Laurent operator

A Laurent operator on the whole integer lattice is a translation-invariant bi-infinite matrix. Under the discrete Fourier transform it becomes multiplication by its symbol, so its $2$-operator norm is the essential supremum of the symbol's modulus.

A Toeplitz operator is a constant-diagonal operator on a one-sided sequence space. It models the restriction of a translation-invariant stencil to a half-line, where the boundary prevents direct diagonalization by the full-lattice Fourier transform.

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.

On the Hardy space of the circle, a continuous symbol gives a Fredholm Toeplitz operator exactly when it never vanishes. Nonvanishing gives a parametrix $T_{1/f}$ modulo compact semicommutators. Write $f(z)=z^n{e}^{h(z)}$, where $n$ is its counterclockwise winding number and $h$ is continuous. Homotopy through nonvanishing symbols and local constancy of the Fredholm index reduce the index to that of the shift $T_{z^n}$, namely $-n$. If $f$ vanishes at a point, normalized Hardy reproducing kernels concentrating there are weakly null unit vectors on which $T_f$ tends to zero, contradicting a Fredholm parametrix.