# Tuesday January 28th

Setup:
Fix an elementary cobordism $(W; M_0, M_1)$, a Morse function $f: W\to [-1, 1]$ with exactly one critical point $p$ with index $\ind_(p) = \lambda$.
This yields a gradient-like vector field $\xi$, and $D_L = W^s(p) = \theset{x\in W \suchthat \lim_{t\to\infty} \psi_x(t) = p}$ the stable manifold.

Theorem (Deformation Retract of Cobordism Onto ??)
:   $W \cong M_0 \union D_L$, a $\lambda$ dimensional disk, is a homotopy equivalence.
    More precisely, there is a deformation retract.

### Proof

Take the characteristic embedding $\phi_L: S^{\lambda - 1} \cross OD^{n-\lambda} \injects M_0$.
We have a cobordism $(W(M_0, \phi_L); M_0, \chi(M_0, \phi_L)) \cong (W; M_0, M_1)$.

Recall that the LHS is constructed via $(M_0 \setminus \phi(S^{\lambda-1} \cross 0)) \cross D_1 \disjoint L_\lambda / \sim$.

Retraction 1:
$W(M_0, \phi_L) \mapsvia{r_t} M_0 \union C$.
We'll construct this retraction.
This follows the green integral curves to retract onto the red.

![Image](figures/2020-01-28-11:19.png)\

Identify $D_L = \theset{(\vector x, \vector 0)} \subset L_\lambda$ in the local picture:

![Image](figures/2020-01-28-11:17.png)\

Define $C = \theset{(\vector x, \vector y) \suchthat \norm{\vector y} \leq \frac 1 {10}}$.

Choose $(Z, c)$ such that $z\in M_0 \setminus \phi_L( S^{\lambda-1} \cross OD^{n-\lambda}  )$ and $c\in [-1, 1]$.
Let $r_t(z, c) = (z, c + t(-1-c))$, note what happens at $t=-1, 1, 0$.

We can parameterize the integral curves in the local picture as $(\vector x/r, r\vector y)$.

So for $(\vector x, \vector y) \in L_\lambda$, we can define


\begin{align*}
r_t(\vector x, \vector y) =
\begin{cases}
(\vector x, \vector y) & \norm{\vector y} \leq \frac 1 {10} \iff (\vector x, \vector y) \in C \\
? & ? \\
(\vector x/ \rho(t), \rho(t) \vector y) & \norm{\vector y} \geq \frac{1}{10}
\end{cases}
.\end{align*}

where $\rho(t)$ is the solution of

\begin{align*}
\rho(t)^2 \norm{\vector y}^2 - \norm{\vector x}^2/\rho(t)^2 = \qty{ \norm{\vector y}^2 - \norm{\vector x}^2  }(1-t) - t \\
\rho(t) \norm{\vector y}^2 \geq \frac{1}{10} \implies \rho(t) \geq \frac{1}{10 \norm{\vector y}}
.\end{align*}

So we define $\rho(t) = \max(\text{positive solutions for the above equation}, \frac{1}{10\norm{\vector y}})$.

Retraction 2:
$M_0 \union C \mapsvia{r_t'} M_0 \union D_L$

![Image](figures/2020-01-28-11:33.png)\

We want the restriction of $r_t'$ to $M_0\setminus C$ to be the identity, so for $(\vector x, \vector y) \in C$ we define

\begin{align*}
r_t'(\vector x, \vector y) =
\begin{cases}
(\vector x, (1-t) \vector y) & \norm{\vector x} \leq 1 \\
(\vector x, \alpha(t)\vector y) & 1 \leq \norm{\vector x} \leq \sqrt{1 + \frac 1 {100}}
\end{cases}
.\end{align*}

![Image](figures/2020-01-28-11:37.png)\

We define $\alpha(t)$ at $t=0$ to be the identity, and at $t=1$ we want $\norm{\alpha(t) \vector y}^2 - \norm{x}^2 = -1$, and solving yields

\begin{align*}
\alpha(t) = (1-t) + t\frac{\sqrt{\norm{x}^2 - 1 }}{\norm{\vector y}}
.\end{align*}

$\qed$

Corollary
: For $M$ a closed smooth $n\dash$manifold with a Morse function $f: M \to \RR$, $M$ is homotopy-equivalent to a CW complex with one $\lambda\dash$cell for each critical point of index $\lambda$.

