--- a/text/smallblobs.tex Sat May 29 08:35:06 2010 -0700
+++ b/text/smallblobs.tex Sat May 29 15:01:43 2010 -0700
@@ -29,25 +29,25 @@
When $\beta$ is a collection of disjoint embedded balls in $M$, we say that a homeomorphism of $M$ `makes $\beta$ small' if the image of each ball in $\beta$ under the homeomorphism is contained in some open set of $\cU$. Further, we'll say a homeomorphism `makes $\beta$ $\epsilon$-small' if the image of each ball is contained in some open ball of radius $\epsilon$.
-On a $1$-blob $b$, with ball $\beta$, $s$ is defined as the sum of two terms. Essentially, the first term `makes $\beta$ small', while the other term `gets the boundary right'. First, pick a one-parameter family $\phi_\beta : \Delta^1 \to \Homeo(M)$ of homeomorphisms, so $\phi_\beta(1,0)$ is the identity and $\phi_\beta(0,1)$ makes the ball $\beta$ small --- in fact, not just small with respect to $\cU$, but $\epsilon/2$-small, where $\epsilon > 0$ is such that every $\epsilon$-ball is contained in some open set of $\cU$. Next, pick a two-parameter family $\phi_{\eset \prec \beta} : \Delta^2 \to \Homeo(M)$ so that $\phi_{\eset \prec \beta}(0,x_1,x_2)$ makes the ball $\beta$ $\frac{3\epsilon}{4}$-small for all $x_1+x_2=1$, while $\phi_{\eset \prec \beta}(x_0,0,x_2) = \phi_\eset(x_0,x_2)$ and $\phi_{\eset \prec \beta}(x_0,x_1,0) = \phi_\beta(x_0,x_1)$. (It's perhaps not obvious that this is even possible --- see Lemma \ref{lem:extend-small-homeomorphisms} below.) We now define $s$ by
-$$s(b) = \restrict{\phi_\beta}{x_0=0}(b) + \restrict{\phi_{\eset \prec \beta}}{x_0=0}(\bdy b).$$
+On a $1$-blob $b$, with ball $\beta$, $s$ is defined as the sum of two terms. Essentially, the first term `makes $\beta$ small', while the other term `gets the boundary right'. First, pick a one-parameter family $\phi_\beta : \Delta^1 \to \Homeo(M)$ of homeomorphisms, so $\phi_\beta(1,0)$ is the identity and $\phi_\beta(0,1)$ makes the ball $\beta$ small --- in fact, not just small with respect to $\cU$, but $\epsilon/2$-small, where $\epsilon > 0$ is such that every $\epsilon$-ball is contained in some open set of $\cU$. Next, pick a two-parameter family $\phi_{\eset \prec \beta} : \Delta^2 \to \Homeo(M)$ so that $\phi_{\eset \prec \beta}(0,x_1,x_2)$ makes the ball $\beta$ $\frac{3\epsilon}{4}$-small for all $x_1+x_2=1$, while $\phi_{\eset \prec \beta}(x_0,0,x_2) = \phi_\beta(x_0,x_2)$ and $\phi_{\eset \prec \beta}(x_0,x_1,0) = \phi_\eset(x_0,x_1)$. (It's perhaps not obvious that this is even possible --- see Lemma \ref{lem:extend-small-homeomorphisms} below.) We now define $s$ by
+$$s(b) = \restrict{\phi_\beta}{x_0=0}(b) - \restrict{\phi_{\eset \prec \beta}}{x_0=0}(\bdy b).$$
Here, $\restrict{\phi_\beta}{x_0=0} = \phi_\beta(0,1)$ is just a homeomorphism, which we apply to $b$, while $\restrict{\phi_{\eset \prec \beta}}{x_0=0}$ is a one parameter family of homeomorphisms which acts on the $0$-blob $\bdy b$ to give a $1$-blob. To be precise, this action is via the chain map identified in Lemma \ref{lem:CH-small-blobs} with $\cV_0$ the open cover by $\epsilon/2$-balls and $\cV_1$ the open cover by $\frac{3\epsilon}{4}$-balls. From this, it is immediate that $s(b) \in \bc^{\cU}_1(M)$, as desired.
