Demonstrating Tangent Space Independence in Manifolds

[Calabi]

Hello I'm french so sorry for the mistake. If we have a manifold and a point p with a card (U, x) defined on on an open set U which contain p, of the manifold, we can defined the tangent space in p by the following equivalence relation : if we have 2 parametered curve : dfinded from $$\mathbb{R}$$ to the manifold, I wright this curve \gamma_{1} and \gamma_{2}. If \gamma_{1}(0) = \gamma_{2}(0) = p and that I know it exist a card (U, x) on p, the 2 curve are linked by the equvalence relation if (x o \gamma_{1})'(0) = (x o \gamma_{2})'(0).
The quotient space is the set of the equivalence class for this relation in p is the tangeant space in p. The quantity (x o \gamma)'(0) for a certain curve permit a bijection from the tangent space to \mathbb{R}^{n}. So the tangent space is a vectorial space. And a vector of this space is a an equivalences class. A set of curve which have the same (x o \gamma)'(0)

How to demosntrate that this definition is independant of the choice of card please?

I try to say that (x o \gamma_{1})'(0) = (x o \gamma_{2})'(0) \Leftrightarrow (x o y^{-1} o y o \gamma_{1})'(0) = (x o y^{-1} o y o \gamma_{2})'(0) but I don't know how to go on.

Thank you in advance and have a nice afternoon:D.

 

[lavinia]

If F: V -> U is the change of coordinates then compose your curves with F and use the Chain Rule.

In your notation,

$xoy−1$
is equal to F

 

[Fredrik]

I'm french so sorry for the mistake.

No problem. The only significant mistake you made was "card". The word is "chart" (or "coordinate system").

I would also recommend that you use the LaTeX code \circ instead of an o, and that you take a quick look at our LaTeX guide: https://www.physicsforums.com/help/latexhelp/

How to demosntrate that this definition is independant of the choice of card please?

You're lucky today. I have answered this before. See posts #12 and #14 here: https://www.physicsforums.com/threads/tangent-space-in-manifolds-how-do-we-exactly-define.685722/

 

[Calabi]

Hello thank you Cartman. I'm going to see your post thanks.

 

[Calabi]

I bet that you perhaps don't have time for this but please it's very important.

(xoγ1)′(0)=(xoγ2)′(0)⇔(xoy−1oyoγ1)′(0)=(xoy−1oyoγ2)′(0)

Could you developped please? I'm sorry to ask which seems basic for you. I'm juste not aware with the notation.
I'm going to check the message Frederik post. And I keep you in touch if I have news.

Thank you in advance and have a nice afternoon.

 

[Fredrik]

Lavinia is just telling you to use the chain rule, as I did in the first calculation in post #12 in the thread I linked to.

When the functions are defined on subsets of $\mathbb R$, the chain rule is written as $(f\circ g)'(x)=f'(g(x))g'(x)$. When they're defined on subsets of $\mathbb R^n$ rather than $\mathbb R$, it takes the form
$$(f\circ g)_{,i}(x)=\sum_{j=1}^n f_{,j}(g(x))g^j{}_{,i}(x).$$ If you don't like the comma notation for partial derivatives, you can write
$$D_i(f\circ g)(x)=\sum_{j=1}^n D_j f(g(x))D_i g^j(x).$$ I strongly recommend that you don't use notations like $\frac{\partial}{\partial x^i}$ for partial derivatives of the kind encountered in a calculus course, because this looks too much like the partial derivatives in differential geometry, which are defined using the partial derivative concept from calculus.

