Pushforward of the vector field $\dfrac{d}{dx}$ by exponential, i.e $exp_{*}\dfrac{d}{dx}$ = $x\dfrac{d}{dx}$ on $R_{+}^*$

Question

I'm trying to proof the following statement coming from a book:

"Pushforward of the vector field $\dfrac{d}{dx}$ by exponential, i.e $exp_{*}\dfrac{d}{dx}$ = $x\dfrac{d}{dx}$ on $R_{+}^*$"

The only statement I have in my possession is that for $\phi$ a diffeomorphism between $M \rightarrow N$ , and $X$ a vector field, (which is a section of $TM$), on has the following property:

$\forall x$ in $N$

$(\phi_{*}X)(x) = d_{\phi^{-1}(x)}\phi$ . $X(\phi^{-1}(x))$

Hence, I tried applying the formula, $exp_{*}\dfrac{d}{dx}(x)$ = $(d_{\exp^{-1}(x)}\exp)$ . $\dfrac{d}{dx}(\exp^{-1}(x))$ = $d_{\ln(x)}\exp$ . $\dfrac{d}{dx}(\ln(x))$
= $d_{\ln(x)}\exp$ . $\dfrac{1}{x}$ = $exp(1/x)$

But if the result I'm trying to prove is right, I should have that $\forall x$ in $R_{+}^*$ $x\dfrac{d}{dx}(x)$ = $x$

Am I missing something more conceptual ? Thank you in advance for explanation

Answer 1

The pushforward of a vector field is well-defined just by a diffeomorphism (or in a Lie group), that is, if $\varphi :M\to N$ is such diffeomorphism then $\varphi _*:\mathfrak{X}(M)\to \mathfrak{X}(N)$, where $\mathfrak{X}(D)$ is the space of smooth vector fields on $D$.

The pushforward is defined by

$$ (\varphi _*X)_{\varphi (p)}f=X_p(f\circ \varphi ) $$

In this case if $\varphi :\mathbb{R}\to (0,\infty ),\, x\mapsto e^x$ you have

$$ \left(\exp_*\frac{\partial}{\partial x}\right)_rf=\left .\frac{\partial}{\partial x}\right|_{\ln r}(f \circ \exp)=(f'\circ \exp)(\ln r)\exp'(\ln r)=r\cdot f'(r)=\left .t\frac{\partial}{\partial t}\right|_{r}f $$

where $\frac{\partial}{\partial t}$ is the canonical basis of $\mathfrak{X}((0,\infty ))$ and $f$ is any smooth function in $(0,\infty )$.

Answer 2

$\def\R{\mathbb{R}}%$@Masacroso's is the answer, but I think we could also use the Fréchet derivative of the map between two copies of $\R$, $Dexp: \R_1\to\R_2$, $Dexp_x=e^x=y$; for an arbitrary $h\in \R_1$, $Dexp_x(h)=e^xh\in \R_2.$

The map derivative, on the other hand, is a pushforward of tangent vectors. We can use what's called canonical identification or natural isomorphism: $g_1: T\R_1\to\R_1$ and $g_2: T\R_2\to\R_2$. The relation formula (see @levap's answer here for example) between the pushforward and the Fréchet derivative of our map $exp$ is

$$exp_{*p}=g_2^{-1}\circ Dexp\circ g_1:T\R_1\to T\R_2$$ And so we have $exp_{*x}(h\tfrac{d}{dx})=[g_2^{-1}\circ Dexp\circ g_1](h\tfrac{d}{dx})=[g_2^{-1}\circ Dexp_x](h)=g_2^{-1}(e^xh)=(e^xh)\tfrac{d}{dy}$,

where $\tfrac{d}{dy}$ is the basis vector on the $T\R_2$. Taking $h=1$ we finally have $$exp_{*x}(\tfrac{d}{dx})=y\tfrac{d}{dy}$$