2

I have a differential $1$-form $\omega = x\mathrm{d}x + x\mathrm{d}y$ and I need to find its Lie derivative along $X = (x+y)\partial_{x} - 2y\partial_{y}$. The first approach is by using Cartan identity: \begin{multline} \imath_{X} \omega = x\mathrm{d}x\left((x+y)\partial_{x} - 2y\partial_{y}\right) + x\mathrm{d}y\left((x+y)\partial_{x} - 2y\partial_{y}\right) = \\ = x(x + y) - 2xy = x^2-xy; \end{multline} \begin{equation} \mathrm{d}(\imath_{X}\omega) = \mathrm{d}\left(x^{2} - xy\right) = 2x\mathrm{d}x - y\mathrm{d}x - x\mathrm{dy} = (2x - y)\mathrm{d}x - x\mathrm{d}y \end{equation} Next \begin{equation} \mathrm{d}\omega = \mathrm{d}x\wedge\mathrm{d}x + \mathrm{d}x\wedge\mathrm{d}y = \mathrm{d}x\wedge\mathrm{d}y; \end{equation} \begin{multline} \imath_{X}(\mathrm{d}\omega) = \imath_{X}(\mathrm{d}x\otimes\mathrm{d}y - \mathrm{d}y\otimes\mathrm{d}x) = \\ = \mathrm{d}x((x+y)\partial_{x} -2y\partial_{y})\mathrm{d}y - \mathrm{d}y((x+y)\partial_{x} -2y\partial_{y})\mathrm{d}x = \\ = (x+y)\mathrm{d}y + 2y\mathrm{d}x \end{multline} Finally, using the Cartan identity \begin{multline} \mathfrak{L}_{X}\omega = \mathrm{d}(\imath_{X}\omega) + \imath_{X}(\mathrm{d} \omega) = (2x - y)\mathrm{d}x - x\mathrm{d}y + (x + y)\mathrm{d}y + 2y\mathrm{d}x = \\ = (2x + y)\mathrm{d}y + y\mathrm{d}y \end{multline} However I’d like to find it using flow definition. As I understand the flow could be found in this manner \begin{gather} \begin{cases}\dot{x} = x + y\\ \dot{y} = -2y\end{cases}\\ \begin{pmatrix}x(t)\\ y(t)\end{pmatrix} = \begin{pmatrix}e^{t} & \frac{e^{t}- e^{-2t}}{3}\\ 0 & e^{-2t}\end{pmatrix}\begin{pmatrix}x(0) \\ y(0)\end{pmatrix} \end{gather} Therefore, flow is \begin{equation} \Phi^{t}(x_{0},y_{0}) = \begin{pmatrix}e^{t} & \frac{e^{t}- e^{-2t}}{3}\\ 0 & e^{-2t}\end{pmatrix} \end{equation}

From here I don’t quite understand how to find a pullback lift $h^{t}(x,y)$ and Lie derivative from here, using definition \begin{equation} \mathfrak{L}_{X}\omega = \lim_{t\to 0} \frac{h^{t}_{(x,y)}(\omega(\Phi^{t}(x,y))) - \omega(x,y)}{t} \end{equation}

And the last thing: is this a correct formula \begin{equation} \mathfrak{L}_{X}\omega = \left(X^{j}\frac{\omega_{i}}{\partial{x}_{j}} + \omega_{j}\frac{\partial X^{i}}{\partial x_{j}}\right)\mathrm{d}x^{i}? \end{equation} As I understand it is a coordinate representation of Cartan identity. I saw it here, however there are some troubles in upper and lower indexes in it and it does not give a correct answer for my example.

Matthew Willow
  • 762

1 Answers1

2

Your flow is:

$$\phi^t(x,y)=\begin{pmatrix}e^{t} & \frac{e^{t}- e^{-2t}}{3}\\ 0 & e^{-2t}\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}e^tx+Ey\\e^{-2t}y\end{pmatrix}$$

where $E=E(t)=\frac{e^t-e^{-2t}}{3}$ to save space.

Then, fixing $t$, the pull back form

\begin{align} (\phi^t)^*\omega &=(e^tx+Ey)d(e^tx+Ey)+(e^tx+Ey)d(e^{-2t}y)\\ &=e^t(e^tx+Ey)dx+(E+e^{-2t})(e^tx+Ey)dy \end{align} Now, you need to compute $$\lim_{t\to 0}\frac{(\phi^t)^*\omega -\omega}{t}$$ This is a direct computation, using $$\lim_{t\to 0}\frac{e^{2t}-1}t=2, \;\;\lim_{t\to 0}E=0,\;\;\lim_{t\to 0}\frac Et=1,\;\;\lim_{t\to 0}\frac{e^{-t}-1}{t}=-1$$ and you will get your desired answer.

In terms of coordinates, $\omega=\omega_idx^i$.

Then,

$$\mathcal L_X\omega=\mathcal L_X\omega_idx^i+\omega_i\mathcal L_X dx^i$$ $$\mathcal L_X\omega_i=X^j\partial_j\omega_i$$ $$\mathcal L_X(dx^i)=\partial_jX^i dx^j$$

So your formula is almost correct, just an index mistake at the end.

FYI, you can work from the definition and you don't need Cartan formula to compute $\mathcal L_X dx^i$: \begin{align} (\mathcal L_X dx^i)\left(\frac{\partial}{\partial x^j}\right)&=\partial_X \left(dx^i\left(\frac{\partial}{\partial x^j}\right)\right)-dx^i\left(\left[X,\frac{\partial}{\partial x^j}\right]\right)\\ &=-dx^i\left(\left[X,\frac{\partial}{\partial x^j}\right]\right)\;\;\;\;\;\text{because }\; dx^i\left(\frac{\partial}{\partial x^j}\right)=\delta^i_j\\ &=\partial_jX^i \end{align}

Y.H. Chan
  • 2,708