We now check that $s$, as defined so far, is a chain map, calculating
\begin{align*}
-\bdy (s(b)) & = \restrict{\phi_\beta}{x_0=0}(\bdy b) + (\bdy \restrict{\phi_{\eset \prec \beta}}{x_0=0})(\bdy b) \\
- & = \restrict{\phi_\beta}{x_0=0}(\bdy b) + \restrict{\phi_\eset}{x_0=0}(\bdy b) - \restrict{\phi_\beta}{x_0=0}(\bdy b) \\
+\bdy (s(b)) & = \restrict{\phi_\beta}{x_0=0}(\bdy b) - (\bdy \restrict{\phi_{\eset \prec \beta}}{x_0=0})(\bdy b) \\
+ & = \restrict{\phi_\beta}{x_0=0}(\bdy b) - \restrict{\phi_\beta}{x_0=0}(\bdy b) + \restrict{\phi_\eset}{x_0=0}(\bdy b) \\
& = \restrict{\phi_\eset}{x_0=0}(\bdy b) \\
& = s(\bdy b)
\end{align*}
Next, we compute the compositions $s \circ i$ and $i \circ s$. If we start with a small $1$-blob diagram $b$, first include it up to the full blob complex then apply $s$, we get exactly back to $b$, at least assuming we adopt the convention that for any ball $\beta$ which is already small, we choose the families of homeomorphisms $\phi_\beta$ and $\phi_{\eset \prec \beta}$ to always be the identity. In the other direction, $i \circ s$, we will need to construct a homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ for $*=0$ or $1$.
-The homotopy $h$ is defined by $$h(b) = \phi_\eset(b)$$ when $b$ is a $0$-blob (here $\phi_\eset$ is a one parameter family of homeomorphisms, so this is a $1$-blob), and $$h(b) = \phi_\beta(b) + \phi_{\eset \prec \beta}(\bdy b)$$ when $b$ is a $1$-blob (here $\beta$ is the ball in $b$, and the first term is the action of a one parameter family of homeomorphisms on a $1$-blob, and the second term is the action of a two parameter family of homeomorphisms on a $0$-blob, so both are $2$-blobs). We then calculate
+The homotopy $h$ is defined by $$h(b) = \phi_\eset(b)$$ when $b$ is a $0$-blob (here $\phi_\eset$ is a one parameter family of homeomorphisms, so this is a $1$-blob), and $$h(b) = \phi_\beta(b) - \phi_{\eset \prec \beta}(\bdy b)$$ when $b$ is a $1$-blob (here $\beta$ is the ball in $b$, and the first term is the action of a one parameter family of homeomorphisms on a $1$-blob, and the second term is the action of a two parameter family of homeomorphisms on a $0$-blob, so both are $2$-blobs). We then calculate
\begin{align*}
-(\bdy h+h \bdy)(b) & = \bdy (\phi_{\beta}(b) + \phi_{\eset \prec \beta}(\bdy b)) + \phi_\eset(\bdy b) \\
- & = \restrict{\phi_\beta}{x_0=0}(b) - \restrict{\phi_\beta}{x_1=0}(b) - \phi_\beta(\bdy b) + (\bdy \phi_{\eset \prec \beta})(\bdy b) + \phi_\eset(\bdy b) \\
- & = \restrict{\phi_\beta}{x_0=0}(b) - b - \phi_\beta(\bdy b) + \restrict{\phi_{\eset \prec \beta}}{x_0=0}(\bdy b) - \phi_\eset(\bdy b) + \phi_\beta(\bdy b) + \phi_\eset(\bdy b) \\
- & = \restrict{\phi_\beta}{x_0=0}(b) - b + \restrict{\phi_{\eset \prec \beta}}{x_0=0}(\bdy b) \\
+(\bdy h+h \bdy)(b) & = \bdy (\phi_{\beta}(b) - \phi_{\eset \prec \beta}(\bdy b)) + \phi_\eset(\bdy b) \\
+ & = \restrict{\phi_\beta}{x_0=0}(b) - \restrict{\phi_\beta}{x_1=0}(b) - \phi_\beta(\bdy b) - (\bdy \phi_{\eset \prec \beta})(\bdy b) + \phi_\eset(\bdy b) \\
+ & = \restrict{\phi_\beta}{x_0=0}(b) - b - \phi_\beta(\bdy b) - \restrict{\phi_{\eset \prec \beta}}{x_0=0}(\bdy b) + \phi_\beta(\bdy b) - \phi_\eset(\bdy b) + \phi_\eset(\bdy b) \\
+ & = \restrict{\phi_\beta}{x_0=0}(b) - b - \restrict{\phi_{\eset \prec \beta}}{x_0=0}(\bdy b) \\
& = (i \circ s - \id)(b).
