Is integration on manifolds unique?

Solution 1:

This is a special case of a result from analysis and measure theory. The following two theorems are quoted from section 2.5 of Federer "Geometric Measure Theory." Theorem 2 is a consequence of Theorem 1. The spaces $L^p(X,\mu)$ are the "Lp spaces" of measure theory.

Theorem 1

Let $X$ be a measurable space and let $L$ be a set of measurable functions $X\to\mathbb{R}$ such that $c\in[0,\infty)$ and $f,g\in L$ implies $f+g\in L$ and $cf\in L$ and $\inf\{f,g\}\in L$ and $\inf\{f,c\}\in L$, and provided $f\leq g$, also $g-f\in L$.

(For example, if $X$ is a smooth compact manifold, then $L$ could be the space of continuous functions.)

If $\lambda:L\to\mathbb{R}$ satisfies

1. $\lambda(f+g)=\lambda(f)+\lambda(g)$ for any $f,g\in L$
2. $\lambda(cf)=c\lambda(f)$ for any $c\in[0,\infty)$ and any $f,g\in L$
3. $f\geq g$ implies $\lambda(f)\geq\lambda(g)$
4. if $f_n$ is pointwise increasing and convergent to $g$, then $\lambda(f_n)$ is increasing and convergent to $\lambda(g)$.

then there is a measure $\mu$ on $X$ such that $\lambda(f)=\int_X f\,d\mu$ for all $f\in L$.

Theorem 2

Let $(X,\mu)$ be a measure space, which is the union of countably many measurable subsets with finite measure. Let $\lambda:L^1(X,\mu)\to\mathbb{R}$ be a linear map such that $$\sup\Big\{|\lambda(f)|:f\in L^1(X,\mu)\text{ s.t. }\int_X |f|\,d\mu=1\Big\}<\infty.$$ Then there exists $g\in L^{\infty}(X,\mu)$ such that $$\lambda(f)=\int_X fg\,d\mu\qquad\forall f\in L^1(X,\mu).$$ If $g'$ is any other such function then $g=g'$ almost everywhere with respect to $\mu$.

Application of Theorem 2. Let $M$ be a smooth compact oriented $n$-manifold, and let $T:\Omega^n_{C^\infty}(M)\to\mathbb{R}$ be a linear map. Let $\omega$ be a smooth nonvanishing $n$-form on $M$. Suppose that $$\underbrace{\sup\Big\{|T(\nu)|:\nu\in\Omega^n_{C^\infty}(M)\text{ s.t. }\int_M \Big|\frac{\nu}{\omega}\Big|\omega\leq 1\Big\}}_{c}<\infty.$$ Define a measure $\mu$ on $M$ by $\mu(A)=\int_A\omega$, and define $\lambda:C^\infty(M)\to\mathbb{R}$ by $\lambda(f)=T(f\omega).$ If $f\in C^\infty(M)$ has $\int_M |f|\omega=1$ then one has $|T(f\omega)|\leq c$. So (see wikipedia "continuous linear extension") $\lambda$ extends to a map $L^1(M,\mu)\to\mathbb{R}$, and $$\sup\Big\{|T(f)|:f\in L^1(M,\mu)\text{ s.t. }\|f\|_{L^1(M,\mu)}\leq 1\Big\}<\infty.$$ According to Theorem 2, there exists $g\in L^\infty(M,\mu)$ such that $\lambda(f)=\int_M fg\omega$ for all $f\in L^1(M,\mu)$. So, for any smooth $n$-form $\nu$, one has $$T(\nu)=T\Big(\frac{\nu}{\omega}\omega\Big)=\int_M \frac{\nu}{\omega}g\omega=\int_M g\nu.$$ By taking $\nu$ to be supported in arbitrarily small regions, it should be an easy exercise to show that diffeomorphism-invariance of $T$ implies that $g$ is constant.

On the extra condition. Note that if $\omega'$ is another smooth nonvanishing $n$-form with $\frac{\omega'}{\omega}$ positive everywhere, then $$\Big|\frac{\nu}{\omega'}\Big|\omega'=\Big|\frac{\nu}{\omega'}\Big|\frac{\omega'}{\omega}\omega=\Big|\frac{\nu}{\omega}\Big|\omega$$ and so the sup-condition on $T$ is actually of a topological character, and not dependent on the selection of $\omega.$ I think it is plausible that the sup-condition follows from the diffeomorphism-invariance of $T$ anyway. The idea of a proof could be to use a triangulation to decompose $M$ into small disjoint regions which are mutually diffeomorphic to each other, and to write $\nu$ as the sum of its restrictions to each region. Then $\nu$, using $\omega$ as a reference, could be well-approximated by a step function with supports given by the small regions, with $T$ applied to such a form being nicely comparable with the integral of the form, using linearity and diffeomorphism-invariance