Hypertraces and finite injective algebras

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. New original text: public domain (CC0).

A hypertrace can remember a specified normal trace while it is approximated by finite-rank densities. Retaining that trace is the useful extra condition: a Hilbert–Schmidt vector then controls the product of left and right multiplication, and finite sums of product vectors give completely positive maps through matrix algebras. We first carry out this construction for an algebra with a faithful normal tracial state. We subsequently assemble arbitrary finite algebras from central corners and semifinite algebras from finite corners.

The spectral-projection argument is a second consequence of the same density approximation. It is developed afterwards, together with its noncommuting counterexample. Nearly tracial states are treated in a separate final module, because the finite-model construction no longer needs to obtain its chosen trace from an almost invariant projection state.

1. Conventions and exact foundations

Inner products are linear in the second variable. The symbol $\operatorname{Tr}$ denotes the unnormalized operator trace, and $\|x\|_2^2=\operatorname{Tr}(x^*x)$ for Hilbert–Schmidt operators. For a tracial state $\tau$ on $M$, write $\|x\|_{2,\tau}^2=\tau(x^*x)$. An algebra is finite in the Murray–von Neumann sense: an isometry in it is unitary. This does not assert finite Hilbert-space dimension.

We use the definitions in Finite models of a von Neumann algebra. Semidiscreteness means a net of normal completely positive contractions $S:M\to M_n(\mathbb C)$ and completely positive contractions $R:M_n(\mathbb C)\to M$ for which $RS(x)\to x$ ultraweakly, for every $x\in M$. Injectivity means completely positive extension from operator systems, with the unit value and hence the norm preserved. In particular, extending the identity of a faithfully normally represented algebra gives a ucp retraction $E:B(H)\to M$. The reverse implication from semidiscreteness is proved directly in Section 6 by finite matrix extension and compactness.

The faithful-trace finite-model construction below does not invoke the preceding lesson's commutant criterion. Its matrix maps and their normalization are proved explicitly in Section 5.

Three projection and trace contracts are needed for the general finite and semifinite passages, and for the final nearly tracial-state module:

1. CT. Every finite von Neumann algebra has a positive, faithful, normal, unital, center-linear trace $T:M\to Z(M)$. Thus $T(ab)=T(ba)$. Every projection is an orthogonal sum of monic projections: on its central support a monic projection is one of finitely many equivalent orthogonal projections adding to the central unit.
2. PC. Projection comparison holds. Equivalent projections in a finite algebra have equivalent complements. A finite algebra decomposes centrally into finite homogeneous type I algebras and an algebra with no nonzero abelian projection; the latter admits repeated halving of its unit into equivalent projections. On a finite homogeneous type I component, matrix units identify the algebra with $M_n(Z)$.
3. SF. A finite orthogonal sum of finite projections is finite. In a semifinite algebra every nonzero projection contains a nonzero finite projection.

These foundations are proved in Projection comparison and normal center-valued traces: C05–C06 give comparison and finite complements, D05–D10 give the finite central decomposition, equivalent dyadic partitions and monic decomposition, NC–CT construct the normal faithful center-valued trace, and C10–C12 give the finite-sum and semifinite projection consequences. Here semifinite uses the projection convention: every nonzero projection contains a nonzero finite projection; C12 proves its equivalence to an orthogonal decomposition of the unit into finite projections. No faithful normal state on the whole algebra is assumed.

The concrete series-vector predual and its bounded strong-continuity estimate are proved in H03 of Regular-group operator foundations. Its coordinate Hilbert-space construction applies to any Hilbert space: choose a maximal orthonormal family by Zorn's lemma. H01 gives the orthogonal projection onto its closed span. A nonzero vector in the orthogonal complement could be normalized and added to the family, so maximality makes that span the whole space. Finite orthogonal truncations then identify the space unitarily with the corresponding $\ell^2(I)$ of H00. This also covers the zero space. We use a concrete von Neumann algebra in commutant form $M=\mathcal S''$; H01 proves $M''=M$ in this form.

The scalar, norming and continuous C*-calculus proofs used here are Elementary calculus for infinite products and Infinite tensor products, F01–F08. The finite threshold and spectral-step proofs S01–S06 supply the projection approximation and endpoint cuts used below, and CS01–CS06 supply the character construction. The following constructions supply convex separation, compactness, finite matrix extension, finite-rank trace estimates and the fixed spatial completion. The finite-trace realization is proved in Section 2. All closures and approximation parameters use nets. No separability assumption is imposed.

Convex separation (SEP). The norming and extension form of Hahn–Banach is proved in Infinite tensor products, F01. Here is its needed convex consequence. If a nonempty convex set has zero in its Banach-space weak closure, then zero is in its norm closure.

