# Noncommutative Poincaré inequalities

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.)

There are results in this direction, usually phrased in terms of quantum Markov semigroups (QMS). Let $$\mathcal{M}$$ be a von Neumann algebra with a normal faithful tracial state $$\tau$$. A quantum Markov semigroup is a weak$$^\ast$$ continuous family $$(P_t)$$ of unital completely positive maps from $$\mathcal{M}$$ to $$\mathcal{M}$$. Assume further that the maps $$P_t$$ satisfy $$\tau((P_t x)y)=\tau(x P_t y)$$.
If $$\mathcal{L}$$ is the generator of $$(P_t)$$, the carré du champ and iterated carré du champ are defined as \begin{align*} \Gamma(x,y)&=\frac 12 (x^\ast\mathcal{L}(y)+\mathcal{L}(x)^\ast y-\mathcal{L}(x^\ast y))\\ \Gamma_2(x,y)&=\frac 1 2(\Gamma(x,\mathcal{L}(y))+\Gamma(\mathcal{L}(x),y)-\mathcal{L}\Gamma(x,y)). \end{align*} The QMS $$(P_t)$$ is said to satisfy an $$L^p$$-Poincaré inequality if $$\|x\|_p\leq C_p\max\{\|\Gamma(x,x)^{1/2}\|_p,\Gamma(x^\ast,x^\ast)\|_p\},\qquad x\in(\ker\mathcal{L})^\perp.$$ By a result of Junge and Zeng [1], $$\Gamma_2(x,x)\geq \lambda\Gamma(x,x)$$ implies the $$L^p$$-Poincaré inequality for $$p\geq 2$$ with constant $$C_p\sim \frac{p}{\sqrt{\lambda}}$$. Examples include
• the heat semigroup on a compact Riemannian manifolds with Ricci curvature bounded below by $$K$$ (with $$\lambda=K$$),
• the QMS on the group von Neumann algebra of the free group given by $$P_t \lambda_g=e^{-t|g|}\lambda_g$$, where $$|g|$$ is the length of $$g$$ as a reduced word in the generators and their inverses (with $$\lambda=1$$),
• the QMS generated by the number operator on $$q$$-Gaussian algebras.
Derivations fit into this picture because whenever $$\delta\colon \mathcal{M}\to \mathcal{M}$$ is a derivation that satisfies $$\delta(x^\ast)=\delta(x)^\ast$$, the operator $$\mathcal{L}=\delta^\dagger\delta$$ generates a $$\tau$$-symmetric QMS. In this case, the carré du champ operator is given by $$\Gamma(x,y)=\delta(x)^\ast\delta(y)$$, so that the $$L^p$$-Poincaré inequality takes the form from the question. However, the usual trace on the algebra of all bounded operators on an infinite-dimensional Hilbert space is not finite. I think the extension of these results to the semi-finite case should be doable, but I don't think anyone has written it down in detail.