blob: comparison text/hochschild.tex

equal deleted inserted replaced

-:675f53735445
+:1d76e832d32f
 So far we have provided no evidence that blob homology is interesting in degrees
 greater than zero.
 In this section we analyze the blob complex in dimension $n=1$.
 We find that $\bc_*(S^1, \cC)$ is homotopy equivalent to the
-Hochschild complex of the 1-category $\cC$. (Recall from \S \ref{sec:example:traditional-n-categories(fields)} that a $1$-category gives rise to a $1$-dimensional system of fields; as usual, talking about the blob complex with coefficients in a $n$-category means first passing to the corresponding $n$ dimensional system of fields.)
+Hochschild complex of the 1-category $\cC$.
+(Recall from \S \ref{sec:example:traditional-n-categories(fields)} that a
+$1$-category gives rise to a $1$-dimensional system of fields; as usual,
+talking about the blob complex with coefficients in a $n$-category means
+first passing to the corresponding $n$ dimensional system of fields.)
 Thus the blob complex is a natural generalization of something already
 known to be interesting in higher homological degrees.
 It is also worth noting that the original idea for the blob complex came from trying
 to find a more ``local" description of the Hochschild complex.
 \begin{thm} \label{hochthm}
 The blob complex $\bc_*(S^1; C)$ on the circle is homotopy equivalent to the
 usual Hochschild complex for $C$.
 \end{thm}
-This follows from two results. First, we see that
+This follows from two results.
+First, we see that
 \begin{lem}
 \label{lem:module-blob}%
 The complex $K_*(C)$ (here $C$ is being thought of as a
 $C$-$C$-bimodule, not a category) is homotopy equivalent to the blob complex
-$\bc_*(S^1; C)$. (Proof later.)
+$\bc_*(S^1; C)$.
+(Proof later.)
 \end{lem}
 Next, we show that for any $C$-$C$-bimodule $M$,
 \begin{prop} \label{prop:hoch}
 The complex $K_*(M)$ is homotopy equivalent to $\HC_*(M)$, the usual
 \ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there
 is a quasi-isomorphism
 $$\cP_*(M) \iso \coinv(F_*).$$
 %
 Observe that there's a quotient map $\pi: F_0 \onto M$, and by
-construction the cone of the chain map $\pi: F_* \to M$ is acyclic. Now
+construction the cone of the chain map $\pi: F_* \to M$ is acyclic.
-construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by
+Now construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by $i+j$.
-$i+j$. We have two chain maps
+We have two chain maps
 \begin{align*}
 \cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\
 \intertext{and}
 \cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \coinv(F_j).
 \end{align*}
-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.
+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.
 In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free.
-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
+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
 $$\cP_*(M) \quismto \coinv(F_*).$$
 %If $M$ is free, that is, a direct sum of copies of
 %$C \tensor C$, then properties \ref{item:hochschild-additive} and
 %\ref{item:hochschild-free} determine $\HC_*(M)$. Otherwise, choose some
 %compute every homology group of $\HC_*(M)$; we already know $\HH_0(M)$
 %(it's just coinvariants, by property \ref{item:hochschild-coinvariants}),
 %and higher homology groups are determined by lower ones in $\HC_*(K)$, and
 %hence recursively as coinvariants of some other bimodule.
-Proposition \ref{prop:hoch} then follows from the following lemmas, establishing that $K_*$ has precisely these required properties.
+Proposition \ref{prop:hoch} then follows from the following lemmas,
+establishing that $K_*$ has precisely these required properties.
 \begin{lem}
 \label{lem:hochschild-additive}%
 Directly from the definition, $K_*(M_1 \oplus M_2) \cong K_*(M_1) \oplus K_*(M_2)$.
 \end{lem}
 \begin{lem}
 In particular, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$.
 We want to define a homotopy inverse to the above inclusion, but before doing so
 we must replace $\bc_*(S^1)$ with a homotopy equivalent subcomplex.
 Let $J_* \sub \bc_*(S^1)$ be the subcomplex where * does not lie on the boundary
-of any blob.  Note that the image of $i$ is contained in $J_*$.
+of any blob.
+Note that the image of $i$ is contained in $J_*$.
 Note also that in $\bc_*(S^1)$ (away from $J_*$)
 a blob diagram could have multiple (nested) blobs whose
 boundaries contain *, on both the right and left of *.
 We claim that $J_*$ is homotopy equivalent to $\bc_*(S^1)$.
 where there are no labeled points
 in $N_\ep$, except perhaps $*$, and $N_\ep$ is either disjoint from or contained in
 every blob in the diagram.
