Differential forms on vector spaces vs. on manifolds

Question

A differential 1-form on a vector space $V$ is given by $$dx_i:V\rightarrow \mathbb{R}$$ $$dx_i(\textbf{v})=v_i$$ where $\textbf{v}\in V$. A differential 1-form on a manifold $M$ is given by $$dx_i:T_pM\rightarrow \mathbb{R}$$ My question is, is it still the case that $dx_i(\textbf{v})=v_i$ for $\textbf{v}\in T_pM$?

Answer 1

Differential forms on vector spaces are a special case of differential forms on manifolds. Indeed, take a real, finite dimensional vector space $V$, say of dimension $n$. A choice of basis for $V$ is the same thing as an identification of $V$ with $\mathbb{R}^n$, which we can interpret as a choice of an atlas for $V$ given by a single chart.

What leads to confusion here is the fact that for any $v\in V$, we have $T_vV\cong\mathbb{R}^n\cong V$ canonically (as we have fixed a basis), and often people don't make any distinction between what lives in the manifold $V$ and what is instead a tangent vector. This for example happens in physics, where often one has the impression that positions and velocities live in the same place, which is absolutely not true.

Now what you call $dx_i$ is a differential form on $V$, which in particular gives us a linear map $$(dx_i)_v:T_vV\longrightarrow\mathbb{R}$$ by sending a tangent vector $\hat{v}\in T_vV$ at $v$ to its $i$-th component. In more general manifolds, you can do something similar only locally after a choice of local coordinates.

Answer 2

There are many questions at play here. Firstly, there are no differential forms on a vector space (I'll explain what I mean by this maybe-not-strictly-true statement below). The 1-form you mentioned above is a linear functional.

A differential $k$-form on a manifold $M$ is a map $\omega$ which takes in a point $p \in M$ on the manifold and returns an alternating $k$-linear form on the tangent space $T_pM$. That is, $\omega(p) : V \times \cdots \times V \to \mathbb{R}$ is itself a $k$-linear form. The map $\omega$ itself is not an alternating $k$-linear form; you must evaluate it at a point $p$ on a manifold. Then it becomes a $k$-linear form on the tangent space at that point.

So then, coming around to your question, there's a few complications. First, what is the choice of basis? You are moving from tangent space to tangent space with a differential $k$-form, and each time, you need to pick a new basis. Your definition of $dx_i$ makes sense, for instance, if $V = \mathbb{R}^n$; then $dx_i(v) = v\cdot e_i = v_i$. Note that in this case, $dx_i$ is a linear functional (there is no need to appeal to concepts like differential forms on a single vector space). But I am not sure if it makes sense to talk about it on $M$, a general manifold, since a choice of basis is needed to make sense of this.