 Solution

Let $C=[A,B]$ be trace class and set $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$. The integral identity $e^{B}A{e}^{-B}-A={\int }_0^1{e}^{sB}[B,A]e^{-sB}ds$ is trace-norm convergent. It shows that this difference is trace class. The analogous Duhamel identity then makes $F(t)-I$ trace class and differentiable in that norm. Hence $F(t)\in G(H)$.

Compute the right logarithmic derivative without taking separate traces of $A$:Trace invariance under bounded invertible similarity and the integral identity give $\mathrm{Tr}(\dot{F}{F}^{-1})=-{\int }_0^1\mathrm{Tr}[B,A]ds=\mathrm{Tr}[A,B]$. Cyclicity with a trace-class factor makes this equal to the trace in part (b). As $F(0)=I$, integrating the scalar differential equation givesThis Fredholm determinant of an exponential commutator can be nontrivial in infinite dimension: it is not legitimate to set the trace of a trace-class commutator to zero when $A,B$ themselves need not be trace class.