# January 21st

## Elementary Cobordism

Recall that an elementary cobordism is a cobordism that has a Morse function with exactly one critical point.

Definition (Handles)
: An $n\dash$dimensional $\lambda\dash$handle is a copy of $D^{\lambda} \times D^{n-\lambda}$ which is attached to $\bd M^n$ via an embedding $\phi: \bd D^\lambda \cross D^{n-\lambda} \injects \bd M$.

Example
:   Let $\lambda = 1, n = 2, n-\lambda = 1$ and take $M^2 = D^2$ and we attach $D^1 \cross D^1$.
    Note that there's not necessarily a smooth structure on the resulting manifold, so we can "smooth corners":

    ![Image](figures/2020-01-21-11:09.png)\

Example
: ![Image](figures/2020-01-21-11:12.png)\

> Note: the above is just a homeomorphism.

Definition (Surgery)
:   Let $M$ be an $n-1$ dimensional smooth manifold, and $\rho: S^{\lambda-1} \cross D^{n-\lambda} \injects M^{n-1}$ be an embedding.

    Then noting that $\bd D^{n-\lambda} = S^{n-\lambda-1}$, consider the space

    \begin{align*}
    X(M, \phi) &= (M\setminus \rho(S^{\lambda-1} \cross \theset{0}) ) \cross (D^\lambda \cross S^{n-\lambda-1}) \\
    & / \generators{ \rho(u, tv) \sim (tu, v) \suchthat t\in (0, 1), \forall u\in S^{\lambda-1}, \forall v\in S^{n-\lambda-1} }
    ,\end{align*}

    where we note that we can parameterize $D^{n-\lambda} = tv$ where $v$ is a point on the boundary.

    Note that this accomplishes the goal of smoothing, and is referred to as **surgery** (of type $\lambda, n-\lambda$) on $M$ along $\phi$.

Example
: ![Image](figures/2020-01-21-11:30.png)

Example
:   $n-1 = 3$ and $\lambda = 2$ implies $\lambda-1 = 1$, and take $\rho: S^1 \cross D^2 \to S^3$, which has image a tubular neighborhood of a knot.
    Then $\phi(S^1 \cross\theset{0}) = K$ for some knot, and $(S^3\setminus K) \disjoint (D^2 \cross S^1) / \cdots$.
    Then note that $\bd \phi(\theset{u} \cross D^2) = \theset{u}\cross S^1$, which no longer bounds a disk since we have removed the core of tube.

    ![Image](figures/2020-01-21-11:42.png)\

Theorem (Cobordism and Morse Function Induced by Surgery)
:   Suppose $M' = X(M, \rho)$ is obtained from $M$ by surgery of type $\lambda$.
    Then there exists an elementary cobordism $(W; M, M')$ with a Morse function $f: W \to [-1, 1]$ with only one index $\lambda$ critical point.

Example
:   Let $M=S^1$ and $\lambda = 1$.

    ![Image](figures/2020-01-21-11:49.png)\

### Proof: Surgeries Come From Cobordisms With Special Morse Functions

Write $\RR^n = \RR^\lambda \cross \RR^{n-\lambda}$, and $(\vector x, \vector y) \in \RR^n$.

Then
\begin{align*}
L_\lambda
= \{(\vector x, \vector y) \suchthat -1
\leq -\norm{\vector x}^2 +\norm{\vector y}^2
\leq 1,~ \norm{\vector x}\norm{\vector y}
< \sinh(1) \cosh(1)\}
.\end{align*}

![Image](figures/2020-01-21-11:54.png)\

The left boundary is given by $\bd_L: \norm{\vector y}^2 - \norm{\vector x}^2 = -1$, and there is a map

\begin{align*}
S^{\lambda-1} \cross D^{n-\lambda} &\mapsvia{\text{diffeo}} \bd_L \\
(u, tv) &\mapsto (u\cosh(t), v\sinh(t)) \quad t\in [0, 1)
,\end{align*}

which is clearly invertible.

The right boundary is given by $\bd_R: \norm{\vector y}^2 - \norm{\vector x}^2 = 1$, and there is a map

\begin{align*}
S^{\lambda-1} \cross D^{n-\lambda} &\mapsvia{\text{diffeo}} \bd_L \\
(tu, v) &\mapsto (u\sinh(t), v\cosh(t))
.\end{align*}

In the above picture, we can consider the orthogonal trajectories, which are given by $y^2 - x^2 = c$, which has gradient $(-x, y)$ and $xy = c$ which has gradient $(y, x)$, so these are orthogonal.

Recall that near a point $p\in M$, the morse function has the form $f(\vector x, \vector y) = f(p) - \norm{\vector x}^2 + \norm{\vector y}^2$ with a gradient-like vector field given by $\xi = (-\vector x, \vector y)$.

The orthogonal trajectories will generally be of the form $\norm{\vector x}\norm{\vector y} = c$, which we can parameterize as $t \mapsto (t\vector x, \frac 1 t \vector y)$.

**Construction of $W$**:

Take
\begin{align*}
W(M, \phi)
&= ( (M\setminus \phi(S^{\lambda-1} \cross \theset{0})) \cross D^1 )\disjoint L_\lambda  \\
& / \generators{ \phi(u, tv) \cross c \sim (\vector x, \vector y) \suchthat \norm{\vector x}^2 - \norm{\vector y}^2 = c, (\vector x, \vector y) \in \text{orth. traj. starting from} (u\cosh(t), v\sinh(t)) }
.\end{align*}

![Image](figures/2020-01-21-12:10.png)\

This amounts to closing up in the following two ways:

![Image](figures/2020-01-21-12:14.png)\

This has two boundaries: when $c = -1$, we obtain $M$, and $c=1$ yields $X(M, \phi)$.
The Morse function is given by $f: W(M, \phi) \to [-1, 1]$ where

\begin{align*}
\begin{cases}
f(z, c) = c & z \in M\setminus \phi(S^{\lambda - 1} \cross \theset{0}), c\in D^1\\
f(\vector x, \vector y) = \norm{\vector x}^2 - \norm{\vector y}^2 & (\vector x, \vector y) \in L_\lambda
\end{cases}
.\end{align*}