Geometric understanding of differential forms.

I would like to understand differential forms more intuitively. I have yet to find a book which explains how the use of the exterior product in differential forms ties into the geometrical significance of it. Most books briefly introduce the exterior algebra and then go on to prove theorems about differential forms without ever geometrically motivating anything, and this does not help me to truly understand them. Could someone explain or give a reference to a text explaining differential forms more geometrically. I would very much like to have a deep understanding of them.

Thanks very much


A differential k-form on an n-manifold can be visualised as a “density” of (n – k) submanifolds. In $\mathbb{R}^3$ a 3-form is a point density. The form $dx \wedge dy \wedge dz$ is a uniform point density such that there is 1 full point in a unit cube. The integral $\int_S dx \wedge dy \wedge dz$ measures the number of points inside $S$, which is equal to the volume of $S$. The form $f dx \wedge dy \wedge dz$ is a point density with density $f(x,y,z)$ at that point.

The form $dx$ is a density of planes of constant $x$ (i.e. $yz$-planes) such that there is 1 full plane in a unit of $x$. In other words, the line segment $(t,0,0)$ for $t=a$ to $t=b$ crosses $b – a$ planes (note that the orientation matters). As with a point density, it’s not that we cross a plane discretely at $x=0, x=1, x=2$, rather, the $yz$-planes are distributed uniformly along the $x$-axis. For a curve $\gamma : [0,1] \rightarrow \mathbb{R^3}$ we can count the number of $dx$ planes it crosses. This is denoted $\int_\gamma dx = x(\gamma(1)) – x(\gamma(0))$. A general form $\alpha = A dx + B dy + C dz$ can be visualised as a density of surfaces with general orientation. The integral $\int_\gamma \alpha$ counts the number of times the curve $\gamma$ pierces a surface in the surface density. In other words, it counts the intersections of $\gamma$ with the surface density.

For a function $f : \mathbb{R}^3 \rightarrow \mathbb{R}$ the form $df$ represents surfaces of constant $f$. Viewing $x$ as a function that gives the $x$ coordinate for a point, you can see that $dx$ corresponds to planes of constant $x$. If $r = \sqrt{x^2 + y^2 + z^2}$ then $dr$ are the surfaces of constant $r$, i.e. spheres centered at the origin.

A 2-form $dx \wedge dy$ represents lines in the $z$-direction. Lines in the $z$-direction are formed by the intersection of a plane of constant $x$ and a plane of constant $y$, i.e. lines in the $z$-direction are lines of constant $x$ and $y$. For functions $f$ and $g$, the form $df \wedge dg$ represents curves of constant $f$ and $g$. For two general 1-forms $\alpha$ and $\beta$, which represent densities of surfaces, the form $\alpha \wedge \beta$ represents the density of curves formed by the intersection of those surfaces. The form $dx \wedge dy \wedge dz$ is the point density of intersecting the lines $dx \wedge dy$ in the $z$-direction with the $xy$-planes $dz$.

Given a parameterised surface $A : \mathbb{R}^2 \rightarrow \mathbb{R}^3$, the integral $\int_A \alpha$ of a 2-form $\alpha$ is the number of times the lines of $\alpha$ intersect the surface $A$.

The operation $d$ forms the boundary of the density of curves/surfaces/volumes. For example, of $\alpha$ is a 2-form representing a collection of curves, $d\alpha$ represents the collection of endpoints of those curves. We can understand the formula $d(df) = 0$; the curves $df$ of constant $f$ have no endpoints. The boundary of a density of volumes is a density of surfaces, the boundary of a density of surfaces is a density of curves, the boundary of a density of curves is a density of points, the boundary of a density of points is zero. Let’s understand the form $x dy \wedge dz$. Visualize this as small lines of constant $y,z$ i.e. lines in the x direction. The lines have density $x$ near a point $x,y,z$. This can only be accomplished if the line density has a collection of boundary points. As we move further along $x$, the density of lines in the $x$ direction gets higher, and those lines have to start somewhere. Indeed, we see that $d(x dy \wedge dz) = dx \wedge dy \wedge dz$. The collection of lines $x dy \wedge dz$ has a uniform density of start points. On the other hand, $y dy \wedge dz$ has no net start points. This is just a collection of lines in the $x$ direction that gets denser as we move in the $y$ direction, but those lines still go on from $x = -\infty$ to $x = \infty$. Indeed, we see that $d(y dy \wedge dz) = 0$.

