# Tuesday February 11th

## Cancellation

The setup:
$f: W \to [0, 1]$ a morse function with $\crit(f) = \theset{p, q}$ with $\ind(p) = \lambda$ and $\ind(q) = \lambda + 1$, with a gradient-like vector field $\xi$ such that there exists a *single* flow line $T$ from $p$ to $q$.

Lemma (Modifying Gradient-Like Vector Fields)
:   There exists a gradient-like vector field $\xi'$ for $f$ such that

    1. $T$ is still the single flow line from $p$ to $q$.
    2. $\xi'$ is *standard* in a neighborhood $U_T$ of $T$, i.e. there exists $h: U_T \to \RR^n$ such that
      \begin{align*}
      h(p) &= (0, 0, \cdots, 0) \\
      h(q) &= (1, 0,\cdots, 0)
      .\end{align*}
      $h(T)$ is contained in the $x\dash$axis, and
      $$
      h_* \xi' = (V(x_1), -x_2, \cdots, -x_{\lambda+1}, x_{\lambda+2}, \cdots, x_n)
      ,$$
      where $V$ satisfies the property that near 0 and 1, $\abs{V'} = 1$:

      ![Image](figures/2020-02-11-11:08.png)\

### Proof

Let $\eta = V(x_1)$ from above.
Define the following vector field:

\begin{align*}
F(\vector x) = f(p) + 2 \int_0^{x_1} v(t) ~dt -x_2^2 - x_3^2 - \cdots x_{\lambda+1}^2 \cdots + x_n^2
.\end{align*}

Then $\eta$ is gradient-like for $F$, and we can pick $v(t)$ such that

\begin{align*}
F(1, 0, \cdots, 0) &=
f(p) + 2\int_0^1 v(t) ~dt = f(q) \\
\implies \int_0^1 v(t) &= \frac{1}{2}\qty{ f(q) - f(p)  }
.\end{align*}

We know that $v(t) = t$ near $(0, 0, \cdots, 0)$, and since $\int_0^1 t~dt = 2\qty{ \frac{x_1^2}{2} }$, we have

\begin{align*}
F(\vector x)
&= f(p) + 2\int_0^{x_1} t~dt + \cdots + x_n^2 = f(p) + x_1^2 - x_2^2 - \cdots - x_{\lambda+1}^2 + \cdots + x_n^2 \\
\implies \eta(\vector x)
&= (x_1, -x_2, \cdots, -x_{\lambda+1}, x_{\lambda+2}, \cdots x_n)
.\end{align*}

Then there exists a neighborhood $\tilde U_1$ of $p$ and $h_1: \tilde U_1 \to \RR^n$ such that $\tilde h_1(p) = (0, 0, \cdots, 0)$ with $F \circ \tilde h_1 = f$ and $\tilde h_{1*} = \eta$.


![Image](figures/2020-02-11-11:21.png)\

Similarly, near $(1, 0, \cdots, 0)$ we have $v(t) = 1 -t$ and since $\int_0^1 v(t)~dt = f(q) - f(p)$, we have

\begin{align*}
F(\vector x) &= f(p) + 2\int_0^1 v(t) ~dt + 2\int_1^{x_1} (1-t) ~dt - \cdots + x_n^2 \\
&= f(q) - (x_1 - 1)^2 - x_2^2 - \cdots + x_n^2
,\end{align*}

and there exists a neighborhood $\tilde U_2$ of $q$ and $\tilde h_2: \tilde U_2 \to \RR^n$ such that $\tilde h_2(q) = (1, 0, \cdots, 0)$, $F\circ \tilde h_2 = f$, and $\tilde h_{2*} \xi = \eta$.

So pick $(\tilde U_1,\tilde h_1)$ and $(\tilde U_2, \tilde h_2)$ such that $\tilde U_1 \intersect \tilde U_2 = \emptyset$ and $\tilde h_1(\tilde U_1) \intersect \tilde h_2(\tilde U_2) = \emptyset$.

Pick $$a_1 < f(p) < b_1 < b_2 < f(q) < a_2$$ such that
$$f\inv [a_1, b_1] \intersect T \subset \tilde U_1\quad\text{and}\quad f\inv [b_2, a_2] \intersect T \subset \tilde U_2$$
and set $p_i = f\inv(b_i) \intersect T$.

