Diffeomorphism invariance, Lie derivative

Question

There is written in the Hamilton's Ricci flow book about Lie Derivative this:

The Lie derivative, which measures the infinitesimal lack of  
diffeomorphism invariance of a tensor with respect to a 1-parameter group of  
diffeomorphisms generated by a vector field, has the following properties: 

(1) If $f$ is a function, then $\cal L_x f = Xf$,

(2) If $Y$ is a vector field, then ${\cal L_X} Y = [X,Y]$.

I do not understand what intuitively is the "diffeomorphism invariance".

Answer 1

To make things simple, let $X,Y$ be vector fields on $\mathbb{R}^n$. Then $\mathcal{L}_XY$ gives the instantaneous rate of change of $Y$ in the direction of the flow $\phi_t$ in which $X$ induces. You can show that,

$$ \mathcal{L}_XY(p) = [X,Y](p) = \lim_{t \to 0} \frac{Y( \phi_t(p)) - Y(p)}{t}$$

Here $\{\phi_t\}$ is the $1$-parameter group of differomorphisms. By definition of the limit, the lie derivative is just asking, ``how does $Y$ change in the direction of the velocity for the phase flow?". Letting $t$ be very small we have,

$$ \mathcal{L}_XY(p) t \approx Y(\phi_t(p))-Y(p)$$

i.e the lie derivative tells us also how $Y$ varies with respect to the flow i.e. under the coordinate change $\phi_t$, how different is $Y(\phi_t(p))$ and $Y(p)$? Hence it is measuring the invariance of $Y$ (which is a tensor) w.r.t the diffeomorphism $\phi_t$.