# 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.

• This seems a lot like "On the Visualisation of Differential Forms": citeseerx.ist.psu.edu/viewdoc/… . I'd never heard of this picture before today (your answer and that link), except that I independently discovered part of it yesterday. Commented Aug 18, 2018 at 2:30
• I'm sure it was well known at some point. This visual interpretation is common when teaching vector calculus / electromagnetism, and differential forms are just the generalisation of that to n dimensions. Strangely though, Cartan, the inventor of differential forms, never mentions any visual interpretation in his book. I've heard from several people who teach differential forms that they were unaware of this interpretation. It's a bit of a mystery where the visual intuition got lost. Perhaps Cartan was aware of it, but considered it too obvious to mention, or perhaps it got lost before Cartan. Commented Nov 23, 2018 at 16:25
• This is a very nice answer. But I'd like to gently push back on the implication that this visual, geometric interpretation has been "lost." In fact, these perspectives are crucial to topology (via Poincare duality), to geometric measure theory (via the study of currents), and to algebraic geometry (via intersection theory), all of which are active areas of research. Commented May 4, 2021 at 13:26
• There surely is some kind of geometric intuition there, but I find the intuition quite vague. For example, it's clear that the wedge product is somehow related to intersection, but precisely how isn't described, and as far as I know, pictures are not drawn or described. Contrast this with physics books about electromagnetism: the pictures are there, and the intuition is clear and quantitative. Each electron has definite amount of field lines emanating from it, and that's why the flux through a surface is equal to the amount of electrons enclosed. Commented May 5, 2021 at 12:16
• It's on my todolist :D Commented Feb 27, 2022 at 18:04

There are grave issues with the common thinking inherent to differential forms. For instance, in vector calculus, you can consider a vector field

$$\mathbf G(\mathbf r) = \frac{\mathbf r}{4\pi r^3}$$

You can meaningfully integrate the divergence of this vector field,

$$\int_{M} \nabla \cdot \mathbf G \, dV$$

and if your region contains the origin, you get 1, else 0.

In differential forms language, you have as your only hammer the exterior derivative. Instead of considering vector fields like you're used to from vector calculus, you have to convert $\mathbf G$ to a 2-form $\omega$.

$$\omega = \frac{z \, dx \wedge dy + y \, dz \wedge dx + x \, dy \wedge dz}{4\pi r^3}$$

And then you do an integral like

$$\int_{M} d\omega$$

as is usually done. But this is the big problem with learning differential forms when coming from vector calculus. You have to take all the stuff you're familiar with and turn it upside down. You have to take vector fields and turn them into two-forms to take meaningful divergences; you have to convert them to one-forms to take meaningful curls (though this is much less peculiar).

What's actually going on here can be explained in a different formalism called geometric calculus. It enforces that when we integrate over geometric regions (surfaces, volumes, and so on), the integration measure itself not only contains magnitude formation (the area of the surface, the size of the volume) but direction information, too. So instead of $dV$, we integrate over $d\mathbf V$, a 3-vector measure. Since all 3-vectors in 3d are scalar multiples of each other, we call this simply $\mathbf I \, dV$.

We can meaningfully integrate vector field divergences this way without arbitrarily converting them to two-forms. Rather, the "need" to do so follows from algebra.

$$\int_M (\nabla \cdot \mathbf G) (\mathbf I \, dV)$$

The 3-vector $\mathbf I$ turns directions to orthogonal planes and points to volumes. I submit to you, for reasons too complex to get into, that $(\nabla \cdot \mathbf G) \mathbf I = \nabla \wedge [\mathbf G \mathbf I]$ using the underlying rules of Clifford algebra that geometric calculus is built on top of. It is $\mathbf G \mathbf I$ that is the 2-form $\omega$, and it is $\nabla \wedge$ that truly represents, for all objects, the exterior derivative.

What this says geometrically is that there is an equivalence between the following: taking the divergence of a vector field (which yields a scalar) and converting that vector field to its orthogonal planes, finding the derivative orthogonal to all of those planes, and then converting the resulting volume back to a scalar. That is the nature of the divergence as seen in differential forms, and it becomes very clear using geometric calculus.

