# Prove directional derivative of a vector-valued function is independent of choice of basis

I'm trying to figure out this problem from Tu, Differential Geometry: Connections, Curvature and Characteristic Classes (problem 20.2)

We all know how the directional derivative of an $$\mathbb{R}$$-valued function on a manifold $$M$$ is defined. Let $$f : M \rightarrow \mathbb{R}$$ be smooth, and let $$X_p$$ be a tangent vector to $$M$$ at $$p$$. Then $$X_p f$$ is the directional derivative of $$f$$ at $$p$$ in the direction $$X_p$$. For let $$c(t)$$ be a smooth curve in $$M$$, with $$c(0) = p$$ and $$c^{'} (0) = X_p$$. Then $$X_p f = \frac{d}{dt}|_{t = 0} f(c(t)).$$

Now Tu in his book extends this to vector-valued functions $$f : M \rightarrow V$$ where $$V$$ is some finite-dimensional vector space. Let $$v_1, \ldots, v_n$$ be a basis for $$V$$, and write $$f = \sum f^{i} v_i$$ for some smooth real-valued functions $$f^{i} : M \rightarrow \mathbb{R}$$. Then define $$X_p f := \sum (X_p f^{i}) v_i.$$

Problem: Show the above definition is independent of the choice of the basis.

Attempt: I let $$u_1, \ldots, u_n$$ be another basis for $$V$$. Then $$v_i = \sum_j a_i^{j} u_j$$ for some coefficients $$a_{i}^{j}$$. Then $$X_p f = \sum_i (X_p f^{i}) (\sum_j a_{i}^{j} u_j) = \sum_{ij} a_{i}^{j} (X_p f^{i}) u_j.$$ Now I'm not really sure how to proceed. Does this already conclude the proof?

$$\sum_j(X_p\tilde {f}{}^j)u_j$$ where $$f=\sum_j\tilde f{}^ju_j$$. To find the $$\tilde f{}^j$$ just plug in the equations for the $$v_i$$ into
$$f=\sum_i f^iv_i$$
• But can I bring the coefficients $a_{i}^{j}$ inside $(X_p f^{i})$? Don't I have to worry about the Leibniz rule then? Are the coefficients $a_{i}^{j}$ numbers or functions? Jul 6, 2019 at 7:09
• The $a^j_i$ are numbers so you can safely bring them inside. Jul 6, 2019 at 7:43