# Vector Field Flow

In the last post, we finally explained how to differentiate a smooth map between manifolds, justifying the phrase a manifolds is a space on which one can do calculus. We have only scratched the surface. There is much more to say about manifolds. Before we can go deeper, we need a better understanding of vector fields. Let $$X$$ be a manifold. Recall that a vector field is a section of the tangent bundle $$TX$$. If $$x^1,\dots,x^d$$ are coordinates on $$U \subseteq X$$, then the vector fields on $$U$$ look like $f^i \frac{\partial}{\partial x^i}$ (we are using the Einstein summation convention). One way to construct vector fields on $$X$$ is by describing them on coordinate patches and checking that they change coordinates correctly. For example, consider the coordinates $$\theta : S^1 \backslash \{ (1,0) \} \to (0,2 \pi)$$ and $$\phi : S^1 \backslash \{ (-1,0) \} \to (-\pi,\pi)$$.

Then $$d \phi / d \theta = 1$$ so $$d / d \theta = d / d \phi$$ where both vector fields are defined. Therefore we have a globally defined vector field which we call $$d / d \theta$$. If you need many coordinate charts to cover your manifold, then constructing vector fields in this way can be very time consuming.

### Local group actions

Consider the manifold $$\mathbb{R}$$. The addition map $$\mathbb{R} \times \mathbb{R} \to \mathbb{R}, (x,y) \mapsto x + y$$ is smooth. That means that $$\mathbb{R}$$ is a Lie group. Lie groups are the most important objects in differential geometry and we will study them in general later, but for now let us just focus on the real line. If $$X$$ is a manifold, we can talk about smooth group actions $$\mathbb{R} \times X \to X$$. Remember that group actions satisfy the following laws: $s \cdot (t \cdot x) = (s + t) \cdot x \qquad 0 \cdot x = x$ Often in differential geometry, asking for group actions defined for all $$t \in \mathbb{R}$$ is too much. Instead, we talk about local group actions $$(-\epsilon,\epsilon) \times X \to X$$ which are defined for $$t$$ in an open neighborhood of the identity. We still require the same laws as a total group action, but only when the equations make sense. Every local group action of $$\mathbb{R}$$ on $$X$$ gives a vector field defined by $$x \mapsto t \cdot x$$.

For example, consider the action $$\mathbb{R} \times S^1 \to S^1$$ defined by $$t \cdot (\cos \theta, \sin \theta) = (\cos (\theta + t),\sin(\theta + t))$$. In coordinates, the action is given by $$t \cdot \theta = \theta + t$$. Since $\frac{d}{dt}_{t=0} (\theta + t) = 1$ it follows that the vector field corresponding to this action is $$d/d\theta$$. Therefore, we have constructed $$d/d\theta$$ globally by defining a group action rather than by patching local vector fields together.

### Every vector field comes from a local group action

Constructing vector fields from local group actions is nice. The amazing fact is that every vector field can be constructed in this way. This is our second example of an integrability result (the first example was the inverse function theorem). Let $$V$$ be a vector field on the manifold $$X$$. For each $$p \in X$$, we want to construct a curve $$\gamma_p(t) \in X$$ defined for $$t \in (-\epsilon,\epsilon)$$ such that $\gamma_p'(t) = V(\gamma_p(t)) \qquad \gamma_p(0) = p$ Choose a coordinate chart $$x^1,\dots,x^d$$ around $$p$$ such that $$p = (0,\dots,0)$$. If we write $$V = f^i \frac{\partial}{\partial x^i}$$, then we are faced with the following initial value problem: ${\bf x}'(t) = {\bf f}({\bf x}(t)) \qquad {\bf x}(0) = 0$ Since $$f$$ is smooth, the Picard-Lindelof theorem from functional analysis tells us that there is a unique solution defined for $$t \in (-\epsilon,\epsilon)$$. These curves give us a map $$(-\epsilon,\epsilon) \times X \to X$$ defined by $$(t,x) \mapsto \gamma_x(t)$$. By definition, $$0 \cdot x = \gamma_x(0) = x$$. Also, since solutions are unique, we must have $\gamma_{\gamma_x(s)}(t) = \gamma_x(t+s)$ which translates to $$t \cdot (s \cdot x) = (t+s) \cdot x$$. This local group action is often called the flow of $$V$$. We shall often use the notation $$L_{t,V}(x) = t \cdot x$$. In words, $$L_{t,V}(x)$$ flows the point $$x$$ along the vector field $$V$$ for $$t$$ seconds.

