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

2012-08-30T13:09:13Z

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?

Does the algebra of bounded variation functions have a "noncommutative geometric" meaning and generalization? - MathOverflow

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