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CAR Algebras, Majorana Generators, and Clifford Comparison

1 · Prerequisites

2 · Summary

This page fixes the complex CAR relations and the plus- $Q$ Clifford convention. It proves the finite-mode CAR algebra is $M_{2^{n}} (C)$ , gives the inverse Majorana maps, and constructs CAR C*-algebras over arbitrary complex Hilbert spaces as directed limits of finite-dimensional matrix algebras.

For the standard countable mode set, the occupation representation is irreducible, and composing it with particle-hole exchange gives an inequivalent irreducible representation. This proves inequivalent irreducible representations may occur. The abstract Hilbert-space Fock existence proof retains Countable Choice for the Hilbert tensor/Fock construction and adjoints; the countable occupation proof is a separate choice-free branch. Neither branch classifies all representations or proves a QFT spin-statistics theorem.

The norm completion defining the CAR algebra over an arbitrary Hilbert space assumes Countable Choice. The countable-mode CAR completion stays choice-free because its explicit rational finite-matrix subalgebra is countable and dense.

Reading route

Read clifford-universal-algebras-and-pbw first for the plus- $Q$ Clifford quotient, its universal property, and the ordered PBW basis. This is a page-level reading route; each item lists its actual logical suppliers.

Local results

Finite-Mode Complex CAR Algebra defines the algebraic universal presentation.
Finite-Dimensional CAR Algebra over an Inner-Product Space, Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra, and Finite-Subspace CAR Inclusions Are Isometric extend the finite matrix model and isometric embeddings to every finite-dimensional subspace.
CAR C*-Algebra over a Complex Hilbert Space and CAR over a Hilbert Space Has the Universal Inductive-Limit Norm define and prove the universal inductive-limit norm over an arbitrary complex Hilbert space. Fermionic Fock Space over a Complex Hilbert Space and Fock Creation and Annihilation Operators Satisfy the CAR supply its Fock representation under Countable Choice by explicit normalized-wedge and pairing transport from the canonical mathematical result Bosonic and fermionic Fock operators with exact domains.
Square-Summable Complex Families Form a Hilbert Space proves the occupation-space $ℓ^{2}$ completion is Hilbert without Countable Choice, which keeps the explicit countable representation branch choice-free.
Finite-Mode CAR and the Complex Clifford Algebra via Majorana Generators proves the mutually inverse Majorana/Clifford maps with $γ_{2 j - 1} = a_{j} + a_{j}^{†}$ and $γ_{2 j} = i (a_{j}^{†} - a_{j})$ .
Finite CAR Occupation Operators Give All Matrix Units constructs the finite occupation representation and all matrix units; The Finite-Mode CAR Algebra Is a Full Matrix Algebra proves the universal finite C*-completion is the full matrix algebra.
Canonical Finite-Mode CAR Inclusions Are Isometric, Countably Infinite-Mode CAR C*-Algebra, and The Countable CAR Algebra Has the Universal Inductive-Limit Norm give the explicit standard countable construction and its choice-free occupation representation.
The Countable-Mode CAR Fock Representation Is Irreducible, Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR, and Countable-Mode CAR Has Inequivalent Irreducible Representations establish the explicit pair of inequivalent irreducible representations.

The companion page contains checked one- and two-mode calculations and the countable-mode counterexample.

Companion: car-algebras-majorana-generators-and-clifford-comparison-examples.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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Fermionic Fock Space over a Complex Hilbert Space

Definition

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), and let $H$ be a complex Hilbert space. For each $k \geq 0$ , let $Λ^{k} H$ be the Hilbert exterior power with the Gram-determinant inner product and wedge convention of Hilbert exterior powers and induced operators, with $Λ^{0} H = C$ . The algebraic direct sum $X_{0} = ⨁_{k \geq 0}^{a l g} Λ^{k} H$ , where distinct particle-number summands are orthogonal and the inner product on a finite sum is the sum of the component inner products, is an inner-product space. Its norm completion exists under Countable Choice by Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences and has its compatible Banach-space and Hilbert-space structures by The metric completion of a normed space carries a unique compatible Banach-space structure and The norm completion of an inner-product space is a Hilbert space. The fermionic Fock space over $H$ is this Hilbert completion. The vacuum vector $Ω$ is the image of $1 \in Λ^{0} H$ in the completion.

On finite sums of decomposable wedges define the creation rule $C (f) ξ = f \land ξ$ for $f \in H$ . This rule is initially defined on a dense subspace. Its bounded extension and the CAR relations for $C (f)$ and its Hilbert-space adjoint are proved in Fock Creation and Annihilation Operators Satisfy the CAR.

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Finite-Mode Complex CAR Algebra

Definition

Fix an integer $n \geq 0$ . Let $F_{n}$ be the complex vector space with basis all finite words in the symbols $a_{1}, \dots, a_{n}, a_{1}^{†}, \dots, a_{n}^{†}$ , including the empty word $1$ . Give $F_{n}$ concatenation as multiplication and the conjugate-linear involution that reverses each word and exchanges $a_{j}$ with $a_{j}^{†}$ . Thus $F_{n}$ is the free unital complex $*$ -algebra on $a_{1}, \dots, a_{n}$ .

For $x, y$ in an algebra write ${x, y} = x y + y x$ . Let $J_{n}$ be the smallest two-sided $*$ -ideal in $F_{n}$ containing

${a_{i}, a_{j}}, {a_{i}^{†}, a_{j}^{†}}, {a_{i}, a_{j}^{†}} - δ_{i j} 1 (1 \leq i, j \leq n) .$

Here a two-sided $*$ -ideal is a complex linear subspace closed under left and right multiplication and under $*$ . It exists as the intersection of all such ideals containing these relations; the collection is nonempty because it contains $F_{n}$ . The relations are $*$ -stable, since taking adjoints exchanges the first two families and changes the third relation with indices $i, j$ to the same relation with indices $j, i$ . Define the algebraic finite-mode CAR $*$ -algebra by ${C A R}_{n}^{a l g} = F_{n} / J_{n}$ , with generators denoted by the same symbols. Its quotient map gives the universal property: every tuple in a unital complex $*$ -algebra satisfying these relations induces a unique unital $*$ -homomorphism from ${C A R}_{n}^{a l g}$ . For $n = 0$ there are no generators or relations and ${C A R}_{0}^{a l g} = C$ .

This definition is algebraic; the finite-mode universal C*-completion is identified in The Finite-Mode CAR Algebra Is a Full Matrix Algebra.

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Square-Summable Complex Families Form a Hilbert Space

Statement

For every set $I$ , the complex square-summable family space $ℓ^{2} (I, C)$ with its canonical norm and inner product from Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ is complete. Thus it is a complex Hilbert space. Its finitely supported families are dense, and the coordinate vectors form an orthonormal basis. This result uses no choice principle.

Facts & Assumptions

Given: A set $I$ and a Cauchy sequence $(x^{(m)})_{m \in N}$ in $ℓ^{2} (I, C)$ .

[F1]

For $x \in ℓ^{2} (I, C)$ , $∥ x ∥_{2}^{2} = \sup_{F \subseteq I, F finite} \sum_{i \in F} ∣ x_{i} ∣^{2}$ , and the canonical inner product is defined by the absolutely summable family $\sum_{i \in I} x_{i} \overline{y_{i}}$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[F2]

The complex plane is complete: every Cauchy sequence in $C$ has a unique limit (The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts).

[F3]

A complex Hilbert space is a complete complex inner-product space (Hilbert space).

[A1]

Replacement forms the image of a set under a uniquely defined class function (The Axiom Schema of Replacement: for each formula $φ$ , if $φ$ defines a class function on $A$ then its image on $A$ is a set).

Proof

1.1givenA1F1F2construct

For each $i \in I$ , the singleton finite-sum bound gives $∣ x_{i}^{(m)} - x_{i}^{(n)} ∣ \leq ∥ x^{(m)} - x^{(n)} ∥_{2}$ , so $(x_{i}^{(m)})_{m}$ is Cauchy in $C$ . By [F2] it has a unique limit $x_{i}$ . These uniquely determined coordinates form a function $x : I \to C$ by [A1]; no choices are made.

1.2F1choosealgebra

Let $x \in ℓ^{2} (I, C)$ and $ε > 0$ . By [F1], $s = ∥ x ∥_{2}^{2}$ is the supremum of its finite coordinate sums, so there is a finite $F \subseteq I$ with $\sum_{i \in F} ∣ x_{i} ∣^{2} > s - ε^{2} / 4$ . For every finite $G \subseteq I ∖ F$ , disjointness gives $\sum_{i \in G} ∣ x_{i} ∣^{2} \leq s - \sum_{i \in F} ∣ x_{i} ∣^{2} < ε^{2} / 4$ . Taking the supremum over such $G$ yields $∥ x - x ∣_{F} ∥_{2} \leq ε / 2 < ε$ , proving finite-support families are dense. The coordinate vectors have inner products $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ by the canonical inner product in [F1], and their finite span is exactly the finitely supported families, so they form an orthonormal basis.

2.1F1step 1.1algebra

Since $(x^{(m)})$ is Cauchy, choose $N$ such that $∥ x^{(m)} - x^{(n)} ∥_{2} < 1$ whenever $m, n \geq N$ . Then $∥ x^{(m)} ∥_{2} \leq ∥ x^{(N)} ∥_{2} + 1$ for $m \geq N$ ; the finitely many earlier norms have a finite maximum, so there is $M < \infty$ with $∥ x^{(m)} ∥_{2} \leq M$ for every $m$ . For any finite $F \subseteq I$ , coordinate convergence and continuity of the finite sum give $\sum_{i \in F} ∣ x_{i} ∣^{2} = \lim_{m} \sum_{i \in F} ∣ x_{i}^{(m)} ∣^{2} \leq M^{2}$ . Taking the supremum over finite $F$ in [F1] shows $x \in ℓ^{2} (I, C)$ .

