Nigel Hitchen, in his notes on differential manifolds gives a definition of the derivative of a smooth map between two manifolds that appears to assume the the following assertion is self-evident:

Given two manifolds $M$ and $N$, a smooth map $F: M \rightarrow N$, and a tangent vector $X_a$ at the point $a\in M$, the map $X'_{F(a)}: C^\infty(N) \rightarrow C^\infty(\mathbb{R})$ defined by $$X'_{F(a)}(f)=X_a(f\circ F)$$ for $f\in C^{\infty}(N)$ is a tangent vector at $F(a)$.

I see that $X'(1) = 0$ but how about the Leibnitz rule?

  • $\begingroup$ That follows from the fact that $X_a$ safisfies the Leibniz rule. $\endgroup$
    – user98602
    Dec 29 '15 at 23:15
  • $\begingroup$ Perhaps I'm misunderstanding something - which is highly likely. But if $f : N \to \mathbb R$ and $F : M \to N$ then what does $F \circ f$ mean? We do $f$ first and then $F$. Hence $(F \circ f) : N \to \mathbb{R} \ \ (?=?) \ \ M \to N$. Did you mean to write $f \circ F$? This makes more sense $(f \circ F) : M \to N \to \mathbb R$. $\endgroup$ Dec 29 '15 at 23:25
  • $\begingroup$ @Flyby: Yes, he meant $f \circ F$. $\endgroup$
    – user98602
    Dec 29 '15 at 23:27
  • $\begingroup$ Thanks, I meant $f \circ F$. It makes no sense as I wrote it. $\endgroup$ Dec 29 '15 at 23:28
  • $\begingroup$ Also, I meant $X'(const) = 0$ $\endgroup$ Dec 29 '15 at 23:30

If $\{f,g\} \subset C^\infty(N)$ then $\{f\circ F,g\circ F, fg\circ F\} \subset C^{\infty}(M)$ and $fg\circ F = (f\circ F)(g\circ F)$, so $X'_{F(a)}$ inherits the Leibnitz rule from $X_a$.

  • $\begingroup$ This $\{f, g\}$ notation is extremely confusing as it's also used for the Poisson bracket. $\endgroup$
    – Pedro
    Dec 31 '15 at 19:16

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