A question on defining vectors as equivalence classes

  • #1
I understand that a tangent vector, tangent to some point [itex]p[/itex] on some [itex]n[/itex]-dimensional manifold [itex]\mathcal{M}[/itex] can defined in terms of an equivalence class of curves [itex][\gamma][/itex] (where the curves are defined as [itex]\gamma: (a,b)\rightarrow U\subset\mathcal{M}[/itex], passing through said point, such that [itex]\gamma (0)= p[/itex]), under the equivalence relation $$\gamma_{1} \sim \gamma_{2} \iff \left(\varphi\circ\gamma_{1}\right)'(0)= \left(\varphi\circ\gamma_{2}\right)'(0)$$
where [itex](U,\varphi )[/itex] is some coordinate chart such that [itex]\varphi :U\rightarrow\mathbb{R}^{n}[/itex], with [itex]\varphi (p)= x= \lbrace x^{\mu}\rbrace[/itex].

Am I correct in assuming that this definition relies on the fact that the directional derivative of a function is independent of the curve one chooses to parametrise it by? If so, is this the correct way to prove it?

Let [itex]f:\mathcal{M}\rightarrow\mathbb{R}[/itex] be a differential function of class [itex]C^{k}[/itex] and let [itex]\gamma_{1}: (a,b)\rightarrow U[/itex] and [itex]\gamma_{2}: (a,b)\rightarrow U[/itex] be two curves, parametrised by [itex]t[/itex] and [itex]s[/itex], respectively, both passing through the point [itex]p\in U\subset\mathcal{M}[/itex] such that [itex]\gamma_{1} (0)=p= \gamma_{2} (0)[/itex]. Furthermore, suppose that $$\left(\varphi\circ\gamma_{1}\right)'(0)= \left(\varphi\circ\gamma_{2}\right)'(0)$$ (via the coordinate chart as defined above).

We have then, that the directional derivative of the function [itex]f[/itex] through the point [itex]p\in U\subset\mathcal{M}[/itex] is given by $$\frac{df}{dt}\Biggr\vert_{t=0}= \frac{d(f\circ\gamma_{1})}{dt}\Biggr\vert_{t=0} = \frac{\partial f (p)}{\partial x^{\mu}}\left(\varphi\circ\gamma_{1}\right)'(0) = \frac{\partial f (p)}{\partial x^{\mu}}\left(\varphi\circ\gamma_{2}\right)'(0) = \frac{d(f\circ\gamma_{2})}{ds}\Biggr\vert_{s=0}$$
As such, the directional derivative of [itex]f[/itex] at [itex]p\in U\subset\mathcal{M}[/itex] is independent of the curve it's parametrised by.

Given this we can define the a tangent vector [itex]\dot{q}[/itex] at a point [itex]q\in\mathcal{M}[/itex] as the equivalence class of curves passing though the point [itex]q\in\mathcal{M}[/itex] (as defined earlier). The tangent space to [itex]\mathcal{M}[/itex] at the point [itex]q\in\mathcal{M}[/itex] is then defined in the following manner $$\lbrace\dot{q}\rbrace = \lbrace [\gamma] \;\vert \quad\gamma (0)=q\rbrace$$ [itex]\dot{q}[/itex] then acts on functions [itex]f[/itex] (as defined earlier) to produce the directional derivative of [itex]f[/itex] at the point [itex]q[/itex] in the direction of [itex]\dot{q}[/itex] as follows $$\dot{q}[f] =\frac{d(f\circ\gamma)}{dt}\Biggr\vert_{t=0}$$
Would this be correct? (I'm deliberately using the notation [itex]\dot{q}[/itex] for the tangent vectors as I'm approaching it from a physicist's point of view, with the aim of motivating the phase space for Lagrangian dynamics, and explicitly showing why [itex]q[/itex] and [itex]\dot{q}[/itex] can be treated as independent variables in the Lagrangian).

From this, can one then prove that the definition of a tangent vector as an equivalence class of curves is independent of coordinate chart.

