author | kevin@6e1638ff-ae45-0410-89bd-df963105f760 |
Fri, 05 Jun 2009 16:10:37 +0000 | |
changeset 70 | 5ab0e6f0d89e |
parent 69 | d363611b1f59 |
child 74 | ea9f0b3c1b14 |
permissions | -rw-r--r-- |
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%!TEX root = ../blob1.tex |
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In this section we analyze the blob complex in dimension $n=1$ |
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and find that for $S^1$ the blob complex is homotopy equivalent to the |
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Hochschild complex of the category (algebroid) that we started with. |
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\nn{need to be consistent about quasi-isomorphic versus homotopy equivalent |
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in this section. |
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since the various complexes are free, q.i. implies h.e.} |
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Let $C$ be a *-1-category. |
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Then specializing the definitions from above to the case $n=1$ we have: |
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\begin{itemize} |
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\item $\cC(pt) = \ob(C)$ . |
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\item Let $R$ be a 1-manifold and $c \in \cC(\bd R)$. |
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Then an element of $\cC(R; c)$ is a collection of (transversely oriented) |
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points in the interior |
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of $R$, each labeled by a morphism of $C$. |
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The intervals between the points are labeled by objects of $C$, consistent with |
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the boundary condition $c$ and the domains and ranges of the point labels. |
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\item There is an evaluation map $e: \cC(I; a, b) \to \mor(a, b)$ given by |
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composing the morphism labels of the points. |
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Note that we also need the * of *-1-category here in order to make all the morphisms point |
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the same way. |
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\item For $x \in \mor(a, b)$ let $\chi(x) \in \cC(I; a, b)$ be the field with a single |
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point (at some standard location) labeled by $x$. |
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Then the kernel of the evaluation map $U(I; a, b)$ is generated by things of the |
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form $y - \chi(e(y))$. |
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Thus we can, if we choose, restrict the blob twig labels to things of this form. |
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\end{itemize} |
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We want to show that $\bc_*(S^1)$ is homotopy equivalent to the |
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Hochschild complex of $C$. |
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Note that both complexes are free (and hence projective), so it suffices to show that they |
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are quasi-isomorphic. |
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In order to prove this we will need to extend the blob complex to allow points to also |
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be labeled by elements of $C$-$C$-bimodules. |
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Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$. |
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We define a blob-like complex $K_*(S^1, (p_i), (M_i))$. |
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The fields have elements of $M_i$ labeling $p_i$ and elements of $C$ labeling |
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other points. |
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The blob twig labels lie in kernels of evaluation maps. |
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(The range of these evaluation maps is a tensor product (over $C$) of $M_i$'s.) |
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Let $K_*(M) = K_*(S^1, (*), (M))$, where $* \in S^1$ is some standard base point. |
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In other words, fields for $K_*(M)$ have an element of $M$ at the fixed point $*$ |
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and elements of $C$ at variable other points. |
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We claim that |
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\begin{thm} \label{hochthm} |
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The blob complex $\bc_*(S^1; C)$ on the circle is quasi-isomorphic to the |
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usual Hochschild complex for $C$. |
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\end{thm} |
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This follows from two results. First, we see that |
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\begin{lem} |
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\label{lem:module-blob}% |
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The complex $K_*(C)$ (here $C$ is being thought of as a |
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$C$-$C$-bimodule, not a category) is homotopy equivalent to the blob complex |
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$\bc_*(S^1; C)$. (Proof later.) |
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\end{lem} |
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Next, we show that for any $C$-$C$-bimodule $M$, |
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\begin{prop} \label{prop:hoch} |
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The complex $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual |
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Hochschild complex of $M$. |
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\end{prop} |
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\begin{proof} |
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Recall that the usual Hochschild complex of $M$ is uniquely determined, |
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up to quasi-isomorphism, by the following properties: |
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\begin{enumerate} |
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\item \label{item:hochschild-additive}% |
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$HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$. |
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\item \label{item:hochschild-exact}% |
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An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
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exact sequence $0 \to HC_*(M_1) \into HC_*(M_2) \onto HC_*(M_3) \to 0$. |
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\item \label{item:hochschild-coinvariants}% |
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$HH_0(M)$ is isomorphic to the coinvariants of $M$, $\coinv(M) = |
