Finding the flow lines of a given vector field on $S^2$

Question

Good morning!

I have a quick question which I couldn't find the answer to.

Say we have a vector field $\eta : S^2 \to TS^2$ which takes a point $(x,y,z) \in S^2$ to the tangent vector $(-y, x, 0)$.

I am trying to figure out the flow lines of $\eta$, ie, fixing $(x,y,z)$, I'm trying to find a curve $\gamma$ through $(x,y,z$) such that $\gamma'(0) = (-y, x, 0)$.

I was just thinking to set: $\gamma : t \mapsto (x - yt, y + xt, z)$, which does satisfy the property above... but I want it to be a curve in $S^2$, and I do not know how to enforce that, because unless $x \ne y$ or $t = 0$, $||\gamma(t)||^2 \ne 1$.

Answer 1

In addition to Lee Mosher's answer here is another way of finding the flow line of this particular vector field: You wish to find $\gamma\colon (-\varepsilon,\varepsilon) \to \mathbb{S}^2$ such that $\gamma(0)=p_0=(x_0,y_0,z_0)^T$ and $$ \begin{align*} \dot{\gamma}(0) = \begin{pmatrix} -y_0 \\ x_0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} = L_z \cdot p_0. \end{align*} $$ Now why did I write this as a matrix-vector multiplication? The reason is the following:
The special orthogonal group $\mathrm{SO}(3)$ operates on $\mathbb{S}^2$ by left-multiplication. Its Lie algebra $\mathfrak{so}(3)$ is the set of skew-symmetric $(3\times 3)$-matrices. One basis is given by $$ \begin{align*} \mathfrak{so}(3) =\mathrm{span}\{L_x,L_y,L_z\} = \mathrm{span} \left\{ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \right\}. \end{align*} $$ Here $L_x, L_y, L_z$ are the infinitesimal generators of rotations about the $x$-, $y$- and $z$-axis.
(I wrote earlier that $\mathfrak{so}(3)$ is isomorphic to $\mathbb{R}^3$. This is not only the case as same-dimensional vector spaces but as Lie algebras. This means that the Lie bracket needs to transform naturally. Choosing the above basis you can directly compute that $\mathfrak{so}(3)$ with the Lie bracket $[A,B]_{\mathfrak{so}(3)}=AB-BA$ is isomorphic to the Lie algebra $\mathbb{R}^3$ with the Lie bracket given by the cross product, i.e. $[X,Y]_{\mathbb{R}^3}=X\times Y$. Note that this is a special feature of $\mathbb{R}^3$ and doesn't hold in different dimensions!)

You can now use the exponential map $\exp\colon \mathfrak{so}(3)\to \mathrm{SO}(3), A \mapsto \sum_{k=0}^{\infty} \tfrac{1}{k!}A^k$ to explicitly calculate the rotations around the corresponding axes about an angle $\theta$ as $R_i(\theta)=\exp(\theta L_i)$ for $i\in\{x,y,z\}$. For example, $$ \begin{align*} R_z(\theta)= \exp(\theta L_z) = \begin{pmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}. \end{align*} $$ You can check here directly that $\dot{R_z}(0) = L_z$. More generally, this is because the differential of the exponential map at the origin of the Lie algebra is the identity map.

The flow line of your vector field (which corresponds to $L_z$) is then given by the action of $\mathrm{SO}(3)$ on $\mathbb{S}^2$, i.e. left-multiplication: $$ \begin{align*} \gamma\colon (-\pi, \pi) \to \mathbb{S}^2, \theta \mapsto R_z(\theta)\cdot p_0= \begin{pmatrix} \cos(\theta)x_0-\sin(\theta)y_0 \\ \sin(\theta)x_0+\cos(\theta)y_0 \\ z_0 \end{pmatrix} \end{align*} $$ which is of course a parametrisation of the circle of latitude in Lee Mosher's answer :)

Edit: You might notice that if you calculate the first-order Taylor approximation of the flow line which I wrote down then you will get the one that you wrote down in your original question. Maybe some geometric intuition here is that the exponential map of the group action guarantees that you "stay on the manifold" which is being acted on (here $\mathbb{S}^2$). Linear approximation of the exponential map will make you "leave your manifold" straight away. Of course, this only makes sense if your manifold is embedded in some higher-dimensional ambient manifold. I hope that makes sense.

Answer 2

Let's consider the point $p_0 = (x_0,y_0,z_0)$ at which the tangent vector is $\eta(p_0)=(-y_0,x_0,0)$.

Noticing that the $z$-coordinate of $\eta(p_0)$ is zero, one sees that $\eta(p_0)$ is parallel to the $xy$-plane, and in fact $\eta(p_0)$ is tangent to the horizontal plane $P_{z_0} = \{(x,y,z_0) \mid x,y \in \mathbb R\}$. The same is true for each $p \in S^2 \cap P_{z_0}$: the tangent vector $\eta(p)$ is tangent to the plane $P_{z_0}$; it follows that $\eta(p)$ is tangent to the "latitude circle" $S^2 \cap P_{z_0}$. But this proves that the latitude circle $S^2 \cap P_{z_0}$ is the flow line. And this argument is independent of $z_0$. Thus, the foliation of $S^2$ by flow lines is identical to the folation by latitude circles (with the north and south poles as stationary points of the flow).