# Thursday February 6th

Cancellation:
Let $f: W \to I$ be Morse, $\crit(f) = \theset{p, q}$ with $f(p) < f(q)$ and $\ind(p) = \lambda,~\ind(q) = \lambda+1$.
Let $\xi$ be its gradient-like vector field, then $S_R^c(p) \intersect S_L^c(q) = \pt$, so there exists a unique integral curve $T$ from $p$ to $q$.

In this situation $W$ is diffeomorphic to the product cobordism.


We will show

Theorem (1, Modifying Vector Fields)
: We can change $\xi$ in a compact neighborhood of $T$ to get a nonvanishing vector field $\xi'$ for which the integral curves originate at $M_0$ and end at $M_1$.

Example
:   \hfill
    ![Image](figures/2020-02-06-11:11.png)\

Moreover, it takes a particularly nice standard form, described in the following way:

Proposition
:   There exists a neighborhood $U_T$ and a coordinate chart $h: U_T \to \RR^n$ such that

    1. $h(p) = (0, \cdots, 0)$ and $h(q) = (1, 0, \cdots, 0)$.

    2. $h_* \xi = (V(x_1), -x_2, -x_3, \cdots, -x_{\lambda+1}, x_{\lambda+2}, \cdots, x_n)$.

    3. $V(x)$ is smooth and positive over $(0, 1)$with $V(0) = V(1) = 0$, and $V(x) < 0$ everywhere else.

    4. (Minor) $\abs{V'(0)} = \abs{V'(1)} = 1$.

    Thus  we have
    \begin{align*}
    (x_1, -x_2, \cdots, -x_{\lambda+1}, x_{\lambda+2}, \cdots, x_n) &\quad \text{near } p \\
    (-x_1, -x_2, \cdots, -x_{\lambda+1}, x_{\lambda+2}, x_n) &\quad \text{near } q
    .\end{align*}

![Image](figures/2020-02-06-11:20.png)\

### Proof of Theorem

*Step 1:*

Consider $(U(x_1, \rho), -x_2, -x_3, \cdots, -x_{\lambda+1}, x_{\lambda+2}, \cdots, x_n)$ where $\rho(\vector x) = (x_2^2 + x_3^2 + \cdots x_n^2)^{1/2}$, which measures the distance between the two curves above.
Some facts:

1. $U(x_1, \phi)$ is equal to $V(x_1)$ outside of a compact neighborhood of $h(T)$ in $h(U_T)$.

2. $U(x_1, 0) < 0$ for all $x_1$.

Then $\xi' = h_*(u, -x_2, \cdots, x_n)$ in $U_T$ and $\xi' = \xi$ everywhere else.
Thus $\xi'$ is nowhere vanishing.

*Step 2:*

We want to pick $U'$ such that $T\subset U' \subset \bar U \subset U_T$ where $\bar U$ is a compact set such that any trajectory of $\xi$ that exits $U$ never re-enters $U'$.

![Image](figures/2020-02-06-11:35.png)\

Suppose such a $U'$ does not exist.
Then there exist sequences of points $\theset{s_k}, \theset{r_k} \subset U$ and $\theset{t_k} \subset W\setminus U$ all on the same integral curves $\gamma_k$ such that $\theset{s_k} \to T$ and $\theset{r_k} \to T$.
Since $W\setminus U$ is compact, $\theset{t_k}$ has a limit point $A$.
Then consider $\psi_A(t)$, which are integral curves that originate from $M_0$ and end on $M_1$.

![Image](figures/2020-02-06-11:39.png)\

Then there exists a neighborhood $A$ such that for each $a\in A$, the integral curves (half trajectories) containing $a$ originate on $M_0$.
Moreover, for $k$ large enough, all $t_k$ are in $A$.
The union of all of these half trajectories has a positive distance from $T$, so there is a small enough $U$ disjoint from these trajectories, so $\theset{s_k} \not\to T$, a contradiction.

$\qed$

We now consider the flow lines of $\xi'$ in $h(U_T)$.
We have
$$
\dd{x_1}{t} = u(x_1, \phi), \dd{x_2}{t} = -x_2, \cdots, \dd{x_n}{t} = x_n
.$$
Thus $x_2 = x_2^0 e^{-t}, \cdots, x_n = x_n^0 e^t$.

