# Tuesday January 14th

## Existence of Morse Functions

Notation
: Let $F(M; \RR)$ be the space of smooth functions from $M$ to $\RR$ with the $C^2$ topology.

Theorem (Morse Functions Are Dense)
: Morse functions form an open dense subset of $F(M; \RR)$ in the $C^2$ topology.

Recall that the $C^2$ topology is defined by noting that $F(M, \RR)$ is an abelian group under addition, so we'll define open sets near the zero function and define open sets around $f$ by translation.
(I.e., if $N$ is an open neighborhood of 0, then $N+f$ is an open neighborhood of $f$.)

So we'll define a base of open sets around 0.
Take a finite cover of $M$, say by coordinate systems $\theset{U_\alpha}$.
Then let $h_\alpha: U_\alpha \to \RR^n$.
Now (exercise) we can find a compact refinement $C_\alpha \subset U_\alpha$ with each $C_\alpha$ compact and $\union_\alpha C_\alpha = M$.
We can now define $f_\alpha \definedas f \circ h_\alpha\inv$ for any $f: M \to \RR$

\begin{center}
\begin{tikzcd}
U_\alpha \arrow[rr, "h_\alpha"] \arrow[dd, "\restrictionof{f}{U_\alpha}"] &  & \RR^n \arrow[lldd, "f_\alpha"] \\
&  &                                \\
C_\alpha                                                                  &  &
\end{tikzcd}
\end{center}

Now for each $\delta > 0$, define

\begin{align*}
N(\delta) = \theset{
f: M\to \RR \suchthat \begin{cases}
\abs{f_\alpha(p)} < \delta \\
\abs{\dd{f_\alpha}{x_i} } < \delta \\
\abs{ \frac{\partial^2 f_\alpha}{\partial x_i \partial x_j}  } < \delta
\end{cases} \quad \forall p\in h_\alpha(C_\alpha),~\forall \alpha
}
.\end{align*}


Corollary
: $f + N(\delta)$ (for all $\delta$) is a basis for open neighborhoods around $f$.

Lemma
: This topology does not depend on the choice of $\theset{U_\alpha, h_\alpha}$.

Proof
: See Milnor 2.

Lemma (1)
: Let $f: U \to \RR$ be a $C^2$ map for $U \subseteq \RR^n$.
  For "almost all" linear maps $L: \RR^n \to \RR$, $f + L$ has only nondegenerate critical points.

Almost all: Note that $\hom(\RR^n, \RR) \cong \RR^n$, so the complement of the set of such maps has measure zero in $\RR^n$.

Proof
:   Consider $X = U \cross \hom(\RR^n, \RR)$, which contains a subspace $M = \theset{(x, L) \suchthat \partial_x (f + L) = 0}$, i.e. $x$ is a critical point of $f$.
    If $\partial_x f + L = 0$, then $L = -\partial_x f$.
    We thus obtain an identification of $M$ with $U$ by sending $x\in U$ to $(x, -\partial_x f)$.

    There is also a projection onto the second component, where $(x, L) \mapsto L$.
    So let $\pi: X \to \hom(\RR^n, \RR)$ be this projection; then there is a map $\tilde \pi: U \to \hom(\RR^n, \RR)$ given by $x \mapsto \partial_x f$.
    Note that $f + L$ has a *degenerate* critical point iff there is an $x\in U$ such that $\partial_x (f + L) = 0$ (or equivalently $L = -\partial_x f$), and the second derivative of $f+L$ is zero.
    Since $L$ is linear, this says that the matrix $\qty{ \frac{\partial f^2}{\partial x_i \partial x_j} }(x)$ is singular.
    But this says $x$ is a critical point for $\tilde \pi$.

    This happens iff $\tilde \pi(x) = -\partial_x f = L$, so $L$ is a critical value for $\tilde \pi$.
    Thus $f+L$ has a degenerate critical point $\iff L$ is a critical value for $\tilde \pi$.

    Now Sard's theorem applies: if $g: M^n \to \RR^n$ is a map from any manifold to $\RR^n$ that is $C^1$, then the set of critical values of $g$ in $\RR^n$ has measure zero.

    Thus the set of critical values of $\tilde \pi$ has measure zero, and thus for almost all $L$, $f+L$ has no degenerate critical points.

Summary: Consider the map of first derivatives.
It has a critical point whenever the 2nd derivative is singular, which is exactly the nondegeneracy condition.

Lemma (2)
:   Let $K \subset U \subset \RR^n$ with $K$ compact and $U$ open, and let $f: U \to \RR$ have only nondegenerate critical points.
    Then there exists a $\delta > 0$ such that every $g: U \to \RR$ that is $C^2$ which satisfies

    1. $\abs{ \dd{f}{x_i}(p) - \dd{g}{x_i}(p) } < \delta$, and
    1. $\abs{ \frac{\partial x}{\partial x_i \partial x_j}(p) - \frac{\partial g}{\partial x_i \partial x_j}(p) } < \delta$

    for all $i, j$ and $p\in K$ has only nondegenerate critical points.

Proof
:   Define $\abs{df} = \sqrt{ \abs{\dd{f}{x_1}}^2 + \cdots + \abs{\dd{f}{x_n}}^2 }$.
    Now note that $S(f) = \abs{df} + \abs{\det\qty{ \frac{\partial f^2}{\partial x_i \partial x_j} }} \geq 0$.
    This is an equality iff both terms are zero, and the first term is zero iff $x$ is a critical point, while the second term is zero iff $x$ is degenerate.

