# Clarification on the meaning of dx in the integral and differential setting

I have some questions regarding differential forms, notation, and integrals.

i would like to point out that I read the other post in this forums regarding this issue with no succes.

I have always been working with expressions such as : \begin{align} \int fdx \end{align} Considering $dx$ as a placeholder and an indication on the variable for which f need to be integrated. the riemannian integral says that taking the limite of a defined sum turns $\Delta x$ Into $dx$. This definition never satisfied me. So I turned into lebesgue integral where $d\mu$ represent the measure that will provide a volume of the domain for which f need to be integrated. I was a little more happy with this ( despite the fact that lebesgue is a nice generalisation of riemann )

Then I was introduced to differential geometry and read that the infamous $dx$ represent the differential of the coordinate function x. Then I learned how to integrate differential form where the integral of a form is defined to be: \begin{align} \int \omega = \int f dx\wedge dy = \int f dx dy \end{align} ( i wrote the case of 2 forms to display the transition between the integral of the form and the integral it translate to ) At that point I was completly lost. My exposure to differential was that $df$ is a functional that takes a vector and return a scalar: \begin{align} df(X) = Xf \end{align}

Then I read that differential can be interpreted as infinitesimals which doesn't really make sense to me. I see the operation $Xf$ as the directional derivative of the function f in the X direction. $df$ if mainly for me a change of perspective where instead of varying the function $f$ we vary the vector $X$ for the directial derivative.

I also studied n-forms which made the whole buisness more confusing as my introduction to n-forms was that it an element of the antisymmetric tensor product space. So basically an n-form is for me a (0-n) tensor that is antisymmetric.

Finally I have been exposed to interpreting $dx$ as the differential of the coordinate function on a manifold. Which is a form that happen to be a basis member of the cotangent space.

So here are my questions :

1) How should we interpret $dx$ in the riemannian integral. As a placeholder ? A differential ? Something else ?

2) If it is ok to treat it as a differential, why do we do this ? What the differential is supposed to represent in the framework of integration since for me it is just a change of perspective on the directional derivative or an element of the tensor product space.

3) are they any relation between the measure used in lebesgue integration and differential forms ?

4) My main problem here is to reconcile all of the things I learned about integration and differential forms. I want to be able to glance at $dx$ in any situation without being afraid of the meaning of it.

Thank you for any pointers or answers.

• Very interesting and well formulated question about this polysemic notation. There are still other uses of "dx" like in infinitesimals of nonstandard analysis, not to speak of the powerful-yet-not-always-rigorous our physicist colleagues use differential notation :). Have a look also to this reference – Jean Marie Dec 22 '16 at 8:50
• I apologize if that induce any inconvenience... – user149705 Dec 22 '16 at 15:47
• @user149705: No need to apologize. This is a good question and I was only expecting that it would have attracted more attention than it really has. This lack of interest is probably due to the fact that similar questions have been already asked (this is a type of question that shows up on MathSE every several months), and to the fact that a significant part of the world is making plans for Christmas and holidays. Not a good time to ask questions! – Alex M. Dec 22 '16 at 22:03
• Aha that's true ! I've been thinking and reading a ton of material online and just couldn't wait any longer... – user149705 Dec 22 '16 at 22:56

1) How should we interpret dx in the Riemann integral. As a placeholder ? A differential ? Something else ?

The correct answer from differential geometry is that for the single-variable Riemann integral, $dx$ represents the $1$-form that is given as the differential of the coordinate function $x$. In particular, it is good to mentally distinguish this from the common notation $dx$ for Lebesgue integration in the $x$-variable, since the Lebesgue integral is not, strictly speaking, a complete generalization of the Riemann integral. (Namely, the Lebesgue integral does not distinguish orientations: integrating from $0$ to $1$ vs. $1$ to $0$ is the same in Lebesgue integration, but introduces a sign in Riemann integration.) Practically speaking, it just tells you what variable you are integrating in.

2) If it is ok to treat it as a differential, why do we do this ? What the differential is supposed to represent in the framework of integration since for me it is just a change of perspective on the directional derivative or an element of the tensor product space.

A $1$-form $\theta$ on a manifold $M$ can be thought of as a function that evaluates the length of local line segments, possibly up to a scaling factor. When $M = \mathbb{R}$ and $\theta = dx$, this is an exact analogy: if $X = \partial_x$ is the coordinate tangent vector, and we think of $X$ having length $1$, then $dx(X) = 1$; and if we scale $X$ to $\lambda X$, so that $\lambda X$ has length $\lambda$, then $dx(\lambda X) = \lambda$. We can equate a line segment in $\mathbb{R}$ with the tangent vector $X$ with the same length and starting at the left endpoint of the segment, so $dx$ in this sense evaluates the length of lines.

More generally, we may have curvilinear coordinates on the manifold, for example if $M = S^1$ with (local) angular coordinates $\theta$. Then the length-evaluation analogy is no longer exact, but holds in an approximate sense: $d\theta$ can be thought of as evaluating the length of infinitesimally small segments starting at $p$ and going in the tangent direction $\partial_\theta$.

Even more generally, a $k$-form $\theta^1\wedge\cdots\wedge\theta^k$ can be thought of as evaluating the volume of infinitesimal $k$-dimensional parallelepipeds. Again, when $M = \mathbb{R}^n$ and $\theta = \theta^1\wedge\cdots\wedge\theta^k = dx^1\wedge\cdots\wedge dx^k$, this analogy is exact: put in $k$ tangent vectors based at the same point, and $\theta$ returns precisely the volume of the corresponding parallelepiped spanned by those $k$ vectors. This also explains the role of the alternating property: if two of the spanning vectors are linearly dependent, then the $k$ vectors fail to span a full $k$-dimensional parallelepiped, and therefore the span has zero $k$-dimensional volume.

