An abstract characterization of line integrals

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 $$I(omega,c)inmathbb{R}$$ to every pair of a smooth $$1$$-form $$omega$$ on a smooth manifold $$M$$ with boundary and smooth path $$ccolon[0,1]to M$$ such that:

• (adjunction) if $$fcolon Mto N$$ is a smooth map of smooth manifolds with boundary, $$omega$$ is a smooth $$1$$-form on $$N$$, and $$ccolon[0,1]to M$$ is a smooth path in $$M$$, then$$I(f^*omega,c)=I(omega,fcirc c)$$
• (normalisation) if $$M=[0,1]$$, $$mathbf 1colon[0,1]rightarrow[0,1]$$ 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 $$ccolon[0,1]to M$$ 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 $$int_0^1left(int_0^xg(y)mathrm dyright)h(x)mathrm dx$$, and by demanding the same adjunction relation you get a way to define a double line integral $$I(omega_1omega_2,c)$$ for all pairs of smooth $$1$$-forms $$omega_1,omega_2$$ on a manifold $$M$$ with boundary, and all smooth paths $$ccolon[0,1]to M$$. For more details, see the works of Kuo-Tsai Chen, who was the first to develop this theory systematically

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 $$ccolon[0,1]to M$$ such that $$dot c(t)neq0$$ for all $$tin[0,1]$$. I think this restriction could probably be removed with enough effort.

Theorem:

For any manifold $$M$$, line integration is the unique function $$IcolonOmega^1(M)timesmathcal C_Mtomathbb R$$ 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 $$omegainOmega^1(M)$$. Here $$c_1c_2$$ denotes the composite path, defined by$$c_1c_2(t)=begin{cases}c_1(2t)&0leq tleq1/2\c_2(2t-1)&1/2leq tleq1.end{cases}$$
• (additivity in the $$1$$-form) We have $$I(omega_1+omega_2,c)=I(omega_1,c)+I(omega_2,c)$$ for all $$omega_1,omega_2inOmega^1(M)$$ and all $$cinmathcal C_M$$.
• (locality) If $$omegainOmega^1(M)$$ satisfies $$omega_{c(t)}(dot c(t))=0$$ for all $$tin[0,1]$$, then $$I(omega,c)=0$$.
• (exact forms) If $$fcolon Mtomathbb R$$ 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 $$0leq k<2^N$$, 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 $$omegainOmega^1(M)$$. Then there exists a smooth function $$fcolon Mtomathbb R$$ such that $$omega_{c(t)}(dot c(t))=mathrm df_{c(t)}(dot c(t))$$ for all $$0.

Proof: The pullback $$c^*omega$$ is a smooth $$1$$-form on $$[0,1]$$, hence is $$mathrm df_0$$ for some smooth $$f_0colon[0,1]tomathbb R$$. We want to show that $$f_0$$ extends to a smooth map $$fcolon Mtomathbb R$$ (i.e. $$f_0=fcirc c$$).

To do this, we first extend $$c$$ to a smooth map $$ccolon(-epsilon,1+epsilon)to M$$ 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 $$ccolon(-epsilon,1+epsilon)hookrightarrow M$$ is an embedding. The tubular neighbourhood theorem then implies that the embedding $$c$$ extends to an embedding $$tilde ccolon(-epsilon,1+epsilon)times (-1,1)^{d-1}hookrightarrow M$$, 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 $$f_0colon(-epsilon,1+epsilon)tomathbb R$$, and then extend this again to a smooth function $$f_0colon(-epsilon,1+epsilon)times(-1,1)^{d-1}tomathbb R$$. Multiplying by an appropriate bump function if necessary, we may assume that $$f_0$$ vanishes outside $$(-frac12epsilon,1+frac12epsilon)times(-frac12,frac12)^{d-1}$$.

We’ve now constructed an extension $$f=f_0circtilde c^{-1}$$ 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 $$Omega^1(M)timesmathcal C_Mtomathbb R$$ which satisfy the given conditions. We need to show that $$I(omega,c)=I'(omega,c)$$ for all $$omegainOmega^1(M)$$ and all immersed paths $$c$$.

To do this, suppose first that $$c$$ is embedded. By Lemma 2 we can choose a smooth map $$fcolon Mtomathbb R$$ such that $$omega_{c(t)}(dot c(t))=mathrm df_{c(t)}(dot c(t))$$ for all $$cin[0,1]$$. 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 $$0leq k<2^N$$. 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 $$I(omega,c)=0$$.