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 \begin{align*} {\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) \\ & \quad \times \exp (- \operatorname {tr}_{\mathcal {H}}(AB)). \end{align*} The product formula for the case of higher regularized Fredholm determinants ${\det }_{\mathcal {H},k}(I_{\mathcal {H}} - A)$, $A \in \mathcal {B}_k(\mathcal {H})$, $k \in \mathbb {N}$, $k \geqslant 2$, does not seem to be easily accessible and hence this note aims at filling this gap in the literature.