 

[Calabi]

Hello, so if I have (x o \gamma_{1})'(0)=(x o \gamma_{2})'(0), I can developed by using the chaine rule, sorry but here I use the notations d rondes I'll stop later promise :

\frac{ \part (x o y^{-1} o y o \gamma_{1} )_{i} }{\part t}(0) = \sum_{k=1}^{n} \frac{\part (x o y^{-1})_{i}}{\part y^{k}}(( y o \gamma_{1})(0)) \frac{\part (y o \gamma_{1})_{k}}{\part t}(0) = \frac{ \part (x o y^{-1} o y o \gamma_{2} )_{i} }{\part t}(0) = \sum_{k=1}^{n} \frac{\part (x o y^{-1})_{i}}{\part y^{k}}(( y o \gamma_{2})(0)) \frac{\part (y o \gamma_{2})_{k}}{\part t}(0). I see that ther's the jacobian here of the charts changing. Who's defined because of the C^{k} differntial manifold strucuture.

Then what can I do please? I don't see the same definition with this other charts I introduce.

I recall that my first goal was to proove that if I defined the tangent space with a certain charts, with the curve the bijection with \mathbb{R}^{n}, I can do the same things with anothers charts.

So it's equivalence to say that the vectroial space structure, which we have show by the bijection with \mathbb{R}^{n} with a certain charts, is independant of the choice of charts.

Thank you in advance and have a nice afternoon.

 

[Fredrik]

You need to fix your LaTeX. It's \partial, not \part, and \circ, not o.

 

[Calabi]

Well I get what I have to do, that was not so complicate, so if I have (x o \gamma_{1})'(0)=(x o \gamma_{2})'(0), I can developed by using the chaine rule, sorry but here I use the notations d rondes I'll stop later promise :
(y o \gamma_{1})'(0) =
\frac{ \partial (y o x^{-1} o x o \gamma_{1} )_{i} }{\partial t}(0) = \sum_{k=1}^{n} \frac{\partial (y o x^{-1})_{i}}{\partial x^{k}}(( y o \gamma_{1})(0)) \frac{\partial (x o \gamma_{1})_{k}}{\partial t}(0) =

\sum_{k=1}^{n} \frac{\partial (y o x^{-1})_{i}}{\partial x^{k}}(( y o \gamma_{1})(0)) \frac{\partial (x o \gamma_{2})_{k}}{\partial t}(0) = \frac{ \partial (y o x^{-1} o x o \gamma_{2} )_{i} }{\partial t}(0)

= (y o \gamma_{2})'(0).

So if 2 curve are equivalent in a certain charts they are in others charts to. So it's independante of the choice of charts.

Which I can Right : (y o \gamma_{1})'(0) = \frac{\partial y}{\partial x}(x o \gamma_{1})'(0)= \frac{\partial y}{\partial x}(x o \gamma_{2})'(0) = (y o \gamma_{2})'(0)..

Where \frac{\partial y}{\partial x} is the jacobian of the charts changing.

Is it True or False please?

Which canonic base we choose with a certain charts please?

I'd like to check if the jacobian is a matrix passage between 2 base each associataed to a charts.

Thank you in advance and have a nice morning.

 

[Fredrik]

Which I can Right : (y o \gamma_{1})'(0) = \frac{\partial y}{\partial x}(x o \gamma_{1})'(0)= \frac{\partial y}{\partial x}(x o \gamma_{2})'(0) = (y o \gamma_{2})'(0)..

Where \frac{\partial y}{\partial x} is the jacobian of the charts changing.

Is it True or False please?

True.

Which canonic base we choose with a certain charts please

The ordered basis for the tangent space $T_pM$ associated with the chart $x$ is $\big(\frac{\partial}{\partial x^1}\!\big|_p,\dots,\frac{\partial}{\partial x^n}\!\big|_p\big)$. For each $i\in\{1,\dots,n\}$, $\frac{\partial}{\partial x^i}\!\big|_p$ is defined by
$$\frac{\partial}{\partial x^i}\bigg|_p f =(f\circ x^{-1})_{,i}(x(p))$$ for all smooth $f:M\to\mathbb R$.

