Vector fields and the tangent bundle.

We have seen how to define tangent vectors at a point of a manifold. In many problems we are interested in vector fields, that is a choice of vector at every point of a manifold. We can think of this in the following manner. Take the union of all the tangent spaces, denote it by

$$TM = \bigcup_{x \in M} T_xM$$

and call it the tangent bundle to $M$. There is an important map $\pi\colon TM \to M$ called the projection that sends a vector $X \in T_xM$ to the point $\pi(X)=x$ at which it is located. A vector field is a map $X \colon M \to TM$ with the special property that $X(x) \in T_xM$. This property can be also written as $\pi\circ X =$   id$_M$, that is $\pi(X(x)) = x$. Such a map $X \colon M \to TM$ is called a section of the projection map $\pi$. We want to consider smooth vector fields and as we already have a notion of smooth function between manifolds the simplest way to define smooth vector fields is to make $TM$ a manifold. To do this involves a construction that we will use again later so we ill state it in more general form that immediately necessary.

Let $E$ be a set with a surjection $\pi \colon E \to M$ where $M$ is a manifold. Denote by $E_x$ the fibre of $E$ over $x$, that is the set $\pi^{-1}(x)$. Let $V$ be a finite dimensional vector space. Assume that we can cover $M$ by co-ordinate charts $\{(U_\alpha , \psi_\alpha \}_{\alpha \in I}$ such that for every $\alpha \in I$ and for every $x \in U_\alpha$ there is a bijection

$$\phi_a(x) \colon E_x \to V$$

such that the map

$$\begin{array}{ccc} U_\alpha \cap U_\b & \to & GL(V) \\ x & \mapsto & \phi_a(x)\circ \phi_b(x)^{-1} \end{array}$$

is smooth where $GL(V)$ is the group of all linear isomorphisms of $V$. Then it is possible to make $E$ a manifold as follows. We define bijections

$$\begin{array}{ccccc} \chi_a &\colon &\pi^{-1}(U)& \to &U \times V\\ & & x & \mapsto &(\pi(x), \phi_a(\pi(x))v) \end{array}$$

To make these into charts we should really identify $V$ with some $\mathbb{R}^k$ but we will not bother to do that. To check compatibility we note that $\chi_\alpha (U_\alpha \cap U_\b ) = U_\alpha \cap U_\b\times V$ which is open in $\mathbb{R}^n \times V$. Likewise for $\chi_\b (U_\alpha \cap U_\b )$. Then the map we want to check is smooth is the map

$$U_\alpha \cap U_\b\times V \to U_\alpha \cap U_\b\times V$$

which sends $(x, v)$ to $(x, \phi_a(x)\circ \phi_b(x)^{-1}v)$ and this is smooth and invertible. By interchanging $\alpha$ and $b$ we deduce that this map is a diffeomorphism. Hence we have made $E$ into a manifold. Notice that with this manifold structure the map $\chi_\alpha$ is a diffeomorphism, as the co-ordinate charts of a manifold are diffeomorphisms. Notice also that each $E_x$ is a vector space from Lemma ??. Moreover it easy to check that the addition and scalar multiplication are smooth. Define a section of $\pi \colon E \to M$ to be a smooth map $s \colon M \to E$ which satisfies $s(x) \in E_x$ for all $x \in M$. If $s$ is such a section then on restriction to $U_\alpha$ we can define a map $s_a\colon U \to V$ by $s_\alpha (x) = \phi_\alpha (x)(s(x))$. The $s_a$ are clearly smooth. The converse is also true if $s$ is any map and the $s_a$ defined in this way are smooth then $s$ is smooth.

Consider now the case of the tangent bundle. Let $(U_\alpha, \psi_\alpha)$ be a co-ordinate chart on $M$. Then $V =R^n$ and $\phi_a(x) = d\psi_a(x)$. The condition we require to hold is that the map

$$x \mapsto d\phi_\b (\psi^{-1}(x)) \circ d\psi_\alpha ^{-1}(x) = d(\psi_\b\circ\psi^{-1}_a) (x) = d_i(\psi^j_b\circ\psi^{-1}_\alpha )(x)$$

is smooth. But this is just the Jacobian matrix of partial derivatives which depends smoothly on $x$.

We can now define

Definition B.1   A smooth vector field on a manifold $M$ is a smooth section of the tangent bundle.

To understand what it means to be smooth in terms of co-ordinates recall the definition of $d\psi(x)$. We have the co-ordinate vector fields $(\partial/\partial\psi^i)(x)$ for $i = 1, \dots, n$. Then

$$d\psi(x)(\frac{\partial\phantom{\psi^i}}{\psi^i})(x) = e^i$$

where $e^i$ is the standard basis vector of $\mathbb{R}^n$. So clearly this is a smooth map so that the co-ordinate vector fields are smooth.

More generally if $X$ is a vector field we can write it as

$$X(x) = \sum_{i=1}^n X^i(x) \frac{\partial\phantom{\psi^i}}{\psi^i}(x)$$

for any $x \in U$, and functions $X^i\colon U \to \mathbb{R}$. Then

$$d\psi(x)(X(x)) = (X^1(x), \dots, X^n(x)).$$

This proves

Proposition B.1   Let$X$ be a vector field on a manifold $M$. Then if $X$ is smooth and $(U, \psi)$ is a co-ordinate chart then if we let

$$X(x) = \sum_{i=1}^n X^i(x) \frac{\partial\phantom{\psi^i}}{\psi^i}(x)$$

the functions $X^i\colon U \to \mathbb{R}$ are smooth. Conversely if $X$ is a vector field and we can cover $M$ with co-ordinate charts $(U, \psi)$ such that the corresponding $X^i\colon U \to \mathbb{R}$ are smooth then $X$ is smooth.