110 \begin{prop} |
110 \begin{prop} |
111 The complex $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual |
111 The complex $K_*(M)$ is quasi-isomorphic to $HC_*(M)$, the usual |
112 Hochschild complex of $M$. |
112 Hochschild complex of $M$. |
113 \end{prop} |
113 \end{prop} |
114 \begin{proof} |
114 \begin{proof} |
115 %First, since we're working over $\Complex$, note that saying two complexes are quasi-isomorphic simply means they have isomorphic homologies. |
115 Recall that the usual Hochschild complex of $M$ is uniquely determined, |
116 %\todo{We really don't want to work over $\Complex$, though; it would be nice to talk about torsion!} |
116 up to quasi-isomorphism, by the following properties: |
117 |
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118 Recall that the usual Hochschild complex of $M$ is uniquely determined, up to quasi-isomorphism, by the following properties: |
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119 \begin{enumerate} |
117 \begin{enumerate} |
120 \item \label{item:hochschild-additive}% |
118 \item \label{item:hochschild-additive}% |
121 $HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$. |
119 $HC_*(M_1 \oplus M_2) \cong HC_*(M_1) \oplus HC_*(M_2)$. |
122 \item \label{item:hochschild-exact}% |
120 \item \label{item:hochschild-exact}% |
123 An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
121 An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
124 exact sequence $0 \to HC_*(M_1) \into HC_*(M_2) \onto HC_*(M_3) \to 0$. |
122 exact sequence $0 \to HC_*(M_1) \into HC_*(M_2) \onto HC_*(M_3) \to 0$. |
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123 \item \label{item:hochschild-coinvariants}% |
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124 $HH_0(M)$ is isomorphic to the coinvariants of $M$, $\coinv(M) = |
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125 M/\langle cm-mc \rangle$. |
125 \item \label{item:hochschild-free}% |
126 \item \label{item:hochschild-free}% |
126 $HC_*(C\otimes C)$ (the free $C$-$C$-bimodule with one generator) is |
127 $HC_*(C\otimes C)$ (the free $C$-$C$-bimodule with one generator) is contractible; that is, |
127 quasi-isomorphic to the 0-step complex $C$. |
128 quasi-isomorphic to its $0$-th homology (which in turn, by \ref{item:hochschild-coinvariants}, is just $C$) via the quotient map $HC_0 \onto HH_0$. |
128 \item \label{item:hochschild-coinvariants}% |
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129 $HH_0(M)$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. |
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130 \end{enumerate} |
129 \end{enumerate} |
131 (Together, these just say that Hochschild homology is `the derived functor of coinvariants'.) |
130 (Together, these just say that Hochschild homology is `the derived functor of coinvariants'.) |
132 We'll first recall why these properties are characteristic. |
131 We'll first recall why these properties are characteristic. |
133 |
132 |
134 Take some $C$-$C$ bimodule $M$, and choose a free resolution |
133 Take some $C$-$C$ bimodule $M$, and choose a free resolution |
135 \begin{equation*} |
134 \begin{equation*} |
136 \cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0. |
135 \cdots \to F_2 \xrightarrow{f_2} F_1 \xrightarrow{f_1} F_0. |
137 \end{equation*} |
136 \end{equation*} |
138 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 |
137 We will show that for any functor $\cP$ satisfying properties |
139 $HC_i(F_j)$, with $i,j \geq 0$, graded by $i+j$. |
138 \ref{item:hochschild-additive}, \ref{item:hochschild-exact}, |
140 |
139 \ref{item:hochschild-coinvariants} and \ref{item:hochschild-free}, there |
141 Observe that we have two chain maps |
140 is a quasi-isomorphism |
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141 $$\cP_*(M) \iso \coinv(F_*).$$ |
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142 % |
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143 Observe that there's a quotient map $\pi: F_0 \onto M$, and by |
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144 construction the cone of the chain map $\pi: F_* \to M$ is acyclic. Now |
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145 construct the total complex $\cP_i(F_j)$, with $i,j \geq 0$, graded by |
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146 $i+j$. We have two chain maps |
142 \begin{align*} |
147 \begin{align*} |
143 HC_i(F_j) & \xrightarrow{HC_i(\pi)} HC_i(M) \\ |
148 \cP_i(F_*) & \xrightarrow{\cP_i(\pi)} \cP_i(M) \\ |
144 \intertext{and} |
149 \intertext{and} |
145 HC_i(F_j) & \xrightarrow{HC_0(F_j) \onto HH_0(F_j)} \operatorname{coinv}(F_j). |
