Proving \alpha^i = v x^i on tangent vector space T_pM"

In summary, the conversation is about proving a property for tangent vector spaces on a differentiable manifold using the concept of inner product spaces. The approach involves defining an inner product on the tangent space and using it to prove the property. The idea of using coordinates and curves is also discussed.
  • #1
QuarkHead
8
0
Let me first confess this a copy/paste of a question I asked on another forum; I trust it's not against the rules.

Let [tex]M[/tex] be a [tex]C^{\infty}[/tex] manifold, and, for some neighbourhood [tex]U\ni p \subsetneq M[/tex] let there be local coordinates [tex]x^i [/tex] such that [tex]p=(x^1,\,x^2,...,x^n)[/tex]

Suppose that [tex]T_pM[/tex] is a tangent vector space at [tex]p[/tex], and define a coordinate basis for [tex]T_pM[/tex] as [tex]\frac{\partial}{\partial x^i}[/tex].

By modeling on "ordinary" linear algebra, suppose that any [tex]v \in T_pM = \sum\nolimits_ i \alpha^i \frac{\partial}{\partial x^i}[/tex], where the [tex]\{\alpha^i\}[/tex] are scalar.

I want to prove that [tex]\alpha^i = v x^i[/tex].

My thoughts, based on inner product spaces...

Suppose [tex]V[/tex] is a vector space with inner products. Let the set [tex]\{e_j\} [/tex] denote the basis vectors. Then any [tex]v \in V[/tex] can be expressed as [tex]v = \sum \nolimits_j a^j e_j[/tex], where the [tex]\{a^j\}[/tex] are scalar.

Now the inner product of an arbitrary basis vector with an arbitrary vector will be denoted by [tex](v,e_i) = \sum \nolimits_j(a^j e_j, e_i)=\sum \nolimits_j a^j( e_j, e_i)[/tex] (since inner products are bilinear) hence [tex] (v, e_i) = \sum \nolimits_j a^j(e_j,e_i) = \sum \nolimits_j a^j \delta _{ij} = a^i[/tex]

This looks promising, except we don't have an IP on [tex]T_pM[/tex], and moreover, the [tex]\{x^i\}[/tex] are coordinates, not a basis!

Where do I go from here? I tried the simple operation [tex]vx^i = \sum \nolimits_i \alpha^j \frac{\partial}{\partial x^j}x^i = \alpha^j\delta_{ij}= \alpha^i[/tex] but I am told this is no proof

Any thoughts out there? Have I effed up somewhere?
 
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  • #2
This is not really my cup of tea, but can't you simply define the inner product? After all the tangent space is just a linear vector space like Rn, just define a map [tex](e_i, e_j) \mapsto \delta_{ij}[/tex] and extend it linearly. Would that solve your problem?

Also I don't really get the last remark. Isn't [itex]\partial_i = \frac{\partial}{\partial x^i}[/itex] by definition the basis dual to [itex]x_i[/itex], such that [itex]\partial_i x^j \equiv \delta_{ij}[/itex]?

You might want to ask someone more knowledgeable, but that's my thought on it.
 
  • #3
Well,

[tex]
vx^i = \sum \nolimits_i \alpha^j \frac{\partial}{\partial x^j}x^i = \alpha^j\delta_{ij}= \alpha^i
[/tex]

it right, and is the classical way to do it that you will find in every book existing on differentiable manifold theory, provided you define properly what you mean by that x^i being differentiated...
 
  • #4
I thank you both for your responses.
quasar987 said:
Well,

[tex]
vx^i = \sum \nolimits_i \alpha^j \frac{\partial}{\partial x^j}x^i = \alpha^j\delta_{ij}= \alpha^i
[/tex]

it right, and is the classical way to do it that you will find in every book existing on differentiable manifold theory
I was told that this is sufficient to prove uniqueness, but not existence. Maybe I was told wrong; again I thank you both
 
  • #5
I have no idea about rigour, but I think is it has nothing to do with inner products. It starts by defining a manifold as something on which you can put coordinates. Coordinates are functions from the manifold to the real numbers. Since we already put coordinate functions, it's no problem if we put other functions F. Then we introduce curves which are maps from the real numbers into the manifold (say like time t along a trajectory). Then whatever functions on the manifold will change along the curve parameter F(t), for which we can just do normal differentiation. The partial derivative of x wrt x is the special case of differentiating a coordinate function along a coordinate curve.
 

FAQ: Proving \alpha^i = v x^i on tangent vector space T_pM"

1. What is the Tangent Space Question?

The Tangent Space Question is a mathematical concept in differential geometry that asks, given a point on a manifold, how many independent directions can one move in while staying on the manifold at that point.

2. Why is the Tangent Space Question important in science?

The Tangent Space Question is important because it helps us understand the local behavior of a manifold and its curvature at a specific point. This has applications in fields such as physics, engineering, and computer science, where the shape and local properties of a surface or space are crucial for understanding and solving problems.

3. How is the Tangent Space Question related to differential geometry?

The Tangent Space Question is a fundamental concept in differential geometry as it is used to define the tangent space at a point on a manifold. The tangent space is a vector space that approximates the manifold at that point and is a useful tool for studying its properties.

4. Can the Tangent Space Question be applied to real-world problems?

Yes, the Tangent Space Question has many real-world applications. For example, it is used in robotics to determine the movement of a robot arm in a specific direction, in computer graphics to create smooth surfaces in 3D modeling, and in physics to understand the curvature of space-time in general relativity.

5. Are there any open questions or problems related to the Tangent Space Question?

Yes, there are ongoing research efforts to better understand and generalize the Tangent Space Question for more complex manifolds. Additionally, there are open questions about the relationship between the tangent space and other structures on a manifold, such as the normal space and the cotangent space.

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