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M/\langle cm-mc \rangle$. |
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\item \label{item:hochschild-free}% |
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$HC_*(C\otimes C)$ is contractible. |
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(Here $C\otimes C$ denotes |
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the free $C$-$C$-bimodule with one generator.) |
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That is, $HC_*(C\otimes C)$ is |
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quasi-isomorphic to its $0$-th homology (which in turn, by \ref{item:hochschild-coinvariants}, is just $C$) via the quotient map $HC_0 \onto HH_0$. |
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\end{enumerate} |
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(Together, these just say that Hochschild homology is `the derived functor of coinvariants'.) |
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We'll first recall why these properties are characteristic. |
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Take some $C$-$C$ bimodule $M$, and choose a free resolution |
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\begin{equation*} |
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\cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0. |
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\end{equation*} |
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We will show that for any functor $\cP$ satisfying properties |
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\ref{item:hochschild-additive}, \ref{item:hochschild-exact}, |
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97 |
\ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there |
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98 |
is a quasi-isomorphism |
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99 |
$$\cP_*(M) \iso \coinv(F_*).$$ |
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100 |
% |
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101 |
Observe that there's a quotient map $\pi: F_0 \onto M$, and by |
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102 |
construction the cone of the chain map $\pi: F_* \to M$ is acyclic. Now |
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103 |
construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by |
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104 |
$i+j$. We have two chain maps |
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105 |
\begin{align*} |
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106 |
\cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\ |
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107 |
\intertext{and} |
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108 |
\cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \coinv(F_j). |
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109 |
\end{align*} |
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110 |
The cone of each chain map is acyclic. In the first case, this is because the `rows' indexed by $i$ are acyclic since $HC_i$ is exact. |
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111 |
In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free. |
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112 |
Because the cones are acyclic, the chain maps are quasi-isomorphisms. Composing one with the inverse of the other, we obtain the desired quasi-isomorphism |
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113 |
$$\cP_*(M) \quismto \coinv(F_*).$$ |
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114 |
|
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115 |
%If $M$ is free, that is, a direct sum of copies of |
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116 |
%$C \tensor C$, then properties \ref{item:hochschild-additive} and |
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117 |
%\ref{item:hochschild-free} determine $HC_*(M)$. Otherwise, choose some |
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118 |
%free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we |
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119 |
%have a short exact sequence $0 \to K \into F \onto M \to 0$, and hence a |
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120 |
%short exact sequence of complexes $0 \to HC_*(K) \into HC_*(F) \onto HC_*(M) |
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121 |
%\to 0$. Such a sequence gives a long exact sequence on homology |
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122 |
%\begin{equation*} |
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123 |
%%\begin{split} |
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124 |
%\cdots \to HH_{i+1}(F) \to HH_{i+1}(M) \to HH_i(K) \to HH_i(F) \to \cdots % \\ |
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125 |
%%\cdots \to HH_1(F) \to HH_1(M) \to HH_0(K) \to HH_0(F) \to HH_0(M). |
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126 |
%%\end{split} |
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127 |
%\end{equation*} |
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128 |
%For any $i \geq 1$, $HH_{i+1}(F) = HH_i(F) = 0$, by properties |
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129 |
%\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so |
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130 |
%$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}. |
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131 |
% |
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132 |
%This tells us how to |
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133 |
%compute every homology group of $HC_*(M)$; we already know $HH_0(M)$ |
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134 |
%(it's just coinvariants, by property \ref{item:hochschild-coinvariants}), |
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135 |
%and higher homology groups are determined by lower ones in $HC_*(K)$, and |
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136 |
%hence recursively as coinvariants of some other bimodule. |
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137 |
|
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138 |
Proposition \ref{prop:hoch} then follows from the following lemmas, establishing that $K_*$ has precisely these required properties. |
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139 |
\begin{lem} |
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140 |
\label{lem:hochschild-additive}% |
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141 |
Directly from the definition, $K_*(M_1 \oplus M_2) \cong K_*(M_1) \oplus K_*(M_2)$. |
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142 |
\end{lem} |
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143 |
\begin{lem} |
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144 |
\label{lem:hochschild-exact}% |
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145 |
An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
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146 |
exact sequence $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$. |
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147 |
\end{lem} |
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148 |
\begin{lem} |
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149 |
