How to visualize the gradient as a one-form?

Question

I just finished reading the proof that the gradient is a covariant vector or a one-form, but I am having a difficult time visualizing this. I still visualize gradients as vector fields instead of the level sets associated with dual vectors. How to visualize the gradient as a one-form?

Answer 1

To clear up some confusion in the comments: when Carroll refers to the gradient of a function $f$ as a $1$-form he probably intends to refer to the exterior derivative $df$ of $f$. This is a $1$-form containing all of the information which is contained in the gradient, but which can be defined in the absence of a (pseudo-)Riemannian metric. In particular it contains precisely the data needed to differentiate $f$ with respect to any tangent vector; the result tells you how fast $f$ is changing in a particular direction at a particular point, which is what the gradient is supposed to do.

In the presence of a (pseudo-)Riemannian metric, one can identify $1$-forms and vector fields, and the vector field corresponding to $df$ along this identification is what people usually call the gradient or gradient vector field of $f$. Unlike $df$, it depends on a choice of (pseudo-)Riemannian metric. It's unusual and imprecise to refer to $df$ itself as the gradient.

Answer 2

It depends on how you like to visualize $1$-forms, but I'll assume familiarity with the "parallel planes in the tangent space" visualization:

Given a function $f$ on your $n$-manifold $M$ and a point $p \in M$, we often denote the differential of $f$ at $p$ by $df_p$ (I actually prefer the less common but more "functorial" notation $T_p f$ but I'll stick to the classical notation here.) If $df_p \neq 0$, then at least near $p$, the level set $N := f^{-1}(c)$, $c := f(p)$, is a hypersurface ($(n - 1)$-manifold) through $p$, and we can think of the tangent space $T_p N$ as a hyperplane in $T_p M$.

We can identify this hyperplane readily: We can write any vector $X$ tangent to $N$ at $p$ as $X = \gamma'(0)$ for some curve $\gamma: I \to N$. Then, evaluating $df$ on $X$ gives $$df(X) = X \cdot f = \gamma'(0) \cdot f = (f \circ \gamma)'(0).$$ On the other hand, $\gamma$ is a curve in $N$, which is the set in $M$ on which $f$ takes the value $c$, so $f \circ \gamma$ is the constant function with value $c$, and $$df(X) = (f \circ \gamma)'(0) = 0.$$ Thus (counting dimensions gives that) the kernel of $df$---that is, the purple plane through the origin in the above diagram---is precisely $T_p N$. So, we can visualize the $1$-form $df_p$ as the parallel planes in $T_p M$ parallel to the hyperplane $T_p N$ tangent to the level set of $f$ through $p$.

Replacing the function $f$ with $\lambda f$ for some constant $\lambda > 1$ gives a differential $\lambda df_p$ at $p$, which has the same kernel, but more closely spaced planes, which is hence characteristic of "steeper" functions $f$.