5 |
5 |
6 So far we have provided no evidence that blob homology is interesting in degrees |
6 So far we have provided no evidence that blob homology is interesting in degrees |
7 greater than zero. |
7 greater than zero. |
8 In this section we analyze the blob complex in dimension $n=1$. |
8 In this section we analyze the blob complex in dimension $n=1$. |
9 We find that $\bc_*(S^1, \cC)$ is homotopy equivalent to the |
9 We find that $\bc_*(S^1, \cC)$ is homotopy equivalent to the |
10 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.) |
10 Hochschild complex of the 1-category $\cC$. |
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11 (Recall from \S \ref{sec:example:traditional-n-categories(fields)} that a |
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12 $1$-category gives rise to a $1$-dimensional system of fields; as usual, |
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13 talking about the blob complex with coefficients in a $n$-category means |
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14 first passing to the corresponding $n$ dimensional system of fields.) |
11 Thus the blob complex is a natural generalization of something already |
15 Thus the blob complex is a natural generalization of something already |
12 known to be interesting in higher homological degrees. |
16 known to be interesting in higher homological degrees. |
13 |
17 |
14 It is also worth noting that the original idea for the blob complex came from trying |
18 It is also worth noting that the original idea for the blob complex came from trying |
15 to find a more ``local" description of the Hochschild complex. |
19 to find a more ``local" description of the Hochschild complex. |
65 \begin{thm} \label{hochthm} |
69 \begin{thm} \label{hochthm} |
66 The blob complex $\bc_*(S^1; C)$ on the circle is homotopy equivalent to the |
70 The blob complex $\bc_*(S^1; C)$ on the circle is homotopy equivalent to the |
67 usual Hochschild complex for $C$. |
71 usual Hochschild complex for $C$. |
68 \end{thm} |
72 \end{thm} |
69 |
73 |
70 This follows from two results. First, we see that |
74 This follows from two results. |
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75 First, we see that |
71 \begin{lem} |
76 \begin{lem} |
72 \label{lem:module-blob}% |
77 \label{lem:module-blob}% |
73 The complex $K_*(C)$ (here $C$ is being thought of as a |
78 The complex $K_*(C)$ (here $C$ is being thought of as a |
74 $C$-$C$-bimodule, not a category) is homotopy equivalent to the blob complex |
79 $C$-$C$-bimodule, not a category) is homotopy equivalent to the blob complex |
75 $\bc_*(S^1; C)$. (Proof later.) |
80 $\bc_*(S^1; C)$. |
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81 (Proof later.) |
76 \end{lem} |
82 \end{lem} |
77 |
83 |
78 Next, we show that for any $C$-$C$-bimodule $M$, |
84 Next, we show that for any $C$-$C$-bimodule $M$, |
79 \begin{prop} \label{prop:hoch} |
85 \begin{prop} \label{prop:hoch} |
80 The complex $K_*(M)$ is homotopy equivalent to $\HC_*(M)$, the usual |
86 The complex $K_*(M)$ is homotopy equivalent to $\HC_*(M)$, the usual |
112 \ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there |
118 \ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there |
113 is a quasi-isomorphism |
119 is a quasi-isomorphism |
114 $$\cP_*(M) \iso \coinv(F_*).$$ |
120 $$\cP_*(M) \iso \coinv(F_*).$$ |
115 % |
121 % |
116 Observe that there's a quotient map $\pi: F_0 \onto M$, and by |
122 Observe that there's a quotient map $\pi: F_0 \onto M$, and by |
117 construction the cone of the chain map $\pi: F_* \to M$ is acyclic. Now |
123 construction the cone of the chain map $\pi: F_* \to M$ is acyclic. |
118 construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by |
124 Now construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by $i+j$. |
119 $i+j$. We have two chain maps |
125 We have two chain maps |
120 \begin{align*} |
126 \begin{align*} |
121 \cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\ |
127 \cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\ |
122 \intertext{and} |
128 \intertext{and} |
123 \cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \coinv(F_j). |
129 \cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \coinv(F_j). |
124 \end{align*} |
130 \end{align*} |
125 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. |
131 The cone of each chain map is acyclic. |
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132 In the first case, this is because the `rows' indexed by $i$ are acyclic since $\HC_i$ is exact. |