3.1F1step 1.1step 2.1algebra

Given $ε > 0$ , choose $N$ so that $∥ x^{(m)} - x^{(n)} ∥_{2} < ε / 2$ for all $m, n \geq N$ . Fix $m \geq N$ . For every finite $F \subseteq I$ , coordinate convergence yields $\sum_{i \in F} ∣ x_{i}^{(m)} - x_{i} ∣^{2} = \lim_{n} \sum_{i \in F} ∣ x_{i}^{(m)} - x_{i}^{(n)} ∣^{2} \leq ε^{2} / 4$ . Taking the supremum over finite $F$ gives $∥ x^{(m)} - x ∥_{2} \leq ε / 2 < ε$ . Hence $x^{(m)} \to x$ in $ℓ^{2} (I, C)$ and every Cauchy sequence converges.

4.1F1F3step 3.1∎

By [F1], $ℓ^{2} (I, C)$ is an inner-product space, and step 3.1 proves it is complete. It is therefore a Hilbert space by [F3]. The proof used no family of choices.

Remarks

The proof works for arbitrary index sets, including $I = \emptyset$ , and does not require a countable enumeration or Countable Choice.

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Finite-Dimensional CAR Algebra over an Inner-Product Space

Definition

Let $F$ be a finite-dimensional complex inner-product space, with the pairing linear in its first variable and conjugate-linear in its second. The algebraic CAR $*$ -algebra over $F$ , denoted ${C A R}^{a l g} (F)$ , is generated by symbols $c (f)$ , one for each $f \in F$ , subject to complex linearity in $f$ and the relations

${c (f), c (g)} = 0, {c (f)^{*}, c (g)^{*}} = 0, {c (f)^{*}, c (g)} = ⟨ g, f ⟩ 1 (f, g \in F),$

where ${x, y} = x y + y x$ . Precisely, take the free unital complex $*$ -algebra on symbols $c (f)$ , with formal adjoints, and quotient by the two-sided $*$ -ideal generated by these relations and the linearity relations. The quotient has the universal property that every complex-linear family $C (f)$ in a unital complex $*$ -algebra satisfying the displayed relations induces a unique unital $*$ -homomorphism from ${C A R}^{a l g} (F)$ .

If $e_{1}, \dots, e_{n}$ is an orthonormal basis of $F$ , the elements $a_{j} : = c (e_{j})^{*}$ and $a_{j}^{†} : = c (e_{j})$ satisfy the finite-mode relations of Finite-Mode Complex CAR Algebra. The algebraic definition does not choose a basis; its finite-dimensional matrix realization is established in Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra.

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Finite CAR Occupation Operators Give All Matrix Units

Statement

For $n \geq 0$ , let $P_{n}$ be the set of subsets of ${1, \dots, n}$ , and let $F_{n} = ℓ^{2} (P_{n})$ with orthonormal basis $e_{S}$ for $S \in P_{n}$ . For $1 \leq j \leq n$ define operators on this finite-dimensional Hilbert space by

$c_{j} e_{S} = {\begin{cases} 0 & j \notin S, \\ (- 1)^{∣ {s \in S : s < j} ∣} e_{S ∖ {j}} & j \in S, \end{cases} c_{j}^{†} e_{S} = {\begin{cases} (- 1)^{∣ {s \in S : s < j} ∣} e_{S \cup {j}} & j \notin S, \\ 0 & j \in S . \end{cases}$

Then $c_{j}^{†}$ is the Hilbert-space adjoint of $c_{j}$ , and $a_{j} \mapsto c_{j}$ , $a_{j}^{†} \mapsto c_{j}^{†}$ defines a $*$ -representation $π_{n} : {C A R}_{n}^{a l g} \to End (F_{n})$ . Put $N_{j} = c_{j}^{†} c_{j}$ and $P_{\emptyset} = \prod_{j = 1}^{n} (I - N_{j})$ , with the empty product equal to $I$ . For $S = {i_{1} < \dots < i_{p}}$ and $T = {j_{1} < \dots < j_{q}}$ , set

$E_{S, T} = c_{i_{1}}^{†} \dots c_{i_{p}}^{†} P_{\emptyset} c_{j_{q}} \dots c_{j_{1}},$

with empty operator products equal to $I$ . Then $E_{S, T} e_{U} = δ_{T, U} e_{S}$ . Consequently the image of $π_{n}$ is all of $End (F_{n}) ≅ M_{2^{n}} (C)$ .

Facts & Assumptions

Given: An integer $n \geq 0$ , the finite-mode algebra ${C A R}_{n}^{a l g}$ , and the finite set $P_{n}$ .

[F1]

The algebra ${C A R}_{n}^{a l g}$ is generated by $a_{j}, a_{j}^{†}$ subject to ${a_{i}, a_{j}} = 0$ , ${a_{i}^{†}, a_{j}^{†}} = 0$ , and ${a_{i}, a_{j}^{†}} = δ_{i j} 1$ (Finite-Mode Complex CAR Algebra).

Proof

1.1givenconstructalgebra

On basis vectors, $c_{j}^{†}$ reverses exactly the same basis transition as $c_{j}$ and has the same real sign. Hence its matrix is the transpose-conjugate matrix of $c_{j}$ . Each of $c_{j}^{2}$ and $(c_{j}^{†})^{2}$ vanishes because after one allowed toggle, the second toggle of the same type is forbidden. On a basis vector containing $j$ , $c_{j}^{†} c_{j}$ is the identity and $c_{j} c_{j}^{†}$ is zero; when $j$ is absent these roles reverse. Thus ${c_{j}, c_{j}^{†}} = I$ .

2.1givenF1step 1.1algebra

Fix $j < k$ . Whether the $j$ -toggle is allowed depends only on membership of $j$ , and whether the $k$ -toggle is allowed depends only on membership of $k$ . Therefore, if either toggle is forbidden, both orders of the two distinct toggles give zero. If both are allowed, performing the $j$ -toggle first changes the number of occupied indices below $k$ by one, while the $k$ -toggle does not change the number below $j$ . The signs in the displayed formulas therefore differ by $- 1$ when the order is reversed. This proves every distinct-mode anticommutator among $c_{j}, c_{j}^{†}, c_{k}, c_{k}^{†}$ is zero. Along with step 1.1 these are exactly [F1], so the specified map extends through the quotient to a $*$ -representation $π_{n}$ .

2.2givenstep 1.1constructalgebra

The operator $N_{j}$ is diagonal in the basis $e_{S}$ , with eigenvalue $1$ when $j \in S$ and $0$ otherwise. Hence the commuting product $P_{\emptyset}$ is the projection onto $C e_{\emptyset}$ . For $T = {j_{1} < \dots < j_{q}}$ , the right-hand product $c_{j_{q}} \dots c_{j_{1}}$ removes the elements of $T$ in increasing order; on $e_{T}$ each sign is $+ 1$ , and on $e_{U}$ with $T ⊊ U$ its result is a scalar multiple of $e_{U ∖ T}$ . If $T \subseteq̸ U$ it gives zero. The middle projection therefore kills every $e_{U}$ except $e_{T}$ , and sends $e_{T}$ to $e_{\emptyset}$ . The left-hand creation product sends $e_{\emptyset}$ to $e_{S}$ , again with sign $+ 1$ because its indices are applied from largest to smallest. Thus $E_{S, T} e_{U} = δ_{T, U} e_{S}$ .

3.1step 2.2algebra∎

The operators $E_{S, T}$ are all standard matrix units relative to the basis $(e_{S})_{S \in P_{n}}$ , so their linear span is $End (F_{n})$ . Each lies in the $*$ -algebra generated by the $c_{j}, c_{j}^{†}$ , proving surjectivity. For $n = 0$ , the sole basis vector is $e_{\emptyset}$ , $P_{\emptyset} = I$ , and $E_{\emptyset, \emptyset} = I$ , so the same conclusion holds.

Remarks

The construction uses only the finite occupation basis. It does not import an infinite-particle-domain argument or a choice principle.

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Fock Creation and Annihilation Operators Satisfy the CAR

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space with inner product linear in the first variable, and let $F_{-} (H)$ be its fermionic Fock space. The creation rule $C (f) ξ = f \land ξ$ extends to a bounded operator on $F_{-} (H)$ with $∥ C (f) ∥ = ∥ f ∥$ . Writing $A (f) = C (f)^{*}$ for its adjoint, one has, for all $f, g \in H$ ,

${C (f), C (g)} = 0, {A (f), A (g)} = 0, {A (f), C (g)} = ⟨ g, f ⟩ I .$

Thus $f \mapsto C (f)$ is a bounded complex-linear CAR family in the convention of Finite-Dimensional CAR Algebra over an Inner-Product Space.

Facts & Assumptions

Given: Countable Choice, a complex Hilbert space $H$ , its fermionic Fock space, and vectors $f, g \in H$ .

[F1]

The Fock space is the Hilbert completion of the orthogonal direct sum of the Hilbert exterior powers; the vacuum is $1 \in Λ^{0} H$ (Fermionic Fock Space over a Complex Hilbert Space).

[F2]

The local tensor powers use the product pairing, their antisymmetrizers are the signed permutation averages $A_{n}$ , and $x_{1} \land \dots \land x_{n} = \sqrt{n!} A_{n} (x_{1} \otimes \dots \otimes x_{n})$ . Wedges of decomposable vectors have inner product $⟨ x_{1} \land \dots \land x_{k}, y_{1} \land \dots \land y_{k} ⟩ = \det [⟨ x_{i}, y_{j} ⟩]$ (Hilbert exterior powers and induced operators).

