author | scott@6e1638ff-ae45-0410-89bd-df963105f760 |
Mon, 30 Jun 2008 20:20:52 +0000 | |
changeset 21 | b7812497643a |
parent 19 | ea489bbccfbf |
child 28 | f844cffa5c03 |
permissions | -rw-r--r-- |
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rearranging the Hochschild section. Splitting things up into lemmas, and explaining why those lemmas are what we need.
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1 |
In this section we analyze the blob complex in dimension $n=1$ |
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and find that for $S^1$ the homology of the blob complex is the |
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Hochschild homology of the category (algebroid) that we started with. |
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4 |
\nn{or maybe say here that the complexes are quasi-isomorphic? in general, |
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5 |
should perhaps put more emphasis on the complexes and less on the homology.} |
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6 |
|
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7 |
Notation: $HB_i(X) = H_i(\bc_*(X))$. |
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8 |
|
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9 |
Let us first note that there is no loss of generality in assuming that our system of |
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fields comes from a category. |
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11 |
(Or maybe (???) there {\it is} a loss of generality. |
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12 |
Given any system of fields, $A(I; a, b) = \cC(I; a, b)/U(I; a, b)$ can be |
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13 |
thought of as the morphisms of a 1-category $C$. |
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14 |
More specifically, the objects of $C$ are $\cC(pt)$, the morphisms from $a$ to $b$ |
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15 |
are $A(I; a, b)$, and composition is given by gluing. |
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16 |
If we instead take our fields to be $C$-pictures, the $\cC(pt)$ does not change |
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and neither does $A(I; a, b) = HB_0(I; a, b)$. |
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18 |
But what about $HB_i(I; a, b)$ for $i > 0$? |
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19 |
Might these higher blob homology groups be different? |
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20 |
Seems unlikely, but I don't feel like trying to prove it at the moment. |
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21 |
In any case, we'll concentrate on the case of fields based on 1-category |
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22 |
pictures for the rest of this section.) |
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23 |
|
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24 |
(Another question: $\bc_*(I)$ is an $A_\infty$-category. |
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25 |
How general of an $A_\infty$-category is it? |
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26 |
Given an arbitrary $A_\infty$-category can one find fields and local relations so |
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27 |
that $\bc_*(I)$ is in some sense equivalent to the original $A_\infty$-category? |
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28 |
Probably not, unless we generalize to the case where $n$-morphisms are complexes.) |
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29 |
|
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30 |
Continuing... |
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31 |
|
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32 |
Let $C$ be a *-1-category. |
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33 |
Then specializing the definitions from above to the case $n=1$ we have: |
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34 |
\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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37 |
Then an element of $\cC(R; c)$ is a collection of (transversely oriented) |
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38 |
points in the interior |
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39 |
of $R$, each labeled by a morphism of $C$. |
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40 |
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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42 |
\item There is an evaluation map $e: \cC(I; a, b) \to \mor(a, b)$ given by |
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43 |
composing the morphism labels of the points. |
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44 |
Note that we also need the * of *-1-category here in order to make all the morphisms point |
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45 |
the same way. |
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46 |
\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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47 |
point (at some standard location) labeled by $x$. |
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48 |
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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50 |
Thus we can, if we choose, restrict the blob twig labels to things of this form. |
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51 |
\end{itemize} |
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52 |
|
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53 |
We want to show that $HB_*(S^1)$ is naturally isomorphic to the |
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54 |
Hochschild homology of $C$. |
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55 |
\nn{Or better that the complexes are homotopic |
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or 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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59 |
%Given an interval (1-ball) so labeled, there is an evaluation map to some tensor product |
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%(over $C$) of $C$-$C$-bimodules. |
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%Define the local relations $U(I; a, b)$ to be the direct sum of the kernels of these maps. |
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%Now we can define the blob complex for $S^1$. |
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63 |
%This complex is the sum of complexes with a fixed cyclic tuple of bimodules present. |
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64 |
