Calculate the Lie derivative using definition

Question

Definition: If $X$ and $Y$ are vector fields on a smooth manifold $M$, the Lie derivative of $Y$ in the direction of $X$ is the vector field $$ \mathcal{L}_X Y := \frac{d}{dt}\bigg\vert_{t=0}(F^X_{-t})_\ast Y. $$ Here $F^X$ denotes the flow of $X$.

Using this definition, calculate the Lie derivative $\mathcal{L}_X Y$ where $X = \frac{\partial}{\partial x}$ and $Y = x \frac{\partial}{\partial y}$ (here $x,y$ are the standard coordinates on $\Bbb{R}^2$).

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My attempt:

A previous exercise asked to calculate this via the Lie bracket, to which I found that $[X,Y] = \frac{\partial}{\partial y}$, so I know the Lie derivative must equal this as well.

I calculated the flow of $X$ as being $F^X: \Bbb{R}^2 \times \Bbb{R} \to \Bbb{R}: ((x,y),t) \mapsto (x+t,y)$. Hence at a time$-t$ we get $F^X_{-t} = (x-t,y)$. But when I calculate the derivative of this map, I just get that the derivative, $(F^X_{-t})_\ast = \operatorname{Id}$. But this is independent of $t$, so that would mean $\mathcal{L}_X Y = 0$, which is not the correct answer.

Where did I go wrong here? Is the flow correct? What about the derivative? Am I interpreting the notation incorrectly?

Thanks in advance!

Answer 1

Write down $$\lim_{t\to 0}\frac{(F_{-t})_*(Y(F_t(x,y)))-Y(x,y)}t=\lim_{t\to 0}\frac{(F_{-t})_*((x+t)\partial/\partial x_2)-(x\,\partial/\partial x_2)}t.$$

Answer 2

Let $X,Y$ be vector fields on $M$. We compute (in general) that $\mathcal L_XY=[X,Y]$, where $[X,Y]=XY-YX$ is the Lie bracket of them.

In your case you can just put $X =\frac{\partial}{\partial x}$ and $Y = x \frac{\partial}{\partial y}$ and get the desired result.

Here is the proof:

Suppose $f\in C^\infty(M),\,p\in M.$ (i.e., smooth function on $M$ that maps to $\mathbb R$)

Note that by definition of vector field, we should have

$$((F^X_{-t})_*Y)_pf=Y_{F^X_t(p)}(f\circ F^X_{-t})=(F^X_{t})^*(Y(f\circ F^X_{-t}))(p)$$

To be more precise, $(Y(f\circ F^X_{-t}))$ is a smooth map and pullback of it under $(F^X_{t})$ is just the usual composition of two maps, which corresponds to the middle one.

As for the first equality, actually this could be viewed as definition of pushforward of vector fields, since we defined that, for a diffeomorphism $F:N\to M,$ the pushforward of a tangent vector $X_p \in T_pN$ is given by $ F_{*,p}:T_pN \rightarrow T_{F(p)}M,$ where $$F_*X\colon q\mapsto(F_*X)_q = dF_{F^{-1}(q)}(X_{F^{-1}(q)}).$$ In our case $F^X_{-t}$ is a flow along $X$ so it is a (local) diffeomorphism by definition, and $q=p$, corresponding to our notations.

(I think this is the desired answer to what you confused so far, but I will write down the rest of proof for those who may have the same question.)

We have \begin{align} (\mathcal L_XY)f&=\lim_{t\to0}\frac{((F^X_{-t})_*)Y)f-Yf}{t}\\ &=\lim_{t\to0}\frac{(F^X_{t})^*(Y(f\circ F^X_{-t}))-Yf}{t}\\ &=\lim_{t\to0}\frac{(F^X_{t})^*(Y(f\circ F^X_{-t}))-Y(f\circ F^X_{-t})+Y(f\circ F^X_{-t})-Yf}{t}\\ &=X(Yf)-Y(Xf)=[X,Y]f. \end{align} Note that, as we just know, $(Y(f\circ F^X_{-t}))$ is a smooth map, may write it as $G$ for convenience, then we have $$\lim_{t\to0}\frac{(F^X_{t})^*(Y(f\circ F^X_{-t}))-Y(f\circ F^X_{-t})}{t}=\lim_{t\to0}\frac{(F^X_{t})^*G-G}{t}=\frac{d}{dt}\bigg\vert_{t=0}G\circ(F^X_{t})$$ which becomes $dG\left(\frac{d}{dt}\bigg\vert_{t=0}F^X_{t}\right)=dG(X)=XG,$ and the other term is similar, by using chain rule.