23
$\begingroup$

Let $\omega$ be an $n$-form on an oriented $n$-manifold $M$. To integrate $\omega$, we choose an atlas $(O_\alpha, (x^1_\alpha,\dots, x^n_\alpha))_\alpha$ for $M$ and a partition of unity $\phi_\alpha$ subordinate to the atlas. Then we write $\omega|_{O_\alpha} = f_\alpha \mathrm{d}x^1 \wedge \dots \wedge \mathrm{d}x^n$ and define $\int_M \omega = \sum_\alpha \int_{O_\alpha} \phi_\alpha f_\alpha dx^1\cdots dx^n$, where now the "d"'s represent the Lebesgue measure rather than the exterior derivative of differential forms. Then we show that the result doesn't depend on the choice of atlas or partition of unity.

Is there an alternate definition that avoids the coordinates? It seems to me that one should be able to define integration of a differential form in a coordinate-independent way and then derive the above formula as a consequence.

It's not actually the partition of unity that bugs me the most. What really puzzles me is the way we use coordinates to "magically" transform our differential form into a measure. This transformation doesn't depend on a choice of coordinates, so why should we have to use coordinates to describe it?

$\endgroup$
17
  • 2
    $\begingroup$ Why do you believe it should be definable without coordinates? For convenience, suppose $M$ is compact. Consider linear operators from the space of $n$-forms to $\mathbb{R}$. Of course, $\int_M$ is one such operator; but so is $2 \int_M$. How would you distinguish between the two? $\endgroup$
    – Zhen Lin
    Nov 15, 2014 at 16:50
  • $\begingroup$ One way to do it is to require that $\int_{\mathbb{R}^n} f\mathrm{d}x^1 \wedge \dots \wedge \mathrm{d}x^n = \int_{\mathbb{R}^n} fdx^1 \cdots dx^n$ (and ask that $\int_M$ be natural with respect to pullback of forms). So it would suffice to have a coordinate-free way of defining the $n$-form $\mathrm{d}x^1 \wedge \dots \wedge \mathrm{d}x^n$. This should be possible because you can't recover the coordinates from the form. $\endgroup$
    – tcamps
    Nov 15, 2014 at 17:09
  • $\begingroup$ But how would you distinguish the $n$-form $\mathrm{d} x^1 \wedge \cdots \wedge \mathrm{d} x^n$ from $2 \, \mathrm{d} x^1 \wedge \cdots \wedge \mathrm{d} x^n$? $\endgroup$
    – Zhen Lin
    Nov 15, 2014 at 17:11
  • $\begingroup$ I don't know. It should be possible in principle because it's certainly possible to distinguish them when you allow yourself coordinates - one of them is the wedge of the differentials of the coordinates while the other is not -- but the full power of the coordinates is not used: we can have $\mathrm{d}x^1 \wedge \dots \wedge \mathrm{d}x^n = \mathrm{d}x'^1 \wedge \dots \wedge \mathrm{d}x'^n$ for different coordinate systems $x,x'$. $\endgroup$
    – tcamps
    Nov 15, 2014 at 17:23
  • $\begingroup$ I suppose we can equally ask how to distinguish the Lebesgue measure among other measures on $\mathbb{R}^n$ -- even among Haar measures -- without coordinates. $\endgroup$
    – tcamps
    Nov 15, 2014 at 17:24

5 Answers 5

34
$\begingroup$

There is a coordinate free approach to integration in the book Global Calculus by Ramanan, Chapter 3. In particular, the change of variable formula is deduced from abstract nonsense in Corollary 2.9. Unfortunately, I don't really understand the abstract nonsense carried out earlier. I cannot even precisely pinpoint where the "magic" is happening. I will try to summarize my understanding of what he does.

Roughly, Ramanan constructs the sheaf of densities as a subsheaf of the sheaf of Borel measures in Def. 2.6. To me, this definition is a bit unclear, because it uses a flat homomorphism from top dimensional forms to Borel measures even though it is remarked in Rem. 2.3 that this does not in general exist.

He claims that it is obvious that one has a canonical isomorphism of the sheaf of densities with the tensor product of top dimensional differential forms and the orientation sheaf (i.e. with the usual definition of densities found for example on Wikipedia. I cannot confirm that this is obvious, but if it is obvious, this suggests that the magic is happening somewhere else.