Suppose instead that its norm closure $K$ misses zero. Choose $\varepsilon>0$ for which $U=K+B(0,\varepsilon)$ misses zero. Choose $u_0\in U$, and let $V=U-u_0$. This is open, convex and contains zero. Its gauge

is finite and nonnegative, positively homogeneous and subadditive. Finiteness follows from a ball about zero in $V$; convexity proves subadditivity by adding the two scaled representatives. Moreover $V=\{p<1\}$: convexity proves one inclusion, and openness permits a small outward scaling of each point for the other. For $z=-u_0$, we have $p(z)\ge1$. The real linear functional $tz\mapsto t$ is dominated by $p$, for positive $t$ by homogeneity and for negative $t$ by nonnegativity. Hahn–Banach extends it to a real linear $g\le p$. A ball in $V$ bounds $p(x)$ by a constant times $\|x\|$; applying this to $x$ and $-x$ proves continuity of $g$.

For $u\in U$, $g(u)=g(u-u_0)-1<0$. Since $k+B(0,\varepsilon)\subset U$, this gives $g(k)\le-\varepsilon\|g\|<0$ for every $k\in K$. In a complex space, $G(x)=g(x)-ig(ix)$ is a bounded complex linear functional with real part $g$. Thus this inequality also separates zero in the Banach weak topology, a contradiction. Translating proves the same separation for any point outside a closed convex set.

For the weak* topology of a dual space, a neighbourhood excluding a convex set is specified by finitely many real and imaginary parts of evaluations. Their joint map takes values in a finite-dimensional real space and sends the excluded point outside the closure of the convex image. The preceding separation argument there supplies a real linear separator; pulling it back gives a real combination of those evaluations. For hermitian functionals, each real part is evaluation at the selfadjoint part of the tested element, and each imaginary part at its imaginary selfadjoint part. Their real combination is therefore evaluation at one selfadjoint element. This is the weak* separation used below.

Products and dual balls (COMPACT). Closed scalar disks are compact by Elementary calculus for infinite products, R02. Closed subsets inherit the finite-open-cover property by adding their open complement to a cover. We now prove the arbitrary-product consequence.

A proper filter is a family of subsets closed under finite intersections and supersets, containing no empty set. Every proper filter extends to a maximal one by Zorn's lemma: the union of a chain is still proper. Such an ultrafilter contains either a set or its complement. Indeed, if adding a set would remain proper, maximality puts that set in the filter; otherwise some filter member is disjoint from it, and its complement is already in the filter.

In a compact space, the closures of the members of an ultrafilter have the finite intersection property, so they have a common point $x$. Every open neighbourhood of $x$ belongs to the ultrafilter: otherwise its closed complement would belong and contain $x$, a contradiction. Conversely, an open cover without a finite subcover has complements generating a proper filter. An ultrafilter extending it cannot converge anywhere, since it contains the complement of one covering neighbourhood at every proposed limit. Thus compactness is equivalent to convergence of every ultrafilter.

For an ultrafilter on a product of nonempty compact spaces, inverse images under a coordinate projection give an ultrafilter on that coordinate. Choose one limit in each coordinate. A basic product neighbourhood imposes only finitely many coordinate conditions, whose inverse images have intersection in the original ultrafilter. It therefore converges to the chosen tuple. The product is compact. A product with an empty factor is empty and compact; the product with no factors is a singleton.

For a complex normed space $E$, embed the closed radius-$C$ ball of $E^*$, $C\ge0$, in

The equations for additivity and complex homogeneity define a closed subset. A tuple satisfying them is exactly a linear functional bounded by $C\|x\|$. The subspace topology is pointwise convergence on $E$, hence the weak* topology. The ball is therefore weak* compact. On a unital C*-algebra the states form the closed subset of its dual unit ball specified by $f(1)=1$ and $f(a^*a)\ge0$ for all $a$, so they too are weak* compact.

For clarity, compactness supplies subnets, not necessarily subsequences. The closures of the tails of a net in a compact space have the finite intersection property. A common point meets every neighbourhood on every tail. Use triples consisting of a neighbourhood, a tail index and a chosen later index whose net value lies in that neighbourhood. Order them by shrinking the neighbourhood and advancing both indices. They are directed: after combining two neighbourhoods, choose a hit beyond both chosen indices and both tail indices. Projection to the chosen index is order preserving and eventually beyond every original index. It defines a subnet converging to the common point.

