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Let $M$ be a smooth manifold (endowed with a Riemann structure, if useful). If $\omega \in \Omega^1 (M)$ is a smooth $1$-form and $c : [0,1] \to M$ is a smooth curve, one defines the line integral of $\omega$ along $c$ as

$$I(\omega, c) = \int _0 ^1 \omega_{c(t)} (\dot c (t)) \ \mathrm d t \ .$$

This is clear, but this is just a formula that does not give any insight into the innards of the concept.

Is is possible to define the concept of line integral by purely abstract properties?

To give two analogies, the algebraic tensor product is defined by some universal property, and then it is shown that it exists and is essentially unique. Similarly, the Haar measure on locally-compact groups is defined as a regular, positive measure, that is invariant under (left) translations, and then is shown to exist and be essentially unique. Do you know of any similar approach for line integrals?


To clarify: if $\mathcal C$ is the space of smooth curves in $M$, I am attempting to understand line integration as a map $I : \Omega ^1 (M) \times \mathcal C \to \mathbb R$ uniquely determined by some properties: what are these properties? For sure, linearity in the first argument is among them. What else is needed?

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  • $\begingroup$ Are you asking for something like the following: Let $\Omega$ be the set of maps $I: C([0,1],M)$ satisfying an appropriate list of universal properties. Then, for each $I \in \Omega$, there exists a $1$-form $\omega$ such that, for each continuous $C: [0,1] \rightarrow M$, $I(C)$ is the line integral of $\omega$ along $C$. $\endgroup$
    – Deane Yang
    Jul 16, 2020 at 14:34
  • $\begingroup$ I don't think that I want "there exists a $1$-form $\omega$"; I believe that I want "for all $1$-forms $\omega$" instead. Alexander Betts' answer below seems extremely close to what I had in mind, but I'll leave the question open for several more days, to give it a chance to collect as many useful answers as possible. $\endgroup$
    – Alex M.
    Jul 16, 2020 at 15:24
  • $\begingroup$ My point is that I believe you can define the universal properties so that you do not even need to assume that what you are integrating $1$-forms along each curve $C$. The fact that the integrand is a $1$-form is already implied by the universal properties, if properly chosen. I think what I am proposing is even more general than what you are asking for. $\endgroup$
    – Deane Yang
    Jul 16, 2020 at 19:28
  • $\begingroup$ Since I do not fully get your ideas, I have edited my question to hopefully make clearer what I am looking for. $\endgroup$
    – Alex M.
    Jul 17, 2020 at 8:00

2 Answers 2

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I'll suggest here another possible characterisation, expanding on a suggestion of the OP in one of the comments. Again, this is an assertion that certain known properties of line integration characterise it uniquely; this doesn't provide a "new" construction of line integration. Unlike my previous answer, here all the action takes place on the one manifold $M$.

To avoid various technicalities, I'm going to redefine $\mathcal C_M$ to be the set of immersed paths, i.e. smooth paths $c\colon[0,1]\to M$ such that $\dot c(t)\neq0$ for all $t\in[0,1]$. I think this restriction could probably be removed with enough effort.

Theorem:

For any manifold $M$, line integration is the unique function $I\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$ satisfying the following properties:

  • (additivity in the path) Suppose that $c_1$ and $c_2$ are two immersed paths that are composable, i.e. all the derivatives $c_1^{(i)}(1)=c_2^{(i)}(0)$. Then $I(\omega,c_1c_2)=I(\omega,c_1)+I(\omega,c_2)$ for all $\omega\in\Omega^1(M)$. Here $c_1c_2$ denotes the composite path, defined by$$c_1c_2(t)=\begin{cases}c_1(2t)&0\leq t\leq1/2\\c_2(2t-1)&1/2\leq t\leq1.\end{cases}$$
  • (additivity in the $1$-form) We have $I(\omega_1+\omega_2,c)=I(\omega_1,c)+I(\omega_2,c)$ for all $\omega_1,\omega_2\in\Omega^1(M)$ and all $c\in\mathcal C_M$.
  • (locality) If $\omega\in\Omega^1(M)$ satisfies $\omega_{c(t)}(\dot c(t))=0$ for all $t\in[0,1]$, then $I(\omega,c)=0$.
  • (exact forms) If $f\colon M\to\mathbb R$ is smooth, then $I(\mathrm df,c)=f(c(1))-f(c(0))$.

The proof uses two lemmas.

