 \begin{align*} |
166 \begin{align*} |
166 \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)+ \\ |
167 \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)+ \\ |
167 & \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) + \\ |
168 & \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) + \\ |
168 & \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) |
169 & \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) |
169 \end{align*} |
170 \end{align*} |
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171 We will show that the first and third rows cancel, and that the second row gives with $s(\bdy b)$. |
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172 |
170 Now $$\sum_{\substack{i_1 = 1 \\ i_1 \not\in \rest(i)}}^{k} (-1)^{\sigma(i_1\rest(i))} b_{i_1\rest(i)} = (-1)^{\sigma(\rest(i))} \bdy (b_{\rest(i)})$$ |
173 Now $$\sum_{\substack{i_1 = 1 \\ i_1 \not\in \rest(i)}}^{k} (-1)^{\sigma(i_1\rest(i))} b_{i_1\rest(i)} = (-1)^{\sigma(\rest(i))} \bdy (b_{\rest(i)})$$ |
171 |
174 |
172 |
175 |
173 On the other hand, we have |
176 On the other hand, we have |
174 \begin{align*} |
177 \begin{align*} |
175 s(\bdy b) & = \sum_{q=1}^k (-1)^{q+1} s(b_q) \\ |
178 s(\bdy b) & = \sum_{q=1}^k (-1)^{q+1} s(b_q) \\ |
176 & = \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). |
179 & = \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). |
177 \end{align*} |
180 \end{align*} |
178 \todo{to be continued...} |
181 \todo{to be continued...} |
179 |
182 |
180 Finally, we need to check that $dh+hd=i\circ s$. \todo{} |
183 Finally, the calculation that $\bdy h+h \bdy=i\circ s - \id$ is very similar, and we omit it. |
181 \end{proof} |
184 \end{proof} |