Geometric calculus is a natural and powerful extension to the vector calculus you probably already know. The notation is a lot more similar, surely, while still being able to fully express everything you would need to do in higher dimensions. It is called geometric calculus for a reason, as many of the main proponents try to emphasize, as much as possible, the geometric nature of the underlying algebra and the calculus.

If you become interested in this field, the canonical reference is Clifford Algebra to Geometric Calculus by Hestenes and Sobczyk. It's a slightly older text, but it contains good material on the relationship between multivector functions and differential forms. A more modern treatment is Vector and Geometric Calculus by Macdonald, which aims to be a more elementary text. Geometric Algebra for Physicists by Doran and Lasenby contains many applications and a good chapter on geometric calculus that condenses Hestenes and Sobczyk into something a little more approachable.

• Clifford algebra is beautiful and powerful, but a lot to ask of someone at the freshman/sophomore level learning multivariable calculus. Of course, a good first course should emphasize all the correspondences between vector fields and differential forms in $\mathbb R^3$. Commented Jul 10, 2013 at 22:38
• For such an audience I would recommend Macdonald's book (well, both his books, the first being Linear and Geometric Algebra, which introduces more of the algebraic aspects). They are lean and efficient at what they do and in maintaining solid contact with traditional matrix algebra, vector algebra, vector calculus, and other more traditional approaches. Commented Jul 10, 2013 at 22:40

I'll also toot my own horn and add my own book Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds, which I wrote when my students couldn't penetrate Hubbard and Hubbard :) Differential forms come about by adding layers of calculus to the algebra of determinants, which compute oriented $k$-dimensional volumes.

I would also suggest Harley Flanders's Differential Forms with Applications to the Physical Sciences (now in Dover). At a higher level, especially if you're interested in differential geometry and physics, look at a surprisingly concrete exposition, R.W.R. Darling's Differential Forms and Connections.

• Excellent reference on the use of Darling, this textbook got me through my first semester in Grad school and beyond, I adore it (notwithstanding the typos / errors!)
– user284001
Commented May 5, 2021 at 8:25

Here are a couple of references that you might find helpful.

Vector Calculus, Linear Algebra, and Differential Forms by Hubbard and Hubbard. This book is written for undergraduates and integrates (pardon the pun) the study of multivariable calculus, linear algebra, and finishes with differential forms. If you'd like intuition for the geometry of forms in $\mathbb{R}^3$, without too much high-level symbol-pushing, this is a good place to look.

The Geometry of Differential Forms, by Morita, is a monograph which starts with basic definitions and proceeds to describe the utility of differential forms in various contexts, including (if my memory serves) Hodge theory and bundle-valued forms.

One way (that I discovered accidentally reading a book on exterior algebras) to understand differential forms more intuitively is to see them as linear functionals instead of multilinear antisymmetric maps, what can be done if we see the tangent space of a manifold also as an exterior algebra. In short: I translate the abstraction of differential forms to an abstraction of multivectors in the tangential space, what is to me a very natural and intuitive way to think the notion of volume.

In the following if $$V$$ is a vector space of dimension $$n$$ then $$\bigwedge V:=\bigwedge V^0\oplus \bigwedge V^1\oplus \ldots \oplus \bigwedge V^n$$ is an exterior algebra on $$V$$.

Let $$M$$ an smooth manifold of dimension $$n$$, then every homogeneous differential form $$\omega\in \bigwedge T^*M$$ of $$k$$ degree induces a linear functional in $$\bigwedge^k TM$$ by $$\omega(x_1\wedge x_2\wedge \ldots \wedge x_n):=\omega (x_1,x_2,\ldots ,x_n)$$, by example any bilinear map $$a\wedge b$$, for $$a,b\in \bigwedge^1 T^*M$$, can be seen now as a linear functional on $$\bigwedge^2 TM$$ using the assignment $$a\wedge b(c\wedge d):=a\wedge b(c,d)$$.

