217 where there are no labeled points |
217 where there are no labeled points |
218 in $N_\ep$, except perhaps $*$, and $N_\ep$ is either disjoint from or contained in |
218 in $N_\ep$, except perhaps $*$, and $N_\ep$ is either disjoint from or contained in |
219 every blob in the diagram. |
219 every blob in the diagram. |
220 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
220 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
221 |
221 |
222 We define a degree $1$ chain map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. Let $x \in L_*^\ep$ be a blob diagram. |
222 We define a degree $1$ map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. Let $x \in L_*^\ep$ be a blob diagram. |
223 \nn{maybe add figures illustrating $j_\ep$?} |
223 \nn{maybe add figures illustrating $j_\ep$?} |
224 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 |
224 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 |
225 of $x$ to $N_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$, |
225 of $x$ to $N_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$, |
226 write $y_i$ for the restriction of $z_i$ to $N_\ep$, and let |
226 write $y_i$ for the restriction of $z_i$ to $N_\ep$, and let |
227 $x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin N_\ep$, |
227 $x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin N_\ep$, |
240 \end{proof} |
240 \end{proof} |
241 |
241 |
242 \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
242 \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
243 We now prove that $K_*$ is an exact functor. |
243 We now prove that $K_*$ is an exact functor. |
244 |
244 |
245 Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules |
245 As a warm-up, we prove |
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246 that the functor on $C$-$C$ bimodules |
246 \begin{equation*} |
247 \begin{equation*} |
247 M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M) |
248 M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M) |
248 \end{equation*} |
249 \end{equation*} |
249 is exact. For completeness we'll explain this below. |
250 is exact. |
250 |
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251 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
251 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
252 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor, so $\hat{f}(a \tensor k \tensor b) = a \tensor f(k) \tensor b$, and similarly for $\hat{g}$. |
252 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor, so |
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253 \[ |
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254 \hat{f}(\textstyle\sum_i a_i \tensor k_i \tensor b_i) = |
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255 \textstyle\sum_i a_i \tensor f(k_i) \tensor b_i , |
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256 \] |
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257 and similarly for $\hat{g}$. |
253 Most of what we need to check is easy. |
258 Most of what we need to check is easy. |
254 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. |
259 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. |
255 If $\sum_i (a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ 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 |
260 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 |
256 $e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$, and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$. |
261 $e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$, and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$. |
257 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 $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$. |
262 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 $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$. |
258 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 $\hat{g}$. |
263 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 $\hat{g}$. |
259 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. |
264 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. |
260 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 |
265 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 |
283 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
288 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
284 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
289 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
285 |
290 |
286 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 $*$, |
291 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 $*$, |
287 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. |
292 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. |
288 There is a quotient map $\pi: M \to \coinv{M}$, and the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; if $y \in K_1(M)$, the blob in $y$ either contains $*$ or does not. If it doesn't, then |
293 |
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294 There is a quotient map $\pi: M \to \coinv{M}$. |
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295 We claim that the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; |
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296 i.e.\ that $\pi(\ev(\bd y)) = 0$ for all $y \in K_1(M)$. |
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297 There are two cases, depending on whether the blob of $y$ contains the point *. |
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298 If it doesn't, then |
289 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 |
299 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 |
290 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 |
300 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 |
291 $\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$ |
301 $\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$ |
292 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 $*$, |
302 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 $*$, |
293 and there are labels $c_i$ at the labeled points outside the blob. We know that |
303 and there are labels $c_i$ at the labeled points outside the blob. We know that |
330 \end{align*} |
340 \end{align*} |
331 \caption{Defining $s_\ep$.} |
341 \caption{Defining $s_\ep$.} |
332 \label{fig:sy} |
342 \label{fig:sy} |
333 \end{figure} |
343 \end{figure} |
334 |
344 |
335 Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. |
345 Define a degree 1 map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. |
336 Let $x \in K_*^\ep$ be a blob diagram. |
346 Let $x \in K_*^\ep$ be a blob diagram. |
337 If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $N_\ep$ to |
347 If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $N_\ep$ to |
338 $x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $N_\ep$. |
348 $x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $N_\ep$. |
339 If $*$ is contained in a twig blob $B$ with label $u = \sum z_i$, $j_\ep(x)$ is obtained as follows. |
349 If $*$ is contained in a twig blob $B$ with label $u = \sum z_i$, $j_\ep(x)$ is obtained as follows. |
340 Let $y_i$ be the restriction of $z_i$ to $N_\ep$. |
350 Let $y_i$ be the restriction of $z_i$ to $N_\ep$. |