This is a question on how (or if) people in the community think about the Poincaré inequality in noncommutative geometry. In geometry, the Poincaré 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 Poincaré 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 Poincaré 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.)