1 %!TEX root = ../blob1.tex |
1 %!TEX root = ../blob1.tex |
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3 \section{Hochschild homology when $n=1$} |
3 \section{Hochschild homology when $n=1$} |
4 \label{sec:hochschild} |
4 \label{sec:hochschild} |
5 |
5 |
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6 So far we have provided no evidence that blob homology is interesting in degrees |
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7 greater than zero. |
6 In this section we analyze the blob complex in dimension $n=1$ |
8 In this section we analyze the blob complex in dimension $n=1$ |
7 and find that for $S^1$ the blob complex is homotopy equivalent to the |
9 and find that for $S^1$ the blob complex is homotopy equivalent to the |
8 Hochschild complex of the category (algebroid) that we started with. |
10 Hochschild complex of the category (algebroid) that we started with. |
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11 Thus the blob complex is a natural generalization of something already |
10 \nn{initial idea for blob complex came from thinking about...} |
12 known to be interesting in higher homological degrees. |
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13 |
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14 It is also worth noting that the original idea for the blob complex came from trying |
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15 to find a more ``local" description of the Hochschild complex. |
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12 \nn{need to be consistent about quasi-isomorphic versus homotopy equivalent |
17 \nn{need to be consistent about quasi-isomorphic versus homotopy equivalent |
13 in this section. |
18 in this section. |
14 since the various complexes are free, q.i. implies h.e.} |
19 since the various complexes are free, q.i. implies h.e.} |
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36 |
41 |
37 We want to show that $\bc_*(S^1)$ is homotopy equivalent to the |
42 We want to show that $\bc_*(S^1)$ is homotopy equivalent to the |
38 Hochschild complex of $C$. |
43 Hochschild complex of $C$. |
39 Note that both complexes are free (and hence projective), so it suffices to show that they |
44 Note that both complexes are free (and hence projective), so it suffices to show that they |
40 are quasi-isomorphic. |
45 are quasi-isomorphic. |
41 In order to prove this we will need to extend the blob complex to allow points to also |
46 In order to prove this we will need to extend the |
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47 definition of the blob complex to allow points to also |
42 be labeled by elements of $C$-$C$-bimodules. |
48 be labeled by elements of $C$-$C$-bimodules. |
43 |
49 |
44 Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$. |
50 Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$. |
45 We define a blob-like complex $K_*(S^1, (p_i), (M_i))$. |
51 We define a blob-like complex $K_*(S^1, (p_i), (M_i))$. |
46 The fields have elements of $M_i$ labeling $p_i$ and elements of $C$ labeling |
52 The fields have elements of $M_i$ labeling $p_i$ and elements of $C$ labeling |
221 is exact. For completeness we'll explain this below. |
227 is exact. For completeness we'll explain this below. |
222 |
228 |
223 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
229 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
224 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
230 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
225 Most of what we need to check is easy. |
231 Most of what we need to check is easy. |
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232 \nn{don't we need to consider sums here, e.g. $\sum_i(a_i\ot k_i\ot b_i)$ ?} |
226 If $(a \tensor k \tensor b) \in \ker(C \tensor K \tensor C \to K)$, to have $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$, we must have $f(k) = 0 \in E$, which implies $k=0$ itself. If $(a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, clearly |
233 If $(a \tensor k \tensor b) \in \ker(C \tensor K \tensor C \to K)$, to have $\hat{f}(a \tensor k \tensor b) = 0 \in \ker(C \tensor E \tensor C \to E)$, we must have $f(k) = 0 \in E$, which implies $k=0$ itself. If $(a \tensor e \tensor b) \in \ker(C \tensor E \tensor C \to E)$ is in the image of $\ker(C \tensor K \tensor C \to K)$ under $\hat{f}$, clearly |
227 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
234 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
228 If $\hat{g}(a \tensor e \tensor b) = 0$, then $g(e) = 0$, so $e = f(\widetilde{e})$ for some $\widetilde{e} \in K$, and $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
235 If $\hat{g}(a \tensor e \tensor b) = 0$, then $g(e) = 0$, so $e = f(\widetilde{e})$ for some $\widetilde{e} \in K$, and $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
229 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 $\hat{g}$. |
236 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 $\hat{g}$. |
230 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. |
237 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. |