Lie derivative of vector fields on $S^1$

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. $Y$ 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 $$S^1$$. $$L_X(Y) := \lim\limits_{t \rightarrow 0} \frac{dF^{-t}(Y(F^t(\theta))) - Y(\theta)}{t} = (\sin^2{\theta}, - \sin{\theta}\cos{\theta}) = -\sin{\theta} \partial_{\theta}$$

Questions

$$1.~~$$ 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. 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.
• 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 $\partial_\theta$. – geometry_geek Jul 19 '19 at 11:24
• @geometry_geek the Lie derivative of vector fields is anti-symmetric so you should get $$L_YX = -L_XY.$$ – user7530 Jul 20 '19 at 20:55