diff -r 675f53735445 -r 1d76e832d32f text/hochschild.tex --- a/text/hochschild.tex Fri Jun 04 17:00:18 2010 -0700 +++ b/text/hochschild.tex Fri Jun 04 17:15:53 2010 -0700 @@ -7,7 +7,11 @@ 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. @@ -67,12 +71,14 @@ 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 $M$, @@ -114,17 +120,19 @@ $$\cP_*(M) \iso \coinv(F_*).$$ % 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 -construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by -$i+j$. We have two chain maps +construction the cone of the chain map $\pi: F_* \to M$ is acyclic. +Now construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by $i+j$. +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 @@ -150,7 +158,8 @@ %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, $K_*(M_1 \oplus M_2) \cong K_*(M_1) \oplus K_*(M_2)$. @@ -185,7 +194,8 @@ 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 *. @@ -219,10 +229,13 @@ 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 -of $x$ to $N_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$, +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 +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 $y_i - s(y_i)$. @@ -256,14 +269,24 @@ \] 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. -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 -$e_i$ 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. +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. +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 +$e_i$ 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 @@ -275,32 +298,44 @@ \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 -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 +direct sum of copies of this sort of functor. +The direct sum is indexed by +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 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. +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 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 -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 +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 +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 $*$, -and there are labels $c_i$ at the labeled points outside the blob. We know that +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 $*$, +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*} @@ -310,7 +345,8 @@ \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) . @@ -340,7 +376,8 @@ \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 @@ -355,7 +392,8 @@ %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 $s_\ep(y)$. @@ -512,7 +550,8 @@ \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} @@ -529,7 +568,8 @@ \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.