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Question: On $\mathbf{R}^2$, let $X=y \frac{\partial}{\partial x}$ and let $Y=x \frac{\partial}{\partial y}$, with corresponding flows given by $\phi_t(x, y)=(x+t y, y)$ and $\psi_t(x, y)=(x, y+t x)$. Then show that the Lie derivative, $\mathcal L_XY=-x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$

The definition given in the book,

Definition 4.7.1. Let $T$ be a tensor field of type $(r, s)$ on $\mathbf{R}^n$ and let $X$ be a vector field with flow $\phi_t: \mathbf{R}^n \rightarrow \mathbf{R}^n$. The Lie derivative of $T$ with respect to $X$, denoted by $\mathcal{L}_X T$, is the $(r, s)$-tensor defined as $$ \mathcal{L}_X T=\left.\frac{d}{d t}\right|_{t=0}\left(\phi_t^* T\right) . $$

But when we have a ambient space like for this example then do we need the flow to define the Lie derivative? Inspired from this M.SE answer, I give a try below:
$$ \begin{align} \mathcal L_XY&=\mathcal L_X\left(x\frac{\partial}{\partial y}\right)\\ &=[\mathcal L_X(x)]\frac{\partial}{\partial y}+x[\mathcal L_X\left(\frac{\partial}{\partial y}\right)]\\ &=[y\frac{\partial}{\partial x}(x)]\frac{\partial}{\partial y}+x[y\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}\right)]\\ &=y\frac{\partial}{\partial y}+? \end{align} $$

but here I guess $\mathcal L_X(\frac{\partial}{\partial y})$ doesn't make sense! Or I am not expert enough to guess the meaning.

falamiw
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  • Apply the definition you just put up there! – Ted Shifrin Feb 06 '24 at 18:17
  • Okay, the book doesn't show any computation for pullback explicitly. here is mentioned two flow, then which flow I should pick? Like for $\phi_t^*\left(x\frac{\partial}{\partial y}\right)$ But isn't pullback only defined on forms? Here it is a vector field @TedShifrin – falamiw Feb 06 '24 at 18:25
  • Right. The notation is explained in the book. For a contravariant tensor, you push forward by the flow. – Ted Shifrin Feb 06 '24 at 19:20
  • You should explicitly mention which book you are referring to. "The book" is a bit vague – Didier Feb 06 '24 at 19:31
  • There is a typo in your expression for $L_XY$: it should end with $y\frac{\partial}{\partial y}$, note $x\frac{\partial}{\partial y}$. – Didier Feb 06 '24 at 19:49
  • Thank you @Didier fixed the typo. Can you help me with another question related with covariant derivative. Actually, I was looking for resource where things are explained with example. but couldn't manage. Let me know if you have one where connections, curvature, bundles are explained with example. Thanks again. – falamiw Feb 07 '24 at 20:12
  • This book by R.W.R. Darling may be helpful to you. – Ted Shifrin Feb 08 '24 at 16:28

1 Answers1

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Just use the definition that you are given. The flow of $X$ is $\phi_t(x,y) = (x+ty,y)$. The Lie derivative $\mathcal{L}_XY$ is given by $$ L_XY = \left.\frac{d}{dt}\right|_{t=0} (\phi_t)^*Y, $$ and for $p\in \Bbb R^2$, the pullback $(\phi_t)^*Y$ at $p$ is given by $$ ((\phi_t)^*Y)_p = {(d\phi_t)_p}^{-1} \left(Y_{\phi_t(p)}\right). $$ Let us proceed step by step. Fix $t,x$ and $y$. Then $\phi_t(x,y) = (x+ty,y)$. Hence, $Y_{\phi_t(x,y)} = (x+ty)\frac{\partial}{\partial y}$. Now, $d\phi_t(x,y)$ is the linear isomorphism whose matrix in the basis $\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\}$ is given by $$ \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}, $$ and it follows that its inverse ${(d\phi_t)_{(x,y)}}^{-1}$ is the linear isomorphism whose matrix in the same basis is $$ \begin{pmatrix} 1 & -t \\ 0 & 1 \end{pmatrix}. $$ Hence, we have $$ ((\phi_t)^*Y)_{(x,y)} = (x+ty)\left(-t\frac{\partial}{\partial x} + \frac{\partial}{\partial y} \right) = -tx\frac{\partial}{\partial x} + ty\frac{\partial}{\partial y} -t^2y \frac{\partial}{\partial y}. $$ Its derivative with respect to $t$ is $$ \frac{d}{dt} ((\phi_t)^*Y)_{(x,y)} = -x \frac{\partial}{\partial x} +y\frac{\partial}{\partial y} -2ty\frac{\partial}{\partial x}. $$ Evaluating at $t=0$ gives $$ (\mathcal{L}_XY)_{(x,y)} = -x\frac{\partial}{\partial x} +y\frac{\partial}{\partial y}, $$ which is the desired result.

As an instructive exercise, you can mimic this proof to compute $\mathcal{L}_YX$. You should find the opposite of the above result.

Didier
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  • Thank you very much for your detail answer. One last confusion, why the inverse ${(d\phi_t)p}^{-1}$ in the $((\phi_t)^*Y)_p = {(d\phi_t)_p}^{-1} \left(Y{\phi_t(p)}\right)$. I ask this because isn't it the inverse of the Jacobian? I guess it by jacobian only but the inverse thing make me confused, @Didier – falamiw Feb 06 '24 at 19:37
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    @falamiw This is the definition of the pullback of a vector field by a diffeomorphism, that's it. More conceptually, you are pulling a vector that is tangent at the point $\phi_t(x,y)$ back at the original point $(x,y)$. The linear map $(d\phi_t)_{(x,y)}$ sends vectors tangent at $(x,y)$ to vectors tangent at $\phi_t(x,y)$, so it is not the right linear map to use. Meanwhile, its inverse goes in the reverse direction, which is what you are looking for. This may seems arbitrary on $\Bbb R^2$, but once you will work on abstract manifolds, you'll see that it's the only thing that makes sense. – Didier Feb 06 '24 at 19:44