math214:02-05

$$\gdef\Q{\mathbb Q} \gdef\P{\mathbb P} \gdef\RM{\backslash}$$

We will do some basic definitions of submanifolds. Then we talk about Whitney Embedding Theorem

(1) Let $M, N$ be two smooth manifolds. A ** smooth embedding of $M$ into $N$ ** is a smooth immersion $F: M \to N$ that is also a topological embedding, i.e., a homeomorphism onto its image $F(M) \subset N$ where $F(M)$ is equipped with the subspace topology.

** Example and Non-example of smooth embeddings**

- $S^1 \into \R^2 $. We view $S^1 = \R / 2\pi \Z$, and the map is $\theta \mapsto (\cos(\theta), \sin(\theta))$. (Yes)
- $(0, 2\pi) \into \R^2$. $\theta \mapsto (\cos(\theta), \sin(\theta))$. (Still yes)
- The figure eight curve. one maps $(-\pi, \pi)$ to $\R^2$ as a figure eight, $$ \beta(t) = (\sin(2t), \sin(t) )$$, we see that $\lim_{t \to \pm \pi} \beta(t) = (0,0) = \beta(0)$. Hence it is example that an injective smooth immersion is not a smooth embedding.
- Another example is the dense curve on a two torus. Pick an irrational number $c \in \R - \Q$. Then we can consider the map $\phi: \R \to T^2 = \R^2 / \Z^2$, $x \mapsto [(x, cx)]$.

**Prop: ** if $F: M \to N$ is an injective smooth immersion, and $M$ is compact, then $F$ is a smooth embedding.

(2) A closely related terminology is **an embedded submanifold**. Let $M$ be a smooth manifold, an embedded submanifold $S$ is a subset $S \subset M$ such that

- $S$ is a (topological) manifold in the subspace topology.
- $S$ is endowed with a smooth structure with respect to which $i: S \into M$ a smooth embedding.

The codimension of $S$ in $M$ is $co\dim_M S = \dim M - \dim S$. An open subset of $M$ can be viewed as a codimension-0 submanifold, and is called an 'open submanifold' of $M$.

(3) **Def:** An embedded submanifold $S \subset M$ is **properly embedded**, if the inclusion $i: S \into M$ is a proper map.

- Recall the definition of 'proper map': a continuous map of topological spaces $f: M \to N$ is proper if inverse image of compact sets are compact. Note that image of compact sets under $f$ are always compact, since you can pullback open cover to open cover.
- The inclusion $f: (-1, 1) \into \R$ is not a proper map, since $f^{-1}([0,2]) = [0,1)$ is not compact in $(-1,1)$. (nonetheless, $[0,1) \subset (-1,1)$ is closed)
- The inclusion $f: \R \into \R^2, f(x) = (x,0) $ is a proper map.

Intuitively speaking, a properly embedded submanifold does not have “loose ends”.

**Prop:** An embedded submanifold $S \subset M$ is properly embedded if and only if $S$ is a closed subset of $M$.

(4) local $k$-slice. Let $S$ be an embedded submanifold of $M$, of dimension $k$. Then for any point $p \in S$, we can find a coordinate chart $(U, (x_1, \cdots, x_n))$ of $p$ in $M$, such that $S \cap U = \{q \in U: x_{k+1}(q) = \cdots = x_n(q) = 0\}$. Such a coordinate is called a slice coordinate.

For simplicity, I will only prove the compact version.

Theorem: Every smooth compact manifold of dimension $n$ admit a proper smooth embedding into $\R^{2n+1}$.

Let $M$ be a smooth compact manifold of dimension $n$.

** Step 1: ** Show that one can embed $M$ into $\R^N$ for a large enough $N$. Let $M$ be covered by finitely many coordinate charts $\{(U_1,\varphi_i)\}_{i=1}^m$. Let $\{f_i\}$ be a partition of unity of $M$ subject to the cover $\{U_i\}$, i.e. $f_i \geq 0$, $\sum_i f_i = 1$ and $supp(f_i) \subset U_i$. We then define a map
$$ \Phi: M \to \R^{nm + m} , \quad p \mapsto (\varphi_1(p) f_1(p), \cdots, \varphi_m(p) f_m(p); f_1(p), \cdots, f_m(p))$$
First, we note that $\Phi$ is well-defined and smooth, indeed $\varphi_i(p)f_i(p): U \to \R^n$ can be viewed as a function on $M$ by extension by zero. Next, we note that $\Phi$ is a smooth immersion. Suppose not, and $0 \neq v \in T_p M$ is in the kernel of $d\Phi_p$, then assume $f_j(p) \neq 0$, we would have $d f_j(p)(v) = 0$, then $d (\varphi_j \cdot f_j)(v) = f_j(p) d (\varphi_j)(v)$, since $d (\varphi_j)$ is a bijection, and $f_j(p) \neq 0$, we have $d (\varphi_j \cdot f_j)(v)\neq 0$, this contradicts with $d \Phi_p(v) = 0$. Hence $\Phi$ is a smooth immersion. Since $M$ is compact, we have $\Phi$ a proper embedding.

** Step 2: ** Show that, if $\Phi: M \to \R^N$ is any smooth embedding, and if $N > 2n+1$, then we can find quotient map $\pi_v : \R^N \to \R^N / (\R \cdot v) \cong \R^{N-1}$, such that $\pi_v \circ \Phi: M \to \R^{N-1}$ is a smooth embedding.

Since $M$ is compact, hence smooth embedding corresponds to injective immersion. We identify $M$ as a embedded submanifold of $\R^N$. Let $[v] \in \R \P^{N-1}$ be the line containing $v$. Then

- $\pi_v|_M$ is injective, if and only if for any distinct points $p,q \in M$, $p-q \notin [v]$.
- $\pi_v|_M$ is an immersino, if and only if for any $p \in M$, $T_p M \subset T_p \R^N$ does not contain vector $v$.

To show that such nice $v$ exists, we will consider all possible directions $[v] \in \R\P^{N-1}$, and use Sard theorem to say the bad directions are negligible. Let $\Delta_M \subset M \times M$ be the diagonal. And let $M \subset TM$ as the zero section. Then we have two maps $$ \kappa: (M \times M) \RM \Delta_M \to \R \P^{N-1}, \quad (p,q) \mapsto [p-q] $$ $$ \tau: TM \RM M \to \R \P^{N-1}, \quad (p, w) \mapsto [w] $$ For both map, the source manifold is $2n$-dimensional, and the target manifold is $N-1$ dimensional, by assumption $N-1 > 2n$, hence the image of $\kappa$ and $\tau$ are the singular value set, hence negligible.

Then, we can repeat this step iteratively, until $N = 2n+1$, then we are done. QED.

math214/02-05.txt · Last modified: 2020/02/05 23:22 by pzhou