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 Answer 1


(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




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):


  • $\begingroup$ 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? $\endgroup$
    – B.Hueber
    Commented Jan 21, 2020 at 13:29
  • 1
    $\begingroup$ Yes, it is indeed the composition of the two functions $Y(f\circ\phi^{-1}_t)$ and $\phi_t$. $\endgroup$
    – Balloon
    Commented Jan 21, 2020 at 13:52

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