Positive functionals and their Hilbert quotients (GNS). A positive linear functional $f$ on a unital C*-algebra obeys Cauchy–Schwarz for $\langle a,b\rangle_f=f(a^*b)$, by expanding its positive quadratic form. The zero diagonal case follows by varying the phase and magnitude of the scalar coefficient. The calculus and order estimates of Infinite tensor products, F06–F08 give

Thus $\|f\|=f(1)$, including the case $f(1)=0$. The zero vectors of this form are a linear subspace, by Cauchy–Schwarz. The inequality

makes left multiplication descend to the quotient and bounded by $\|a\|$. Complete it using F01. The formula for the form shows that the adjoint is multiplication by $a^*$; multiplication and the unit are preserved. This gives a bounded *-representation with cyclic vector $[1]$, and $f(a)=\langle[1],a[1]\rangle$. For a state that vector has norm one. A norm-dense unital *-subalgebra has dense quotient vectors, since $\|[a]-[b]\|\le\|a-b\|$.

Completely positive extension into a matrix algebra (FEXT). Let $S$ be a unital selfadjoint subspace of a unital C*-algebra $D$, and let $\Phi:S\to M_n$ be completely positive. It has an extension $\widetilde\Phi:D\to M_n$ with the same unit value.

First, a positive scalar functional $f$ on $S$, with $f(1)=c$, extends positively to $D$. On the real selfadjoint part, $-\|s\|1\le s\le\|s\|1$ gives $|f(s)|\le c\|s\|$. Here $f(s)$ is real, since $s+\|s\|1$ and $1$ are positive elements of $S$. Real Hahn–Banach extends this functional to $D_{\rm sa}$ with the same norm and value at one. If $a\ge0$ is nonzero, then $\|1-a/\|a\|\|\le1$, so the extended value at $a$ is nonnegative. Complexification gives a positive extension to $D$, with norm $c$ by GNS above. When $c=0$ all these functionals are zero.

Matrix orders can be taken in a faithful concrete realization of $D$. Such a realization follows from the scalar extension and GNS arguments just given. For $0\ne b\ge0$, its spectral radius is $\|b\|$ by F04 and F08. Evaluation of a polynomial in $b$ at that spectral value is bounded by its operator norm, by F03–F06. It extends to a state on the closure of the polynomials: positivity follows by approximating each square $a^*a$ with polynomial squares. Positive scalar extension gives a state of $D$ taking value $\|b\|$ at $b$. Take the Hilbert direct sum of the GNS representations for all states of $D$. For $b=a^*a$, a state with $f(b)=\|a\|^2$ gives $\|\pi_f(a)[1]\|=\|a\|$. The direct sum is therefore isometric and faithful, as each summand is contractive. Its matrix algebras are concrete C*-algebras. Any two faithful realizations give the same matrix norms and orders: apply the automatic contractivity in F05 to the entrywise *-isomorphism and its inverse, and use F08 for positivity. The zero algebra uses the zero Hilbert space.

Apply scalar extension to the operator system $M_n(S)\subset M_n(D)$ and the positive functional

Its positivity follows by testing the positive matrix $[\Phi(s_{ij})]$ on the vector with components $e_1,\ldots,e_n$. Its value at the identity is $F_\Phi(1)=\operatorname{Tr}(\Phi(1))$, equal to $n$ when $\Phi$ is unital. Let $F$ be its positive extension. Define

This agrees with $\Phi$ on $S$, including its unit value. To check complete positivity, take $[a_{kl}]\ge0$ in $M_m(D)$ and vectors $\xi_k\in\mathbb C^n$. The matrix

is a scalar compression of $[a_{kl}]$, hence positive in $M_n(D)$. Its value under $F$ is exactly $\sum_{k,l}\langle\xi_k,\widetilde\Phi(a_{kl})\xi_l\rangle$. This proves positivity at every matrix level.

The norm bound follows directly for any completely positive map on an operator system. For $\|s\|\le1$, the matrix $\begin{bmatrix}1&s\\s^*&1\end{bmatrix}$ is positive: it is $VV^*$ for $V=\begin{bmatrix}1&0\\s^*&(1-s^*s)^{1/2}\end{bmatrix}$ in the containing matrix algebra. Applying complete positivity and the positive-form Cauchy–Schwarz inequality gives

Consequently $\|\Phi\|=\|\Phi(1)\|$. This applies to the extension and to every matrix amplification. Thus a cpc map extends cpc, and a ucp map extends ucp. This finite-target construction uses scalar positive extension and matrix compressions, with no dilation prerequisite.

