Fermions, Fock space and quasi-free factors

Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. Self-checked by the writing AI. New original text: public domain (CC0).

Anticommuting generators can produce ordinary matrix algebras. Parity strings explain how: they absorb the signs between successive modes. Positive occupation densities on these finite algebras fit together into a state for every covariance. A reduced doubled exterior space then realizes its completed cyclic representation and proves that its generated von Neumann algebra is a factor.

The proofs use finite Hilbert tensor products, positive square roots, finite matrix algebras and completion of a cyclic inner-product space. Their exact constructions are supplied in Sections 9–12 below and ITP Section 10. Section 6 constructs bounded finite-mode approximants directly; its constant-state conclusions use the completed product model and modular graph in ITP Section 11, and the finite-corner trace construction in Sections 18–20 below. The AF comparison is AF-algebras, Section 4 and Example 5.3. No classification theorem is used in the factor or approximation proofs.

Inner products are linear in the second variable. Accordingly, creation $c(f)$ is linear in $f$, and annihilation $a(f)=c(f)^*$ is conjugate-linear. Our CAR convention is

The occupation covariance is $\omega(a(f)^*a(g))=\langle g,Tf\rangle$. Thus the empty vacuum has $T=0$. Using a linear creation generator instead changes the placement of adjoints and the covariance parameter. These conventions must be changed together.

The one-particle Hilbert space $K$ may be arbitrary. We explicitly distinguish finite dimension, separable infinite dimension and a general directed finite-mode approximation. The finite-mode, state and factor phenomena are compared with the selected freely accessible CAR materials below.

1. Exterior space and bounded operators

For $n\ge1$, let $U_\sigma$ permute tensor coordinates according to the left action of the symmetric group, using $\sigma^{-1}$ on indices. Define

The group law gives $A_n^2=A_n$: in the double sum each resulting permutation occurs $n!$ times with its sign. The adjoint of $U_\sigma$ is $U_{\sigma^{-1}}$, whose sign is the same, so $A_n^*=A_n$. Thus these are orthogonal projections.

For an orthonormal basis $(e_j)$, indexed by an arbitrarily ordered set, put

The vectors $[e_{j_1},\ldots,e_{j_n}]$, with $j_1<\cdots<j_n$, form an orthonormal basis of $\Lambda^nK$. Distinct increasing tuples have disjoint permutation orbits, while a tuple with repeated entries has zero antisymmetrization. Expanding the tensor basis proves completeness. More generally,

This follows by expanding $A_n$ and taking the product of the coordinate inner products.

The Hilbert direct sum

is the fermionic Fock space; $\Omega=1\in\Lambda^0K$ is its vacuum. If $\dim K=d<\infty$, then

The unscaled exterior product of homogeneous vectors is $\eta\wedge\zeta=A_{p+q}(\eta\otimes\zeta)$, for degrees $p,q$. It is associative: both iterated products equal the full antisymmetrization of the three tensor factors, since full antisymmetrization absorbs a preceding antisymmetrization of any consecutive block. It defines a product on the algebraic sum of finite-degree vectors.

On that sum define

In particular $c(f)[g_1,\ldots,g_n]=[f,g_1,\ldots,g_n]$. For a unit vector $f$, choose a basis beginning with $f$. On the occupation basis, $c(f)$ carries a vector not containing $f$ to the corresponding vector containing $f$, up to its sign, and kills those already containing $f$. It is an isometry on the first closed span and zero on its orthogonal complement. Therefore $c(f)$ is bounded with norm one. Linearity and scaling give $\|c(f)\|=\|f\|$.

The adjoint has the contraction formula

To verify it, take inner products with arbitrary product vectors and expand the determinant (2) along its first row. Alternation gives $\{c(f),c(g)\}=0$. Formula (4) gives

on the occupation core. Adjoints give the other relations in (1). Boundedness extends all identities to $\mathcal F(K)$.

Lemma 1.1. The Fock CAR representation is irreducible.