### Lie Derivatives

Now that we know every vector field comes from a local group action of $$\mathbb{R}$$, we can start doing some tricky things with vector fields. Let $$X$$ be a manifold, $$V$$ a vector field and $$f : X \to \mathbb{R}$$ a smooth function. Then $V(f) = \lim_{t \to 0} \frac{L_{t,V}^* f -f}{t}$ where $$L_{t,V}^* f(x) = f(L_{t,V}(x)) = f(t \cdot x)$$. This formula is not new. We already discussed it in the tangent spaces post. Motivated by this, we can do something which is new. Let $$U,V$$ be vector fields. Then define $\mathcal{L}_V U = \lim_{t \to 0} \frac{U - L_{t,V}'(U) }{t}$ This is a vector field on $$X$$ called the Lie derivative of $$U$$ along $$V$$. We can compute $$\mathcal{L}_V U$$ as follows: $(\mathcal{L}_V U)_x(f) = \lim_{t \to 0} \frac{U_x(f) - L'_{t,V}(U_{L_{-t,V}x})(f)}{t} = \lim_{t \to 0} \frac{U_x(f)-U_{L_{-t,V}x}L^*_{t,V} f}{t} =$ $\lim_{t \to 0} \frac{U_x(f)-U_{L_{-t,V}x}(f) + U_{L_{-t,V}x}(f) - U_{L_{-t,V}x}L^*_{t,V} f}{t} = \lim_{t \to 0} \frac{Uf(x) - L_{-t,V}^* Uf(x)}{t} + \lim_{t \to 0} \frac{U_{L_{-t,V}x}(f - L_{t,V}^*f)}{t} = V_xUf - U_x Vf$ More cleanly, $$\mathcal{L}_V U = VU - UV = [V,U]$$. This makes sense because the commutator of two derivations is a derivation. We call $$[V,U]$$ the Lie bracket. If $$x^1,\dots,x^d$$ are coordinates on $$X$$ and $V = f^i \frac{\partial}{\partial x^i} \qquad U = g^j \frac{\partial}{\partial x^j}$ then $[V,U] = \left( f^i \frac{\partial g^k}{\partial x^i} - g^j \frac{\partial f^k}{\partial x^j} \right) \frac{\partial}{\partial x^k}$

### What does the Lie bracket mean?

It is tempting to say things like $$[V,U]$$ is the rate of change of $$U$$ in the direction of $$V$$. This is not correct. The vector $$[V,U]_x$$ is not determined by $$V_x$$ and $$U$$. You need to know both $$V$$ and $$U$$ in an open neighborhood of $$x$$. In fact, without further structure, there is no way to define the rate of change of $$U$$ along $$V$$ in a coordinate independent way. What $$[V,U]$$ really measures is the obstruction to the local group actions for $$V$$ and $$U$$ commuting with each other. We will make this precise when we start thinking about Lie theory, but we can understand it intuitively now. From the definition of the Lie derivative, we have $[tV,sU] = sU - L'_{t,V}(sU) \qquad \mod t^2,s^2.$ This equation is encoded in the following cartoon:

### Frobenius's Theorem

Let $$X$$ be a manifold and $$E \subseteq TX$$ a rank $$r$$ sub bundle of the tangent bundle. We call the coordinate chart $$x^1,\dots,x^d$$ on $$U$$ adapted to $$E$$ if $$\partial / \partial x_1, \dots, \partial / \partial x_r$$ are a basis for $$E$$. The Frobenius theorem says that we can cover $$X$$ with coordinate patches adapted to $$E$$ if and only if $$E$$ is closed under the Lie bracket. This is the third integrability result we have encountered! The Frobenius theorem will be important later, but its role in this post is to demonstrate that the Lie bracket contains interesting geometric information about $$X$$. The proof is by induction on $$r$$.