Borel measures can be integrated by definition (by pairing with the constant unit function), so viewing densities as a subsheaf of Borel measures, one can integrate densities. On the other hand, densities on $\mathbb{R}^n$ have the form $f dx_1 \dots dx_n$. I don't know where he proves that the integral of this density agrees with the Lebesgue integral of $f$.

$\endgroup$
0
7
$\begingroup$

I've got a partial answer that in particular addresses Zhen Lin's objections. It requires relating integration in different dimensions in two different ways.

We rely on two principles:

  1. External product: $\int_{X\times Y} \omega \boxtimes \eta = (\int_X \omega) \cdot (\int_Y \eta)$, whenever $\omega$ is a top-dimensional form on $X$ and $\eta$ is a top-dimensional form on $Y$. By $\omega \boxtimes \eta$ I mean the external product, $\omega\boxtimes \eta = \pi_X^*(\omega)\wedge \pi_Y^*(\eta)$.
  2. Stokes' theorem: $\int_{\partial \Omega} \omega = \int_{\Omega} \mathrm{d}\omega$

Suppose we've gotten as far as agreeing that for 1-dimensional integrals, $\int_{\Omega} f \mathrm{d}x = c\int_{\Omega} f dx$ for some scalar $c$. Then consider the area of the unit square $\int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y$.

  • Using (1), $\int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y = (\int_I \mathrm{d}x)(\int_I \mathrm{d}y) = (c\int_I dx)(c\int_I dy) = c^2$
  • Using (2), $\int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y = \int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y = \int_{\partial (I\times I)} x\mathrm{d}y = c \int_{\partial (I\times I)} x dy = c$ (using that $\mathrm{d}(x\mathrm{d}y) = \mathrm{d}x \wedge \mathrm{d}y$)

So we have $c^2 = c$, and so $c = 0$ or $c=1$. Of course $c = 0$ can be eliminated as a degenerate case. From this we can conclude that $\int f \mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^n = \int f dx^1\cdots dx^n$ by approximating $f$ by polynomials, say (since the integral of a monomial can be integrated by the external product rule, and so the integral of a polynomial can be calculated by linearity from there. Some continuity principle is needed.), and using pullback and linearity principles we can derive the value of the integral in general, say by the partition of unity argument.

Edit Here's a way to derive the condition $\int f\mathrm{d}x = c\int f dx$ from a weaker assumption. Assume that $\int_{\Omega} f\mathrm{d}x = \int_{\Omega} fg dx$ for some function $g$, where $\Omega \subseteq \mathrm{R}^n$ is the closure of a bounded open subdomain of $\mathbb{R}^n$, although we only need subdomains of $I=[0,1]$, the unit interval. (Plausibly this can be concluded from some general continuity and naturality conditions on the integration operator.) Consider the integral $\int_{[0,t]} \mathrm{d}x$:

  • By the assumption, $\int_{[0,t]} \mathrm{d}x = \int_{[0,t]} g(x) dx = G(t)$ where $G$ is the antiderivative of $g$ (with $G(0) = 0$), using the ordinary fundamental theorem of calculus.

  • Let $\phi: I \to [0,t]$ be the multiply-by-$t$ map. By pullback, $\int_{[0,t]} \mathrm{d}x = \int_I \phi^*(\mathrm{d}x) = \int_I t\mathrm{d}x = t G(1)$.

So $G(t) = tc$ where $c = G(1)$. By differentiating, $g(t) = c$, and we have $\int f \mathrm{d}x = c \int f dx$ as desired.