Let $U_1$ and $U_2$ be closed neighborhood of the arc $p \to p_1$ in $\tilde U_1 \intersect f\inv[a_1, b_1]$ and $q \to p_2$ in $\tilde U_2 \intersect f\inv[b_2, a_2]$.

Let $h_i = \tilde h_i \mid_{U_i}$.
Then $\xi$ yields a diffeomorphism $\psi: f\inv(b_1) \to f\inv(b_2)$.

![Image](figures/2020-02-11-11:37.png)\

Fix a small neighborhood $\lambda$ of $h_1(p_1)$ in $h_1(f\inv(b_1) \intersect U_1)$, following the flow lines of $\eta$ yields a diffeomorphism $\phi: V_1 \to V_2$ where $V_2$ is a sufficiently small neighborhood of $h_2(p_2)$ in $h_2(f\inv(b_2) \intersect U_2)$, the following diagram commutes:

\begin{center}
\begin{tikzcd}
f\inv(b_1) \arrow[rr, "\psi"]                &  & f\inv(b_2)                \\
                                             &  &                           \\
V_1 \arrow[rr, "\phi"] \arrow[uu, "h_1\inv"] &  & V_2 \arrow[uu, "h_2\inv"]
\end{tikzcd}
\end{center}


**If** for $V_1$ small enough we have $h_2\inv \circ \phi \circ h_1$ restricted to $h_1\inv(V_1)$ is equal to $\psi$, then we can extend $(h_1, h_2)$ to a diffeomorphism $h$ from $U_1 \intersect U_0 \intersect U_2$, where $U_0$ is a small neighborhood of $p_1 p_2$ such that it preserves the trajectories and level sets.
We then obtain $h_* \eta = K \eta$, where $K$ is some positive function.

We can extend $K$ to a positive smooth function over $W$, which yields $\xi' \definedas \frac 1 K \xi$ and thus $h_* \xi' = \eta$.
So $\xi'$ is a gradient-like vector field for $f$.

In case the above inequality does *not* hold, we can use an isotopy to change $\psi \to \psi'$ and change $\xi \to \xi'$ in $f\inv[a, b]$ such that the integral curves of $\xi'$ induce $\psi'$.
So find an isotopy such that $\psi'$ is equal to $h_2\inv \phi h_1$ near $p_1$, and furthermore $\psi'(S_R^{b_1}(p))$ intersects $S_L^{b_2}(q)$ transversely in $p_2$, i.e. $p_2$ is the only intersection point.

We can do this last step locally.
Let $\phi' = h_2\inv \phi h_1$, then $(\phi')\inv \psi: \tilde V_1 \injects \tilde V_1$ for some small neighborhood $\tilde V_1 \subset V_1$ containing $p_1$.
Note that if
$S_R^{b_1}(p) \transverse S_L^{b_1}(q)$
at $p_1$, then
$(\phi')\inv\psi S_R^{b_1}(p) \transverse (\phi')\inv\psi S_L^{b_1}(q)$
at $p_1$, and so
$S_R^{b_1}(p) \transverse (\phi')\inv\psi S_L^{b_1}(q)$.

Theorem (?)
:   Identify $\RR^n = \RR^a \oplus \RR^b$ where $a+b = n$.
    Suppose that $h: \RR^n \injects \RR^n$ is an orientation-preserving embedding such that $h(0) = 0$ with $h(\RR^a) \transverse \RR^b$ with *intersection number* $+1$ at $\theset{\vector 0}$.

    Then there exists a smooth isotopy $h_t: \RR^n \to \RR^n$ such that

    1. $h_0 = h$,
    2. $h_t(0) = 0, h_t(x) = x$ for $x$ outside of a neighborhood $N$ of zero,
    3. $h_1 = \id$ in a smaller neighborhood of zero in $N$
    4. $h_1(\RR^a) \intersect \RR^b = \theset{\vector 0}$.

Rough idea: modify $h(\RR^1)$ in a neighborhood of $0$:

![Image](figures/2020-02-11-12:13.png)\

Next time: a second cancellation theorem.
Suppose $W, M_0, M_1$ are simply connected and $2\leq \lambda < \lambda+1 \leq n-3$.
If $$S_R^c(p) \cdot S_L^c(q) = \pm 1,$$ then $W \cong M_0 \cross [0, 1]$ are diffeomorphic.

We'll briefly review the intersection number later.
Also: homological intersection number.