[F3]

The inner product on $H$ is linear in its first variable and conjugate-linear in its second (Real and complex inner-product spaces and their induced length).

[A1]

Countable Choice is assumed for the canonical Hilbert tensor/Fock construction and Hilbert-space adjoints (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Schwartz/distribution spaces and Hilbert Fock conventions, The Hilbert-space adjoint of a bounded operator).

[F4]

Under [A1], every bounded operator on a Hilbert space has a bounded adjoint, and $∥ T^{*} ∥ = ∥ T ∥$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F5]

Under Countable Choice, the canonical fermionic Fock creation and annihilation operators are bounded adjoints with norms $∥ f ∥$ and satisfy the CAR on the whole Fock space, using the second-variable-linear pairing $⟨ f, g ⟩_{2}$ (Bosonic and fermionic Fock operators with exact domains).

[F6]

The canonical Fock space is the square-summable direct sum of antisymmetric tensor sectors; its pairing is $⟨ f, g ⟩_{2} = ⟨ g, f ⟩$ , its vacuum is $(1, 0, \dots)$ , and sector creation is $c^{†} (f) ξ = \sqrt{n + 1} A_{n + 1} (f \otimes ξ)$ (Schwartz/distribution spaces and Hilbert Fock conventions).

[F7]

Completed Hilbert tensor products have the product pairing and dense elementary tensors; completions are identified by the unique surjective linear isometry fixing the dense algebraic tensor product (Hilbert tensor construction: inner product, completion and product bases).

Proof

1.1givenA1F1F2F3F5F6F7construct

Reverse the canonical second-variable pairing as in [F6], so all tensor pairings are linear in the first variable. The identity on elementary tensors preserves the product pairings of [F2, F7]; by their dense spans and completion uniqueness it extends, degree by degree, to a surjective linear isometry between the local and canonical tensor powers. It intertwines every permutation on the dense elementary tensors and hence the finite averages $A_{n}$ , so it restricts to an isometry $J_{n}$ of their antisymmetric sectors; set $J_{0} = I_{C}$ . Their orthogonal direct sum is an isometry on finite sector sums. The local Fock space is its completion by [F1], and finite sector sums are dense in the canonical square-summable Fock space by [F5, F6]. Thus it extends to a surjective linear isometry $J$ between the Fock spaces, taking the local vacuum to the canonical vacuum.

2.1F1F2F6step 1.1algebra

The permutation averages satisfy $A_{n + 1} (I \otimes A_{n}) = A_{n + 1}$ : applying any permutation of the last $n$ factors before $A_{n + 1}$ multiplies it by that permutation's sign, and the signed average cancels this sign. Hence, for $n \geq 1$ , on a decomposable $n$ -wedge $ξ = x_{1} \land \dots \land x_{n}$ , the normalization in [F2] gives $c^{†} (f) J_{n} ξ = \sqrt{n + 1} \sqrt{n!} A_{n + 1} (f \otimes A_{n} (x_{1} \otimes \dots \otimes x_{n})) = J_{n + 1} (f \land ξ)$ by [F6]. For $n = 0$ , both sides send the scalar vacuum to $f$ . Thus the local creation rule on its dense decomposable subspace is $J^{- 1} c^{†} (f) J$ .

3.1givenA1F1F4F5F6step 1.1step 2.1algebra∎

By [F5] and step 2.1, $C (f)$ extends to $J^{- 1} c^{†} (f) J$ with norm $∥ f ∥$ . The adjoint identity and uniqueness in [F4] give $A (f) = J^{- 1} c (f) J$ . Transporting the canonical CAR of [F5] through $J$ gives the two zero anticommutators and ${A (f), C (g)} = ⟨ f, g ⟩_{2} I = ⟨ g, f ⟩ I$ by [F6]. Creation is complex-linear in $f$ on the dense wedge subspace and therefore after bounded extension. This includes $f = 0$ and $H = {0}$ , whose Fock space retains the vacuum line.

Remarks

This proof transports the canonical published mathematical Fock result through the explicit normalization and pairing identification; it does not reprove that supplier. The annihilation map $f \mapsto A (f)$ is conjugate-linear under the stated first-variable-linear inner-product convention. The creation map $f \mapsto C (f)$ is complex-linear.

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Finite-Mode CAR and the Complex Clifford Algebra via Majorana Generators

Statement

Let $n \geq 0$ , let $V = C^{2 n}$ with ordered basis $e_{1}, \dots, e_{2 n}$ , and put $Q (\sum_{r} z_{r} e_{r}) = \sum_{r} z_{r}^{2}$ . Use the plus- $Q$ convention of Universal Clifford algebra of a quadratic space. Give ${C l}^{+} (V, Q)$ the conjugate-linear anti-involution that fixes each $e_{r}$ ; it is defined by conjugating coefficients, reversing words, and fixing the basis generators. Then there is a $*$ -isomorphism

${C l}^{+} (V, Q) ≅ {C A R}_{n}^{a l g}$

specified by

$e_{2 j - 1} ⟼ γ_{2 j - 1} = a_{j} + a_{j}^{†}, e_{2 j} ⟼ γ_{2 j} = i (a_{j}^{†} - a_{j}) .$

The inverse sends $a_{j}$ to $(e_{2 j - 1} + i e_{2 j}) / 2$ and $a_{j}^{†}$ to $(e_{2 j - 1} - i e_{2 j}) / 2$ . These are finite-mode algebra maps; they assert no relation between CAR and CCR.

Facts & Assumptions

Given: An integer $n \geq 0$ , the finite-mode CAR $*$ -algebra, and the complex quadratic space $(V, Q)$ with its ordered basis and plus- $Q$ convention.

[F1]

In ${C A R}_{n}^{a l g}$ , ${a_{i}, a_{j}} = 0$ , ${a_{i}^{†}, a_{j}^{†}} = 0$ , and ${a_{i}, a_{j}^{†}} = δ_{i j} 1$ ; its quotient presentation gives the unique $*$ -homomorphism for every tuple satisfying these relations (Finite-Mode Complex CAR Algebra).

[F2]

The plus- $Q$ Clifford algebra is the quotient with canonical generators satisfying ${\overset{ˉ}{v}}^{2} = Q (v) 1$ (Universal Clifford algebra of a quadratic space).

[F3]

Its polarized relation is $\overset{ˉ}{v} \overset{ˉ}{w} + \overset{ˉ}{w} \overset{ˉ}{v} = 2 B (v, w) 1$ (Polarized relations in a Clifford algebra).

[F4]

Every linear map into a unital associative $k$ -algebra satisfying $f (v)^{2} = Q (v) 1$ extends uniquely to a unital algebra map from $C l (V, Q)$ (Universal property of the Clifford algebra).

Proof

1.1givenF2constructalgebra

On $T_{C} (V)$ define $e_{r}^{*} = e_{r}$ , conjugate scalars, and reverse each tensor word. For $v = \sum_{r} z_{r} e_{r}$ , let $\overline{v} = \sum_{r} \overline{z_{r}} e_{r}$ . Then the adjoint of the defining relation $j (v)^{2} - Q (v) 1$ is $j (\overline{v})^{2} - \overline{Q (v)} 1 = j (\overline{v})^{2} - Q (\overline{v}) 1$ , another defining relation. The generated two-sided ideal is therefore $*$ -stable, so this operation descends to the stated anti-involution on the Clifford quotient.

1.2F1algebra

By [F1], $γ_{2 j - 1}^{*} = γ_{2 j - 1}$ and $γ_{2 j}^{*} = γ_{2 j}$ . The same-mode relations give $γ_{2 j - 1}^{2} = a_{j} a_{j}^{†} + a_{j}^{†} a_{j} = 1$ and $γ_{2 j}^{2} = - (a_{j}^{†} - a_{j})^{2} = a_{j}^{†} a_{j} + a_{j} a_{j}^{†} = 1$ . Also ${γ_{2 j - 1}, γ_{2 j}} = i {a_{j} + a_{j}^{†}, a_{j}^{†} - a_{j}} = 0$ . For distinct modes every term in the expanded anticommutator vanishes by [F1]. Thus ${γ_{r}, γ_{s}} = 2 δ_{r s} 1$ .

2.1givenF4step 1.2construct

For $v = \sum_{r} z_{r} e_{r}$ , linearity and step 1.2 give $(\sum_{r} z_{r} γ_{r})^{2} = \sum_{r} z_{r}^{2} 1 + \sum_{r < s} z_{r} z_{s} {γ_{r}, γ_{s}} = \sum_{r} z_{r}^{2} 1 = Q (v) 1$ . Consequently [F4] gives a unique unital algebra map $Φ : {C l}^{+} (V, Q) \to {C A R}_{n}^{a l g}$ with $Φ (e_{r}) = γ_{r}$ . Since each $γ_{r}$ is self-adjoint, $Φ$ preserves $*$ .

2.2givenF1F3step 1.1construct

Put $b_{j} = (e_{2 j - 1} + i e_{2 j}) / 2$ and $b_{j}^{†} = (e_{2 j - 1} - i e_{2 j}) / 2$ in the Clifford algebra. By [F3], ${e_{r}, e_{s}} = 2 δ_{r s} 1$ . Expanding the three CAR anticommutators for the $b_{j}, b_{j}^{†}$ therefore gives ${b_{i}, b_{j}} = 0$ , ${b_{i}^{†}, b_{j}^{†}} = 0$ , and ${b_{i}, b_{j}^{†}} = δ_{i j} 1$ . The anti-involution from step 1.1 gives $b_{j}^{*} = b_{j}^{†}$ . The quotient definition in [F1] now yields a unique $*$ -homomorphism $Ψ : {C A R}_{n}^{a l g} \to {C l}^{+} (V, Q)$ with these generator images.

