Concrete example of the Lie derivative of a one-form

It is completely correct.

The given formula for the Lie derivative of a one-form follows from Cartan's identity:

$$\mathcal{L}_X\alpha = i_X(d\alpha) + d(i_X\alpha).$$

Let's compute $\mathcal{L}_X\alpha$ using this identity and check we get the same result.

First, we can compute the exterior derivative of $\alpha$:

$$d\alpha = 2ydy\wedge dx + 2xdx\wedge dy = (2x-2y)dx\wedge dy.$$

Using the formula $i_X(\beta\wedge\gamma) = (i_X\beta)\wedge\gamma + (-1)^{|\beta|}\beta\wedge(i_X\gamma)$, we can compute the interior product of $d\alpha$:

\begin{align*} i_X(d\alpha) &= (2x - 2y)i_X(dx)\wedge dy - (2x - 2y)dx\wedge i_X(dy)\\ &= (2x - 2y)dx(X)dy - (2x - 2y)dy(X)dx\\ &= (2x - 2y)dx\left(\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y}\right)dy - (2x-2y)dy\left(\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y}\right)dx\\ &= (2x-2y)dx\left(\frac{\partial}{\partial x}\right)dy - (2x-2y)(xy)dy\left(\frac{\partial}{\partial y}\right)dx\\ &= (2xy^2 - 2x^2y)dx + (2x-2y)dy. \end{align*}

For the second term, the interior product of $\alpha$ is

\begin{align*} i_X\alpha &= y^2i_X(dx) + x^2i_X(dy)\\ &= y^2dx(X) + x^2dy(X)\\ &= y^2dx\left(\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y}\right) + x^2dy\left(\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y}\right)\\ &= y^2dx\left(\frac{\partial}{\partial x}\right) + x^2(xy)dy\left(\frac{\partial}{\partial y}\right)\\ &= y^2 + x^3y \end{align*}

and taking the exterior derivative, we have

$$d(i_X\alpha) = 2ydy + 3x^2ydx + x^3dy = 3x^2ydx + (x^3 + 2y)dy.$$

Combining, we see that

\begin{align*} \mathcal{L}_X\alpha &= i_X(d\alpha) + d(i_X\alpha)\\ &= (2xy^2 - 2x^2y)dx + (2x - 2y)dy + 3x^2ydx + (x^3 + 2y)dy\\ &= (2xy^2 + x^2y)dx + (2x+x^3)dy \end{align*}

which is the same result you obtained.

Tags:

Analysis

Differential Geometry

Differential Forms

Lie Derivative

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