Does the algebra of bounded variation functions have a “noncommutative geometric” meaning and generalization?

According to Gelfand-Naimark theory, $C^*$-algebras of continuous functions $\mathcal{C}^0(X,\mathbb{C})$ on a compact Hausdorff topological space completely capture its topology. Furthermore, every commutative $C^*$-algebra arises in this way. In noncommutative geometry (NCG), noncommutative $C^*$-algebras generalize the notion of compact (locally compact, if the algebra is possibly non unital) Hausdorff topological space.

Given a measure space $(X,\mu)$, the algebra $L^\infty (X,\mu)$ of essentially bounded functions on $X$ is a von Neumann algebra which completely describes the "measure theory" on $(X,\mu)$. In NCG, noncommutative von Neumann algebras are considered, which somehow generalize measure theory to the NC setting.

I learn from this wikipedia entry that a certain "chain rule" holds for the space $\mathrm{BV}(\Omega)$ of bounded variation functions on an open subset $\Omega\subseteq\mathbb{R}^n$, making it an algebra, and even a Banach algebra.

I would like to know:

1) Which geometric aspect of $\Omega$ -if any- is completely described by $\mathrm{BV}(\Omega)$ ?

2) Which is -if there is any- the "right" NCG generalization of the $\mathrm{BV}(\Omega)$ algebra?

-
I don't know of any work done on $BV(\Omega)$ itself - note that its Gelfand dual is even bigger than the Stone-Cech compactification. Still, the question reminds me of a related construction called the Higson compactification. The Higson compactification of a proper metric space $X$ is the Gelfand dual of the space of continuous functions on $X$ whose variation vanishes at infinity. It plays an important role in the descent principle which relates the Novikov conjecture to considerations in geometric group theory and metric geometry, for instance. –  Paul Siegel Aug 30 '12 at 14:47
If you're interested, you can learn more in Roe's "Lectures on Coarse geometry" as well as in several papers by Dranishnikov and Ferry. –  Paul Siegel Aug 30 '12 at 14:49
This probably doesn't help with your question, but something similar was one of the things I had in mind when I asked this question mathoverflow.net/questions/90435/… –  Yemon Choi Aug 30 '12 at 22:16
In one direction quantum probabilists do use a noncommutative quadratic variation. But I don't think this is the right kind of noncommutative (in Gelfand-Naimark senseof your question); they consider operator-valued functions, but the domain of the function is still just $\mathbb{R}$. –  Ollie Margetts Aug 31 '12 at 9:18