3.1step 2.1step 2.2algebra∎

On each generator, $Φ Ψ (a_{j}) = (γ_{2 j - 1} + i γ_{2 j}) / 2 = a_{j}$ and $Φ Ψ (a_{j}^{†}) = (γ_{2 j - 1} - i γ_{2 j}) / 2 = a_{j}^{†}$ ; likewise $Ψ Φ (e_{2 j - 1}) = e_{2 j - 1}$ and $Ψ Φ (e_{2 j}) = e_{2 j}$ . The two composites are identities because each algebra is generated by the displayed generators and both maps preserve its operations. Thus $Φ$ and $Ψ$ are inverse $*$ -isomorphisms.

Remarks

The scalar-product pairing convention is irrelevant to these generator relations: the modes are indexed by an orthonormal basis and the CAR use only $δ_{i j}$ . The sign is fixed by $e_{r}^{2} = + 1$ .

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The Finite-Mode CAR Algebra Is a Full Matrix Algebra

Statement

For every integer $n \geq 0$ , the universal finite-mode CAR $*$ -algebra ${C A R}_{n}^{a l g}$ of Finite-Mode Complex CAR Algebra is $*$ -isomorphic to $End (F_{n}) ≅ M_{2^{n}} (C)$ under the occupation representation of Finite CAR Occupation Operators Give All Matrix Units, where $End (F_{n})$ denotes all complex-linear endomorphisms under composition. Give this matrix algebra the operator norm and conjugate-transpose involution; it is a C*-algebra in the sense of C star algebra. Every unital $*$ -representation on a nonzero Hilbert space is unitarily identified with the matrix action on the occupation coordinate tensored with the identity on one Hilbert space, and is therefore isometric. Consequently the universal C*-norm, the supremum of operator norms over these representations, equals the matrix norm, so the universal C*-completion is the same full matrix algebra.

Facts & Assumptions

Given: An integer $n \geq 0$ , the finite-mode algebra ${C A R}_{n}^{a l g}$ , the ordered basis of $C^{2 n}$ used for its Clifford comparison, and the occupation representation on $F_{n}$ .

[F2]

The Majorana maps give a $*$ -isomorphism ${C l}^{+} (C^{2 n}, Q) ≅ {C A R}_{n}^{a l g}$ (Finite-Mode CAR and the Complex Clifford Algebra via Majorana Generators).

[F3]

For a $2 n$ -dimensional complex quadratic space and a fixed ordered basis, the plus- $Q$ Clifford algebra has the $2^{2 n}$ ordered square-free monomials as a basis (PBW basis and exterior associated graded for a Clifford algebra).

[F4]

The occupation representation maps onto $End (F_{n})$ , with $E_{S, T} e_{U} = δ_{T, U} e_{S}$ (Finite CAR Occupation Operators Give All Matrix Units).

[F5]

For $m \geq 1$ , $M_{m} (C)$ with conjugate transpose and the operator norm is a unital complex C*-algebra (Finite matrix algebras with the operator norm are C star algebras).

[F6]

A complex C*-algebra is a complete complex normed associative algebra with a conjugate-linear involution reversing multiplication and satisfying $∥ x^{*} x ∥ = ∥ x ∥^{2}$ (C star algebra).

Proof

1.1givenF2F3algebra

By [F2] the vector-space dimension of ${C A R}_{n}^{a l g}$ is the dimension of ${C l}^{+} (C^{2 n}, Q)$ . Applying [F3] with $2 n$ basis vectors gives $\dim_{C} {C A R}_{n}^{a l g} = 2^{2 n} = 4^{n}$ , including $n = 0$ .

2.1givenF4step 1.1algebra

The occupation representation in [F4] is a surjective $*$ -homomorphism onto $End (F_{n})$ . Since $P_{n}$ has $2^{n}$ elements, $\dim_{C} End (F_{n}) = (2^{n})^{2} = 4^{n}$ . By step 1.1, the source also has dimension $4^{n}$ , so the kernel has dimension zero. Thus the occupation representation is a $*$ -isomorphism.

3.1F4step 2.1construct

Put $m = 2^{n}$ and identify ${C A R}_{n}^{a l g}$ with $M_{m} (C)$ by step 2.1. In any unital $*$ -representation $ρ$ on a nonzero Hilbert space $K$ , the projections $P_{S} = ρ (E_{S, S})$ are pairwise orthogonal and sum to $I$ . Set $K_{0} = P_{\emptyset} K$ . The map $U_{S} = ρ (E_{S, \emptyset}) ∣_{K_{0}} : K_{0} \to P_{S} K$ is unitary because the matrix-unit relations give $U_{S}^{*} U_{S} = I_{K_{0}}$ and $U_{S} U_{S}^{*} = P_{S}$ . Hence $U : ⨁_{S \in P_{n}} K_{0} \to K$ , $U ((ξ_{S})_{S}) = \sum_{S} U_{S} ξ_{S}$ , is unitary and carries each $ρ (E_{S, T})$ to the standard matrix unit on the subset coordinate with identity multiplicity on $K_{0}$ .

4.1F5step 3.1algebra

The space $K_{0}$ is nonzero: if $y \in P_{S} K$ , then $y = ρ (E_{S, \emptyset}) ρ (E_{\emptyset, S}) y$ , and $ρ (E_{\emptyset, S}) y \in K_{0}$ ; if $K_{0} = 0$ , this forces every $P_{S} K = 0$ , contradicting $\sum_{S} P_{S} = I$ and $K \neq {0}$ . Under the unitary in step 3.1, $A = (A_{S, T})$ acts on $⨁_{S} K_{0}$ as $A \otimes I_{K_{0}}$ . For each vector $(ξ_{T})_{T}$ , its finitely many coordinates span a finite-dimensional subspace of $K_{0}$ ; expand them in a finite orthonormal basis and apply the scalar matrix norm bound to obtain $∥ A \otimes I_{K_{0}} ∥ \leq ∥ A ∥$ . Conversely, for a unit vector $η \in K_{0}$ , vectors $(x_{S} η)_{S}$ give $∥ A \otimes I_{K_{0}} ∥ \geq ∥ A ∥$ after taking the supremum over unit $x \in C^{m}$ . Thus every unital representation is isometric.

5.1givenF5F6step 2.1step 4.1algebra∎

The identity representation on $C^{m}$ is one of the representations in the universal supremum, and step 4.1 shows every unital representation has the same matrix operator norm. The universal norm is therefore finite and equals the matrix operator norm. By [F5] and the C*-axioms in [F6], this is a complete C*-norm, so the algebraic isomorphism of step 2.1 already identifies the universal C*-completion with $M_{m} (C)$ . For $n = 0$ , $m = 1$ and the same argument gives $C$ .

Remarks

The proof uses only finite-dimensional matrix units and finite Gram–Schmidt. No choice axiom or infinite-dimensional representation classification is used.

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Canonical Finite-Mode CAR Inclusions Are Isometric

Statement

For $n \geq 0$ , let $ι_{n} : {C A R}_{n} \to {C A R}_{n + 1}$ be the unique unital $*$ -homomorphism sending each of the first $n$ annihilation and creation generators to the generator with the same index. Here ${C A R}_{n}$ is the universal finite-mode C*-algebra of The Finite-Mode CAR Algebra Is a Full Matrix Algebra. Then $ι_{n}$ is isometric and injective. Under the occupation-basis identifications $F_{n + 1} ≅ F_{n} \otimes C^{2}$ , it is represented by $A \mapsto A \otimes I_{2}$ .

Facts & Assumptions

Given: An integer $n \geq 0$ and the finite-mode C*-algebras ${C A R}_{n}, {C A R}_{n + 1}$ .

[F1]

The finite-mode C*-algebra is universal for the finite CAR relations and is represented faithfully on $F_{n}$ as $M_{2^{n}} (C)$ (The Finite-Mode CAR Algebra Is a Full Matrix Algebra).

[F2]

In the occupation basis, the first $n$ annihilation and creation operators act by the toggle formulas in the finite-mode matrix-unit construction (Finite CAR Occupation Operators Give All Matrix Units).

[F3]

The algebraic finite-mode presentation has the quotient universal property for tuples satisfying the CAR (Finite-Mode Complex CAR Algebra).

Proof

1.1givenF1F3construct

The first $n$ generators in ${C A R}_{n + 1}$ satisfy the $n$ -mode CAR relations. The quotient universal property in [F3], followed by the finite-mode C*-completion in [F1], therefore gives the unique unital $*$ -homomorphism $ι_{n}$ with the stated generator values.

1.2givenF2algebra

Identify $e_{S} \otimes e_{ε}$ with $e_{S \cup ({n + 1} if ε = 1)}$ , where $S \subseteq {1, \dots, n}$ and $ε \in {0, 1}$ . For $j \leq n$ , the number of occupied indices below $j$ is unchanged by the last occupation bit. The formulas in [F2] therefore give $c_{j}^{(n + 1)} = c_{j}^{(n)} \otimes I_{2}$ and $(c_{j}^{†})^{(n + 1)} = (c_{j}^{†})^{(n)} \otimes I_{2}$ . By multiplication and linearity, the representation of $ι_{n} (x)$ is $π_{n} (x) \otimes I_{2}$ for every $x \in {C A R}_{n}$ .