$\endgroup$
9
  • $\begingroup$ Stokes' theorem is usually proved by appealing to the coordinate definition of integration (c.f. Warner's book). $\endgroup$
    – Neal
    Nov 17, 2014 at 16:29
  • $\begingroup$ Of course the usual proof of Stokes' theorem has to use the usual definition of integration. I'm looking to reverse the usual logic and derive the usual definition from other plausible axioms. But I agree that it would be nice to do this with axioms that are not so heavy-duty. Another point is that one would like to derive the integration formula not just for differential forms, but for more general densities as defined here. These do not satisfy Stokes' theorem. $\endgroup$
    – tcamps
    Nov 18, 2014 at 0:07
  • $\begingroup$ In the first part of the argument, I used a 2-dimensional instance of Stokes' Theorem above, but actually it's much simpler to use a 1-dimensional version, i.e. the fundamental theorem of calculus for differential forms: $\int_I \mathrm{d}f = \int_{\partial I} f$. It's a natural normalization condition to ask that $\int_{\Omega} f = \sum_{x \in \Omega} f(x)$ when $\Omega$ is zero-dimensional and positively oriented. If $\int_I \mathrm{d}x = c \int_I dx = c$, then we have $c = \int_{\partial I} x = x(1)-x(0) = 1$. $\endgroup$
    – tcamps
    Nov 20, 2014 at 1:00
  • $\begingroup$ In fact, the normalization condition on 0-forms follows from the external product hypothesis, since $1_{\mathrm{pt}} \boxtimes \omega = \omega$ (where $1_{\mathrm{pt}}$ is the function on a point which takes the value $1$). We should also hypothesize that switching orientations should switch the sign of the integral. $\endgroup$
    – tcamps
    Nov 21, 2014 at 17:14
  • $\begingroup$ I can't understand "Suppose we've gotten as far as agreeing that for 1-dimensional integrals, $\int_{\Omega} f \mathrm{d}x = c\int_{\Omega} f dx$". Doesn't this already imply $c=0$ or $c=1$, without appealing to exterior product? Or are the two integral operators meant to be distinct? The left-hand is a generic operator, the right-hand is standard Lebesgue or something? $\endgroup$
    – ziggurism
    Jun 27, 2018 at 17:50
4
$\begingroup$

It's probably worth trying to write down a precise statement. I think many variations should be possible.

Notation:

  • Let $Man$ be the category of compact oriented smooth manifolds $(M,o_M)$ with boundary, with orientation-preserving smooth embeddings as morphisms. Here $o_M$ is an orientation form, well-defined up to multiplication by an everywhere-positive function. The category $Man$ is symmetric monoidal under cartesian product $\times$. The unit is the positively-oriented one point manifold $\bullet$. If $M_1,M_2 \to M$ are maps which are jointly surjective and $M_1 \cap M_2$ is lower-dimensional, then say that $M = M_1 \cup M_2$ is a partition.

  • Let $\iota Man \subset Man$ be the symmetric monoidal subcategory with the same objects whose morphisms are the orientation-preserving diffeomorphisms.

  • Let $Vect$ be category of real vector spaces equipped with a closed cone of "nonnegative" vectors. A morphism is a linear map preserving nonnegativity. The category $Vect$ is symmetric monoidal under tensor product $\otimes$. The unit is $\mathbb R$ with the usual positive cone.

  • Let $\Omega^{topdim}: Man^{op} \to Vect$ be defined by $(M^n,o_M) \mapsto \Omega^n(M)$. A form is in the positive cone if it is of the form $fo_M$ where $f$ is an everywhere-positive smooth function. The functor $\Omega^{topdim}$ is lax symmetric-monoidal via the maps $\Omega^{topdim}(\bullet) \cong \mathbb R$ and $\boxtimes: \Omega^{topdim}(M) \otimes \Omega^{topdim}(N) \to \Omega^{topdim}(M \times N)$.

  • Let $\mathbb R$ also denote the functor $Man^{op} \to Vect$ which is constant at $\mathbb R$. This functor is lax symmetric monoidal in a natural way.

Definition:

  1. An integral operator is a monoidal natural transformation $\int: \Omega^{topdim}|_{\iota Man} \to \mathbb R$ such that $\int_M \omega = \int_{M_1} i_1^\ast \omega + \int_{M_2} i_2^\ast \omega$ for every partition $M = M_1 \cup M_2$ (where $i_j: M_j \to M$ is the inclusion).

  2. An integral operator $\int$ is said to satisfy Stokes' theorem if $\int_M \mathrm{d} \omega = \int_{\partial M} \omega$ for every $M \in Man$ and $\omega \in \Omega^{topdim}(\partial M)$.

