
diff -r e0b304e6b975 -r e1d24be683bb text/hochschild.tex
--- a/text/hochschild.tex	Wed Oct 28 00:54:35 2009 +0000
+++ b/text/hochschild.tex	Wed Oct 28 02:44:29 2009 +0000
@@ -3,11 +3,16 @@
 \section{Hochschild homology when $n=1$}
 \label{sec:hochschild}
 
+So far we have provided no evidence that blob homology is interesting in degrees 
+greater than zero.
 In this section we analyze the blob complex in dimension $n=1$
 and find that for $S^1$ the blob complex is homotopy equivalent to the 
 Hochschild complex of the category (algebroid) that we started with.
+Thus the blob complex is a natural generalization of something already
+known to be interesting in higher homological degrees.
 
-\nn{initial idea for blob complex came from thinking about...}
+It is also worth noting that the original idea for the blob complex came from trying
+to find a more ``local" description of the Hochschild complex.
 
 \nn{need to be consistent about quasi-isomorphic versus homotopy equivalent
 in this section.
@@ -38,7 +43,8 @@
 Hochschild complex of $C$.
 Note that both complexes are free (and hence projective), so it suffices to show that they
 are quasi-isomorphic.
-In order to prove this we will need to extend the blob complex to allow points to also
+In order to prove this we will need to extend the 
+definition of the blob complex to allow points to also
 be labeled by elements of $C$-$C$-bimodules.
 
 Fix points $p_1, \ldots, p_k \in S^1$ and $C$-$C$-bimodules $M_1, \ldots M_k$.
@@ -223,6 +229,7 @@
 Suppose we have a short exact sequence of $C$-$C$ bimodules $$\xymatrix{0 \ar[r] & K \ar@{^{(}->}[r]^f & E \ar@{->>}[r]^g & Q \ar[r] & 0}.$$
 We'll write $\hat{f}$ and $\hat{g}$ for the image of $f$ and $g$ under the functor.
 Most of what we need to check is easy.
+\nn{don't we need to consider sums here, e.g. $\sum_i(a_i\ot k_i\ot b_i)$ ?}
 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
 $e$ is in the image of the original $f$, so is in the kernel of the original $g$, and so $\hat{g}(a \tensor e \tensor b) = 0$.
 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 $a \tensor e \tensor b = \hat{f}(a \tensor \widetilde{e} \tensor b)$.




