I am trying to do a simple calculation of $L_X(Y)$ and $L_Y(X)$ over the manifold $M = S^1$ where $X = \partial_{\theta} = (-\sin{\theta}, \cos{\theta})$ and $Y = \cos{\theta}\partial_\theta$ using the local coordinates $\phi(\theta) = (\cos{\theta}, \sin{\theta})$. In particular I am self-studying from Peter Peterson's Riemannian Geometry and want to really understand what the Lie derivative is of $2$ vector fields from the difference quotient definition with local flow. <span class=$Y$ has a source and a sink."> I have the following. In local coordinates, the local flow defined by $X$ is $F^t(\theta) = (\cos{(\theta + t)} , \sin{(\theta + t)})$ and is a curve on $S^1$. $$L_X(Y) := \lim\limits_{t \rightarrow 0} \frac{dF^{-t}(Y(F^t(\theta))) - Y(\theta)}{t} = (\sin^2{\theta}, - \sin{\theta}\cos{\theta}) = -\sin{\theta} \partial_{\theta}$$


$1.~~$ I am having a hard time picturing what it means for $L_X(Y) = 0$ at the rightmost and leftmost points of the circle. Relatedly, how does the vector (1,0) relate to $Y$ "flowing along $X$" at the top point of $S^1$? I get that these derivatives are really about curves in $T_pM$. In particular, the difference quotient is fixing a point $p$ on $S^1$, flowing along $X$, taking the vector given by $Y$, and pulling it back to $T_pM$, thus tracing a curve in $T_pM$ as $t$ varies.

$2.~~$ I am having trouble calculating the local flow of $Y$ to ultimately calculate $L_Y(X)$. I have tried the following: Let $G^t(\theta) = (\cos(\eta(t, \theta)), \sin(\eta(t, \theta)))$. Then $\dot{G} = (-\sin(\eta)\dot{\eta}, \cos(\eta)\dot{\eta}) = \cos(\eta) (-\sin(\eta), \cos(\eta)) = Y(G^t(\theta))$ where the dot is differentiation w.r.t. $t$. This gives $\dot{\eta} = \cos(\eta)$. I'm not sure how to proceed. I also know that $G^0(\theta) = (\cos\theta,\sin\theta)$, and $G^t(\pm \frac{\pi}{2}) = (0, \pm 1)$. A hint on how to proceed would be greatly appreciated. Separation of variables (treating $\theta$ as constant)/an ODE solver seems to give rather unpleasant solutions that I can't easily use to plug into $X$. Is thinking of these things as literal vectors as opposed to the derivation viewpoint hindering me?


1 Answer 1

  1. Yes, exactly, though since your manifold is one-dimensional these curves may be hard to visualize, since $T_pM$ is just isomorphic to $\mathbb{R}$ and the curve just moves back and forth on the line. For example at $\theta=0$, you can visualize $T_pM$ as a vertical tangent line passing though the right point of your circle. Then consider how $$f(t) = dF^{-t} \circ Y \circ F^t(0)$$ behaves as a function of $t$: at $t=0$ you get the unit vector pointing up, and for $t$ slightly negative or slightly positive the vector points up and is slightly less than unit length. It's clear then that $f$ has a critical point at $t=0$ (the "curve" $f(t)$ has a cusp where it changes from moving up to moving down) which agrees with your computation that $L_XY=0$. $$$$ Similarly at $\theta=\pi/2$ you can imagine $T_pM$ as the horizontal tangent line, and as $t$ varies form being slightly negative to slightly positive, $f(t)$ varies from a vector pointing to the left along the line to vanishing at $t=0$ to pointing right; the derivative of $f$ is now negative in agreement with your answer $L_XY = -\partial_\theta$. $$$$ Notice that since $X$ is constant, all you're really doing here is computing the covariant derivative of $Y$: $L_XY = \nabla_{X(\theta)} Y$. To really understand what's happening with the flows I suggest trying a slightly less simple example, e.g. with $X$ and $Y$ two vector fields on the sphere.

  2. I don't think you should expect any particularly nice formula for the flow. Your approach of solving the ODE $$\frac{d\theta}{dt} = \cos\theta$$ on the obvious chart seems right to me. You can get a (not particularly pleasant) closed-form solution by separation of variables, and Wolfram Alpha can't do any better either.

  • $\begingroup$ Thanks! Short of a solution, I am thinking about some properties for $L_Y(X)$. Am I correct that $L_Y(X) = 0$ at $\theta = \pm \pi/2$? What about the sign of $L_Y(X)$ at $\theta = 0, \pi$?. If my understanding is correct, I shouldn't be able to associate a sign to a lie derivative in general, but since this is a 1-manifold, I mean the sign of the coefficient in front of $\partial_\theta$. $\endgroup$ Jul 19, 2019 at 11:24
  • $\begingroup$ @geometry_geek the Lie derivative of vector fields is anti-symmetric so you should get $$L_YX = -L_XY.$$ $\endgroup$
    – user7530
    Jul 20, 2019 at 20:55

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