Chapter 4: Vector Fields

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Chapter 4: Vector Fields

Contents
Contents
 1.  Vector fields
 2.  Pushforward of a vector field
 3.  Lie brackets
 4.  Integral curves and flows

We have been familiar with some vector fields, such as an electric field E=(E _ x,E _ y,E _ z) which is a continuous function to \mathbb R^3 and can be visualized by attaching an arrow to each point. Similarly, a vector filed X on a manifold M is the assignment of a tangent vector X _ p\in T _ pM to each point p\in M.
1. Vector fields We now formally define a vector field.

Sometimes such a vector field is called a rough vector field. We are more interested in the smooth one, in which TM is treated as a smooth manifold by Propostion 11 in Chapter 2. Let (U,(x^i)) be a chart. We can write the value of X:M\to TM at any p\in U in terms of coordinate basis vectors X _ p=X^i(p)\frac{\partial}{\partial x^i}| _ p, and we get n component functions of X, X^i:U\to\mathbb R, in this chart.

An example is given by p\mapsto \frac{\partial}{\partial x^i}| _ p, which is called the i-th coordinate vector field and denoted by \frac{\partial}{\partial x^i}. All of the component functions are constant here.

It turns out that we can tell whether a vector field is smooth by investigating the component functions. This is Proposition 2.

Let \mathcal X(M) be the set of all smooth vector fields on M. If X\in \mathcal X(M) and f:U\to\mathbb R is a smooth function, we can define a new function
\[
Xf:U\to\mathbb R,\quad p\mapsto(Xf)(p)=X _ pf.
\]It is clear that Xf is locally determined. In particular, for any open subset V\subseteq U we have (Xf)| _ V=X(f| _ V).

With this function we have another useful smoothness criterion. This is Proposition 3.

Let (x^i,v^i) be the natural coordinates on \pi^{-1}(U)\subseteq TM in the given chart. Then the coordinate representation of X:M\to TM is \widehat{X}(x)=(x^i,X^i(x)), from which we can see X is smooth on U iff all X^i is smooth on U.

Suppose X is smooth, p\in M contained in a chart (U,(x^i)) and f is a smooth function on M. Then for x\in U we have
\[
Xf(x)=\Big(X^i(x)\frac{\partial}{\partial x^i}\Big| _ x\Big)f=X^i(x)\frac{\partial f}{\partial x^i}(x).
\]Thus Xf is smooth on U. Since p is arbitrary, Xf is smooth on M.

Now if f is a smooth function defined on an open subset U of M, we take a point p\in U. Then we pick a smooth bump function \psi that is equal to 1 in a neighborhood of p and supported in U, and define \tilde{f}=\psi f on M. We have known X\tilde f is smooth on M. Since X\tilde f=Xf in a neighborhood of p, Xf is smooth in this neighborhood. Therefore Xf is smooth on U for p\in U is arbitrary.

Assume now Xf is smooth whenever f is smooth on an open subset of M. We may prove X is smooth. Let (U,(x^i)) be a chart. Then Xx^i=X^j\frac{\partial}{\partial x^j}(x^i)=X^i. By assumption Xx^i is smooth, so the component functions of X are smooth. Thus X is smooth on M.

The next proposition is easy to verify. We define a vector field
\[ fX:M\to TM,\quad p\mapsto(fX) _ p=f(p)X _ p.\]Note that this is not the same as the function Xf.

Given a smooth vector field, consider the map f\mapsto Xf. We know this is a linear map from C^\infty(M) to C^\infty(M). It is also a derivation for we can check it satisfies the linearity and product rule
\[
X(fg)=f\, Xg+g\, Xf.
\]
The next proposition shows that the derivations can be identified with smooth vector fields.

We have seen the "if" direction. Now assume D is a derivation. For any p\in M, we define a function X _ p:C^\infty\to\mathbb R by X _ pf=(Df)(p), which is linear from the linearity of D. It is easy to see this is a tangent vector at p. Thus we have defined a vector field. It is smooth since Xf=Df is smooth for any f\in C^\infty(M).
2. Pushforward of a vector field If X is a smooth vector field on M and F:M\to N is a smooth map. For any p\in M, we can obtain a tangent vector in T _ {F(p)}N by applying the differential of F to X _ p. However this does not generally define a vector field on N. If there is a vector field Y for which we have \mathrm dF _ p(X _ p)=Y _ {F(p)} for all p, we say X,Y are F-related. Such a vector field Y may not exist, but we will see if F is diffeomorphism, there is a unique Y that is F-related to X.

Let's take a look at the relation \mathrm dF _ p(X _ p)=Y _ {F(p)} when acting on a function f. For the left hand side, it is X _ p(f\circ F)=X(f\circ F)(p). For the right hand side, it is (Yf)(F(p))=((Yf)\circ F)(p). Thus, we have

This proposition can be proved if we define a vector field by
\[
(F _ \ast X) _ q=\mathrm dF _ {F^{-1}(q)}(X _ {F^{-1}(q)}).
\]This F _ \ast X:N\to TN is also smooth since F _ \ast X=\mathrm dF\circ X\circ F^{-1}. The uniqueness is easy to see.

