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?
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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?




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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?



An abstract characterization of line integrals
I'll suggest here another possible characterisation, expanding on a suggestion of the OP in [one of the comments](https://mathoverflow.net/questions/365761/an-abstract-characterization-of-line-integrals#comment923647_365779). 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](https://mathoverflow.net/a/365779), 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 $I(\mathrm df,c)=f(c(1))-f(c(0))$.




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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](https://math.stackexchange.com/questions/1059110/every-immersion-is-locally-an-embedding) 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](https://en.wikipedia.org/wiki/Borel%27s_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](https://en.wikipedia.org/wiki/Tubular_neighborhood) 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.




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*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.




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**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 $I(\omega,c)=0$.


