Definition of tangent vector on smooth manifold

  • #1
cianfa72
1,692
192
TL;DR Summary
About the coordinate-free definition of tangent vector on manifold
I would ask for a clarification about the following definition of tangent vector from J. Lee - Introduction to Smooth Manifold. It applies to Euclidean space ##R^n## with associated tangent space ##R_a^n## at each point ##a \in R^n##.

$$D_v\left. \right|_a (f)=D_vf(a)=\left. \frac {df(a + tv)} {dt} \right|_{t=0}$$
From my understanding the above is actually a coordinate-free definition. In other words ##a## inside ##f(a + tv)## is a point in ##R^n## and it is not the tuple of coordinates in some affine basis. The same for ##v##: it is a vector and is not the tuple of vector's components in some vector space basis.

So for example ##a=
\begin{bmatrix}
1 \\
5 \\
3 \\
2
\end{bmatrix} ## is a point in ##R^4## and ##v=
\begin{pmatrix}
2 \\
1 \\
6 \\
4
\end{pmatrix} ## is a vector in ##R^4## with vector space structure.
 
Last edited:
Physics news on Phys.org
  • #2
cianfa72 said:
TL;DR Summary: About the coordinate-free definition of tangent vector on manifold

I would ask for a clarification about the following definition of tangent vector ...
$$D_vf(a)=$$

... The Pantheon of Derivatives ...
 
  • #3
Is there a question here?
 
  • #4
martinbn said:
Is there a question here?
Yes, it is. Does my coordinate-free interpretation of tangent vector as in post #1 actually make sense w.r.t. the Lee definition ?
 
  • #5
cianfa72 said:
Yes, it is. Does my coordinate-free interpretation of tangent vector as in post #1 actually make sense w.r.t. the Lee definition ?
Yes. Why?
 
  • #6
cianfa72 said:
Yes, it is. Does my coordinate-free interpretation of tangent vector as in post #1 actually make sense w.r.t. the Lee definition ?
Yes and no. You only said that "all derivatives are directional derivatives." It is the major part of Weierstraß's notation ##f(x_0+v)=f(x_0)+ D_{x_0}(f)\cdot v + r(v)## (part 1 of my link).

However, you mentioned Riemannian manifolds. They are defined by coordinates as the central part of the concept, so you need a lot more to "eliminate" the coordinates again (part 3 of my link).

Weierstraß's formalism works fine in real (or complex) Euclidean spaces as in your example. General manifolds require a more elaborate handling.
 
  • #7
martinbn said:
Yes. Why?
Because in many text tangent vectors are defined using either implicitely or explicitly a smooth chart in a Atlas. The above from Lee seems to be instead a coordinate indipendent/intrinsic definition (even though at that stage in the book it is limited to Euclidean spaces).
 
  • #8
cianfa72 said:
Because in many text tangent vectors are defined using either implicitely or explicitly a smooth chart in a Atlas. The above from Lee seems to be instead a coordinate indipendent/intrinsic definition (even though at that stage in the book it is limited to Euclidean spaces).
You can use coordinates to define coordinate independent objects. And there are definitions of tangent vectors that do not use coordinates.

I am still don't understand the motivation for your question.
 
  • #9
If you pick a chart and define tangent vectors using it, then you must show that that definition is actually coordinate-indipendent. The kind of definition as in Lee is instead intrinsecally coordinate-free i.e. it defines, let me say from the beginning, an indipendent object.
 
  • #10
cianfa72 said:
If you pick a chart and define tangent vectors using it, then you must show that that definition is actually coordinate-indipendent. The kind of definition as in Lee is instead intrinsecally coordinate-free i.e. it defines, let me say from the beginning, an indipendent object.
And?
 
  • #11
Nothing, I just asked for a confirmation of my understanding of such definition.
 
  • #12
cianfa72 said:
Nothing, I just asked for a confirmation of my understanding of such definition.
But you didn't ask about the the dinition itself nor about your understanding of it. You asked whether it counts as coordinate independent. I am still puzzled. If you understand the definition what difference does it make if it is called this or that!
 
  • #13
martinbn said:
You asked whether it counts as coordinate independent.
That's the point, it is !
 
  • #14
Another point related to this: from the geodesic equation using an affine parameter we get ##g(V,V)=const## where ##V## is the geodesic tangent vector at each point along it. We can check this calculating ##\nabla_V g(V,V)## using the fact the ##\nabla_Vg=0## since the connection is metric compatible.

My doubt is the following: the map ##g(V,V)## is defined only along the geodesic curve and is not off it. Nevertheless the calculation in a coordinate basis actually involves derivatives in directions where the map is not defined.

How is that possibile?
 
Last edited:
  • #15
cianfa72 said:
Another point related to this: from the geodesic equation using an affine parameter we get ##g(V,V)=0## where ##V## is the geodesic tangent vector at each point along it.
Can you edit your post. This makes no sense. You probably mistyped it.
cianfa72 said:
We can check this calculating ##\nabla_V g(V,V)## using the fact the ##\nabla_Vg=0## since the connection is metric compatible.