\label{lem:hochschild-coinvariants}% |
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150 |
$H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
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151 |
\end{lem} |
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152 |
\begin{lem} |
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153 |
\label{lem:hochschild-free}% |
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154 |
$K_*(C\otimes C)$ is quasi-isomorphic to $H_0(K_*(C \otimes C)) \iso C$. |
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155 |
\end{lem} |
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156 |
|
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157 |
The remainder of this section is devoted to proving Lemmas |
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158 |
\ref{lem:module-blob}, |
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159 |
\ref{lem:hochschild-exact}, \ref{lem:hochschild-coinvariants} and |
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160 |
\ref{lem:hochschild-free}. |
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161 |
\end{proof} |
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162 |
|
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163 |
\begin{proof}[Proof of Lemma \ref{lem:module-blob}] |
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164 |
We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$. |
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165 |
$K_*(C)$ differs from $\bc_*(S^1)$ only in that the base point * |
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166 |
is always a labeled point in $K_*(C)$, while in $\bc_*(S^1)$ it may or may not be. |
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167 |
In particular, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$. |
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168 |
|
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169 |
We define a left inverse $s: \bc_*(S^1) \to K_*(C)$ to the inclusion as follows. |
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170 |
If $y$ is a field defined on a neighborhood of *, define $s(y) = y$ if |
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171 |
* is a labeled point in $y$. |
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172 |
Otherwise, define $s(y)$ to be the result of adding a label 1 (identity morphism) at *. |
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173 |
Extending linearly, we get the desired map $s: \bc_*(S^1) \to K_*(C)$. |
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174 |
%Let $x \in \bc_*(S^1)$. |
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175 |
%Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in |
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176 |
%$x$ with $s(y)$. |
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177 |
It is easy to check that $s$ is a chain map and $s \circ i = \id$. |
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178 |
|
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179 |
Let $N_\ep$ denote the ball of radius $\ep$ around *. |
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180 |
Let $L_*^\ep \sub \bc_*(S^1)$ be the subcomplex |
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181 |
spanned by blob diagrams |
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182 |
where there are no labeled points |
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183 |
in $N_\ep$, except perhaps $*$, and $N_\ep$ is either disjoint from or contained in |
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184 |
every blob in the diagram. |
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185 |
Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
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186 |
\nn{what if * is on boundary of a blob? need preliminary homotopy to prevent this.} |
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187 |
|
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188 |
We define a degree $1$ chain map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. Let $x \in L_*^\ep$ be a blob diagram. |
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189 |
\nn{maybe add figures illustrating $j_\ep$?} |
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190 |
If $*$ is not contained in any twig blob, we define $j_\ep(x)$ by adding $N_\ep$ as a new twig blob, with label $y - s(y)$ where $y$ is the restriction |
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of $x$ to $N_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$, |
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write $y_i$ for the restriction of $z_i$ to $N_\ep$, and let |
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$x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin N_\ep$, |
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and have an additional blob $N_\ep$ with label $y_i - s(y_i)$. |
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Define $j_\ep(x) = \sum x_i$. |
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It is not hard to show that on $L_*^\ep$ |
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\[ |
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\bd j_\ep + j_\ep \bd = \id - i \circ s . |
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\] |
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\nn{need to check signs coming from blob complex differential} |
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Since for $\ep$ small enough $L_*^\ep$ captures all of the |
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homology of $\bc_*(S^1)$, |
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it follows that the mapping cone of $i \circ s$ is acyclic and therefore (using the fact that |
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these complexes are free) $i \circ s$ is homotopic to the identity. |
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\end{proof} |
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|
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\begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
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We now prove that $K_*$ is an exact functor. |
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|
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%\todo{p. 1478 of scott's notes} |
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Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules |
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\begin{equation*} |
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M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M) |
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\end{equation*} |
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is exact. For completeness we'll explain this below. |
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|
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Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
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We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
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Most of what we need to check is easy. |
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If $(a \tensor k \tensor b) \in \ker(C \tensor K \tensor C \to K)$, to have $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$, we must have $f(k) = 0 \in E$, which implies $k=0$ itself. If $(a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, clearly |
