The product formula for regularized Fredholm determinants

Abstract

For trace class operators $A,B\in {\mathcal{B}}_1(\mathcal{H})$ ($\mathcal{H}$ a complex, separable Hilbert space), the product formula for Fredholm determinants holds in the familiar form \[ {\det}_{\mathcal{H}} ((I_{\mathcal{H}} - A) (I_{\mathcal{H}} - B)) = {\det}_{\mathcal{H}} (I_{\mathcal{H}} - A) {\det}_{\mathcal{H}} (I_{\mathcal{H}} - B). \] When trace class operators are replaced by Hilbert--Schmidt operators $A,B\in {\mathcal{B}}_2(\mathcal{H})$ and the Fredholm determinant ${\det }_{\mathcal{H}}(I_{\mathcal{H}}-A)$, $A\in {\mathcal{B}}_1(\mathcal{H})$, by the 2nd regularized Fredholm determinant ${\det }_{\mathcal{H},2}(I_{\mathcal{H}}-A)={\det }_{\mathcal{H}}((I_{\mathcal{H}}-A)\exp (A))$, $A\in {\mathcal{B}}_2(\mathcal{H})$, the product formula must be replaced by \[ {\det}_{\mathcal{H},2} ((I_{\mathcal{H}} - A) (I_{\mathcal{H}} - B)) = {\det}_{\mathcal{H},2} (I_{\mathcal{H}} - A) {\det}_{\mathcal{H},2} (I_{\mathcal{H}} - B) \exp(- {\rm tr}(AB)). \] The product formula for the case of higher regularized Fredholm determinants ${\det }_{\mathcal{H},k}(I_{\mathcal{H}}-A)$, $A\in {\mathcal{B}}_k(\mathcal{H})$, $k\in \mathbb{N}$, $k\ge 2$, does not seem to be easily accessible and hence this note aims at filling this gap in the literature.

10 pages, reference added, extended proofs in Sect. 2 a bit