Let $M$ be a $m$-dimensional smooth manifold, $p \in M$ a point and consider smooth curves $\alpha: (-\epsilon,\epsilon) \to M$, $\beta: (-\tilde{\epsilon},\tilde{\epsilon}) \to M$ centered at $p$, meaning $\alpha(0)=\beta(0)=p$.

We define the following equivalence relation on these curves:

$\alpha \sim \beta :\Longleftrightarrow $ for every chart $\varphi: U \to U'$ with $p \in U$ we have $(\varphi \circ \alpha)'(0) = (\varphi \circ \beta)'(0)$.

We call an equivalence class a tangent vector.

Let $\alpha$ be such a curve through $p \in M$ and $f: V \to \mathbb{R}$ a smooth real valued function defined on an open neighborhood $V$ of $p$. Then we call $(f \circ \alpha)'(0)$ the (directional) derivative of $f$ in direction $\alpha$.

My notes say that this derivative only depends on the equivalence class of $\alpha$, so this must mean that if $\alpha \sim \beta$ we have $(f \circ \alpha)'(0) = (f \circ \beta)'(0)$. But I do not understand why this holds.

I know that if $\alpha \sim \beta$ we have $(\varphi \circ \alpha)'(0) = (\varphi \circ \beta)'(0)$ for any chart $\varphi: U \to U'$ defined on an open neighborhood $U$ in $p$. But I do not know how to make the transition from $\varphi$, which maps to an open subset $U' \subset \mathbb{R}^m$, to $f$ which only maps to $\mathbb{R}$.

Could you please explain this problem to me? Thank you!

  • 1
    $\begingroup$ Try the chain rule :) $\endgroup$ – Ted Shifrin Dec 17 '17 at 1:03

Just note that $(f \circ \alpha) = (f \circ \varphi^{-1} \circ \varphi \circ \alpha)$ and $\alpha(0)=\beta(0)=p$ and $(\varphi \circ \alpha)'(0) = (\varphi \circ \beta)'(0)$. So by the chain rule \begin{align} (f \circ \alpha)'(0) &= (f \circ \varphi^{-1} \circ \varphi \circ \alpha)'(0) \\ &= D(f \circ \varphi^{-1})_{\varphi(p)} \circ (\varphi \circ \alpha)'(0)\\ &= D(f \circ \varphi^{-1})_{\varphi(p)} \circ (\varphi \circ \beta)'(0)\\ &= (f \circ \varphi^{-1} \circ \varphi \circ \beta)'(0) \\ &= (f \circ \beta)'(0). \end{align} In fact, this definition of directional derivatives can be extend for all global function $ f \in C^{\infty}(M)$ using bump function argument. See Jeff. Lee or here.


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