Compact translations and twisted torus actions

Translations on any compact Hausdorff abelian group have scalar character fibers. Scalar fixed points also force one-dimensional nonzero fibers for general compact abelian actions, but the products of those fibers may carry nontrivial phases. A concrete two-torus action shows why the circle classification does not extend to arbitrary compact abelian groups.

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

Character fibers on a compact abelian group

Let $G$ be compact Hausdorff abelian, with normalized Haar measure and additive notation. For a continuous character $\chi_p:G\to\mathbb T$, let $(\alpha_sf)(t)=f(t+s),\qquad f\in L^\infty(G). \tag{T1}$ All functions here are Haar equivalence classes. Haar conventions, density of continuous functions in $L^1$, and continuity of $L^1$ translations are supplied by L24. The preadjoint of $\alpha_s$ is $g(t)\mapsto g(t-s)$; as in L102, $L^1$ is exactly the normal functional space of the multiplication algebra on $L^2(G)$. The factorization $g=\xi\overline\eta$ and the multiplier commutant proof there apply verbatim to this finite measure space. Thus (T1) is a point-ultraweakly continuous normal action.

Its average is $E(f)=\int_G\alpha_s(f)\,ds =\left(\int_G f\right)1. \tag{T2}$ To prove the formula without choosing representatives for uncountably many translates, average on the predual. For continuous $g$, Haar invariance gives the equality of continuous functions $\int_G g(t-s)\,ds=(\int_Gg)1$. The Bochner integral is legitimate by compactness and $L^1$ translation continuity, as in L24. Approximate any $g\in L^1$ by continuous functions. Both its averaged translate and $g\mapsto(\int g)1$ are contractions on $L^1$, so the formula extends to all $g$. Taking the adjoint proves (T2).

If $f$ is fixed by all translations, $f=E(f)$, so $f$ is constant. If $\alpha_s(f)=\chi_p(s)f$, multiply by $\overline{\chi_p}$. The product is fixed, and hence constant. Conversely $\chi_p$ itself has that covariance. Therefore $L^\infty(G)_p=\mathbb C\chi_p. \tag{T3}$ The normal projection onto this fiber has the precise formula $P_p f=\int_G\overline{\chi_p(s)}\,\alpha_s(f)\,ds =\left(\int_G f(t)\overline{\chi_p(t)}\,dt\right)\chi_p. \tag{T4}$ Indeed $\chi_p E(f\overline{\chi_p})$ equals the first integral: multiplication by a fixed bounded function is normal, and the integrands agree as equivalence classes. Then use (T2). For $G=\mathbb T^2$, this gives fibers $\mathbb C e^{2\pi i(ms+nt)}$, $(m,n)\in\mathbb Z^2$.

The multiplication phases in scalar fixed-point actions

Now let a compact abelian group act normally and point-ultraweakly continuously on a nonzero von Neumann algebra $M$, with fixed algebra $\mathbb C1$. For any character $p$, the proof of (E4) in L103 uses only multiplication of characters. Thus each nonzero fiber is spanned by a unitary $w_p$, and the set $J$ of appearing characters is a subgroup of the discrete dual. Choose $w_0=1$. Products of the chosen unitaries determine phases $w_pw_q=c(p,q)w_{p+q},\qquad |c(p,q)|=1. \tag{T5}$ Associativity and normalization give $c(p,q)c(p+q,r)=c(q,r)c(p,q+r),\qquad c(p,0)=c(0,p)=1. \tag{T6}$ Replacing $w_p$ by $b_pw_p$, where $|b_p|=1$ and $b_0=1$, replaces $c(p,q)$ by $c'(p,q)=b_pb_q\overline{b_{p+q}}c(p,q). \tag{T7}$ These formulas follow simply by computing the two products. When $J=\mathbb Z$, choosing powers of one generator makes all phases one, as in L103. With two generators, the phase comparing their two orders of multiplication need not disappear: (T7) multiplies $c(p,q)$ and $c(q,p)$ by the same scalar, so their ratio is unchanged.