Proof. A vector annihilated by every $a(f)$ is a multiple of $\Omega$: a nonzero coefficient at a nonempty finite occupation set gives a nonzero coefficient after contraction by one of its basis vectors. If $x$ commutes with all creators and annihilators, then $a(f)x\Omega=0$, so $x\Omega=\lambda\Omega$. Products of creators applied to $\Omega$ span a dense subspace. Commutation makes $x=\lambda1$ on it and hence everywhere. $\square$

2. Finite modes are full matrix algebras

Consider any CAR family on a nonzero representation Hilbert space. For a unit $f$, write $a=a(f)$ and $n=a^*a$. Relations (1) give

Neither $n$ nor $1-n$ is zero: either possibility together with the relations would force $a=0$ and then $1=0$. Hence $\|a\|=1$. Scaling shows

The operators

are matrix units. The two products $aa^*=e_{11}$ and $a^*a=e_{22}$, together with the displayed relations, verify every matrix-unit product and adjoint.

For orthonormal $f_1,\ldots,f_d$, put $a_j=a(f_j)$, $n_j=a_j^*a_j$, and

The $Z_j$ are commuting selfadjoint unitaries. Two anticommutations show that $n_i$ commutes with $a_j,a_j^*$ for $i\ne j$; $Z_j$ anticommutes with $a_j,a_j^*$. Each $b_j$ therefore gives a copy of the above matrix units. The copies for distinct $j$ commute. Indeed, moving the $i$-th annihilator or creator past the $j$-th generator contributes one CAR sign, and moving it past the factor $Z_i$ in the later string contributes a second sign. They cancel. Their diagonal projections commute with all the other copies as well.

Theorem 2.1. For a $d$-dimensional subspace $E\subset K$, the unital C*-algebra generated by its CAR family is $M_{2^d}$.

Proof. Multiply one matrix unit from each commuting system to obtain $E_{\alpha\beta}$, with $\alpha,\beta\in\{1,2\}^d$. These satisfy

Fix an index $o$. The space $Q=E_{oo}H$ is nonzero: if $E_{oo}=0$, then $E_{\alpha o}^*E_{\alpha o}=0$, hence every $E_{\alpha\alpha}=E_{\alpha o}E_{o\alpha}=0$, contradicting their sum. The map

is isometric, because the displayed products make different summands orthogonal and preserve each summand's inner product. It is onto, since every $E_{\alpha\alpha}H=E_{\alpha o}Q$. Under $U$, $E_{\alpha\beta}$ is the usual matrix unit tensored with $1_Q$. Thus for every scalar matrix $X=[x_{\alpha\beta}]$,

The upper bound follows by expanding in an orthonormal basis of $Q$; the lower bound follows by testing $v\otimes\eta$ for a fixed unit $\eta\in Q$ and a maximizing finite-dimensional unit vector $v$. This proves an onto isometric matrix identification, including arbitrary representation multiplicity. Formula (6) recovers the original $a_j$, so the two generated algebras agree. For $d=0$, the algebra is the scalar identity algebra. The complete finite-unit construction is in Sections 10.3–10.4. $\square$

Define $\operatorname{CAR}(K)$ using the concrete Fock family and the identity. The theorem gives a canonical embedded $\operatorname{CAR}(E)$ for every finite-dimensional $E$. The union over $E$ is norm dense: every polynomial uses finitely many vectors, and (5) makes approximation of generator vectors norm continuous.

The finite automorphism assertion used in Exercise 9 also follows from matrix units. If $\beta$ is an automorphism of $M_N$, the corner of $\beta(E_{oo})$ is scalar, so its range in $\mathbb C^N$ has dimension one. Choose a unit vector $\eta$ in that range. The vectors $\beta(E_{\alpha o})\eta$ form an orthonormal basis by the matrix-unit products and diagonal sum. The resulting unitary $U\delta_\alpha=\beta(E_{\alpha o})\eta$ satisfies $\beta(E_{\alpha\beta})=UE_{\alpha\beta}U^*$, hence $\beta=\operatorname{Ad}U$.

Any other CAR family induces isomorphisms of these finite matrix algebras by matching generators. The maps are compatible on overlaps, since they match the same polynomials. Their isometric union map extends to an isomorphism onto the generated C*-algebra. This proves representation independence and the CAR universal property without choosing a norm on a free algebra.

