--- a/text/evmap.tex Sun Sep 19 23:01:49 2010 -0500
+++ b/text/evmap.tex Sun Sep 19 23:11:59 2010 -0500
@@ -279,37 +279,32 @@
\[
h(y) = e(y - r(y)) + c(r(y)) .
\]
-\nn{up to sign, at least}
We must now verify that $h$ does the job it was intended to do.
For $x\in \btc_{ij}$ with $i\ge 2$ we have
-\nn{ignoring signs}
\begin{align*}
- \bd h(x) + h(\bd x) &= \bd(e(x)) + e(\bd x) \\
- &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x) + e(\bd_t x) \\
- &= \bd_b(e(x)) + e(\bd_b x) \quad\quad\text{(since $\bd_t(e(x)) = e(\bd_t x)$)} \\
- &= x .
+ \bd h(x) + h(\bd x) &= \bd(e(x)) + e(\bd x) && \\
+ &= \bd_b(e(x)) + (-1)^{i+1} \bd_t(e(x)) + e(\bd_b x) + (-1)^i e(\bd_t x) && \\
+ &= \bd_b(e(x)) + e(\bd_b x) && \text{(since $\bd_t(e(x)) = e(\bd_t x)$)} \\
+ &= x . &&
\end{align*}
For $x\in \btc_{1j}$ we have
-\nn{ignoring signs}
\begin{align*}
- \bd h(x) + h(\bd x) &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x - r(\bd_b x)) + c(r(\bd_b x)) + e(\bd_t x) \\
- &= \bd_b(e(x)) + e(\bd_b x) \quad\quad\text{(since $r(\bd_b x) = 0$)} \\
- &= x .
+ \bd h(x) + h(\bd x) &= \bd_b(e(x)) + \bd_t(e(x)) + e(\bd_b x - r(\bd_b x)) + c(r(\bd_b x)) - e(\bd_t x) && \\
+ &= \bd_b(e(x)) + e(\bd_b x) && \text{(since $r(\bd_b x) = 0$)} \\
+ &= x . &&
\end{align*}
For $x\in \btc_{0j}$ with $j\ge 1$ we have
-\nn{ignoring signs}
\begin{align*}
- \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(e(x - r(x))) + \bd_t(c(r(x))) +
+ \bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) - \bd_t(e(x - r(x))) - \bd_t(c(r(x))) +
e(\bd_t x - r(\bd_t x)) + c(r(\bd_t x)) \\
- &= x - r(x) + \bd_t(c(r(x))) + c(r(\bd_t x)) \\
+ &= x - r(x) - \bd_t(c(r(x))) + c(r(\bd_t x)) \\
&= x - r(x) + r(x) \\
&= x.
\end{align*}
Here we have used the fact that $\bd_b(c(r(x))) = 0$ since $c(r(x))$ is a $0$-blob diagram, as well as that $\bd_t(e(r(x))) = e(r(\bd_t x))$ \nn{explain why this is true?} and $c(r(\bd_t x)) - \bd_t(c(r(x))) = r(x)$ \nn{explain?}.
For $x\in \btc_{00}$ we have
-\nn{ignoring signs}
\begin{align*}
\bd h(x) + h(\bd x) &= \bd_b(e(x - r(x))) + \bd_t(c(r(x))) \\
&= x - r(x) + r(x) - r(x)\\