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In order to explain to non-experts what a vector field is, one usually describes an assignment of an arrow to each point of space. And this works quite well also when moving to manifolds, where a generalized arrow will be a tangent vector.

My question is: What are similar objects that can help with imagining differential forms?

Positive qualities for such an object would be, for example:

  • it helps justify change-of-coordinate formulas and formulas for the pullback via a function;
  • it is "easily drawable";
  • it helps understand more complicated differential-form-based concepts, e.g. connections, cohomology groups, etc.
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A one-form assigns to each vector tangent to a manifold a real number in a linear way. You may think of a vector tangent to a manifold as being determined by two points on the manifold that are "infinitesimally close", and hence view a 1-form as a function from such infinitesimal pairs of points to the real numbers. This analogy persists in higher dimensions; an n-form is a way of assigning n-points which are mutually infinitesimally close a real number (with the additional assumption that this assignment be antisymmetric). This is more than an analogy. The techniques of synthetic differential geometry allow for a rigorous definition of n-forms this way and it can be proven to line up with the classical definition. The details can be found here:

http://home.imf.au.dk/kock/van00.PDF

Anders Kock has many of his papers on his webpage including some which explain the concept of connections in this language. The advantage of the synthetic approach is that the definitions seem to line up with heuristic arguments often used to think about these objects. Here is the webpage:

http://home.imf.au.dk/kock/

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    $\begingroup$ The definition of Koch speaks of "assigning a number to infinitesimal simplexes, so that the number is zero if the simplex has two coincident vertices".. this sonds like a nice definition to use for cohomology! $\endgroup$
    – Mircea
    Jun 4, 2010 at 19:40
  • $\begingroup$ @Mircea It does not seem to be essentially new. See Grothendieck's infinitesimal cohomology in Dix Exposés, or materials of crystalline cohomology, like Stacks Project. $\endgroup$
    – user20948
    Nov 21, 2019 at 8:51
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The $k$-forms that are easiest to describe are those with $k \in \{0,1,n-1,n\}$. A 0-form on an $n$-manifold is a function. A 1-form on an $n$-manifold, if you imagine it in $n+1$ dimensions, is like an arrangement of shingles on a roof: At each point of the manifold, it defines a directional slope, which as other people have said, is the same as a dual vector on tangent vectors. An $n$-form is a density, i.e., an entity that you can integrate over the manifold. And an $(n-1)$-form is a flux (like, say, describing oil coming out of a well): At each point it has a null tangent direction, and it assigns a non-zero volume to each cross section.

Of course you can think of any $k$-form as a $k$-dimensional flux, and for general values of $k$ you might as well. But when $k$ is 1 or $n-1$, it is somewhat easier to visualize the condition that the form is closed. A 1-form is closed when the shingles locally mesh as the slope of a smooth roof, i.e., the form is locally integrable. An $(n-1)$-form is closed when the flux is locally conservative, which for instance is the case with fluid flow. In fact, theorem: A closed, non-zero $(n-1)$-form is equivalent to a 1-dimensional foliation with a transverse volume structure.

The reason that other values of $k$ are harder is that while you do get an entirely analogous algebraic integrability condition when the form is closed, you might not get the same kind of geometric integrability. A non-zero 1-form has an $(n-1)$-dimensional kernel at each point. (Although the visualization that I suggested is in $n+1$ dimensions, it is also true in $n$ dimensions that these tangent hyperplanes mesh when the 1-form is closed.) A non-zero $(n-1)$-form has a 1-dimensional kernel at each point. But a $k$-form for other values of $k$ doesn't usually have a kernel. (Okay, a maximum rank 2-form in odd dimensions also has a 1-dimensional kernel, and it is equivalent to a 1-foliation with a transverse symplectic structure.)

I have heard the statement that only 1-forms and 2-forms are any good. (Well, that's an overstatement, but they are more important than the others except for maybe $0$ and $n$.) In particular, symplectic forms show up a lot, so it is important to try to imagine them even though by definition they have no kernels. I think of a symplectic form as a calibration for a local complex structure. (Or an almost complex structure, which might be all that exists globally.) I.e., among the different tangent 2-planes of a symplectic $2n$-manifold, the ones that are complex lines have the greatest pairing with the symplectic form, while the ones that are real planes have vanishing pairing, and the pairing minimum is achieved by complex lines with the wrong orientation.