Corollary 2.2. The CAR algebra is simple. For separable infinite-dimensional $K$,

For finite-dimensional $K$ it is $M_{2^{\dim K}}$.

Proof. Suppose a closed two-sided ideal $J$ contains $0\ne x$, and put $r=\|x\|$. Choose a local matrix element $y$ with $\|x-y\|<r/16$. Then

In that finite matrix algebra, choose a rank-one projection $e$ onto an eigenvector of $y^*y$ with largest eigenvalue $t=\|y\|^2>3r^2/4$. Finite-dimensional diagonalization is proved in Section 10.7. Hence $ey^*ye=te$, and $h=ex^*xe\in J\cap e\operatorname{CAR}(K)e$ satisfies $\|h-te\|<r^2/4<t$. The corner has identity $e$; the geometric series

converges in it. Thus $e=h^{-1}h\in J$. Within the same local full matrix algebra, choose partial isometries $v_i$ carrying $e$ to its orthogonal rank-one diagonal projections. Their sum gives $1=\sum_i v_i e v_i^*\in J$. Thus every nonzero closed ideal is the whole algebra. This argument also applies when the chosen local algebra is scalar; the full corner estimates are in Sections 10.8–10.9.

For a countable orthonormal basis, the strings (6) identify the first $d$ modes with the first $d$ tensor factors, compatibly under $x\mapsto x\otimes1$. Their dense union gives (7). For an arbitrary nonseparable $K$, the finite-subspace directed limit remains valid; (7) is asserted here only with a countable basis. $\square$

It is the dimension of $K$ that decides the finite versus infinite case. An infinite-dimensional representation space alone does not: amplifying one mode on $\mathbb C^2\otimes\ell^2$ still generates $M_2\otimes1$.

3. Parity, orthogonal subspaces and automorphisms

For a unitary $u$ on $K$, the family $a(uf)$ satisfies (1). The universal property gives an automorphism

If $u_\lambda\to u$ strongly, then

Telescoping finite products proves convergence on polynomials; their norm density and the automorphisms' isometric bounds prove point-norm convergence on the whole algebra. This applies in particular to the strong* topology on the unitary group.

Let $\gamma=\alpha_{-1}$. Its even and odd spaces are

Every element splits by the bounded projections $(\operatorname{id}\pm\gamma)/2$. Polynomials of even, respectively odd, total generator degree are dense in these spaces.

If $L\perp N$, all generators and adjoint generators from the two spaces anticommute. Moving a homogeneous word of degree $p$ past one of degree $q$ gives the sign $(-1)^{pq}$. By polynomial approximation,

Thus even elements commute across the spaces; two odd elements anticommute.

Proposition 3.1. Multiplication gives faithful spatial tensor identifications

and the analogous identification with the first leg even.

Proof. Take finite-dimensional $E\subset L,F\subset N$, and order the $E$-modes before the $F$-modes in (6). On $E\oplus F$, a generator of the second leg becomes

An even second-leg word has an even number of these strings, so is exactly $1\otimes y$. The first leg is $x\otimes1$. To check the norm in arbitrary faithful representations, diagonalize finite Fock parity. If $f=\dim F>0$, its even and odd occupation spaces both have dimension $2^{f-1}$, and the even algebra is

Every faithful representation contains both central blocks. On each block Theorem 2.1's matrix-unit unitary removes only a nonzero multiplicity space; it leaves the matrix norm unchanged. Thus for $z=\sum_k x_k\otimes y_k$, the tensor norm is the maximum of the two finite matrix norms of $\sum_k x_k\otimes y_k^\pm$, independently of those faithful representations. In the parity-string realization these are exactly the two blocks of $\sum_k x_ky_k$. For $f=0$ the second leg is scalar and the same assertion is immediate. This proves injectivity and the exact spatial norm on every finite tensor algebra, rather than only commutation of its generators.