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$e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
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If $\hat{g}(a \tensor e \tensor b) = 0$, then $g(e) = 0$, so $e = f(\widetilde{e})$ for some $\widetilde{e} \in K$, and $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
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Finally, the interesting step is in checking that any $q = \sum_i a_i \tensor q_i \tensor b_i$ such that $\sum_i a_i q_i b_i = 0$ is in the image of $\ker(C \tensor E \tensor C \to C)$ under $\hat{g}$. |
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For each $i$, we can find $\widetilde{q_i}$ so $g(\widetilde{q_i}) = q_i$. However $\sum_i a_i \widetilde{q_i} b_i$ need not be zero. |
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Consider then $$\widetilde{q} = \sum_i (a_i \tensor \widetilde{q_i} \tensor b_i) - 1 \tensor (\sum_i a_i \widetilde{q_i} b_i) \tensor 1.$$ Certainly |
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$\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further, |
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\begin{align*} |
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\hat{g}(\widetilde{q}) & = \sum_i (a_i \tensor g(\widetilde{q_i}) \tensor b_i) - 1 \tensor (\sum_i a_i g(\widetilde{q_i}) b_i) \tensor 1 \\ |
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& = q - 0 |
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\end{align*} |
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(here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$). |
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|
69 | 234 |
Similar arguments show that the functors |
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\begin{equation} |
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\label{eq:ker-functor}% |
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M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
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\end{equation} |
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are all exact too. Moreover, tensor products of such functors with each |
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other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M) |
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\tensor C \tensor \ker(C \tensor C \to M)$) are all still exact. |
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|
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Finally, then we see that the functor $K_*$ is simply an (infinite) |
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direct sum of copies of this sort of functor. The direct sum is indexed by |
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configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors, |
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with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$, and all other labelled points corresponding |
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to tensor factors of $C$. |
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\end{proof} |
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\begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
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We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
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|
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We define a map $\ev: K_0(M) \to M$. If $x \in K_0(M)$ has the label $m \in M$ at $*$, and labels $c_i \in C$ at the other labeled points of $S^1$, reading clockwise from $*$, |
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we set $\ev(x) = m c_1 \cdots c_k$. We can think of this as $\ev : M \tensor C^{\tensor k} \to M$, for each direct summand of $K_0(M)$ indexed by a configuration of labeled points. |
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There is a quotient map $\pi: M \to \coinv{M}$, and the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; if $y \in K_1(M)$, the blob in $y$ either contains $*$ or does not. If it doesn't, then |
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suppose $y$ has label $m$ at $*$, labels $c_i$ at other labeled points outside the blob, and the field inside the blob is a sum, with the $j$-th term having |
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labeled points $d_{j,i}$. Then $\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \in \ker(\DirectSum_k C^{\tensor k} \to C)$, and so |
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$\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$ |
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Similarly, if $*$ is contained in the blob, then the blob label is a sum, with the $j$-th term have labelled points $d_{j,i}$ to the left of $*$, $m_j$ at $*$, and $d_{j,i}'$ to the right of $*$, |
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and there are labels $c_i$ at the labeled points outside the blob. We know that |
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$$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$ |
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and so |
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\begin{align*} |
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\ev(\bdy y) & = \sum_j m_j d_{j,1}' \cdots d_{j,k'_j}' c_1 \cdots c_k d_{j,1} \cdots d_{j,k_j} \\ |
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& = \sum_j d_{j,1} \cdots d_{j,k_j} m_j d_{j,1}' \cdots d_{j,k'_j}' c_1 \cdots c_k \\ |
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& = 0 |
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\end{align*} |
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where this time we use the fact that we're mapping to $\coinv{M}$, not just $M$. |
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|
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The map $\pi \compose \ev: H_0(K_*(M)) \to \coinv{M}$ is clearly surjective ($\ev$ surjects onto $M$); we now show that it's injective. \todo{} |
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\end{proof} |
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\begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] |
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We show that $K_*(C\otimes C)$ is |
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quasi-isomorphic to the 0-step complex $C$. We'll do this in steps, establishing quasi-isomorphisms and homotopy equivalences |
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$$K_*(C \tensor C) \quismto K'_* \htpyto K''_* \quismto C.$$ |
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|
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Let $K'_* \sub K_*(C\otimes C)$ be the subcomplex where the label of |
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the point $*$ is $1 \otimes 1 \in C\otimes C$. |
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We will show that the inclusion $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism. |
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|
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Fix a small $\ep > 0$. |
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Let $N_\ep$ be the ball of radius $\ep$ around $* \in S^1$. |