150 \cP_*(F_j) & \xrightarrow{\cP_0(F_j) \onto H_0(\cP_*(F_j))} \coinv(F_j). |
146 \end{align*} |
151 \end{align*} |
147 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. |
152 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. |
148 In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free. |
153 In the second case, this is because the `columns' indexed by $j$ are acyclic, since $F_j$ is free. |
149 |
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150 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 |
154 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 |
151 $$HC_*(M) \iso \operatorname{coinv}(F_*).$$ |
155 $$\cP_*(M) \quismto \coinv(F_*).$$ |
152 |
156 |
153 %If $M$ is free, that is, a direct sum of copies of |
157 %If $M$ is free, that is, a direct sum of copies of |
154 %$C \tensor C$, then properties \ref{item:hochschild-additive} and |
158 %$C \tensor C$, then properties \ref{item:hochschild-additive} and |
155 %\ref{item:hochschild-free} determine $HC_*(M)$. Otherwise, choose some |
159 %\ref{item:hochschild-free} determine $HC_*(M)$. Otherwise, choose some |
156 %free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we |
160 %free cover $F \onto M$, and define $K$ to be this map's kernel. Thus we |
182 \label{lem:hochschild-exact}% |
186 \label{lem:hochschild-exact}% |
183 An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
187 An exact sequence $0 \to M_1 \into M_2 \onto M_3 \to 0$ gives rise to an |
184 exact sequence $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$. |
188 exact sequence $0 \to K_*(M_1) \into K_*(M_2) \onto K_*(M_3) \to 0$. |
185 \end{lem} |
189 \end{lem} |
186 \begin{lem} |
190 \begin{lem} |
187 \label{lem:hochschild-free}% |
191 \label{lem:hochschild-coinvariants}% |
188 $K_*(C\otimes C)$ is quasi-isomorphic to the 0-step complex $C$. |
192 $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
189 \end{lem} |
193 \end{lem} |
190 \begin{lem} |
194 \begin{lem} |
191 \label{lem:hochschild-coinvariants}% |
195 \label{lem:hochschild-free}% |
192 $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$, $M/\langle cm-mc \rangle$. |
196 $K_*(C\otimes C)$ is quasi-isomorphic to $H_0(K_*(C \otimes C)) \iso C$. |
193 \end{lem} |
197 \end{lem} |
194 |
198 |
195 The remainder of this section is devoted to proving Lemmas |
199 The remainder of this section is devoted to proving Lemmas |
196 \ref{lem:module-blob}, |
200 \ref{lem:module-blob}, |
197 \ref{lem:hochschild-exact}, \ref{lem:hochschild-free} and |
201 \ref{lem:hochschild-exact}, \ref{lem:hochschild-coinvariants} and |
198 \ref{lem:hochschild-coinvariants}. |
202 \ref{lem:hochschild-free}. |
199 \end{proof} |
203 \end{proof} |
200 |
204 |
201 \begin{proof}[Proof of Lemma \ref{lem:module-blob}] |
205 \begin{proof}[Proof of Lemma \ref{lem:module-blob}] |
202 We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$. |
206 We show that $K_*(C)$ is quasi-isomorphic to $\bc_*(S^1)$. |
203 $K_*(C)$ differs from $\bc_*(S^1)$ only in that the base point * |
207 $K_*(C)$ differs from $\bc_*(S^1)$ only in that the base point * |
204 is always a labeled point in $K_*(C)$, while in $\bc_*(S^1)$ it may or may not be. |
208 is always a labeled point in $K_*(C)$, while in $\bc_*(S^1)$ it may or may not be. |
205 In other words, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$. |
209 In other words, there is an inclusion map $i: K_*(C) \to \bc_*(S^1)$. |
206 |
210 |
207 We define a quasi-inverse \nn{right term?} $s: \bc_*(S^1) \to K_*(C)$ to the inclusion as follows. |
211 We define a left inverse $s: \bc_*(S^1) \to K_*(C)$ to the inclusion as follows. |
208 If $y$ is a field defined on a neighborhood of *, define $s(y) = y$ if |
212 If $y$ is a field defined on a neighborhood of *, define $s(y) = y$ if |
209 * is a labeled point in $y$. |
213 * is a labeled point in $y$. |
210 Otherwise, define $s(y)$ to be the result of adding a label 1 (identity morphism) at *. |
214 Otherwise, define $s(y)$ to be the result of adding a label 1 (identity morphism) at *. |
211 Let $x \in \bc_*(S^1)$. |
215 Let $x \in \bc_*(S^1)$. |
212 Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in |
216 Let $s(x)$ be the result of replacing each field $y$ (containing *) mentioned in |
217 in a neighborhood $B_\ep$ of *, except perhaps *. |
221 in a neighborhood $B_\ep$ of *, except perhaps *. |
218 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
222 Note that for any chain $x \in \bc_*(S^1)$, $x \in L_*^\ep$ for sufficiently small $\ep$. |
219 \nn{rest of argument goes similarly to above} |
223 \nn{rest of argument goes similarly to above} |
