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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?

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    $\begingroup$ 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 '14 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 '14 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 '14 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 '14 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 '14 at 17:24
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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.

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  • $\begingroup$ Stokes' theorem is usually proved by appealing to the coordinate definition of integration (c.f. Warner's book). $\endgroup$
    – Neal
    Nov 17 '14 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 '14 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 '14 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 '14 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 '18 at 17:50
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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.

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  • $\begingroup$ 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 '18 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 '18 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 '18 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 '18 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 '18 at 2:28

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