Now: the meaning of an exterior product on a vector space can be understood as a way to define abstract volumes as if a volume is a kind of vector. Then from this point of view we can see a volume form in $$\bigwedge T^*M$$ as a linear functional on the volume vectors of $$\bigwedge TM$$, where by "volume vector" I mean an element of $$\bigwedge TM$$ of maximum degree.

Then each point of $$M$$ have attached a volume vector at each of it points, that is, if $$\varphi =(\varphi^1, \varphi ^2,\ldots ,\varphi ^n)$$ is a chart on $$U\subset M$$ then $$\frac{\partial}{\partial \varphi^1}\wedge \frac{\partial}{\partial \varphi ^2}\wedge \ldots \wedge \frac{\partial}{\partial \varphi^n}$$ defines a volume vector at each point of $$U$$, and following this point of view if $$\omega\in \bigwedge^n T^*M$$ then the integral $$\int_U\omega$$ can be understood as a "sum" of the scalars $$\omega(\frac{\partial}{\partial \varphi^1}\wedge \frac{\partial}{\partial \varphi ^2}\wedge \ldots \wedge \frac{\partial}{\partial \varphi^n})$$, what, at least to me, seems insanely intuitive :).

• . . n i c e . . Commented Dec 28, 2022 at 17:28

I guess it's already too late though, for someone who landed here in the future, I found this lecture series very useful to grasp the intuition of the differential forms as well as a firm foundation on the theoretical aspects! He explains with very good practices so that we can review if we really get a point of each concept!!

Steps which lead me to more of an understanding:

### 1. Understand Rotation vs Reflection

1. The Orthogonal Group $$O(n)$$ contains all distance preserving transformations of euclidean space. (determinant equal $$\pm 1$$)
2. There are two connected components (sets are connected, if they can not be the union of two open sets), i.e. the two components are both connected sets but disjoints, so $$O(n)$$ itself is not connected - there is a discontinuity.
3. The special orthogonal group $$SO(n)$$ is the connected component containing the identity. It is a normal subgroup and we also refer to these operations as rotation. Its elements have determinant 1.
4. The other connected component are the reflections and have determinant $$-1$$. Since two reflections have determinant $$1$$, they do not form a group.

Since the reflections are not connected to the identity it is not possible to continuously transform an object into its reflection. Objects which are different from their reflections (even modulo rotation - continuous distance preserving operations) are called chiral.

chirality allows us to understand why the multilinear maps we will consider later on should be alternating. I.e. swapping two coordinates flips us into the reflections. And swapping two coordinates will get us back to rotation. If one of the coordinates you swap is the same, then this operation only touches 3 dimension, so even in higher dimension it is possible to visualize how this is only a rotation since the other dimensions remain untouched.

### 2. Understand that determinants measure a (signed) change in volume.

Where the sign indicates whether a reflection took place. The sign is enough, because two reflections are simply a rotation no matter the dimension.

Proofsketch: Let $$A$$ be a linear map. $$\lambda(A \cdot)$$ is still a translation invariant measure, so by uniqueness of the Lebesgue measure (as the only translation invariant measure up to constants) it is just the Lebesgue measure up to a constant. To calculate the constant it is sufficient to consider $$\lambda( A [0,1]^d)$$. For diagonal $$A$$ it is immediately obvious that it should be the product of entries. But with the singular value decomposition we can decompose any general matrix $$A$$ into a positive definite diagonal matrix and distance preserving members of the orthogonal group.

### 3. Understand the set of alternating multilinear maps

Since the set of alternating $$n$$-multilinear maps in $$n$$ dimensions is one dimensional, the determinant is its basis with the property that the determinant of the identity is $$1$$.

Now we understand multilinear maps as a (weighted) measure of volume change. If we were to measure $$k$$-volume in $$n$$-dimensions, we have more possible directions in which we can measure volume, so we stop having only one basis element (the determinant) and instead get multiple. But the principle remains the same. :handwaving: (maybe someone can one up me here)