2.1givenF1step 1.2algebra∎

For any $A \in M_{2^{n}} (C)$ and $(x_{0}, x_{1}) \in C^{2^{n}} \oplus C^{2^{n}}$ , $∥ (A \otimes I_{2}) (x_{0}, x_{1}) ∥_{2}^{2} = ∥ A x_{0} ∥_{2}^{2} + ∥ A x_{1} ∥_{2}^{2} \leq ∥ A ∥^{2} (∥ x_{0} ∥_{2}^{2} + ∥ x_{1} ∥_{2}^{2})$ , so $∥ A \otimes I_{2} ∥ \leq ∥ A ∥$ . Conversely, vectors $(x, 0)$ with $∥ x ∥_{2} \leq 1$ give $∥ A \otimes I_{2} ∥ \geq ∥ A ∥$ . By [F1] and step 1.2, $∥ ι_{n} (x) ∥ = ∥ π_{n} (x) \otimes I_{2} ∥ = ∥ π_{n} (x) ∥ = ∥ x ∥$ . Thus $ι_{n}$ is isometric and therefore injective; the argument includes $n = 0$ .

Remarks

The embeddings use the fixed ordering in which the new mode is appended last. No choice principle is used.

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Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra

Statement

Let $F$ be a finite-dimensional complex inner-product space of dimension $n$ , and let ${C A R}^{a l g} (F)$ be the algebra of Finite-Dimensional CAR Algebra over an Inner-Product Space. For any orthonormal basis of $F$ , the assignment to the $n$ standard finite-mode annihilation and creation generators identifies ${C A R}^{a l g} (F)$ with ${C A R}_{n}^{a l g}$ . It therefore has a unique finite-dimensional universal C*-norm, and its completion is $*$ -isomorphic to $M_{2^{n}} (C)$ . Every unital $*$ -representation of this completion on a nonzero Hilbert space is isometric.

Facts & Assumptions

Given: A finite-dimensional complex inner-product space $F$ , with inner product linear in the first variable, and an orthonormal basis $e_{1}, \dots, e_{n}$ .

[F1]

The algebra ${C A R}^{a l g} (F)$ is the universal unital $*$ -algebra generated complex-linearly by $c (f)$ with ${c (f), c (g)} = 0$ , ${c (f)^{*}, c (g)^{*}} = 0$ , and ${c (f)^{*}, c (g)} = ⟨ g, f ⟩ 1$ (Finite-Dimensional CAR Algebra over an Inner-Product Space).

[F2]

The standard finite-mode algebra is universal for ${a_{i}, a_{j}} = 0$ , ${a_{i}^{†}, a_{j}^{†}} = 0$ , and ${a_{i}, a_{j}^{†}} = δ_{i j} 1$ (Finite-Mode Complex CAR Algebra).

[F3]

The standard finite-mode algebra has universal C*-completion $M_{2^{n}} (C)$ , and every unital $*$ -representation on a nonzero Hilbert space is isometric (The Finite-Mode CAR Algebra Is a Full Matrix Algebra).

[F4]

Every finite-dimensional inner-product space has a finite orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

Proof

1.1givenF1F2F4algebra

A finite orthonormal basis exists by [F4]; fix the basis in Given. For $f = \sum_{j = 1}^{n} λ_{j} e_{j}$ , set $C (f) = \sum_{j} λ_{j} a_{j}^{†}$ in the standard finite-mode algebra. Since the inner product is linear in the first variable, $⟨ g, f ⟩ = \sum_{j} μ_{j} \overline{λ_{j}}$ when $g = \sum_{j} μ_{j} e_{j}$ . Bilinearity of the anticommutator and the finite-mode relations give ${C (f), C (g)} = 0$ , ${C (f)^{*}, C (g)^{*}} = 0$ , and ${C (f)^{*}, C (g)} = \sum_{j} \overline{λ_{j}} μ_{j} 1 = ⟨ g, f ⟩ 1$ . Thus [F1] gives a unital $*$ -homomorphism $Φ : {C A R}^{a l g} (F) \to {C A R}_{n}^{a l g}$ .

2.1givenF1F2step 1.1algebra

Conversely, the elements $c (e_{j})^{*}$ and $c (e_{j})$ in ${C A R}^{a l g} (F)$ satisfy the standard finite-mode relations by [F1] and orthonormality. The universal property [F2] gives $Ψ : {C A R}_{n}^{a l g} \to {C A R}^{a l g} (F)$ . The maps $Φ Ψ$ and $Ψ Φ$ fix, respectively, every standard generator and every $c (e_{j})$ and its adjoint. Since the $e_{j}$ span $F$ , linearity makes them fix all generators; hence the maps are inverse $*$ -isomorphisms.

3.1F3step 2.1algebra∎

By [F3] and step 2.1, the finite-mode universal C*-norm transports to ${C A R}^{a l g} (F)$ and its completion is $M_{2^{n}} (C)$ . The same transport identifies any unital $*$ -representation of this completion with a unital representation of the standard finite-mode algebra. Such a representation is isometric by [F3]. If $n = 0$ , both algebras are $C$ and the identity representation is isometric as well.

Remarks

Changing the orthonormal basis changes the generators by a unitary linear transformation of $F$ . The basis-free universal presentation gives the same algebra, and the transported finite matrix norm is independent of that choice because every unital representation of the finite matrix algebra is isometric.

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Countably Infinite-Mode CAR C*-Algebra

Definition

Let $N = {1, 2, \dots}$ be the countable mode set. For each $n$ take the universal finite-mode C*-algebra ${C A R}_{n}$ whose dense algebraic presentation is Finite-Mode Complex CAR Algebra. Identify each stage with its image in the next under the isometric inclusion of Canonical Finite-Mode CAR Inclusions Are Isometric. The inclusions are compatible, injective, and preserve the C*-norm. Their union ${C A R}_{l o c} = ⋃_{n \geq 0} {C A R}_{n}$ is therefore a normed $*$ -algebra with norm $∥ x ∥ = ∥ x ∥_{{C A R}_{n}}$ whenever $x \in {C A R}_{n}$ ; isometry makes this independent of the stage. Define the countably infinite-mode CAR C-algebra* ${C A R}_{N}$ to be the norm completion of ${C A R}_{l o c}$ in the sense of C star algebra. Its canonical generator $a_{j}$ is the image of the $j$ th finite-mode generator in any stage $n \geq j$ . The C*-axioms pass to the completion because each finite stage satisfies them and multiplication and involution are continuous. The universal property of this inductive-limit construction is proved in The Countable CAR Algebra Has the Universal Inductive-Limit Norm.

The existence of this particular completion needs no choice principle. Let $π_{n} : {C A R}_{n} \to M_{2^{n}} (C)$ be the occupation-basis isomorphism of The Finite-Mode CAR Algebra Is a Full Matrix Algebra, and let $D_{n} \subseteq {C A R}_{n}$ be the inverse image of $M_{2^{n}} (Q (i))$ , where $Q (i) = {a + i b : a, b \in Q}$ . Write $N_{0} = {0, 1, 2, \dots}$ for coding indices, to distinguish it from the positive mode set $N$ . The theorem $Q$ is countably infinite supplies a surjection $N_{0} \to Q$ , and pairing two such indices by $N \times N \approx N$ gives an explicit enumeration of $Q (i)$ . In these coordinates the canonical inclusion is $A \mapsto A \otimes I_{2}$ (Canonical Finite-Mode CAR Inclusions Are Isometric), so $D_{n} \subseteq D_{n + 1}$ . Each finite array of Gaussian rationals is coded by a finite tuple in $N_{0}$ ; repeated pairing, followed by pairing with the stage index, enumerates $D = ⋃_{n \geq 0} D_{n}$ without choosing from a family of stage enumerations. Entrywise approximation of real and imaginary parts by rationals (The rationals embed densely in the reals) shows Gaussian-rational matrices are dense in each finite-dimensional matrix stage: in fixed size $m$ , $∥ A - B ∥_{o p} \leq m \max_{r, s} ∣ a_{r s} - b_{r s} ∣$ . Hence $D$ is dense in ${C A R}_{l o c}$ .

Concretely, form the set of Cauchy sequences in ${C A R}_{l o c}$ , identify two sequences when the norm of their termwise difference tends to zero, and use the limiting norm and termwise algebra operations. Write $\hat{d} ([a_{n}], [b_{n}]) = \lim_{n} ∥ a_{n} - b_{n} ∥$ for the resulting completion metric. Products are well defined because Cauchy sequences are bounded and $∥ x_{m} y_{m} - x_{n} y_{n} ∥ \leq ∥ x_{m} ∥ ∥ y_{m} - y_{n} ∥ + ∥ x_{m} - x_{n} ∥ ∥ y_{n} ∥$ ; the involution is isometric. Thus the completion is a normed $*$ -algebra, and its C*-identity passes to limits. It is complete without Countable Choice: the constant-sequence image of ${C A R}_{l o c}$ is dense in the sequence-class completion, so $D$ is dense there too. Fix the explicit enumeration $q : N_{0} ↠ D$ . For any Cauchy sequence $(x_{m})$ of completion points, let $d_{m}$ be the first $q (r)$ within $2^{- m}$ of $x_{m}$ . The least-index rule defines each $d_{m}$ uniquely. The triangle inequality shows $(d_{m})$ is Cauchy; for $m, n$ sufficiently large, $\hat{d} (x_{m}, d_{n}) \leq \hat{d} (x_{m}, x_{n}) + 2^{- n}$ is arbitrarily small, so the class $[d_{n}]$ is the limit of $(x_{m})$ . Hence this sequence-class construction is a completion of ${C A R}_{l o c}$ and defines ${C A R}_{N}$ choice-free. Its canonical generators and inductive-limit norm are unchanged.

