Intuitive thinking of differential forms in terms of gradient, divergence and curl?

Is there a way to visualize the differential forms -

Like I found that if $\alpha$ is a 1 - form then $d \alpha$ represents curl.

and $\sigma$ is a 2 - form then $d \sigma$ represents the divergence.

and $d^{2} \alpha = 0$ resembles $\nabla \times (\nabla()) = 0$.

How do I visualize the above, is it intuitive?

I will try to give a visual representation of forms and provide intuition into exterior derivatives using Stokes' theorem:

$$\int_{\partial M} \alpha = \int_{M} d\alpha.$$

Differential 1-forms

A differential 1-form $$\alpha$$ is pointwisely a "covector", a function $$\alpha|_p$$ which assigns a real value to each vector tangent to $$p \in M$$. This assignment is linear:

(For simplicity, I use a 2-D manifold as the total space, rather than $$\mathbb R^3$$.)

In the picture above, no covectors are drawn, but only vectors. This is because unlike (column) vectors, covectors are linear functions $$\mathbb R^n\to \mathbb R$$, which are hard to draw.

However, a covector may be regarded as a "row vector" since a linear map of the type $$\mathbb R^n\to \mathbb R$$ is equivalent to a left-multiplication by some $$1\times n$$ matrix. So if needed, one could still draw an arrow (column vector) $$v$$ that represents the function $$f(x) = \langle v, x \rangle$$ or $$v^T$$.

Under this representation, it is possible to draw a 1-form $$\alpha$$ as a vector field (or in this case, a co-vector field). Physically, integrating the co-vector field (1-form) along a curve is identical to measuring the work done by a vector field.

$$\int_\gamma \alpha = \int_\gamma \overset \rightharpoonup \alpha\cdot \overset \rightharpoonup {dr}$$

where $$\overset \rightharpoonup \alpha$$ denotes the vector field "representing" the 1-form $$\alpha$$, and $$\overset \rightharpoonup {dr}$$ is the infinitesimal movement of the curve $$\gamma$$.

Clarification:

Although I draw arrows for covectors, I agree with John Hughes that covectors (dual vectors) should be regraded as functions rather than arrows eventually, to get a really suitable intuition.

Differential 2-forms

Likewise, a 2-form $$\sigma$$ is pointwisely an alternating linear map (which we call a 2-covector) that assigns to each pair of vectors a real number ...

... or we could say that $$\sigma|_p$$ assigns to each oriented parallelogram based at $$p$$ a real number, since an ordered pair of tangent vectors at a same point $$p\in M$$ determines an oriented parallelogram:

The "alternating" property requires that $$\sigma|_p$$ satisfy $$\sigma|_p(v,v)=0$$ for any $$v \in T_p(M)$$. Thus $$\sigma|_p(v_1,v_2)=-\sigma|_p(v_2,v_1)$$, which characterizes the notion of signed area.

Recall that for a 1-form $$\sigma$$, we draw an arrow representing a covector at each point $$p \in M$$. For a fixed $$p \in M$$, this arrow points in the unit direction for which the linear function $$\sigma|_p$$ attains a maximal value. Similarly for 2-forms, we can associate to each $$p \in M$$ an oriented parallelogram (with fixed area) in the direction that attains the maximal value.

The directions of the faces representing 2-covectors form a vector field, which is how people regard a 2-form as a vector field that points the directions in which a parallelogram can get a maximal value, but now we know that that's somewhat different.

Verification:

Let $$v_1:(a_1,b_1,c_1),v_2:(a_2,b_2,c_2)$$ be unit vectors that determine the parallelogram facing the direction that achieves the maximal value under the 2-covector $$\sigma = p\, dy\wedge dz + q\, dz \wedge dx + r \, dx \wedge dy.$$ A calculation reveals that \begin{aligned} \sigma(v_1,v_2) &= p(b_1c_2-b_2c_1) + q(c_1a_2-c_2a_1) + r(a_1b_2-b_2a_1)\\[0.6em] &= (p,q,r) \cdot(v_1 \times v_2). \end{aligned} Hence, the cross product of the two vectors must be parallel to the $$(p,q,r)$$ so that the evaluation $$\sigma(v_1, v_2)$$ become maximal.

Exterior derivative and Exact 2-forms

The exterior derivative of a 1-form $$\alpha$$ is a 2-form $$d\alpha$$. Stokes' theorem states that for a manifold $$M$$ with boundary $$\partial M$$, integrating $$\alpha$$ along $$\partial M$$ is equivalent to integrating $$d\alpha$$ over $$M$$: $$\int_{\partial M} \alpha = \int_{M} d\alpha.$$ So, how do we evaluate the 2-form $$d\alpha$$ over $$M$$? Assume that a pair of vectors $$v_1,v_2$$ determines an infinitesmal oriented parallelogram $$P$$, then we may regard $$d\alpha(v_1,v_2)=d\alpha(P)\approx \int_{P} d\alpha= \int_{\partial P} \alpha.$$

Clarification:

It requires more work to show that this intuition works in some sense (maybe you could skip this part). Let's assume our manifold is just $$\mathbb R^3$$, and prepare a plane $$N\subset \mathbb R^3$$ that contains $$P$$. Then $$d\alpha$$ is a top form when restricted to $$N$$. That is, there exists some real valued function $$f$$ (area density function) such that $$d\alpha|_q=f(q)\,ds\wedge dt,\;\; \forall q\in N$$ where $$(s,t)$$ is an orthonormal coordinate on $$N$$. Then by mean value theorem $$\int_{\partial P} \alpha = \int_P d\alpha\; \underset{q\in P}{\overset{M.V.T}{=\!=\!=}} \; f(q)\cdot \text{Area}(P)\approx f(p)\cdot\text{Area}(P).$$ Write $$v_1=c_{1,1}\vec s+c_{1,2}\vec t\;\;,\;\;v_2=c_{2,1}\vec s+c_{2,2}\vec t.$$ Then \begin{aligned} ...& \approx f(p)\cdot\text{Area}(P) = f(p)\cdot \begin{vmatrix}c_{11}& c_{11}\\ c_{21}& c_{22}\end{vmatrix} = f(p)\cdot (ds\wedge dv)(c_{1,1}\vec s+c_{1,2}\vec t,c_{2,1}\vec s+c_{2,2}\vec t)\\[0.7em] &= f(p)\cdot (ds\wedge dv)(v_1,v_2) = \sigma_p(v_1,v_2).\end{aligned}

However, in general it isn't possible to embed a parallelogram in any manifold $$M$$. Moreover, $$T_p M$$ and $$M$$ are not always assumed to lie in some ambient space. So ... maybe it is better to leave it as a mere intuition.

This also shows that: if the set of arrows representing the 1-form $$\alpha$$ rotates counter-clockwisely in the same plane of the parallelogram, then $$d\alpha$$ assigns a positive value to the parallelogram (curl).

(This picture may be misleading: even if the closed path does not contain the center of rotation, it is still possible for the integral to be positive.)

Differential 3-Forms

A differential 3-form gives each parallelepiped spaned by three vectors a value.

Exterior derivative and Exact 3-forms

Likewise, a two form $$\sigma$$ can may assign values to the boundary of a parallelepiped. This induces a 3-form $$d\sigma$$:

Note that the boundary of the parallelepiped in this example is positively oriented, i.e. every normal vector points outward.

We have shown that a 2-form can evaluate a parallelogram (spanned by $$v_1,v_2$$) via the inner product $$(p,q,r) \cdot (v_1 \times v_2)$$. So, if a vector field representing a 2-form $$\sigma$$ has divergence, then $$d\sigma$$ (integration along the boundary of a small 3D region) is positive.

For simplicity, I drew a polyhedron rather than a 3D region with smooth boundary.

Intuition for $$d^2=0$$

Because the integrals along the edges of the polyhedron cancel out, summing $$d\alpha$$ over each face, i.e. $$d(d\alpha)$$ yields zero.

This is why $$d^2\alpha=0$$.

• Nice diagrams. Where did you get them from? Jul 23, 2019 at 5:48
• @Peter4075 thanks. I drew them by a graphic pad. Jul 24, 2019 at 5:38
• This is the clearest explanation and diagrams I have seen. Thank you! Mar 29, 2020 at 6:52

...and if $f$ is a 0-form (i.e., a real-valued function), then $df$ looks like the gradient.

In fact, the operator $d$ is designed (along with the definition of alternating $k$-forms) to generalize the various notions of "derivative" that already arise for functions and vector fields. As for "visualizing," I don't think I can help except to say that these things are natural generalizations.

There's one small point I want to make: if $\alpha$ is a $1$-form, then $d\alpha$ is a 2-form, i.e., it looks like $$p dx\wedge dy + q dy \wedge dz + r dz \wedge dx$$ If you think of 1-forms as "like vector fields" (they have three components, labelled by $x$, $y$, and $z$, for instance), then although a 2-form is similarly like a vector field, in that it has three components, but the labelling is different. You have to compose with a map that sends $$dx \wedge dy \to dz; dy \wedge dz \to dx; dz\wedge dx \to dy$$ to get back to an "ordinary" vector field. Once you do this, you find that you've got something that looks a lot like the curl.

That intermediate map represents a duality between $k$-forms and $(3-k)$-forms in 3-space. (When you say $d\sigma$ represents the divergence, you're implicitly transforming $dx \wedge dy \wedge dz$ to the constant function $1$, which is the duality from $3$-forms to $0$-forms). In general, $k$-forms and $(n-k)$-forms are "dual" in $n$-dimensional space, and it's tempting to use this duality to think of 2-forms as "really being 1-forms", etc. I advise against this, and suggest you try to develop some intuition for 2-forms as things that consume two tangent vectors, while 1-forms consume 1 tangent vector, etc. It'll pay off in the end.

The other thing to realize is that 1-forms and vectors are different, just the way that the vector space $V$ and the vector space $V^{*}$ of all linear maps from $V$ to $\Bbb R$ are different, even though (for finite dimensions) they're isomorphic. We would call things in $V$ 'vectors' and things in $V^{*}$ "dual vectors", because they consume a vector to produce a number.

In the same way, we have vector fields (which I assume you know about), and 1-forms: at each point in space, a vector field gives you a vector. At each point in space, a 1-form gives you a dual-vector (i.e., something that can consume a vector to give you a number).