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I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from Lebesgue measure theory and is more like a generalization of Riemann integration.

Ok so here's the problem. I have always viewed Lebesgue measure theory as 'solving the issues with Riemann integration'. For example, a big problem with Riemann integration is that the space of Riemann integral functions is not complete. The fact that $L^p$ spaces in the Lebesgue theory are complete seems like a huge improvement on the Riemann situation, and is vital for so many concepts in functional analysis, PDEs, operator theory, and numerical analysis.

So if we then consider differential geometry and integration via differential forms, unless I am misunderstanding something, we lose all the benefits of Lebesgue theory?

It seems like if do lose all those benefits we are in a very bad situation. For example, how are we supposed to rigorously define solution spaces for PDEs if we can't use $L^p$ spaces and thus can't use Sobolev spaces? How can we obtain acceptable convergence of some sequence that may arise during our work if we are operating in this generalized Riemann setting where we lack completeness?

In summary, if differential forms are a generalization of Riemann integration how are we supposed to perform analysis when we no longer have the power and utility of Lebesgue measure theory?

sonicboom
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    What makes you think that differential forms are incompatible with Lebesgue measure and integration theory? – md2perpe Jul 20 '18 at 07:53

1 Answers1

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People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian” in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian” in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.

Suppose that $M$ is a smooth $n$-manifold. By definition, it's locally diffeomorphic to $\mathbb{R}^n$, so that you can define a set $S \subset M$ to be measurable if and only if $x(S \cap U) \subset \mathbb{R}^n$ is Lebesgue measurable for every local coordinate chart $x: U \to x(U) \subset \mathbb{R}^n$. This gives you a $\sigma$-algebra of Lebesgue measurable sets on $M$ that correctly completes the Borel $\sigma$-algebra generated by the open sets on $M$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.

Now, suppose that $M$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $g$—again, the “Riemannian” here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $x : U \to x(U) \subset \mathbb{R}^n$, you can define a measure $\lambda_{g,x}$ on $U$ by setting $$ \lambda_{g,x}(S \cap U) := \int_{x(S \cap U)} \sqrt{\det\left(g\left(\tfrac{\partial}{\partial x^i},\tfrac{\partial}{\partial x^j}\right)\right)} \,d\lambda $$ for any Lebesgue measurable $S \subset M$, where $\lambda$ denotes Lebesgue measure on $\mathbb{R}^n$. By paracompactness of the manifold $M$, one can cover $M$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $\lambda_g$, the Riemannian measure [!] on $M$ with respect to $g$, which is a complete $\sigma$-finite measure on the $\sigma$-algebra of Lebesgue measurable sets in $M$.

Let me now describe the basic properties of $\lambda_g$.

  1. The measure $\lambda_g$ is compatible with calculations in local coordinates, in the precise sense that $\lambda_g(S \cap U) = \lambda_{g,x}(S \cap U)$ for any Lebesgue measurable $S$ and any local coordinate chart $x : U \to x(U) \subset \mathbb{R}^n$.

  2. If $g^\prime$ is any another Riemannian metric, then the Riemannian measures $\lambda_g$ and $\lambda_{g^\prime}$ will be mutually absolutely continuous $\sigma$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $g$ and $g^\prime$.

  3. Suppose that $M$ is orientable, and let $\mathrm{vol}_g \in \Omega^n(M)$ be the Riemannian volume form defined by $g$. Then for any Riemann integrable $f$ on $M$, $$ \int_M f \, \mathrm{vol}_g = \int_M f \,d\lambda_g, $$ so that $\lambda_g$ really is the (completed) Radon measure on $M$ corresponding to the positive functional $C_c(M) \ni f \mapsto \int_M f \, \mathrm{vol}_g$ via the Riesz representation theorem. In other words, integration with respect to $\lambda_g$ really is the “Lebesgue-ification” of integration against the top-degree form $\mathrm{vol}_g$.

Once you've constructed the Riemannian measure on your Riemannian manifold $(M,g)$, the sky is now the limit—you can construct $L^p$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I'm personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman's proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.

P.S. People tend to take the extension of Lebesgue theory from $\mathbb{R}^n$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné's Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn't require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $1$-densities, and the construction of the Riemannian measure $\lambda_g$ is really the construction of the canonical $1$-density $\lvert \mathrm{vol}_g \rvert$ associated to $g$.


ADDENDUM

One can define a measurable $k$-form on $M$ to be a map $\omega : M \to \wedge^k T^\ast M$, such that the following hold.

  1. For every $m \in M$, $\omega(m) \in \wedge^k T^\ast M_m$ (i.e., $\omega$ is a set-theoretic section of $\wedge^k T^\ast M$).
  2. For every local coordinate chart $x : U \to x(U) \subset \mathbb{R}^n$, the pullback $(x^{-1})^\ast \omega : x(U) \to \wedge^k \mathbb{R}^n$ defined by $$ (x^{-1})^\ast\omega := \sum_{i_1 < \cdots < i_k} \omega\left(\tfrac{\partial}{\partial x^{i_1}},\dotsc,\tfrac{\partial}{\partial x^{i_k}}\right) dx^{i_1} \wedge \cdots \wedge dx^{i_k} $$ (with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $\omega(X_1,\dotsc,X_k) : M \to \mathbb{R}$ be measurable (in the above sense) for any smooth vector fields $X_1,\dots,X_k \in \mathfrak{X}(M)$.

