# Thursday January 30th

## Morse Inequality Example

Example:
Consider $f: \CP^n \to \RR$ where (recall) $\CP^n = S^{2n+1} \subset \CC^{n+1} / \sim$ where $\vector z \sim \lambda \vector z$ for all $\abs{\lambda} = 1$ in $\CC\units$, where $f$ is given by $[z_0: \cdots : z_n] \mapsto \sum i \abs{z_i}^2$.

> Note that we can take the coefficients to be any $n+1$ distinct real numbers, here we just take $1, 2, \cdots, n+1$ for simplicity.

Cover $\CP^n$ with $n+1$ coordinate charts $(U_j, h_j)$ where $U_j = \theset{\vector z \suchthat z_j \neq 0}$ and $h_j: U_j \to \RR^{2n}$ is given by first defining $[z_0 : \cdots : z_n] \mapsto (\hat z_j z_0, \hat z_j z_1, \cdots, \hat z_j z_n)$ where $\hat z_j = z_j/\abs{z_j}$.

Denote the image coordinates by $z_k = x_k + i y_k$.
Then define $h_j$ by $[z_0 : \cdots : z_n] \mapsvia{h_j} (x_0, y_0, \cdots, x_{j-1}, y_{j-1}, x_{j+1}, y_{j+1}, \cdots, x_n, y_n)$.

Note that $\abs{z_j} = 1 - \sum_{i\neq j} \abs{z_i}^2$, so this is a one-to-one correspondence (i.e. we can recover the magnitude of $z_j$ from the image point).

So what is $f$ in these coordinates?
We can write

\begin{align*}
f \circ h_j\inv (x_0, y_0, \cdots, \hat x_j, \hat y_j, \cdots x_n, y_n)
&= \sum{i\neq j} i (x_i^2 + y_i^2) + j \abs{z_j}^2 \\
&= \sum_{i\neq j} i(x_i^2 + y_i^2) + j - \sum_{i\neq j} j(x_i^2 + y_i^2) \\
&= j + \sum_{i\neq j} (i-j) (x_i^2 + y_i^2) \\
&= j + (-j)(x_0^2 + y_0^2) + (-j + 1)(x_1^2 + y_1^2) + \cdots + (n-j)(x_n^2 + y_n^2)
.\end{align*})

So what are the critical points?
The derivative is zero iff $x_i = y_i = 0$ for some $i\neq j$.
So there is only one critical point, $p_j = [0: 0 : \cdots : \cdots 1_j :\cdots : 0]$.
Thus there are $n+1$ critical points given by $\mathrm{crit}(f) = \theset{p_0, \cdots, p_n}$.
Using the above equations, we can find that $\ind_f(p_j) = 2j$ (count positive and negative terms).

Note that we had the inequality $b_i(M) \leq \abs{\theset{\text{critical points with index } i}}$.
Noting that $H^i(\CP^n; \FF) = \FF$ for $i=0,2,4,\cdots,2n$ and 0 otherwise, so we have exact equality here.

> Note that there are $\QQ HS$ where the inequality has to be strict, but equality can be obtained with $\CP^n, S^n$, etc.

## Rearrangement

Fix a Morse function $f: W\to [0, 1]$, with $p, q\in \crit(f)$ and $f(p) < f(q)$.
Can we change $f$ to a new Morse function $g$ such that $\crit(g) = \crit(f)$ and $g(p) > g(q)$, where $g = f + \const$ in a neighborhood of $p$ and a neighborhood of $q$?

Note that we obtain elementary cobordisms in each case:

![Image](figures/2020-01-30-11:38.png)

Pick $\xi$ a gradient-like vector field for $f$, which decomposes $W^*(p) = W^s(p) \union W^u(p)$.

