can someone please informally (but intuitively) explain what "differential form" mean? I know that there is (of course) some formalism behind it - definition and possible operations with differential forms, but what is the motivation of introducing and using this object (differential form)? I have heard that they somehow generalize integration, are used for integration of manifolds and can evaluate k-dimensional integrals in n-dimensional space ($k \leq n$), but is it really true and is it the main motivation of introducing this object into mathematics? Thank you for explanation

  • $\begingroup$ How familiar are you with manifolds? Do you know what a vector field on a manifold is? $\endgroup$
    – Plutoro
    Commented Jul 20, 2018 at 23:22
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    $\begingroup$ You might want to check out my lectures on YouTube, starting here. $\endgroup$ Commented Jul 20, 2018 at 23:38
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    $\begingroup$ The idea behind integration is to chop things up into tiny pieces and add up the contributions of the individual pieces. What type of object should we integrate over a manifold? Chop up the manifold into tiny pieces, so that the $i$th piece is approximately a paralellopiped spanned by tangent vectors $v_1,\ldots,v_n$. The thing we integrate should take those tangent vectors as input and return a real number as output. You can see the output should be an alternating multilinear function of those input vectors, for consistency. You just discovered differential forms. $\endgroup$
    – littleO
    Commented Jul 21, 2018 at 1:22
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    $\begingroup$ Take a look at this: math.ucla.edu/~tao/preprints/forms.pdf $\endgroup$
    – Wraith1995
    Commented Jul 21, 2018 at 1:42
  • $\begingroup$ Do you know anything about tensors? The shortest and "tl;dr" version of forms is that they are "extensions" of linear functions. Take this explanation very lightly. The reason why "differential" is added here is most likely because that is usually the object of interest in the maps - smooth manifolds. $\endgroup$
    – IAmNoOne
    Commented Jul 21, 2018 at 9:58

3 Answers 3


To talk about differential forms, first we need to talk about manifolds and vector fields. Informally speaking, a manifold is any space which is locally Euclidean. That is, the area around every point in a manifold "looks like" Euclidean space, but the space as a whole may not be Euclidean. Examples include spheres and tori. More complex examples are varied and interesting, but are difficult to define in an informal setting.

A smooth manifold is a manifold where the Euclidean regions around each point are in some sense "compatible." This means that if the Euclidean regions of two points overlap, I can both Euclidean coordinate systems in that overlap region, and transfer from one to the other in an infinitely differentiable way.

A smooth function on a smooth manifold is a function whose range is the real numbers, and which is infinitely differentiable with respect Euclidean coordinate systems in the Euclidean regions around points in the manifold. The set of all smooth functions on a manifold $M$ is called $C^\infty(M)$

If things are moving a little fast for you, you may some up the last three paragraphs as, "we have spaces that up close look like Euclidean space, and functions on them that are in some sense differentiable."

Now we will talk about a vector at a point on a smooth manifold. This definition is probably going to sound really strange, but it really is the simplest way to define vectors on smooth manifolds. A vector $v$ at a point $x$ in a smooth manifold $M$ is any function whose domain is $C^\infty(M)$ and whose range is $\mathbb R$, and which satisfies the following three properties:

  • $v(f+g)=v(f)+v(g)$
  • $v(\lambda f)=\lambda v(f)$
  • $v(fg)=v(f)g(x)+f(x)v(g)$

Where $f$ and $g$ are smooth functions on $M$ and $\lambda$ is a real number (notice something that looks like the product rule). What does this mean, and how does it relate to vectors as we are used to seeing them? We are used to seeing vectors defined by a collection of components. But the problem with that is that those coordinates depend on the coordinate system we choose to use. The definition I just gave does not. But if you like coordinates, do not worry; we can transfer between these two definitions. If your coordinates are $(x_1,\cdots,x_n)$ and your vector expressed in Euclidean coordinates at a point $x$ is $v=(v_1,\cdots,v_n)$, we can write the vector as an object of the form I defined above by writing $$v=v_1\frac{\partial}{\partial x_1}+\cdots+v_n\frac{\partial}{\partial x_n}.$$ It can act on a function $f$ by differentiation: $$v(f)=v_1\frac{\partial f}{\partial x_1}+\cdots+v_n\frac{\partial f}{\partial x_n}.$$ It is easy to observe that this satisfies each of the three properties above.

A smooth vector field on a smooth manifold is a collection of vectors on a manifold, one at each point, which vary is a smooth (differentiable) way. In other words, it is a function $X$ whose domain and range are $C^\infty(M)$ such that

  • $X(f+g)=X(f)+X(g)$
  • $X(\lambda f)=\lambda X(f)$
  • $X(fg)=X(f)g+fX(g)$

If it is easier for you, it is okay to imagine a vector as a little arrow lying tangent to some surface, and to imagine a vector field as a bunch of such arrows covering the manifold. This is a natural image to think of, but it turns out to be rather unhelpful in practice. But as this is an "informal discussion," go ahead.

We are finally ready to define a differential form.

A differential $k$-form on an $n$ dimensional smooth manifold $M$ is any multilinear function $\omega$ which takes as input $k$ smooth vector fields on $M$, $X_1,\cdots,X_k$ and outputs a scalar function on $M$ so that $$\omega(X_1,\cdots,X_i,\cdots,X_j,\cdots,X_k)=-\omega(X_1,\cdots,X_j,\cdots,X_i,\cdots,X_k).$$ The latter property is called antisymmetry.