The unions of the finite subalgebras are dense in both legs. For the even leg apply $(\operatorname{id}+\gamma)/2$ to a local polynomial approximation. Approximating both entries in a finite sum of simple tensors, with $\|x\otimes y\|=\|x\|\|y\|$, makes these finite tensor algebras dense in the spatial completion. Their compatible isometries therefore extend onto the algebra generated by multiplication: their ranges contain the two legs and have dense polynomial span. Interchanging the order of the legs proves the other assertion. Section 12 supplies the block and completion details. $\square$

Every word can also be put in normal order by using

Move each creator left; each swap either reduces the number of inversions or removes two letters. Repetition terminates and gives a linear combination of

This includes the identity. Their span is a dense *-subalgebra. A bounded functional, and in particular a state, is determined by its values there.

4. A positive state for every covariance

Let $T\in B(K)$ with $0\le T\le1$. For every finite-dimensional $E\subset K$, its compression $S_E=P_ET|_E$ is a positive contraction. Choose an ordered orthonormal eigenbasis of $S_E$, with eigenvalues $p_1,\ldots,p_d$, and use the finite matrix identification of Section 2 to define the positive occupation density

Its diagonal entries are nonnegative and sum to one, including at $p_j=0,1$; hence $\varphi_E$ is a positive state. No global eigenbasis of $T$, and no reducing condition on $E$, is being assumed.

We will first fit these finite states together. For the later cyclic model put $A=(1-T)^{1/2}$, $B=T^{1/2}$. Their commuting positive roots are supplied by the norm-series construction in Sections 8.1–8.2 of the periodic CAR lesson. On the conjugate Hilbert space, inner products satisfy $\langle\bar f,\bar g\rangle=\langle g,f\rangle$. Let

Use ordinary Fock annihilators and creators on $\mathcal F(\mathcal L)$, denoted by $a_{\mathcal L},c_{\mathcal L}$. Define

Both terms are conjugate-linear in $f$. The two legs are orthogonal, so their mixed CAR terms vanish. The remaining scalar anticommutator is

The anticommutator without an adjoint is zero. Thus (12) is a CAR representation. We will identify its vacuum state with the state constructed from the finite densities.

Theorem 4.1. There is exactly one state with the normal-ordered moments

It is invariant under the gauge action $\alpha_{e^{i\theta}1}$, and hence even. These states are called gauge-invariant quasi-free states.

Proof. Compute first in an eigenbasis $h_1,\ldots,h_d$ of $S_E$. A normal word with a repeated creation or annihilation index is zero. A word with different creation and annihilation index sets changes the occupation set, so has zero diagonal expectation under $\rho_E$. If the sets agree, move each list into increasing order. The two sorting signs are also the row and column signs of the diagonal-covariance determinant. The resulting word is $\prod_{j\in I}a(h_j)^*a(h_j)$, whose expectation is $\prod_{j\in I}p_j$. The empty product is one. Multilinearity, with creation linear and annihilation conjugate-linear, proves (13) on $\operatorname{CAR}(E)$ with entries $\langle g_i,S_Ef_j\rangle=\langle g_i,Tf_j\rangle$.

Formula (11) now makes this finite state independent of the chosen eigenbasis. If $E\subset F$, restriction of $\varphi_F$ to $\operatorname{CAR}(E)$ has exactly those same normal moments and is therefore $\varphi_E$. Any two finite spaces lie in their finite sum. Consequently these states define a well-defined positive functional on the directed local union, with $|\varphi(x)|\le\|x\|$ and $\varphi(1)=1$. Extend it by norm continuity to $\operatorname{CAR}(K)$. If local $x_\lambda\to x$ in norm, then $x_\lambda^*x_\lambda\to x^*x$; positivity of each finite density gives $\omega_T(x^*x)\ge0$. This constructs an actual positive state for arbitrary $K$. Uniqueness follows from (11). Under $\alpha_{e^{i\theta}1}$, a normal word acquires the phase $e^{i(n-m)\theta}$; its value in (13) is unchanged. Thus the state is gauge invariant and, at $\theta=\pi$, even.