%If $M$ is a $C$-$C$-bimodule, let $G_*(M)$ denote the summand of $\bc_*(S^1)$ corresponding |
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%to the cyclic 1-tuple $(M)$. |
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66 |
%In other words, $G_*(M)$ is a blob-like complex where exactly one point is labeled |
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%by an element of $M$ and the remaining points are labeled by morphisms of $C$. |
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%It's clear that $G_*(C)$ is isomorphic to the original bimodule-less |
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%blob complex for $S^1$. |
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%\nn{Is it really so clear? Should say more.} |
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71 |
|
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%\nn{alternative to the above paragraph:} |
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73 |
Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$. |
19 | 74 |
We define a blob-like complex $K_*(S^1, (p_i), (M_i))$. |
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75 |
The fields have elements of $M_i$ labeling $p_i$ and elements of $C$ labeling |
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76 |
other points. |
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The blob twig labels lie in kernels of evaluation maps. |
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78 |
(The range of these evaluation maps is a tensor product (over $C$) of $M_i$'s.) |
19 | 79 |
Let $K_*(M) = K_*(S^1, (*), (M))$, where $* \in S^1$ is some standard base point. |
80 |
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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82 |
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\todo{Some orphaned questions:} |
19 | 84 |
\nn{Or maybe we should claim that $M \to K_*(M)$ is the/a derived coend. |
85 |
Or maybe that $K_*(M)$ is quasi-isomorphic (or perhaps homotopic) to the Hochschild |
|
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86 |
complex of $M$.} |
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87 |
|
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88 |
\nn{What else needs to be said to establish quasi-isomorphism to Hochschild complex? |
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89 |
Do we need a map from hoch to blob? |
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90 |
Does the above exactness and contractibility guarantee such a map without writing it |
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91 |
down explicitly? |
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92 |
Probably it's worth writing down an explicit map even if we don't need to.} |
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93 |
|
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94 |
|
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95 |
We claim that |
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96 |
\begin{thm} |
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97 |
The blob complex $\bc_*(S^1; C)$ on the circle is quasi-isomorphic to the |
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98 |
usual Hochschild complex for $C$. |
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99 |
\end{thm} |
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100 |
|
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101 |
This follows from two results. First, we see that |
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102 |
\begin{lem} |
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103 |
\label{lem:module-blob}% |
19 | 104 |
The complex $K_*(C)$ (here $C$ is being thought of as a |
15
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105 |
$C$-$C$-bimodule, not a category) is quasi-isomorphic to the blob complex |
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106 |
$\bc_*(S^1; C)$. (Proof later.) |
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107 |
\end{lem} |
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108 |
|
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109 |
Next, we show that for any $C$-$C$-bimodule $M$, |
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110 |
\begin{prop} |
19 | 111 |
The complex $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual |
15
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112 |
Hochschild complex of $M$. |
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113 |
\end{prop} |
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114 |
\begin{proof} |
19 | 115 |
%First, since we're working over $\Complex$, note that saying two complexes are quasi-isomorphic simply means they have isomorphic homologies. |
116 |
%\todo{We really don't want to work over $\Complex$, though; it would be nice to talk about torsion!} |
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117 |
|
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118 |
Recall that the usual Hochschild complex of $M$ is uniquely determined, up to quasi-isomorphism, by the following properties: |
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119 |
\begin{enumerate} |
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120 |
\item \label{item:hochschild-additive}% |
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121 |
$HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$. |
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122 |
\item \label{item:hochschild-exact}% |
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123 |
An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
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124 |
exact sequence $0 \to HC_*(M_1) \into HC_*(M_2) \onto HC_*(M_3) \to 0$. |
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125 |
\item \label{item:hochschild-free}% |
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126 |
$HC_*(C\otimes C)$ (the free $C$-$C$-bimodule with one generator) is |
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127 |
quasi-isomorphic to the 0-step complex $C$. |
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128 |
\item \label{item:hochschild-coinvariants}% |
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129 |
$HH_0(M)$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. |
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130 |
\end{enumerate} |
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131 |
(Together, these just say that Hochschild homology is `the derived functor of coinvariants'.) |
19 | 132 |
We'll first recall why these properties are characteristic. |
133 |
||
134 |
Take some $C$-$C$ bimodule $M$, and choose a free resolution |
|
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135 |
\begin{equation*} |
19 | 136 |
\cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0. |
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137 |
\end{equation*} |
19 | 138 |
There's a quotient map $\pi: F_0 \onto M$, and by construction the cone of the chain map $\pi: F_j \to M$ is acyclic. Now construct the total complex |
139 |
$HC_i(F_j)$, with $i,j \geq 0$, graded by $i+j$. |
|
140 |
||
141 |
Observe that we have two chain maps |
|
142 |
\begin{align*} |
|
143 |
HC_i(F_j) & \xrightarrow{HC_i(\pi)} HC_i(M) \\ |
|
144 |
\intertext{and} |
|
145 |
HC_i(F_j) & \xrightarrow{HC_0(F_j) \onto HH_0(F_j)} \operatorname{coinv}(F_j). |
|
146 |
\end{align*} |
|
147 |
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. |
|
148 |
In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free. |
|
149 |
||
150 |
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 |
|
151 |
$$HC_*(M) \iso \operatorname{coinv}(F_*).$$ |
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152 |
|
19 | 153 |
%If $M$ is free, that is, a direct sum of copies of |
154 |
%$C \tensor C$, then properties \ref{item:hochschild-additive} and |
|
155 |
%\ref{item:hochschild-free} determine $HC_*(M)$. Otherwise, choose some |
|
156 |
%free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we |
|
157 |
%have a short exact sequence $0 \to K \into F \onto M \to 0$, and hence a |
|
158 |
%short exact sequence of complexes $0 \to HC_*(K) \into HC_*(F) \onto HC_*(M) |
|
159 |
%\to 0$. Such a sequence gives a long exact sequence on homology |
|
160 |
%\begin{equation*} |
|
161 |
%%\begin{split} |
|
162 |
%\cdots \to HH_{i+1}(F) \to HH_{i+1}(M) \to HH_i(K) \to HH_i(F) \to \cdots % \\ |
|
163 |
%%\cdots \to HH_1(F) \to HH_1(M) \to HH_0(K) \to HH_0(F) \to HH_0(M). |
|
164 |
%%\end{split} |
|
165 |
%\end{equation*} |
|
166 |
%For any $i \geq 1$, $HH_{i+1}(F) = HH_i(F) = 0$, by properties |
|
167 |
%\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so |
|
168 |
%$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}. |
|
169 |
% |
|
170 |
%This tells us how to |
|
171 |
%compute every homology group of $HC_*(M)$; we already know $HH_0(M)$ |
|
172 |
%(it's just coinvariants, by property \ref{item:hochschild-coinvariants}), |
|
173 |
%and higher homology groups are determined by lower ones in $HC_*(K)$, and |
|
174 |
%hence recursively as coinvariants of some other bimodule. |
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175 |
|
19 | 176 |
The proposition then follows from the following lemmas, establishing that $K_*$ has precisely these required properties. |
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177 |
\begin{lem} |
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178 |
\label{lem:hochschild-additive}% |
19 | 179 |
Directly from the definition, $K_*(M_1 \oplus M_2) \cong K_*(M_1) \oplus K_*(M_2)$. |
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180 |
\end{lem} |
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181 |
\begin{lem} |
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182 |
\label{lem:hochschild-exact}% |
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183 |
An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
19 | 184 |
exact sequence $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$. |
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185 |
\end{lem} |
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186 |
\begin{lem} |
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187 |
\label{lem:hochschild-free}% |
19 | 188 |
$K_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$. |
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189 |
\end{lem} |
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190 |
\begin{lem} |
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191 |
\label{lem:hochschild-coinvariants}% |
19 | 192 |
$H_0(K_*(M))$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. |
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193 |
\end{lem} |
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194 |
|
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195 |
The remainder of this section is devoted to proving Lemmas |
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196 |
\ref{lem:module-blob}, |
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197 |
\ref{lem:hochschild-exact}, \ref{lem:hochschild-free} and |
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198 |
\ref{lem:hochschild-coinvariants}. |
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199 |
\end{proof} |
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200 |
|
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201 |
\begin{proof}[Proof of Lemma \ref{lem:module-blob}] |
19 | 202 |
We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$. |
203 |
$K_*(C)$ differs from $\bc_*(S^1)$ only in that the base point * |
|
204 |
is always a labeled point in $K_*(C)$, while in $\bc_*(S^1)$ it may or may not be. |
|
205 |
In other words, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$. |
|
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206 |
|
19 | 207 |
We define a quasi-inverse \nn{right term?} $s: \bc_*(S^1) \to K_*(C)$ to the inclusion as follows. |
15
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208 |
If $y$ is a field defined on a neighborhood of *, define $s(y) = y$ if |
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* is a labeled point in $y$. |
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210 |
Otherwise, define $s(y)$ to be the result of adding a label 1 (identity morphism) at *. |
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211 |
Let $x \in \bc_*(S^1)$. |
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212 |
Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in |
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213 |
$x$ with $y$. |
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214 |
It is easy to check that $s$ is a chain map and $s \circ i = \id$. |
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215 |
|
19 | 216 |
Let $L_*^\ep \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points |
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217 |
in a neighborhood $B_\ep$ of *, except perhaps *. |
19 | 218 |
Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
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219 |
\nn{rest of argument goes similarly to above} |
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220 |
\end{proof} |
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221 |
\begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
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222 |
\todo{} |
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223 |
\end{proof} |
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\begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] |