One more remark: The geometric picture of a foliation with a transverse volume structure holds for closed $k$-forms that are also non-zero simple forms (i.e., wedge products of linearly independent 1-forms). I think it's a theorem that any closed $k$-form is locally a sum of closed, simple $k$-forms. If that's correct, then that's also a way to visualize a closed $k$-form, as an algebraic superposition of volumed foliations. $k=1$ and $k=n-1$ are special cases in which every non-zero form is simple.

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    $\begingroup$ I tend to interpret the 'goodness' of the $1$- and $2$-forms as saying that we do not understand $k$-forms for $k\geq3$ well enough :P $\endgroup$ Jun 4, 2010 at 4:59
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    $\begingroup$ Re "any good" remark: 2-forms are curvature forms of a connection, so obviously they are very important; their exterior powers represent higher Chern classes, so they are important as well. 3-forms also arise naturally, for example, there is a canonical invariant 3-form on any compact Lie group, given by $(X,Y,Z)\mapsto K([X,Y],Z)$ on the level of the Lie algebra. J.-L. Brylinski wrote a whole book treating integral 3-forms (3-curvature forms of gerbes). $\endgroup$ Jun 4, 2010 at 15:24
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If you look at Misner-Thorn-Wheeler's book, Gravitation, you will see heroic attempts to draw pictures of differential forms. But I think this is misguided. Not everything can be drawn directly as a picture.

To me, a 1-form is a measuring instrument defining a unit of speed for a vector field. Without a $1$-form, there is no natural way to measure the length of a vector field or the speed of its integral curves. A 1-form is the simplest co-ordinate-free way to do so.

Higher exterior powers of vectors and forms have corresponding but more elaborate explanations.

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    $\begingroup$ Well, the 2-dimensional analogue of a vector field is a field of infinitesimal parallelograms. So a 2-form is an instrument for measuring the area of such an infinitesimal parallelogram. $\endgroup$
    – Deane Yang
    Jun 4, 2010 at 22:22
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    $\begingroup$ A 1-form certainly assigns a number to a vector, but to me it seems kind of misleading to think of this number as the "speed" of the vector because "speed" shouldn't be linear in the vector. Sure, the speed should scale linearly with the vector. But it shouldn't be additive the way a 1-form is. Two vectors pointing in opposite directions should have the same speed, not the opposite speed. A "speed" should not have a codimension-1 hyperplane of vectors in its kernel. And so on. $\endgroup$
    – Tim Campion
    May 8, 2013 at 2:21
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    $\begingroup$ Moreover, the most important subclass of 1-forms are the closed ones, whose integrals along curves are locally path-independent. If you had a notion of "speed" with this property, then its integral should be a distance. Then a "closed speed" gives you distances which are locally path-independent. Very weird. $\endgroup$
    – Tim Campion
    May 8, 2013 at 2:22
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    $\begingroup$ Tim, you make some good points. I shouldn't have said just "speed". It's a "directional speed" but in the following sense: Imagine a bunch of evenly spaced ordered set of parallel planes in space and an object traveling through the planes. The orientation of the planes represents the "direction" in an affine invariant way (i.e., without using the notion of angle or the inner product). Then you can measure how quickly the object passes through the parallel planes. And since the parallel planes are ordered, you can also assign a sign to this speed. $\endgroup$
    – Deane Yang
    May 8, 2013 at 3:36
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    $\begingroup$ See also W Burke's `Applied Differential Geometry' for wonderful pictures. Burke's pictures are basically carefully thought out, clarified, pared-down versions of those in Misner-Thorne-Wheeler. $\endgroup$ Mar 29, 2015 at 22:50
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$p$-forms are what you can integrate over submanifolds of dimension $p$. While this may sound way too operative to be called an intuition, it'll get you quite far.

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    $\begingroup$ It helped me when I realized that $p$-forms are actually integrated over oriented submanifolds of dimension $p$, and those submanifolds may have boundaries. $\endgroup$
    – Axel Boldt
    Feb 20, 2015 at 23:49
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Terry Tao has wonderful article in the PCM on differential forms. I frequently refer to it when I'm trying to get my head around some concept.

http://terrytao.wordpress.com/2007/12/25/pcm-article-differential-forms/

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Since in the question you only mention vector fields and not multivectors, I presume that you only ask about 1-forms, not higher forms :D

Now for 1-forms my personal geometric intuition instrument for 1-forms is this: each linear form $\varphi$ on a vector space determines a foliation of the latter into parallel hyperplanes $\varphi^{-1}(c)$. Call the hyperplane $\varphi^{-1}(0)$ the $\textrm{codirection}$ of $\varphi$ and call the distance between the hyperplanes $\varphi^{-1}(0)$ and $\varphi^{-1}(1)$ the $\textrm{colength}$ of $\varphi$. Clearly each $\varphi$ is uniquely determined by its codirection and colength.