It remains to check the doubled model, independently of this existence argument. Second quantization of $e^{i\theta}1_{K_0}\oplus e^{-i\theta}1_{\overline{K_1}}$ fixes the vacuum and multiplies (12) by $e^{-i\theta}$, so its unbalanced moments vanish. In a balanced word, the creator component of the leftmost adjoint in (12) has zero expectation against the left vacuum. The other component is $a_{\mathcal L}(0\oplus\overline{Bf_1})$. Move it to the right; its scalar anticommutators with the remaining adjoints vanish. Against the generator labelled $g_i$, its scalar term is

Moving past the intervening letters gives the sign $(-1)^{i+1}$. The final annihilator kills the right vacuum. This is the expansion along the first column of (13); induction from the empty word proves all its moments. By uniqueness, this vector state is the finite-density state just constructed. The full contraction induction, including unbalanced words, is in Section 11.2. $\square$

The state with $T=0$ is the ordinary annihilation vacuum. With $T=1$ the representation in (12) is the filled, or anti-Fock, representation. For finite-dimensional $E\subset K$, restriction to $\operatorname{CAR}(E)$ has covariance $P_ET|_E$; $E$ need not reduce $T$. This follows directly by restricting the entries of (13).

The vacuum is cyclic for (12), including when $T$ has a kernel or a one-eigenspace. Indeed the one-particle creation components of its generators and adjoints span densely

Products of such operators applied to the vacuum give the desired wedge of their creation components plus terms of smaller particle degree. Inductively all those smaller-degree terms lie in the closed cyclic subspace; subtracting them puts the wedge there too. By continuity and the occupation basis, these wedges span all of $\mathcal F(\mathcal L)$. On the state quotient, the map $[x]\mapsto\pi_T(x)\Omega$ preserves inner products $\omega_T(x^*y)$, kills exactly the null space, and intertwines multiplication on its whole dense domain. Its extension to the completed quotient is an onto unitary by the just proved cyclicity. Thus (12) with its vacuum is the completed state GNS representation. This also proves the asserted identification at the covariance endpoints; Sections 11.3–11.4 give the full dense-degree and completion estimates.

5. Why the quasi-free representation is a factor

The implication from cyclic irreducibility to purity used in Exercise 8 can be proved on the same completed quotient. If a positive functional $\psi$ satisfies $0\le\psi\le\omega_T$, its form $B([x],[y])=\psi(x^*y)$ is well-defined and bounded there: positive-form Cauchy–Schwarz gives $|B([x],[y])|^2\le\omega_T(x^*x)\omega_T(y^*y)$. Riesz representation produces $0\le D\le1$ with $B(\xi,\eta)=\langle\xi,D\eta\rangle$. The equality $B(\pi_T(a^*)\xi,\eta)=B(\xi,\pi_T(a)\eta)$ for every algebra element $a$ makes $D$ commute with the representation. If it is irreducible, $D=\lambda1$, so $\psi=\lambda\omega_T$. Applied to the dominated functional $t\psi_1$ in a nontrivial state decomposition $\omega_T=t\psi_1+(1-t)\psi_2$, evaluation at $1$ gives $\lambda=t$ and $\psi_1=\psi_2=\omega_T$. Thus the state is pure. The bounded-form and Riesz constructions are supplied in Section 18.5 and ITP Section 10.2; this implication needs no converse criterion.

Let $M_T=\pi_T(\operatorname{CAR}(K))''$ on this cyclic Fock space. The following argument does not require diagonalizing $T$, a separable $K$, or a faithful state.

Split off the endpoint spaces and write

Both $A$ and $B$ preserve this middle space. For $f\in K_{\rm m}$ define the complementary field

Every field in (12), and its adjoint, anticommutes with every $b(f)$ and $b(f)^*$. For the mixed adjoint anticommutator, the only possible scalar terms cancel because

here $AB=BA$. All other terms vanish by orthogonality of the legs and the CAR.

Let $P$ be Fock parity, acting as $(-1)^n$ on $\Lambda^n\mathcal L$. It anticommutes with every annihilator and creator, so each $Pb(f)$ commutes with (12) and its adjoint. Thus

The untwisted families $\pi_T(a(g))$ and $b(f)$ together generate all ordinary Fock fields. On the middle space,

The cancellation again uses $AB=BA$, and the remaining coefficients are $A^2+B^2=1$. On $\ker T$, the first-leg field is already (12); on $\ker(1-T)$, the second-leg creator is already (12). These spaces and the middle space span the active legs $\mathcal L$. Lemma 1.1 therefore makes the joint untwisted family irreducible.