Claims:

  1. The integral operators are in bijection with positive real numbers $c$. The bijection sends $c$ the map $c \int: (M^n,o_M) \mapsto (\omega \mapsto c^n \int_M \omega)$ where $\int_M \omega$ is the usual integration of differential forms.

  2. The usual integration operator (i.e. the case $c=1$ from Claim 1) is the unique integration operator satisfying Stokes' theorem.

Remark: Note that Stoke's Theorem is used only in a very weak sense as a normalization condition. Perhaps some other principle could be substituted here.

Proof Sketch:

  • Existence: use the standard theory of integration of differential forms and verify that these properties hold in that case.

  • The partition property implies a locality property: if a form vanishes in a region, then the integral over that region is zero.

  • Therefore, the value of an integral operator may be computed via a partition of unity, reducing to the case where $M = D^n$ is a disk.

  • It also follows from locality that a integrating a form is given by writing it in coordinates and integrating against some absolutely continuous measure.

  • Deduce as above that this absolutely continuous measure is a scalar multiple of Lebesgue measure.

  • Conclude from monoidalness that the scalar for dimension $n$ is given by $c^n$ where $c$ is the scalar for dimension 1.

$\endgroup$
6
  • $\begingroup$ Compare with the exterior derivative. We can give a coordinate free axiomatization, it is the unique linear operator such that $d^2=0$ and satisfies the graded Leibniz law and $df(v) = v(f)$ (I think this is a normalization condition?). Then we need to prove existence (and uniqueness?) theorem by giving an explicit construction of the operator. We can give a coordinate construction as $d(\omega_I\,dx^I) = \sum(-1)^i\frac{\partial \omega_I}{\partial x^i}dx^i\wedge dx^I$. Then show that this formula transforms correctly under coordinate transformations so defines a global coordinate free object. $\endgroup$
    – ziggurism
    Jun 27, 2018 at 19:48
  • $\begingroup$ Or alternatively give it as $d\omega(u,v) = v(\omega(u))-v(\omega(u))+\omega([u,v])$, which is manifestly coordinate free. $\endgroup$
    – ziggurism
    Jun 27, 2018 at 19:48
  • $\begingroup$ Is it too much to ask for the same thing for the integral operator? A coordinate free axiomatic characterization and a coordinate free construction of the operator? $\endgroup$
    – ziggurism
    Jun 27, 2018 at 19:48
  • $\begingroup$ @ziggurism That's an interesting suggestion, it would be really cool to get a coordinate-free construction! In the case of $\mathrm{d}$, the proof I'm familiar with that the formula you give defines a 2-form (or that $[u,v]$ is a vector field for that matter) involves writing things out in coordinates. So it involves embedding 2-forms into a larger space (the space of all $\mathbb R$-bilinear functions on vector fields), writing down a coordinate-free formula that manifestly makes sense in the larger setting, and then verifying that the result lies in the original space. Maybe imitate this? $\endgroup$
    – tcamps
    Jun 27, 2018 at 21:36
  • $\begingroup$ Hm that's a good point. Maybe even this apparently coordinate free construction has a hidden reliance on a coordinate computation to be well-defined. I don't think we need this for [u,v] though, right? For any commutative algebra, commutator of derivations is a derivation by a purely formal check. Or are you thinking of defining the bracket via the Lie derivative? As for showing that $d\omega$ is $C^\infty$-linear, that also seems to be a purely formal check, with no recourse to local coordinates. Unless I'm overlooking something? $\endgroup$
    – ziggurism
    Jun 28, 2018 at 2:28
2
$\begingroup$

Yes, this is possible. We can think of it on an interval $[a,b]:$ is it possible to define the integral of $f(x)\,dx$ without using coordinates?

Well, a Riemann sum is this: $$\sum_{i=1}^n f(a+i(b-a)/n)\frac{b-a}{n}$$ right? This seems to use coordinates. Well, this is a right hand Riemann sum, but there is also a left hand Riemann sum. Of course, you can actually choose any point between the left and right endpoints to evaluate $f\,.$ But you can also take the average of the function at the left and right endpoints, etc. They all converge to the integral. Call the summands you can use in the Riemann sum "approximations".

