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

[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

I choose to parametrize the ellipse using

[tex]x=a\cos[\theta][/tex]

[tex]y=b\sin[\theta][/tex]

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

[tex]\frac{d}{d\theta} = -\frac{a}{b} y \frac{\partial}{\partial x} + \frac{b}{a} x \frac{\partial}{\partial y}[/tex]

Next I compute the Lie brackets

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

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

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

[tex]

\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}

[/tex]

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

[tex]\tan[\phi] = \frac{y}{x}[/tex]

But the derivative dy/dx of the ellipse at say the point x = y is [itex]dy/dx = -b^2/a^2[/itex], and so the "outward" normal to the ellipse has a gradient of [itex]a^2/b^2[/itex]. Now, can we expect the [itex]\partial/\partial x[/itex] 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.)

[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

I choose to parametrize the ellipse using

[tex]x=a\cos[\theta][/tex]

[tex]y=b\sin[\theta][/tex]

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

[tex]\frac{d}{d\theta} = -\frac{a}{b} y \frac{\partial}{\partial x} + \frac{b}{a} x \frac{\partial}{\partial y}[/tex]

Next I compute the Lie brackets

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

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

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

[tex]

\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}

[/tex]

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

[tex]\tan[\phi] = \frac{y}{x}[/tex]

But the derivative dy/dx of the ellipse at say the point x = y is [itex]dy/dx = -b^2/a^2[/itex], and so the "outward" normal to the ellipse has a gradient of [itex]a^2/b^2[/itex]. Now, can we expect the [itex]\partial/\partial x[/itex] 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.)

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