# Push forward and Lie-bracket

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?

• At the end, it should be $\psi_*X(\psi_*Y(f)) - \psi_*Y(\psi_*X(f))$. – Malkoun Nov 8 '19 at 19:52
• Yes exactly.... – Udalricus.S. Nov 8 '19 at 20:02
• 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. – ZxJx Nov 8 '19 at 20:04
• I second @ZxJx's comment. Keep track of the basepoint! – Malkoun Nov 8 '19 at 20:06

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.

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 $$F_*[X_1,X_2]=[Y_1,Y_2]=[F_*X_1,F_*X_2].$$