# Is integration on manifolds unique?

I asked this question on math.stackexchange.com, but haven't received an answer, so I thought I'd brave the waters here.

Suppose $M$ is a smooth oriented compact connected $m$-dimensional manifold and let $A^m(M)$ denote the set of smooth exterior differential $m$-forms on $M$. We can integrate members of $A^m(M)$, and the mapping $\phi\mapsto\int_M\phi$ has several properties:

• It's not identically $0$.
• It's linear.
• It's symmetric.

By "it's symmetric", I mean that if $f:M\to M$ is an orientation-preserving diffeomorphism, then $\int_M \phi = \int_M f^*\circ \phi$. I think that expresses the idea of symmetry I have in mind. Basically, all points on $M$ are equivalent as far as integration is concerned.

My question is: Do these properties uniquely determine the space $A^m(M)$ and the usual definition of integration? More precisely, is there some smooth compact connected $m$-manifold $M$, a space $S$ of smooth sections of the tensor bundle of $M$, a large family $F$ of diffeomorphisms of $M$, and a non-trivial linear functional $\lambda:S\to\mathbb{R}$ such that $\lambda( \phi) = \lambda( f^*\circ \phi)$ for all $\phi\in S$ and $f\in F$, but is not just integration of exterior $m$-forms, up to constant scale factor?

My motivation for asking this question: I suspect almost everyone who's ever learned about integration of differential forms on manifolds has wondered why those particular definitions were chosen. The answer is probably some variant of, "Because those definitions work." But must it be this way? Are they the only definitions that make integration work?

• This is a comment since we just answer that integration is determined (not the space $A^m(M)$). Fix a nonvanishing $\phi_0 \in A^m(M)$ and define its integral to be whatever you like, we need to show everything else is determined by it. By Moser's "On the volume elements of a manifold" any other volume form is given by $\alpha f^* \circ \phi_0$ for some $\alpha \in \mathbb R$ and a diffeomorphism $f$. So, the integration is determined on volume forms by symmetry and scaling. Then note all $m$ forms can be written as the difference between two volume forms. – Tim Carson Aug 1 '17 at 5:02
• Richard Palais has a result along these lines. Peter Michor will probably show up soon and explain the story. – Ben McKay Aug 1 '17 at 7:43

Here's a somewhat abstract but pretty satisfying way to see that integration of compactly supported densities on a not-necessarily orientable and not-necessarily compact $n$-manifold $M$ is unique.

This point of view is explained in Remark 3.3.10 in Kashiwara & Schapira's Sheaves on Manifolds.

Let $t: M \to 1$ be the terminal map to a point, and recall that the category of sheaves of $\mathbb{R}$-vector spaces on a point is just the familiar category of vector spaces.

Recall also that the pushforward with proper supports along the terminal map $t$ is the compactly supported global sections functor; i.e. $$t_!=\Gamma_c(M; -).$$

The counit $\epsilon: Rt_! t^! \Rightarrow 1$ of the Poincare-Verdier duality adjunction $Rt_! \dashv t^!$ is a generalization of integration.

Since adjoints, like other universal objects, are unique up to unique isomorphism, it follows that integration is unique in this sense.

To see how the counit here generalizes integration, apply it to the real numbers (or complex numbers) and take the $0$th cohomology to recover integration. In a bit more detail, notice that

$$H^0 Rt_! t^! \mathbb{R}=H^0 Rt_! \mathrm{or}_M[n] =H^0 R \Gamma_c(M; \mathrm{or}_M[n]) = H^n_c(M; \mathrm{or}_M).$$

So the counit mentioned above induces the integration map in compactly supported sheaf cohomology: $$\int_M: H^n_c(M; \mathrm{or}_M) \to \mathbb{R}.$$

The densities (or $n$-forms if you pick an orientation for an orientable manifold) appear when you take the usual de Rham resolution of the orientation sheaf $\mathrm{or}_M$, thus using de Rham cohomology to compute sheaf cohomology.

For $p: E \to M$ a topological submersion, we can use Poincare-Verdier duality, the relative orientation sheaf $\mathrm{or}_{E/M}$, and a somewhat similar approach to define fibre integration of "relative" densities.

• Thanks. I haven't studied sheaf theory, but for this argument to satisfactorily answer the question, it has to imply that integration can only be defined in a diffeomorphism-invariant way for orientable manifolds, only for m-forms, and only for orientation-preserving diffeomorphisms. Does this approach pick off all these facets? To me, the focus on $m$-forms and orientability seems to appear out of thin air in the theory of manifolds and I'm looking for a theorem that basically states, "If you want integration, you need $m$-forms and orientability". – Will Nelson Aug 1 '17 at 6:27
• You actually don't need orientability for integration; on non-orientable manifolds you integrate densities, and this is the more fundamental notion. If the manifold is indeed orientable, a choice of orientation is just a choice of trivialization of the orientation sheaf. The differential forms appear when you take the de Rham resolution of the orientation sheaf. Diffeomorphism invariance just follows from homeomorphism invariance of sheaf cohomology. – ಠ_ಠ Aug 1 '17 at 6:35
• @WillNelson: The bundle of densities is the tensor product of the bundle of top-degree forms and the orientation bundle. So diffeomorphisms induce isomorphisms of bundles of densities and integrals of densites are invariant under all diffeomorphisms, not just orientation-preserving ones. 2-forms on a Möbius strip are not densities because the orientation bundle is nontrivial. – Dmitri Pavlov Aug 1 '17 at 21:37
• @WillNelson You can define the density bundle as Dmitri Pavlov, or as the natural bundle associated to the frame bundle for the 1-dimensional representation $\mathrm{GL}(\mathbb{R}^n) \to \mathrm{GL}(\mathbb{R})$ given by $g \mapsto |\det(g)|^{-1}$. Densities are smooth sections of this bundle. Being a natural bundle, diffeomorphisms preserve it. For an elementary introduction to densities, see p. 428 in chapter 16 of John Lee's Smooth Manifolds. – ಠ_ಠ Aug 1 '17 at 22:49
• @WillNelson: There are plenty of books on manifolds that treat integration using densities, for instance, Ramanan's Global Calculus. Remark 3.2.7 in Ramanan identifies the sheaf of densities with the sheaf of infinitely differentiable Borel measures, which is close in spirit to what you want. – Dmitri Pavlov Aug 2 '17 at 8:22