Flowing a vector field along a diffeomorphism

Question

For simplicity I am happy to work on Euclidean space $\mathbb{R}^d$. Let $X$ be a vector field on $\mathbb{R}^d$, which I think of as a first-order differential operator (that is, I would write $Xf$ for $X\cdot \nabla f$). Let $\varphi:\mathbb{R}^d\to\mathbb{R}^d$ be a diffeomorphism, and write $\varphi$ for the pullback map on functions

$$ \varphi^*f(x) = f(\varphi(x)). $$

I would like to know the expression for the differential operator $$ \varphi^* X (\varphi^*)^{-1}. $$

I think that $X(\varphi^{-1})^*$ is the pushforward of $X$ by $\varphi^{-1}$, but honestly I get confused once I apply the chain rule too many times and I am not sure how to do things correctly.

Answer 1

It clarifies to fix a smooth function $f \in \mathcal{C}^{\infty}(\mathbf{R}^{d})$ and a point $p \in \mathbf{R}^{d}$.

1. Apply $(\phi^{*})^{-1}$. $$ ((\phi^{*})^{-1}f)(p) = ((\phi^{-1})^{*}f)(p) = f(\phi^{-1}(p)). $$

2. Apply $X$. $$ (X(\phi^{*})^{-1}f)(p) = \left( df_{\phi^{-1}(p)} \circ d(\phi^{-1})_{p} \right)(X_{p}). $$

3. Apply $\phi^{*}$. $$ (\phi^{*}X(\phi^{*})^{-1}f)(p) = (X(\phi^{*})^{-1}f)(\phi(p)) = \left( df_{p} \circ d(\phi^{-1})_{\phi(p)} \right)(X_{\phi(p)}). $$

Finally, $$ \left( df_{p} \circ d(\phi^{-1})_{\phi(p)} \right)(X_{\phi(p)}) = df_{p}(d(\phi^{-1})_{\phi(p)}(X_{p})) = df_{p}\left( (\phi^{-1})_{*}X_{p} \right) = \left( (\phi^{-1})_{*}X_{p} \right)f. $$ In sum, $$ \phi^{*}X(\phi^{*})^{-1} = (\phi^{-1})_{*}X, $$ as you suspect.