Finite-rank trace estimates (FP). A finite-rank operator $d$ and its adjoint have a common finite-dimensional reducing support. Define its trace on that support and extend by zero. The value is independent of enlargement. Expanding into rank-one operators proves $\operatorname{Tr}(dA)=\operatorname{Tr}(Ad)$ for every bounded $A$. A finite polar decomposition $d=v|d|$, or its singular-vector expansion, gives

For the upper bound write $d=\sum_j s_j\theta_{u_j,w_j}$ with orthonormal singular vectors and $s_j\ge0$; each trace term is $s_j\langle w_j,A u_j\rangle$, whose modulus is at most $s_j$. The contraction $A=v^*$ attains the sum. This includes the zero operator. Finite-rank trace functionals are finite series-vector functionals of H03, so (F1) is also their exact predual norm.

The Hilbert–Schmidt inner product on finite-rank operators is $\langle d,e\rangle_2=\operatorname{Tr}(d^*e)$. Finite matrix Cauchy–Schwarz gives its norm inequality. Expanding in an orthonormal basis of the finite initial support gives $\|Ad\|_2\le\|A\|\|d\|_2$; taking adjoints gives $\|dA\|_2\le\|A\|\|d\|_2$. If $d,e$ are finite rank and $de=v|de|$, then

All these calculations occur on finite supports; no decomposition of an arbitrary compact operator is used.

The fixed spatial completion (SP). For arbitrary Hilbert spaces, the coordinate realizations H00–H01 define the Hilbert tensor product as $\ell^2(I\times J)$, with elementary coordinate tensors total. Equivalently it completes finite tensors in their product inner product. On a finite tensor, orthonormalize its finitely many second vectors. The identity $\|\sum_j\xi_j\otimes\eta_j\|^2=\sum_j\|\xi_j\|^2$ for orthonormal $\eta_j$ proves that $a\otimes1$ is bounded by $\|a\|$; exchanging the factors treats $1\otimes b$. Product vectors give the reverse inequality, by choosing unit vectors whose image norms approach the two operator norms. Therefore

including zero factors. The rank-one map

is unitary: a finite trace calculation gives the same inner products, and finite rank-one sums are dense on both sides. Under it, $a\otimes\bar b$ acts by $d\mapsto adb^*$, with the bounds just proved.

Take the closure of the algebraic operators $\sum_i a_i\otimes\bar b_i$ on this one $H\otimes\bar H$. It is a concrete unital C*-algebra, denoted $C$. The algebraic action is faithful. To see this, express a tensor using linearly independent first operators $a_1,\ldots,a_m$. Their vector coefficients span the dual of that finite-dimensional space: otherwise a nonzero linear combination would have all matrix coefficients zero. Finite combinations of those coefficients isolate each $a_i$; if the tensor acts as zero, each resulting second operator is zero. This proves injectivity.

In this lesson $\|z\|_{\min}$ is the operator norm of $z$ in this faithful spatial completion. Every use below has this fixed trace Hilbert space. Through COM's linear map $JbJ\mapsto\bar b$, this is the indicated spatial completion of $M\odot M'$. It is exactly the original concrete spatial norm on $H\otimes H$: the map $U:\bar H\to H$, $U(\bar\xi)=J\xi$, is a linear unitary, and $U\bar bU^*=JbJ$. Consequently $1\otimes U$ intertwines the two displayed algebraic actions and preserves their operator norms. No comparison with another trace representation is required.

For any faithful concrete unital C*-algebra $D\subset B(K)$, finite convex mixtures of its vector states are weak* dense in its state space. If not, weak* SEP supplies a selfadjoint $a\in D$ and a state whose value exceeds every unit-vector expectation. With $t=\sup_{\|\xi\|=1}\langle\xi,a\xi\rangle$, we have $a\le t1$ as an operator, hence in $D$ by F08, so no state can exceed $t$. This contradiction proves density. A vector state changes in functional norm by at most $(\|\xi\|+\|\eta\|)\|\xi-\eta\|$ when its vector changes; this follows by expanding the two scalar products. Thus any total family of finite product vectors gives the same convex density after normalization.

Normal trace factorization from CT. If $\tau$ is a normal scalar tracial state on a finite algebra, then $\tau=\tau\circ T$. To verify the implication we consume from CT, take a monic projection $p$ with $n$ equivalent orthogonal copies summing to a central projection $z$. Traciality gives $\tau(p)=\tau(z)/n$, and center-linearity and traciality give $T(p)=z/n$. Consequently $\tau$ and $\tau\circ T$ agree on monic projections. Normality and the orthogonal monic decomposition give agreement on every projection. Uniform spectral approximation then gives agreement on every selfadjoint element, and linearity on all elements. Both normality and the full monic-decomposition contract are used in this argument. The identity for singular tracial states is proved separately in Section 8.

