Lecture Notes

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====== 2020-02-07, Friday ======

A vector field $X$ on a smooth manifold $M$, is an assignment to each $p \in M$ an element $X(p) \in T_p M$. We say $X$ is smooth, if for any smooth function $f \in C^\infty(M)$, the derivative $X(f)$ is also a smooth function. 

**Exercise** : Show that $X$ is smooth if and only if the following is true:  for any coordinate patch $(U, (x_1, \cdots, x_n))$, if we expand the vector field $X$ as
$$ X = \sum_{i=1}^n X^i \frac{\d}{\d x_i},  $$
then the coefficients function $X^i: U \to \R$ are smooth functions. 

Example: on $\R^n$, given any smooth function $f: \R^n \to \R$, we can consider the gradient vector field
$$ \nabla f = \sum_{i} \d_i f \d_i. $$
The zero of $\nabla f$ are called critical points of $f$. **Warning: ** for general smooth manifold, we do not have the notion of a gradient vector field for $f$, unless we are given a metric tensor. 

===== Commutator of Vector fields =====
Given two vector fields, $X, Y$, we can define its commutator as follows, for any smooth function $f$, we define 
$$ [X, Y]_p f = X_p Y_p (f) - Y_p X_p (f) $$
It is an exercise to check that this $[X, Y]_p$ is indeed a derivation, hence $[X, Y]_p \in T_p M$. 
Concretely, if we have coordinates $u_i$, we have $[\d_i, \d_j]=0$. 
$$ [ \sum_i X^i \d_i , \sum_j Y^j \d_j ] = \sum_{i,j} X^i \d_i Y_j \d_j - Y_j \d_j X^i \d_i. $$

It is an exercise to check that $[ [X, Y], Z] + [ [Y, Z], X ] + [ [ Z, X], Y ] = 0$. Hence the space of vector fields forms a ** Lie algebra **. 

===== Integral Curve =====
Let $X$ be a smooth vector field on $M$. Let $p \in M$ be any points. And integral curve of $X$ through $p$ is a map 
$$ \gamma: (a, b) \mapsto M $$
such that $0 \in (a,b)$ and $\gamma(0) = p$, and $\dot \gamma(t) = X_{\gamma(t)}$ for all $t \in (a,b)$. 

If we work in coordinate near $p$, then finding an integral curve through $p$ is equivalent to solving an ODE. By the fundamental theorem of ODE, there exists an $\epsilon>0$, such that we have an integral curve for $t \in (-\epsilon, \epsilon)$ through $p$. 

===== Flow =====
Given an integral curve through a point $p$, we can define the motion of $p$ for some small interval of time $t$. If we consider the motion of all the points, we get a flow on $M$. However, there is subtlety that the flow may not exist for arbitrary long time. 

We can define flow on a manifold without talking about a vector field. We first consider an easy case. 

** Global Flow **
A smooth **global flow** on $M$, or a one-parameter group action on $M$, is a smooth map 
$$ \theta: \R \times M \to M$$
such that for each $t \in \R$, we define $\theta_t(p) = \theta(t, p)$ for all $p \in M$. Then, we require the group law $\theta_t \circ \theta_s = \theta_{t+s}$ for all $t, s \in \R$. 

For each $p \in M$, we define $\theta^p: \R \to M$ by $\theta^p(t) = \theta(t,p)$. The image of this curve is the orbit of $p$. 

We may define a vector field $V$ on $M$, such that its value at $p$ is $V_p = \dot \theta^p(0)$. $V$ is called the ** generator ** of the flow $\theta$. 

** Flow that does not exists for all time **
In general, we do not have a smooth map $\R \times M \to M$. Instead, we replace $\R \times M$ by an open set $\gdef\dcal{\mathcal D} \dcal \subset \R \times M$, such that for each $p \in M$, the set $\dcal^p = \{t \in \R, (t, p) \in \dcal \}$ is an interval containing $0$. Such a $\dcal$ is called a **flow domain**. 

A flow on $M$ is a map $\theta: \dcal \to M$ where $\dcal$ is a flow domain, and $\theta(0, p) = p$ and for each $p\in M$, $\theta_s(\theta_t(p)) = \theta_{s+t}(p)$ when the maps are defined. 

** Theorem (Fundamental Theorem on Flows) ** Let $V$ be a smooth vector field on $M$. Then there exists a unique **maximal** flow $\theta: \dcal \to M$ whose infinitesimal generator is $V$. 

Sketch of proof: We can define a set $\dcal = \{(t,p): p \in M, \z{flow exists for time $t$} \}$. Then one need to prove that $\dcal$ is open. This is done using the group law of $\theta$ and ODE theorems.
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