 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$.
-We define a degree $1$ map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. Let $x \in L_*^\ep$ be a blob diagram.
+We define a degree $1$ map $j_\ep: L_*^\ep \to L_*^\ep$ as follows.
+Let $x \in L_*^\ep$ be a blob diagram.
 \nn{maybe add figures illustrating $j_\ep$?}
-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
+If $*$ is not contained in any twig blob, we define $j_\ep(x)$ by adding
-of $x$ to $N_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$,
+$N_\ep$ as a new twig blob, with label $y - s(y)$ where $y$ is the restriction
+of $x$ to $N_\ep$.
+If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$,
 write $y_i$ for the restriction of $z_i$ to $N_\ep$, and let
 $x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin N_\ep$,
 and have an additional blob $N_\ep$ with label $y_i - s(y_i)$.
 Define $j_\ep(x) = \sum x_i$.
 	\hat{f}(\textstyle\sum_i a_i \tensor k_i \tensor b_i) =
 						\textstyle\sum_i a_i \tensor f(k_i) \tensor b_i ,
 \]
 and similarly for $\hat{g}$.
 Most of what we need to check is easy.
-Suppose we have $\sum_i (a_i \tensor k_i \tensor b_i) \in \ker(C \tensor K \tensor C \to K)$, assuming without loss of generality that $\{a_i \tensor b_i\}_i$ is linearly independent in $C \tensor C$, and $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$. We must then have $f(k_i) = 0 \in E$ for each $i$, which implies $k_i=0$ itself.
+Suppose we have $\sum_i (a_i \tensor k_i \tensor b_i) \in \ker(C \tensor K \tensor C \to K)$,
-If $\sum_i (a_i \tensor e_i \tensor b_i) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, again by assuming the set  $\{a_i \tensor b_i\}_i$ is linearly independent we can deduce that each
+assuming without loss of generality that $\{a_i \tensor b_i\}_i$ is linearly independent in $C \tensor C$,
-$e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$, and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$.
+and $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$.
-If $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$, then each $g(e_i) = 0$, so $e_i = f(\widetilde{e_i})$ for some $\widetilde{e_i} \in K$, and $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$.
+We must then have $f(k_i) = 0 \in E$ for each $i$, which implies $k_i=0$ itself.
-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 $\hat{g}$.
+If $\sum_i (a_i \tensor e_i \tensor b_i) \in \ker(C \tensor E \tensor C \to E)$
-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.
+is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$,
+again by assuming the set  $\{a_i \tensor b_i\}_i$ is linearly independent we can deduce that each
+$e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$,
+and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$.
+If $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$, then each $g(e_i) = 0$, so $e_i = f(\widetilde{e_i})$
+for some $\widetilde{e_i} \in K$, and $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$.
+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 $\hat{g}$.
+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.
 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
-$\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further,
+$\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$.
+Further,
 \begin{align*}
 \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 \\
 & = q - 0
 \end{align*}
 (here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$).
 Similar arguments show that the functors
 \begin{equation}
 \label{eq:ker-functor}%
 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M)
 \end{equation}
-are all exact too. Moreover, tensor products of such functors with each
+are all exact too.
+Moreover, tensor products of such functors with each
 other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M)
 \tensor C \tensor \ker(C \tensor C \to M)$) are all still exact.
 Finally, then we see that the functor $K_*$ is simply an (infinite)
-direct sum of copies of this sort of functor. The direct sum is indexed by
+direct sum of copies of this sort of functor.
-configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors,
+The direct sum is indexed by
-with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$ (depending on whether they contain a marked point $p_i$), and all other labelled points corresponding
+configurations of nested blobs and of labels; for each such configuration, we have one of
+the above tensor product functors,
+with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor}
+or $\ker(C^{\tensor k} \to C)$ (depending on whether they contain a marked point $p_i$), and all other labelled points corresponding
 to tensor factors of $C$ and $M$.
 \end{proof}
 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}]
 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$.
-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 $*$,
+We define a map $\ev: K_0(M) \to M$.
-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.
+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 $*$,
+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.
 There is a quotient map $\pi: M \to \coinv{M}$.
 We claim that the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$;
 i.e.\ that $\pi(\ev(\bd y)) = 0$ for all $y \in K_1(M)$.
 There are two cases, depending on whether the blob of $y$ contains the point *.