Suppose that [itex](U,\varphi )[/itex] and [itex](V, \psi )[/itex] are two coordinate charts such that [itex]U \cap V \neq\emptyset[/itex] and let [itex]p\in U \cap V[/itex]. Let [itex]\gamma_{1}[/itex] and [itex]\gamma_{2}[/itex] be two coordinate curves (as defined previously) such that [itex]\gamma_{1} (0)=p=\gamma_{2} (0)[/itex]. It follows from the chain rule, that $$\left(\psi\circ\gamma_{1}\right)^{\prime}(0)=\left((\psi\circ\varphi^{-1})\circ (\varphi\circ\gamma_{1})\right)^{\prime}(0) \qquad\qquad\qquad\qquad \\ = \left(\psi\circ\varphi^{-1}\right)^{\prime}(\varphi (p))\left(\varphi\circ\gamma_{1}\right)^{\prime} (0) \qquad \\ = \left(\psi\circ\varphi^{-1}\right) ^{\prime}(\varphi (p))\left(\varphi\circ\gamma_{2}\right)^{\prime} (0) \qquad \\ = \left((\psi\circ\varphi^{-1})\circ (\varphi\circ\gamma_{2})\right)^{\prime}(0) \qquad\quad \\ = \left(\psi\circ\gamma_{2}\right)^{\prime}(0)\qquad\qquad\qquad\qquad\quad$$
As such, if the equivalence relation holds in one coordinate chart [itex](U,\varphi )[/itex] then it holds in any other (as [itex](V, \psi )[/itex] was chosen arbitrarily, other than it overlap with [itex](U, \varphi )[/itex] in the neighbourhood of [itex]p\in \mathcal{M}[/itex]).

Would this be correct?

Apologies in advance for the long-windedness of this post, just keen to check my understanding.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Looks reasonable. I assume you are assuming that ##\gamma_1## and ##\gamma_2## are in the same equivalence class based on ##(\varphi \circ \gamma_1)'(0) = (\varphi \circ \gamma_2)'(0)## in your last part, this was not obvious to me on first reading.
 
  • #3
are in the same equivalence class based on (φ∘γ1)′(0)=(φ∘γ2)′(0)
Yes, I am. Sorry I didn't make it very clear.
Is what I've put correct in general though?
 
  • #4
177
61
We have then, that the directional derivative of the function [itex]f[/itex] through the point [itex]p\in U\subset\mathcal{M}[/itex] is given by $$\frac{df}{dt}\Biggr\vert_{t=0}= \frac{d(f\circ\gamma_{1})}{dt}\Biggr\vert_{t=0} = \frac{\partial f (p)}{\partial x^{\mu}}\left(\varphi\circ\gamma_{1}\right)'(0) = \frac{\partial f (p)}{\partial x^{\mu}}\left(\varphi\circ\gamma_{2}\right)'(0) = \frac{d(f\circ\gamma_{2})}{ds}\Biggr\vert_{s=0}$$
As such, the directional derivative of [itex]f[/itex] at [itex]p\in U\subset\mathcal{M}[/itex] is independent of the curve it's parametrised by.
Hi Don't panic!
I am not sure I understand you here. What is ##\frac{\partial f(p)}{\partial x^\mu}##?

You should go through local coordinate chart: define ##\tilde f := f\circ \varphi^{-1}##, ##\tilde\gamma_{1,2} = \varphi\circ \gamma_{1,2}##. Then $$f\circ\gamma_{1,2} = \tilde f\circ\tilde\gamma_{1,2}. $$ Condition ##\gamma_1 \sim\gamma_2## means that ##\tilde\gamma_1'(0) = \tilde\gamma_2'(0)##, which immediately implies that ##(f\circ \gamma)'(0) = (\tilde f\circ \tilde \gamma)'(0)## does not depend on the choice of the representative ##\gamma## (just use chain rule).

Probably before doing this calculation you should prove that equivalence of ##\gamma##s does not depend on a choice of ##\varphi##, you've done this at the end of your post. Then the directional derivative does not depend on the choice of ##\varphi##, because it is just derivative of ##f\circ\gamma##, no ##\varphi## is involved here (but to compute the derivative, we need to involve ##\varphi## ).
 