A twisted regular algebra and its faithful trace

Fix $\theta\in\mathbb R$ and write $e(r)=e^{2\pi ir}$. On $J=\mathbb Z^2$, set $c_\theta((m,n),(p,q))=e(\theta np). \tag{T8}$ For $a=(m,n)$, $b=(p,q)$, $d=(r,k)$, the two exponents in (T6) are $\theta(np+(n+q)r)$ and $\theta(qr+n(p+r))$, which agree. The normalization is immediate.

On $H=\ell^2(J)$, define $\lambda(a)\delta_b=c_\theta(a,b)\delta_{a+b}, \qquad M_\theta=\{\lambda(a):a\in J\}''. \tag{T9}$ Each $\lambda(a)$ permutes the orthonormal basis with modulus-one coefficients, so is unitary. The cocycle identity gives $\lambda(a)\lambda(b)=c_\theta(a,b)\lambda(a+b). \tag{T10}$ In particular $\lambda(a)^*=\overline{c_\theta(a,-a)}\lambda(-a)$; the same expression is both a left and a right inverse because $c_\theta(a,-a)=c_\theta(-a,a)$.

For $b\in J$, also define $\rho(b)\delta_a=c_\theta(a,b)\delta_{a+b}. \tag{T11}$ These are unitaries. Applying both $\lambda(d)\rho(b)$ and $\rho(b)\lambda(d)$ to $\delta_a$, equality of their coefficients is precisely $c_\theta(a,b)c_\theta(d,a+b)=c_\theta(d,a)c_\theta(d+a,b)$. Thus every $\rho(b)$ commutes with $M_\theta$. Since $\rho(b)\delta_0=\delta_b$, the vector $\delta_0$ is separating: if $x\delta_0=0$ for $x\in M_\theta$, then $x\delta_b=\rho(b)x\delta_0=0$ for all $b$, and hence $x=0$.

It follows that $\tau(x)=\langle x\delta_0,\delta_0\rangle \tag{T12}$ is a faithful normal state. For positivity, $\tau(x^*x)=\|x\delta_0\|^2$, which also proves faithfulness. Normality is that of a vector coefficient. It is a trace: on two generators, both $\tau(\lambda(a)\lambda(b))$ and $\tau(\lambda(b)\lambda(a))$ vanish unless $a+b=0$; in that case their coefficients agree by the preceding inverse identity. Extend by bilinearity to finite sums. The finite sums form a unital *-algebra with bicommutant $M_\theta$; BD1–5 gives bounded ultraweak approximation of every element by such sums. For one fixed finite sum, separate ultraweak continuity of multiplication extends the trace identity in the other variable; then fix that arbitrary element and extend in the first variable. Separate continuity follows directly from the vector-series tests in CP4, by moving the fixed bounded factor to a test vector. No joint continuity of multiplication is used.

The GNS map for this trace is concretely $\Lambda_\tau(x)\mapsto x\delta_0$: it preserves inner products by (T12), and its range contains every $\delta_a=\lambda(a)\delta_0$. It therefore extends to a unitary onto $H$. The $\lambda(a)$ form a complete orthonormal family in this trace Hilbert space.

A faithful ergodic two-torus action

For $s=(s_1,s_2)\in\mathbb T^2$, define $D_s\delta_{(m,n)}=e(ms_1+ns_2)\delta_{(m,n)}. \tag{T13}$ This unitary group is strongly continuous: approximate a vector in $\ell^2(J)$ by a finite-support vector; continuity on the finite support is elementary, and the remaining difference is at most twice the tail norm. Direct calculation gives $D_s\lambda(m,n)D_s^*=e(ms_1+ns_2)\lambda(m,n). \tag{T14}$ Consequently $\alpha_s=\operatorname{Ad}D_s$ preserves $M_\theta$. Conjugation is normal by the vector-series description of normal functionals. For each fixed $x$, its individual vector coefficients under conjugation are continuous by strong continuity of $D$. An absolutely summable vector series has uniformly small tails bounded by $\|x\|$ times the sum of products of vector norms, so every normal coefficient is continuous. This proves point-ultraweak continuity. Also $D_s\delta_0=\delta_0$, so $\tau\circ\alpha_s=\tau$.

