Vector fields

 Preliminary exam prep

Here we will have an overview of vector fields and all things related to them. Let $M$ be an $n$-dimensional manifold, and $\pi:M\to TM$ its tangent bundle.

Definition: A vector field is a map $X:M\to TM$ such that $\pi\circ X = \id_M$.

A vector field may also be viewed as a section of the tangent bundle, and smooth vector fields as the space of smooth sections $\Gamma(TM)$. Given a chart $(U,\varphi)$ of $M$ near $p$, we have the pushforward $\varphi_*:T_pM\to T_{\varphi(p)}(\R^n) = \R^n$, where we may assume $\varphi(p)=0$. Given the standard basis $\{e_i\}$ of $\R^n$, we get a basis of $T_pM$ given by
$$\left\{\left.\frac{\dy}{\dy x_i}\right|_p = (\varphi_*)^{-1}(e_i)\right\}_{i=1}^n.$$
Recall that $TM$ may be viewed as the space of derivations, or maps $C^\infty(M)\to \R$ satisfying the Leibniz rule. Then for $p\in M$, we have $X(p):C^\infty(M)\to \R$, so we have $X(p)(f) = X_p(f)\in \R$ for all $f\in C^\infty(M)$. Hence $X_p\in T_pM$, and $X(f)\in C^\infty(M)$. Briefly,
$$\begin{array}{r c l} f\ :\ M & \to & \R, \\ X\ :\ M & \to & TM, \end{array} \hspace{2cm} \begin{array}{r c l} Xf\ :\ M & \to & \R, \\ fX\ :\ M & \to & TM. \end{array}$$

Definition: Given a vector field $X\in \Gamma(TM)$, an integral curve of $X$ is a smooth curve $\gamma:\R \to M$ such that $\gamma'(t) = X_{\gamma(t)}$ for all $t\in \R$.

The domain of $\gamma$ need not be all of $\R$, though any integral curve may be extended to a maximal integral curve, for which the domain can not be made larger. A collection of integral curves for a particular vector field is a flow.


Definition: A flow, or a one paramater group of diffeomorphisms, is a smooth map $\psi:\R\times M\to M$ such that

1. $\psi(t,\cdot)$ is a diffeomorphism of $M$, for all $t$,
2. $\psi(0,\cdot) = \id_M$,
3. $\psi(s+t,\cdot) = \psi(s,\cdot)\circ \psi(t,\cdot)$.

For convenience, we write $\psi_t(p) = \psi(t,p)$, Note that fixing $p\in M$, the map $\psi(\cdot,p)$ is a integral curve. Moreover, flows and vector fields are related uniquely by
$$\left.\frac{d f}{d t} \psi_t(p)\right|_{t=0} = X_p(f).$$
Indeed, if we have a flow $\psi$ and an element $f\in \Hom(T^*_pM,\R)$, this gives us a vector field $X\in \Gamma(TM)$. Conversely, if we have a vector field $X$, by the existence and uniqueness of solutions to first order ordinary differential equations (with boundary conditions), we can find a $\psi$ that satisfies this equality.

Definition: Let $X,Y\in \Gamma(TM)$ and $\psi$ be the associated flow of $X$. The Lie derivative of $Y$ in the direction of $X$, or Lie bracket of $X$ and $Y$, is an element of $\Gamma(TM)$ given by
$$\begin{align*} \left(\mathcal L_XY\right)_p(f) & = \left.\frac{df}{dt}\right|_{t=0}\bigg((\psi_t)_*^{-1}(Y_{\psi_t(p)}(f))\bigg) \\ & = [X,Y]_p(f) \\ & = X_p(Y(f)) - Y_p(X(f)) \end{align*}$$

The Lie derivative has some properties, among them $\mathcal L_X(fY) = X(fY) + f(\mathcal L_XY)$ for any $f\in C^\infty(M)$. If we let $Y$ be the map $M\to TM$ given by
$$\begin{array}{r c l} Y\ :\ M & \to & \Hom(T^*M,\R),\\ p & \mapsto & \left(\begin{array}{r c l} f_p\ :\ C^\infty(M) & \to & \R, \\ g & \mapsto & g(p), \end{array}\right), \end{array}$$
then $Yf = f$, so $\mathcal L_XY = X-X = 0$, and we have $\mathcal L_X f = Xf$.

