Does the algebra of bounded variation functions have a "noncommutative geometric" meaning and generalization? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:24:03Z http://mathoverflow.net/feeds/question/105943 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105943/does-the-algebra-of-bounded-variation-functions-have-a-noncommutative-geometric Does the algebra of bounded variation functions have a "noncommutative geometric" meaning and generalization? Qfwfq 2012-08-30T13:09:13Z 2012-08-30T13:09:13Z <p>According to Gelfand-Naimark theory, <code>$C^*$</code>-algebras of continuous functions $\mathcal{C}^0(X,\mathbb{C})$ on a compact Hausdorff topological space completely capture its topology. Furthermore, every commutative <code>$C^*$</code>-algebra arises in this way. In noncommutative geometry (NCG), noncommutative <code>$C^*$</code>-algebras generalize the notion of compact (<em>locally</em> compact, if the algebra is possibly non unital) Hausdorff topological space.</p> <p>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.</p> <p>I learn from <a href="http://en.wikipedia.org/wiki/Bounded_variation" rel="nofollow">this</a> 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.</p> <p>I would like to know:</p> <blockquote> <p>1) Which geometric aspect of $\Omega$ -if any- is completely described by $\mathrm{BV}(\Omega)$ ?</p> <p>2) Which is -if there is any- the "right" NCG generalization of the $\mathrm{BV}(\Omega)$ algebra?</p> </blockquote>