 

[Calabi]

Well many thanks Frederic. But you anticiped the things : here you present the base but by saying that there an isomorphism between quotient space of the equivalence relation we defined with the curve and all the derivation operators space(who check the Leibniz and linear rules.). who act(bon le mots en français est agir.). on all the scalar field defined on neighbourhood of the point p.

Which base could I choose with a certain charts please? Without the derivation notation.

I thinks the equivalence class of curve whose the quantity (x o \gamma)'(0) check the following relation : (x o \gamma)'(0)_{i} = \delta_{i}^{j}. So it's equats the n uplets (x o \gamma)'(0) is nul exept in j.

And for a certain vector(elements of the quotient/an equivalence class.). with a certain charts the n uplets (x o \gamma)'(0) is all the components of the vector in the base asocitated with the charts I defined.

Is it True or False please?

Thank you in advance and have a nice afternon.

 

[lavinia]

I am reluctant to add anything more here since Frederick has explained everything but may this will help you a little more.

You are really talking about two different ideas. The first is the idea of a directional derivative, the second the idea of independence of coordinate charts.

- Directional derivative.

Given any vector at a point in space, one may define the directional derivative of a function with respect to v. Take any curve whose derivative at the point is the vector,v, then differentiate the function along the curve. It turns out that the derivative of the function is independent of the curve chosen as along as its derivative at the point is the vector v. Thus one may form an equivalence relation for curves that pass through the point by saying that they are equivalent if they have the same derivative at the point.

What you are doing is defining the equivalence relation first before you prove that the derivatives of a function are the same for all equivalent curves. But at some point you must prove that. Otherwise the definition is useless.

- Independence of coordinate chart

Here all you need to do is show that if two curves have the same derivative at a point then the curves obtained by composing them with a diffeomorphism also have the same derivative. A change of coordinates is nothing more than a diffeomorphism. This follows immediately from the Chain Ruke.

 

[Calabi]

Hello Lavina, and thanks. Sorry but the curse I read define the tangent space as the quotient space I defined and then/after he talks about the derivation space. So me I want to first define the base I talk about then see the isomorphism between the 2 space to indentifed them.
If you not agree on that tell me : I'm a beginner and you're the expert. So if it's better just tell me.

By the way what I say about the base is it True or False please?

Thank you in advance and have a nice afternoon.

 

[Fredrik]

But you anticiped the things : here you present the base but by saying that there an isomorphism between quotient space of the equivalence relation we defined with the curve and all the derivation operators space(who check the Leibniz and linear rules.). who act(bon le mots en français est agir.). on all the scalar field defined on neighbourhood of the point p.

Which base could I choose with a certain charts please? Without the derivation notation.

OK, I think I see what you mean. Let $T_pM$ be the tangent space at p, defined as a set of equivalence classes of curves, and let $D_pM$ be the other tangent space at p, defined as a set of derivations at p. I gave you a basis for $D_pM$ when you wanted a basis for $T_pM$. So let's think it through. There must be an n-tuple of curves $(\gamma_1,\dots,\gamma_n)$ such that $([\gamma_1],\dots,[\gamma_n])$ is the ordered basis for $T_pM$ that corresponds to the ordered basis for $D_pM$ that I mentioned in my previous post. The natural choice seems to be to define $\gamma_i$ by $\gamma_i(t)=x^{-1}(te_i)$ for all $t\in\mathbb R$ such that $te_i$ is in the range of $x$.

The isomorphism between $T_pM$ and $D_pM$ is the function $\phi:T_pM\to D_pM$ defined by $\phi([C])(f)=(f\circ C)'(0)$ for all smooth $f:M\to\mathbb R$. It's not hard to show that the right-hand side is independent of what curve C we choose from the equivalence class [C].
$$\phi([C])(f)=(f\circ C)'(0)=(f\circ x^{-1}\circ x\circ C)'(0)=(f\circ x)_{,i}(x(C(0))) (x\circ C)^i{}'(0).$$ For all $D\in [C]$, we can substitute D for C on the right, without changing the value of the right-hand side.