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Finite-Subspace CAR Inclusions Are Isometric

Statement

If $F \subseteq G$ are finite-dimensional complex inner-product subspaces of a Hilbert space $H$ , the generator-preserving map ${C A R}^{a l g} (F) \to {C A R}^{a l g} (G)$ extends to an isometric unital $*$ -homomorphism between their finite-dimensional C*-completions.

Facts & Assumptions

Given: Finite-dimensional subspaces $F \subseteq G \subseteq H$ with the inherited inner products.

[F1]

Each finite-dimensional CAR algebra is the standard finite-mode CAR algebra after a choice of orthonormal basis, and its universal C*-completion is a matrix algebra (Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra).

[F2]

The algebraic CAR presentation is universal for the linear creation generators and their adjoints (Finite-Dimensional CAR Algebra over an Inner-Product Space).

[F3]

The standard inclusion from $n$ to $n + 1$ modes, in the occupation representation, is $A \mapsto A \otimes I_{2}$ and is isometric (Canonical Finite-Mode CAR Inclusions Are Isometric).

[F4]

Every finite-dimensional inner-product space has a finite orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

Proof

1.1givenF2construct

The generators $c_{G} (f)$ for $f \in F$ satisfy the defining relations of ${C A R}^{a l g} (F)$ . Its universal property [F2] therefore defines a unique unital $*$ -homomorphism $ι_{F, G}$ sending $c_{F} (f)$ to $c_{G} (f)$ for every $f \in F$ .

2.1givenF1F3F4step 1.1constructalgebra∎

Let $d = \dim F$ and $m = \dim G$ . Choose an orthonormal basis $e_{1}, \dots, e_{d}$ of $F$ using [F4], and choose a finite vector-space basis $v_{1}, \dots, v_{m}$ of $G$ . Initialize the current orthonormal list by $u_{j} = e_{j}$ for $1 \leq j \leq d$ , and let $s$ denote its current length, initially $d$ . Process $v_{1}, \dots, v_{m}$ in order: set $r_{k} = v_{k} - \sum_{j = 1}^{s} ⟨ v_{k}, u_{j} ⟩ u_{j}$ , where the current list is orthonormal; then $⟨ r_{k}, u_{j} ⟩ = 0$ for every $j$ . If $r_{k} \neq 0$ , append $r_{k} / ∥ r_{k} ∥$ ; otherwise keep the list unchanged. After processing $v_{k}$ , the list spans $F + span {v_{1}, \dots, v_{k}}$ . Thus the resulting finite orthonormal list is a basis of $G$ and retains $e_{1}, \dots, e_{d}$ as its initial vectors. Under these basis identifications in [F1], $ι_{F, G}$ is the standard inclusion of the first $d$ modes into $m$ modes. If $d < m$ , this is the composition of the inclusions from $r$ to $r + 1$ for $d \leq r < m$ ; each is isometric by [F3], so the composition is isometric. If $d = m$ , the map is the identity and is isometric. Thus $ι_{F, G}$ is isometric and in particular injective.

Remarks

Only finite orthonormal bases are extended. No basis of $H$ is selected.

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CAR C*-Algebra over a Complex Hilbert Space

Definition

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for this norm-completion construction, and let $H$ be a complex Hilbert space. Order its finite-dimensional complex subspaces by inclusion. This is a directed system because the span $F + G$ of any two finite-dimensional subspaces is finite-dimensional. To each such $F$ , assign the finite-dimensional CAR C*-algebra of Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra, with the generator-preserving maps of Finite-Subspace CAR Inclusions Are Isometric. Identify each stage with its image under these injective maps and let ${C A R}_{l o c} (H)$ be their algebraic union.

The common finite-stage norm is well defined on this union because every inclusion is isometric. Under Countable Choice, Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences supplies a metric completion and The metric completion of a normed space carries a unique compatible Banach-space structure equips it with the compatible Banach-space structure; this is the norm completion of Completion of a normed space. Define the CAR C-algebra over $H$ **, $C A R (H)$ , to be this norm completion of ${C A R}_{l o c} (H)$ . Addition, multiplication, scalar multiplication, and involution are the compatible finite-stage operations; submultiplicativity and isometry of the involution make them extend continuously to the completion, and the C-identity passes to limits. The identity is nonzero because the zero-dimensional stage is $C$ and every inclusion preserves its unit. For $f \in H$ , write $c (f)$ for the image of the generator in the stage $span {f}$ ; the universal inductive-limit property and its norm are proved in CAR over a Hilbert Space Has the Universal Inductive-Limit Norm.

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Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR

Statement

There is a unique $*$ -automorphism $α$ of ${C A R}_{N}$ satisfying

$α (a_{j}) = a_{j}^{†}, α (a_{j}^{†}) = a_{j} (j \in N) .$

It is involutive: $α^{2} = I$ .

Facts & Assumptions

Given: The countable-mode CAR C*-algebra as the completion of its finite-mode stages and the canonical generators $a_{j}, a_{j}^{†}$ .

[F1]

Each finite stage is the universal finite-mode C*-algebra with the CAR relations and its algebraic quotient presentation (Finite-Mode Complex CAR Algebra, The Finite-Mode CAR Algebra Is a Full Matrix Algebra).

[F2]

The finite stages embed isometrically and compatibly, and their union is dense in ${C A R}_{N}$ (Countably Infinite-Mode CAR C*-Algebra, Canonical Finite-Mode CAR Inclusions Are Isometric).

Proof

1.1givenF1construct

On a finite stage, exchange $a_{j}$ with $a_{j}^{†}$ . The first two CAR families are exchanged, while ${a_{i}^{†}, a_{j}} - δ_{i j} 1$ is the same relation as ${a_{j}, a_{i}^{†}} - δ_{j i} 1$ because $δ_{i j} = δ_{j i}$ . The exchange also respects $*$ , since $(a_{j}^{†})^{*} = a_{j}$ . The quotient universal property in [F1] therefore gives a unital $*$ -homomorphism $α_{n} : {C A R}_{n} \to {C A R}_{n}$ . Applying it twice fixes every generator, so $α_{n}^{2} = I$ and $α_{n}$ is an automorphism.

2.1givenF1F2step 1.1algebra∎

For $m \leq n$ , the maps $α_{n} \circ ι_{m, n}$ and $ι_{m, n} \circ α_{m}$ agree on every generator of ${C A R}_{m}$ , hence agree on that stage by its universal property. The compatible automorphisms therefore define an automorphism $α_{0}$ of the algebraic union. For $x$ in a finite stage, the universal norm there is the supremum of $∥ ρ (x) ∥$ over its unital representations; precomposition by the bijection $α_{n}$ permutes those representations, so $∥ α_{n} (x) ∥ = ∥ x ∥$ . Thus $α_{0}$ is isometric and extends continuously to the completion in [F2]. Its inverse extends from the compatible maps $α_{n}^{- 1} = α_{n}$ , and $α^{2} = I$ because this holds on the dense union.

Remarks

This exchange is a $*$ -algebra automorphism. It makes no statement about implementation by a unitary in the Fock representation.

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The Countable CAR Algebra Has the Universal Inductive-Limit Norm

Statement

Throughout this item, use the positive mode set $N = {1, 2, \dots}$ .

Let ${C A R}_{N}$ be the inductive-limit C*-algebra of Countably Infinite-Mode CAR C*-Algebra. A bounded CAR representation on a nonzero complex Hilbert space $H$ means a sequence of bounded linear operators $T_{j}$ together with specified bounded adjoints $T_{j}^{*}$ , in the sense of The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , satisfying

${T_{i}, T_{j}} = 0, {T_{i}^{*}, T_{j}^{*}} = 0, {T_{i}, T_{j}^{*}} = δ_{i j} I_{H} (i, j \in N) .$

Write $B_{a d} (H)$ for the subalgebra of $B (H)$ consisting of bounded operators that possess bounded adjoints, with that adjoint as involution. For every such representation there is a unique contractive unital $*$ -homomorphism $ρ : {C A R}_{N} \to B_{a d} (H)$ with $ρ (a_{j}) = T_{j}$ . The inductive-limit norm on the algebraic union ${C A R}_{l o c}$ equals the universal norm

$∥ x ∥_{u} = \sup_{ρ} ∥ ρ (x) ∥, x \in {C A R}_{l o c},$

where the supremum is over these representations. The norm is finite and the family of representations is nonempty; a concrete representation is the occupation-space representation on $ℓ^{2} (P_{f i n} (N))$ constructed in the proof. Its coordinate vectors $e_{S}$ form an orthonormal basis, so finite-support vectors are dense.

Facts & Assumptions

Given: The finite and countably infinite-mode CAR algebras, a nonzero complex Hilbert space $H$ with inner product linear in the first variable, and a bounded CAR representation $(T_{j}, T_{j}^{*})_{j \geq 1}$ .

[F1]

${C A R}_{N}$ is the completion of the increasing union ${C A R}_{l o c}$ of the finite-mode C*-algebras under their isometric inclusions (Countably Infinite-Mode CAR C*-Algebra, Canonical Finite-Mode CAR Inclusions Are Isometric).

[F2]

Every finite-mode C*-algebra is $M_{2^{n}} (C)$ , and every unital $*$ -representation on a nonzero Hilbert space is isometric and has the matrix-unit amplification form proved there (The Finite-Mode CAR Algebra Is a Full Matrix Algebra).

[F3]

The finite occupation operators are given on subset basis vectors by toggling one mode with sign $(- 1)^{∣ {s \in S : s < j} ∣}$ , and they satisfy the finite CAR (Finite CAR Occupation Operators Give All Matrix Units).

