# Tuesday, March 3rd

> Lecture by Weiwei..

:::{.remark}
Recall 2nd Cancellation Theorem:
With some conditions on $(W, V_0, V_1)$ with only two critical points and $S_R \cdot S_L' = \pm 1$, then $W = V_0 \cross [0, 1]$.
Our goal: show that if $W$ is simply connected and $H_*(W, V) = 0$, then $W = V \cross [0, 1]$.
:::

:::{.remark}
Cancellation of C.P., with the assumption of vanishing homology:
Recall that if $\bd W = X$, then the induced orientation 

1. An $(n-1)\dash$frame $\{ \xi_1, \cdots, \xi_{n+1} \}$ is positive iff $\{ \nu, x_i\}$ is positive in $W$, where $\nu$ is the outward normal.

2. $H_n(W, X) \to H_{n-1}(X)$ sends $[W] \to [X]$ (fundamental classes).
:::

:::{.theorem title="Main theorem"}
: Let $(W, V_0, V_1)$ be a triad of dimension $n$  with a Morse function $f: M\to \RR$ such that 

- All critical points have index $\lambda$ and are on the same level.[^same_level_reminder]

- $\xi$ is a gradient-like vector field of $f$
- $2\leq \lambda \leq n-2$ and $W$ is connected

Then given an integral basis of $H_{\lambda}(W, W)$ there exists a Morse function $f'$ with a gradient-like vector field $\xi'$ such that $f = f'$ in a neighborhood of $V \union V'$, the critical points of $f$ and $f'$ are still on the same level, and the $D_L$S of $\xi'$ determine the given basis.

[^same_level_reminder]: 
Same level: $f(x_i) = p$, a single $p\in \RR$.

:::

:::{.remark}
The idea here is cut a cobordism into pieces, then pick handles that are nullhomologous in the larger manifold, but form a basis of relative homology, and we choose bases in higher indices such that their boundaries are the basis of lower indices (yielding algebraic cancellation).

See diagram
:::

:::{.remark title="Algebraic lemma"}
See diagram.
Write 
\[
W &= c_0 c_1 \cdots c_n \\
W_\lambda &= c_0 c_1 \cdots c_\lambda \\
W_{-1} &= V
.\]

Define $C_\lambda = H_\lambda(W_\lambda, W_{\lambda - 1}) = H_*(W_\lambda, W_{\lambda - 1})$, then $\del_\lambda: C_\lambda \to C_{\lambda - 1}$.
We have
\[
H_*(W_{\lambda - 1}, W_{\lambda - 2}) \to H_*(W_{\lambda}, W_{\lambda - 2}) \to H_*(W_\lambda, W_{\lambda - 1}) \mapsvia{\del_\lambda} H_{* - 1}(W_{\lambda - 1}, W_{\lambda - 2})
.\]

:::

:::{.theorem title="?"}
$\del^2 = 0$, so this yields a chain complex and $H_\lambda(C_*) = H_{\lambda}(W, V)$.
:::

:::{.remark}
See diagram, need to set up different homology because total homology doesn't "see" handle attachment, and this allows us to do intersection theory.

\begin{tikzcd}
                                        &  & o                                         &  &                                               &  &   \\
{H_{\lambda + 1}(\lambda + 1, \lambda)} &  & {H_\lambda(W_{\lambda}, W_{\lambda - 2})} &  & {H_\lambda(W_{\lambda + 1}, W_{\lambda - 2})} &  & 0 \\
                                        &  &                                           &  &                                               &  &   \\
                                        &  & {H_\lambda(W_\lambda, W_{\lambda - 1})}   &  &                                               &  &   \\
                                        &  &                                           &  &                                               &  &   \\
                                        &  & {H_\lambda (\lambda - 1, \lambda - 2)}    &  &                                               &  &  
\end{tikzcd}


We want $H_\lambda(C_*) = H_\lambda(\lambda + 1, \lambda - 2) = H_\lambda(W, V)$.
:::

> End of class..