A completely positive retraction is bimodular (MD). Here is the short dilation argument, to make this use of complete positivity explicit. For a ucp map $\Phi:D\to B(K)$, put on $D\odot K$ the positive semidefinite form

Complete positivity gives positivity of the form, and its scalar Cauchy–Schwarz inequality makes its zero vectors a linear subspace. Left multiplication by $a$ is bounded by $\|a\|$, since the matrix $[b_i^*(\|a\|^2-a^*a)b_j]$ is positive. Quotient by the zero vectors and complete. This gives a representation $\pi$ and an isometry $V\xi=1\otimes\xi$, with $\Phi(a)=V^*\pi(a)V$. If $\Phi(a^*a)=\Phi(a)^*\Phi(a)$, then

The equality for $aa^*$ gives the same result for $a^*$. Hence $\pi(a)V=V\Phi(a)$ and $V^*\pi(a)=\Phi(a)V^*$, so $\Phi(ax)=\Phi(a)\Phi(x)$ and $\Phi(xa)=\Phi(x)\Phi(a)$. For a retraction fixing a C*-subalgebra both equalities hold for every element of that subalgebra. This is the standard dilation and multiplicative-domain argument; Arveson's paper supplies its freely accessible primary construction.

The same dilation construction for a completely positive map that need not be unital gives $V^*V=\Phi(1)$, with bounded $V$. Thus $\|\Phi\|=\|\Phi(1)\|$, and applying the construction at matrix levels gives the complete norm bound. This also proves the norm assertion used below for subunital maps.

2. A hypertrace with a prescribed trace

For a unital C*-algebra $A\subseteq B(H)$, a hypertrace is a state $\sigma$ on $B(H)$ for which $\sigma(ax)=\sigma(xa)$, for $a\in A$, $x\in B(H)$. This is equivalent to invariance under every $\operatorname{Ad}(u)(x)=uxu^*$, $u\in\mathcal U(A)$. For the converse apply invariance to $xu$, giving $\sigma(ux)=\sigma(xu)$, and span the algebra by unitaries.

We will use the following explicit spanning formula. If $x\ne0$, let $r=\|x\|$, $a=(x+x^*)/(2r)$, $b=(x-x^*)/(2ir)$. The selfadjoint contractions $a,b$ give unitaries $u=a+i(1-a^2)^{1/2}$, $v=b+i(1-b^2)^{1/2}$, and

The absolute coefficient sum is $2r$. The zero element needs no such decomposition.

The normal trace realization (REP). Let $M$ have its concrete predual $Q$ from H03, and let $\tau$ be its faithful normal tracial state. Here normal functionals are the ultraweakly continuous functionals of that concrete predual. Complete $M$ for $\langle x,y\rangle=\tau(x^*y)$ to obtain $H$. Positivity, traciality and the C*-order inequality give

Thus left multiplication $L_a$ and right multiplication $R_b$ are bounded; $R_bR_c=R_{cb}$. Their adjoints are $L_{a^*}$ and $R_{b^*}$. The left representation is faithful, since $L_a1=0$ forces $\tau(a^*a)=0$. It is isometric by F05 applied to it and F07 for an injective *-homomorphism.

The canonical map $Q\to M^*$ is isometric: H03 gives $Q^*=M$, and the norming Hahn–Banach formula of F01 identifies the norm of a vector of $Q$ with its supremum against the unit ball of $M$. Its image is therefore norm closed. Fixed multiplication preserves $Q$, since pulling back a series-vector functional replaces its two vectors by bounded-operator images.

For $x,y\in M$, the functional $a\mapsto\langle x1,L_a y1\rangle=\tau(x^*ay)$ belongs to $Q$. Approximate arbitrary $\xi,\eta\in H$ by vectors of $M1$. The bound $\|L_a\|\le\|a\|$ makes the corresponding functionals converge in norm on $M$, so $a\mapsto\langle\xi,L_a\eta\rangle$ belongs to $Q$, with norm at most $\|\xi\|\|\eta\|$. A series of such coefficients with summable norm products converges in $Q$. Hence $L$ is ultraweakly continuous.

We next prove directly that its image is a von Neumann algebra. For $C\ge0$, let

Every Hilbert weak test of $x1$ is a coefficient just proved to belong to $Q$. COMPACT therefore makes $K_C$ weakly compact. The Hilbert weak topology is Hausdorff, so a compact subset is closed: for a point outside it, separate that point from each compact-set point by disjoint neighbourhoods, and select finitely many of the latter to cover the compact set. Intersect the corresponding neighbourhoods of the outside point. Thus $K_C$ is also norm closed. It is convex and bounded by $C$.

