Spectral fibers of circle translation

Write $\mathbb T=\mathbb R/\mathbb Z$, equip it with normalized Haar measure, and put $u(s)=e^{2\pi is}$. Translation on $L^\infty(\mathbb T)$ has precisely one scalar degree of freedom in each integer frequency. We prove this as a statement about equivalence classes, with explicit normal projections and convergent Fourier approximations.

Original expression: CC0-1.0 to the extent of rights held.

Translation and its Fourier coefficients

Let $M=L^\infty(\mathbb T)$, acting by multiplication on $L^2(\mathbb T)$, and define $(\alpha_t f)(s)=f(s+t),\qquad \widehat f(n)=\int_{\mathbb T}f(s)e^{-2\pi ins}\,ds. \tag{C1}$ Translation preserves null sets, so the definition is independent of the representative. It preserves multiplication, adjoints and the essential supremum norm, and $\alpha_{-t}$ is its inverse.

We record why this is a von Neumann algebra in the stated representation. The map $f\mapsto M_f$ is isometric: the upper norm bound is immediate, and indicators of sets where $|f|>r$ give the lower bound for every $r<\|f\|_\infty$. If $B\in B(L^2(\mathbb T))$ commutes with all bounded multipliers, let $h=B1$. Then $Bg=gh$ for every bounded $g$. For a measurable set $A$, $\|1_Ah\|_2=\|B1_A\|_2\leq\|B\|\mu(A)^{1/2}.$ If $|h|>\|B\|+\epsilon$ on a set of positive measure, this inequality on that set is impossible. Thus $h\in L^\infty$ and $\|h\|_\infty\leq\|B\|$. Bounded truncations are dense in $L^2$, so $B=M_h$ on the whole Hilbert space. The multiplier algebra is therefore its own commutant; commutation equations are weak-operator closed, so it is a von Neumann algebra. This proof works on any finite measure space.

Under the pairing $\langle f,g\rangle=\int fg$, its preadjoint is $g(r)\mapsto g(r-t)$. This follows by Haar substitution. L24, translations proves norm continuity of this action on $L^1$, hence point-ultraweak continuity of $\alpha$. For the multiplication representation the same pairing describes all normal functionals: a vector coefficient is $\int f\xi\overline\eta$, and Cauchy–Schwarz puts its density in $L^1$; every $g\in L^1$ factors as $g=\xi\overline\eta$ with $\xi,\eta\in L^2$, by taking square roots of $|g|$ and its measurable phase. Summable vector coefficients have $L^1$ sums, by CP4–6. Thus these statements concern the actual normal topology.

Haar substitution also gives $\widehat{\alpha_t f}(k)=e^{2\pi ikt}\widehat f(k). \tag{C2}$ If $\alpha_t f=e^{2\pi int}f$ for every $t$, then $(e^{2\pi ikt}-e^{2\pi int})\widehat f(k)=0$. When $k\ne n$, take $t=1/(2(k-n))$ modulo one to make the two phases unequal. Hence all coefficients except the $n$-th vanish. Since $f\in L^2(\mathbb T)$, completeness of the orthonormal family $u^k$, proved in CC0, gives $M_n:=\{f:\alpha_t f=e^{2\pi int}f\text{ for every }t\} =\mathbb C u^n. \tag{C3}$ The reverse inclusion is immediate from (C1). In particular the fixed algebra is the scalars.

Normal projections and bounded Fejér approximation

By AT5, the scalar integral $P_n f=\int_{\mathbb T}e^{-2\pi int}\alpha_t(f)\,dt \tag{C4}$ defines a normal contraction, where the integral is evaluated against normal functionals. Changing $t$ to $t+h$ shows $\alpha_h(P_nf)=e^{2\pi inh}P_nf$, and (C4) is the identity on $M_n$. Applying the normal coefficient functional $\widehat{\phantom f}(n)$ and using (C2) proves $P_nf=\widehat f(n)u^n. \tag{C5}$ Thus $P_n$ is a normal rank-one projection and $P_nP_k=0$ for $n\ne k$. No pointwise identity simultaneously outside one exceptional null set for every translation is needed.

For $N\geq1$, use the kernels already proved in CC0: $F_N(t)=\frac1N\left|\sum_{j=0}^{N-1}e^{2\pi ijt}\right|^2 =\sum_{|j|<N}\left(1-\frac{|j|}{N}\right)e^{2\pi ijt}. \tag{C6}$ They are nonnegative, have integral one, and their integrals outside every neighborhood of zero tend to zero. Consequently $Q_Nf=\int_{\mathbb T}F_N(t)\alpha_t(f)\,dt =\sum_{|n|<N}\left(1-\frac{|n|}{N}\right)\widehat f(n)u^n, \qquad \|Q_Nf\|_\infty\leq\|f\|_\infty. \tag{C7}$ The equality follows from the finite expansion and the symmetry of the coefficients.

These approximants converge to $f$ in $L^1$ and $L^2$, and ultraweakly in $L^\infty$. For $p=1,2$, translation is continuous in $L^p$, by L24 and CC0. For a class $h\in L^p$, bounded-map compatibility and the integral norm inequality give $\|Q_Nh-h\|_p\leq \int F_N(t)\|\alpha_t h-h\|_p\,dt. \tag{C8}$ Near zero the integrand's norm factor is arbitrarily small; elsewhere it is at most $2\|h\|_p$, and the kernel mass tends to zero. The same proof with the continuous scalar function $t\mapsto\omega(\alpha_t f)$ gives ultraweak convergence for $f\in M$. For a continuous $f$, uniform continuity makes the estimate valid in the supremum norm as well. Thus (C7) supplies both uniqueness of Fourier coefficients and a bounded family of trigonometric polynomial approximations.

One unitary generates the multiplication algebra

Let $M_z$ be multiplication by $z(s)=e^{2\pi is}$. Its integer powers are the trigonometric monomial multipliers. For any $f\in L^\infty(\mathbb T)$, (C7) gives polynomials $Q_Nf$ with the common supremum bound $\|f\|_\infty$, converging to $f$ in $L^2$. For bounded $g$, $\|(Q_Nf-f)g\|_2\leq\|g\|_\infty\|Q_Nf-f\|_2\longrightarrow0. \tag{C8a}$ For arbitrary $g\in L^2$, approximate by bounded truncations and use the common operator bound $2\|f\|_\infty$ for the difference. Thus $M_{Q_Nf}\to M_f$ strongly. Every such multiplier belongs to $W^*(M_z)$, since bicommutants are strongly closed: a strong limit commutes with each fixed operator commuting with the approximants. Conversely the multiplier algebra is a von Neumann algebra containing $M_z$, by the commutant proof above. We conclude $W^*(M_z)=\{M_f:f\in L^\infty(\mathbb T)\}. \tag{C8b}$ The coefficient calculation after (C1) also proves that this identification and its inverse preserve the ultraweak topology.

Reversing the direction of translation

If instead $\beta_t f(s)=f(s-t)$, then $\beta_t=\alpha_{-t}$. Its degree-$n$ fiber is therefore $\mathbb C u^{-n}$, with projection $P_n^\beta f=\widehat f(-n)u^{-n}. \tag{C9}$ This follows by substituting $-t$ in (C4), or directly from (C2). It explains why the sign of a fiber must be tied to the stated action and pairing.

These computations give the translation model in Takesaki, Theory of Operator Algebras II, Exercise XI.1.2, p.330.