263 \begin{equation} |
263 \begin{equation} |
264 \label{eq:ker-functor}% |
264 \label{eq:ker-functor}% |
265 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
265 M \mapsto \ker(C^{\tensor k} \tensor M \tensor C^{\tensor l} \to M) |
266 \end{equation} |
266 \end{equation} |
267 are all exact too. Moreover, tensor products of such functors with each |
267 are all exact too. Moreover, tensor products of such functors with each |
268 other and with $C$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M) |
268 other and with $C$ or $\ker(C^{\tensor k} \to C)$ (e.g., producing the functor $M \mapsto \ker(M \tensor C \to M) |
269 \tensor C \tensor \ker(C \tensor M \to M)$) are all still exact. |
269 \tensor C \tensor \ker(C \tensor C \to M)$) are all still exact. |
270 |
270 |
271 Finally, then we see that the functor $K_*$ is simply an (infinite) |
271 Finally, then we see that the functor $K_*$ is simply an (infinite) |
272 direct sum of this sort of functor. The direct sum is indexed by |
272 direct sum of copies of this sort of functor. The direct sum is indexed by |
273 configurations of nested blobs and positions of labels; for each such configuration, we have one of the above tensor product functors, |
273 configurations of nested blobs and of labels; for each such configuration, we have one of the above tensor product functors, |
274 with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor}, and all other labelled points corresponding |
274 with the labels of twig blobs corresponding to tensor factors as in \eqref{eq:ker-functor} or $\ker(C^{\tensor k} \to C)$, and all other labelled points corresponding |
275 to tensor factors of $C$. |
275 to tensor factors of $C$. |
276 \end{proof} |
276 \end{proof} |
277 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
277 \begin{proof}[Proof of Lemma \ref{lem:hochschild-coinvariants}] |
278 \todo{} |
278 We show that $H_0(K_*(M))$ is isomorphic to the coinvariants of $M$. |
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279 |
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280 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 $*$, |
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281 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. |
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282 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 |
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283 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 |
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284 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 |
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285 $\ev(\bdy y) = 0$, because $$C^{\tensor \ell_1} \tensor \ker(\DirectSum_k C^{\tensor k} \to C) \tensor C^{\tensor \ell_2} \subset \ker(\DirectSum_k C^{\tensor k} \to C).$$ |
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286 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 $*$, |
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287 and there are labels $c_i$ at the labeled points outside the blob. We know that |
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288 $$\sum_j d_{j,1} \tensor \cdots \tensor d_{j,k_j} \tensor m_j \tensor d_{j,1}' \tensor \cdots \tensor d_{j,k'_j}' \in \ker(\DirectSum_{k,k'} C^{\tensor k} \tensor M \tensor C^{\tensor k'} \tensor \to M),$$ |
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289 and so |
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290 \begin{align*} |
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291 \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} \\ |
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292 & = \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 \\ |
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293 & = 0 |
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294 \end{align*} |
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295 where this time we use the fact that we're mapping to $\coinv{M}$, not just $M$. |
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296 |
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297 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. \todo{} |
279 \end{proof} |
298 \end{proof} |
280 \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] |
299 \begin{proof}[Proof of Lemma \ref{lem:hochschild-free}] |
281 We show that $K_*(C\otimes C)$ is |
300 We show that $K_*(C\otimes C)$ is |
282 quasi-isomorphic to the 0-step complex $C$. We'll do this in steps, establishing quasi-isomorphisms and homotopy equivalences |
301 quasi-isomorphic to the 0-step complex $C$. We'll do this in steps, establishing quasi-isomorphisms and homotopy equivalences |
283 $$K_*(C \tensor C) \quismto K'_* \htpyto K''_* \quismto C.$$ |
302 $$K_*(C \tensor C) \quismto K'_* \htpyto K''_* \quismto C.$$ |