Then just as a vector field is specification of a direction and of a length at each point, a 1-form is a specification of a codirection and of a colength at each point.

Intuitively, while a vector field tells us in which direction to move and at what speed from each point, a 1-form tells us parallelly to which hyperplane to foliate and with what density at each point.

This by the way is in accordance with the fact that connections form an affine space corresponding to the linear space of 1-forms. A connection in these terms is the same thing except we do not specify which of the hyperplanes is the zeroth one: to specify a codirection and a colength, one does not need to specify a hyperplane, it suffices to know the set of parallel hyperplanes and the ratio of the distance between $\varphi^{-1}(c)$ and $\varphi^{-1}(c')$ and of $c-c'$ for some distinct $c$ and $c'$. Thus we can view a connection as a specification of an $\textrm{affine codirection}$ (that is, a class of parallel hyperplanes) and of an $\textrm{affine colength}$, that is, of a ratio as above.

I admit this is not mathematics, but then the question was primarily about intuition, right?

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90% of the time, all you need to know about k-forms is that they are something that turns a k-tuple of vector fields into a function on your manifold. (Aka they eat vector fields and give functions.)

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    $\begingroup$ 47.5% of all statistics is made up on the spot! (FTR: I agree with the other 95% of your answer) $\endgroup$ Jun 4, 2010 at 4:09
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Well, since you consider satisfying to draw a vector field using the usual drawing of a vector, one interpretation of your question is how can one draw a linear form. I learned a nice way to do that from Patrick Massot: just draw the kernel and the level of the value $1$. These two geometric object carry all the information about the linear form. Below are quick sketches of what one can easily do on a board.

enter image description here

enter image description here

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    $\begingroup$ Since everyone knows how vector addition and multiplication with scalars looks in terms of the standard pictures, it's a nice exercise to figure out the analogous pictures for covectors. $\endgroup$ Mar 30, 2015 at 8:16
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One-forms are coordinates on the tangent space at a given point. Given a basis of a n-dimensional vector space, the dual basis is just the nth coordinates functions on the vector space in the given basis. Of course, if the basis of vector depends on the point, its dual basis also changes as we move the point on the manifold.

n-forms in n-dimensional space are a way to compute volume, or more generally the extensive quantities that appears in physics: mass, electric charge etc. The "value" of the n-form at a point is the infinitesimal mass located at the point.

Explaining forms of intermediate degree is a bit tricky. I think that some physical intuition can help again. Given a thin surface of non-homogeneous material in 3-dimensional ambient space, we want to describe how the mass is distributed on the surface. At each point of the surface, we want a 2-dimensional element of "volume". This infinitesimal mass depends on the point chosen on the surface. The distribution of 2-dimensional mass can be described by a 2-form. That form retains two informations: the way the tangent plane to the surface is oriented in space (which is completely determined by the normal vector to the surface), and the infinitesimal mass value.

Arguably, such an explanation suggests that a p-form can be "integrated" so as to give some p-dimensional manifold, which is completely wrong. Still it provides some intuition for p-forms, which are meant to be integrated on manifolds, and it can be a starting point for a more elaborate answer.

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  • $\begingroup$ A 2-form in a 3-manifold does not retain the information of the orientation of the tangent plane to the surface. It rather provides a mass density (or charge density, to account for the sign) for each surface. $\endgroup$ Jul 7, 2010 at 3:37
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I'm sure Mircea you've already thought of what I'm about to say, but if you cheat a little (viz. you assume you're manifold has a metric) you can always think of cotangent vectors as tangent vectors which act on others by scalar product.

I guess as one imagines the wedging of tangent vectors as (higher dimensional) parallelograms one can apply a similar trick.

p.s. Another idea might be to view cotangent vectors as their kernels, so hyperplanes in tangent space. This gives a picture of the projectivisation of the cotangent space. Maybe this trick can be extended as well to differential forms?