126 In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free. |
133 In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free. |
127 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 |
134 Because the cones are acyclic, the chain maps are quasi-isomorphisms. |
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135 Composing one with the inverse of the other, we obtain the desired quasi-isomorphism |
128 $$\cP_*(M) \quismto \coinv(F_*).$$ |
136 $$\cP_*(M) \quismto \coinv(F_*).$$ |
129 |
137 |
130 %If $M$ is free, that is, a direct sum of copies of |
138 %If $M$ is free, that is, a direct sum of copies of |
131 %$C \tensor C$, then properties \ref{item:hochschild-additive} and |
139 %$C \tensor C$, then properties \ref{item:hochschild-additive} and |
132 %\ref{item:hochschild-free} determine $\HC_*(M)$. Otherwise, choose some |
140 %\ref{item:hochschild-free} determine $\HC_*(M)$. Otherwise, choose some |
217 where there are no labeled points |
227 where there are no labeled points |
218 in $N_\ep$, except perhaps $*$, and $N_\ep$ is either disjoint from or contained in |
228 in $N_\ep$, except perhaps $*$, and $N_\ep$ is either disjoint from or contained in |
219 every blob in the diagram. |
229 every blob in the diagram. |
220 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
230 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
221 |
231 |
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. |
232 We define a degree $1$ map $j_\ep: L_*^\ep \to L_*^\ep$ as follows. |
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233 Let $x \in L_*^\ep$ be a blob diagram. |
223 \nn{maybe add figures illustrating $j_\ep$?} |
234 \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 |
235 If $*$ is not contained in any twig blob, we define $j_\ep(x)$ by adding |
225 of $x$ to $N_\ep$. If $*$ is contained in a twig blob $B$ with label $u=\sum z_i$, |
236 $N_\ep$ as a new twig blob, with label $y - s(y)$ where $y$ is the restriction |
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237 of $x$ to $N_\ep$. |
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238 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 |
239 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 $B \setmin N_\ep$, |
240 $x_i$ be equal to $x$ on $S^1 \setmin B$, equal to $z_i$ on $B \setmin N_\ep$, |
228 and have an additional blob $N_\ep$ with label $y_i - s(y_i)$. |
241 and have an additional blob $N_\ep$ with label $y_i - s(y_i)$. |
229 Define $j_\ep(x) = \sum x_i$. |
242 Define $j_\ep(x) = \sum x_i$. |
230 |
243 |
254 \hat{f}(\textstyle\sum_i a_i \tensor k_i \tensor b_i) = |
267 \hat{f}(\textstyle\sum_i a_i \tensor k_i \tensor b_i) = |
255 \textstyle\sum_i a_i \tensor f(k_i) \tensor b_i , |
268 \textstyle\sum_i a_i \tensor f(k_i) \tensor b_i , |
256 \] |
269 \] |
257 and similarly for $\hat{g}$. |
270 and similarly for $\hat{g}$. |
258 Most of what we need to check is easy. |
271 Most of what we need to check is easy. |
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. |
272 Suppose we have $\sum_i (a_i \tensor k_i \tensor b_i) \in \ker(C \tensor K \tensor C \to K)$, |
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 |
273 assuming without loss of generality that $\{a_i \tensor b_i\}_i$ is linearly independent in $C \tensor C$, |
261 $e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$, and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$. |
274 and $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$. |
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 $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$. |
275 We must then have $f(k_i) = 0 \in E$ for each $i$, which implies $k_i=0$ itself. |
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 $\hat{g}$. |
276 If $\sum_i (a_i \tensor e_i \tensor b_i) \in \ker(C \tensor E \tensor C \to E)$ |
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. |
277 is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, |
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278 again by assuming the set $\{a_i \tensor b_i\}_i$ is linearly independent we can deduce that each |
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279 $e_i$ is in the image of the original $f$, and so is in the kernel of the original $g$, |
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280 and so $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$. |