[F4]

A complex Hilbert space is complete for its induced norm (Hilbert space), and its pairing is linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

[F5]

$B (H)$ denotes bounded linear operators on $H$ , with operator norm $∥ S ∥ = \sup_{∥ x ∥ \leq 1} ∥ S x ∥$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F6]

The operators possessing bounded adjoints form the unital $*$ -subalgebra $B_{a d} (H)$ : the pairing identity gives $(S + T)^{*} = S^{*} + T^{*}$ , $(z S)^{*} = \overline{z} S^{*}$ and $(S T)^{*} = T^{*} S^{*}$ , all bounded; uniqueness and conjugation give $(S^{*})^{*} = S$ and $I^{*} = I$ . The specified adjoint $S^{*}$ is characterized by $⟨ S x, y ⟩ = ⟨ x, S^{*} y ⟩$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ ).

[F7]

$∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F8]

For every set $I$ , the canonical complex square-summable family space $ℓ^{2} (I, C)$ is a Hilbert space, its finitely supported families are dense, and its coordinate vectors form an orthonormal basis (Square-Summable Complex Families Form a Hilbert Space).

Proof

1.1givenF1F2construct

Restrict the given family to its first $n$ modes. The quotient universal property of Finite-Mode Complex CAR Algebra gives the finite algebraic $*$ -homomorphism on generators, and [F2] identifies it with a contractive unital $*$ -representation $ρ_{n} : {C A R}_{n} \to B (H)$ . If $m \leq n$ , the maps $ρ_{n} \circ ι_{m, n}$ and $ρ_{m}$ agree on all generators of ${C A R}_{m}$ , so uniqueness of the finite quotient map gives $ρ_{n} \circ ι_{m, n} = ρ_{m}$ .

1.2givenF1F2F3F4F8construct

Let $P_{f i n} (N)$ be all finite subsets of $N$ , and let $F_{\infty} = ℓ^{2} (P_{f i n} (N))$ with coordinate vectors $e_{S}$ . By [F8], this is a complex Hilbert space, the vectors $e_{S}$ are an orthonormal basis, and their finite span is dense. Define $c_{j} e_{S} = 0$ if $j \notin S$ , and $c_{j} e_{S} = (- 1)^{∣ {s \in S : s < j} ∣} e_{S ∖ {j}}$ if $j \in S$ ; define $c_{j}^{†}$ by the reverse toggle with the same sign. Each map is a partial isometry on the basis and extends boundedly, and the coefficient of each reverse basis transition shows $c_{j}^{†}$ is its adjoint. For any finite set of indices, choose $n$ containing them; under the decomposition of the basis into first- $n$ and tail subsets, the first $n$ operators act as the finite occupation representation tensored with the identity. Hence the CAR hold for every pair of indices. The finite universal property gives compatible contractive representations of all ${C A R}_{n}$ on $F_{\infty}$ , which extend to a representation $π_{\infty}$ of the inductive limit.

2.1givenF1F4F5F6F7step 1.1constructalgebra

Define $ρ_{0}$ on ${C A R}_{l o c}$ by $ρ_{0} (x) = ρ_{n} (x)$ whenever $x \in {C A R}_{n}$ . Compatibility in step 1.1 makes this well defined, and contractivity at each stage gives $∥ ρ_{0} (x) ∥ \leq ∥ x ∥$ . The space $B (H)$ is complete: for an operator-norm Cauchy sequence $(S_{k})$ , each $(S_{k} x)$ converges in $H$ ; its pointwise limit $S x$ is linear and bounded because $(∥ S_{k} ∥)$ is bounded, and taking the pointwise limit in the uniform Cauchy bound shows $∥ S_{k} - S ∥ \to 0$ . Thus $ρ_{0}$ extends uniquely and contractively to the completion. Products pass to limits by $∥ A B ∥ \leq ∥ A ∥ ∥ B ∥$ . For any bounded operator $A$ with a specified adjoint, [F6] and [F7] give $∥ A^{*} y ∥ = \sup_{∥ x ∥ \leq 1} ∣ ⟨ A^{*} y, x ⟩ ∣ = \sup_{∥ x ∥ \leq 1} ∣ ⟨ A x, y ⟩ ∣ \leq ∥ A ∥ ∥ y ∥$ , hence $∥ A^{*} ∥ \leq ∥ A ∥$ . Conjugating [F6] shows $A$ is the adjoint of $A^{*}$ ; uniqueness gives $(A^{*})^{*} = A$ , so applying the same inequality to $A^{*}$ yields $∥ A^{*} ∥ = ∥ A ∥$ . Therefore if $S_{k} \to S$ , the adjoints $S_{k}^{*}$ are Cauchy and converge to some $R$ ; passing to the limit in [F6] gives $⟨ S x, y ⟩ = ⟨ x, R y ⟩$ , so $R$ is the adjoint of $S$ . Thus the extension takes values in $B_{a d} (H)$ , preserves $*$ , and is a unital $*$ -homomorphism into this subalgebra.

2.2givenF2step 1.2algebra

Under the basis bijection $P_{f i n} (N) ≅ P_{n} \times P_{f i n} ({n + 1, n + 2, \dots})$ , the first $n$ toggle signs depend only on the first subset coordinate because every tail index is larger than every mode $j \leq n$ . Thus the restriction of $π_{\infty}$ to ${C A R}_{n}$ is the finite matrix representation acting identically on each tail coordinate. The tail contains the empty subset, so it is nonzero; by the isometric amplification statement in [F2], $∥ π_{\infty} (x) ∥ = ∥ x ∥_{{C A R}_{n}}$ for every $x \in {C A R}_{n}$ .

3.1givenF1F2step 2.1step 1.2step 2.2algebra∎

For any bounded CAR representation $σ$ and $x \in {C A R}_{n}$ , its restriction is contractive by [F2], so $∥ σ (x) ∥ \leq ∥ x ∥_{{C A R}_{n}}$ . Step 2.2 supplies a representation attaining this norm. Therefore the supremum over all representations equals the inductive-limit norm on each finite stage. Every element of ${C A R}_{l o c}$ lies in some finite stage, so the universal norm is finite and equals the inductive-limit norm on the entire union. Step 1.2 proves the representing family is nonempty. The extension in step 2.1 is unique because ${C A R}_{l o c}$ is dense.

Remarks

No choice axiom is used. A representation is supplied together with its bounded adjoint operators; the proof does not assert that every bounded operator on every Hilbert space has an adjoint in the absence of additional choice principles.

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The Countable-Mode CAR Fock Representation Is Irreducible

Statement

Throughout this item, use the positive mode set $N = {1, 2, \dots}$ .

Let $π_{F}$ be the countable-mode occupation representation of ${C A R}_{N}$ on $F_{\infty} = ℓ^{2} (P_{f i n} (N))$ constructed in The Countable CAR Algebra Has the Universal Inductive-Limit Norm. Then $π_{F}$ is irreducible: its only closed invariant subspaces are ${0}$ and $F_{\infty}$ . In addition, the common kernel of all annihilators $c_{j}$ is the vacuum line $C e_{\emptyset}$ , whereas the common kernel of all creators $c_{j}^{†}$ is ${0}$ .

Facts & Assumptions

Given: The countable-mode CAR algebra, its occupation representation $π_{F}$ , and the orthonormal basis $(e_{S})_{S \in P_{f i n} (N)}$ .

[F1]

The operators $c_{j}, c_{j}^{†}$ toggle membership of $j$ in $S$ with sign $(- 1)^{∣ {s \in S : s < j} ∣}$ , are adjoints, satisfy the CAR, and extend to $π_{F}$ (The Countable CAR Algebra Has the Universal Inductive-Limit Norm).

[F2]

The coordinate vectors $e_{S}$ form an orthonormal basis of $F_{\infty}$ , and their finite span is dense (The Countable CAR Algebra Has the Universal Inductive-Limit Norm).

Proof

1.1givenF1F2algebra

Write $ξ = \sum_{S} ξ_{S} e_{S} \in F_{\infty}$ and suppose $c_{j} ξ = 0$ for every $j$ . For any nonempty finite $S$ , take its least element $j$ . The coefficient of $e_{S ∖ {j}}$ in $c_{j} ξ$ is $(- 1)^{∣ {s \in S : s < j} ∣} ξ_{S} = ξ_{S}$ , and no other input subset maps to this output under $c_{j}$ . Hence $ξ_{S} = 0$ for every nonempty $S$ , so $ξ = ξ_{\emptyset} e_{\emptyset}$ . Conversely every annihilator kills $e_{\emptyset}$ .

1.2givenF1F2choose

If $ξ \neq 0$ , choose a finite subset $S$ with $ξ_{S} \neq 0$ and let $j$ be the least positive integer not in $S$ . The coefficient of $e_{S \cup {j}}$ in $c_{j}^{†} ξ$ is $(- 1)^{∣ {s \in S : s < j} ∣} ξ_{S} \neq 0$ , and this output has the unique input $S$ . Thus $c_{j}^{†} ξ \neq 0$ , proving that no nonzero vector is killed by every creator.

1.3givenF1F2choose

Let $K \subseteq F_{\infty}$ be a nonzero closed subspace invariant under $π_{F} ({C A R}_{N})$ . Choose $0 \neq ξ \in K$ and, by [F2], a finite subset $S = {j_{1} < \dots < j_{q}}$ with $ξ_{S} \neq 0$ . If $q = 0$ , put $η = ξ$ ; otherwise put $η = c_{j_{q}} \dots c_{j_{1}} ξ$ . In either case $η \in K$ . The operator used is the identity when $q = 0$ and the displayed annihilator product otherwise. On basis vectors it vanishes unless $S \subseteq U$ , and then gives a sign times $e_{U ∖ S}$ ; hence only $U = S$ contributes to its empty-set coefficient, so $η_{\emptyset} = \pm ξ_{S} \neq 0$ .

