diff -r aac9fd8d6bc6 -r ea489bbccfbf text/hochschild.tex --- a/text/hochschild.tex Thu Jun 26 17:56:20 2008 +0000 +++ b/text/hochschild.tex Fri Jun 27 04:24:25 2008 +0000 @@ -71,18 +71,18 @@ %\nn{alternative to the above paragraph:} Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$. -We define a blob-like complex $F_*(S^1, (p_i), (M_i))$. +We define a blob-like complex $K_*(S^1, (p_i), (M_i))$. The fields have elements of $M_i$ labeling $p_i$ and elements of $C$ labeling other points. The blob twig labels lie in kernels of evaluation maps. (The range of these evaluation maps is a tensor product (over $C$) of $M_i$'s.) -Let $F_*(M) = F_*(S^1, (*), (M))$, where $* \in S^1$ is some standard base point. -In other words, fields for $F_*(M)$ have an element of $M$ at the fixed point $*$ +Let $K_*(M) = K_*(S^1, (*), (M))$, where $* \in S^1$ is some standard base point. +In other words, fields for $K_*(M)$ have an element of $M$ at the fixed point $*$ and elements of $C$ at variable other points. \todo{Some orphaned questions:} -\nn{Or maybe we should claim that $M \to F_*(M)$ is the/a derived coend. -Or maybe that $F_*(M)$ is quasi-isomorphic (or perhaps homotopic) to the Hochschild +\nn{Or maybe we should claim that $M \to K_*(M)$ is the/a derived coend. +Or maybe that $K_*(M)$ is quasi-isomorphic (or perhaps homotopic) to the Hochschild complex of $M$.} \nn{What else needs to be said to establish quasi-isomorphism to Hochschild complex? @@ -101,19 +101,19 @@ This follows from two results. First, we see that \begin{lem} \label{lem:module-blob}% -The complex $F_*(C)$ (here $C$ is being thought of as a +The complex $K_*(C)$ (here $C$ is being thought of as a $C$-$C$-bimodule, not a category) is quasi-isomorphic to the blob complex $\bc_*(S^1; C)$. (Proof later.) \end{lem} Next, we show that for any $C$-$C$-bimodule $M$, \begin{prop} -The complex $F_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual +The complex $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual Hochschild complex of $M$. \end{prop} \begin{proof} -First, since we're working over $\Complex$, note that saying two complexes are quasi-isomorphic simply means they have isomorphic homologies. -\todo{We really don't want to work over $\Complex$, though; it would be nice to talk about torsion!} +%First, since we're working over $\Complex$, note that saying two complexes are quasi-isomorphic simply means they have isomorphic homologies. +%\todo{We really don't want to work over $\Complex$, though; it would be nice to talk about torsion!} Recall that the usual Hochschild complex of $M$ is uniquely determined, up to quasi-isomorphism, by the following properties: \begin{enumerate} @@ -129,47 +129,67 @@ $HH_0(M)$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. \end{enumerate} (Together, these just say that Hochschild homology is `the derived functor of coinvariants'.) -We'll first explain why these properties are characteristic. Take some -$C$-$C$ bimodule $M$. 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 -free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we -have a short exact sequence $0 \to K \into F \onto M \to 0$, and hence a -short exact sequence of complexes $0 \to HC_*(K) \into HC_*(F) \onto HC_*(M) -\to 0$. Such a sequence gives a long exact sequence on homology +We'll first recall why these properties are characteristic. + +Take some $C$-$C$ bimodule $M$, and choose a free resolution \begin{equation*} -%\begin{split} -\cdots \to HH_{i+1}(F) \to HH_{i+1}(M) \to HH_i(K) \to HH_i(F) \to \cdots % \\ -%\cdots \to HH_1(F) \to HH_1(M) \to HH_0(K) \to HH_0(F) \to HH_0(M). -%\end{split} +\cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0. \end{equation*} -For any $i \geq 1$, $HH_{i+1}(F) = HH_i(F) = 