My doubt is the following: the map ##g(V,V)## is defined only along the geodesic curve and is not off it. Nevertheless the calculation in a coordinate basis actually involves derivatives in directions where the map is not defined.

How is that possibile?
 
  • #16
Yes sorry, ##g(V,V)## is constant along the geodesic curve. My question is about why one is allowed to take directional derivatives along the curve when the map ##g(V,V)## is not defined off the curve.
 
  • #17
cianfa72 said:
Yes sorry, ##g(V,V)## is constant along the geodesic curve. My question is about why one is allowed to take directional derivatives along the curve when the map ##g(V,V)## is not defined off the curve.
The directional derivative depends only on the values along an integral curve of the vector in the direction you take the derivative of. So you can deferentiate quantities that are defined only along the curve. The general notion is induced connection along a submanifold.
 
  • #18
martinbn said:
The directional derivative depends only on the values along an integral curve of the vector in the direction you take the derivative of. So you can deferentiate quantities that are defined only along the curve.
But in a coordinate chart the directional derivative operator along a curve ##\frac {D} {d\lambda} ()## is defined as ##a^i {\partial_i}()##. Hence derivatives of the function along the coordinate basis are actually involved (i.e. the function the directional derivative operator acts on is required to be defined in an open neighborhood of each point along the curve).
 
Last edited:
  • #19
If you cannot figour it out why dont you look ot up! Lee's book on Riemannian manifold has that at the end of the section on connections. Where he discusses vector fields along curves and their derivatives along the curve.
 
  • #20
martinbn said:
Lee's book on Riemannian manifold has that at the end of the section on connections. Where he discusses vector fields along curves and their derivatives along the curve.
I took a look at Lee' book chapter 4. He introduces the concept of estendibile vector field starting from a vector field defined along a curve. I believe the same extension can be applied also for functions defined only along a curve.
 
  • #21
cianfa72 said:
TL;DR Summary: About the coordinate-free definition of tangent vector on manifold

I would ask for a clarification about the following definition of tangent vector from J. Lee - Introduction to Smooth Manifold. It applies to Euclidean space ##R^n## with associated tangent space ##R_a^n## at each point ##a \in R^n##.

$$D_v\left. \right|_a (f)=D_vf(a)=\left. \frac {df(a + tv)} {dt} \right|_{t=0}$$
From my understanding the above is actually a coordinate-free definition. In other words ##a## inside ##f(a + tv)## is a point in ##R^n## and it is not the tuple of coordinates in some affine basis. The same for ##v##: it is a vector and is not the tuple of vector's components in some vector space basis.

So for example ##a=
\begin{bmatrix}
1 \\
5 \\
3 \\
2
\end{bmatrix} ## is a point in ##R^4## and ##v=
\begin{pmatrix}
2 \\
1 \\
6 \\
4
\end{pmatrix} ## is a vector in ##R^4## with vector space structure.
Hmm. I would say that is stretching things. There are two problems here. First, you are only dealing with ##\mathbb{R}^n##. The second is that for something to be coordinate free it shouldn't rely on a coordinate chart around the point ##a##.

Honestly, I don't think the question makes much sense until you talke about tangent vectors on a general manifold.
 
  • #22
jbergman said:
There are two problems here. First, you are only dealing with ##\mathbb{R}^n##. The second is that for something to be coordinate free it shouldn't rely on a coordinate chart around the point ##a##.
Indeed, there is no chart involved. ##\mathbb R^n## is an affine Euclidean space. ##a## and ##v## are respectively a point in ##\mathbb R^n## and a vector in the translation vector space (so ##a + tv## is well-defined and coordinate-free).

jbergman said:
Honestly, I don't think the question makes much sense until you talke about tangent vectors on a general manifold.
Yes, the step further is define tangent vectors as derivation on smooth functions defined on the manifold.
 

Similar threads

A About computing the tangent space at 1 of certain lie groups
    • Differential Geometry
Replies
4
Views
1K
I Differential operator vs one-form (covector field)
    • Differential Geometry
Replies
10
Views
642
I The Road to Reality - exercise on scalar product
    • Differential Geometry
Replies
2
Views
514
I Tangent space basis vectors under a coordinate change
    • Differential Geometry
Replies
12
Views
3K
I Definition of tangent space: why germs?
    • Differential Geometry
Replies
6
Views
6K
I Tangent vector basis and basis of coordinate chart
    • Differential Geometry
Replies
6
Views
3K
I Differential k-form vs (0,k) tensor field
    • Differential Geometry
Replies
8
Views
2K
A A claim about smooth maps between smooth manifolds
    • Differential Geometry
Replies
20
Views
2K
I Maps with the same image are actually different curves?
    • Differential Geometry
Replies
21
Views
2K
A Hilbert spaces and kets "over" manifolds
    • Differential Geometry
Replies
10
Views
2K

Hot Threads

Recent Insights

Back
Top