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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.

mavavilj
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1 Answers1

3

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.

Matematleta
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  • The first derivative is at $g'(a)$ right? – mavavilj Sep 28 '18 at 22:31
  • Also I don't really see how you apply chain rule. Is $v$ really the only thing containing $x_i$s? – mavavilj Sep 28 '18 at 22:32
  • Also where do you use $v_a=v^i e_i|_a$? – mavavilj Sep 28 '18 at 22:33
  • The first derivative is the $definition$ of $g'$ at $0$. But $g$ may also de written as $g=f\circ h$ where $h(t)=a+tv.$ Then, $g'(t)=\sum\frac{\partial f_i}{\partial x_i}(a+tv)\cdot \frac{d h_i}{dt}(t)$. I did not use $v_a=v^i e_i|_a$ at all, only the fact that $f$ is differentiable at $a$. I am of course, using the standard coordinate charts on the Euclidean spaces. – Matematleta Sep 28 '18 at 22:47
  • Is it? I think that when $t$ disappears then one is left with $f$s at $a$: https://en.wikipedia.org/wiki/Derivative#Rigorous_definition – mavavilj Sep 28 '18 at 22:49
  • My calculation is correct: it's a standard argument. It seems to apply to your problem. I suspect the $v_a$ refer to an arbitrary coordinate chart at $a$ in some abstract manifold, and what they are saying is that if one uses the standard chart, then you get the stated result, which is the one which I proved. – Matematleta Sep 28 '18 at 22:53
  • But why, because it's the same formula as in the above link, yet you say it's at $0$. But it's perhaps, because you defined $g$ as function of $t$. Although I didn't mention there being such function. – mavavilj Sep 28 '18 at 22:55
  • Also here the dir. deriv. is at $a$ not $t$: https://math.boisestate.edu/~zteitler/teaching/2013C/slides/slides_2013C_275_10-07_gradient-directional-derivatives.pdf. Here it's at $0$ http://www.utdallas.edu/~jwz120030/Teaching/PastCoursesUMBC/M423S10/M423L2B.pdf. – mavavilj Sep 28 '18 at 23:01
  • It's $a+tv$. I corrected the typo.$a$ and $v$ are fixed and $g$ is a function of $t$. $g(t)=a+tv$ while $g(0)=a$. So by $definition$ the derivative of $g$ at $0$ is $\lim_{t\to 0}\frac{g(t)-g(0)}{t}=\lim_{t\to 0}\frac{f(a+tv)-f(a)}{t}$ – Matematleta Sep 28 '18 at 23:02
  • I am pretty sure my calculation is correct. Defining $g$ as above is a trick, and it gets you to the result quickly. But I will happily delete my answer if you wish. – Matematleta Sep 28 '18 at 23:04
  • I think it's okay, but I'm unsure about where/whether you apply the $e_i$ part. – mavavilj Sep 28 '18 at 23:05
  • It's also possible that my version could be correct, since it's essentially the definition for partial derivative (multiplied by $v_i$) at $x_i$: https://en.wikipedia.org/wiki/Partial_derivative#Basic_definition – mavavilj Sep 28 '18 at 23:07
  • I am not an expert in the theory of manifolds, but I suspect the definition of derivative is for an abstract manifold with some coordinate chart $(v_a)$ about $a\in M$. The gist of the exercise is that if one uses the standard chart on a Euclidean space, then the result is the one I calculated. If you have another proof, I'd be glad to check it out. – Matematleta Sep 28 '18 at 23:07
  • Here's a solution that suggest that my version is correct (as well): http://homepages.math.uic.edu/~jwood/top/M549revnotes1.pdf p.2 – mavavilj Sep 28 '18 at 23:15
  • what is "your" version? You are quoting other people's proofs. I am not sure what you are even saying. Sorry. – Matematleta Sep 28 '18 at 23:55
  • "What I read is that this means that ..." in my question. – mavavilj Sep 29 '18 at 09:30
  • I have Lee's book somewhere. I will look up the statement and get back to you. – Matematleta Sep 29 '18 at 13:58
  • @mavavilj I wrote out the calculation in detail. – Matematleta Sep 29 '18 at 14:49