Why does $D_v f(a)=\lim_{t \rightarrow 0} \frac{f(a+tv)-f(a)}{t}$ reduce to $v^i \frac{df}{d x^i}(a)$ when $v_a = v^i e_i |_a$?

Question

Why does $D_v f(a)=D v|_a f=\lim_{t \rightarrow 0} \frac{f(a+tv)-f(a)}{t}$ reduce to

$$v^i \frac{df}{d x^i}(a)$$

when $v_a = v^i e_i |_a$?

What I read is that this means that

$$v_i D_v f(a)=v_i \lim_{t \rightarrow 0} \frac{f(a+te_i)-f(a)}{t}$$

But I'm unsure as to what's the "phenomenon" here. And whether I really understand what the notation $e_i|_a$ is supposed to mean.

Is this perhaps the partial derivative of the particular variable $x_i$? Since $a+te_i$ appears to increment $h$ only on that particular coordinate.

This appears in Lee's Introduction to Smooth Manifolds p. 52.

Answer 1

Here is what Lee is saying:

Fix a point in $\mathbb R^n$ and choose a geometric tangent vector $v_a\in \mathbb R^n.\ v_a$ is called a "geometric vector" because it specifies a direction in $\mathbb R^n.$ I will henceforth write $v$ for $v_a$ to simplify notation. (Lee retains the $v_a$ nomenclature).

Now, $v$ induces a map $\tilde v:C^{\infty}(\mathbb R^n)\to \mathbb R$ defined by

$\tilde v(f)=\frac{d}{dt}\big |_{t=0}f(a+tv).$ This formula is just the $definition$ of the directional derivative of $f$ (in the direction specified by $v$) at the point $a$. That is, $\tilde v(f)=\frac{d}{dt}\big |_{t=0}f(a+tv)=D_vf(a).$

Setting $g(t)=f(a+tv)$, we have,

$\tag 1\ g'(0)=\lim_{t \rightarrow 0} \frac{f(a+tv)-f(a)}{t}\ =\frac{d}{dt}\big |_{t=0}f(a+tv)=D_vf(a)=\tilde v(f).$

That is, taking the derivative of $g$ at $t=0$ is precisely $\tilde v(f).$

But...to actually do the calculation, we have to specify a coordinate system.

If we adopt the standard basis on $\mathbb R^n$, then in particular $v=(v_1,\cdots, v_n)=\sum^n_{i=1} v_ie_i:=v^ie_i\big |_a$ in the Einstein notation.

Now, we can do the calculation:

We have shown that $D_vf(a)=g'(0).$ On the other hand, $g$ is a compostion, i.e. $g(t)=f(h(t))$ where $h(t)=a+vt$. All that remains now is to apply the chain rule:

$g'(t)=\sum^n_{i=1}\frac{\partial f}{\partial x_i}(a+tv)\cdot\frac{dh}{dt}(t)=\sum^n_{i=1}v_i\frac{\partial f}{\partial x_i}(a+tv)$

and evaluate at $t=0:$

$\tag2 g'(0)=\sum v_i\frac{\partial f_i}{\partial x_i}(a)=v^i\frac{\partial f_i}{\partial x_i}(a)$

the last equality is just a changeover to the Einstein notation.

Now combine $(1)$ and $(2)$ to see that this is precisely Lee's result.