blob: comparison text/hochschild.tex

equal deleted inserted replaced

-:aac9fd8d6bc6
+:ea489bbccfbf
 %blob complex for $S^1$.
 %\nn{Is it really so clear?  Should say more.}
 %\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.
+Let $K_*(M) = K_*(S^1, (*), (M))$, 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 $*$
+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.
+\nn{Or maybe we should claim that $M \to K_*(M)$ is the/a derived coend.
-Or maybe that $F_*(M)$ is quasi-isomorphic (or perhaps homotopic) to the Hochschild
+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?
 Do we need a map from hoch to blob?
 Does the above exactness and contractibility guarantee such a map without writing it
 \end{thm}
 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.
+%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!}
+%\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}
 \item \label{item:hochschild-additive}%
 $HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$.
 quasi-isomorphic to the 0-step complex $C$.
 \item \label{item:hochschild-coinvariants}%
 $HH_0(M)$ 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
+We'll first recall why these properties are characteristic.
-$C$-$C$ bimodule $M$. If $M$ is free, that is, a direct sum of copies of
-$C \tensor C$, then properties \ref{item:hochschild-additive} and
+Take some $C$-$C$ bimodule $M$, and choose a free resolution
-\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 F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0.
-\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
+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
-\ref{item:hochschild-additive} and \ref{item:hochschild-free}, and so
+$HC_i(F_j)$, with $i,j \geq 0$, graded by $i+j$.
-$HH_{i+1}(M) \iso HH_i(F)$. For $i=0$, \todo{}.
+Observe that we have two chain maps
-This tells us how to
+\begin{align*}
-compute every homology group of $HC_*(M)$; we already know $HH_0(M)$
+HC_i(F_j) & \xrightarrow{HC_i(\pi)} HC_i(M) \\
-(it's just coinvariants, by property \ref{item:hochschild-coinvariants}),
+\intertext{and}
-and higher homology groups are determined by lower ones in $HC_*(K)$, and
+HC_i(F_j) & \xrightarrow{HC_0(F_j) \onto HH_0(F_j)} \operatorname{coinv}(F_j).
-hence recursively as coinvariants of some other bimodule.
+\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 proposition then follows from the following lemmas, establishing that $F_*$ has precisely these required properties.
+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
+$$HC_*(M) \iso \operatorname{coinv}(F_*).$$
+%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 $K_*$ 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
 \ref{lem:module-blob},
 \ref{lem:hochschild-exact}, \ref{lem:hochschild-free} and
 \ref{lem:hochschild-coinvariants}.
 \end{proof}
 \begin{proof}[Proof of Lemma \ref{lem:module-blob}]
-We show that $F_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$.
+We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$.
-$F_*(C)$ differs from $\bc_*(S^1)$ only in that the base point *
+$K_*(C)$ 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.
+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 $i: F_*(C) \to \bc_*(S^1)$.
+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 *.
 Let $x \in \bc_*(S^1)$.
 Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in
 $x$ 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~$*$
 labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$.
 (See Figure xxxx.)
 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.
+Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows.
-Let $x \in F^\ep_*$ be a blob diagram.
+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.
 Let $y_i$ be the restriction of $z_i$ to $B_\ep$.
 Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin B_\ep$,
 and have an additional blob $B_\ep$ with label $y_i - s_\ep(y_i)$.
 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
+where $\sigma_\ep : K_*^\ep \to K_*^\ep$ is given by replacing the restriction $y$ of each field
-mentioned in $x \in F^\ep_*$ with $s_\ep(y)$.
+mentioned in $x \in K_*^\ep$ with $s_\ep(y)$.
-Note that $\sigma_\ep(x) \in F'_*$.
+Note that $\sigma_\ep(x) \in K'_*$.
-If $j_\ep$ were defined on all of $F_*(C\otimes C)$, it would show that $\sigma_\ep$
+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 $F'_* \to F_*(C\otimes C)$.
+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
+If $x$ is a cycle in $K_*(C\otimes C)$, then for sufficiently small $\ep$ we have
-$x \in F_*^\ep$.
+$x \in K_*^\ep$.
-(This is true for any chain in $F_*(C\otimes C)$, since chains are sums of
+(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
+Then $x$ is homologous to $s_\ep(x)$, which is in $K'_*$, so the inclusion map
-$F'_* \sub F_*(C\otimes C)$ is surjective on homology.
+$K'_* \sub K_*(C\otimes C)$ is surjective on homology.
-If $y \in F_*(C\otimes C)$ and $\bd y = x \in F'_*$, then $y \in F^\ep_*$ for some $\ep$
+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.
+Let $K''_* \sub K'_*$ be the subcomplex of $K'_*$ where $*$ is not contained in any blob.
-We will show that the inclusion $i: F''_* \to F'_*$ is a homotopy equivalence.
+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.
 For $G'_*$ the proof is the same as in (\ref{bcontract}), except that the splitting
 $G'_0 \to H_0(G'_*)$ concentrates the point labels at two points to the right and left of $*$.
 Thus any cycle lies in the image of the normal blob complex of a disjoint union
 of two intervals, which is contractible by (\ref{bcontract}) and (\ref{disjunion}).
 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
+Next we construct a degree 1 map (homotopy) $h: K'_* \to K'_*$ such that
-for all $x \in F'_*$ we have
+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 $h_0 = 0$.
+Since $K'_0 = K''_0$, we can take $h_0 = 0$.
-Let $x \in F'_1$, with single blob $B \sub S^1$.
+Let $x \in K'_1$, with single blob $B \sub S^1$.
-If $* \notin B$, then $x \in F''_1$ and we define $h_1(x) = 0$.
+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 $B$.
 So $x' \in G'_l$ for some $l \le k$.
 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.
 \nn{need to say something else in degree zero}
 \end{proof}

changeset 19	ea489bbccfbf
parent 15	7340ab80db25
child 28	f844cffa5c03