So, what is the motivation behind such an object? As far as I know, the most important application of differential forms is, by far, integration on manifolds. There may have been some other reason for their initial discovery and definition, but this is what they are used for. When you think of integration, you think of calculating area and volume. In an $n$-dimensional manifold, we may want to measure the volume or area of any submanifold of $n$ with any dimension less than or equal to $n$. Given a coordinate system on $M$, a differential $k$-form tells us how to measure $k$-dimensional volume according to that coordinate system. Assuming you have taken a multivariable calculus course, you probably remeber seeing a picture of a spherical coordinate volume element.

Spherical coordinates volume element

Pictures like this one also give us some idea about how differential forms work. The picture labels infinitesimal changes in the $\theta$, $\phi$ and $r$ directions, and shows how we can calculate the infinitesimal volume swept out by these changes. Instead, we could consider three vector fields which at each point have vectors tangent to the $\theta$, $\phi$ and $r$ directions respectively. A differential 3-form would combine these three vector fields into the same volume element.

Why the antisymmetry? The antisymmetry allows us to consider orientations. Again, if you have studied multivariable calculus, you know that when integrating over a surface in 3 dimensional space, it is usually important to note which direction normal vectors to the surface point. But if our surface sits in a space of 4 dimensions or more, there is not a unique normal direction at each point, so we instead use the order of our coordinates to determine orientation. If we switch two coordinates we switch orientations. We will get the same result from integration except the sign will be reversed.

I hope I answered some of your questions.

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    $\begingroup$ Wish I had more than just an upvote to give. Great answer! $\endgroup$ Commented Jul 21, 2018 at 5:01
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    $\begingroup$ What a beautiful answer! Wow. This helped me a lot. Thank you. $\endgroup$
    – Alapan Das
    Commented Mar 29, 2020 at 14:23
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    $\begingroup$ I don't think anyone could have given the questioner a better answer then that if they tried. : ) $\endgroup$ Commented Dec 20, 2020 at 9:28
  • $\begingroup$ Thanks for the great answer! The part I'm stuck on is: your new definition of a vector looks like a directional derivative. This makes sense because there should be a 1-1 relationship between vectors and directional derivatives, is that right? And so in the case of your vector field, $X: C^\infty(M) \rightarrow C^\infty(M)$ looks like it could send $f$ to its directional (along the directions given by the vector field) derivatives. Is this an acceptable way to understand it? If not, why not? $\endgroup$
    – 900edges
    Commented Jan 19, 2022 at 13:40
  • $\begingroup$ @900edges Yes, a vector is like a directional derivative at a point (but it is scaled depending on the magnitude of the vector), and a vector field $X$ maps a smooth function $f$ to another smooth function $g$ such that $g(x)$ is the (scaled) directional derivative of $f$ at $x$ in the direction of $X(x)$. $\endgroup$
    – Plutoro
    Commented Jan 19, 2022 at 15:16

One day Élie Cartan was having a lot of Absinthe and suddenly he had a strange idea: "What if I take the change of variable formula and represent it without that annoying Jacobian determinant?".

First Mr. Cartan expressed, as usual, the integrand and the region with new coordinate function $X' = \Phi(X)$. Then he formally expressed the differentials with new differentials $$dx'= \sum \frac{\partial x'}{\partial x^*}dx$$

Now he could extract the Jacobian determinant if he treated differentials in a very bizarre way,$ dx \cdot dy = -dy \cdot dx$, just like with matrix columns in determinant. Hence, he could get rid of the Jacobian determinant $\det [J]$ and encode that information into the differentials: $$ \begin{align}\det [J]d(x_1)∧...∧d(x_n)&=\det[J]Det[x_1,..., x_n]=\det[J*X] \\ &=\det[X']=\det[x'_1,..., x'_n]=d(x'_1)∧...∧d(x'_n) \end{align}$$

Which gives,

$$ \to ∫f(X')d(x'_1)∧...∧d(x'_n)= ∫f((ϕ(X)) \det[J_ϕ]d(x_1)∧...∧d(x_n)$$

As a true Frenchman Cartan quickly axiomatized and generalized his powerful invention, and left future generations wondering what the Hell's going on...

From gem of a YouTube comment in the comments of this lecture by Ted Shifrin

Related MO answer

  • $\begingroup$ After going through differential forms about a few months on and off, I have concluded that this comment was possibly the most sensible answer $\endgroup$ Commented Jun 25, 2021 at 22:31
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    $\begingroup$ Thanks for transcribing and conserving this wonderful gem. I cannot stop making other people aware of it. Wish I knew who that anonymous person was who made that comment in youtube 8 years ago. $\endgroup$
    – Kurt G.
    Commented Jun 24, 2023 at 21:17

Unfortunately, there is a ton of formalism behind differential forms, and an "intuitive" explanation of them for a non-specialist is probably too much to ask. I myself have spent a reasonable amount of time thinking about what differential forms "are," trying to derive them from first principles, or trying to understand them as generalizations of this-or-that object in elementary calculus.

At this point, however, I think that the best way to approach the daunting concept of differential forms is to realize that differential forms are defined to be the thing that makes Stokes' Theorem true. In other words, you can approach understanding forms in two different ways:

  1. You can try to understand differential forms first, and interpret Stokes' Theorem as a profound result about forms, or
  2. You can realize that the Fundamental Theorem of Calculus, Green's Theorem, Divergence Theorem, Kelvin-Stokes, etc. are all tantalizingly similar, and you can then try to understand why forms are the correct abstraction that makes all of these theorems special cases of the same general Stokes' Theorem.

I think that the second approach is more enlightening, and leads to less banging your head against the wall trying to understand "why" the definitions are what they are. So I would advise first thoroughly understanding all of the theorems I listed above, and then, once you've convinced yourself that they are all "the same" in some mysterious way you can't quite put your finger on, try picking up a book on differential forms that doesn't start with manifolds as a prerequisite, e.g. Spivak's Calculus on Manifolds. Understanding determinants very well probably won't hurt either.


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