The product formula for regularized Fredholm determinants: two new proofs

Abstract

For an $m$-summable operator $A$ in a separable Hilbert space the higher regularized Fredholm determinant $\det\nolimits_m(I+A)$ generalizes the classical Fredholm determinant. Recently, Britz et al presented a proof of a product formula \[ \det\nolimits_m\bigl( (I+A)\cdot(I+B) \bigr) = \det\nolimits_m (I+A) \cdot \det\nolimits_m (I+B) \cdot \exp\operatorname{Tr}\bigl({X_m(A,B)}\bigr), \] where $X_m(A,B)$ is an explicit polynomial in $A,B$ with values in the trace class operators. If $m=1$ then $X_1(A,B)=0$, hence the formula generalizes the classical determinant product formula. One of the purposes of this note is to present two very simple alternative proofs of the formula. The first proof is a priori analytic and makes use of the fact that $z\mapsto \det\nolimits_m(I+zA)$ is holomorphic, while the second proof is completely algebraic. The algebraic proof has, in our opinion, some interesting aspects in its own about the trace and commutators. Secondly, we extend the above mentioned formula to several factors \[ \det\nolimits_m\Bigl( \prod_{l=1}^r (I+A_l) \Bigr) =\left( \prod_{l=1}^r \det\nolimits_m (I+A_l) \right) \cdot \exp\operatorname{Tr}\bigl({X_{m,r}(A_r,\ldots,A_r)}\bigr). \] The latter is more than just a straightforward generalization as we will gain more insights into the combinatorics behind it. Also we will present an algebraized version of the analytic proof in the language of formal power series. The upshot is that the two identities are just combinatorial in nature.