Theorem 5.1. $M_T$ is a factor for every positive contraction $T$.

Proof. Parity preserves $M_T$, since it sends each generator to its negative, and hence preserves its center. If a central $x$ is even, it commutes with $P$. It also commutes with every $Pb(f)$, because $x\in M_T$ and (15) holds. These two commutations imply that $x$ commutes with every $b(f)$. Centrality supplies commutation with (12). Irreducibility of the joint family makes $x$ scalar.

Every central element splits into its even and odd parts under $x\mapsto PxP$. If the odd part of the center were nonzero, one of its real or imaginary parts would give a nonzero selfadjoint odd central $z$. Its square is even and central, hence a scalar $c1$, with $c=\|z\|^2>0$. Rescale so that $z^2=1$. For any odd central $w$, the product $zw$ is even and scalar, so $w$ is a scalar multiple of $z$. The center would therefore be exactly $\mathbb C1+\mathbb Cz$.

Let $V_\theta=e^{i\theta}1_{K_0}\oplus e^{-i\theta}1_{\overline{K_1}}$ and let $U_\theta=\Gamma(V_\theta)$ be its degreewise second quantization. On a finite occupation vector with $r$ first-leg and $s$ second-leg entries, $U_\theta$ has scalar $e^{i(r-s)\theta}$. Continuity on finite linear combinations and their density, with $\|U_\theta\|=1$, prove strong continuity on the entire Fock space, without a countability assumption on its basis. The defining creator formula gives $U_\theta\pi_T(a(f))U_\theta^*=e^{-i\theta}\pi_T(a(f))$, so conjugation preserves $M_T$ and its center; it commutes with parity. It is concretely normal, since conjugation transforms each summand in a square-summable vector-coefficient functional into another such summand.

For the normalized odd $z$, conjugation must therefore give $U_\theta zU_\theta^*=\varepsilon(\theta)z$, where $\varepsilon(\theta)\in\{1,-1\}$: the odd central space is one-dimensional, and selfadjointness and $z^2=1$ force this sign. Strong continuity at zero makes $\varepsilon(\theta)=1$ for all sufficiently small $|\theta|$. Indeed, for a fixed nonzero $\xi$, the alternative sign would make $\|(U_\theta zU_\theta^*-z)\xi\|=2\|\xi\|$. For any real $\theta$, choose an integer $n$ with $\theta/n$ in that neighbourhood. The group law gives $\varepsilon(\theta)=\varepsilon(\theta/n)^n=1$. But $V_\pi=-1$ on both active legs, so $U_\pi=P$, which sends the odd $z$ to $-z$. This contradiction removes the odd center. The even center was scalar, so $\mathcal Z(M_T)=\mathbb C1$. Section 13 supplies the full endpoint and complementary-field verification. $\square$

The twisted fields were used only to control the center. We have not assumed, or needed, that they generate the whole commutant.

6. Finite-dimensional approximation and product states

Theorem 6.1. For separable $K$, every representation of $\operatorname{CAR}(K)$ generates an AFD von Neumann algebra. In particular every $M_T$ is an AFD factor. For arbitrary $K$, the same proof gives an increasing directed family of finite-dimensional subalgebras with strong* dense union.

Proof. For any representation $\pi$ put $e=\pi(1)$ and

On $eH$, Theorem 2.1 gives a finite full matrix algebra; the complementary block is scalar. Thus every $D_E$ is a finite-dimensional unital algebra on the full representation space. The matrix-unit coefficients recover its finitely many entries, so a WOT limit of elements of $D_E$ again lies in $D_E$. These are finite-dimensional von Neumann subalgebras. They increase with $E$. Polynomial norm density and continuity of $\pi$ give $A_0''=\pi(\operatorname{CAR}(K))''$; adjoining $1-e$ adds no commutation condition, since a commutant already commutes with $e$.