Now, suppose that $N$ is an oriented $k$-dimensional submanifold of $M$ (compact and without boundary, for simplicity), and let $x : U \to x(U) \subset \mathbb{R}^n$ be a local coordinate chart of $M$, such that $x(N \cap U) = V_{x,N} \times \{0\}$ for some open $V_{x,N} \subset \mathbb{R}^k$, and such that restriction of $x$ to a diffeomorphism $N \cap U \to V_{x,N}$ is orientation-preserving. Then we can define $$ \int_{N \cap U} \omega := \int_{V_{x,N}} \omega\left(\tfrac{\partial}{\partial x^{1}},\dotsc,\tfrac{\partial}{\partial x^{k}}\right) d\lambda_{\mathbb{R}^k} $$ whenever the Lebesgue integral on the right-hand side exists (with $\lambda_{\mathbb{R}^k}$ the Lebesgue measure on $\mathbb{R}^k$). We can then define $\omega$ to be integrable on $N$ whenever it's integrable in this way on $N \cap U$ for any suitable local coordinate chart $x : U \to \mathbb{R}^n$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $\int_N \omega$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way.

Physor
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Branimir Ćaćić
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  • There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $\int_M f \, \mathrm{vol}_g = \int_M f \,d\lambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...) – Bob Jul 20 '18 at 09:13
  • The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $\int_M f \mathrm{vol}_g \geq 0$ for $f \geq 0$, and in particular, so that $\mathrm{Vol}_g(M) := \int_M 1 \,d\lambda_g = \int_M 1 \mathrm{vol}_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $\lambda_g$). – Branimir Ćaćić Jul 20 '18 at 09:21
  • In fact, if you like, the Riemannian volume form is *defined* so that $\int_M f \mathrm{vol}_g = \int_M f \,d\lambda_g$ in exactly the same way that the standard orientation on $\mathbb{R}^n$ is defined so that $\int_U f \,dx^1 \wedge \cdots \wedge dx^n = \int_U f \,d\lambda$. – Branimir Ćaćić Jul 20 '18 at 09:24
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    I see, so the point is that you pick a canonical orientation... thanks :) – Bob Jul 20 '18 at 09:25
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    The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals). –  Jul 20 '18 at 13:18
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    Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms? – sonicboom Jul 20 '18 at 14:19
  • @Hurkyl I've quickly sketched integrating a measurable $k$-form over a compact oriented $k$-dimensional submanifold. – Branimir Ćaćić Jul 20 '18 at 14:28
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    @sonicboom Anything billing itself as an introduction to "global analysis" will fit the bill, indeed, any textbook that needs to introduce Sobolev spaces on manifolds for any reason (e.g., to study the spectral theory of some Laplace-type operator or to talk about index theory of geometric elliptic operators, particularly the Dirac operator). As an example, check out Section 10.2 of [these notes by Nicolaescu.](https://www3.nd.edu/~lnicolae/Lectures.pdf). – Branimir Ćaćić Jul 20 '18 at 14:31
  • Be aware, though, that the moment you're happy to work with oriented manifolds, most people just define Lebesgue integration on a Riemannian manifold *as* integration with respect to the Radon measure corresponding, via the Riesz representation theorem, to integration against the Riemannian volume form. – Branimir Ćaćić Jul 20 '18 at 14:32
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    If your manifold is nonorientable (or even just nonoriented, because while it may be orientable you may not have picked one of the two orientations), then you can work with pseudoforms instead of ordinary forms. This is actually what you always do; an orientation allows you to convert between forms and pseudoforms, but even without it, Riesz lets you convert between top-rank pseudoforms and absolutely continuous Radon measures. – Toby Bartels Dec 09 '19 at 15:45
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    It's a mistake to say that this about a manifold being locally *homeomorphic* to $\mathbb{R}^n$. This suggests that you can use any topological manifold and define measurable sets using any continuous coordinate chart. You need (at least) a *differentiable* manifold (and use only differentiable coordinate charts). The reason is that homeomorphisms don't always respect measurability (one example is the sum of the Cantor function and the identity function). But *diffeomorphisms* (whether you mean smooth maps with smooth inverses or merely differentiable maps with differentiable inverses) do. – Toby Bartels Dec 09 '19 at 16:05
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    (Actually, my previous comment should say *continuously* differentiable, that is $C^1$. Although I'm not sure; I can't think of a differentiable function with a differentiable inverse that is not absolutely continuous, so maybe differentiability is enough. Regardless, it would be unusual; $C^1$ is the usual notion of differentiability used with manifolds. And if we want to allow unusual notions for the sake of finding the weakest condition that allows us to define measurability, then we'd use *absolutely continuous* manifolds: only require the transition functions to be absolutely continuous.) – Toby Bartels Dec 09 '19 at 16:10
  • @TobyBartels, thank you for your comments and your correction—I totally sleepwalked into that. – Branimir Ćaćić Dec 09 '19 at 19:55
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    Great answer indeed! To address @sonicboom's request, I suggest also §10 of Bourbaki's "Variétés differentielles et analytiques", which bears probably some relation to Dieudonné's treatment. – Riccardo Pengo Apr 09 '21 at 08:37