220 \end{proof} |
224 \end{proof} |
221 \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
225 \begin{proof}[Proof of Lemma \ref{lem:hochschild-exact}] |
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226 \todo{p. 1478 of scott's notes} |
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227 Essentially, this comes down to the unsurprising fact that the functor on $C$-$C$ bimodules |
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228 \begin{equation*} |
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229 M \mapsto \ker(C \tensor M \tensor C \xrightarrow{c_1 \tensor m \tensor c_2 \mapsto c_1 m c_2} M) |
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230 \end{equation*} |
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231 is exact. For completeness we'll explain this below. |
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232 |
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233 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$ |
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234 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor. |
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235 Most of what we need to check is easy. |
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236 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 $f(k) = 0 \in E$, so |
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237 be $k=0$ 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 |
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238 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$. |
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239 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 $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$. |
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240 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 $\hat{g}$. |
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241 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. |
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242 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 |
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243 $\widetilde{q} \in \ker(C \tensor E \tensor C \to E)$. Further, |
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244 \begin{align*} |
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245 \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 \\ |
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246 & = q - 0 |
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247 \end{align*} |
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248 (here we used that $g$ is a map of $C$-$C$ bimodules, and that $\sum_i a_i q_i b_i = 0$). |
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249 |
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250 Identical arguments show that the functors |
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251 \begin{equation*} |
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252 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
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253 \end{equation*} |
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254 are all exact too. |
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255 |
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256 Finally, then \todo{explain why this is all we need.} |
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257 \end{proof} |
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258 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
222 \todo{} |
259 \todo{} |
223 \end{proof} |
260 \end{proof} |
224 \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] |
261 \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] |
225 We show that $K_*(C\otimes C)$ is |
262 We show that $K_*(C\otimes C)$ is |
226 quasi-isomorphic to the 0-step complex $C$. |
263 quasi-isomorphic to the 0-step complex $C$. |
231 |
268 |
232 Fix a small $\ep > 0$. |
269 Fix a small $\ep > 0$. |
233 Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$. |
270 Let $B_\ep$ be the ball of radius $\ep$ around $* \in S^1$. |
234 Let $K_*^\ep \sub K_*(C\otimes C)$ be the subcomplex |
271 Let $K_*^\ep \sub K_*(C\otimes C)$ be the subcomplex |
235 generated by blob diagrams $b$ such that $B_\ep$ is either disjoint from |
272 generated by blob diagrams $b$ such that $B_\ep$ is either disjoint from |
236 or contained in each blob of $b$, and the two boundary points of $B_\ep$ are not labeled points of $b$. |
273 or contained in each blob of $b$, and the only labeled point inside $B_\ep$ is $*$. |