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  • $\begingroup$ I agree about 1-forms and 1-vectors, but I'm not sure of the dualtity between k-forms and k-vectors, for k different from 0,1,n-1,n (if n is the dimension of the manifold)... my intuition would then be that the dual of k-forms are just "simple k-vector"-fields. To justify that, observe that a form should be something that integrates on a submanifold, and the vector $e_1\wedge e_2+e_3\wedge e_4$ in $\mathbb R^4$ is not representing the tangent of a submanifold. $\endgroup$
    – Mircea
    Jun 4, 2010 at 18:51
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As a possible aid toward improving "geometric imagination of differential forms," I include below three somewhat random but suggestive images:


      DifferentialForms3
      Sources: Image #1. Image #2. Image #3.


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Perhaps it is useful to read the discussion given here. It deals with cohomology, but the intuition behind forms is also discussed.

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Certain special $k$-forms on an $n$-manifold $M$ do lend themselves to visualization.

For simplicity, let me assume that $M$ comes with a Riemannian metric.

Now suppose that we have a field $\phi$ of oriented $k$-planes and a scalar field (i.e. real-valued function) $s$ on $M$. The metric and the orientation induce $k$-forms on the $k$-planes in $\phi$. (I'll take $k$-forms on a $k$-plane as trivial to visualize, or nearly so.)

Given a $k$-tuple $T$ of tangent vectors at a point $p$, one can project the vectors in $T$ to $\phi(p)$, the $k$-plane over $p$, compute the volume (according to the $k$-form induced by the metric) of the resulting parallelepiped, then use the orientation of $\phi(p)$ to get a sign and finally scale by $s(p)$.

The linearity of projection makes this recipe yield an alternating multilinear function of the vectors in $T$. Thus the given data determine a $k$-form in a way roughly as easy to picture as a vector field.

General differential $k$-forms arise as linear combinations of these special ones. The usual coordinate representation of $k$-forms works exactly this way, albeit in that case the $k$-plane fields derive from the coordinate system alone, independently from the $k$-form one hopes to visualize.

That leaves the question of how much better we do (at visualization) if we adapt the $k$-plane fields to the $k$-form we want to see. In general, we dod not do much better. For large $n$, and for $k$ intermediate between $0$ and $n$, the dimension of the Grassmann variety of $k$-planes in $n$-space simply doesn't supply nearly enough degrees of freedom for significantly reducing the number of summands in the linear combination. And correct me if I err here, but I don't see anything that would make a representation with a minimal number of summands in any way canonical, even generically. But perhaps one can say something more in low dimensional cases, say $2$-forms on a $4$-manifold. A dimension count suggests two fields of $2$-planes should suffice.

Let me add that by much the reasoning above, we get the projective Grassmann coordinates on the Grassmann variet $Gr(n,k)$ of $k$-planes in $n$-space. A differential $k$-form then determines a hyperplane section of the Grassmann variety. The hyperplane section "knows" the differential form up to a scalar. (We have here an generalization of regarding a (co)vector as a magnitude and a direction.) So up to a scalar field, we can regard a differential form as a bundle of Grassmann variety hyperplane sections. That may not make it easier to "see" differential forms, but it does translate them into something geometrical, canonical and coordinate free.

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I'd like to comment on some of the ideas here: You should ask yourself whether you want to make the (implicit) assumption that your manifold is already endowed with a volume form $\omega$. If the answer is "yes", then this gives you an isomorphism between $k$-vectors (antisymmetric tensors of degree $k$ of tangent vectors) and $n-k$-forms (where $n$ is the dimension of $M$) by inserting the $k$-vector into $\omega$. From this you get your interpretation: $n-k$-forms describe orientation and length/area/volume (choose depending on dimension) of a small $k$-dimensional cube. This isomorphism between $k$-vectors and $n-k$-forms essentially is the one that lets us describe the orientation of a flat surface in 3 dimensions by one arrow perpendicular to it, or by two non-parallel arrows inside of it.

But what if you do not want to make this assumption? Then you have to decide which one of forms and tangent vectors are the "arrow perpendicular to it" and which are the "arrow inside of it". I'd say that the tangent vectors are "arrows inside of it" and forms are "arrows perpendicular to it". For example, the differential of a smooth function are the arrows perpendicular to the levelsets of the function, while the tangent vectors of a curve are the arrows "along" the curve, which becomes "inside" the curve if we zoom in so that the curve becomes just a line.

Whatever version you choose, it might always be good to keep in mind that our intuition likes to implicitly apply the isomorphism between forms and vectors, so we have to be careful if this is not allowed.

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I personally learned a lot from trying to formulate and visualize the analogue of differential geometry over $GF(2)$, also known as differential logic. There is an exposition here.

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