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Lie Brackets and Parallelograms

  1. Jul 24, 2014 #1
    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!
     
  2. jcsd
  3. Jul 24, 2014 #2
    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.
     
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