Lemma 1: Let $c$ be an immersed path. Then there is a non-negative integer $N$ such that for all $0\leq k<2^N$, the restriction $c|_{[2^{-N}k,2^{-N}(k+1)]}$ of $c$ to the interval $[2^{-N}k,2^{-N}(k+1)]$ is an embedding.

Proof (outline): This follows from the standard fact that an immersion is locally an embedding (see e.g. this MO question), and that $[0,1]$ is compact.

Lemma 2: Let $c$ be an embedded path in $M$ and $\omega\in\Omega^1(M)$. Then there exists a smooth function $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $0<t<1$.

Proof: The pullback $c^*\omega$ is a smooth $1$-form on $[0,1]$, hence is $\mathrm df_0$ for some smooth $f_0\colon[0,1]\to\mathbb R$. We want to show that $f_0$ extends to a smooth map $f\colon M\to\mathbb R$ (i.e. $f_0=f\circ c$).

To do this, we first extend $c$ to a smooth map $c\colon(-\epsilon,1+\epsilon)\to M$ for some $\epsilon>0$. This is possible by Borel's Lemma, which says that we can choose smooth maps $(-\epsilon,0]\to M$ and $[1,1+\epsilon)\to M$ having the same higher-order derivatives at $0$ and $1$ as $c$, respectively.

Decreasing $\epsilon$ if necessary, we may even assume that $c\colon(-\epsilon,1+\epsilon)\hookrightarrow M$ is an embedding. The tubular neighbourhood theorem then implies that the embedding $c$ extends to an embedding $\tilde c\colon(-\epsilon,1+\epsilon)\times (-1,1)^{d-1}\hookrightarrow M$, where $d=\dim(M)$. In other words, we have $c(t)=\tilde c(t,0,\dots,0)$ for all $t$.

We now extend $f_0$ as follows. By Borel's Lemma again, we may extend $f_0$ to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\to\mathbb R$, and then extend this again to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\times(-1,1)^{d-1}\to\mathbb R$. Multiplying by an appropriate bump function if necessary, we may assume that $f_0$ vanishes outside $(-\frac12\epsilon,1+\frac12\epsilon)\times(-\frac12,\frac12)^{d-1}$.

We've now constructed an extension $f=f_0\circ\tilde c^{-1}$ of $f$ on the open neighbourhood $\mathrm{im}(\tilde c)$ of the image of $c$. Moreover, we've ensured that this extension has compact support (it vanishes outside a compact subspace), so we can extend $f$ to all of $M$ by specifying that it is $0$ outside $\mathrm{im}(\tilde c)$. This yields the desired $f$. This proves Lemma 2.


Proof of Theorem: We show unicity. Let $I$ and $I'$ be two functions $\Omega^1(M)\times\mathcal C_M\to\mathbb R$ which satisfy the given conditions. We need to show that $I(\omega,c)=I'(\omega,c)$ for all $\omega\in\Omega^1(M)$ and all immersed paths $c$.

To do this, suppose first that $c$ is embedded. By Lemma 2 we can choose a smooth map $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $c\in[0,1]$. Using additivity in the $1$-form, locality and the condition about exact forms, we find that $I(\omega,c)=I(\mathrm df,c)=f(1)-f(0)$. Since the exact same argument applies to $I'$, we have $I(\omega,c)=I'(\omega,c)$.

Now let us deal with the general case. By Lemma 1 we can choose a non-negative integer $N$ such that $c|_{[2^{-N}k,2^{-N}(k+1)]}$ is an embedded path for all $0\leq k<2^N$. A repeated application of the additivity property implies that $I(\omega,c)=\sum_{k=0}^{2^N-1}I(\omega,c|_{[2^{-N}k,2^{-N}(k+1)]})$ and similarly for $I'$. Since we already know that $I$ and $I'$ agree on embedded paths, we obtain that $I(\omega,c)=I'(\omega,c)$, as desired. This concludes the proof.


