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$\textbf{Proposition:}$ Suppose $F: M \to N$ is a smooth map between manifolds with out without boundary, let $X \in \mathcal{X}(M)$ and $Y \in \mathcal{Y}(N)$. Then $X$ and $Y$ are $F$-related if and only if for every smooth valued function $f$ defined on an open subset of $N$, $$ X \left ( f \circ F \right)= \left(Yf \right) \circ F$$.

My question concerns the following statement in the proof of the above proposition,

For any $p \in M$, and any smooth real-valued $f$ defined in a nieghborhood of $F(p)$ we have, $$X \left ( f \circ F \right ) (p) = X_p \left ( f \circ F \right) = dF_p(X_p) f$$

How does one go about showing the following equality, $$X_p \left ( f \circ F \right) = dF_p(X_p) f ? $$ It looks like it could be an application of the chain rule but I am not sure...

Yuugi
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2 Answers2

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That's just the definition of the differential of $F$ at $p$, $dF_p$. Take a look at page 55:

If $F\colon M\to N$ is a smooth map, for each $p\in M$, $dF_p\colon T_pM\to T_{F(p)}N$ is defined as follows: If $v\in T_pM$, and $f\in C^\infty(N)$, then $dF_p(v)\in T_{F(p)}(N)$ is the derivation that acts as $$ dF_p(v)(f)=v(f\circ F). $$

Since $X\colon M\to TM$ is a vector field, $X_p\in T_pM$, and that's your $v$ in this case.

Ben West
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The answer here is actually "by definition". If you have a smooth map $F: X \rightarrow Y$ then $$dF_p: T_p X \rightarrow T_{F(p)}Y$$ How is this defined? Take $V \in T_p X$ and send it to the tangent vector $dF_p V \in T_{F(p)}Y$, now that is just notation, but how does this act on a function? $$ dF_p (V) f := V (f\circ F)$$ See the paragraph on `differential of a smooth function' in Lee's book (p. 55 in my copy). Hope that helps!

Tim Weelinck
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  • Page number was useful in addition to your answer. Thank you. – Yuugi Feb 26 '15 at 01:44
  • Possibly not appropriate fort the comment section, bt my class is not actually using Lee, and I am not entirely sure on how to interpret $V(f \circ F)$. If $V \in T_p(X)$, I am imagining $V$ as a tangent vector at $p$, that is some column vector with real components.. – Yuugi Feb 26 '15 at 01:52
  • Ah, well I can strongly recommend using Lee's book to learn about these things. Although you can also try taking a look into Loring W. Tu's book. Anyway, imagining $V$ as a column vector only makes sense if you know what the basis of the vector space is. You can think of this basis in terms of derivations, or germs of functions. The latter giving an easy way to actually compute the derivative, but this is too much for a comment. I recommend you reading a good introduction (such as Tu or Lee) on the tangent space to a manifold, and computing some concrete examples. – Tim Weelinck Feb 26 '15 at 02:33