math214:01-29

# 2020-01-29, Wednesday

$$\gdef\d\partial, \gdef\t\tilde$$

We first introduce the notion of a tangent bundle. Next, we mention that a smooth map $f: M \to N$ induces a global differential $df: TM \to TN$. This finishes up Ch 3. Next, we introduce some more terminologies in Ch 4, Immersion and Submersion.

### Tangent Bundle

Last time, we defined the 'setwise' tangent bundle as $$TM = \bigsqcup_{p \in M} T_p M.$$ Elements of $TM$ is a pair $(p, v)$ where $p \in M$ and $v \in T_p M$. This $TM$ has a natural projection map $$\pi : TM \to M, \quad (p,v) \mapsto p$$.

Now, we equip $TM$ with a smooth manifold structure by equip it with a coordinate charts. If $(U, \varphi)$ is a chart for $M$, we will define a chart $(\t U, \t \varphi)$ on $TM$, where $\t U := \pi^{-1}(U)$, and $$\t \varphi: \t U \mapsto \varphi(U) \times \R^n \subset \R^n \times \R^n.$$ Let $u^i = x^i \circ \varphi: U \to \R$ be components of $\varphi$, recall that $\{ \frac{\d}{d u^i}(p) \}_{i=1}^n$ is a basis for $T_p M$, thus we may decompose $v \in T_p M$ as $$v = \sum_{i=1}^n v^i \frac{\d}{d u^i}(p).$$ Thus, we can define $\t \varphi$ as $$\t \varphi: (p, v) \mapsto (u^1, \cdots, u^n; v^1, \cdots, v^n).$$

Suppose we have two such charts $(U_1, \varphi_1)$ and $(U_2, \varphi_2)$, we want to check that on $\t U_1 \cap \t U_2$, the transition map $$\t \varphi_2 \circ \t \varphi_1^{-1}: \t \varphi_1( \t U_1 \cap \t U_2) \to \t \varphi_2( \t U_1 \cap \t U_2)$$ is a diffeomorphism. The base coordinates $u^i$ changes according to $\varphi_2 \circ \varphi_1^{-1}$, which is a diffeomorphism, the fiber coordinates $v^i$ changes as $$v^i_2 = \sum_{j=1}^n \frac{\d u_2^i }{\d u_1^j } v^j_1 \quad \Leftarrow \quad v = \sum_i v_1^i \frac{\d }{\d u_1^i} = \sum_i v_2^i \frac{\d }{\d u_2^i}$$ Since the Jacobian matrices $\frac{\d u_2^i }{\d u_1^j }$ are smooth and invertible, we see $\t \varphi_2 \circ \t \varphi_1^{-1}$ is smooth. It is not hard to verify its inverse is smooth as well. Thus, we have produced from each coordinate chart $(U, \varphi)$ on $M$ a coordinate chart $(\t U, \t \varphi)$ on $TM$, thus giving $TM$ a smooth manifold structure. See Lee p63 for details.

### Global differential

Last time, we defined pointwise linear map $df(p): T_p M \to T_{f(p)} N$. Put them together, we get a global differential $df: TM \to TN$. To verify that this map is a smooth map, we just need to work in local coordinate chart, and verify that the corresponding map between charts is smooth.

## Immersion and submersions

### Implicit Function Theorem, Inverse Function Theorem

We recall the following results from calculus.

Inverse Function Theorem. If $f: \R^n \to \R^n$ is smooth, $f(0)=0$, and $df(0)$ is invertible, then there exists a neighborhood $U$ of $0$, such that $f|_U$ is invertible with smooth inverse.

Implicit Function Theorem. Let $f: \R^n_x \times \R_y^m \to \R_z^m$ be a smooth function, where subscripts $x,y,z$ are coordinates on the corresponding factors. Assume that $f(0,0) = 0$, and the (partial) Jacobian matrix $\frac{\d f}{\d y} \vert_{(0,0)}$ is invertible. Then, there exists a neighborhood $0 \in U \subset \R^n_x$, and a smooth map $g:U \to \R^m_y$, such that $g(0)=0$ and $f(x,g(x))=0$ for all $x \in U$.

### Local property of $f$ is determined by $df$

The rank of a map at point $p$ is the rank of the linear map $df(p): T_p M \to T_{f(p)} N$. More concretely, if we put coordinate charts around $p$ and $f(p)$, then $df(p)$ is represented as the Jacobian matrix.

We only consider constant rank map, that is $r=rank(df(p))$ is constant on $M$. There are several situtations

• If $r = \dim M$, i.e. $df$ is injective, we call $f$ an immersion.
• If $r = \dim N$, i.e. $df$ is surjective, we call $f$ an submersion.
• If $r = \dim M = \dim N$, $df$ is bijective, we call $f$ a local diffeomorphism.

Note that these are local properties of $f$.

In general, we have the rank theorem (Lee Thm 4.12, p81), which says $f$ can factorize as a submersion followed by an immersion $$\gdef\into\hookrightarrow \gdef\onto\twoheadrightarrow$$ $$f: M \onto Z \into N$$