This question can look like a duplicate of this one, but it's kind of different. I'm trying to relate some geometrical meanings I've seem in some books to the definition of differential forms in $\mathbb{R}^n$ as mappings $p \mapsto \omega(p)\in \Lambda^k(\mathbb{R}^n_{\phantom{n}p})$.

Differential forms seems to be object with high geometrical importance. However, I'm failing to grasp what they really represent. Many books, mainly on Physics, try to give one geometrical interpretation for differential forms as "families of surfaces" such that the value on a vector is the number of surfaces the vector crosses.

This confuses me a little. Why do this interpretation makes any sense? I mean, if I want to construct an object with this geometrical property, why it should be a function associating skew-symmetric tensors to each point in space?

Also, vector fields are easy to understand. We know what each vector is at each point, we picture as a small arrow, and we know that they can describe things with directions, they can describe rates of change being derivatives, and so on. Now this geometrical interpretation they give does not allow us to picture differential forms at points, just the association at each point.

My understanding was the following as I see: Differential forms replace the classical $dx$, $dA$, $dV$ and so on, that were considered infinitesimal objects. My idea is that in that case, $\omega(p)$ would represent just a small patch of the surfaces $\omega$ represents and because of that, we could think of $\omega$ really relating to those infinitesimal objects. I'm unsure of this intuition, and I can't see how this would lead us towards the rigorous definition of differential forms.

So, what's the true geometrical meaning of differential forms and how this meaning implies that the algebraic definition we give is a good one?

Thanks very much in advance!

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2 Answers2


A 2-form is a function that eats a parallelogram (technically it eats 2 vectors, which you should think of as spanning a parallelogram) and spits out a number proportional to its area. A 3-form eats a parallelepiped (the 3-dimensional analog of a parallelogram) and spits out a number proportional to its volume. A 4-form eats a 4-dimensional parallelotope and spits out a number proportional to its hypervolume. A 1-form eats a line segment (which you can think of as a 1-dimensional parallelogram) and spits out a number proportional to its length. A 0-form eats a single point (which you can think of as a 0-dimensional parallelogram) and spits out a number, though there's nothing for it to be proportional to since a point has no extension in space. I think you get the picture. In general an n-form eats n vectors, which you should think of as spanning an n-dimensional parallelotope, and spits out a number proportional to its hypervolume.

Usually books that teach differential forms obscure this. They will define an n-form as a "real-valued multilinear, skew-symmetric function of n vectors". But it means the same thing. Multilinearity and skew-symmetry = output is proportional to length/area/volume/hypervolume. The determinant, which is used to compute the volume of a parallelepiped (and its higher and lower dimensional analogs), has the same two properties.

So why do we require forms to have this property? Well it's just because it's needed for integration. Imagine a curve you want to integrate over. The first step is to approximate it with line segments. Then you apply some function to each line segment in order to get a number. You need that number to shrink as the size of the line segment shrinks otherwise the sum won't converge. Think about it, if the output of the function was independent of the length of the input, then as more segments were added to the approximation the sum would just shoot up to infinity. Now think of a surface you want to integrate over. You can approximate it with parallelograms, imagine the scales of an armadillo. Then for each parallelogram you apply some function that spits out a number. We need the numbers to shrink as the scales do so the sum actually converges. If you want to integrate over some 3-dimensional volume, approximate it with parallelepipeds and again evaluate a function for each parallelepiped. The output of this function needs to shrink with its input for the sum to converge. These functions that we integrate over curves/surfaces/volumes/hypervolumes are forms.

Now let me explain why you write forms as linear combinations of elementary forms. It has to do with the generalized Pythagorean theorem, which I'll just call the GPT. In the same way that the length of a line segment is equal to the sum of the squared lengths of its projections onto the various coordinate axes, the area of an arbitrary parallelogram is equal to the sum of the squared areas of its projections onto the various coordinate planes. And the volume of a parallelepiped is equal to the sum of the squared volumes of its projections onto the various 3-dimensional subspaces. And so on. So the Pythagorean theorem applies to more than just line segments.