We call this F _ \ast X the pushforward of X by F.
3. Lie brackets If X,Y are smooth vector fields on M. The function f\mapsto YXf is not necessarily a vector field. However, a surprising fact is that XY-YX is a vector field.

We prove some of these propositions. To show [X,Y] is a vector field it suffices to show it is a derivation.

\begin{align*}
& [ X , Y ] (fg)=X(f Y g+g Y f)-Y(f X g+g X f) \\
={} & X f Y g+f X Y g+X g Y f+g X Y f-Y f X g-f Y X g-Y g X f-g Y X f \\
={} & f X Y g+g X Y f-f Y X g-g Y X f = f [ X , Y ] g+g [ X , Y ] f .
\end{align*}

Proposition 10 is obvious, for mixed partial derivatives of smooth functions commute. For the Jacobi identity, let \sum be the cyclic sum and compute:
\begin{align*}
& \sum[X,[Y,Z]]f=\sum (X[Y,Z]f-[Y,Z]Xf) \\
={} & \sum (XYZf-XZYf-YZXf+ZYXf)=0.
\end{align*}
To prove Proposition 13, we use Proposition 6 and
\begin{gather*}
X _ {1} X _ {2}(f \circ F)=X _ {1}((Y _ {2} f) \circ F)=(Y _ {1} Y _ {2} f) \circ F , \\
X _ 2X _ 1(f\circ F)=(Y _ 2Y _ 1f)\circ F,
\end{gather*}which implies
\[
[X _ 1,X _ 2](f\circ F)=(Y _ 1Y _ 2f)\circ F-(Y _ 2Y _ 1f)\circ F=([Y _ 1,Y _ 2]f)\circ F.
\]

4. Integral curves and flows The integral curves of a vector field are smooth curves whose velocity at each point is the given vector field at that point.

If 0\in J, we call \gamma(0) the starting point of \gamma. A integral curve is maximal if its domain cannot be extended to a larger interval.

For example, consider the vector field W=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y} on \mathbb R^2. If \gamma(t)=(x(t),y(t)) is a smooth curve, the condition for it to be an integral curve, \gamma'(t)=W _ {\gamma(t)}, is
\[
x'(t) \frac{\partial}{\partial x}\Big| _ {\gamma(t)}+y'(t) \frac{\partial}{\partial y}\Big| _ {\gamma(t)}=x(t) \frac{\partial}{\partial y}\Big| _ {\gamma(t)}-y(t) \frac{\partial}{\partial x}\Big| _ {\gamma(t)}.
\]Therefore we need to solve the system of ordinary differential equations
\begin{align*}
x'(t) & =-y(t), \\
y'(t) & =x(t).
\end{align*}The solution is x(t)=a\cos t-b\sin t, y(t)=a\sin t+b\cos t. Thus if a,b are neither zero it is an integral starting at p=(a,b) and obtained by rotating this point counterclockwise about the origin at angle t.

In this example, we also have \gamma _ s(\gamma _ t(p))=\gamma _ {s+t}(p).

\Rightarrow: F\circ\gamma is an integral curve because
\[
(F\circ\gamma)'(t)=\mathrm dF _ {\gamma(t)}(\gamma'(t))=\mathrm dF _ {\gamma(t)}(X _ {\gamma(t)})=Y _ {F(\gamma(t))}.
\]\Leftarrow: Given p\in M. It can be shown there is an integral curve \gamma:(-\varepsilon,\varepsilon)\to M of X starting at p (the existence of which will be discussed later). Then F\circ\gamma is an integral curve of Y starting at F(p), so it is straightforward to see that Y _ {F(p)} =\mathrm dF _ p(X _ p), which shows X,Y are F-related.

Flow

From the example above, we can see in order to find integral curves given a vector field a system of ODE need to be solved. From the knowledge of ODE, it is said that such a system always has a unique solution.

The uniqueness means if z satisfies the same differential equation, then the domain of z is a subset of (a(p _ 0),b(p _ 0)) and z agrees with y in it.

Next we study the dependence of an integral curve on its initial point. The function y is now written as y(t,q). The condition for y to be an integral curve starting at p is \frac{\partial}{\partial t}(t,q)=f(y(t,q)), with initial condition y(0,q)=q. Again we need the following theorem from ODE:

From this theorem, given a smooth vector field X and a point p in a chart U, there is a unique F(t,q), as an integral curve of X, starting at p for each q in some neighborhood of p. We usually write F _ t(q) for F(t,q).

As functions of t, both F _ t(F _ s(q)) and F _ {t+s}(q) are integral curves starting at F _ s(q). By the uniqueness, we have
\[
F _ t(F _ s(q))=F _ {t+s}(q).
\]We take the following definition.

Given a smooth vector field X, a unique F is determined, called a local flow generated by X. The function F _ t(q) of t is called a flow line of the local flow. Each flow line is an integral curve of X. If (-\varepsilon,\varepsilon)=\mathbb R, the flow is called a global flow, and in this case the vector field is called a complete vector field.