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Let $K_*^\ep \sub K_*(C\otimes C)$ be the subcomplex |
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generated by blob diagrams $b$ such that $N_\ep$ is either disjoint from |
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or contained in each blob of $b$, and the only labeled point inside $N_\ep$ is $*$. |
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%and the two boundary points of $N_\ep$ are not labeled points of $b$. |
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For a field $y$ on $N_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ |
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labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$. |
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(See Figure \ref{fig:sy}.) Note that $y - s_\ep(y) \in U(N_\ep)$. We can think of |
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$\sigma_\ep$ as a chain map $K_*^\ep \to K_*^\ep$ given by replacing the restriction $y$ to $N_\ep$ of each field |
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appearing in an element of $K_*^\ep$ with $s_\ep(y)$. |
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Note that $\sigma_\ep(x) \in K'_*$. |
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292 |
\begin{figure}[!ht] |
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\begin{align*} |
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y & = \mathfig{0.2}{hochschild/y} & |
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s_\ep(y) & = \mathfig{0.2}{hochschild/sy} |
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\end{align*} |
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\caption{Defining $s_\ep$.} |
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\label{fig:sy} |
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\end{figure} |
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300 |
|
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Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. |
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Let $x \in K_*^\ep$ be a blob diagram. |
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If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $N_\ep$ to |
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$x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $N_\ep$. |
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If $*$ is contained in a twig blob $B$ with label $u = \sum z_i$, $j_\ep(x)$ is obtained as follows. |
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Let $y_i$ be the restriction of $z_i$ to $N_\ep$. |
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Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin N_\ep$, |
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and have an additional blob $N_\ep$ with label $y_i - s_\ep(y_i)$. |
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Define $j_\ep(x) = \sum x_i$. |
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\nn{need to check signs coming from blob complex differential} |
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Note that if $x \in K'_* \cap K_*^\ep$ then $j_\ep(x) \in K'_*$ also. |
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312 |
|
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313 |
The key property of $j_\ep$ is |
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\eq{ |
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\bd j_\ep + j_\ep \bd = \id - \sigma_\ep. |
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316 |
} |
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317 |
If $j_\ep$ were defined on all of $K_*(C\otimes C)$, this would show that $\sigma_\ep$ |
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is a homotopy inverse to the inclusion $K'_* \to K_*(C\otimes C)$. |
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319 |
One strategy would be to try to stitch together various $j_\ep$ for progressively smaller |
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$\ep$ and show that $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$. |
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321 |
Instead, we'll be less ambitious and just show that |
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$K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. |
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323 |
|
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If $x$ is a cycle in $K_*(C\otimes C)$, then for sufficiently small $\ep$ we have |
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$x \in K_*^\ep$. |
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326 |
(This is true for any chain in $K_*(C\otimes C)$, since chains are sums of |
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327 |
finitely many blob diagrams.) |
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328 |
Then $x$ is homologous to $s_\ep(x)$, which is in $K'_*$, so the inclusion map |
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$K'_* \sub K_*(C\otimes C)$ is surjective on homology. |
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If $y \in K_*(C\otimes C)$ and $\bd y = x \in K'_*$, then $y \in K_*^\ep$ for some $\ep$ |
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331 |
and |
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332 |
\eq{ |
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333 |
\bd y = \bd (\sigma_\ep(y) + j_\ep(x)) . |
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334 |
} |
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335 |
Since $\sigma_\ep(y) + j_\ep(x) \in F'$, it follows that the inclusion map is injective on homology. |
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336 |
This completes the proof that $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. |
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337 |
|
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338 |
Let $K''_* \sub K'_*$ be the subcomplex of $K'_*$ where $*$ is not contained in any blob. |
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339 |
We will show that the inclusion $i: K''_* \to K'_*$ is a homotopy equivalence. |
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340 |
|
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First, a lemma: Let $G''_*$ and $G'_*$ be defined the same as $K''_*$ and $K'_*$, except with |
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342 |
$S^1$ replaced some (any) neighborhood of $* \in S^1$. |
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343 |
Then $G''_*$ and $G'_*$ are both contractible |
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and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence. |
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For $G'_*$ the proof is the same as in (\ref{bcontract}), except that the splitting |
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$G'_0 \to H_0(G'_*)$ concentrates the point labels at two points to the right and left of $*$. |
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347 |
For $G''_*$ we note that any cycle is supported \nn{need to establish terminology for this; maybe |
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in ``basic properties" section above} away from $*$. |
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Thus any cycle lies in the image of the normal blob complex of a disjoint union |