 If it doesn't, then
-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
+suppose $y$ has label $m$ at $*$, labels $c_i$ at other labeled points outside the blob,
-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
+and the field inside the blob is a sum, with the $j$-th term having
+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
 $\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$
-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 $*$,
+Similarly, if $*$ is contained in the blob, then the blob label is a sum, with the
-and there are labels $c_i$ at the labeled points outside the blob. We know that
+$j$-th term have labelled points $d_{j,i}$ to the left of $*$, $m_j$ at $*$, and $d_{j,i}'$ to the right of $*$,
+and there are labels $c_i$ at the labeled points outside the blob.
+We know that
 $$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$
 and so
 \begin{align*}
 \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} \\
 & = \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 \\
 & = 0
 \end{align*}
 where this time we use the fact that we're mapping to $\coinv{M}$, not just $M$.
-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.
+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.
 This is equivalent to showing that
 \[
 	\ev\inv(\ker(\pi)) \sub \bd K_1(M) .
 \]
 The above inclusion follows from
 As shown in Figure \ref{fig:hochschild-1-chains}, $cm - mc$ lies in $\ev(\bd K_1(M))$.
 \end{proof}
 \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}]
 We show that $K_*(C\otimes C)$ is
-quasi-isomorphic to the 0-step complex $C$. We'll do this in steps, establishing quasi-isomorphisms and homotopy equivalences
+quasi-isomorphic to the 0-step complex $C$.
+We'll do this in steps, establishing quasi-isomorphisms and homotopy equivalences
 $$K_*(C \tensor C) \quismto K'_* \htpyto K''_* \quismto C.$$
 Let $K'_* \sub K_*(C\otimes C)$ be the subcomplex where the label of
 the point $*$ is $1 \otimes 1 \in C\otimes C$.
 We will show that the inclusion $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism.
 generated by blob diagrams $b$ such that $N_\ep$ is either disjoint from
 or contained in each blob of $b$, and the only labeled point inside $N_\ep$ is $*$.
 %and the two boundary points of $N_\ep$ are not labeled points of $b$.
 For a field $y$ on $N_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$
 labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$.
-(See Figure \ref{fig:sy}.) Note that $y - s_\ep(y) \in U(N_\ep)$.
+(See Figure \ref{fig:sy}.)
+Note that $y - s_\ep(y) \in U(N_\ep)$.
 Let $\sigma_\ep: K_*^\ep \to K_*^\ep$ be the chain map
 given by replacing the restriction $y$ to $N_\ep$ of each field
 appearing in an element of  $K_*^\ep$ with $s_\ep(y)$.
 Note that $\sigma_\ep(x) \in K'_*$.
 \begin{figure}[t]
 \mathfig{0.6}{hochschild/2-chains-0}
 \end{equation*}
 \begin{equation*}
 \mathfig{0.4}{hochschild/2-chains-1} \qquad \mathfig{0.4}{hochschild/2-chains-2}
 \end{equation*}
-\caption{The 0-, 1- and 2-chains in the image of $m \tensor a \tensor b$. Only the supports of the 1- and 2-blobs are shown.}
+\caption{The 0-, 1- and 2-chains in the image of $m \tensor a \tensor b$.
+Only the supports of the 1- and 2-blobs are shown.}
 \label{fig:hochschild-2-chains}
 \end{figure}
 \begin{figure}[t]
 \begin{equation*}
 \caption{One of the 2-cells from Figure \ref{fig:hochschild-2-chains}.}
 \label{fig:hochschild-example-2-cell}
 \end{figure}
 In degree 2, we send $m\ot a \ot b$ to the sum of 24 ($=6\cdot4$) 2-blob diagrams as shown in
-Figure \ref{fig:hochschild-2-chains}. In Figure \ref{fig:hochschild-2-chains} the 1- and 2-blob diagrams are indicated only by their support.
+Figure \ref{fig:hochschild-2-chains}.
+In Figure \ref{fig:hochschild-2-chains} the 1- and 2-blob diagrams are indicated only by their support.
 We leave it to the reader to determine the labels of the 1-blob diagrams.
 Each 2-cell in the figure is labeled by a ball $V$ in $S^1$ which contains the support of all
 1-blob diagrams in its boundary.
 Such a 2-cell corresponds to a sum of the 2-blob diagrams obtained by adding $V$
 as an outer (non-twig) blob to each of the 1-blob diagrams in the boundary of the 2-cell.

changeset 342	1d76e832d32f
parent 321	76c301fdf0a2
child 400	a02a6158f3bd