Lecture Notes

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====== 2020-02-12, Wednesday ======

===== Tubular Neighborhood Theorem =====
Let $M \subset \R^n$ be a embedded submanifold. Let $NM$ be the normal bundle of $M$, defined by
$$ NM= \{ (x, v) \in T\R^n \mid x \in M, v \perp w, \forall w \in T_x M \}. $$
Thm: $NM$ is an embedded $n$-dimensional submanifold of $T\R^n$.
Let $M_0 =\{(x,v) \in  NM| v = 0\} \subset NM$ be the zero section. 

There is a shift map: 
$$E: NM \to \R^n, \quad (x,v) \mapsto x+v. $$ 
One can check that for any $x \in M_0$, $dE_x: T_x NM \to T_x \R^n$ is an isomorphism. Hence $E$ is a diffeomorphism in a neighborhood of $x$. Since $x$ is arbitrary, we have $E$ is a diffeomorphism in a neighborhood $V \supset M_0$. More precisely, we want to constraint the shape of $U$. We define

**Tubular neighborhood of $M$ in $\R^n$** is a neighborhood $U$ of $M$, that is the diffeomorphic image under $E$ of an open subset of the form 
$$ V = \{(x,v) \in NM : |v| < \delta(x) \}$$
for some positive continuous function $\delta: M \to \R$. 

See Thm 6.24 for a proof of the existence of a tubular neighborhood. Mainly, one try to construct the 'radius' function $\delta$ and show that it is continuous. 

We define a map $r: U \to M$, that is identity map when restrict to $M$, such a map is called a retraction. 
$r: = \pi \circ E^{-1}$, where $\pi: NM \to M$ is the projection map. 




===== Transversality =====
Let $V$ be a finite dimensional vector space, $W_1, W_2$ be two subspaces, we say $W_1$ and $W_2$ are transversal, if $W_1 + W_2 = V$. 

Consider two embedded submanifolds $N_1, N_2$ in $M$. Suppose $\dim M =m, \dim N_1 = n_1, \dim N_2 = n_2$. If $n_1 + n_2 < m$, then intuition tells us that $N_1$ and $N_2$ 'generically' does not intersect(no intersection is a special case of transverse intersection). If $n_1 + n_2 \geq m$, then 'generically' $N_1$ and $N_2$ should intersects transversally, that is, for any $p \in N_1 \cap N_2$, we should have  $T_p N_1 + T_p N_2 = T_p M$. But what does 'generically' mean? Let's make it more precise.

Let's consider a more general setting. Let $S \subset M$ be an embedded submanifold, and let $F: N \to M$ be any smooth map. We say that **$F$ is transverse to $S$**, if for every $x \in F^{-1}(S)$, we have
$$ T_{F(x)} S + DF_x (T_x N) = T_{F(x)} M. $$

To understand the following theorem, we need to have 
  * Theorem 5.8 (Local Slice Criterion for Embedded Submanifolds) Any embedded submanifold $S$ has slice coordinate chart. 
  * Theorem 5.12 (Constant-Rank Level Set Theorem) : If $F: M \to N$ has constant rank, then for any $x \in F(M)$, $F^{-1}(x)$ is a properly embedded submanifold of $M$. 

**Theorem (6.30)**: (a) If $F: N \to M$ is a smooth map between smooth manifolds, $S \subset M$ embedded manifold, and $F$ is transverse to $S$, then $F^{-1}(S)$ is an embedded submanifold of $N$ with the same codimension as $S$ in $M$ \\
(b) If $S', S$ are two embedded submanifolds, that intersects transversely, then $S \cap S'$ is an embedded submanifold of $M$, with $co\dim_M (S \cap S') = co\dim_M S + co\dim_M S'$.  

Sketch of proof: (b) follows from (a) by taking $F = \iota: S' \into M$, and $S' \cap S \into S' \into M$,  composition of embedding is embedding. 

To prove (a), we take a point $x \in S \cap F(N)$, and take slice coordinate chart $(U, (x_1, \cdots, x_n))$ of $M$ adapted to $S$, that is,  $S \cap U$ can be written as $\{x_{k+1}=0, \cap x_n=0\}$ where $k =\dim S$. Denote $\varphi=(x_{k+1}, \cdots, x_n): U \to \R^{n-k}$. Then, consider $(\varphi \circ F): F^{-1}(U) \to \R^{n-k}$, thus $(\varphi \circ F)^{-1}(0) = F^{-1}(U \cap S)$. One want to show that $(\varphi \circ F)$ has $0$ a regular value, i.e for any $x \in (\varphi \circ F)^{-1}(0)$, $D(\varphi \circ F)_x: T_x N \to T_0 \R^{n-k}$ is surjective. That is guaranteed by transversality condition. $\Box$


Next, one consider the parametrix transversality theorem. 

Let $S, M, N$ be smooth manifolds. A smooth family of maps $F_s: N \to M$ parametrized by $s \in S$ is a smooth map $F: N \times S \to M$, whereas $F_s(-) = F(-, s)$. 

** Theorem (6.35) ** Let $X \subset M$ be an embedded submanifold, and let $F: N \times S \to M$ be a smooth map. If $F$ is transversal to $X$, then for almost every $s \in S$, $F_s: N \to M$ is transverse to $X$. 

Proof: Let $W=F^{-1}(X)$. Let $\pi: N \times S \to S$, and $\pi|_W$ its restriction to $W$. By Sard theorem, the critical value of $\pi|_W$ is negligible. Suffice to prove that for any point  $s \in S$ away from the critical value $\pi|_W$, $F_s$ is transversal to $X$. (remember that if $X \cap F_s(N) = \emptyset$, it is still transverse intersection). 

Let's do some simple dimension counting. Suppose $codim_M X = k$, then $codim_{N \times S} W = k$. Let $\dim N = n$, $\dim S = d_s$. If $k > n$, then $\dim W = d_s + n - k < d_s$, then for most $s$, $F_s(N) \cap X = \emptyset$ and we are done. 

Now assume $k \leq n$. We can see that $s \in S$ is a critical value, if and only if $N \times \{s\}$ intersects $W$ non-transversely (hence intersection is non-empty). $\Box$. 

** Theorem 6.36 (Transversality Homotopy Theorem). **
Suppose $M$ and $N$ are smooth manifolds and $X \subset􏰌 M$ is an embedded submanifold. Every smooth map $f:  N \to M $ is homotopic to a smooth map $g: N \to M$ that is transverse to $X$.

Sketch of proof: We are going to let $S=B^k$, $k$-dimensional unit open ball, and build a smooth map $F: N \times S \to M$ that is a submersion, hence transverse to $X$. First, we use Whitney embedding theorem to embed $M$ to $\R^k$, $\iota: :M \to \R^k$. Create tubular neighborhood $U$ of $M$ in $\R^k$. Then, we will wiggle the composition $\iota\circ f: N \to \R^k$ in all the direction, within $U$, then project back to $M$.
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