If my definition of $\gamma_i$ is appropriate, we should have $\phi([\gamma_i])=\frac{\partial}{\partial x^i}\!\big|_p$ for all $i$. So let's verify that we do. We have $\phi([\gamma_i])=(f\circ\gamma_i)'(0)$, and $f\circ\gamma_i$ is the map that takes $t$ to $(f\circ\gamma_i)(t)=f(\gamma_i(t))=f(x^{-1}(te_i))$, so we have
$$\phi([\gamma_i])=(f\circ\gamma_i)'(0)=\lim_{t\to 0}\frac{f(x^{-1}(te_i))-f(x^{-1}(0))}{t} =(f\circ x^{-1})_{,i}(0)= (f\circ x^{-1})_{,i}(x(p)) =\frac{\partial}{\partial x^i}\bigg|_p f.$$

I thinks the equivalence class of curve whose the quantity (x o \gamma)'(0) check the following relation : (x o \gamma)'(0)_{i} = \delta_{i}^{j}. So it's equats the n uplets (x o \gamma)'(0) is nul exept in j.

And for a certain vector(elements of the quotient/an equivalence class.). with a certain charts the n uplets (x o \gamma)'(0) is all the components of the vector in the base asocitated with the charts I defined.

Is it True or False please?

I'm thinking that the $i$th component of $(x\circ\gamma_j)'(0)$ is $(x^i\circ\gamma_j)'(0)$, and that $x^i\circ\gamma_j$ is the map that takes $t$ to $(x^i\circ\gamma_j)(t)=x^i(\gamma_j(t))=x^i(x^{-1}(te_j)) =(te_j)^i =t\delta^i_j$. This implies that $(x\circ\gamma_j)'(0)=\delta^i_j$. So if I understand you correctly, you got it right.

 

[Calabi]

Yeah and the jacobian is the matrix passage betwwen this 2 basis. All ight so it's answe to all my question. Well thanks to all of you. Lavina and Fredrik.

Of course now I want to study the derivation space in p. If I have question about that could I continue here please? Exept if it surcharge the thead. However it's the same context. Thank you in advance and have a nice afternoon.

 

[Fredrik]

Of course now I want to study the derivation space in p. If I have question about that could I continue here please? Exept if it surcharge the thead.

You can continue here or start a new thread. It doesn't matter much.

 

[Calabi]

Oh hell I've made mistake :
so if I have (x o \gamma_{1})'(0)=(x o \gamma_{2})'(0), I can developed by using the chaine rule, sorry but here I use the notations d rondes I'll stop later promise :
(y o \gamma_{1})'(0) =
\frac{ \partial (y o x^{-1} o x o \gamma_{1} )_{i} }{\partial t}(0) = \sum_{k=1}^{n} \frac{\partial (y o x^{-1})_{i}}{\partial x^{k}}(( x o \gamma_{1})(0)) \frac{\partial (x o \gamma_{1})_{k}}{\partial t}(0) =
\sum_{k=1}^{n} \frac{\partial (y o x^{-1})_{i}}{\partial x^{k}}(( x o \gamma_{2})(0)) \frac{\partial (x o \gamma_{2})_{k}}{\partial t}(0) = \frac{ \partial (y o x^{-1} o x o \gamma_{2} )_{i} }{\partial t}(0)
= (y o \gamma_{2})'(0).
So if 2 curve are equivalent in a certain charts they are in others charts to. So it's independante of the choice of charts.
Which I can Right : (y o \gamma_{1})'(0) = \frac{\partial y}{\partial x}(x o \gamma_{1})'(0)
= \frac{\partial y}{\partial x}(x o \gamma_{2})'(0) = (y o \gamma_{2})'(0)..
Where \frac{\partial y}{\partial x} is the jacobian of the charts changing.
Is it True or False please?I reverse a x with a y.

Good afternoon.

 

[Fredrik]

Yes, that looks better. I didn't see the mistake before.