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This is a problem that comes from Jost's Riemannian Geometry and Geometric Analysis (Chapter 1, Exercise 16).

We consider the constant vector field $X(x)=a$ for all $x\in\mathbb{R}^{n+1}$. We obtain a vector field $\tilde{X}(x)$ on $\mathbb{S}^n$ by projecting $X(x)$ onto $T_x\mathbb{S}^n$ for $x\in\mathbb{S}^n$. Determine the corresponding flow on $\mathbb{S}^n$.

I've been stuck on this for a while, and haven't been sure exactly what to try. This is the partial work that I have so far:

Let $\mathbf{1}_{n+1}$ be the $(n+1)\times(n+1)$ identity matrix. Note that the projection onto the tangent space $T_x\mathbb{S}^n$ is simply given by $$P_x:=\mathbf{1}_{n+1}-xx^\top:\mathbb{R}^{n+1}\to T_x\mathbb{S}^n.$$ Now we see that $$\tilde{X}(x)=P_x(X(x))=P_x(a)=(\mathbf{1}_{n+1} - xx^\top)a = a - xx^\top a = a - \langle{a,x}\rangle x\in T_x\mathbb{S}^n.$$ So for any $p\in\mathbb{S}^n$ we are looking for a solution to the following ODE $$\begin{cases}\dot\gamma(t)=\tilde{X}(\gamma(t))=a-\langle{a,\gamma(t)}\rangle\gamma(t),\\\gamma(0)=p.\end{cases}$$ We have two fixed point solutions given by $\gamma_{\pm}(t)=\pm\frac{a}{\|a\|}.$ I'm pretty sure (although I haven't shown this yet) that if we start from any point that isn't a fixed point that the flow will tend to $\frac{a}{\|a\|}$.

At first I thought that the flow might be related to the projection of the flow of $X$ in $\mathbb{R}^{n+1}$, i.e. take $\Gamma:\mathbb{R}\to\mathbb{R}^{n+1}$ given by $\Gamma(t)=p + ta$, and then see if $\gamma(t):=\frac{\Gamma(t)}{\|\Gamma(t)\|}$ solves the ODE. Unfortunately, it doesn't solve it on $\mathbb{S}^1\subseteq\mathbb{R}^2$ (although a reparameterization would probably do the trick in this case since $\mathbb{S}^1$ is $1$-dimensional and the direction of the flow seems to be correct), and in any higher dimensions it doesn't even flow in the right direction. So I'm pretty sure this idea is incorrect, and I wasn't able to get it to work at all.

The second idea I had was to consider the geodesic starting at $p$ and in the direction $\dot\gamma(0)=a-\langle{a,p}\rangle p\in T_p\mathbb{S}^n$, and maybe reparameterize it if need be. Unfortunately, this also didn't work and I wasn't able to figure out whether the reparameterization idea would work here.

Other than that, I've been pretty stuck and I'm not sure if I'm supposed to be able to just be able to solve this ODE directly or if there is some geometric insight that I can use to determine this flow. Any hints and explanations would be extremely appreciated! I've been working on this for a long time, and this is the only question in this chapter I haven't been able to figure out.

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    $\begingroup$ Have you thought about what it might look like if $a$ is the vector from the origin to the North Pole? $\endgroup$ – Steve D Jul 20 '17 at 5:51
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First, choose cartesian coordinates $(x_0,x_1,...,x_n)$ on $\mathbb{R}^{n+1}$ such that $\vec{a} = (a,0,0,...,0)$, $a\in \mathbb{R}$, and such that $\mathbb{S}^n$ is given by $\sum_{k=0}^n x_k^2 = 1$. Then, by symmetry argument (i.e. an eventual rotation around the $\vec{a}$ axis), all you have to do is solve the equation EDO : $$\frac{d \gamma_0(t)}{d t} = a - a \gamma_0^2(t)$$ $$\frac{d \gamma_1(t)}{d t} = -a\gamma_0(t)\gamma_1(t)$$ on the half-circle $\{(x_0,x_1,0,0,...,0)|x_0^2 + x_1^2 = 1, x_1 \geq 0\}$. You can solve the first equation. Then you can solve the second using the solution of the first. Now take the solution on the half circle and act on it using a $O(n)$ group action preserving the $\vec{a}$-axis and sending the equator $\mathbb{S}^{n-1}$ of $\mathbb{S}^n$ to itself. Now you have all the integral curves you were looking for (which then implies the flow).

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  • $\begingroup$ Instead of just solving it on the half-circle, couldn't we solve it on the the entire circle $\{(x_0,x_1,0,\dots,0)~:~x_0^2 + x_1^2=1\}$ and then consider the actions of $SO(n+1,R)$ which preserves the $\vec{a}$-axis on these solutions to obtain the flow map? Or is there something I'm missing regarding why you chose the half-circle and $GL(n+1,R)$ actions? $\endgroup$ – pomegranate Jul 20 '17 at 15:33
  • $\begingroup$ The reason I'm not so sure about the half-circle is that when we are solving it on the circle $S^1\subseteq R^2$ I'm not sure what $GL(2,R)$ action would allow us to transform our solutions on the half-circle to those on the entire circle. Furthermore, it seems like we can solve the ODE directly with any initial conditions on the circle $\{x_0^2 + x_1^2 = 1\}$. $\endgroup$ – pomegranate Jul 20 '17 at 15:36
  • $\begingroup$ Oh ! Sorry I didn't mean $GL(n,\mathbb{R})$ but $O(n)$ (we need to preserve the sphere), I'll edit that. Also, the solution on both side of the circle isn't necessary because the solution on the left side of the circle is given by reflection of the one on the right. Note that you don't need the full $O(n+1)$ group but just the $O(n)$ subgroup sending the equator to itself. $\endgroup$ – Noé AC Jul 20 '17 at 16:09
  • $\begingroup$ Great! That all makes sense! I made the mistake of thinking that the half-circle you were considering had the constraint that $x_0\geq 0$, instead of $x_1\geq 0$. $\endgroup$ – pomegranate Jul 20 '17 at 18:56

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