So let's look at the example of a 1-form that eats line segments embedded in 3-dimensional space. In general it's gonna look like $adx + bdy + cdz$ (if you forgot, $dx$, $dy$, and $dz$ are just functions that eat a line segment and spit out its projections on the x axis, y axis, and z axis respectively). All that's happening is you're taking the dot product of a vector $(a,b,c)$ with another vector $(dx,dy,dz)$ which equals the projection of $(a,b,c)$ onto $(dx,dy,dz)$ times the length of $(dx,dy,dz)$ (the length of $(dx,dy,dz)$ is $\sqrt{dx^2 + dy^2 + dz^2}$ ie the length of the line segment by the GPT). In other words $adx + bdy + cdz$ is literally just another way of writing: (projection of $(a,b,c)$ onto $(dx,dy,dz)$) times (length of the line segment). Since the length of the line segment is a factor in this product, the function is obviously proportional to the length of the line segment. Any 1-form can be written like this.

Another example: A 2-form that eats parallelograms embedded in 3-dimensional space is gonna have the form $a(dx \wedge dy) + b(dx \wedge dz) + c(dy \wedge dz)$ (if you forgot, $dx \wedge dy$, $dx \wedge dz$, and $dy \wedge dz$ are just functions that eat parallelograms and spit out the areas of their projections on the xy, xz, and yz planes respectively). So this is just another way of writing the dot product of $(a,b,c)$ and $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$ which is just the projection of $(a,b,c)$ onto $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$ times the length of $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$ (which is $\sqrt{(dx \wedge dy)^2 + (dx \wedge dz)^2 + (dy \wedge dz)^2}$ ie the area of the parallelogram by the GPT). In other words the linear combination is just equal to: (projection of $(a,b,c)$ onto $(dx \wedge dy, dx \wedge dz, dy \wedge dz)$) times (area of the parallelogram). Which is clearly a function proportional to the area of the parallelogram.

Another example: A 2-form that eats parallelograms in the plane. It has the general form $a(dx \wedge dy)$. You only need one term because $dx \wedge dy$ already gives you the area of the parallelogram. In the same way $dx$ gives you the length of your line segment if you're only in 1 dimension. It's only when you're in a dimension higher than the dimension of the line segment/parallelogram/parallelepiped/parallelotope that you're gonna have to invoke the GPT ie have a linear combination of multiple elementary forms.

So hopefully you see that differential forms are actually very simple objects. They're merely generalized integrands. Other things in exterior calculus like the exterior derivative, the generalized stokes theorem, etc are similarly very simple when explained properly.

edit: a slightly cleaned up version of this post with some pictures can be found here: https://simplermath.wordpress.com/2020/02/13/understanding-differential-forms/

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  • Greater explanation! – Allawonder Nov 05 '19 at 02:08
  • can i have a reference for the generalized Pythagorean theorem? Thank you – karhas Jun 07 '20 at 20:45
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    @karhas Read the paper titled "An n-dimensional Pythagorean Theorem" by William J. Cook. Everyone should learn this because it really clarifies a lot of otherwise obtuse math. For example think of computing a determinant using the expansion by minors technique, with the generalized pythagorean theorem you can see that the whole process is just a recursive application of the basic base x height formula for parallelograms. – someguy67 Jun 08 '20 at 15:31
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    The connection between the Pythagorean theorem and expressing a general k-form as a linear combination of basis forms is unclear to me still. The n-dimensional Pythagorean theorem referenced relates to sums of the *squares* of the (n − 1)- dimensional volumes of the projections of a parallelotope. However, the basis forms measure the volumes of the projections themselves, rather than the squares. Could you further clarify the connection you have in mind? – Noah M Jul 30 '20 at 07:21
  • @NoahM you have to interpret the linear combination of basis forms as a dot product. so adx + bdy for example is (a,b) dot (dx,dy) which geometrically is the projection of (a,b) onto (dx,dy) times the length of (dx,dy). the length of (dx,dy) is sqrt(dx^2 + dy^2) that's where the sum of squares comes from, and so by the GPT (or whatever you want to call it) you have the length of the line segment/area of the parallelogram/volume of the parallelepiped. i explained it in the post, but maybe i did it poorly. or maybe im misunderstanding your question. – someguy67 Jul 30 '20 at 15:09
  • i wrote this blog post a while ago, it might be a little more coherent: https://simplermath.wordpress.com/2020/02/13/understanding-differential-forms/ – someguy67 Jul 30 '20 at 15:12

Caveat: This answer is (judiciously!) incomplete, and makes no pretense of giving the One True Geometric Meaning of differential forms. Also, there are so many conventions (regarding spaces and their duals, index placement, and vectors/covectors versus vector fields/$1$-forms) that it's impossible to be notationally and terminologically consistent with other sources you may have read.