Suppose $$r = 1$$. Then $$E \subseteq TX$$ is a line bundle. Fix $$p \in X$$. Then we can choose coordinates $$x^1,\dots,x^d$$ around $$p \in U \subseteq X$$ such that $$E \lvert_U$$ is spanned by the nowhere vanishing vector field $$V$$. Consider the local action $$a : \mathbb{R} \times U \dashrightarrow U$$ generated by $$V$$ (we use the dashed arrow to indicate that $$a$$ is only defined in an open neighborhood of $$\{ 0 \} \times U$$). The jacobian $$a'_{(0,0)}$$ has $$V_p \not= 0$$ in the first column and a $$d \times d$$ identity matrix in the last $$d$$ columns. By permuting the $$x^i$$, we can assume that the first $$d$$ columns of $$a'_{(0,0)}$$ are invertible. The inverse function theorem implies that $$t,x_1,\dots,x_{d-1}$$ are coordinates in a small neighborhood of $$p$$ via the map $$(t,x_1,\dots,x_{d-1}) \mapsto a(t,x_1,\dots,x_{d-1},0)$$. Then $$V = \partial / \partial t$$ in this coordinate chart. Since our choice of $$p$$ was arbitrary, we can cover $$X$$ with charts adapted to $$E$$.

Now assume that $$E \subseteq TX$$ is a rank $$r$$ sub bundle. Fix $$p \in X$$. We can choose an open subset $$p \in U \subseteq X$$ and vector fields $$V_1,\dots,V_r$$ on $$U$$ which are a basis for $$E$$ at each point in $$U$$. By shrinking $$U$$, we can choose coordinates $$x^1,\dots,x^d$$ around $$p$$ such that $$V_r = \partial / \partial x_1$$. For $$j < r$$ we have $V_j = f^i_j \frac{\partial}{\partial x^i}$ By changing basis on $$E$$ (doing column operations), we can assume that $$f_j^1 = 0$$ for each $$j < r$$. Let $$F$$ be the sub bundle on $$U$$ spanned by $$V_1,V_2,\dots,V_{r-1}$$. Since $$E$$ is closed under the Lie bracket, $$f_j^1 = 0$$ implies that $$F$$ is closed under the Lie bracket. By induction, we can shrink $$U$$ to a smaller neighborhood of $$p$$ and find coordinates $$y^i$$ such that for $$j < r$$ we have $$V_j = \partial / \partial y^j$$ and $V_r = f^i \frac{\partial}{\partial y^i}$ By changing basis on $$E$$ again, we have $$f^i = 0$$ for $$i < r$$. If we shrink $$U$$ to a smaller neighborhood of $$p$$ and permute $$y^{r},y^{r+1},\dots,y^d$$, we can assume that $$f^r$$ does not vanish on $$U$$. If we redefine $$V_r = V_r / f^r$$, then $V_r = \frac{\partial}{\partial y^r} + \sum_{i > r} f^i \frac{\partial}{\partial y^i}$ which implies $[V_j,V_r] = \sum_{k > r} \frac{\partial f^k}{\partial y^j} \frac{\partial}{\partial y^k}$ Since $$E$$ is integral, this Lie bracket is a section of $$E$$ which implies that it equals $$0$$. Therefore $$f^i$$ only depends on $$y^r,y^{r+1},\dots,y^d$$. By the $$r=1$$ case, we can shrink $$U$$ to a smaller neighborhood of $$p$$ and replace the coordinates $$y^r,y^{r+1},\dots,y^d$$ with coordinates $$z^r,z^{r+1},\dots,z^d$$ so that $$V_r = \partial / \partial z^r$$. This completes the proof!

If $$E$$ is closed under the Lie bracket, we call it integrable. The Frobenius theorem says that if $$E$$ is an integrable subbundle, then it is the union of the tangent bundles of interleaved sub manifolds. This should be geometrically obvious in the case of a nowhere vanishing vector field, but it is harder to visualize in the higher dimensional case.

### Exercises

1. Write down the local group actions for the following vector fields:
• $$x d/dx$$
• $$x^2 d/dx$$
• $$x \partial / \partial y - y \partial / \partial x$$
2. The sub bundle spanned by $$U = \partial/ \partial x + y \partial / \partial z$$ and $$V = \partial / \partial y$$ is not integrable. Can you see why geometrically?