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Let $\psi$ be a diffeomorphism in $\mathbb{R}^{n}$. I have to proove that for two vector fields $X,Y\in\mathfrak{X}(\mathbb{R}^{n})$ the following property holds

$\psi_{\ast}[X,Y] = [\psi_{\ast}X,\psi_{\ast}Y]$,

where $[\cdot,\cdot]$ denotes the Lie-Bracket of vector fields.

First of all, I review some definitions:

(1) In this context, vector fields are viewed as derivations:

$X:C^{\infty}(\mathbb{R}^{n})\to C^{\infty}(\mathbb{R}^{n})$, where $X$ is linear and fulfills the Leibniz rule.

(2) For a vector $v\in T_{p}\mathbb{R}^{n}$, here also viewed as a derivation $v:C^{\infty}(\mathbb{R}^{n})\to \mathbb{R}$, the push foward is defined via:

$\psi_{\ast}v(f):=v(f\circ\psi)$,

where $f\in C^{\infty}(\mathbb{R}^{n})$

Now to the proof:

$\psi_{\ast}[X,Y](f)=[X,Y](f\circ\psi)=X(Y(f\circ\psi)) - Y(X(f\circ\psi)) = X(\psi_{\ast} Y(f)) - Y(\psi_{\ast} X(f))$

But now I don`t no how to continue the proof.....Is there an error in my calculation?

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    $\begingroup$ At the end, it should be $\psi_*X(\psi_*Y(f)) - \psi_*Y(\psi_*X(f))$. $\endgroup$ – Malkoun Nov 8 '19 at 19:52
  • $\begingroup$ Yes exactly.... $\endgroup$ – Udalricus.S. Nov 8 '19 at 20:02
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    $\begingroup$ I think it might help to see why Malkoun's statement is true, by proving the result for a diffeomorphism between two manifolds. Mostly, keep track of what the domain/codomains are for the maps involved. $\endgroup$ – ZxJx Nov 8 '19 at 20:04
  • $\begingroup$ I second @ZxJx's comment. Keep track of the basepoint! $\endgroup$ – Malkoun Nov 8 '19 at 20:06
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There is a definition issue here--you have never actually defined what $\psi_*X$ means. You have defined what $\psi_*v$ means when $v$ is a tangent vector at a single point, but not what $\psi_*X$ means when $X$ is an entire vector field. The correct definition of $\psi_*X$ to use is not $\psi_*X(f)=X(f\circ\psi)$ but rather $$\psi_*X(f)=X(f\circ\psi)\circ \psi^{-1}.$$ To see that this makes sense, let $v$ be the value of $X$ at a point $p$. Then $\psi_*v(f)=v(f\circ\psi)$ defines a tangent vector $\psi_*v$ not at the point $p$ but at the point $\psi(p)$, since $v(f\circ\psi)$ depends on the values of $f\circ\psi$ near $p$ and thus on the values of $f$ near $\psi(p)$. Thus, $\psi_*v$ should be defined to be the value of the vector field $\psi_*X$ at $\psi(p)$, not at $p$. To get the value of $\psi_*X$ at $p$, you have to precompose with $\psi^{-1}$, hence the formula above.

Using this correct formula, you should have no difficulty verifying that $\psi_*[X,Y] = [\psi_*X,\psi_*Y]$, the key point being that when you compute $\psi_*X(\psi_*Yf))$ you get $$X((\psi_*Yf)\circ\psi)\circ\psi^{-1}=X((Y(f\circ \psi)\circ \psi^{-1})\circ\psi)\circ\psi^{-1}=X(Y(f\circ\psi))\circ\psi^{-1}$$ with the inner $\psi^{-1}$ and $\psi$ cancelling out.

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I don't think the result is all that easy. Following Lee (cf. Introduction to Smooth Manifolds), we can prove something more general, from which your formula follows immediately. First, note that $X:C^{\infty}(\mathbb{R}^{n})\to C^{\infty}(\mathbb{R}^{n})$ is a map that sends $f$ to $Xf:\mathbb R^n\to \mathbb R:p\mapsto X_pf.$ Then, if $F:M\to N$ is a diffeomorpshism, $X\in \mathfrak X(M),\ Y\in \mathfrak X(N)$, say that $X$ and $Y$ are $F$ -related $\Leftrightarrow (F_*)_p(X_p)=Y_{F(p)},$ which is to say $F_*X=Y.$

To prove your claim, in steps, we have

$(1).\ X$ and $Y$ are $F$-related $\Leftrightarrow X(f\circ F)=(Yf)\circ F $, which is true simply because

$$X_p(f\circ F)=((F_*)_pX_p)f\ \text{and}\ (Yf)\circ F(p)=Y_{F(p)}f.$$

$(2).\ $ if $X_1,X_2\in \mathfrak X(M),\ Y_1,Y_2\in \mathfrak X(N)$ then if $X_1$ and $Y_1$ are $F$-related and $X_2$ and $Y_2$ are $F$-related, then $[X_1,X_2]$ is $F$ -related to $[Y_1,Y_2].$ This is a direct calculation. Using $(1)$ twice in each of the following, we have

$$X_1X_2(f\circ F)=X_1(Y_2f\circ F)=(Y_1Y_2f)\circ F$$

and

$$X_2X_1(f\circ F)=X_2(Y_1f\circ F)=(Y_2Y_1f)\circ F,$$

which implies that

$$([X_1,X_2])f\circ F=(Y_1Y_2f)\circ F-(Y_2Y_1f)\circ F=([Y_1,Y_2]f)\circ F,$$

and the result now follows from another application of $(1).$

$(3).$ Unwinding all this, we have shown that $F_*[X_1,X_2]=[Y_1,Y_2]=[F_*X_1,F_*X_2].$

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