Finite $N=2$ rest-fiber algebra with zero central charge and strict bound

For the displayed subset-basis action, removing two distinct labels in opposite orders gives opposite signs, while removing the same label twice gives zero; hence $\{c_j,c_k\}=0$. For $j\mathrm{\ne }k$, inserting $k$ and removing $j$ in opposite orders also gives opposite signs. For $j=k$, if $j\mathrm{\notin }S$ then $c_j{c}_j^{†}\mathrm{\mid }S⟩=\mathrm{\mid }S⟩$ and $c_j^{†}{c}_j\mathrm{\mid }S⟩=0$, while if $j\in S$ the two terms reverse roles. Thus $\{c_j,c_k^{†}\}={\delta }_{jk}I$ on every basis vector; taking adjoints gives $\{c_j^{†},c_k^{†}\}=0$.

Scaling by $\sqrt{2M}$ gives $\{Q_{\alpha }^I,(Q_{\beta }^J{)}^{†}\}=2M{\delta }^I{}_J{\delta }_{\alpha \beta }I$ and all same-chirality anticommutators zero. These are exactly the rest-frame relations in [F2] with $Z^{IJ}=0$ and $P_0=-M$.

Since $Z=0$, its operator norm is $\mathrm{\parallel }Z{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=0<M$. By [F1], the $N=2$ mass/central-charge equality condition is $M=\mathrm{\mid }z\mathrm{\mid }=\mathrm{\parallel }Z{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$, which fails for these rest-frame algebra data. This conclusion concerns only the finite module and does not assert that it is a BPS sector of a full Poincare representation.

The verification concerns only a finite rest-frame module and uses no choice from a family; [A1] is retained as the inherited interface assumption of the BPS definition. The data supply no Lorentz action, full unitary Poincare representation, or physical particle, so the example makes no claim about the spectrum of any QFT.