Lie dragging a vector around the ellipse

[wandering.the.cosmos]

I've been trying to understand exactly how the Lie derivative parallel transports a vector, by working out an explicit example: Lie dragging $\partial/\partial x$ at (a,0) on the x-y plane anticlockwise along the ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

I choose to parametrize the ellipse using

$$x=a\cos[\theta]$$
$$y=b\sin[\theta]$$

so that we have the tangent vector along the ellipse to be

$$\frac{d}{d\theta} = -\frac{a}{b} y \frac{\partial}{\partial x} + \frac{b}{a} x \frac{\partial}{\partial y}$$

Next I compute the Lie brackets

$$\left[\frac{d}{d\theta},\frac{\partial}{\partial x}\right] = \frac{b}{a} \frac{\partial}{\partial y}$$
$$\left[\frac{d}{d\theta},\frac{\partial}{\partial y}\right] = -\frac{a}{b} \frac{\partial}{\partial x}$$

From this I conclude that parallel transporting $\partial/\partial x$ from (a,0) through an angle $\theta$ along the ellipse yields

$$\exp\left[\theta \pounds_{d/d\theta}\right] \frac{\partial}{\partial x} = \sum_{n=0}^\infty \left( \frac{b}{a}\frac{(-1)^n}{(2n+1)!} \theta^{2n+1} \frac{\partial}{\partial y} + \frac{(-1)^n}{(2n)!} \theta^{2n} \frac{\partial}{\partial x} \right) \\ = \frac{b}{a} \sin[\theta] \frac{\partial}{\partial y} + \cos[\theta] \frac{\partial}{\partial x} \\ = \frac{y}{a} \frac{\partial}{\partial y} + \frac{x}{a} \frac{\partial}{\partial x}$$

It seems this implies that the parallel transported $\partial/\partial x$ gives a vector that is inclined at an angle $\phi$ such that

$$\tan[\phi] = \frac{y}{x}$$

But the derivative dy/dx of the ellipse at say the point x = y is $dy/dx = -b^2/a^2$, and so the "outward" normal to the ellipse has a gradient of $a^2/b^2$. Now, can we expect the $\partial/\partial x$ to remain perpendicular to the ellipse as we lie drag it on its circumference? Why or why not? It does not seem to be the case, as this computation shows. If so, how should we understand what parallel transportation on an ellipse means? (Of course, I could have made an error somewhere, and would appreciate any corrections.)

 

[Doodle Bob]

Now, can we expect the $\partial/\partial x$ to remain perpendicular to the ellipse as we lie drag it on its circumference? Why or why not? It does not seem to be the case, as this computation shows. If so, how should we understand what parallel transportation on an ellipse means? (Of course, I could have made an error somewhere, and would appreciate any corrections.)

I can't really speak to your computations which at first glance look kosher.
But I can say that one would *not* expect the image of d/dx under the elliptical flow to remain perpendicular to the ellipse at all points. The flow of a vector field will preserve the angles of vectors (or even orthogonality) only if it preserves (up to a nonzero function) the Riemannian metric - which in this case is the Euclidean inner product. i.e. the Lie derivative of the metric with respect to the given vector field should be a multiple of that metric. I don't think that a vector field which is tangent along a noncircular ellipse will satisfy that -- or least the one that you have chosen.

Furthermore, don't confuse parallel translation along a curve with transportation by the flow of a vector field tangent to that curve. Parallel transport *will* by definition preserve angles, the dynamic flow (or rather a dynamic flow -- there are infinitely many after all because there are infinitely many ways to extend a vector field along a curve to a neighborhood abotu that curve) tangent to the curve will not necessarily.

 

[Doodle Bob]

Regarding parallel translation...

Since we're dealing with a flat 2-dimensional Riem. manifold here (The Eucldiean plane), calculating the parallel transport of a vector around the ellipse is easy.

In your case, we have d/dx at the point (a,0). It is of unit length and perpendicular to the tangent vector of the ellipse there and pointing "outside" the filled ellipse. The parallel transport of this vector about the curve will have the very same properties at each given point, i.e. unit length and normal to the curve and pointing outside of the filled ellipse (its relative orientation with respect to the tangent vector will not change). Thus, it will be the outward pointing normal vector field of the ellipse.

Things change, of course, if the metric is not flat (say, hyperbolic 2-space), then we get nontrivial holonomy so that, when we parallel-transport a vector all the way around a closed curve, we often end up with a different vector at the place of origin. In the case of hyperbolci 2-space, the difference between the original vector and its wrap-around parallel-transport is related to the total area enclosed by the curve.

Hope that helps.

 

[pmb_phy]

Note: Lie dragging a vector takes place on a one of the curves which forms a congruence which fills the manifold so keep in mind that the ellipse is part of this congruence.

Pete

 

[wandering.the.cosmos]

$$\left[\frac{d}{d\theta},\frac{\partial}{\partial x}\right] = \frac{b}{a} \frac{\partial}{\partial y}$$
$$\left[\frac{d}{d\theta},\frac{\partial}{\partial y}\right] = -\frac{a}{b} \frac{\partial}{\partial x}$$

$$\exp\left[\theta \pounds_{d/d\theta}\right] \frac{\partial}{\partial x} = \frac{y}{a} \frac{\partial}{\partial y} + \frac{x}{a} \frac{\partial}{\partial x}$$

$$\tan[\phi] = \frac{y}{x}$$

I made a sign error with my Lie brackets -- both of them should read

$$\left[\frac{d}{d\theta},\frac{\partial}{\partial x}\right] = -\frac{b}{a} \frac{\partial}{\partial y}$$
$$\left[\frac{d}{d\theta},\frac{\partial}{\partial y}\right] = \frac{a}{b} \frac{\partial}{\partial x}$$

But that'd mean that

$$\exp\left[\theta \pounds_{d/d\theta}\right] \frac{\partial}{\partial x} = -\frac{y}{a} \frac{\partial}{\partial y} + \frac{x}{a} \frac{\partial}{\partial x}$$

And hence it seems that $\partial/\partial x$ would start dipping in the negative y direction as it's being dragged along the ellipse? If so this is quite far from my expectation that Lie dragging gives us a way of parallel transporting w/o using a metric.

 

[Doodle Bob]

If so this is quite far from my expectation that Lie dragging gives us a way of parallel transporting w/o using a metric.

That would be because the two processes are very different. The Lie process for example depends *completely* on how you extend the tangent vector of the curve away from the curve. You chose to extend it by creating the family of parallel ellipses. But you could have also chosen a family of conic sections with some other property, say, sharing a directrix. This would have resulted in a different family of curves hence a different vector field to "Lie drag" along and hence a different Lie transportation than the one you ended up with.

As I stated before, some vector fields do result in isometries, but you have to choose them wisely. As noted by Pete, your ellipse vector field will generate non-isometric transformations on the plane. I believe in this case they would have to foliate to circles and lines.