![Image](figures/2020-01-30-11:40.png)

Lemma
:   Let $f: W \to I$ be a Morse function with 2 critical points $p, q$ and $\xi$ a gradient-like vector field for $f$ such that $W^*(p) \intersect W^*(q) = \emptyset$.
    Then for any two points $a, b \in I$, there exists a Morse function $g$ such that:

    1. $\xi$ is gradient-like for $g$,
    2. $\crit(g) = \crit(f)$, with $g(p) = a$ and $g(q) = b$.
    3. $g = f$ near $M_0$ and $M_1$, and $g= f+\const$ in some neighborhood of $p$, and some neighborhood of $q$.


So this is stronger: we can modify our Morse function to take on any two real numbers.

*Idea of proof:*
We want the two purple sections here, since we want to modify $p$ and $q$ separately:

![Image](figures/2020-01-30-11:45.png)\

### Proof of Lemma

We can find a $\mu: M_0 \to I$ such that $\mu \equiv 0$ near $S_L^P$ and $\mu \equiv 1$ near $S_L^q$.
So extend $\mu$ to $\bar \mu: W \to I$ such that $\bar \mu$ is constant over the integral curves of $\xi$ and $\bar\mu \equiv 0$ near $W^*(p)$ and $\mu \equiv 1$ near $W^*(q)$.

> Here the integral curves are green:

![Image](figures/2020-01-30-11:49.png)\

Let $g(z) = G(f(z), \bar \mu(z))$ where $G: I \cross I \to I$ will be defined as follows:

1. Fix a $y\in W$ to $\bar \mu$ is constant, then $G(\wait, y): I \to I$ is increasing (since $f$ is increasing) and surjective, i.e. $\dd{G}{x} > 0$ everywhere.
2. $G(x, y) = x$ whenever $x$ is near 0 or 1.
3. $\dd{G}{x}(x, 0) = 1$ for $x$ near $f(p)$ and $\dd{g}{x}(x, 1) = 1$ for $x$ near $f(q)$.

> Note that $a = g(p) G(f(p), 0)$ and $b = g(q) = G(f(q), 1)$, and the slope should be constant near $a, b$.

We get something like the following graphs:

![Image](figures/2020-01-30-12:03.png)\

$\qed$

When can such a function exist?
I.e. is this a relatively strong condition?
If $f(p) < f(q)$, it is possible that $W^u(p) \intersect W^s(q) \neq \emptyset$:

![Image](figures/2020-01-30-12:07.png)\


Theorem (Modifying a Vector Field to Separate W's)
: If $f(p) < f(q)$ and $\ind(p) \geq \ind(q)$, then it is possible to change $\xi$ in a neighborhood of $f\inv(x)$ for some $f(p) < c < f(q)$ such that $W_p^u \intersect W_q^s = \emptyset$.

Main Idea:

![Image](figures/2020-01-30-12:10.png)\

Note that $W_p^u \intersect W_q^s = \emptyset$ iff $S_R^c(p) \intersect S_L^c(q) = \emptyset$, where $S_R^C(p) = W_p^u \intersect f\inv(c)$ and $S_L^c(q) = W_q^s \intersect f\inv(c)$.

We have the implication
\begin{align*}
\begin{cases}
\dim W = n \\
\ind(p) = \lambda \\
\ind(q) = \lambda'
\end{cases}
\implies
\begin{cases}
\dim f\inv(c) = n-1\\
\dim S_R^c(p) = n - \lambda - 1 \\
\dim S_L^c(q) = \lambda' - 1
\end{cases}
.\end{align*}

and thus
$$
\dim S_R^c(p) + \dim S_L^c(q) = n - \lambda - \lambda'  2 < n-1 = \dim f\inv(c)
.$$

Lemma (1)
: If $M^m, N^n \subset V^v$ are smooth submanifolds and $m+n < v$ then there exists a diffeomorphism $h: V\to V$ which is isotopic to the identity such that $h(M) \intersect N = \emptyset$.

Idea:
We'll use this new diffeomorphism to modify the vector field $\xi$ to make things disjoint.