Confusion about the notation of directional derivative

by Oliver G   Last Updated August 10, 2019 03:20 AM

From An Introduction to Manifolds by Tu:

$(1)$ Let $D_v = \sum v^i\frac{\partial}{\partial x^i}|_p$ where $v = [v^1, \dots, v^n]$ is a vector in $\Bbb R^n$ and $p = (p^1, \dots, p^n)$ a point in $\Bbb R^n.$ Then $D_v$ is a map that sends a function to a number $D_vf$.

Let $C^{\infty}_p$ be the set of all germs of $f$ at $p$, where a germ is an equivalence class of pairs $(f, U)$ where $U$ is a neighborhood of $p$ in $\Bbb R^n$ and $(f,U) $ is similar to $(g, V)$ if and only if there's an open subset $W \subset U \cap V$ containing $p$ such that $f=g$ when restricted to $W$. Each $f$ is a $C^{\infty}$ function.

Tu then writes:

For each tangent vector at a point $p$ in $\Bbb R^n$, the directional derivative at $p$ gives a map of real vector spaces $$ (2)\space D_v: C^{\infty}_p \rightarrow \Bbb R.$$

$D_v$ is $\Bbb R$-linear and satisfies the Leibniz rule $$D_v(fg) = (D_vf)g(p) + f(p)D_vg.$$

To me, this looks like Tu's giving two different definitions in $(1)$ and $(2)$. The first is a function from a space of functions, and the second is a function from the set of all germs at $p$.

But if he's using the second definition, how is $D_v(f)$ defined? $D_v$ should map an equivalence class $[(f,U)]$ to the real numbers, but I don't understand what the operation is on equivalence classes or how to show this equality on equivalence classes.



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