The radial vector field $x_i \partial/\partial x_i$

Question

In the Wikipedia article on Lie derivative we read that $\mathcal{L}_X \omega$ (where $X$ is a vector field and $\omega$ is a 1-form) evaluates the change of $\omega$ along the flow defined by the vector field $X$. Then $\mathcal{L}_X \omega = 0$ would mean that $\omega$ remains constant along the flow defined by the vector field $X$.

Suppose $X=\sum _i x_i \frac{\partial}{\partial x_i}$ and $\eta$ is a 1-form such, that $\mathcal{L}_X \eta = 0$. What is the meaning of this...? Is $X$ actually the position vector? Then it would mean $\eta$ doesn't change anywhere in $\mathbb R^3$? Does it mean $\eta=\mbox{const}$ in $\mathbb R^3$ ?

The question is:

-What is the meaning of the field $X$? How would you describe it?

-What does it mean for $\eta$ to be an invariant?

Answer 1

We can prove a general result, using a calculation that Lee does on p. 323 in his book on smooth manifolds. The significance of a zero Lie derivative is, paraphrasing Lee, that the flow $\theta_t$ pulls $\eta$ back to itself wherever it is defined. So $\eta$ is constant along the flow of the vector field $X$. Here is a sketch of the proof:

If $M$ is a smooth manifold, and $X$ a smooth vector field on $M$, fix $t$ in the flow domain of $X$ (which in your specific case, is $\mathbb R$. We have the flow of $X$, namely $\theta_t:M\to M$ which induces the pullback on forms: $(\theta_t^*\eta)_pX_p=\eta_{\theta_t(p)}(d\theta_t X_p)=d(\theta_t^*)_p\eta_{\theta_t(p)}$

The definition of the Lie derivative is $(\mathscr L_X\eta))_p=\frac{d}{dt}|_{t=0}(\theta^*_t\eta)_p$.

Now, using properties of the differential and of the flow, we make a change of variables and calculate

$\frac{d}{dt}|_{t=t_0}(\theta^*_t\eta)_p=\frac{d}{dt}|_{t=t_0}d(\theta_t^*)_p\eta_{\theta_t(p)}=$

$\frac{d}{ds}|_{s=0}d(\theta_{s+t_0}^*)_p\eta_{\theta_{s+t_0}(p)}=d(\theta_{t_0}^*)_p\frac{d}{ds}|_{s=0}d(\theta_{s}^*)_{\theta_{t_0}(p)}(\eta_{\theta_{s}(\theta_{t_0}(p))})=$

$d(\theta_{t_0}^*)_p((\mathscr L_X\eta)_{\theta_{t_0}(p)})$.

This calculation shows that if the Lie derivative is zero, then $(\theta^*_t\eta)_p$ is constant. And since $(\theta^*_0\eta)_p=\eta_p$ , this means that $\eta$ is invariant under pullback by $\theta_t.$

Answer 2

The flow of $X$ is $\phi_t(x_1,..,x_n)=(e^tx_1,..,e^tx_n)$, we deduce that $\eta$ is invariant by positive homothetic map.