1

In this article https://en.wikipedia.org/wiki/Tangent_space, under the section title Basis of the tangent space at a point, it says "... Then for every tangent vector v\in TpM, one has...." followed by an equation for a tangent vector v. But I don't see enough reasoning/explanation to justify the equation.

The same topic is also discussed in this article https://bjlkeng.github.io/posts/manifolds/. Similarly, it says right before Equation (10), "So every tangent vector v\in TpM, we have..." followed by Eq (10). Again, I don't see why.

I completely understand Eqs. (8) and (9), showing the derivative of function f over curve gamma(t) as the inner product of tangent vector v and the gradient of f, i.e., it is the directional derivative of f along the direction of v.

However, what I do not understand is why the LHS of Eq. (8) becomes a tangent vector v in Eq. (10), when function f is dropped on the RHS. In other words, how can a tangent vector on the LHS be the same as the weighted sum of n differential operators (treated as basis of tangent space) on the RHS?

Ruye
  • 31
  • Do you have a good grasp of the tangent space conceptually? The expression for $v$ is simply the components of the vector in the direction of each basis vector. Do you understand why the basis vectors are defined that way? – John Douma Apr 15 '22 at 19:36

1 Answers1

1

I think I have some idea about my own question above after reading this post: https://math.stackexchange.com/questions/3330025/why-are-the-partial-derivatives-a-basis-of-the-tangent-space#:~:text=The%20tangent%20space%20is%20viewed%20as%20the%20space,they%20are%20identified%20as%20directional%20derivative%20operators.%20Share

My realization is that the partials $\partial/\partial x_i$ are not the basis of the tangent space $T_p(R^n)$, instead, they are the basis of a different space $D_p(R^n)$, an isomorphic mapping of the tangent space $T_p(R^n)$.

In other words, any tangent vector $v\in T_p(R^n)$ can be mapped to $D_v\in D_p(R^n)$, which can be expressed as a linear combination of the partials $\partial/\partial x_i$ as the basis.

Please correct me if I am wrong.

Ruye
  • 31
  • There are many different ways to define the tangent space and$$T_p\mathbb R^n=D_p(\mathbb R^n)$$is one of them. – Filippo Apr 15 '22 at 19:33
  • In fact, there is a characterization of the tangent space that defines it up to an isomorphism. This is explained here. – Filippo Apr 15 '22 at 19:35