0$, by properties -\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so -$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}. +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 +$HC_i(F_j)$, with $i,j \geq 0$, graded by $i+j$. + +Observe that we have two chain maps +\begin{align*} +HC_i(F_j) & \xrightarrow{HC_i(\pi)} HC_i(M) \\ +\intertext{and} +HC_i(F_j) & \xrightarrow{HC_0(F_j) \onto HH_0(F_j)} \operatorname{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. +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 +$$HC_*(M) \iso \operatorname{coinv}(F_*).$$ -This tells us how to -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. +%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 +%free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we +%have a short exact sequence $0 \to K \into F \onto M \to 0$, and hence a +%short exact sequence of complexes $0 \to HC_*(K) \into HC_*(F) \onto HC_*(M) +%\to 0$. Such a sequence gives a long exact sequence on homology +%\begin{equation*} +%%\begin{split} +%\cdots \to HH_{i+1}(F) \to HH_{i+1}(M) \to HH_i(K) \to HH_i(F) \to \cdots % \\ +%%\cdots \to HH_1(F) \to HH_1(M) \to HH_0(K) \to HH_0(F) \to HH_0(M). +%%\end{split} +%\end{equation*} +%For any $i \geq 1$, $HH_{i+1}(F) = HH_i(F) = 0$, by properties +%\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so +%$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}. +% +%This tells us how to +%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. -The proposition then follows from the following lemmas, establishing that $F_*$ has precisely these required properties. +The proposition then follows from the following lemmas, establishing that $K_*$ has precisely these required properties. \begin{lem} \label{lem:hochschild-additive}% -Directly from the definition, $F_*(M_1 \oplus M_2) \cong F_*(M_1) \oplus F_*(M_2)$. +Directly from the definition, $K_*(M_1 \oplus M_2) \cong K_*(M_1) \oplus K_*(M_2)$. \end{lem} \begin{lem} \label{lem:hochschild-exact}% An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an -exact sequence $0 \to F_*(M_1) \into F_*(M_2) \onto F_*(M_3) \to 0$. +exact sequence $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$. \end{lem} \begin{lem} \label{lem:hochschild-free}% -$F_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$. +$K_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$. \end{lem} \begin{lem} \label{lem:hochschild-coinvariants}% -$H_0(F_*(M))$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. +$H_0(K_*(M))$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. \end{lem} The remainder of this section is devoted to proving Lemmas @@ -179,12 +199,12 @@ \end{proof} \begin{proof}[Proof of Lemma \ref{lem:module-blob}] -We show that $F_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$. -$F_*(C)$ differs from $\bc_*(S^1)$ only in that the base point * -is always a labeled point in $F_*(C)$, while in $\bc_*(S^1)$ it may or may not be. -In other words, there is an inclusion map $i: F_*(C) \to \bc_*(S^1)$. +We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$. +$K_*(C)$ differs from $\bc_*(S^1)$ only in that the base point * +is always a labeled point in $K_*(C)$, while in $\bc_*(S^1)$ it may or may not be. +In other words, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$. -We define a quasi-inverse \nn{right term?} $s: \bc_*(S^1) \to F_*(C)$ to the inclusion as follows. +We define a quasi-inverse \nn{right term?