Proof of Isomorphism between Tangent Space and the Vector Space of all Derivations

Question

In his book An Introduction to Manifolds (Second Edition) on page no. 13, Loring Tu proves the following theorem.

Theorem 2.2 The linear map $\phi: T_P(\mathbb{R}^n) \rightarrow \mathscr{D}_p(\mathbb{R}^n)$ defined by $v \mapsto D_v = \sum_{i} v^i \left.\frac{\partial}{\partial x^i}\right|_p$ is an isomorphism of vector spaces.

Here $\mathscr{D}_p(\mathbb{R}^n)$ is the set of all derivations at $p$ which is, of course, a vector space.

In the proof of surjectivity, he uses Taylor's theorem with remainder to expand a smooth function $f: V \rightarrow \mathbb{R}$ defined on a star-convex neighborhood $V$ of the point $p$:

$$ f(x) = f(p) + \sum_{i} (x^i - p^i) g_i(x), \quad g_i(p) = \frac{\partial f}{ \partial x^i}(p). $$

Then he applies $D$ on both sides and uses $D(f(p)) = 0$ and $D(p^i) = 0$ to arrive at the following equaiton:

$$ Df(x) = \sum (Dx^i) g_i(p) + \sum (p^i - p^i) Dg_i(x) = \sum (Dx^i) \frac{\partial f}{ \partial x^i}(p), $$ where $D$ is a derivation at point $p$.

My Confusion

I can't arrive at this equation.

My Attempts

\begin{eqnarray} f(x) &=& f(p) + \sum_{i} (x^i - p^i) \, g_i(x) \\ \Rightarrow Df(x) &=& D(f(p)) + D\left(\sum_{i} (x^i - p^i) \, g_i(x)\right) \\ \Rightarrow Df(x) &=& \sum_{i} (x^i - p^i)D\left(g_i(x)\right) + \sum_{i} D\left((x^i - p^i)\right) g_i(x) \\ \Rightarrow \left[Df(x)\right]_p &=& \sum_{i} \left[(x^i - p^i)\right]_p \,\,\left[D\left(g_i(x)\right)\right]_p + \sum_{i} \left[D\left((x^i - p^i)\right)\right]_p \,\,\left[g_i(x)\right]_p \\ \Rightarrow Df(p) &=& 0 + \sum_{i} \left[D(x^i)\right]_p \,\,\left[g_i(x)\right]_p \\ \Rightarrow Df(p) &=& \sum_{i} D(p^i) g_i(p) \\ \Rightarrow Df(p) &=& \sum_{i} D(p^i) \frac{\partial f}{\partial x^i}(p). \end{eqnarray}

A change of variables $p \longleftrightarrow x$ yields, $$Df(x) = \sum_{i} D(x^i) \frac{\partial f}{\partial x^i}(\color{red} x)$$ whereas Tu writes, $$Df(x) = \sum_{i} D(x^i) \frac{\partial f}{\partial x^i}(\color{red}p).$$

I don't see where I have made the mistake.

Answer 1

Tu is right. The derivation $D$ assigns to each $f \in C^\infty_p$ a real number. However, in my opinion, it may be confusing to write $Df(x)$ instead of $Df$. In fact, the number $Df \in \mathbb R$ does not depend on any special value $x$. Writing $Df(x)$ simply indicates that the function $f$ has $x$ as its variable. If one wants to include $x$ by all means to the notation, I would prefer to write $D[f(x)]$.

A completely exact approach would be this. Define functions $$c : V \to \mathbb R, c(x) = f(p) ,$$ $$c^i : V \to \mathbb R, c^i(x) = p^i ,$$ $$\pi^i : V \to \mathbb R, \pi^i(x) = x^i.$$ $\pi^i$ is the projection onto the $i$-th coordinate. Then we get $$f = c + \sum (\pi^i - c^i) g_i$$ and $$Df = Dc + \sum D((\pi^i - c^i)g_i) = \sum D(\pi^i - c^i)g_i(p) + \sum (\pi^i(p) -c^i(p))D(g_i) \\ = \sum(D\pi^i -Dc^i)g_i(p) = \sum D\pi^i g_i(p) = \sum D\pi^i \frac{\partial f}{ \partial x^i}(p) .$$ But now note that one simply writes $D\pi^i = Dx^i$, that is, the coordinate projections are denoted by $x^i$. This is common use in the literature.