This is a question on how (or if) people in the community think about the PoincarĂ© inequality in noncommutative geometry. In geometry, the PoincarĂ© inequality (when it exists) gives a bound on a function on a manifold $M$ in terms of its derivative; loosely written: $$ ||u||_{L^p(M)}\leq c||\nabla u||_{L^p(M)} $$ where (of course) u is in the complement of the kernel of $\nabla$ (or, one could subtract the average value from $u$ in the formula above). The PoincarĂ© inequality is very useful in analysis, and I'm looking in the literature for a noncommutative counterpart.

In non-commutative geometry, a common way to introduce differential calculus on an operator algebra is to define the derivation $A\to dA=[A,F]$ for some fixed operator $F$ (perhaps, from a Fredholm module). Now, in what context (and for which $A$) does it hold that
$$
||A||_{L^p}\leq c||dA||_{L^p}
$$

where $||A||^p_{L^p}=\operatorname{tr}|A|^p$ is the $p$-norm with respect to a trace on the operator algebra? More precisely (since $d$ has a non-trivial kernel), is there a natural subspace (of the operator algebra), perhaps the complement of the kernel, on which the PoincarĂ© inequality holds?. Anyway, this might not be the right way to phrase it in noncommutative geometry, so I'm happy if someone can point me in right direction where I can find references to these types of questions.

(Some preliminary remarks: IF the operator $d$ is bounded and IF the kernel of the operator $d$ is complemented and IF the range of $d$ is closed, then there exists a bounded inverse on the complement of the kernel, which implies that $d$ is bounded from below as in the inequality above (then $c$ is more or less one over the norm of the inverse). Of course, this smells like $d$ is a Fredholm operator.)