I'm trying to teach myself about differential forms. I understand that a $k$-form “eats” $k$ vectors to give a number (eg a 1-form eats a single vector to give a number). This is a nice, vivid mental image to help me picture what's going on. However, it's only just dawned on me (this may well be a false dawn) that this “eating” of vectors can happen in one of two ways.

First, (keeping things simple) a 1-form $\omega$ can act on a vector $v$:$$\omega\left(v\right)=\omega_{i}v^{i}.$$This works because of the dual basis definition$$\boldsymbol{e}^{i}\left(\boldsymbol{e}_{j}\right)=\delta_{j}^{i},$$where $\boldsymbol{e}^{i}$ and $\boldsymbol{e}_{j}$ are, respectively, the basis 1-forms and basis vectors.

Second, a $k$-form can be integrated over a $k$-dimensional manifold $M$. For example, a 1-form $\omega$ can be integrated over a curve $C=\mathbf{X}\left(t\right)$ with $a\leq t\leq b$

$$\intop_{C}\omega=\int_{a}^{b}\omega\left(\mathbf{X}^{\prime}\left(t\right)\right)dt.$$Both methods give a number, but the integration version of “eating” only makes use of vector components (the $\mathbf{X}^{\prime}\left(t\right)$ bits), not the basis vectors.

Is this a reasonable way of understanding what “eating” vectors means? Thank you.

EDIT - 25 June 2019

On reflection, I messed up. My premise was false, meaning this is a non-question. There is no separate integration method of differential forms “eating” vectors. The 1-form $\omega$ acts on the vectors $\mathbf{X}^{\prime}\left(t\right)$ because of the dual basis relationship $dx^{i}\left(\mathbf{e}_{j}\right)=\delta_{j}^{i}$. I was confused.

  • 1
    $\begingroup$ A differential form assigns a $k$-tensor to each point on a $k$-manifold. It is the $k$-tensor assigned to a particular point on a manifold which eats tangent vectors. If the manifold is $\mathbb R^k$, then a differential form assigns a $k$-tensor to each point in $\mathbb R^k$. It seems like the concept of a $k$-tensor, and distinguishing between the concepts of a differential form and a $k$-tensor, might be missing from your explanation. $\endgroup$
    – littleO
    Jun 21, 2019 at 8:08
  • 1
    $\begingroup$ Intuitively, to integrate a $k$-form $\omega$ over a $k$-manifold $M$, we chop up the manifold into tiny pieces, in such a way that the $i$th piece is (approximately) a parallelepiped spanned by tangent vectors $v_{i,1},\ldots,v_{i,k} \in T_{p_i}M$. The contribution of the $i$th piece of $M$ to the integral is computed by feeding the tangent vectors $v_{i,1},\ldots, v_{i,k}$ into the $k$-tensor $\omega(p_i)$. We sum up the contributions of all the individual pieces to compute the integral $\int_M \omega$. $\endgroup$
    – littleO
    Jun 21, 2019 at 8:18
  • $\begingroup$ @littleO Thanks for that. I $knew$ I'd get the details wrong. But in terms of the big picture, can I safely assume there are, broadly speaking, two ways differential forms consume vectors? $\endgroup$
    – Peter4075
    Jun 21, 2019 at 8:26
  • $\begingroup$ The integral of a 1-form over a curve in a manifold is definitely not denoted $\int_M \omega$, especially when $\dim(M) > 1$! $\endgroup$
    – KCd
    Jun 21, 2019 at 15:28
  • $\begingroup$ @KCd - whoops. I meant to say a 1-dimensional manifold $M$. I've corrected my question accordingly using $C$ instead of $M$. Hope that it's now clearer. $\endgroup$
    – Peter4075
    Jun 21, 2019 at 16:20

1 Answer 1


There are a few properties of differential forms that you didn't mention. First and foremost an important quality they enjoy is being anti-symmetric. Anti-symmetry of course cannot be seen when only dealing with one forms, you have to look at two forms and up.

What is anti-symmetry?
Well consider the function $$q(x,y)=x-y$$ Now if I exchange x and y I will pick up a negative sign for my function. $$q(y,x)=-q(x,y)$$ In the case of two variables, there are only two rearrangements of the input, namely doing nothing and exchanging x with y.
This concept is rooted in Abstract Algebra and is related to the Symmetric Group.
The symmetric group is a collection of mappings from $$\{1,...,n\}\rightarrow \{1,...n\}$$ Or all possible rearrangements of n letters.
I don't want to get very much into group theory here but this is what's going on.

Differential forms interact with members of the Symmetric Group denoted: $$S_n$$ By satisfying the following relation: $$\forall \sigma \in S_n, \quad \omega \in \Lambda^k(M)\\ \sigma(\omega)=(-1)^m \omega$$ Where m is the number of transpositions in the permutation. Also called the signature or sign.

The key here is that differential k-forms are a special kind of tensor. The kind of tensor that interacts with $S_n$ in the manner described.

What else can I say about them..

-Well they allow us to define the determinant of a linear map for one. I'm not going to run through the construction unless you want me to but it is certainly worth looking at.

-They give rigorous meaning to the cross product in three dimensions and higher.

-They provide the tools for defining div, grad and curl. Which are the result of applying a special map called the exterior derivative to a zero form, 1 form and 2 form.
Again worth looking at.

-I think most importantly they provide a theory of integration and of course Stoke's Theorem. When you study generalized Stokes you will see that the Fundamental Theorem of Calculus, Green's Theorem, Divergence Theorem and classical Stoke's Theorem are all the same but deal with different flavors of forms.

Let me know if you have any further questions, I left most details out.
An excellent reference is 'A Geometric Guide to Differential Forms' by David Bachman.

In answer to your question:

The situation is as follows. You have a curve in say $R^3$ and you parameterize the curve. Now at each point along the curve you have a Tangent Space. What is a Tangent Space? It's a vector space attached to the curve where all the tangent vectors live. In your mind picture it as nothing more than the tangent line.

In our case the vector space is one dimensional. But in differential geometry, when we study surfaces, the dimension of the surface is characterized exactly as the dimension of its tangent space.

Now you have a one-form defined along this curve. When we integrate the one form the idea of partitioning the domain as we do for Riemann Integration is the same. We partition the curve and as we do in each partition interval we choose a tangent space and have the one form act on a vector in that space. Then you add them up to get your number.

So the long winded answer is when you get down to it, integrating a form and evaluating a form are two different things. You build the integral of a form by evaluation and summing. So I think in essence the two cases are different but of course I could be wrong. Maybe someone else could elaborate more on this.

I can help with wedge product too if you want.

Unfortunately or fortunately, however you want to look at it. Tensors are algebraic objects and to really understand them you need some algebra which of course takes some time.

Hope this helps.

  • $\begingroup$ Thanks for that, though I'm afraid a lot of it was above my head. I have come across anti-symmetry when looking at the wedge product. Hate to sound like a stuck record but coming back to my original question, have I correctly identified two different ways that differential forms act on vectors to produce numbers? $\endgroup$
    – Peter4075
    Jun 21, 2019 at 12:14
  • $\begingroup$ Not a problem, tensors and forms are difficult to grasp at first. When you integrate a form you are still feeding it vectors! $\endgroup$ Jun 21, 2019 at 14:47
  • $\begingroup$ I messed up. There aren't two methods of forms "eating" vectors, just the one. Please see my edit. $\endgroup$
    – Peter4075
    Jun 25, 2019 at 7:10

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