If $x$ is an eigenvector of degree $a=(m,n)$, then $D_sx\delta_0=e(ms_1+ns_2)x\delta_0. \tag{T15}$ In the coordinate expansion of $x\delta_0$, every coefficient at $b\ne a$ vanishes: vary a coordinate $s_j$ for which $b_j\ne a_j$ to separate the two characters. Thus $x\delta_0$ is a scalar multiple of $\delta_a$. Since $\delta_0$ is separating, $x$ is the same scalar multiple of $\lambda(a)$. Conversely (T14) supplies every such eigenvector. Hence $(M_\theta)_a=\mathbb C\lambda(a)\quad(a\in\mathbb Z^2), \qquad M_\theta^\alpha=\mathbb C1. \tag{T16}$ Testing (T14) at $a=(1,0)$ and $(0,1)$ shows that the kernel is zero. The action is faithful and ergodic.

Finally, put $U=\lambda(1,0)$ and $V=\lambda(0,1)$. Equation (T10) gives $U^mV^n=\lambda(m,n),\qquad VU=e(\theta)UV. \tag{T17}$ The power formulas hold also for negative integers, because the cocycle is one on either coordinate subgroup. If $\theta\notin\mathbb Z$, (T17) shows $M_\theta$ is noncommutative: $UV$ is unitary, and $e(\theta)-1\ne0$. This includes every irrational $\theta$. Thus a faithful ergodic $\mathbb T^2$-action need not be conjugate to translation on the commutative algebra $L^\infty(\mathbb T^2)$.

The zero-phase model is ordinary translation

When $\theta=0$, define a unitary $\mathcal F:\ell^2(\mathbb Z^2)\longrightarrow L^2(\mathbb T^2), \qquad \mathcal F\delta_{(m,n)}=z_1^mz_2^n. \tag{T18}$ Completeness follows from the circle basis in CC0 and its tensor-product statement, with the product integration convention in L24. On basis vectors, $\mathcal F\lambda(m,n)\mathcal F^*$ is multiplication by $z_1^mz_2^n$.

These multipliers generate the entire multiplication algebra. For $f\in L^\infty(\mathbb T^2)$, average translations against $F_N(s_1)F_N(s_2)$. The result $f_N$ is a finite trigonometric polynomial, satisfies $\|f_N\|_\infty\leq\|f\|_\infty$, and tends to $f$ in $L^2$. For the last assertion, the operators are $L^2$ contractions by the integral norm bound; they converge on each basis vector by the explicit Fejér coefficients, and hence on all $L^2$ by density. Multiplication by $f_N$ tends strongly to multiplication by $f$: for bounded $g$, use $\|(f_N-f)g\|_2\leq\|g\|_\infty\|f_N-f\|_2$; for general $g\in L^2$, approximate by bounded truncations and use the common bound $2\|f\|_\infty$. The multiplication algebra is its own commutant by the finite-measure proof in L102, so it is strongly closed. This proves $\mathcal F M_0\mathcal F^*=L^\infty(\mathbb T^2) \quad\hbox{in its multiplication representation}. \tag{T19}$ Under $\mathcal F$, $D_s$ is translation by $s$, as follows first on the Fourier basis and then by continuity. Conjugating a multiplier therefore translates its function. Thus the construction contains ordinary torus translation at zero phase and noncommutative scalar-fiber actions at nonintegral phase, with the difference recorded exactly by (T17).

This develops the compact-group and two-torus question in Takesaki, Theory of Operator Algebras II, Exercise XI.1.4, p.330.