Product integral for differential forms

Question

Consider the following expression $$e^{\int_{\mathcal{M}}\Omega}$$ where $\Omega$ is the top form on the $m$-dimensional manifold $\mathcal{M}.~~$ How appropriate is it to represent the above expression as a product integral via $$e^{\int_{\mathcal{M}}\Omega} = \prod_{p\in \mathcal{M}} e^{\Omega}$$ where $p$ is a point in $\mathcal{M}$ .

Answer 1

Let us consider a Volterra-type of product integral (original proposal) $$\prod_{a}^b(1+f(x)dx) = \exp\left({\int_{a}^b f(x)dx}\right)$$ Now, we can interpret $f(x)dx$ as a top-form on a 1-D manifold and therefore, with this understanding, we can see that $$e^{f(x)dx} = 1 + f(x)dx$$ Using this equality, we can now write $$\prod_{a}^b(1+f(x)dx) = \prod_{a}^b \exp(f(x)dx)= \exp\left({\int_{a}^b f(x)dx}\right)$$ As one can see. it is now very intuitive to justify the second equality and in turn also justify Volterra's original proposal. Now, we may propose a generalization of the above to higher dimensions (specifically $m$-dimensions) and write $$e^{\int_{\mathcal{M}}\Omega} = \prod_{p \in \mathcal{M}}e^{\Omega(p)}$$ for a top-form $\Omega$ on the $m$-dimensional manifold $\mathcal{M}$ where $$e^{\Omega(p)} = 1+\Omega(p)$$ which is exactly what I proposed in my question. Hence, my proposal may be interpreted as a higher dimensional version of a Volterra-type product integral. I chose to work on the complete manifold $\mathcal{M}$ but I think it is very easy to generalize to an integration range over an open set $U \in \mathcal{M}$, just like in Volterra's original proposal.