= Toeplitz exponential determinant identity
{title2=$\det(T(e^f)T(e^{-f}))=\exp\sum_{n>0}nf_nf_{-n}$}
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 $T(e^f)T(e^{-f})=e^Ae^Be^{-A}e^{-B}$. The trace of $[A,B]$ is $\sum_{n>0}nf_nf_{-n}$ because the shift relation gives $\operatorname{Tr}[(S^*)^m,S^n]=n\delta_{mn}$. Smoothness makes the series converge in trace norm, so the <Fredholm determinant of an exponential commutator> applies.
Back to article page