Remark: Vector fields are 1-forms, or elements of $\mathcal A^0_M(TM) = \Gamma(TM\otimes \bigwedge^0T^*M) = \Gamma(TM)$. We may generalize the definition above to consider the Lie derivative $\mathcal L_X\omega$ of a differential $k$-form $\omega$ . Note that a differential $k$-form takes in $k$ vector fields and gives back a smooth function $M\to \R$. With this in mind, we may define new operations on vector fields:
$$\begin{align*} (\mathcal L_X\omega)(Y_1,\dots,Y_k) & = \mathcal L_X(\omega(Y_1,\dots,Y_k)) - \sum_{i=1}^k\omega(Y_1,\dots,\mathcal L_XY_i,\dots,Y_k) \\ (d\omega)(Y_1,\dots,Y_{k+1}) & = \sum_{i=1}^{k+1}(-1)^{i-1}Y_i(\omega(Y_1,..,\widehat{Y_i},..,Y_{k+1})) + \sum_{j>i=1}^{k+1}(-1)^{i+j}\omega([Y_i,Y_j],Y_1,..,\widehat{Y_i},..,\widehat{Y_j},..,Y_{k+1}) \\ (i_X\omega)(Y_1,\dots,Y_{k-1}) & = \omega(X,Y_1,\dots,Y_{k-1}) \end{align*}$$

The last is the interior product. All three are related by Cartan's formula $\mathcal L_X\omega = d(i_X\omega)+i_X(d\omega)$:
$$\begin{align*} (\mathcal L_{Y_1}\omega)(Y_2,\dots,Y_{k+1}) & = Y_1(\omega(Y_2,\dots,Y_{k+1})) - \sum_{i=2}^{k+1}\omega(Y_2,\dots,[Y_1,Y_i],\dots,Y_k) \\ & = Y_1(\omega(Y_2,\dots,Y_{k+1})) - \sum_{i=2}^{k+1}(-1)^i\omega([Y_1,Y_i],Y_2,\dots,\widehat{Y_i},\dots,Y_k) \\ (d(i_{Y_1}\omega))(Y_2,\dots,Y_{k+1}) & =  \sum_{i=2}^{k+1}(-1)^iY_i(\omega(Y_1,..,\widehat{Y_i},..,Y_{k+1})) - \sum_{j>i=2}^{k+1}(-1)^{i+j}\omega([Y_i,Y_j],Y_1,..,\widehat{Y_i},..,\widehat{Y_j},..,Y_{k+1})\\ (i_{Y_1}(d\omega))(Y_2,\dots,Y_{k+1}) & = (d\omega)(Y_1,\dots,Y_{k+1}) \\ & = \sum_{i=1}^{k+1}(-1)^{i-1}Y_i(\omega(Y_1,..,\widehat{Y_i},..,Y_{k+1})) + \sum_{j>i=1}^{k+1}(-1)^{i+j}\omega([Y_i,Y_j],Y_1,..,\widehat{Y_i},..,\widehat{Y_j},..,Y_{k+1}) \end{align*}$$

Remark: The action of a $k$-differential form on a $k$-vector field is given by $$\left(dx_1\wedge \cdots \wedge dx_k\right)\left(\frac\dy{\dy y_1},\dots,\frac\dy{\dy y_p}\right) = \det\begin{bmatrix} dx_1\frac\dy{\dy y_1} & dx_1\frac{\dy}{\dy y_2} & \cdots & dx_1\frac\dy{\dy y_p} \\ dx_2\frac\dy{\dy y_1} & dx_2\frac{\dy}{\dy y_2} & \cdots & dx_2\frac\dy{\dy y_p} \\ \vdots & \vdots & \ddots & \vdots \\ dx_p\frac\dy{\dy y_1} & dx_p\frac{\dy}{\dy y_2} & \cdots & dx_p\frac\dy{\dy y_p} \end{bmatrix} = \det\left(dx_i\frac\dy{\dy y_j}\right).$$ This may be generalized to get a map $\wedge^k T^*M \oplus \Gamma(TM)^{\oplus \ell} \to \bigwedge^{k-\ell}T^*M$, for $\ell\leqslant k$. For example, given a basis $x,y$ of our space $M$, $$(dx\wedge dy)\left(x\frac\dy{\dy x} + y \frac\dy{\dy y}\right) = dx\left(x\frac\dy{\dy x} + y \frac\dy{\dy y}\right)dy - dy\left(x\frac\dy{\dy x} + y \frac\dy{\dy y}\right)dx = x\ dy - y\ dx.$$ When $\ell=1$, this is just the interior product.

References: Lee (Introduction to smooth manifolds, Chapter 8), Hitchin (Differentiable manifolds, Chapter 3)