Let $T$ commute with all right multiplications, put $C=\|T\|$ and $\xi=T1$. For $b\in M$, use the polar decomposition $b=v|b|$ proved for every concrete von Neumann algebra in Regular-group operator foundations, T04a, and put $d=|b|^{1/2}$. The selfadjoint operator $R_d$ commutes with $T$, so

Indeed both $vd1$ and $d1$ have squared norm $\tau(|b|)$, since $v^*v$ is the support of $|b|$. On the other hand,

The upper bound is tracial Cauchy–Schwarz applied to $\tau(dv^*xd)$, and the lower bound is attained at $x=Cv$. The support function of $K_C$ is $C$-Lipschitz in its Hilbert normal. Since $M1$ is dense, (R1)–(R2) imply $\operatorname{Re}\langle\eta,\xi\rangle\le\sup_{k\in K_C}\operatorname{Re}\langle\eta,k\rangle$ for every $\eta\in H$.

This places $\xi$ in $K_C$. To prove that implication, a nonempty closed convex Hilbert subset has a closest point: the parallelogram identity applied to a minimizing sequence and its midpoints makes the sequence Cauchy. If $k$ were closest to $\xi$ with $\xi\ne k$, minimality along the segment to any $h\in K_C$ would give $\operatorname{Re}\langle\xi-k,h-k\rangle\le0$. But the support inequality in the direction $\xi-k$ gives the opposite strict inequality at $\xi$, since $\operatorname{Re}\langle\xi-k,\xi\rangle=\operatorname{Re}\langle\xi-k,k\rangle+\|\xi-k\|^2$. This is a contradiction.

Consequently $T1=x1$ for some $x\in M$, $\|x\|\le C$. For $y\in M$, commutation gives $T(y1)=R_yT1=xy1=L_x(y1)$; density proves $T=L_x$. We have proved

H01 makes the right hand side a concrete von Neumann algebra.

The inverse map also preserves the ultraweak topology. The functionals $a\mapsto\tau(b^*a)$, $b\in M$, are norm dense in $Q$. If their closed span were proper, the norming Hahn–Banach theorem and $Q^*=M$ would give a nonzero annihilator $a$. Taking $b=a$ would give $\tau(a^*a)=0$, contradicting faithfulness. Each such functional is the vector coefficient $L_a\mapsto\langle b1,L_a1\rangle$ on $L(M)$. Isometry of $L$ preserves the functional norms, and H03 makes normal functionals on its concrete image norm closed. Thus every $f\in Q$ transports to a normal functional on $L(M)$. This proves ultraweak continuity of $L^{-1}$.

Proposition 2.2. Let $M$ be finite and injective, and let $\tau$ be a faithful normal tracial state. Its left representation on $H=L^2(M,\tau)$ has a hypertrace extending $\tau$.

Proof. Extend the identity of the left representation to a ucp map $E:B(H)\to M$, using injectivity. The dilation computation MD makes $E$ bimodular. For $a\in M$, $x\in B(H)$,

Thus $\sigma=\tau\circ E$ is the required state, and $\sigma(a)=\tau(a)$. Normality of $E$ and $\sigma$ is not needed. $\square$

3. Finite densities retain both invariance and the chosen trace

The estimate for square roots is proved first, because it is used before any spectral projection is selected.

Lemma 1.1 (Powers–Størmer). For positive finite-rank operators $h,k$,

Proof. Restrict to the finite-dimensional sum of the ranges of $h$ and $k$, which reduces both selfadjoint operators. Put $d=h-k$, $s=h+k$, and $v=\operatorname{sign}(d)$, zero on the kernel. Then

The first bound is a finite-matrix calculation. Diagonalize the selfadjoint matrix $D=h^2-k^2$, with eigenvalues $\lambda_j$ and orthonormal eigenvectors $\xi_j$. Since $\|v\|\le1$,

To bound the trace below, write $d=d_+-d_-$, with supports $p_+,p_-$. Compression gives

For positive matrices $A,B$, $\operatorname{Tr}(AB)=\operatorname{Tr}(B^{1/2}AB^{1/2})\geq0$. Apply this to the two displayed positive differences and $d_+,d_-$. Adding gives

The two bounds prove the claim, also when one or both operators are zero. This sign-support proof is the finite-matrix specialization of the primary Powers–Størmer argument read in Anantharaman–Popa, Theorem 7.3.7. $\square$

Trace-controlled density approximation (DEN). Suppose $N\subseteq B(H)$ is faithfully normally represented, $\tau$ is a normal state on $N$, and a hypertrace $\sigma$ restricts to $\tau$. For finite sets $U\subseteq\mathcal U(N)$, $B\subseteq N$, and numbers $\alpha,\beta>0$, there is a positive finite-rank operator $h$ such that

where $\rho_d(x)=\operatorname{Tr}(dx)$. One can simultaneously require $\|h-u^*hu\|_2<\beta$ for $u\in U$.