In summary, on an $n$-manifold

  1. An $k$ form represents a density of $(n – k)$ submanifolds.
  2. The wedge product represents intersection of densities.
  3. The $d$ calculates the boundary.
  4. The integral $\int_M \alpha$ of a $k$-form along a $k$-submanifold computes the number of intersection points of the $k$-submanifold with the density of $(n – k)$ submanifolds represented by the form.

We can also intuitively understand the general Stokes theorem $\int_M d\alpha = \int_{\partial M} \alpha$. Let’s consider a function $f$ on $\mathbb{R}^2$. The form $df$ represents the countours of constant $f$ (think of a contour plot), with a density such that there are net $b – a$ contours between the countour $f(x,y) = a$ and $f(x,y) = b$. Now consider a curve $\gamma$ with endpoints. The integral $\int_\gamma df$ calculates the number of contours crossed by $f$. The integral $\int_{\partial \gamma} f$ is just $f(\gamma(1)) – f(\gamma(0))$, i.e. $f$ evaluated along the boundary of $\gamma$ (with the appropriate orientation). We can see that these two integrals are equal: the number of countours crossed by $\gamma$ is precisely the height difference between the start and endpoint.

Now let $V$ be a volume in $\mathbb{R}^3$ with boundary $\partial V$, and let $\alpha$ be a 2-form representing a density of curves. The integral $\int_V d\alpha$ counts the number of endpoints sitting inside $V$. The integral $\int_{\partial V} \alpha$ counts the number of curves piercing through the boundary $\partial V$. You can intuitively understand that these are equal: the curve emanating from an endpoint must either have its other endpoint inside $V$, in which case this part of the curve density does not contribute as one endpoint is positive and the other negative, or the curve has its other endpoint outside $V$, in which case it must pierce its boundary.

The Stokes theorem for a surface in $\mathbb{R}^3$ is a bit tricky to describe with words, but drawing a picture will convince you.

You can also use this picture to understand the pullback, Poincaré lemma, the formula for $d(\alpha \wedge \beta)$, the degree formula, Poincaré duality, and so on.

One last point: why is there this weird inversion of dimension that a $k$-form represents a density of $(n – k)$ submanifolds? Why not use $r$-vectors to represent densities of $r$ submanifolds rather than $n – r$ covectors (which is what differential forms are). In particular, why not use a normal vector field to represent a density of curves? The reason is that $r$ vectors do not have the best transformation properties. A point density on the plane $\mathbb{R}^2$ is something that assigns a real number to an area. The infinitessimal version of this is that for a 2-form $\alpha$ the quantity $\alpha(x,y)(u,v)$ counts the number of points in a small parallelogram spanned by the vectors $u,v$ at the point $x,y$. If we deform the plane then the parallelogram of the vectors will deform with it, and $\alpha(x,y)(u,v)$ stays constant. If we used $r$ vectors rather than $r$ covectors, we would only be invariant under isometries rather than general diffeomorphisms. Suppose we have a vector field on the plane and a curve $\gamma$ in the plane. There is no basis independent way to say how many times the curve intersects with the vector field. This question only makes sense up to some scale factor. Forms have a natural scale attached to them, because a form $df$ naturally eats a tangent vector $y'(t)$ as $df(\gamma(t))(\gamma'(t))$. To do this with the vector field you’d have to choose a basis/coordinate system.

I hope that helps.

Source : Link , Question Author : Markus , Answer Author : Jules

Leave a Comment