Normal abelian representations over a given measure space

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. Self-checked by the writing AI. Original text: CC0 1.0.

This chapter constructs the diagonal field for a normal representation of an already specified $L^\infty$ algebra. It supplies the concrete starting step in comparing representations that agree on a given abelian subalgebra.

1. Exact assumptions and inputs

Let $(X,\Sigma,\mu)$ be a sigma-finite measure space, let $H$ be a separable Hilbert space, and let $$\rho:L^\infty(X,\mu)\longrightarrow B(H)$$ be a unital normal star representation. Normality means preservation of suprema of bounded increasing nets of positive elements. The representation need not be faithful. If it is faithful, the field constructed below is nonzero almost everywhere. A zero Hilbert space is allowed when the representation is zero.

The exact inputs are the Hilbert space basis and orthogonal-projection theorems; monotone convergence, density of bounded measurable functions in $L^2$ of a finite measure, and the positive Radon–Nikodym theorem on a sigma-finite measure space; and the definition and elementary orthogonal-coordinate construction of measurable fields and direct integrals. The measure inputs are proved in Measure and Hilbert space tools. No spectral multiplicity assertion is assumed.

2. Countably many reducing cyclic summands

There is a finite or countable orthogonal decomposition $$H=\bigoplus_{j\in J}H_j,\qquad H_j=\overline{\rho(L^\infty(X,\mu))\xi_j},$$ where the $\xi_j$ are nonzero and each $H_j$ reduces $\rho$.

Proof. Let $(\eta_n)$ be a countable norm-dense sequence in $H$. Proceed through its terms. At stage $n$, take the orthogonal component of $\eta_n$ in the complement of the sum of the cyclic summands already chosen. If it is nonzero, call it $\xi_j$ and add its cyclic subspace. A subspace generated by a vector under a star algebra is invariant under both the algebra and its adjoints, so it is reducing. The already chosen orthogonal sum and its complement are also reducing; the new cyclic subspace therefore lies in that complement. At the end, every $\eta_n$ lies in the closed sum: at its stage its old projection lies in the earlier sum and its remaining component either vanishes or lies in the newly added summand. Density gives the asserted equality. At most one summand is added per stage. $\square$

3. A cyclic summand as a weighted scalar space

For each $j$, set $$\nu_j(E)=\langle\rho(1_E)\xi_j,\xi_j\rangle\quad(E\in\Sigma).$$ This is a finite positive measure absolutely continuous with respect to $\mu$. Indeed, disjoint partial unions give an increasing sequence of indicator functions with supremum the indicator of their union. Normality of $\rho$, followed by the normal vector functional, gives countable additivity. Also $\nu_j(X)=\|\xi_j\|^2$, and a $\mu$-null indicator represents zero in $L^\infty$, so its $\nu_j$ measure is zero. By positive Radon–Nikodym there is a measurable $h_j\ge0$, finite almost everywhere, such that $$\nu_j(E)=\int_Eh_j\,d\mu,\qquad \int_Xh_j\,d\mu=\|\xi_j\|^2.$$ Changing $h_j$ on a measurable null set, make it finite everywhere, and put $S_j=\{h_j>0\}$.

For bounded measurable functions $f,g$, the representation identities give $$\langle\rho(f)\xi_j,\rho(g)\xi_j\rangle =\langle\rho(\overline g f)\xi_j,\xi_j\rangle =\int_X f\overline g\,d\nu_j.$$ The last equality holds first for simple functions by the definition of $\nu_j$, and then for bounded measurable functions by uniform simple approximation and continuity on both sides. Thus $f\mapsto\rho(f)\xi_j$ is an isometry from the bounded functions modulo $\nu_j$-null equality into $H_j$. Bounded functions are dense in $L^2(X,\nu_j)$ by truncation and simple approximation. The isometry consequently extends to a unitary $$V_j:L^2(X,\nu_j)\longrightarrow H_j.$$ It is onto because its range contains the dense cyclic subspace and an isometry from a complete Hilbert space has closed range. For $a\in L^\infty(X,\mu)$, it intertwines multiplication by $a$ with $\rho(a)|_{H_j}$, first on bounded functions and then on all of $L^2(X,\nu_j)$ by density. Absolute continuity ensures that the multiplication depends only on the $\mu$-equivalence class of $a$.