19 | 225 |
We show that $K_*(C\otimes C)$ is |
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226 |
quasi-isomorphic to the 0-step complex $C$. |
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227 |
|
19 | 228 |
Let $K'_* \sub K_*(C\otimes C)$ be the subcomplex where the label of |
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229 |
the point $*$ is $1 \otimes 1 \in C\otimes C$. |
19 | 230 |
We will show that the inclusion $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism. |
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231 |
|
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232 |
Fix a small $\ep > 0$. |
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233 |
Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$. |
19 | 234 |
Let $K_*^\ep \sub K_*(C\otimes C)$ be the subcomplex |
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235 |
generated by blob diagrams $b$ such that $B_\ep$ is either disjoint from |
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236 |
or contained in each blob of $b$, and the two boundary points of $B_\ep$ are not labeled points of $b$. |
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237 |
For a field (picture) $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ |
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238 |
labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$. |
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239 |
(See Figure xxxx.) |
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240 |
Note that $y - s_\ep(y) \in U(B_\ep)$. |
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241 |
\nn{maybe it's simpler to assume that there are no labeled points, other than $*$, in $B_\ep$.} |
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242 |
|
19 | 243 |
Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. |
244 |
Let $x \in K_*^\ep$ be a blob diagram. |
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245 |
If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $B_\ep$ to |
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246 |
$x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $B_\ep$. |
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247 |
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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248 |
Let $y_i$ be the restriction of $z_i$ to $B_\ep$. |
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249 |
Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin B_\ep$, |
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250 |
and have an additional blob $B_\ep$ with label $y_i - s_\ep(y_i)$. |
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251 |
Define $j_\ep(x) = \sum x_i$. |
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252 |
\nn{need to check signs coming from blob complex differential} |
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253 |
|
19 | 254 |
Note that if $x \in K'_* \cap K_*^\ep$ then $j_\ep(x) \in K'_*$ also. |
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255 |
|
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256 |
The key property of $j_\ep$ is |
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257 |
\eq{ |
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258 |
\bd j_\ep + j_\ep \bd = \id - \sigma_\ep , |
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259 |
} |
19 | 260 |
where $\sigma_\ep : K_*^\ep \to K_*^\ep$ is given by replacing the restriction $y$ of each field |
261 |
mentioned in $x \in K_*^\ep$ with $s_\ep(y)$. |
|
262 |
Note that $\sigma_\ep(x) \in K'_*$. |
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263 |
|
19 | 264 |
If $j_\ep$ were defined on all of $K_*(C\otimes C)$, it would show that $\sigma_\ep$ |
265 |
is a homotopy inverse to the inclusion $K'_* \to K_*(C\otimes C)$. |
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266 |
One strategy would be to try to stitch together various $j_\ep$ for progressively smaller |
19 | 267 |
$\ep$ and show that $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$. |
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268 |
Instead, we'll be less ambitious and just show that |
19 | 269 |
$K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. |
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270 |
|
19 | 271 |
If $x$ is a cycle in $K_*(C\otimes C)$, then for sufficiently small $\ep$ we have |
272 |
$x \in K_*^\ep$. |
|
273 |
(This is true for any chain in $K_*(C\otimes C)$, since chains are sums of |
|
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274 |
finitely many blob diagrams.) |
19 | 275 |
Then $x$ is homologous to $s_\ep(x)$, which is in $K'_*$, so the inclusion map |
276 |
$K'_* \sub K_*(C\otimes C)$ is surjective on homology. |
|
277 |
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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278 |
and |
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279 |
\eq{ |
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280 |
\bd y = \bd (\sigma_\ep(y) + j_\ep(x)) . |
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281 |
} |
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282 |
Since $\sigma_\ep(y) + j_\ep(x) \in F'$, it follows that the inclusion map is injective on homology. |
19 | 283 |
This completes the proof that $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. |
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284 |
|
19 | 285 |
Let $K''_* \sub K'_*$ be the subcomplex of $K'_*$ where $*$ is not contained in any blob. |
286 |
We will show that the inclusion $i: K''_* \to K'_*$ is a homotopy equivalence. |
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287 |
|
19 | 288 |
First, a lemma: Let $G''_*$ and $G'_*$ be defined the same as $K''_*$ and $K'_*$, except with |
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289 |
$S^1$ replaced some (any) neighborhood of $* \in S^1$. |
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290 |
Then $G''_*$ and $G'_*$ are both contractible |
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291 |
and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence. |
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292 |
For $G'_*$ the proof is the same as in (\ref{bcontract}), except that the splitting |
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293 |
$G'_0 \to H_0(G'_*)$ concentrates the point labels at two points to the right and left of $*$. |
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294 |
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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296 |
Thus any cycle lies in the image of the normal blob complex of a disjoint union |
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297 |