> Proof: See Milnor's book (Morse Theory).

## Morse Inequalities

Let $M$ be a closed smooth manifold and $f: M\to \RR$ Morse, and fix $F$ a field.
Let $b_i(M)$ denote the $i$th Betti number, which is $\rank H_i(M; F)$.

Weak Morse Inequalities:

1. $b_i(M) \leq$ the number of index $i$ critical points, denoted $c_i(M)$.
2. $\xi(M) = \sum (-1)^i c_i(M)$.

Strong Morse Inequalities:
$b_i(M) - b_{i-1}(M) + \cdots \pm b_0(M) \leq c_i - c_{i-1} + c_{i-2} \cdots \pm c_0$.

Lemma
: The weak inequalities are consequences of the strong ones.

Proof (implying (1))
: We have $b_i - \cdots \leq c_i - \cdots$ and separately $b_{i-1} - \cdots \leq c_{i-1} \cdot$, and adding these inequalities yields $b_i \leq c_i$.

Proof (implying (2))
: To obtain the equality, multiply through by a negative sign.
  For $i> n$, we have $b_{i-1} - b_{i-2} + \cdots = c_{i-1} - c_{i-2} + \cdots$, where the LHS is $\pm \chi(M)$ and the RHS has matching signs.

### Proof of Strong Morse Inequalities

Suppose $f(p_1) < \cdots < f(p_k)$.
We can select points $a_i$ such that $a_0 < f(p_1) < a_1 < \cdots$.
Let $M_i = f\inv [a_0, a_i]$; we then have $\emptyset \definedas M_0 \subset M_1 \subset \cdots M_k = M$.

Using excision, we have

\begin{align*}
H_j(M_i, M_{i-1})
&= H_j( f\inv[a_{i-1}, a_i], f\inv(a_{i-1}) ) \\
&=
\begin{cases}
\FF & j = \ind(p_i) = \lambda_i \\
0 & \text{otherwise}
\end{cases}
.\end{align*}

So $b_j(M_{i}, M_{i-1}) = 1$ iff $j = \lambda_i$, and 0 otherwise.

Lemma (Sublemma)
:   Define $S_i \definedas b_i(X, Y) - b_{i-1}(X, Y) + \cdots$, i.e. the LHS of the strong Morse inequality.
    Then $S_i$ is *subadditive*, i.e. if $X \supset Y \supset Z$ then $S_i(X, Z) \leq S_i(X, Y) + S_i(Y, Z)$.

This implies the strong inequality, since
\begin{align*}
S_i(M, \emptyset) \leq S_i(M_k, M_{k-1}) + S_i(M_{k-1}, M_{k-2}) + \cdots + S_i(M_1, M_0)
.\end{align*}
The RHS here equals $\sum_{j=1}^k T_i(M_j, M_{j-1}) = T_i(M)$, where $T_i(M) = c_i - c_{i-1} + \cdots$.

Write down the relative homology exact sequence:

\begin{center}
\begin{tikzcd}
{H_{i+1}(X, Y)} \arrow[rr, "\partial_i"] &  & {H_i(Y, Z)} \arrow[rr, "f_i"] &  & {H_i(X, Z)} \arrow[rr, "g_i"] &  & {H_i(X, Y)} \arrow[lllldd, "\partial_{i-1}"] \\
&  &                               &  &                               &  &                                              \\
&  & {H_{i-1}(Y, Z)}               &  &                               &  &
\end{tikzcd}
\end{center}

then
\begin{align*}
\rank(\del_i) = \dim \ker(f_i) = b_i(Y,Z) - \rank(f_i) = b_i(Y,Z) - b_i(X, Z) + \rank(g_i) = \cdots = S_i(Y,Z) - S_i(X, Z) + S_i(X, Y) > 0
.\end{align*}
since ranks are positive.