  • #5
I am not sure I understand you here. What is ∂f(p)∂\frac{\partial f(p)}{\partial x^\mu}
I am not sure I understand you here. What is ∂f(p)∂\frac{\partial f(p)}{\partial x^\mu}?

You should go through local coordinate chart: define f~:=fφ−1\tilde f := f\circ \varphi^{-1}, γ~1,2=φγ1,2\tilde\gamma_{1,2} = \varphi\circ \gamma_{1,2}. Then
fγ1,2=f~∘γ~1,2.​
Sorry, that's what I had meant, just laziness on my part that I was abusing notation.

Probably before doing this calculation you should prove that equivalence of γ\gammas does not depend on a choice of φ\varphi, you've done this at the end of your post
Have I proven this correctly?

Then the directional derivative does not depend on the choice of φ\varphi, because it is just derivative of fγf\circ\gamma, no φ\varphi is involved here (but to compute the derivative, we need to involve φ\varphi )
I assume this is how we can claim that the definition of a tangent vector to some point on a manifold is independent of coordinates, i.e. [tex]\frac{df\circ\gamma (t)}{dt}\Biggr\vert_{t=0}= \frac{d (\varphi\circ\gamma (t))^{\mu}}{dt}\frac{\partial}{\partial x^{\mu}}\Biggr\vert_{t=0}f[/tex] is coordinate independent as the left-hand side of this equation makes no reference to any particular coordinate chart [itex]\varphi[/itex]?!

Thanks for taking a look at it all.
 
  • #6
177
61
Have I proven this correctly?
Yes, you calculation showing that the equivalence of paths does not depend on the choice of coordinate chart is correct.

I assume this is how we can claim that the definition of a tangent vector to some point on a manifold is independent of coordinates, i.e. [tex]\frac{df\circ\gamma (t)}{dt}\Biggr\vert_{t=0}= \frac{d (\varphi\circ\gamma (t))^{\mu}}{dt}\frac{\partial}{\partial x^{\mu}}\Biggr\vert_{t=0}f[/tex] is coordinate independent as the left-hand side of this equation makes no reference to any particular coordinate chart [itex]\varphi[/itex]?!
Yes, exactly. But I think right hand side here should be $$\frac{\partial (f\circ\varphi^{-1})}{\partial x^{\mu}}\Biggm\vert_{x=\varphi(\gamma(0))} \frac{d (\varphi\circ\gamma (t))^{\mu}}{dt} \Biggm\vert_{t=0} $$
 
  • #7
mathwonk
Science Advisor
Homework Helper
2020 Award
11,099
1,302
my apologies for the following post, which i see now just reproduces a few lines in the original one, with which i thus completely agree. i just couldn't believe it could take so long to write this down, so i didn't read the whole post until now. the argument is entirely contained already in lines -9 through -5 of post #1.

[repetitious]:

If f,g are curves through p, i.e. smooth maps from the real line equaling p when t = 0, and if F,G are coordinate systems near p, then assume f,g are equivalent in coordinate system G, i.e. that (Gf)'(0) = (Gg)'(0). Then we have (Ff)'(0) = (FG^(-1)Gf)'(0) = (by chain rule) (FG^(-1))'(Gf)'(0) = (by hypothesis) (FG^(-1))'(Gg)'(0) = (chain rule) (FG^(-1)Gg)'(0) = (Fg)'(0). Hence f,g are also equivalent by coordinate system F. qed.
 
Last edited:
  • #8
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
Don't panic, you may be interested in my posts here. Unfortunately it takes a very long time to load the LaTeX on that page.

I don't like the notations in post #1 much. In particular, I don't like the notation ##\frac{\partial f(p)}{\partial x^\mu}## for ##(f\circ\varphi^{-1})_{,\mu}(\phi(p))##. It would be better to write ##\frac{\partial f(p)}{\partial \phi^\mu}##. If you prefer to see an ##x## in the denominator, just call the coordinate system ##x##. That's what I do.
 
