Does the Lie Bracket Always Close the Parallelogram as Described?

[Pond Dragon]

I finally finished my big summer research project. Reviewing how it went, it is clear to me that I lack an understanding of many structures on smooth manifolds. I decided to pull out my old copy of Spivak's A Comprehensive Introduction to Differential Geometry, Volume 1.

I'm currently working through the concept of the Lie derivative, which I don't actually remember from before. I decided that, to build understanding, I'd do a simple example on $\mathbb{RP}^3$.

Fix a chart $x:\{(p^1,p^2,p^3,p^4)\in\mathbb{R}^4~\vert~p^4\neq 0\}/\sim\,\to\mathbb{R}^3,~[p^1,p^2,p^3,p^4]\mapsto (\frac{p^1}{p^4},\frac{p^2}{p^4},\frac{p^3}{p^4})$, where $\sim$ is an equivalence relation defined by $v \sim \lambda v$, for $v\in\mathbb{R}^4$ and $\lambda\in\mathbb{R}$. This notation is used in Manfredo do Carmo's Riemannian Geometry. Consider vector fields $X=\frac{\partial}{\partial x^1}$ and $Y=x^2\frac{\partial}{\partial x^1}+x^1\frac{\partial}{\partial x^3}$.

I calculated $[X,Y]=\frac{\partial}{\partial x^3}$, and calculated the flows as $$\phi^X_t(p)=x^{-1}(t+x^1(p),x^2(p),x^3(p)) \\ \phi^Y_t(p)=x^{-1}(tx^2(p)+x^1(p),x^2(p),tx^1(p)+x^3(p)).$$

This makes the composition $\phi^Y_{-t}\circ\phi^X_{-t}\circ\phi^Y_t\circ\phi^X_t(p)=x^{-1}(x^1(p),x^2(p),t^2(1-x^2(p))+x^3(p))$. The flow $\phi^{[X,Y]}_t(p)=x^{-1}(x^1(p),x^2(p),t+x^3(p))$ completes this "parallelogram" thing that Spivak talks about by the composition $$\phi^{[X,Y]}_{-t^2(1-x^2(p))}\circ\phi^Y_{-t}\circ\phi^X_{-t}\circ\phi^Y_t\circ\phi^X_t(p)=p.$$

Questions: Does this always happen? Can we then think of the flow of the Lie bracket as something that will close this "parallelogram"? If so, in what way?

Thank you!

 

[Pond Dragon]

I made a small mistake.

$$\phi^Y_{-t}\circ\phi^X_{-t}\circ\phi^Y_t\circ\phi^X_t(p)=x^{-1}(x^1(p),x^2(p),t^2+x^3(p)).$$

I apologize.