 Toeplitz exponential determinant identity

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.