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281 If $\hat{g}(\sum_i a_i \tensor e_i \tensor b_i) = 0$, then each $g(e_i) = 0$, so $e_i = f(\widetilde{e_i})$ |
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282 for some $\widetilde{e_i} \in K$, and $\sum_i a_i \tensor e_i \tensor b_i = \hat{f}(\sum_i a_i \tensor \widetilde{e_i} \tensor b_i)$. |
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283 Finally, the interesting step is in checking that any $q = \sum_i a_i \tensor q_i \tensor b_i$ |
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284 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 $\hat{g}$. |
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285 For each $i$, we can find $\widetilde{q_i}$ so $g(\widetilde{q_i}) = q_i$. |
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286 However $\sum_i a_i \widetilde{q_i} b_i$ need not be zero. |
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 |
287 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 |
266 $\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further, |
288 $\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. |
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289 Further, |
267 \begin{align*} |
290 \begin{align*} |
268 \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 \\ |
291 \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 \\ |
269 & = q - 0 |
292 & = q - 0 |
270 \end{align*} |
293 \end{align*} |
271 (here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$). |
294 (here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$). |
273 Similar arguments show that the functors |
296 Similar arguments show that the functors |
274 \begin{equation} |
297 \begin{equation} |
275 \label{eq:ker-functor}% |
298 \label{eq:ker-functor}% |
276 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
299 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
277 \end{equation} |
300 \end{equation} |
278 are all exact too. Moreover, tensor products of such functors with each |
301 are all exact too. |
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302 Moreover, tensor products of such functors with each |
279 other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M) |
303 other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M) |
280 \tensor C \tensor \ker(C \tensor C \to M)$) are all still exact. |
304 \tensor C \tensor \ker(C \tensor C \to M)$) are all still exact. |
281 |
305 |
282 Finally, then we see that the functor $K_*$ is simply an (infinite) |
306 Finally, then we see that the functor $K_*$ is simply an (infinite) |
283 direct sum of copies of this sort of functor. The direct sum is indexed by |
307 direct sum of copies of this sort of functor. |
284 configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors, |
308 The direct sum is indexed by |
285 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 |
309 configurations of nested blobs and of labels; for each such configuration, we have one of |
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310 the above tensor product functors, |
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311 with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} |
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312 or $\ker(C^{\tensor k} \to C)$ (depending on whether they contain a marked point $p_i$), and all other labelled points corresponding |
286 to tensor factors of $C$ and $M$. |
313 to tensor factors of $C$ and $M$. |
287 \end{proof} |
314 \end{proof} |
288 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
315 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
289 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
316 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
290 |
317 |
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 $*$, |
318 We define a map $\ev: K_0(M) \to M$. |
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. |
319 If $x \in K_0(M)$ has the label $m \in M$ at $*$, and labels $c_i \in C$ at the other |
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320 labeled points of $S^1$, reading clockwise from $*$, |
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321 we set $\ev(x) = m c_1 \cdots c_k$. |
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322 We can think of this as $\ev : M \tensor C^{\tensor k} \to M$, for each direct summand of |
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323 $K_0(M)$ indexed by a configuration of labeled points. |
293 |
324 |
294 There is a quotient map $\pi: M \to \coinv{M}$. |
325 There is a quotient map $\pi: M \to \coinv{M}$. |