A differential $k$-form may be viewed as defining (at each point) a "measuring device for $k$-dimensional oriented volume elements". Loosely, for example, if you view a vector field as a velocity field, then (to coin a phrase) a $1$-form may be viewed as a "(vector-valued) speedometer field".

To give this heuristic principle a precise interpretation in $\mathbf{R}^3$, let $\mathbf{e}_i$ denote the Cartesian frame fields (i.e., the vector fields whose values at each point are the standard basis of $\mathbf{R}^3$); $dx^i$ the (dual) coordinate $1$-forms; $\omega_i$ smooth functions; and $a_i = \omega_i(p)$ the value of $\omega_i$ at a point $p$. A $1$-form $$ \omega = \omega_1\, dx^1 + \omega_2\, dx^2 + \omega_3\, dx^3 $$ defines the linear functional $\omega(p) = a_1\, dx^1 + a_2\, dx^2 + a_3\, dx^3$ on the vector space $T_p\mathbf{R}^3 \simeq \mathbf{R}^3$. If $X = X^1 \mathbf{e}_1 + X^2 \mathbf{e}_2 + X^3 \mathbf{e}_3$ is a vector field, then $$ \omega(X)(p) = \sum_{i,j=1}^3 a_i X^j dx^i(\mathbf{e}_j) = a_1 X^1 + a_2 X^2 + a_3 X^3 $$ may be viewed as the "measurement": $a_1$ times the first component of $X$ plus $a_2$ times the second component plus $a_3$ times the third component.

Analogously, if $\omega_{ij}$ are smooth functions and $a_{ij} = \omega_{ij}(p)$, the $2$-form $$ \omega = \omega_{23}\, dx^2 \wedge dx^3 + \omega_{31}\, dx^3 \wedge dx^1 + \omega_{12}\, dx^1 \wedge dx^2 $$ defines a linear functional $\omega(p) = a_{23}\, dx^2 \wedge dx^3 + a_{31}\, dx^3 \wedge dx^1 + a_{12}\, dx^1 \wedge dx^2$ on the space $\bigwedge^2(\mathbf{R}^3)$ of "oriented $2$-plane elements". If $X = X^{23} \mathbf{e}_2 \wedge \mathbf{e}_3 + X^{31} \mathbf{e}_3 \wedge \mathbf{e}_1 + X^{12} \mathbf{e}_1 \wedge \mathbf{e}_2$, then $\omega(X)(p)$ may be viewed as the "measurement": $a_{23}$ times the projection of $X$ on the $(x_2, x_3)$-plane plus $a_{31}$ times the projection of $X$ on the $(x_3, x_1)$-plane plus $a_{12}$ times the projection of $X$ on the $(x_1, x_2)$-plane.

Note that the components of a vector field $X$ are the projections of $X$ onto the coordinate axes, so the two preceding interpretations are more closely analogous than they may first seem.

Andrew D. Hwang
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    +1. I think this is as good an answer as any for an intuitive explanation of forms. – Mathemagician1234 Mar 21 '15 at 03:34
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    Thank you I also struggle with the intuition of $n$-forms for $n>1$. While 1-forms are extremely intuitive to me, they're just the linear approximation of a smooth function on each tangent space. But I can't wrap my head around 2-forms. Your explanation does help, I see what you mean about projections. I guess that's why there are no 4-forms in 3 dimensions. – Gregory Grant May 03 '15 at 23:25
  • One thing that adds to my confusion is that I want to relate a 2-form to the second derivative, but that's obviously not the right way to think about these things. – Gregory Grant May 03 '15 at 23:25