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?


1 Answer 1


You are half way there! You still have to show that what you have calculated is equal to

$$\sum_j(X_p\tilde {f}{}^j)u_j $$ where $f=\sum_j\tilde f{}^ju_j$. To find the $\tilde f{}^j$ just plug in the equations for the $v_i$ into

$$f=\sum_i f^iv_i$$

  • $\begingroup$ 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? $\endgroup$
    – Kamil
    Jul 6, 2019 at 7:09
  • $\begingroup$ The $a^j_i$ are numbers so you can safely bring them inside. $\endgroup$
    – Claire
    Jul 6, 2019 at 7:43

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