--- a/blob1.tex Tue Mar 03 23:27:22 2009 +0000
+++ b/blob1.tex Thu Mar 12 19:53:43 2009 +0000
@@ -52,7 +52,7 @@
% \DeclareMathOperator{\pr}{pr} etc.
\def\declaremathop#1{\expandafter\DeclareMathOperator\csname #1\endcsname{#1}}
-\applytolist{declaremathop}{pr}{im}{gl}{ev}{coinv}{tr}{rot}{Eq}{obj}{mor}{ob}{Rep}{Tet}{cat}{Maps}{Diff}{sign}{supp}{maps};
+\applytolist{declaremathop}{pr}{im}{gl}{ev}{coinv}{tr}{rot}{Eq}{obj}{mor}{ob}{Rep}{Tet}{cat}{Maps}{Diff}{sign}{supp};
@@ -764,8 +764,7 @@
In the other direction, any blob diagram on $X\du Y$ is equal (up to sign)
to one that puts $X$ blobs before $Y$ blobs in the ordering, and so determines
a pair of blob diagrams on $X$ and $Y$.
-These two maps are compatible with our sign conventions \nn{say more about this?} and
-with the linear label relations.
+These two maps are compatible with our sign conventions.
The two maps are inverses of each other.
\nn{should probably say something about sign conventions for the differential
in a tensor product of chain complexes; ask Scott}
@@ -778,7 +777,7 @@
we have a splitting $s: H_0(\bc_*(B^n, c)) \to \bc_0(B^n; c)$
of the quotient map
$p: \bc_0(B^n; c) \to H_0(\bc_*(B^n, c))$.
-\nn{always the case if we're working over $\c$}.
+For example, this is always the case if you coefficient ring is a field.
Then
\begin{prop} \label{bcontract}
For all $c \in \cC(\bd B^n)$ the natural map $p: \bc_*(B^n, c) \to H_0(\bc_*(B^n, c))$
@@ -794,20 +793,20 @@
In other words, add a new outermost blob which encloses all of the others.
Define $h_0 : \bc_0(B^n; c) \to \bc_1(B^n; c)$ by setting $h_0(x)$ equal to
the 1-blob with blob $B^n$ and label $x - s(p(x)) \in U(B^n; c)$.
-\nn{$x$ is a 0-blob diagram, i.e. $x \in \cC(B^n; c)$}
\end{proof}
-(Note that for the above proof to work, we need the linear label relations
-for blob labels.
-Also we need to blob reordering relations (?).)
+Note that if there is no splitting $s$, we can let $h_0 = 0$ and get a homotopy
+equivalence to the 2-step complex $U(B^n; c) \to \cC(B^n; c)$.
-(Note also that if there is no splitting $s$, we can let $h_0 = 0$ and get a homotopy
-equivalence to the 2-step complex $U(B^n; c) \to \cC(B^n; c)$.)
-
-(For fields based on $n$-cats, $H_0(\bc_*(B^n; c)) \cong \mor(c', c'')$.)
+For fields based on $n$-categories, $H_0(\bc_*(B^n; c)) \cong \mor(c', c'')$,
+where $(c', c'')$ is some (any) splitting of $c$ into domain and range.
\medskip
+\nn{Maybe there is no longer a need to repeat the next couple of props here, since we also state them in the introduction.
+But I think it's worth saying that the Diff actions will be enhanced later.
+Maybe put that in the intro too.}
+
As we noted above,
\begin{prop}
There is a natural isomorphism $H_0(\bc_*(X)) \cong A(X)$.
@@ -815,21 +814,6 @@
\end{prop}
-% oops -- duplicate
-
-%\begin{prop} \label{functorialprop}
-%The assignment $X \mapsto \bc_*(X)$ extends to a functor from the category of
-%$n$-manifolds and homeomorphisms to the category of chain complexes and linear isomorphisms.
-%\end{prop}
-
-%\begin{proof}
-%Obvious.
-%\end{proof}
-
-%\nn{need to same something about boundaries and boundary conditions above.
-%maybe fix the boundary and consider the category of $n$-manifolds with the given boundary.}
-
-
\begin{prop}
For fixed fields ($n$-cat), $\bc_*$ is a functor from the category
of $n$-manifolds and diffeomorphisms to the category of chain complexes and
@@ -837,10 +821,6 @@
\qed
\end{prop}
-\nn{need to same something about boundaries and boundary conditions above.
-maybe fix the boundary and consider the category of $n$-manifolds with the given boundary.}
-
-
In particular,
\begin{prop} \label{diff0prop}
There is an action of $\Diff(X)$ on $\bc_*(X)$.
@@ -857,7 +837,7 @@
Let $X$ be an $n$-manifold, $\bd X = Y \cup (-Y) \cup Z$.
Gluing the two copies of $Y$ together yields an $n$-manifold $X\sgl$
with boundary $Z\sgl$.
-Given compatible fields (pictures, boundary conditions) $a$, $b$ and $c$ on $Y$, $-Y$ and $Z$,
+Given compatible fields (boundary conditions) $a$, $b$ and $c$ on $Y$, $-Y$ and $Z$,
we have the blob complex $\bc_*(X; a, b, c)$.
If $b = -a$ (the orientation reversal of $a$), then we can glue up blob diagrams on
$X$ to get blob diagrams on $X\sgl$:
@@ -876,6 +856,8 @@
The above map is very far from being an isomorphism, even on homology.
This will be fixed in Section \ref{sec:gluing} below.
+\nn{Next para not need, since we already use bullet = gluing notation above(?)}
+
An instance of gluing we will encounter frequently below is where $X = X_1 \du X_2$
and $X\sgl = X_1 \cup_Y X_2$.
(Typically one of $X_1$ or $X_2$ is a disjoint union of balls.)
@@ -886,9 +868,8 @@
Note that we have resumed our habit of omitting boundary labels from the notation.
-\bigskip
-\nn{what else?}
+
\section{Hochschild homology when $n=1$}
\label{sec:hochschild}
--- a/text/gluing.tex Tue Mar 03 23:27:22 2009 +0000
+++ b/text/gluing.tex Thu Mar 12 19:53:43 2009 +0000
@@ -249,7 +249,7 @@
\nn{give examples: $A(J^n) = \bc_*(Z\times J)$ and $A(J^n) = C_*(\Maps(J \to M))$.}
-\todo{the motivating example $C_*(\maps(X, M))$}
+\todo{the motivating example $C_*(\Maps(X, M))$}
@@ -406,7 +406,7 @@
use graphical/tree point of view, rather than following Keller exactly
\item define blob complex in $A_\infty$ case; fat mapping cones? tree decoration?
\item topological $A_\infty$ cat def (maybe this should go first); also modules gluing
-\item motivating example: $C_*(\maps(X, M))$
+\item motivating example: $C_*(\Maps(X, M))$
\item maybe incorporate dual point of view (for $n=1$), where points get
object labels and intervals get 1-morphism labels
\end{itemize}