# Concrete example of the Lie derivative of a one-form

Let $\alpha$ be a one-form and $X$ a vector field. For example take:

\begin{align*} \alpha &= y^2 dx + x^2 dy\\ &\\ X &= \frac{\partial}{\partial x}+xy \frac{\partial}{\partial y}. \end{align*}

I'm trying to understand how the Lie derivative works on concrete examples. In particular I want to apply the formula:

$$\mathcal{L}_X\alpha = \left(X^j\frac{\partial \alpha_i}{\partial \phi^j} + \alpha_j \frac{\partial X^j}{\partial \phi^i}\right)d\phi^i.$$

If I understand the definitions correctly, it should be:

\begin{align*} \mathcal{L}_X\alpha =&\ \left(1\frac{\partial y^2}{\partial x} + y^2 \frac{\partial 1}{\partial x}\right)dx +\left(xy\frac{\partial y^2}{\partial y} + x^2 \frac{\partial xy}{\partial x} \right)dx\\ &\ +\left(1\frac{\partial x^2}{\partial x} + y^2 \frac{\partial 1}{\partial y}\right)dy + \left(xy\frac{\partial x^2}{\partial y} + x^2 \frac{\partial xy}{\partial y}\right)dy\\ =&\ (0+0)dx+(2xy^2+x^2y)dx+(2x+0)dy+(0+x^3)dy\\ =&\ (2xy^2+x^2y)dx+(2x+x^3)dy. \end{align*}

Is this correct at all?

If not, I'd appreaciate any tip or hint as to what I'm doing wrong.

• what is this formula that is being applied here? Could you please point me to some reference? – gen May 30 at 10:29

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.

• I've tried to work it out, but I don't get how you calculate $i_X(d\alpha)$. Could you please elaborate it a bit? Regardless of that, thanks a lot for the answer, it has been very helpful! – G. Schiele Aug 5 '16 at 23:08
• I have added some details about computing the interior product. – Michael Albanese Aug 6 '16 at 14:05

I'm late for the party, but here's another solution: fixed$\renewcommand\vec[1]{{\boldsymbol #1}}$ $\vec{X} \in \mathfrak{X}(\Bbb R^2)$, the derivation $L_{\vec{X}}$ defined on functions and vector fields by $L_{\vec{X}}f \doteq \vec{X}(f)$ and $L_{\vec{X}}\vec{Y} \doteq [\vec{X},\vec{Y}]$ extends to tensor fields of any rank by imposing the Leibniz rule, and commuting with contractions. It'll boil down to Cartan's magic formula again if you think (maybe not so) hard about it, but we can use $${\rm d}\alpha(\vec{X},\vec{Y}) = \vec{X}(\alpha(\vec{Y}))-\vec{Y}(\alpha(\vec{X}))-\alpha([\vec{X},\vec{Y}])$$too. So: \begin{align} (L_{\vec{X}}\alpha)(\vec{Y})&\doteq L_{\vec{X}}(\alpha(\vec{Y})) - \alpha(L_{\vec{X}}\vec{Y}) \\ &= \vec{X}(\alpha(\vec{Y})) - \alpha([\vec{X},\vec{Y}]) \\ &= {\rm d}\alpha(\vec{X},\vec{Y})+\vec{Y}(\alpha(\vec{X}))\\ &= {\rm d}\alpha(\vec{X},\vec{Y}) + {\rm d}(\alpha(\vec{X}))(\vec{Y}).\end{align}So we compute $$\alpha = y^2{\rm d}x+x^2{\rm d}y \implies {\rm d}\alpha = 2(x-y){\rm d}x\wedge {\rm d}y \implies {\rm d}\alpha(\vec{X},\cdot) = 2(x-y)({\rm d}y - xy\;{\rm d}x),$$and $$\alpha(\vec{X}) = y^2 + x^3y \implies {\rm d}(\alpha(\vec{X})) = 3x^2y\,{\rm d}x + (x^3+2y){\rm d}y.$$Summing: $$L_{\vec{X}}\alpha = (x^2y+2xy^2){\rm d}x + (x^3+2x){\rm d}y,$$as wanted.