The meaning of $\omega$ = df?

Question

I'm reading Algebraic Topology by William Fulton and I am a little lost about 1-forms, which I assume to be the number of differentials according to the number of dimensions (correct me if I'm wrong. Since a 1-form is $\omega = A \, dx + B \, dy$, for

$$\omega = df$$

would that mean that the derivative of $f$ is in the 1-form?

Thanks for the help.

Answer 1

A smooth map $f: M \to N$ of manifolds (you can also think real space as an example) induces a map on tangent bundles. This is the generalization of the idea from multivariable calculus that total derivative at a point is a linear map. In particular, a smooth map $M \to \mathbb{R}$ defines a linear map from each tangent space to $\mathbb{R}$, that is, a 1-form. Explicitly, in local coordinates, we can write this as

$$ df = \sum_i \frac{\partial f}{\partial x_i} dx_i$$

Answer 2

A $1$-form is just a map that assigns to each point in your manifold a linear functional on the tangent space at that point, denoted $T_pM$. That is, for $p\in M$ we have that $\omega(p):=\omega_p\in T_p^\ast M$.

For simplicity, if you'd like just replace $M=\mathbb{R}^n$ so that $T^\ast_pM=T_pM=\mathbb{R}^n$ and so a $1$-form just eats $n$-tuples and spits out an element of $(\mathbb{R}^n)^\ast$

Assuming $ \ f:U\to\mathbb{R}$ is smooth, for some open $U\subset\mathbb{R}^n$, then $\omega=df$ is just a specific one form. It's the one defined by, for $p\in U$, we have $(df)_p\in (\mathbb{R}^n)^\ast$ where $(df)_p:T_pU=\mathbb{R}^n\to \mathbb{R}$ is defined by sending $V\mapsto V\cdot \left.\frac{\partial f}{\partial x}\right|_p$ the dot product of the vector $n$-tuple $V$ and the $n$-tuple $\left.\frac{\partial f}{\partial x}\right|_p$, otherwise known as the directional derivative of $f$ at $p$ in the direction $V$.