\end{align*}
@@ -88,21 +88,21 @@
\newcommand{\length}[1]{\operatorname{length}(#1)}
We've finally reached the point where we can define a map $s: \bc_*(M) \to \bc^{\cU}_*(M)$, and then a homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ so that $dh+hd=i\circ s$. We have
-$$s(b) = \sum_{i} ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i)$$
-where the sum is over sequences $i=(i_1,\ldots,i_m)$ in $\{1,\ldots,k\}$, with $0\leq m \leq k$, $i(b)$ denotes the increasing sequence of blob configurations
+$$s(b) = \sum_{i} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i)$$
+where the sum is over sequences $i=(i_1,\ldots,i_m)$ in $\{1,\ldots,k\}$, with $0\leq m \leq k$, $\sigma(i)$ is something to do with $i$, $i(b)$ denotes the increasing sequence of blob configurations
$$\beta_{(i_1,\ldots,i_m)} \prec \beta_{(i_2,\ldots,i_m)} \prec \cdots \prec \beta_{()},$$
and, as usual, $b_i$ denotes $b$ with blobs $i_1, \ldots, i_m$ erased. We'll also write
-$$s(b) = \sum_{m=0}^{k} \sum_{\length{i}=m} ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i),$$
+$$s(b) = \sum_{m=0}^{k} \sum_{\length{i}=m} (-1)^{\sigma(i)} \ev(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor b_i),$$
where we arrange the sum according to the length of $i$.
The homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ is similarly given by
-$$h(b) = \sum_{i} ev(\phi_{i(b)}, b_i).$$
+$$h(b) = \sum_{i} (-1)^{\sigma(i)} \ev(\phi_{i(b)}, b_i).$$
Before completing the proof, we unpack this definition for $b \in \bc_2(M)$, a $2$-blob. We'll write $\beta$ for the underlying balls (either nested or disjoint).
Now $s$ is the sum of $5$ terms, split into three groups depending on with the length of the sequence $i$ is $0, 1$ or $2$. Thus
\begin{align*}
-s(b) & = \restrict{\phi_{\beta}}{x_0 = 0}(b) + \\
- & \quad + \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_1) + \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_2) + \\
- & \quad + \restrict{\phi_{\eset \prec \beta_1 \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_2 \prec \beta}}{x_0 = 0}(b_{12}).
+s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(b) + \\
+ & \quad + (-1)^{\sigma(1)} \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_1) + (-1)^{\sigma(2)} \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_2) + \\
+ & \quad + (-1)^{\sigma(12)} \restrict{\phi_{\eset \prec \beta_2 \prec \beta}}{x_0 = 0}(b_{12}) + (-1)^{\sigma(21)} \restrict{\phi_{\eset \prec \beta_1 \prec \beta}}{x_0 = 0}(b_{12}).
\end{align*}
As in the $k=1$ case, the first term, corresponding to $i(b) = \eset$, makes the all balls in $\beta$ $\cV_1$-small. However, if this were the only term $s$ would not be a chain map, because we have no control over $\restrict{\phi_{\beta}}{x_0 = 0}(\bdy b)$. This necessitates the other terms, which fix the boundary at successively higher codimensions.
@@ -116,26 +116,28 @@
\nn{lots of signs are wrong ...}
\begin{align*}
-\bdy s(b) & = \restrict{\phi_{\beta}}{x_0 = 0}(\bdy b) + \\
- & \quad + \restrict{\phi_{\beta}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1}}{x_0 = 0}(b_1) + \restrict{\phi_{\beta}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2}}{x_0 = 0}(b_2) + \\
- & \quad - \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) + \\
- & \quad + \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_1}}{x_0 = 0}(b_{12}) + \\
- & \quad + \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_2}}{x_0 = 0}(b_{12}), \\
+\bdy s(b) & = (-1)^{\sigma()} \restrict{\phi_{\beta}}{x_0 = 0}(\bdy b) + \\
+ & \quad + (-1)^{\sigma(1)} \left( \restrict{\phi_{\beta}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1}}{x_0 = 0}(b_1) - \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) \right) + \\
+ & \quad + (-1)^{\sigma(2)} \left( \restrict{\phi_{\beta}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2}}{x_0 = 0}(b_2) - \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) \right) + \\
+ & \quad + (-1)^{\sigma(12)} \left( \restrict{\phi_{\beta_2 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_2}}{x_0 = 0}(b_{12}) \right) + \\
+ & \quad + (-1)^{\sigma(21)} \left( \restrict{\phi_{\beta_1 \prec \beta}}{x_0 = 0}(b_{12}) - \restrict{\phi_{\eset \prec \beta}}{x_0 = 0}(b_{12}) + \restrict{\phi_{\eset \prec \beta_1}}{x_0 = 0}(b_{12}) \right), \\
\intertext{while}
s(\bdy(b)) & = s(b_1) - s(b_2) \\
- & = \restrict{\phi_{\beta_1}}{x_0=0}(b_1) + \restrict{\phi_{\eset \prec \beta_1}}{x_0=0}(b_{12}) - \restrict{\phi_{\beta_2}}{x_0=0}(b_2) - \restrict{\phi_{\eset \prec \beta_2}}{x_0=0}(b_{12}) .
