So I have studied quite a bit of differential geometry and am comfortable with differential forms and related notions. Of late however, I have been spending time seeing if I actually understand the objects/concepts that I am using all the time. This has led me to investigate whether I understand what a differential form is. For our purposes here, let us suppose that everything is smooth.
Originally, I had been satisfied with considering a differential form as a smooth section of the cotangent bundle, or as a basis vector for the cotangent space. This means however that for each point $p \in M$, where $M$ is a smooth manifold, a differential form $$\omega = \sum_{i_1 \cdots i_n =1}^n f_{i_1 \cdots i_n} dx^{i_1} \wedge \cdots \wedge dx^{i_n}$$ assigns to $p$ a corresponding linear functional, call it $\omega_p$.
I want to make sure that I'm not just symbol pushing and actually understand this properly. Take for example, the 1-form $\omega = f(x) dx$, where $f$ is smooth. Then $\omega$ maps $p$ to the linear functional $\omega_p = f(p)dx$? I don't see how $f(p)dx$ is a linear functional.
Best Answer
I think you may have forgotten what the symbol $dx$ means. Here $x$ is one of the coordinate functions in a chart at $p$, so it is a smooth function from some neighborhood of $p$ to $\mathbb{R}$. Then $dx$ is by definition the total derivative of $x$: that is, it is the functional on the tangent space at $p$ which takes a tangent vector $v$ at $p$ and outputs the directional derivative of the coordinate function $x$ with respect to $v$. Explicitly, if your local coordinates are $(x_1,\dots,x_n)$ with $x=x_i$ for some $i$ and you think of $v$ as a vector $(a_1,\dots,a_n)\in\mathbb{R}^n$ using these local coordinates, then $dx(v)$ will just be $a_i$. The functional $f(p)dx$ is then just this functional $dx$ multiplied by the scalar $f(p)\in\mathbb{R}$.