If you are integrating a $n$-form on a $n$-dimensional manifold (i.e. a volume form), then the idea is the same as integration in Euclidean space. In Euclidean space, you break things up into tiny $n$-dimensional cubes, then approximate the integral by sampling your function on each cube, forming the Riemann sum, and then letting the size of the cubes go to $0$. On a manifold, the idea of a cube as a subset of the manifold doesn't make much sense, but we can define cubes in the tangent space and they take the role of our infinitesimally small cubes. The volume form then tells us that we know how to find infinitesimal volumes (namely, by working in local coordinates, so everything reduces to Euclidean space), and therefore we can form Riemann sums. If you carefully examine the definition of the integral of a volume form over a manifold, this is what actually happens.

All of this is the geometric interpretation of differential forms, which is sometimes unfortunately buried under the entire gigantic algebraic machinery of exterior algebras and cotangent bundles. The whole point of this construction is to provide a practical notion of integration on manifolds that does not depend on how local coordinates are chosen. In practice, nearly every actual numerical calculation (including) that needs to be done on a manifold must be done by choosing local coordinates, performing the calculation in local coordinates (and therefore, essentially, in Euclidean space, where things are known), then piecing the information from each coordinate neighborhood together into a cohesive whole. Because differential forms transform in the correct way under a change in local coordinates, and have a natural interpretation as evaluating volumes, they work nicely for defining integration on manifolds, and even gives us the change of variables formula pretty much for free.

3) Are there any relations between the measure used in Lebesgue integration and differential forms ?

If $M$ is an orientable manifold, and $\theta$ is a volume form, then integrating characteristic functions against $\theta$ almost defines a measure on $M$, save for the fact that the integral of a volume form need not be non-negative and therefore $\theta$ may assign negative volumes to sets. However, after taking absolute values, one obtains what is called a pseudo-form, or volume element, and this gives a measure on $M$. (In fact, one does not need a globally defined volume form to do this in general, and thus this can be done on non-orientable manifolds.) Since there are infinitely many distinct volume forms, this does not give rise to a single canonical measure.

However, if we restrict to a local coordinate neighborhood $(U,x)$, then (identifying $U$ with an open subset of $\mathbb{R}^n$) we have Lebesgue measure on $U$, which comes from integrating the volume form $dx^1\wedge\cdots\wedge dx^n$. Since any two volume forms only differ by multiplication by a nonvanishing function, we have that $\theta = fdx^1\wedge\cdots\wedge dx^n$ where $f$ is smooth and nonvanishing. In this sense, integration against $\theta$ is locally just a rescaled version of Lebesgue integration, which is a concrete but rather weak connection (since you could be off by a completely arbitrary nonvanishing smooth function). This is not entirely surprising, becaues Lebesgue measure is also not an entirely arbitrary measure: it is a very special (essentially unique) measure that respects the symmetries of Euclidean space. With some additional work, one can make additional and more precise connections by studying pullbacks of forms and measures, which may be where you want to head next.

• So I understand the whole thing as being separate notion depending on the context. We start by defining a Riemann integral on euclidean spaces as a limit of sums. Then we define an integration procedure on manifold by treating the basis vectors in the tangent space as principal direction where we can measure their length if we define a volume form ( I though we could measure length only on Riemannan manifold equipped with a metric ? ). – user149705 Dec 22 '16 at 15:41
• Then we reduce the integral of a form as the Riemann integral. So if I defined a notion of integration independent of chart representation by some crazy arbitrary rules, Am I to say this is a valid form of integration as long as I used it in all calculation ? After all the Riemann integral represent intuitively the area under a curve provided we have a unit of length. Here I feel like on a manifold we arbitrarily choose a volume form, and then define an arbitrarily rule to integrate independent of the chart representation.. – user149705 Dec 22 '16 at 15:41
• Also in euclidean space we have a natural isomorphism between tangent vectors and points in $\mathbb{R^n}$ so the form evaluated on a tangent vector represent the length of the vector in $\mathbb{R^n}$. How does that play out with the metric ( that also provide a length for vectors ), Do we force our volume form to produce the same results ? Also could you answer (or edit your answer) to respond to my confusion regarding the equivalence between what you explained and the fact that $df$ if the directional derivative of f while keeping the entry Vector $X$ varying ? Thanks ! – user149705 Dec 22 '16 at 15:46
• Also If I may add, I am not easy with the circular definition of : The integrand in a Riemann integral is to be considered as a differential form, but then to do actual computation this reduce to the Riemann integral... – user149705 Dec 22 '16 at 16:50
• 1. A Riemannian metric gives you a canonical way to measure lengths, and also gives rise to a canonical volume form. An arbitrary $k$-form is a non-canonical way of assigning $k$-volumes (essentially because the pairing between a vector space and its dual is non-canonical unless there is an inner product). Consequently it may be extremely unsatisfactory: for example, the $1$-form $dx$ on $\mathbb{R}^2$ evaluates only the length of a tangent vector in the $x$-direction, and completely ignores the $y$-direction, with the consequence that this only forms a pseudometric on the tangent space. co: – Gyu Eun Lee Dec 22 '16 at 20:58