Multiplication by $\sqrt{h_j}$ is a unitary $$W_j:L^2(X,\nu_j)\longrightarrow L^2(S_j,\mu).$$ The isometry follows from the Radon–Nikodym integral formula, first for simple nonnegative functions and then by monotone convergence. For $b\in L^2(S_j,\mu)$, the measurable function $b/\sqrt{h_j}$ on $S_j$, extended by zero off $S_j$, belongs to $L^2(X,\nu_j)$ and maps to $b$. This proves surjectivity. The unitary $W_j$ commutes with every bounded scalar multiplier. Combining these unitaries identifies $\rho$ with multiplication on $\bigoplus_jL^2(S_j,\mu)$.

4. The measurable Hilbert field and the full intertwiner

Let $(e_j)_{j\in J}$ be the standard orthonormal basis of $\ell^2(J)$, and define $$K(x)=\overline{\operatorname{span}}\{e_j:x\in S_j\},\qquad s_j(x)=1_{S_j}(x)e_j.$$ The Gram coefficients $\langle s_j(x),s_k(x)\rangle=\delta_{jk}1_{S_j}(x)$ are measurable, and the $s_j(x)$ have dense span in $K(x)$. The measurable-field construction therefore makes $K(x)$ a measurable field with this fundamental sequence. The zero space occurs exactly when no $S_j$ contains $x$.

A section $b$ is measurable precisely when its scalar coordinates $b_j(x)=\langle b(x),e_j\rangle$, zero off $S_j$, are measurable. Indeed, these are its scalar products with the fundamental sections, and every measurable section is determined by these coordinates. Conversely, measurable coordinates whose values form a square-summable family at each point give a section: finite coordinate sums are measurable and converge pointwise in norm, which is the defining closure property of this field. Consequently $$T:\int_X^\oplus K(x)\,d\mu(x)\longrightarrow\bigoplus_jL^2(S_j,\mu),\qquad T(b)=(b_j)_j$$ is a unitary. The norm identity is $$\int_X\|b(x)\|^2,d\mu =\int_X\sum_j|b_j(x)|^2,d\mu =\sum_j\int_{S_j}|b_j(x)|^2,d\mu.$$ Here the last equality is simply monotone convergence for the increasing finite coordinate sums. For surjectivity, choose measurable representatives of an element $(b_j)_j$ of the direct sum. The displayed sum is finite, so $\sum_j|b_j(x)|^2<\infty$ outside one measurable null set; set all coordinates to zero on that set. The resulting measurable section lies in the direct integral and maps to the given element. This also proves surjectivity without any choice of uncountably many representatives. The case $J=\varnothing$ gives the zero space.

Let $V=\bigoplus_jV_j$ and $W=\bigoplus_jW_j$. The unitary $$U=T^{-1}WV^{-1}:H\longrightarrow\int_X^\oplus K(x)\,d\mu(x)$$ satisfies, for every $a\in L^\infty(X,\mu)$, $$U\rho(a)U^*=M_a,$$ where $(M_ab)(x)=a(x)b(x)$. All three constituent unitaries intertwine the scalar multipliers, so the equality holds for every $a$, rather than just a chosen countable stock.

If $\rho$ is faithful, put $E=X\setminus\bigcup_jS_j$. Multiplication by $1_E$ vanishes on every scalar summand, hence $\rho(1_E)=0$. Faithfulness gives $1_E=0$ in $L^\infty(X,\mu)$, so $\mu(E)=0$ and $K(x)\ne0$ almost everywhere. Conversely, if $\bigcup_jS_j$ is conull and $M_a=0$, then $a=0$ almost everywhere on each $S_j$. To see this, multiply $a$ by the indicators of the finite-measure pieces of a sigma-finite exhaustion of $S_j$, which belong to $L^2(S_j,\mu)$. Each such product must vanish. Countable union over the pieces and over $j$ gives $a=0$ almost everywhere on $X$. Thus this field realization is faithful exactly when it is nonzero almost everywhere. $\square$

For two faithful normal representations on separable Hilbert spaces, this construction puts both over the same given base $(X,\mu)$. Their other operators can then be decomposed whenever they commute with the diagonal algebra, by the complete diagonal-commutant theorem in Decomposable operators and the diagonal algebra. This conclusion assumes the specified $L^\infty$ base and does not construct a centre decomposition for an arbitrary abstract von Neumann algebra.

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