Here is the bounded approximation step. The finite-vector amplification proof of the bicommutant theorem in ITP Section 10.5 makes $A_0$ WOT dense in $A_0''$. Apply it simultaneously to the finite lists $x\xi_j$ and $x^*\xi_j$. Their image set as $x$ runs over $A_0$ is a real linear subspace of a finite Hilbert direct sum. If the desired tuple were outside its norm closure, orthogonal projection would separate it by a real continuous functional. That functional is a finite sum of WOT coefficients of $x$ and $x^*$, hence is WOT continuous as a real functional. This would contradict WOT density. Consequently every $z\in A_0''$ has a net $x_\lambda\in A_0$ tending strongly* to it.

To retain an exact norm bound, use

The geometric inverse stays in the same complete local algebra as $x$, and

Thus $F(x)$ is a contraction. For $R_z=(1+z^*z)^{-1}$, the inverse identity in Section 14.2 gives

The same inequality for adjoints proves strong* continuity at each fixed $z$, even for nets with no common norm bound. If $y\in A_0''$ is a contraction, put $s=(1-y^*y)^{1/2}$ and $G(y)=y(1+s)^{-1}$. Commuting multiplication gives $1+G(y)^*G(y)=2(1+s)^{-1}$, hence $F(G(y))=y$. Choose the preceding local strong* net toward $G(y)$; its images under $F$ are local contractions tending strongly* to $y$. Scaling proves bounded strong* density at every radius.

For separable $K$, choose finite-dimensional $E_n$ increasing to a dense subspace. Their generator union is norm dense by (5), so the same proof uses the increasing sequence $D_{E_n}$. This is AFD by a sequence of finite algebras; the approximating operators can still require nets when the representation space is nonseparable. For arbitrary $K$, all finite subspaces give the asserted directed family. When $e=0$, all the $D_E$ are scalar; when $H=0$, the assertion is immediate. Factorality for $\omega_T$ is Theorem 5.1. The complete approximation argument is in Section 14. $\square$

If $K=\bigoplus_jK_j$ and $T$ reduces each $K_j$, (13) proves the full finite-block product identity

To see it, put each local polynomial in normal order and expand into basis vectors in its own block. Cross-block contractions vanish. A nonzero determinant term requires equal numbers of creators and annihilators in every block; then each block word is even. Regrouping those balanced blocks contributes no sign, and the determinant splits into their determinants. If some block is unbalanced, both sides vanish. Density extends (17) to arbitrary local elements. Thus reducing spectral blocks give product states. Merely pairwise independence would be weaker than (17).

For a finite-dimensional subspace $E$, use the eigenvalues $p_1,\ldots,p_d$ of its compression $S_E=P_ET|_E$. The occupation basis gives the density matrix

Here $\rho_T$ denotes the density of the restriction to $\operatorname{CAR}(E)$ in this finite basis. The entire diagonal normal-word computation was used in Theorem 4.1; it proves (18) even when $E$ does not reduce $T$. When a family of spaces does reduce $T$, (17) additionally gives compatibility with its block product.

For a countable eigenbasis the compatible Jordan–Wigner identifications (7), together with (18), give a normal GNS identification

For completeness, set $\rho_j=\operatorname{diag}(1-p_j,p_j)$, let $s_j$ be its support projection, and take the finite Hilbert space

with left matrix action. Its cyclic span is all $H_j$: every supported column matrix equals $x\rho_j^{1/2}$, by inversion only on the positive diagonal entries. This includes the rank-one endpoint supports. Complete the increasing finite tensor spaces under the embeddings $\xi\mapsto\xi\otimes\Omega_{n+1}$. For local $x,y$, (18) gives the exact inner product

The map from the CAR cyclic quotient $[x]$ to $x\Omega$ is therefore isometric, respects all compatible local embeddings, and intertwines every generator. Local cyclicity gives dense range in the pointed completion, so its extension is an onto unitary from the whole completed GNS space. By boundedness and norm density it intertwines every CAR element. Taking commutants twice gives (19) on every element of the generated von Neumann algebras, with inverse. Both directions are concretely normal: conjugation carries a square-summable vector-coefficient functional to the series with both vector lists conjugated, preserving their sums of squared norms. Sections 15.3–15.8 give the complete bridge, including finite and endpoint cases.