+ & = \restrict{\phi_{\beta_1}}{x_0=0}(b_1) - \restrict{\phi_{\eset \prec \beta_1}}{x_0=0}(b_{12}) - \restrict{\phi_{\beta_2}}{x_0=0}(b_2) + \restrict{\phi_{\eset \prec \beta_2}}{x_0=0}(b_{12}) .
\end{align*}
\nn{that does indeed work, modulo signs}
We need to check that $s$ is a chain map, and that \todo{} the image of $s$ in fact lies in $\bc^{\cU}_*(M)$. Calculate
\begin{align*}
-\bdy(s(b)) & = \sum_{m=0}^{k-1} \sum_{\length{i}=m} \ev\left(\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^m \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) \\
- & = \sum_{m=0}^{k-1} \sum_{\length{i}=m} \ev\left(\sum_{i' \prec_1 i} \pm \restrict{\phi_{i'(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^m \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{i \prec_1 i'} \pm b_{i'}\right) \\
-\intertext{and telescoping the sum}
- & = \sum_{m=0}^{k-2} \left(\sum_{\length{i}=m} (-1)^m \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \sum_{i \prec_1 i'} \pm b_{i'}\right) \right) + \left(\sum_{\length{i}=m+1} \ev\left(\sum_{i' \prec_1 i} \pm \restrict{\phi_{i'(b)}}{x_0 = 0} \tensor b_i\right) \right) + \\
- & \qquad + (-1)^{k-1} \sum_{\length{i}=k-1} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \sum_{i \prec_1 i'} \pm b_{i'}\right) \\
- & = (-1)^{k-1} \sum_{\length{i}=k-1} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \sum_{i \prec_1 i'} \pm b_{i'}\right)
+\bdy(s(b)) & = \sum_{m=0}^{k} \sum_{\length{i}=m} (-1)^{\sigma(i)} \ev\left(\bdy(\restrict{\phi_{i(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0} \tensor \bdy b_i\right) \\
+ & = \sum_{m=0}^{k} \sum_{\length{i}=m}(-1)^{\sigma(i)} \ev\left(\sum_{p=1}^m \pm \restrict{\phi_{(i\setminus i_p)(b)}}{x_0 = 0})\tensor b_i\right) + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k \pm b_{i \cup \{q\}}\right) \\
+\intertext{which we telescope as}
+ & = \ev \left( \restrict{\phi_\beta}{x_0=0} \tensor \sum_{q=1}^k \pm b_{\{q\}}\right) + \\
+ & \qquad + \sum_{m=1}^{k-1} \sum_{\length{i}=m} \Bigg( \sum_{q=1}^{m+1} \sum_{\substack{i^+ \\ i = i^+ \setminus i^+_q}} (-1)^{\sigma(i^+)} \ev\left(\sum_{p=1}^m \pm \restrict{\phi_{i(b)}}{x_0 = 0})\tensor b_{i^+}\right) + \\
+ & \qquad \qquad + (-1)^{\sigma(i) + m} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k \pm b_{i \cup \{q\}}\right)\Bigg) \\
+ & \qquad + (-1)^k \sum_{\length{i}=k}(-1)^{\sigma(i)} \ev\left(\restrict{\phi_{i(b)}}{x_0 = 0}\tensor \sum_{\substack{q=1 \\ q \not\in i}}^k \pm b_{i \cup \{q\}}\right)\\
\end{align*}
+\todo{to be continued...}
Finally, we need to check that $dh+hd=i\circ s$. \todo{}
\end{proof}