237 For a field (picture) $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ |
274 %and the two boundary points of $B_\ep$ are not labeled points of $b$. |
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275 For a field $y$ on $B_\ep$, let $s_\ep(y)$ be the equivalent picture with~$*$ |
238 labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$. |
276 labeled by $1\otimes 1$ and the only other labeled points at distance $\pm\ep/2$ from $*$. |
239 (See Figure xxxx.) |
277 (See Figure \ref{fig:sy}.) Note that $y - s_\ep(y) \in U(B_\ep)$. We can think of |
240 Note that $y - s_\ep(y) \in U(B_\ep)$. |
278 $\sigma_\ep$ as a chain map $K_*^\ep \to K_*^\ep$ given by replacing the restriction $y$ to $B_\ep$ of each field |
241 \nn{maybe it's simpler to assume that there are no labeled points, other than $*$, in $B_\ep$.} |
279 appearing in an element of $K_*^\ep$ with $s_\ep(y)$. |
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280 Note that $\sigma_\ep(x) \in K'_*$. |
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281 \begin{figure}[!ht] |
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282 \begin{align*} |
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283 y & = \mathfig{0.2}{hochschild/y} & |
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284 s_\ep(y) & = \mathfig{0.2}{hochschild/sy} |
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285 \end{align*} |
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286 \caption{Defining $s_\ep$.} |
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287 \label{fig:sy} |
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288 \end{figure} |
242 |
289 |
243 Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. |
290 Define a degree 1 chain map $j_\ep : K_*^\ep \to K_*^\ep$ as follows. |
244 Let $x \in K_*^\ep$ be a blob diagram. |
291 Let $x \in K_*^\ep$ be a blob diagram. |
245 If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $B_\ep$ to |
292 If $*$ is not contained in any twig blob, $j_\ep(x)$ is obtained by adding $B_\ep$ to |
246 $x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $B_\ep$. |
293 $x$ as a new twig blob, with label $y - s_\ep(y)$, where $y$ is the restriction of $x$ to $B_\ep$. |
248 Let $y_i$ be the restriction of $z_i$ to $B_\ep$. |
295 Let $y_i$ be the restriction of $z_i$ to $B_\ep$. |
249 Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin B_\ep$, |
296 Let $x_i$ be equal to $x$ outside of $B$, equal to $z_i$ on $B \setmin B_\ep$, |
250 and have an additional blob $B_\ep$ with label $y_i - s_\ep(y_i)$. |
297 and have an additional blob $B_\ep$ with label $y_i - s_\ep(y_i)$. |
251 Define $j_\ep(x) = \sum x_i$. |
298 Define $j_\ep(x) = \sum x_i$. |
252 \nn{need to check signs coming from blob complex differential} |
299 \nn{need to check signs coming from blob complex differential} |
253 |
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254 Note that if $x \in K'_* \cap K_*^\ep$ then $j_\ep(x) \in K'_*$ also. |
300 Note that if $x \in K'_* \cap K_*^\ep$ then $j_\ep(x) \in K'_*$ also. |
255 |
301 |
256 The key property of $j_\ep$ is |
302 The key property of $j_\ep$ is |
257 \eq{ |
303 \eq{ |
258 \bd j_\ep + j_\ep \bd = \id - \sigma_\ep , |
304 \bd j_\ep + j_\ep \bd = \id - \sigma_\ep. |
259 } |
305 } |
260 where $\sigma_\ep : K_*^\ep \to K_*^\ep$ is given by replacing the restriction $y$ of each field |
306 If $j_\ep$ were defined on all of $K_*(C\otimes C)$, this would show that $\sigma_\ep$ |
261 mentioned in $x \in K_*^\ep$ with $s_\ep(y)$. |
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262 Note that $\sigma_\ep(x) \in K'_*$. |
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263 |
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264 If $j_\ep$ were defined on all of $K_*(C\otimes C)$, it would show that $\sigma_\ep$ |
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265 is a homotopy inverse to the inclusion $K'_* \to K_*(C\otimes C)$. |
307 is a homotopy inverse to the inclusion $K'_* \to K_*(C\otimes C)$. |
266 One strategy would be to try to stitch together various $j_\ep$ for progressively smaller |
308 One strategy would be to try to stitch together various $j_\ep$ for progressively smaller |
267 $\ep$ and show that $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$. |
309 $\ep$ and show that $K'_*$ is homotopy equivalent to $K_*(C\otimes C)$. |
268 Instead, we'll be less ambitious and just show that |
310 Instead, we'll be less ambitious and just show that |
269 $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. |
311 $K'_*$ is quasi-isomorphic to $K_*(C\otimes C)$. |