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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.

Split a smooth symbol into its strictly negative and strictly positive Fourier modes. Their Toeplitz operators $A,B$ exponentiate exactly to the Toeplitz operators of the corresponding exponentials, and $T(e^f)T(e^{-f})=e^A{e}^B{e}^{-A}{e}^{-B}$. The trace of $[A,B]$ is ${\sum }_{n>0}n{f}_n{f}_{-n}$ because the shift relation gives $\mathrm{Tr}[(S^{\ast }{)}^m,S^n]=n{\delta }_{mn}$. Smoothness makes the series converge in trace norm, so the Fredholm determinant of an exponential commutator applies.