Proof. Finite-rank normal states are convex and weak* dense in the state space of $B(H)$, by SP. Each finite mixture of vector states has a positive finite-rank trace-one density. Formula (F1) identifies the norm of every finite-rank density difference with its $\|\cdot\|_1$ norm. These facts hold in arbitrary Hilbert dimension.

For a finite unitary set $V$, consider the convex set of tuples

where $\rho$ ranges over finite-rank normal states. Restriction is normal because the representation is normal. Its Banach-space weak tests are tuples in $B(H)^{|V|}\oplus N$. Weak* approximation to $\sigma$, its invariance, and $\sigma|_N=\tau$ show that zero is in the weak closure. SEP makes its weak and norm closures equal. In the sum norm we may therefore choose $d\geq0$, finite rank, $\operatorname{Tr}(d)=1$, with

With $h=\sqrt d$, Lemma 1.1 gives $\|h-v^*hv\|_2<\sqrt\delta$, while $\|h\|_2=1$. Include in $V$ the unitaries in (U) for all nonzero elements of $B$, as well as $U$. Let $C$ be the largest absolute coefficient sum for those decompositions, with $C=0$ if $B$ has no nonzero element, and set $C_0=\max(1,C)$. Choose $0<\delta<\min(\alpha,\beta^2/C_0^2)$. Unitary invariance of the Hilbert–Schmidt norm and the triangle inequality give (DEN). If the required sets are empty or consist of zero elements, the same tuple argument with no unitary coordinates still gives the restriction estimate; the zero commutators impose no condition. The construction is the trace-retaining convexity method in the complete proof of Anantharaman–Popa, Theorem 10.2.9. The square-root tolerance is chosen explicitly because its displayed estimate controls the square of the Hilbert–Schmidt error. $\square$

For later comparison with spectral projections, a projection state has a particularly transparent invariance bound. If $p\ne0$ has finite rank $r$, put $\rho_p(x)=\operatorname{Tr}(pxp)/r$. For a unitary $u$, its conjugated state's density is $u^*pu/r$. If $q=u^*pu$, then

Indeed, the finite-rank bound (F2) gives $\|p-q\|_1\leq2\sqrt r\|p-q\|_2$; (F1) and unitary invariance give the displayed state estimate. DEN retains the trace without needing to select such a projection.

4. The spatial scalar estimate and all central tests

Throughout this section $M$ is finite and injective with a faithful normal tracial state $\tau$, and is represented on $H=L^2(M,\tau)$. These are the hypotheses for Lemma 4.1 and Theorem 4.2 as well as (SC). The rule $Jx=x^*$ on $M$ extends to an antiunitary involution, because $\tau(xx^*)=\tau(x^*x)$.

The commutant in this representation (COM). Right multiplication $JbJ$, given by $\xi\mapsto\xi b^*$, commutes with the left algebra. For completeness, let $a\in M'$. Antiunitarity, commutation and the adjoint identity give, for $z\in M$,

Thus $JaJ1=a^*1$, or $Ja1=a^*1$. For $a,b\in M'$, it follows that

Use this identity once with $ac$ instead of $a$, and once with $c$, to obtain $(bJaJ)JcJ1=(JaJb)JcJ1$. These vectors are dense: $M'1$ contains $(JMJ)1=M1$. Hence $JaJ\in M''=M$. Together with the already verified right-multiplication inclusion this proves $M'=JMJ$. This is the complete elementary finite-trace commutant method in Peterson, Proposition 6.4.12; no general-weight commutation theorem is used.

The linear faithful representation of $M'$ on $\bar H$ sending $JbJ$ to $\bar b$ preserves products and adjoints: both maps are conjugate linear in $b$. SP supplies its fixed spatial completion $C$, the Hilbert–Schmidt identification (S2), and the action $d\mapsto adb^*$.