  • #9
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,460
2,625
The way I had seen it is not to mention coordinates at all, and instead define:

If [itex]\gamma_1 : R \rightarrow \mathcal{M}[/itex], and [itex]\gamma_2 : R \rightarrow \mathcal{M}[/itex], then [itex]\gamma_1 \sim \gamma_2[/itex] if
  1. [itex] \gamma_1(0) = \gamma_2(0)[/itex]
  2. For all smooth functions [itex]f : \mathcal{M} \rightarrow R[/itex]: [itex]\frac{d}{dt} f(\gamma_1(t)) = \frac{d}{dt} f(\gamma_2(t))[/itex] at [itex]t=0[/itex]
 
  • #10
177
61
The way I had seen it is not to mention coordinates at all, and instead define:

If [itex]\gamma_1 : R \rightarrow \mathcal{M}[/itex], and [itex]\gamma_2 : R \rightarrow \mathcal{M}[/itex], then [itex]\gamma_1 \sim \gamma_2[/itex] if
  1. [itex] \gamma_1(0) = \gamma_2(0)[/itex]
  2. For all smooth functions [itex]f : \mathcal{M} \rightarrow R[/itex]: [itex]\frac{d}{dt} f(\gamma_1(t)) = \frac{d}{dt} f(\gamma_2(t))[/itex] at [itex]t=0[/itex]
To define what is a smooth function you still need the coordinates.
 
  • #11
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,460
2,625
To define what is a smooth function you still need the coordinates.
Sort of. The definition of a "manifold" is in terms of a collection of patches of [itex]R^N[/itex], together with overlaps and rules relating overlapping patches. Within a given patch, "smooth" means infinitely differentiable in the ordinary sense of functions on [itex]R^N[/itex]. Similarly, the notion of a smooth parametrized path just means that within a patch, [itex]\gamma(t)[/itex] is an infinitely differentiable function from [itex]R[/itex] to [itex]R^N[/itex]. I suppose you could think of the little patches of [itex]R^N[/itex] as a pre-existing coordinate system for that section of the manifold, but it doesn't have to have any relationship to the coordinates that people are actually using (other than the fact that the map between them must be differentiable).

But the specification of the patches (maybe they're called "charts") is part of what it means to be a differentiable manifold. So associated with the manifold is a set of smooth scalar fields and parametrized paths. Then the definition of tangent vector doesn't need to mention coordinates.
 
  • #12
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,460
2,625
I would say that it's backwards to say that you need a coordinate system in order to specify what it means for a function to be "smooth". You need a notion of "smooth" in order to say what's a coordinate system. A coordinate system on a patch [itex]\mathcal{M}'[/itex] is defined to be a smooth, invertible map from [itex]\mathcal{M}'[/itex] to [itex]R^N[/itex].
 
  • #13
177
61
I suppose you could think of the little patches of RNR^N as a pre-existing coordinate system for that section of the manifold,
Yes, exactly, in this thread word "coordinates" (or coordinate charts) was often used for what you (and many books) call just "patches".
 
  • #14
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
I would say that it's backwards to say that you need a coordinate system in order to specify what it means for a function to be "smooth". You need a notion of "smooth" in order to say what's a coordinate system. A coordinate system on a patch [itex]\mathcal{M}'[/itex] is defined to be a smooth, invertible map from [itex]\mathcal{M}'[/itex] to [itex]R^N[/itex].
You need smoothness of functions on subsets of ##\mathbb R^n## to define smooth manifolds and their coordinate systems, and then you need both that first smoothness concept and those coordinate systems to define smoothness of functions on smooth manifolds. ##f## is said to be smooth if ##f\circ x^{-1}## is smooth for all ##x##.
 
  • #15
22,089
3,294
To define what is a smooth function you still need the coordinates.
Not necessarily, you can always introduce them axiomatically. For example: diffeological spaces.
 

Related Threads on A question on defining vectors as equivalence classes

Replies
7
Views
5K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
6K
Replies
11
Views
5K
Replies
1
Views
3K
Replies
7
Views
3K
Replies
6
Views
4K
Replies
1
Views
7K
Replies
3
Views
2K
Top