Ok so if we could find a subspace of these approximation without using coordinates then we will have solved the problem. Here's how to do it: Consider the set of maps $$g:[a,b]\times[a,b]\to \mathbb{R}$$ such that $g(x,x)=0$ for all $x\,,$ and $$dg(x,y)\vert_{y=x}=f\,dx,$$ where $d$ is differentiating in the second component — in coordinates this is just saying that $\frac{\partial}{\partial y}g(x,y)\vert_{y=x}=f(x)$ (this implies $g(x,y)\approx f(x)(y-x)$ when $y-x\approx 0$). We've specified this set of maps without using coordinates, and these are our subspace of approximations. You can use any one of these functions to do the Riemann sum, ie. $$\lim_{n\to\infty}\sum_{i=1}^n g(x_{i-1},x_{i})=\int_a^bf\,dx\,,$$ where $\{x_i\}_{i=0}^n$ are the points in your partition. You might ask: how do we know this set is nonempty? Well you can check that $g(x,y)=f(y)(y-x)\,,g(x,y)=f(x)(y-x)$ are both in the set.

This whole story generalizes to manifolds, so we can define integrals of differential forms coordinate-free. The construction is naturally described using Lie groupoids.

$\endgroup$
5
  • $\begingroup$ I'm a little unclear on how the partitioning of the interval is done without coordinates, even in dimension $1$. If $S$ is a $1$-manifold, what defines an admissible partition $\{x_i\}_{i=0}^n$? $\endgroup$
    – M W
    Aug 26, 2023 at 9:08
  • $\begingroup$ In one dimension a partition is just a choice of points on your manifold, so there is no need to choose cooordinates. You would take a limit over all partitions (they form a directed set, ordered by subdivision). $\endgroup$
    – JLA
    Aug 26, 2023 at 18:49
  • $\begingroup$ That makes some sense. But how do we know what order the points are in? We could in principle get them backwards, right? $\endgroup$
    – M W
    Aug 26, 2023 at 18:55
  • $\begingroup$ @MW Yes that's right, you have to pick an orientation to integrate on manifolds, but this doesn't require coordinates. If your interval is $[a,b]$ then just saying $b>a$ defines an orientation. $\endgroup$
    – JLA
    Aug 26, 2023 at 19:01
  • 1
    $\begingroup$ I think I see now, was sort of worried about the circle, but I guess even there it’s ok since orientation + discreteness gives well defined notion of what comes directly after what. Anyway it’s a very nice construction. $\endgroup$
    – M W
    Aug 26, 2023 at 19:18
2
$\begingroup$

[Update. Having thought about this even more I think it makes sense to expand on my original remarks.]

I've been thinking about this question for a bit, and I've realized a few things.

Basis-free vs chart-free

First, we have clarify what we mean by "coordinate free". A choice of coordinates entails a choice of a local chart, i.e., parametrization by a vector space, and a choice of a (ordered) basis for that vector space.

So we can ask for basis-free approaches, or more ambitiously, chart-free approaches.

Implicit vs explicit

We have to temper our expectations a little bit, as the definition of a manifold defines it in terms of charts. Moreover, even if we have no privileged basis, the definition of an $n$-dimensional vector space is that it has a basis of cardinality $n$. Thus we are going to be implicitly using facts about charts and bases no matter what we do.

(A clever topologist might propose circumventing the second point by defining charts to be parametrized by arbitrary topological vector spaces of topological dimension $n$, but they would struggle mightily to establish much useful about these spaces without accidentally proving the existence of a basis along the way.)

However, what we can ask for are approaches which avoid explicitly invoking charts and/or bases.

There are two approaches that I think work well to this end - a more concrete one that is basis-free but invokes charts, and a more abstract one that eschews either.

Concrete basis-free approach using charts

We will avoid explicit invocation of a basis for the integral, but as per previous remarks, we must use some fact about finite dimensionality somewhere under the hood, and we do so by establishing, via well known linear algebra facts, that the signed volume of an $n$-vector is independent of its representation.