} $s: \bc_*(S^1) \to K_*(C)$ to the inclusion as follows. If $y$ is a field defined on a neighborhood of *, define $s(y) = y$ if * is a labeled point in $y$. Otherwise, define $s(y)$ to be the result of adding a label 1 (identity morphism) at *. @@ -193,25 +213,25 @@ $x$ with $y$. It is easy to check that $s$ is a chain map and $s \circ i = \id$. -Let $G^\ep_* \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points +Let $L_*^\ep \sub \bc_*(S^1)$ be the subcomplex where there are no labeled points in a neighborhood $B_\ep$ of *, except perhaps *. -Note that for any chain $x \in \bc_*(S^1)$, $x \in G^\ep_*$ for sufficiently small $\ep$. +Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. \nn{rest of argument goes similarly to above} \end{proof} \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] \todo{} \end{proof} \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] -We show that $F_*(C\otimes C)$ is +We show that $K_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$. -Let $F'_* \sub F_*(C\otimes C)$ be the subcomplex where the label of +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 $i: F'_* \to F_*(C\otimes C)$ is a quasi-isomorphism. +We will show that the inclusion $i: K'_* \to K_*(C\otimes C)$ is a quasi-isomorphism. Fix a small $\ep > 0$. Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$. -Let $F^\ep_* \sub F_*(C\otimes C)$ be the subcomplex +Let $K_*^\ep \sub K_*(C\otimes C)$ be the subcomplex generated by blob diagrams $b$ such that $B_\ep$ is either disjoint from or contained in each blob of $b$, and the two boundary points of $B_\ep$ are not labeled points of $b$. For a field (picture) $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ @@ -220,8 +240,8 @@ Note that $y - s_\ep(y) \in U(B_\ep)$. \nn{maybe it's simpler to assume that there are no labeled points, other than $*$, in $B_\ep$.} -Define a degree 1 chain map $j_\ep : F^\ep_* \to F^\ep_*$ as follows. -Let $x \in F^\ep_*$ be a blob diagram. +Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. +Let $x \in K_*^\ep$ be a blob diagram. If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $B_\ep$ to $x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $B_\ep$. If $*$ is contained in a twig blob $B$ with label $u = \sum z_i$, $j_\ep(x)$ is obtained as follows. @@ -231,41 +251,41 @@ Define $j_\ep(x) = \sum x_i$. \nn{need to check signs coming from blob complex differential} -Note that if $x \in F'_* \cap F^\ep_*$ then $j_\ep(x) \in F'_*$ also. +Note that if $x \in K'_* \cap K_*^\ep$ then $j_\ep(x) \in K'_*$ also. The key property of $j_\ep$ is \eq{ \bd j_\ep + j_\ep \bd = \id - \sigma_\ep , } -where $\sigma_\ep : F^\ep_* \to F^\ep_*$ is given by replacing the restriction $y$ of each field -mentioned in $x \in F^\ep_*$ with $s_\ep(y)$. -Note that $\sigma_\ep(x) \in F'_*$. +where $\sigma_\ep : K_*^\ep \to K_*^\ep$ is given by replacing the restriction $y$ of each field +mentioned in $x \in K_*^\ep$ with $s_\ep(y)$. +Note that $\sigma_\ep(x) \in K'_*$. -If $j_\ep$ were defined on all of $F_*(C\otimes C)$, it would show that $\sigma_\ep$ -is a homotopy inverse to the inclusion $F'_* \to F_*(C\otimes C)$. +If $j_\ep$ were defined on all of $K_*(C\otimes C)$, it would show that $\sigma_\ep$ +is a homotopy inverse to the inclusion $K'_* \to K_*(C\otimes C)$. One strategy would be to try to stitch together various $j_\ep$ for progressively smaller -$\ep$ and show that $F'_*$ is homotopy equivalent to $F_*(C\otimes C)$. +$\ep$ and show that $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$. Instead, we'll be less ambitious and just show that -$F'_*$ is quasi-isomorphic to $F_*(C\otimes C)$. +$K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. -If $x$ is a cycle in $F_*(C\otimes C)$, then for sufficiently small $\ep$ we have -$x \in F_*^\ep$. -(This is true for any chain in $F_*(C\otimes C)$, since chains are sums of +If $x$ is a cycle in $K_*(C\otimes C)$, then for sufficiently small $\ep$ we have +$x \in K_*^\ep$. +(This is true for any chain in $K_*(C\otimes C)$, since chains are sums of finitely many blob diagrams.) -Then $x$ is homologous to $s_\ep(x)$, which is in $F'_*$, so the inclusion map -$F'_* \sub F_*(C\otimes C)$ is surjective on homology. -If $y \in F_*(C\otimes C)$ and $\bd y = x \in F'_*$, then $y \in F^\ep_*$ for some $\ep$ +Then $x$ is homologous to $s_\ep(x)$, which is in $K'_*$, so the inclusion map +$K'_* \sub K_*(C\otimes C)$ is surjective on homology. +If $y \in K_*(C\otimes C)$ and $\bd y = x \in K'_*$, then $y \in K_*^\ep$ for some $\ep$ and \eq{ \bd y = \bd (\sigma_\ep(y) + j_\ep(x)) . } Since $\sigma_\ep(y) + j_\ep(x) \in F'$, it follows that the inclusion map is injective on homology. -This completes the proof that $F'_*$ is quasi-isomorphic to $F_*(C\otimes C)$. +This completes the proof that $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. -Let $F''_* \sub F'_*$ be the subcomplex of $F'_*$ where $*$ is not contained in any blob. -We will show that the inclusion $i: F''_* \to F'_*$ is a homotopy equivalence. +Let $K''_* \sub K'_*$ be the subcomplex of $K'_*$ where $*$ is not contained in any blob. +We will show that the inclusion $i: K''_* \to K'_*$ is a homotopy equivalence. -First, a lemma: Let $G''_*$ and $G'_*$ be defined the same as $F''_*$ and $F'_*$, except with +First, a lemma: Let $G''_*$ and $G'_*$ be defined the same as $K''_*$ and $K'_*$, except with $S^1$ replaced some (any) neighborhood of $* \in S^1$. Then $G''_*$ and $G'_*$ are both contractible and the inclusion $G''_* \sub G'_*$ is a homotopy equivalence. @@ -278,20 +298,20 @@ Actually, we need the further (easy) result that the inclusion $G''_* \to G'_*$ induces an isomorphism on $H_0$. -Next we construct a degree 1 map (homotopy) $h: F'_* \to F'_*$ such that -for all $x \in F'_*$ we have +Next we construct a degree 1 map (homotopy) $h: K'_* \to K'_*$ such that +for all $x \in K'_*$ we have \eq{ - x - \bd h(x) - h(\bd x) \in F''_* . + x - \bd h(x) - h(\bd x) \in K''_* . } -Since $F'_0 = F''_0$, we can take $h_0 = 0$. -Let $x \in F'_1$, with single blob $B \sub S^1$. -If $* \notin B$, then $x \in F''_1$ and we define $h_1(x) = 0$. +Since $K'_0 = K''_0$, we can take $h_0 = 0$. +Let $x \in K'_1$, with single blob $B \sub S^1$. +If $* \notin B$, then $x \in K''_1$ and we define $h_1(x) = 0$. If $* \in B$, then we work in the image of $G'_*$ and $G''_*$ (with respect to $B$). Choose $x'' \in G''_1$ such that $\bd x'' = \bd x$. Since $G'_*$ is contractible, there exists $y \in G'_2$ such that $\bd y = x - x''$. Define $h_1(x) = y$. The general case is similar, except that we have to take lower order homotopies into account. -Let $x \in F'_k$. +Let $x \in K'_k$. If $*$ is not contained in any of the blobs of $x$, then define $h_k(x) = 0$. Otherwise, let $B$ be the outermost blob of $x$ containing $*$. By xxxx above, $x = x' \bullet p$, where $x'$ is supported on $B$ and $p$ is supported away from $B$. @@ -299,12 +319,12 @@ Choose $x'' \in G''_l$ such that $\bd x'' = \bd (x' - h_{l-1}\bd x')$. Choose $y \in G'_{l+1}$ such that $\bd y = x' - x'' - h_{l-1}\bd x'$. Define $h_k(x) = y \bullet p$. -This completes the proof that $i: F''_* \to F'_*$ is a homotopy equivalence. +This completes the proof that $i: K''_* \to K'_*$ is a homotopy equivalence. \nn{need to say above more clearly and settle on notation/terminology} -Finally, we show that $F''_*$ is contractible. +Finally, we show that $K''_*$ is contractible. \nn{need to also show that $H_0$ is the right thing; easy, but I won't do it now} -Let $x$ be a cycle in $F''_*$. +Let $x$ be a cycle in $K''_*$. The union of the supports of the diagrams in $x$ does not contain $*$, so there exists a ball $B \subset S^1$ containing the union of the supports and not containing $*$. Adding $B$ as a blob to $x$ gives a contraction.