Beta independence of the pure-point supersymmetric heat supertrace

By [F1], $Q\mathrm{/}\sqrt{\lambda }$ is a linear bijection between the finite-dimensional positive-energy parity eigenspaces, so $m_{\lambda }^{+}:=\dim {\mathcal{H}}_{\lambda }^{+}=\dim {\mathcal{H}}_{\lambda }^{-}=:m_{\lambda }^{-}$ for every $\lambda \in \mathrm{\Sigma }$. On any supplied orthonormal basis of ${\mathcal{H}}_{\lambda }^{\pm }$, [F2] gives $⟨T_{\beta }e,e⟩=e^{-\beta \lambda }$ for each basis vector. Therefore the positive-energy contribution to the signed basis sum at this fixed $\lambda$ is $e^{-\beta \lambda }(m_{\lambda }^{+}-m_{\lambda }^{-})=0$.

By [F2], the full diagonal family is absolutely summable. For the partition into its energy blocks, the absolute value of each block sum is at most the sum of the absolute values in that block; hence the family of block sums is also absolutely summable. To see that its sum equals the original sum, fix $\epsilon >0$ and choose a finite set of basis vectors whose omitted absolute tail is below $\epsilon$. Only finitely many energy blocks meet that finite set, and the omitted absolute tail bounds the difference between the full sum and the sum of those complete block sums. Letting $\epsilon$ decrease to zero proves that grouping by energy preserves the sum. Step 1.1 cancels every positive-energy block; each zero-energy even basis vector contributes $1$ and each zero-energy odd basis vector contributes $-1$. The remaining finite sum is $m_0^{+}-m_0^{-}$, independent of $\beta$. The proof uses no trace cyclicity and no positive-energy eigenvector outside the given discrete decomposition.