At $T=\tfrac12 1$, every finite restriction is normalized matrix trace. A finite matrix trace is unique: traciality makes all diagonal matrix-unit values equal and makes every off-diagonal value zero. Norm density therefore makes $\omega_{1/2}$ the unique tracial state on the CAR algebra. Its vector extension $\tau(x)=\langle\Omega,x\Omega\rangle$ on $M_{1/2}$ is tracial on every pair $x,y$: choose uniformly bounded local strong* nets for both; the estimate

and its adjoint counterpart pass their finite trace identity to the joint directed limit. The vector functional is concretely normal. It is faithful: if $\tau(x^*x)=0$, then for $y\in M$,

Cyclicity gives $x=0$. The displayed inequality does not assert positivity of $x^*xyy^*$: traciality rewrites its value as $\tau(xyy^*x^*)$, and $0\le xyy^*x^*\le\|y\|^2xx^*$. If a partial isometry had initial $1$ and proper final projection $q<1$, traciality would give $\tau(1-q)=0$, contradicting faithfulness. Thus the factor is finite.

To justify its type when $K$ is infinite-dimensional, suppose it had a minimal projection $q$. Positive parts and their support projections make $qMq=\mathbb Cq$. Every nonzero projection $r$ then contains a projection equivalent to $q$: factor contact gives $0\ne rMq$, and polar decomposition of such an element has initial projection $q$ and final projection below $r$. A maximal orthogonal family of copies of $q$ has at most $1/\tau(q)$ members, since each has the same strictly positive trace. Its complement must vanish by the preceding contact argument. Partial isometries between this finite family give matrix units; every corner between two members is scalar because $qMq=\mathbb Cq$. Hence $M$ would be a finite full matrix algebra. This contradicts its local subalgebras $M_{2^d}$ for arbitrarily large $d$. An abelian projection in a factor would have scalar corner, because that corner is itself a factor, and hence would be minimal. There are therefore no nonzero abelian projections, while $1$ is finite. This is exactly type $\mathrm{II}_1$. The projection, contact and finite-matrix arguments are fully supplied in Section 16. They use no separability of $K$. In the separable infinite-dimensional case (19) is exactly the tracial infinite product of $M_2$.

For $T=p1$, $0<p<1$, on separable infinite-dimensional $K$, we now prove directly that

Use (19) with $\rho=\operatorname{diag}(1-p,p)$, and denote the pointed product algebra by $M$. Its left and right finite matrix orbits of $\Omega=\bigotimes\rho^{1/2}$ are dense, since $\rho$ is invertible. Let $X_j=E_{12}^{(j)}$ be the left matrix in coordinate $j$; in the CAR model it is the parity-corrected mode $b_j$ of (6), not the odd mode with its prefix string. Right multiplication in that coordinate by $\rho^{-1/2}E_{12}\rho^{1/2}$ gives an operator $R_j\in M'$ with

For $u\in M$, choose a left-local $z$ with $\|(u-z)\Omega\|<\varepsilon$. Beyond its finite support, $z$ commutes with $X_j$, so

On every right-local $r\Omega$, the commutator equals $r[u,X_j]\Omega$. These vectors are dense, and $\|[u,X_j]\|\le2\|u\|$; therefore $[u,X_j]\to0$ strongly on the entire Hilbert space. The same argument for $X_j^*$ gives the adjoint limit. On pairs of finite-stage vectors, the untouched reference tail gives

Their norm-one bounds extend these coefficient limits to arbitrary vectors.

Suppose $M$ had a nonzero finite projection $e$. The corner $eMe$ is a finite factor on the separable space $eH$. The complete finite-factor construction in Sections 18–19 supplies a faithful concretely normal tracial state $\vartheta$ on this corner, with $\vartheta(e)=1$; no trace on all of $M$ is assumed. Put $B_j=eX_je$. The strong commutator limits with $u=e$ give