Remark:

If one only cares about the integrals of closed 1-forms, then this whole setup can be significantly simplified: one doesn't need to restrict to immersed paths, and one can replace the locality condition above with the more natural condition:

  • (locality') If $\omega$ vanishes on an open neighbourhood of the image of $c$, then $I(\omega,c)=0$.
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    $\begingroup$ Give me some time, please, to digest it! It seems that you get linearity ($I(a \omega, c) = a I(\omega, c)$) for free, which I find strange. $\endgroup$
    – Alex M.
    Jul 17, 2020 at 12:28
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I don't know if it's exactly what you're looking for, but line integration is the unique way to assign a real number $I(\omega,c)\in\mathbb{R}$ to every pair of a smooth $1$-form $\omega$ on a smooth manifold $M$ with boundary and smooth path $c\colon[0,1]\to M$ such that:

  • (adjunction) if $f\colon M\to N$ is a smooth map of smooth manifolds with boundary, $\omega$ is a smooth $1$-form on $N$, and $c\colon[0,1]\to M$ is a smooth path in $M$, then$$I(f^*\omega,c)=I(\omega,f\circ c)$$
  • (normalisation) if $M=[0,1]$, $\mathbf 1\colon[0,1]\rightarrow[0,1]$ is the identity path, and $\omega=g(x)\mathrm{d}x$ for a smooth function $g$, then $I(\omega,\mathbf 1)=\int_0^1g(x)\mathrm{d}x$, where the integral denotes the usual Riemann line integral.

(To show this characterises line integration uniquely, apply the adjunction formula to the smooth path $c\colon[0,1]\to M$ to show that $I(\omega,c)=I(c^*\omega,\mathbf 1)=\int_0^1c^*\omega$.)


Remark:

This is the approach that one takes when defining iterated integration of a sequence $\omega_1,\dots,\omega_n$ of $1$-forms along a path $c$. For instance, we know how to double-integrate over the interval $[0,1]$: the double-integral of $g(x)\mathrm dx$ and $h(x)\mathrm dx$ is $\int_0^1\left(\int_0^xg(y)\mathrm dy\right)h(x)\mathrm dx$, and by demanding the same adjunction relation you get a way to define a double line integral $I(\omega_1\omega_2,c)$ for all pairs of smooth $1$-forms $\omega_1,\omega_2$ on a manifold $M$ with boundary, and all smooth paths $c\colon[0,1]\to M$. For more details, see the works of Kuo-Tsai Chen, who was the first to develop this theory systematically

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  • $\begingroup$ This seems very close (at least in spirit) to what I had in mind. What I do not understand, though, is the domain of definition of $I$. If $\mathcal M$ is the class of smooth manifolds with boundary, it seems that the domain of $I$ is the disjoint union of $\Omega^1(M) \times \mathcal C _M$, where $M \in \mathcal M$ and $\mathcal C _M$ is the set of smooth curves $: [0,1] \to M$. Weird, though... $\endgroup$
    – Alex M.
    Jul 16, 2020 at 13:52
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    $\begingroup$ You're right that the domain of $I$ is this funny disjoint union, but perhaps this isn't such a strange domain to get. We probably want to think of $I$ really as a collection of functions $I_M\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$, one for each $M\in\mathcal M$, and we're saying that this whole collection is uniquely determined by some compatibility condition with respect to smooth maps, plus a normalisation condition when $M=[0,1]$. $\endgroup$ Jul 16, 2020 at 14:53
  • $\begingroup$ (ctd) These sorts of structures are actually quite common: if we view $I_M$ as a map $\mathcal C_M\to\mathrm{Hom}(\Omega^1(M),\mathbb R)$, then this "compatibility with smooth maps" condition is just saying that these $I_M$ are the components of a natural transformation (in the sense of category theory). And natural transformations crop up all over the place. $\endgroup$ Jul 16, 2020 at 14:53
  • $\begingroup$ I think that I can come up with version better suited to my needs, that does away with the disjoint union: property (1) should hold for smooth maps $f : M \to M$ (yes, not $N$), and property (2) should be replaced by the condition that $I(\mathrm d \varphi, c) = \varphi(c(1)) - \varphi(c(0))$ for every smooth $\varphi : M \to \mathbb R$. This means that we may work with a single, fixed manifold $M$. Also, one may take it now to be boundaryless (which is what I need). (The boundary was needed by the case $M = [0,1]$ which becomes irrelevant in this approach.) $\endgroup$
    – Alex M.
    Jul 16, 2020 at 15:20
  • $\begingroup$ Glad to hear it! Though I would be a little surprised if those two conditions uniquely determine $I$ in general. Your second condition only pins down line integrals of exact $1$-forms; it's unclear how this helps calculate integrals of inexact forms (non-closed forms seem especially difficult). $\endgroup$ Jul 16, 2020 at 15:40

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