295 We claim that the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; |
326 We claim that the composition $\pi \compose \ev$ is well-defined on the quotient $H_0(K_*(M))$; |
296 i.e.\ that $\pi(\ev(\bd y)) = 0$ for all $y \in K_1(M)$. |
327 i.e.\ that $\pi(\ev(\bd y)) = 0$ for all $y \in K_1(M)$. |
297 There are two cases, depending on whether the blob of $y$ contains the point *. |
328 There are two cases, depending on whether the blob of $y$ contains the point *. |
298 If it doesn't, then |
329 If it doesn't, then |
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 |
330 suppose $y$ has label $m$ at $*$, labels $c_i$ at other labeled points outside the blob, |
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 |
331 and the field inside the blob is a sum, with the $j$-th term having |
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332 labeled points $d_{j,i}$. |
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333 Then $\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \in \ker(\DirectSum_k C^{\tensor k} \to C)$, and so |
301 $\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$ |
334 $\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$ |
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 $*$, |
335 Similarly, if $*$ is contained in the blob, then the blob label is a sum, with the |
303 and there are labels $c_i$ at the labeled points outside the blob. We know that |
336 $j$-th term have labelled points $d_{j,i}$ to the left of $*$, $m_j$ at $*$, and $d_{j,i}'$ to the right of $*$, |
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337 and there are labels $c_i$ at the labeled points outside the blob. |
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338 We know that |
304 $$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$ |
339 $$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$ |
305 and so |
340 and so |
306 \begin{align*} |
341 \begin{align*} |
307 \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} \\ |
342 \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} \\ |
308 & = \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 \\ |
343 & = \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 \\ |
309 & = 0 |
344 & = 0 |
310 \end{align*} |
345 \end{align*} |
311 where this time we use the fact that we're mapping to $\coinv{M}$, not just $M$. |
346 where this time we use the fact that we're mapping to $\coinv{M}$, not just $M$. |
312 |
347 |
313 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. |
348 The map $\pi \compose \ev: H_0(K_*(M)) \to \coinv{M}$ is clearly |
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349 surjective ($\ev$ surjects onto $M$); we now show that it's injective. |
314 This is equivalent to showing that |
350 This is equivalent to showing that |
315 \[ |
351 \[ |
316 \ev\inv(\ker(\pi)) \sub \bd K_1(M) . |
352 \ev\inv(\ker(\pi)) \sub \bd K_1(M) . |
317 \] |
353 \] |
318 The above inclusion follows from |
354 The above inclusion follows from |
353 generated by blob diagrams $b$ such that $N_\ep$ is either disjoint from |
390 generated by blob diagrams $b$ such that $N_\ep$ is either disjoint from |
354 or contained in each blob of $b$, and the only labeled point inside $N_\ep$ is $*$. |
391 or contained in each blob of $b$, and the only labeled point inside $N_\ep$ is $*$. |
355 %and the two boundary points of $N_\ep$ are not labeled points of $b$. |
392 %and the two boundary points of $N_\ep$ are not labeled points of $b$. |
356 For a field $y$ on $N_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ |
393 For a field $y$ on $N_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ |
357 labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$. |
394 labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$. |
358 (See Figure \ref{fig:sy}.) Note that $y - s_\ep(y) \in U(N_\ep)$. |
395 (See Figure \ref{fig:sy}.) |
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396 Note that $y - s_\ep(y) \in U(N_\ep)$. |
359 Let $\sigma_\ep: K_*^\ep \to K_*^\ep$ be the chain map |
397 Let $\sigma_\ep: K_*^\ep \to K_*^\ep$ be the chain map |
360 given by replacing the restriction $y$ to $N_\ep$ of each field |
398 given by replacing the restriction $y$ to $N_\ep$ of each field |
361 appearing in an element of $K_*^\ep$ with $s_\ep(y)$. |
399 appearing in an element of $K_*^\ep$ with $s_\ep(y)$. |
362 Note that $\sigma_\ep(x) \in K'_*$. |
400 Note that $\sigma_\ep(x) \in K'_*$. |
363 \begin{figure}[t] |
401 \begin{figure}[t] |