Let $V$ be an $n$-dimensional oriented vector space, $\mu$ a translation invariant measure on $V$, and let $v_1\wedge\cdots\wedge v_n = w_1\wedge\cdots\wedge w_n \in \Lambda^n(V)$, the $n$-th exterior product. Then the signed $\mu$-volumes of the parallelepipeds $(v_1,\cdots,v_n)$ and $(w_1,\cdots,w_n)$ are equal.

This immediately leads to the following definition:

Let $V$, $\mu$ be as above. The signed $\mu$-volume $\operatorname{vol}_\mu(\lambda)$ of an $n$-vector $\lambda\in\Lambda^n(V)$ is given by the signed $\mu$-volume of any (hence every) parallelepiped $(v_1,\cdots,v_n)$ such that $\lambda=v_1\wedge\cdots\wedge v_n$.

Having adopted this notion of signed volume, we may develop the theory of integration without ever touching $\mathbb R^n$, nor any basis, again. For now the integral is uniquely determined by linearity on differential forms and the following two requirements.

  1. For every orientation preserving diffeomorphism $f\colon M\to N$ and $\omega\in \Omega^n(N)$ we have: $$\int_M f^*\omega = \int_{N}\omega\text{.}$$
  2. For every $n$-dimensional oriented vector space $V$, $U\subseteq V$ open, translation invariant measure $\mu$ on $V$, nonzero $n$-vector $\lambda\in\Lambda^n(V)$, and $\omega\in \Omega^n(U)$ we have: $$\int_U \omega = \frac{\int_{U} \langle \omega(p),\lambda \rangle d\mu(p)}{\operatorname{vol}_\mu(\lambda)}\text{.}$$

We do still need to invoke partitions of unity at some point in our development of the theory of forms, but these partitions of unity can be supported on charts parametrized by arbitrary vector spaces, with no privileged bases.

Abstract chart-free approach via Stokes

The preceding approach was still concrete in that it told us, albeit in a basis-free manner, directly how to compute the integral over a chart, from which a partition of unity then shows us how to compute it explicitly in the general case.

We now consider a more abstract approach, via Stoke's theorem, and the natural $\sigma$-algebra of measurable sets that our manifold is equipped with. Although no measure is defined, it is still equipped with a natural sub-algebra of null-sets. Under the hood, these algebras come from local charts, but from them we get a number of interesting properties that can then be discussed without appealing to charts directly.

Of particular use is the fact that, for any locally finite measure $\mu$ on $M$, if $A\subset M$ is measurable, then $\mu(A)$ approximated by $\mu(\overline{U})$, where $U$ is open with smooth boundary. This is inherited from the charts, where it follows from a Vitali covering theorem, for example.

That fact implies that if an integration operator is to linearly and continuously map compactly supported $n$-forms to measures, then it is completely determined by integration over open sets with smooth boundary.

Moreover, from partitions of unity and the local theory of forms, a compactly supported $n$-form can be decomposed into terms of the form $\phi d\nu$, where $\nu$ is an $(n-1)$-form.

The upshot is that an integration operator is completely determined by its action on forms $\phi d\nu$, and with the help of step functions approximating $\phi$, that action is determined via integrals of the form $$\int_U d\nu\text{,}$$ where $U\subseteq M$ is an open subset of smooth boundary.

But this means we can "define" integration inductively, saying in dimension $0$ that integration of a function over a singleton is given by the evaluation map, and in higher dimensions we define $$\int_U d\nu:=\int_{\partial U}\nu\text{,}$$ for every open $U\subseteq M$ with smooth boundary.

Per our remarks, this uniquely determines the integral operator. Now, the question of existence of such an operator is a thornier issue, as the above "definition" might not be a priori well-defined.

Still, this gives a chart-free characterization of the integral as the only integration operator, if one exists, that both satisfies Stokes, and acts as evaluation in dimension $0$.

One could use this "Stokes" approach to characterize the integral, and then establish existence through the previously mentioned basis-free approach using charts.

$\endgroup$
1
  • 1
    $\begingroup$ I've even taught a course defining a manifold where a chart is a map from an open set of affine $n$-space. I also avoided using partitions of unity by assuming that a manifold can be decomposed into smooth images of parallelotopes (I call this a rectangularization, analogous to a triangulization). $\endgroup$
    – Deane
    Aug 26, 2023 at 20:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .