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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 $p$.


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.

math214/02-07.txt · Last modified: 2020/02/06 23:29 by pzhou