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of two intervals, which is contractible by (\ref{bcontract}) and (\ref{disjunion}). |
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351 |
Actually, we need the further (easy) result that the inclusion |
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$G''_* \to G'_*$ induces an isomorphism on $H_0$. |
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353 |
|
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354 |
Next we construct a degree 1 map (homotopy) $h: K'_* \to K'_*$ such that |
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355 |
for all $x \in K'_*$ we have |
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356 |
\eq{ |
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357 |
x - \bd h(x) - h(\bd x) \in K''_* . |
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358 |
} |
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359 |
Since $K'_0 = K''_0$, we can take $h_0 = 0$. |
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360 |
Let $x \in K'_1$, with single blob $B \sub S^1$. |
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361 |
If $* \notin B$, then $x \in K''_1$ and we define $h_1(x) = 0$. |
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If $* \in B$, then we work in the image of $G'_*$ and $G''_*$ (with respect to $B$). |
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363 |
Choose $x'' \in G''_1$ such that $\bd x'' = \bd x$. |
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364 |
Since $G'_*$ is contractible, there exists $y \in G'_2$ such that $\bd y = x - x''$. |
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365 |
Define $h_1(x) = y$. |
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366 |
The general case is similar, except that we have to take lower order homotopies into account. |
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367 |
Let $x \in K'_k$. |
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368 |
If $*$ is not contained in any of the blobs of $x$, then define $h_k(x) = 0$. |
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369 |
Otherwise, let $B$ be the outermost blob of $x$ containing $*$. |
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By xxxx above, $x = x' \bullet p$, where $x'$ is supported on $B$ and $p$ is supported away from $B$. |
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So $x' \in G'_l$ for some $l \le k$. |
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372 |
Choose $x'' \in G''_l$ such that $\bd x'' = \bd (x' - h_{l-1}\bd x')$. |
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373 |
Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$. |
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374 |
Define $h_k(x) = y \bullet p$. |
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375 |
This completes the proof that $i: K''_* \to K'_*$ is a homotopy equivalence. |
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376 |
\nn{need to say above more clearly and settle on notation/terminology} |
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377 |
|
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378 |
Finally, we show that $K''_*$ is contractible. |
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379 |
\nn{need to also show that $H_0$ is the right thing; easy, but I won't do it now} |
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380 |
Let $x$ be a cycle in $K''_*$. |
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381 |
The union of the supports of the diagrams in $x$ does not contain $*$, so there exists a |
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382 |
ball $B \subset S^1$ containing the union of the supports and not containing $*$. |
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383 |
Adding $B$ as a blob to $x$ gives a contraction. |
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384 |
\nn{need to say something else in degree zero} |
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385 |
\end{proof} |
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386 |
|
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387 |
We can also describe explicitly a map from the standard Hochschild |
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388 |
complex to the blob complex on the circle. \nn{What properties does this |
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389 |
map have?} |
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390 |
|
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391 |
\begin{figure}% |
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392 |
$$\mathfig{0.6}{barycentric/barycentric}$$ |
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393 |
\caption{The Hochschild chain $a \tensor b \tensor c$ is sent to |
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394 |
the sum of six blob $2$-chains, corresponding to a barycentric subdivision of a $2$-simplex.} |
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395 |
\label{fig:Hochschild-example}% |
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396 |
\end{figure} |
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397 |
|
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398 |
As an example, Figure \ref{fig:Hochschild-example} shows the image of the Hochschild chain $a \tensor b \tensor c$. Only the $0$-cells are shown explicitly. |
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399 |
The edges marked $x, y$ and $z$ carry the $1$-chains |
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400 |
\begin{align*} |
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401 |
x & = \mathfig{0.1}{barycentric/ux} & u_x = \mathfig{0.1}{barycentric/ux_ca} - \mathfig{0.1}{barycentric/ux_c-a} \\ |
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402 |
y & = \mathfig{0.1}{barycentric/uy} & u_y = \mathfig{0.1}{barycentric/uy_cab} - \mathfig{0.1}{barycentric/uy_ca-b} \\ |
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403 |
z & = \mathfig{0.1}{barycentric/uz} & u_z = \mathfig{0.1}{barycentric/uz_c-a-b} - \mathfig{0.1}{barycentric/uz_cab} |
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404 |
\end{align*} |
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405 |
and the $2$-chain labelled $A$ is |
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406 |
\begin{equation*} |
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407 |
A = \mathfig{0.1}{barycentric/Ax}+\mathfig{0.1}{barycentric/Ay}. |
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408 |
\end{equation*} |
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409 |
Note that we then have |
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410 |
\begin{equation*} |
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411 |
\bdy A = x+y+z. |
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412 |
\end{equation*} |
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413 |
|
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414 |
In general, the Hochschild chain $\Tensor_{i=1}^n a_i$ is sent to the sum of $n!$ blob $(n-1)$-chains, indexed by permutations, |
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415 |
$$\phi\left(\Tensor_{i=1}^n a_i\right) = \sum_{\pi} \phi^\pi(a_1, \ldots, a_n)$$ |
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416 |
with ... (hmmm, problems making this precise; you need to decide where to put the labels, but then it's hard to make an honest chain map!) |