Quantised Calculus for Perturbed Massive Dirac Operator on Noncommutative Euclidean Space

Abstract

In this paper we study quantised calculus for the massive unperturbed Dirac operator ${\mathcal D}_{m}$, m > 0, as well as perturbed Dirac operator ${\mathcal D}_{m}+V$, on the noncommutative Euclidean space . We prove a necessary and sufficient condition for the quantised derivative $i[\operatorname {sgn}({\mathcal D}_m+V), 1\otimes x]$ to belong to the weak Schatten ideal ${\mathcal L}_{d,\infty }$. This extends and generalises earlier results for Dirac operator on the classical Euclidean space.

Appendix: Auxiliary Computation

Lemma A.1

Let e _ij be the N(d) × N(d)-matrix units and m > 0. For any i, j = 1, …N, we have

$$\displaystyle \begin{aligned}\Re\Big( \int_0^\infty ({\mathcal D}_m+i\lambda)^{-1} (e_{ij}\otimes 1)({\mathcal D}_m+i\lambda)^{-1}d\lambda \Big)=(m^2-\Delta_\theta)^{-1/2} \Psi_{ij},\end{aligned}$$

where Ψ _ij is a bounded operator on .

Proof

Since ${\mathcal D}_m^2+\lambda ^2=-1\otimes (\lambda ^2+m^2-\Delta _\theta )$, it follows that ${\mathcal D}_m^2+\lambda ^2$ commutes with e _ij ⊗ 1. Therefore, we can write

$$\displaystyle \begin{aligned} \Re&\big(({\mathcal D}_m+i\lambda)^{-1} (e_{ij}\otimes 1)({\mathcal D}_m+i\lambda)^{-1}\big) \\&=\frac{1}{2}\Big(\frac{{\mathcal D}_m}{\lambda^2+{\mathcal D}_m^2}-i\frac{\lambda}{\lambda^2+{\mathcal D}_m^2}\Big)(e_{ij}\otimes 1)\Big(\frac{{\mathcal D}_m}{\lambda^2+{\mathcal D}_m^2}-i\frac{\lambda}{\lambda^2+{\mathcal D}_m^2}\Big)+ \\&\qquad +\frac{1}{2}\Big(\frac{{\mathcal D}_m}{\lambda^2+{\mathcal D}_m^2}+i\frac{\lambda}{\lambda^2+{\mathcal D}_m^2}\Big)(e_{ij}\otimes 1)\Big(\frac{{\mathcal D}_m}{\lambda^2+{\mathcal D}_m^2}+i\frac{\lambda}{\lambda^2+{\mathcal D}_m^2}\Big) \\&=\frac{{\mathcal D}_m(e_{ij}\otimes 1){\mathcal D}_m-\lambda^2}{(\lambda^2+{\mathcal D}_m^2)^2}. \end{aligned} $$

Therefore, we can compute

$$\displaystyle \begin{aligned} \int_0^\infty&\Re\big(({\mathcal D}_m+i\lambda)^{-1} (e_{ij}\otimes 1)({\mathcal D}_m+i\lambda)^{-1}\big)d\lambda \\&=-{\mathcal D}_m(e_{ij}\otimes 1) {\mathcal D}_m \int_0^\infty\frac{d\lambda}{(\lambda^2+m^2-\Delta_\theta)^{2}}-\int_0^\infty\frac{\lambda^2d\lambda}{(\lambda^2+m^2-\Delta_\theta)^{2}} \\&=-\frac{\pi}{4}{\mathcal D}_m(e_{ij}\otimes 1) {\mathcal D}_m (m^2-\Delta_\theta)^{-3/2}-\frac{\pi}{4}(m^2-\Delta_\theta)^{-1/2}\\ &=-\frac{\pi}4(m^2-\Delta_\theta)^{-1/2}\cdot \Big(\frac{{\mathcal D}_m}{(m^2-\Delta_\theta)^{1/2}}(e_{ij}\otimes 1) \frac{{\mathcal D}_m}{(m^2-\Delta_\theta)^{1/2}}+1\Big). \end{aligned} $$

Since ${\mathcal D}_m(m^2-\Delta _\theta )^{-1/2}$ is bounded, it follows that

$$\displaystyle \begin{aligned}\Psi_{ij}:=-\frac\pi4\Big(\frac{{\mathcal D}_m}{(m^2-\Delta_\theta)^{1/2}}(e_{ij}\otimes 1) \frac{{\mathcal D}_m}{(m^2-\Delta_\theta)^{1/2}}+1\Big),\end{aligned}$$

is bounded operator, as required. □