# Lie derivative of vector fields and flows of vector fields: Is this formula right?

In the lecture we have defined the Lie derivative as $$\mathcal{L}_{X}Y:=\frac{\mathrm{d}}{\mathrm{d}t}\bigg\vert_{t=0}\Phi_{t}^{\ast}Y$$

where $$X,Y\in\mathfrak{X}(\mathcal{M})$$ are vector fields on a manifold $$\mathcal{M}$$ and $$\Phi$$ is the flow of X.

My goal is now to plug in the definition of the pull-back in order to rewrite this formula without the pull-back....I am often confused with the different definition and notations in diffgeo and therefore I would be fine if someone can say if the following is correct:

(1) The push-forward of a tangent vector (viewed as a derivation) is for a function $$f:\mathcal{M}\to \mathcal{N}$$ between two manifolds and a tagent vector $$v\in T_{p}\mathcal{M}$$ defined as $$f_{\ast}v:=\mathrm{d}_{p}f(v)$$ or in other words: for some function $$h\in C^{\infty}(N)$$: $$(f_{\ast}v)(h):=[\mathrm{d}_{p}f(v)](h):=v(h\circ f).$$

(2) For a vector field $$X\in\mathfrak{X}(M)$$, the push-forward is defined pointwise: $$(f_{\ast}X)_{q}:=\mathrm{d}_{f^{-1}(q)}(X_{f^{-1}(q)})$$ for some $$q\in\mathcal{N}$$, where we have to require that f is a diffeomorphism.

(3) Therefore we find with the flow $$\Phi_{t}:\mathcal{M}\to \mathcal{M}$$ for $$p\in\mathcal{M}$$ and $$f\in C^{\infty}(\mathcal{M})$$: $$(\Phi_{t}^{\ast}Y)_{p}(f):=(\Phi^{-1}_{t\ast}Y)_{p}(f)=[\mathrm{d}_{\Phi_{t}(p)}\Phi_{t}^{-1}(Y_{\Phi_{t}(p)})](f)=Y_{\Phi_{t}(p)}(f\circ \Phi_{t}^{-1})$$

(4) Using that $$\Phi^{-1}_{t}=\Phi_{-t}$$ this yields the formula:

$$(\mathcal{L}_{X}(Y))_{p}(f)=\frac{\mathrm{d}}{\mathrm{d}t}\bigg\vert_{t=0}Y_{\Phi_{t}(p)}(f\circ \Phi_{-t})$$

If we now view vector fields as derivation on $$C^{\infty}(\mathcal{M})$$, namely $$X:C^{\infty}(\mathcal{M})\to C^{\infty}(\mathcal{M})$$ instead of $$X:\mathcal{M}\to T\mathcal{M}$$, this can also be written as: $$(\mathcal{L}_{X}(Y))(f)=\frac{\mathrm{d}}{\mathrm{d}t}\bigg\vert_{t=0}Y(f\circ\Phi_{-t})\circ\Phi_{t}$$

Are the steps and the final formula right?

(1) Yes, but since $$f_*$$ and $$df_p$$ (or $$d_pf$$) are two notations for the same object, I would keep only one (and it would be $$df_p$$),

(2) Yes, but this is of course under the condition $$f$$ is a diffeomorphism (else you can't assign a value for $$q$$ with $$f^{-1}(q)=\varnothing$$, or you can't choose between values if $$f^{-1}(q)$$ has at least two elements),

(3) The last equality doesn't make sense: if $$V$$ is a vector field on a manifold $$N$$ and $$\psi:M\to N$$ a map between two manifolds, we won't have $$V_{\psi(p)}f=V_p(f\circ\psi)$$ ($$p$$ is not even a point of $$N$$). There is a specific counter-example in your case (with your notations):

Take $$M=N=\mathbb{R}$$, $$X=\frac{\partial}{\partial t}$$ (so $$\varphi^t(p)=p+t$$) and $$Y=t\frac{\partial}{\partial t}$$. Then

$$Y_{\varphi^t(p)}(f\circ\varphi^{-t})=(p+t)f'(p)$$

and

$$Y_p(f\circ\varphi^{-t}\circ\varphi^t)=f'(p),$$

so any function $$f:\mathbb{R}\to\mathbb{R}$$ with $$f'(p)\neq 0$$ leads to a counter-example.

(4) It becomes correct if you stop one step before in (3):

$$(\mathcal{L}_XY)_pf=\frac{d}{dt}\big|_{t=0}(Y_{\varphi^t(p)}(f\circ\varphi^{-t})).$$

• Thank you for your Remarks..... To your point number (3)......srry, I had some mistakes with my brackets..... What I want to do is the following.....If we view vector fields as derivations, then: $Y(f\circ \Phi^{-1}_{t})( \Phi_{t}(p)) =(Y(f\circ \Phi^{-1}_{t})\circ \Phi_{t}) (p)$....... I correct the error above..... Is it now correct? – Udalricus.S. Jan 21 at 13:29
• Yes, it is indeed the composition of the two functions $Y(f\circ\phi^{-1}_t)$ and $\phi_t$. – Balloon Jan 21 at 13:52