So if $x_i\neq 0$ for some $\lambda+2 \leq i \leq n$, the $\abs{x_i}$ is increasing exponential and thus it will escape $h(U)$.
The corresponding trajectory will then escape $U$, and so it will follow the integral curves of the original $\xi$ and ends at $M_1$.
If $x_{\lambda+2}^0 = \cdots = x_n^0 = 0$, then
$$
\phi(\vector x) = (x_2^2 + \cdots + x_{\lambda+1}^2)^{1/2} = e^{-t}\qty{ \sum (x_i^0)^2 }^{1/2}
.$$
Thus $\phi(x)$ will decrease exponentially.

If it leaves $U$, we are in the previous case.
Otherwise, if it doesn't leave $U$, then there exists an $\eps>0$ such that $u(x_1, \phi) < 0$ for all
$$
N_\eps = \theset{(x,p) \in h(U) \suchthat \phi< \eps}
.$$
Thus there exists a $-\alpha<0$ such that $u<-\alpha$ on $N_\eps$.

For $t$ large enough,
$$
\phi(\vector x(t)) \in N_\eps\implies \dd{x_1}{t} = u(x_1, \phi) < -\alpha
.$$
Thus $x_1(t)< -\alpha t + \const$ for large enough $t$, and as $t$ increases $\vector x(t)$ will go out of $U$.
By the previous argument, it must end at $M_1$.

Thus every integral curve of $\xi$ starts at $M_0$ and ends at $M_1$.

$\qed$

Lemma
:   $\xi'$ gives a diffeomorphism from
    $$
    W' = (M_0 \cross I; M_0 \cross 0, M_1 \cross 1) \to W = (W; M_0, M_1)
    .$$

Proof
:   Take $\pi: W \to M_0$ and follow the integral curves backward.
    Then for all $x\in M_0$, there is a $\tau(x) \in \RR^{\geq 0}$ such that $\psi_X(\tau(x)) \in M_1$.

    So we get a $$\hat \xi = \tau(\pi(q))\inv \xi_q'$$ and we can define $\phi(x, t) = \hat \phi_X(t)$.

Theorem (2, Modifying a Vector Field Away From Critical Points)
: $\xi'$ is a gradient-like vector field for some Morse function $g: W \to I$ such that $g$ has no critical points (since $\xi'$ has no zeros) and $g = f$ near $M_0$ and $M_1$.

### Proof of Theorem

We want to build a $k: M_0 \cross I \to I$ such that the following diagram commutes:

\begin{center}
\begin{tikzcd}
M_0 \cross I \arrow[rr, "k", dashed] \arrow[dd, "\phi"] &  & I \\
                                                        &  &   \\
W \arrow[rruu, "g"]                                     &  &
\end{tikzcd}
\end{center}

This needs to satisfy

1. $k$ is equal to $f_1 \definedas f\circ \phi$ near $M_0 \cross 0$ and $M_0 \cross 1$.

2. $\dd{k}{t} < 0$.

Since $\dd{f_1}{t} > 0$ near $M_0 \cross 0$ and $M_0 \cross 1$, take $\delta > 0$ such that $\dd{f_1}{t} > 0$ on $M_0 \cross [0, \delta)$ and $M_1 \cross (1-\delta, 1]$.

![Image](figures/2020-02-06-12:13.png)\

So pick $\lambda: I \to I$ such that $\lambda \equiv 1$ near $t=0, 1$ and $\lambda \equiv 0$ on $[\delta, 1-\delta]$.

Then pick any positive $\bar K : M_0 \to \RR$, and then take

\begin{align*}
K(x, u) \definedas \int_0^u \lambda(t) \dd{f_1}{t} + (1 - \lambda(t)) ~\bar K(x) ~dt
.\end{align*}

Then
$$
\dd{K}{u} = \lambda(u) \dd{f_1}{u} + (1-\lambda(u)) ~ \bar K(x) > 0
$$
since the first term is positive near $M_0 \cross 1 \text{ or } 0$, and $\bar K$ is positive everywhere.

To see that it satisfies the first property, note that $\int_0^s \dd{f_1}{t} ~dt = f_1$ for $s$ near 0.

To see that property 2, note

\begin{align*}
\int_0^1 \lambda(t) \dd{f_1}{t} ~dt  + \bar K \int_0^1 (1 - \lambda(t)) ~dt = g(x, 1) = f_1(x)\\
\implies \bar K(x) = \frac{ f_1(x) - \displaystyle\int_0^1 \lambda(t) \dd{f_1}{t} ~dt }{\displaystyle\int_0^1 (1 - \lambda(t)) ~dt }
.\end{align*}

$\qed$