0

My question is pretty much summed up in the title. We have a tangent vector $X\in T_pM$ at a point $p\in M$ for a smooth manifold $M$, a diffeomorphism $\varphi:M\to N$ for another smooth manifold $N$ and a function $f\in\mathcal{F}(N)$.

Why do we get that $X(\varphi^*f)=(\varphi_*X)f$ and what happens to the point $p$ in the transformation? Where $\varphi_*,\,\varphi^*$ are the pushforward and the pullback respectively. I came upon this question, when reading this answer.

Harish Chandra Rajpoot
  • 38,565
Wurstcake
  • 152
  • This answer only refers to functions. $X\omega$ is not defined for $X\in T_pM$ and $\omega$ a $k$-form with $k > 0$. – Didier Mar 13 '23 at 18:01
  • @Didier Oh I see. I'll edit the question. I'd still like to know the answer for functions in that case. – Wurstcake Mar 13 '23 at 18:04
  • 1
    While the cited answer does only give the proof for $k=0$, it does indicate that the formula $d(g^* \omega)=g^* d\omega$ is true in general. For a proof of such, see for instance https://math.stackexchange.com/q/1302562/137524 – Semiclassical Mar 13 '23 at 18:07
  • @Semiclassical Yes, the pullback commutes with the exterior differential, but this has nothing to do with the actual question here, which is about the relation between the action of a tangent vector, the pullback, and the pushforward (which makes no sense for $k$-forms with $k > 0$) – Didier Mar 13 '23 at 18:09
  • 3
    Sometimes this is just the definition of $\varphi_$. What is your definition for $\varphi_$? – Arctic Char Mar 13 '23 at 18:09
  • @ArcticChar I took my definition from wikipedia. There it's $\varphi_*X=d\varphi_pX$. – Wurstcake Mar 13 '23 at 18:24
  • 1
    @ArcticChar And as I wrote it, I started to understand, that it follows quite easily from that as well. $(\varphi_X)(f)=(d\varphi_pX)(f)=X(\varphi)(f)=X(f\circ\varphi)=X(\varphi^f)$ Would this be correct? – Wurstcake Mar 13 '23 at 18:33
  • @Wurstcake Usually, $Xf$ is used for real valued functions $f$, and $X\varphi$ would be misleading. That is why we use $\varphi_*X$ etc. ANyway, your proposal is just reading backward my answer – Didier Mar 13 '23 at 19:03

1 Answers1

2

Consider a function $f\colon N\to \Bbb R$, a tangent vector $X\in T_pM$ and a diffeomorphism $\varphi\colon M \to N$. Note that $X\in T_pM$ and $\varphi_*X = d_p\varphi(X) \in T_{\varphi(p)}N$. Then one has \begin{align} X(\varphi^* f) &= \left(d_p(\varphi^*f)\right) X & \text{by definition of the action of $X\in T_pM$},\\ &= d_p(f\circ \varphi) X & \text{by definition of $\varphi^*f$},\\ &= (d_{\varphi(p)}f \circ d_p\varphi)X & \text{by the chain rule},\\ &= d_{\varphi(p)}f(d_p\varphi X) &\text{by associativity},\\ &= d_{\varphi(p)}f(\varphi_*X) & \text{by definition of $\varphi_*X$},\\ &= (\varphi_* X)f & \text{by definition of the action of $\varphi_*X\in T_{\varphi(p)}N$}. \end{align} Of course, this proof relies on the chain rule, so using this to prove the chain rule would be a circular argument. So let's just pretend that the chain rule has been shown, for instance in coordinates.

Didier
  • 20,820