    Since $f$ has only nondegenerate critical points, this inequality is strictly positive on $K$, i.e. $S(f) > 0$.
    Since $K$ is compact, $S(f)$ takes on a positive infimum on $K$, say $\mu$.
    Then $S(f) \geq \mu > 0$ on $K$.

    Thinking of $S$ as defining a norm, the reverse triangle inequality yields

    \begin{align*}
    \abs{ \abs{df} - \abs{dg}  } \leq \abs{df - dg} \leq \sqrt{n\delta^2} \leq \frac \mu 2
    ,\end{align*}

    where we can choose $\delta$ such that $\sqrt{n\delta^2} < \mu$.

    We can also pick $\delta$ small enough such that

    \begin{align*}
    \abs{ \abs{\det J_f} - \abs{\det( J_g )} } \leq \frac \mu 2
    ,\end{align*}

    where $J_f = \qty{ \frac{\partial f}{\partial x_i \partial x_j} }$ is shorthand for the matrix of partial derivatives appearing previously,
    and we just note that picking entries close enough makes the difference of determinant small enough (although there's something to prove there).

    Then

    \begin{align*}
    \abs{df} - \abs{dg} + \abs{\det(J_f)} - \abs{\det{J_g}} < \mu \\
    \implies 0 \leq \abs{df} + \abs{\det(J_f)} - \mu < \abs{dg} + \abs{\det(J_g)}
    ,\end{align*}

    > The second inequality follows from just moving terms in the first inequality.

    which makes the last term strictly positive, and thus nonzero on $K$.
    Then $g$ has no degenerate critical points in $K$.

Proof summary:

1. $\norm{f}_2(x) = 0$ iff $x$ is a degenerate critical point.
2. $\norm{f}_2(x) \geq \mu > 0$ in $K$.
3. We can pick $\delta$ small enough such that $\norm{f}_2 - \norm{g}_2 < \mu$ on $K$.
4. This forces $\norm{g}_2 > 0$ on $K$, so $g$ has *no* nondegenerate critical points on $K$.

## Proof that Morse Functions are Open

We still want to show that Morse functions form an open dense subset.

To see that they form an open set, suppose $f\in F(M, \RR)$ is Morse.
Then take a finite cover of $M$, say $\theset{(U_i, h_i)}_{i=1}^k$.
Pick compact $C_i \subset U_i$ that still covers $M$.

Note that any $g$ satisfying the 2 required conditions where $\abs{f-g} < \delta$ (?), then $g \in N(\delta) + f$.

By lemma 2, there exists a $\delta > 0$ such that every $g\in N_1 \definedas f + N(\delta)$ has only nondegenerate points in $C_1$.
We can pick a $\delta$ similarly to define an $N_i$ for every $i$.
Then taking $N = \intersect_{i=1}^k N_i$, this yields an open neighborhood of $f$ such that every $g \in N$ has only nondegenerate critical points on $C_1 \union C_2 \cdots \union C_k = M$.

$\qed$

## Proof that Morse Functions are Dense

We want to show that this set is dense, so we'll fix some open set and show that there exists a Morse function in it.

Let $f\in N$ for $N$ an open set; we'll then change $f$ gradually to make it Morse.

Convention
: We'll say $f$ is *good* on $S\subset M$ iff $f$ has only nondegenerate critical points in $S$.

Pick a smooth bump function $\lambda: M^n \to [0, 1]$ such that

- $\lambda \equiv 1$ on an open neighborhood of $C_1$, and
- $\lambda \equiv 0$ on an open neighborhood of $M\setminus U_1$.

> Note: we can do this because $C_1 \subset U_1$ is closed, and $M\setminus U_1$ is closed, so we can find disjoint open sets containing each respectively using the fact that $M^n$ is Hausdorff (?).

Now let $f_1 = f + \lambda L$ for some linear function $L:\RR^n \to \RR$, so $f_1 = f + L$ on an open neighborhood of $C_1$.
By Lemma 1, for almost every $L$, $f_1$ is good.

> Note that we need $\lambda$ because $L$ is only defined on $\RR^n$, not on $M$.

Now $f_1 - f = \lambda L$ is supported in $U_1$.
If we pick the coefficients of $L$ small enough, noting that $\lambda$ is bounded, then the first and second derivatives of $f-f_1$ will be bounded, and we can arrange for $f_1 \in f + N(\varepsilon)$ for $\varepsilon > 0$ as small as we'd like.
For $\varepsilon$ sufficiently small, we can arrange for $N(\varepsilon ) \subset N_\delta$ for the finitely many $\delta$s, and so $N(\varepsilon) \subset N$.

By Lemma 2, there exists a neighborhood $N_1 \subseteq N$ containing $f_1$ such that every $g\in N_1$ is good on $C_1$.
Since $f_1 \in N_1$, we can repeat this process to obtain an $f_2 \in N_2 \subseteq N_1$ and so on inductively.
Then since every $g \in N_2$ is good on $C_2$ and $N_2 \subseteq N_1$, every $g\in N_2$ is good on $C_1 \union C_2$.
This yields an $f_k \in N_k \subset N_{k-1} \subset \cdots \subset N_1 \subset N$, so $f_k$ is good on $\union C_i = M$.

$\qed$

Thursday:
We'll show that every pair of critical points can be arranged to take on different values, and then order them.
This yields $f(p_1) < c_1 < f(p_2) < c_2 < \cdots c_{k-1} < f(p_k)$, and since the $c_i$ are regular values, the inverse images $f\inv(c_i)$ are smooth manifolds and we can cut along them.

![Image](figures/2020-01-14-12:22.png)\