Scalar trace bound (SC). Keep this faithful trace representation fixed. For any normal tracial state $\nu$ of $M$, whether faithful or not, and

we have

Proof. REP transports $\nu$ normally to the fixed left image. Injectivity supplies the ucp retraction $E$ there, and MD makes $\nu\circ E$ a hypertrace extending $\nu$. DEN gives positive finite-rank $h$, $\|h\|_2=1$, whose restriction error is below $\alpha$ and whose commutators with the $b_i^*$ have norm below $\beta$. By (S2) its spatial vector state satisfies

Finite-rank trace cyclicity gives

The Hilbert–Schmidt bound in FP makes its modulus at most $\beta\sum_i\|a_i\|$. Consequently

Let $\alpha,\beta\downarrow0$. Empty sums and zero coefficients give zero without division. In particular this applies to the original faithful trace $\tau$. $\square$

Lemma 4.1. For $a_i,b_i\in M$, put

For every $\eta>0$, there is a normal state $\omega$ on $Z$ such that

Proof. Choose the normal state $\omega=\tau|_Z$. By CT, $\nu=\omega\circ T$ is a normal tracial state on $M$. SC in the unchanged faithful trace representation gives

This gives the stated estimate for every $\eta>0$, with the same state $\omega$. $\square$

Theorem 4.2. Under these hypotheses,

Proof. For every normal state $\omega$ of $Z$, CT makes $\nu=\omega\circ T$ a normal tracial state of $M$. SC in the one fixed representation gives $|\omega(T(c))|\le\|z\|_{\min}$.

These normal states detect the norm of $q=T(c)$. If $q\ne0$, F04 gives $\lambda\in\operatorname{sp}(q)$ with $|\lambda|=\|q\|$, since $q$ is normal. Put $h=(q-\lambda)^*(q-\lambda)$. Zero belongs to its spectrum: if $h$ were invertible, normality would make $h^{-1}(q-\lambda)^*$ a two-sided inverse of $q-\lambda$. For $\delta>0$, the continuous positive part $k=(\delta^2-h)_+$ is therefore nonzero, by F06. Let $e\in Z$ be its nonzero support projection from H02. The other positive part of $h-\delta^2$ annihilates $k$ and hence $e$; thus $he\le\delta^2 e$ and $\|(q-\lambda)e\|\le\delta$. Faithfulness gives $\tau(e)>0$, and

is a normal state of $Z$ with $|\omega_e(q)-\lambda|\le\delta$. Letting $\delta\downarrow0$ proves $\|q\|\le\sup_\omega|\omega(q)|$. The opposite inequality is the state norm bound; $q=0$ is immediate. Combining this norm detection with SC proves the result. Every trace test uses the original Hilbert space and tensor norm. $\square$

5. Explicit completely positive matrix models

Theorem 5.1. A finite injective von Neumann algebra with a faithful normal tracial state is semidiscrete.

Proof. We give the matrix maps, including their contractive normalization. The functional

is the vector functional of $1$ for algebraic left-right multiplication. That multiplication is a *-representation on the algebraic tensor product, because its factors commute. Consequently $\Omega(z^*z)=\|\mu(z)1\|^2\geq0$, and $\Omega(1)=1$. Bound (SC) shows that $\Omega$ extends continuously to the fixed concrete spatial completion $C$ of SP. It remains positive there: approximate the square root of a positive element of $C$ by algebraic tensors and pass to their squares.

SP proves weak* density of finite convex mixtures of vector states in this concrete $C$ on $H\otimes\bar H$. Each vector may moreover be approximated in norm by a finite sum $\sum_i\xi_i\otimes\bar\zeta_i$ with $\zeta_i\in M1$. Such vectors are dense, since $M1$ is dense in $H$. Within each sum we may choose the first vectors $\xi_i$ orthonormal by taking an orthonormal basis of their finite-dimensional span and changing the $\zeta_i$'s accordingly. Normalize the vector afterwards; this keeps $\zeta_i\in M1$. Thus finite convex mixtures of these special vector states are still weak* dense.

Their right marginal, regarded as a normal state of $M$, is $f(b)=\rho(1\otimes Jb^*J)$. The map $b\mapsto Jb^*J$ is a linear *-anti-isomorphism, so the marginal is positive. Its normality follows directly by expanding each special product vector: its marginal is $b\mapsto\tau(wb)$, with bounded $w$ given in (M3). REP and fixed multiplication make this a normal functional. Fix finitely many tensors $z_k\in C$. Tuples

form a convex set. They have zero in their Banach-space weak closure, because their weak tests are the listed tensor evaluations and operators $b\in M$, and $\Omega(1\otimes Jb^*J)=\tau(b)$. SEP therefore supplies, for any $\delta>0$, a finite mixture $\rho$ with

Write the mixture as a finite sum of vector functionals associated to $\eta_l=\sum_{i=1}^{n_l}\xi_{li}\otimes\bar\zeta_{li}$, with $\xi_{li}$ orthonormal for each $l$, and incorporate the square roots of mixture weights into $\zeta_{li}\in M$. Then $\sum_l\|\eta_l\|^2=1$. Set $N=\sum_l n_l$. With blocks indexed by $(l,i)$, define