I was doing Lee's smooth manifold Problem 9-19.
Which is stated as follows:
9-19. Let $M$ be $\mathbb{R}^{3}$ with the $z$ -axis removed. Define $V, W \in \mathfrak{X}(M)$ by $$ V=\frac{\partial}{\partial x}-\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial z}, \quad W=\frac{\partial}{\partial y}+\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial z} $$ and let $\theta$ and $\psi$ be the flows of $V$ and $W$, respectively. Prove that $V$ and $W$ commute, but there exist $p \in M$ and $s, t \in \mathbb{R}$ such that $\theta_{t} \circ \psi_{s}(p)$ and $\psi_{s} \circ \theta_{t}(p)$ are both defined but are not equal.
I solve the ODE and gets the solution:
$$\theta_t\circ \psi_s = (p_1+t,p_2+s,\arctan(\frac{s+p_2}{p_1})+p_3-\arctan(\frac{p_2}{p_1}))\\\psi_s\circ\theta_t = (p_1+t,p_2+s,-\arctan(\frac{t+p_1}{p_2}) +p_3 + \arctan(\frac{p_1}{p_2}))$$
Which is obvious not equal,they both defined for all $(\Bbb{R}^3\setminus \{z\} )\times \Bbb{R}$,but it contradict to the theorem 9.44 that vector field commute if and only if flow commute?If I haven't made mistake in the computation.Is my computation correct,it's so hard to compute.Why does it not consistent with the theorem?
The point is that if we solve the ODE taking the first one for example it has 2 different representation,for starting point $p\in \Bbb{R}^3$ such that $p_2 \ne 0$.then the solution is :
$$\theta_t(p) = (p_1+t,p_2,p_3-\arctan((t+p_1)/p_2)+\arctan(p_1/p_2))$$
Such that $\mathcal{D}^{p} = \Bbb{R}$
The problem occurs when $p_2= 0$.In this case the solution of ODE is:
$$\theta_t(p) = (p_1+t,p_2,p_3)$$ as we can see when $t = -p_1$ it will hit the z-axis.
Let's consider another integral curve ,which has similar formula,the point is if $\theta_t(\psi_s)(p)$ If we choose $s$ properly,it will let $\psi_s(p) $ stop at some point that second coordinate is zero.Which means $\theta_t(q)$ at this point will go straight line to hit the z-axis.
Conversely $\theta_t(p)$ will not hit the z-axis,$\mathcal{D}^p = \Bbb{R}$!So it should not commute,as the definition the defining domian should be consistent for two different composition.The rest of the computation is not necessary,but we should hope if choosen $s,t$ nicely, we should have $\psi_s\circ\theta_t = \theta_t\circ \psi_s$
If $\psi_s(p)$ does not stop at the point that second coordinate is zero.The composition will have the form :
$$\theta_t\circ\psi_s(p) = (p_1+t,p_2,p_3-\arctan((t+p_1)/p_2))+\arctan(p_1/p_2)+\arctan((s+p_2)/(t+p_1)) - \arctan(p_2/(p_1+t))) $$
For the other direction we should use a bit $$f(x,y)= \arctan(x)+\arctan(y)-\arctan(\frac{x+y}{1-xy}) = 0$$ the arctan formula so they are really equal to each other at this point.