of two intervals, which is contractible by (\ref{bcontract}) and (\ref{disjunion}). |
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298 |
Actually, we need the further (easy) result that the inclusion |
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299 |
$G''_* \to G'_*$ induces an isomorphism on $H_0$. |
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300 |
|
19 | 301 |
Next we construct a degree 1 map (homotopy) $h: K'_* \to K'_*$ such that |
302 |
for all $x \in K'_*$ we have |
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303 |
\eq{ |
19 | 304 |
x - \bd h(x) - h(\bd x) \in K''_* . |
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305 |
} |
19 | 306 |
Since $K'_0 = K''_0$, we can take $h_0 = 0$. |
307 |
Let $x \in K'_1$, with single blob $B \sub S^1$. |
|
308 |
If $* \notin B$, then $x \in K''_1$ and we define $h_1(x) = 0$. |
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309 |
If $* \in B$, then we work in the image of $G'_*$ and $G''_*$ (with respect to $B$). |
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310 |
Choose $x'' \in G''_1$ such that $\bd x'' = \bd x$. |
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311 |
Since $G'_*$ is contractible, there exists $y \in G'_2$ such that $\bd y = x - x''$. |
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312 |
Define $h_1(x) = y$. |
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313 |
The general case is similar, except that we have to take lower order homotopies into account. |
19 | 314 |
Let $x \in K'_k$. |
15
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315 |
If $*$ is not contained in any of the blobs of $x$, then define $h_k(x) = 0$. |
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316 |
Otherwise, let $B$ be the outermost blob of $x$ containing $*$. |
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317 |
By xxxx above, $x = x' \bullet p$, where $x'$ is supported on $B$ and $p$ is supported away from $B$. |
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318 |
So $x' \in G'_l$ for some $l \le k$. |
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319 |
Choose $x'' \in G''_l$ such that $\bd x'' = \bd (x' - h_{l-1}\bd x')$. |
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320 |
Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$. |
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321 |
Define $h_k(x) = y \bullet p$. |
19 | 322 |
This completes the proof that $i: K''_* \to K'_*$ is a homotopy equivalence. |
15
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323 |
\nn{need to say above more clearly and settle on notation/terminology} |
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324 |
|
19 | 325 |
Finally, we show that $K''_*$ is contractible. |
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326 |
\nn{need to also show that $H_0$ is the right thing; easy, but I won't do it now} |
19 | 327 |
Let $x$ be a cycle in $K''_*$. |
15
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328 |
The union of the supports of the diagrams in $x$ does not contain $*$, so there exists a |
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329 |
ball $B \subset S^1$ containing the union of the supports and not containing $*$. |
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330 |
Adding $B$ as a blob to $x$ gives a contraction. |
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331 |
\nn{need to say something else in degree zero} |
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332 |
\end{proof} |
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333 |
\begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
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334 |
\todo{} |
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335 |
\end{proof} |
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336 |
|
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337 |
We can also describe explicitly a map from the standard Hochschild |
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338 |
complex to the blob complex on the circle. \nn{What properties does this |
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339 |
map have?} |
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340 |
|
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341 |
\begin{figure}% |
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342 |
$$\mathfig{0.6}{barycentric/barycentric}$$ |
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343 |
\caption{The Hochschild chain $a \tensor b \tensor c$ is sent to |
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344 |
the sum of six blob $2$-chains, corresponding to a barycentric subdivision of a $2$-simplex.} |
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345 |
\label{fig:Hochschild-example}% |
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346 |
\end{figure} |
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347 |
|
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348 |
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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349 |
The edges marked $x, y$ and $z$ carry the $1$-chains |
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350 |
\begin{align*} |
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351 |
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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352 |
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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353 |
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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354 |
\end{align*} |
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355 |
and the $2$-chain labelled $A$ is |
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356 |
\begin{equation*} |
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357 |
A = \mathfig{0.1}{barycentric/Ax}+\mathfig{0.1}{barycentric/Ay}. |
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358 |
\end{equation*} |
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359 |
Note that we then have |
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360 |
\begin{equation*} |
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361 |
\bdy A = x+y+z. |
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362 |
\end{equation*} |
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363 |
|
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364 |
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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365 |
$$\phi\left(\Tensor_{i=1}^n a_i\right) = \sum_{\pi} \phi^\pi(a_1, \ldots, a_n)$$ |
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366 |
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!) |