2.1givenF1F2step 1.3choosealgebra

Given $ε > 0$ , use [F2] to choose a finite-support vector $ζ$ with $∥ η - ζ ∥ < ε / 2$ . For each $n$ , put $N_{j} = c_{j}^{†} c_{j}$ and $Q_{n} = \prod_{j = 1}^{n} (I - N_{j})$ . By [F1], $N_{j} e_{T} = 1_{j \in T} e_{T}$ , so the factors commute and $Q_{n}$ is the diagonal projection that kills $e_{T}$ when $T \cap {1, \dots, n} \neq \emptyset$ ; it belongs to $π_{F} ({C A R}_{N})$ and has norm at most one. Choose $n$ at least every element of every nonempty set in the finite support of $ζ$ (take $n = 1$ if there are none), so $Q_{n} ζ = ζ_{\emptyset} e_{\emptyset}$ . Then $∥ Q_{n} η - η_{\emptyset} e_{\emptyset} ∥ \leq ∥ Q_{n} (η - ζ) ∥ + ∣ ζ_{\emptyset} - η_{\emptyset} ∣ < ε$ : the first term is bounded by $∥ η - ζ ∥$ since $Q_{n}$ is a contraction, and the second by the same norm since a coordinate functional on an orthonormal basis has norm one. Thus for every $ε > 0$ there is $Q_{n} η \in K$ within $ε$ of $η_{\emptyset} e_{\emptyset}$ . Closedness and $η_{\emptyset} \neq 0$ give $e_{\emptyset} \in K$ .

3.1F1F2step 2.1algebra∎

For every finite $S$ , a finite creator product sends $e_{\emptyset}$ to $e_{S}$ up to a sign by [F1]. Invariance gives every basis vector in $K$ , and their finite span is dense by [F2]; hence $K = F_{\infty}$ . This proves there is no nonzero proper closed invariant subspace, so $π_{F}$ is irreducible.

Remarks

The mode choices are explicit (a nonzero coordinate in step 1.3 and least-index choices in steps 1.1–1.2). Step 2.1 uses finite-support density to approximate one vector at a time; it does not select projections onto arbitrary closed subspaces or invoke Riesz representation, orthogonal-projection existence, or a choice axiom.

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CAR over a Hilbert Space Has the Universal Inductive-Limit Norm

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space, and let $C A R (H)$ be the completion of its directed union of finite-dimensional CAR algebras from CAR C*-Algebra over a Complex Hilbert Space. Let $K \neq {0}$ be a complex Hilbert space, and let $C : H \to B (K)$ be a complex-linear family of bounded operators satisfying

${C (f), C (g)} = 0, {C (f)^{*}, C (g)^{*}} = 0, {C (f)^{*}, C (g)} = ⟨ g, f ⟩ I_{K} (f, g \in H) .$

There is a unique unital isometric $*$ -homomorphism $ρ : C A R (H) \to B (K)$ with $ρ (c (f)) = C (f)$ . For each $x$ in the algebraic union, its inductive-limit norm equals the supremum of $∥ ρ (x) ∥$ over all such families on nonzero Hilbert spaces. This family is nonempty: the fermionic Fock representation $C (f) \mapsto C (f)$ of Fock Creation and Annihilation Operators Satisfy the CAR attains the norm on every finite stage.

Facts & Assumptions

Given: A complex Hilbert space $H$ , a nonzero complex Hilbert space $K$ , and a complex-linear CAR family $C : H \to B (K)$ .

[F1]

$C A R (H)$ is the completion of the directed union of its finite-dimensional CAR C*-algebras, with generator-preserving isometric inclusions (CAR C*-Algebra over a Complex Hilbert Space, Finite-Subspace CAR Inclusions Are Isometric).

[F2]

Every finite-dimensional CAR C*-algebra is a full matrix algebra, and every unital $*$ -representation on a nonzero Hilbert space is isometric (Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra).

[F3]

Fermionic Fock space has a nonzero vacuum vector (Fermionic Fock Space over a Complex Hilbert Space).

[F4]

Fock creation operators and their adjoints satisfy the stated CAR relations (Fock Creation and Annihilation Operators Satisfy the CAR).

[F5]

If $Y$ is Banach, then $B (X, Y)$ is Banach for the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach).

[F6]

$B (K)$ denotes bounded linear operators with operator norm $∥ T ∥ = \sup_{∥ x ∥ \leq 1} ∥ T x ∥$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A1]

Countable Choice is retained for the Hilbert tensor/Fock construction and Hilbert-space adjoints in the Fock representation (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F7]

Under Countable Choice, bounded Hilbert-space adjoints satisfy $∥ T^{*} ∥ = ∥ T ∥$ and $(S T)^{*} = T^{*} S^{*}$ (Hilbert-adjoint identities).

[F8]

Operator norms satisfy $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ (Composition satisfies |ST|\le|S|,|T|).

[F9]

A C*-norm is submultiplicative and its involution is isometric (C star algebra).

Proof

1.1givenF1F2construct

For each finite-dimensional $F \subseteq H$ , the restriction $C ∣_{F}$ satisfies the defining relations of ${C A R}^{a l g} (F)$ . Its universal property gives a unital $*$ -homomorphism $ρ_{F}$ from that algebra to $B (K)$ . These homomorphisms agree on inclusions because each sends every $c (f)$ to $C (f)$ . By [F2], every $ρ_{F}$ is isometric since $K \neq {0}$ . They therefore define an isometric $*$ -homomorphism $ρ_{0}$ on the algebraic union.

2.1F1F5F6step 1.1construct

Since $K$ is Hilbert, it is Banach; applying [F5] with $X = Y = K$ shows $B (K)$ is complete for its operator norm [F6]. The isometry $ρ_{0}$ sends Cauchy sequences to Cauchy sequences, so for $x \in C A R (H)$ choose a local sequence $x_{m} \to x$ and define $ρ (x) = \lim_{m} ρ_{0} (x_{m})$ . Isometry makes this limit independent of the chosen sequence and proves that $ρ$ is a linear isometry extending $ρ_{0}$ .

3.1F1F7F8F9step 2.1algebra

Let $x, y \in C A R (H)$ and choose local sequences $x_{m} \to x$ and $y_{m} \to y$ . Since $(x_{m})$ is bounded, [F9] gives $∥ x_{m} y_{m} - x y ∥ \leq ∥ x_{m} ∥ ∥ y_{m} - y ∥ + ∥ x_{m} - x ∥ ∥ y ∥ \to 0$ ; [F8] gives the same estimate for $ρ_{0} (x_{m}) ρ_{0} (y_{m}) \to ρ (x) ρ (y)$ . Because $ρ_{0} (x_{m} y_{m}) = ρ_{0} (x_{m}) ρ_{0} (y_{m})$ , limits prove multiplicativity. Also [F9] gives $x_{m}^{*} \to x^{*}$ , while [F7] and $ρ_{0} (x_{m}) \to ρ (x)$ give $ρ_{0} (x_{m})^{*} \to ρ (x)^{*}$ . Since $ρ_{0} (x_{m}^{*}) = ρ_{0} (x_{m})^{*}$ , taking limits proves $ρ (x^{*}) = ρ (x)^{*}$ . The common unit is preserved on the dense union. Any other continuous extension agrees there and hence equals $ρ$ .

4.1A1F2F3F4step 1.1step 2.1step 3.1∎

By [F3] the Fock space is nonzero, and under [A1] the creation and adjoint operators in [F4] give one of the stated CAR families. Applying the extension construction of steps 1.1, 2.1 and 3.1 to this family gives its representation of $C A R (H)$ . Every such representation restricts on a finite stage to a unital representation of that matrix algebra, so [F2] makes its norm on each local element equal the stage norm. Thus the global family is nonempty and the supremum of represented norms equals the inductive-limit norm for every local element.

Remarks

The directed finite-dimensional CAR system and its finite-stage norms use finite-dimensional subspaces and finite orthonormal bases. The theorem retains Countable Choice for its completion and Hilbert/Fock construction context, including the bounded Hilbert adjoints in the Fock representation that witnesses nonemptiness of the global representation family. This is an abstract CAR C*-algebra statement and supplies no QFT locality or spin-statistics theorem.

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Countable-Mode CAR Has Inequivalent Irreducible Representations

Statement

Let $π_{F}$ be the countable-mode occupation representation on $F_{\infty}$ and let $α$ be the particle-hole automorphism of Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR. Then $π_{F}$ and $π_{h} : = π_{F} \circ α$ are irreducible representations of ${C A R}_{N}$ , and they are not unitarily equivalent. Here two representations are unitarily equivalent if a unitary operator between their Hilbert spaces intertwines their actions. This proves existence of inequivalent irreducible representations; it does not classify all representations.

Facts & Assumptions

Given: The countable-mode CAR algebra, the occupation representation $π_{F}$ , and the particle-hole automorphism $α$ .

[F1]

The occupation representation is irreducible, and no nonzero vector is killed by every creator $c_{j}^{†}$ (The Countable-Mode CAR Fock Representation Is Irreducible).

[F2]

The particle-hole map $α$ is a surjective $*$ -automorphism with $α (a_{j}) = a_{j}^{†}$ (Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR).

Proof

1.1givenF1F2algebra

Since $α$ is a $*$ -automorphism, $π_{h} = π_{F} \circ α$ is a representation. If a closed subspace were invariant under $π_{h} ({C A R}_{N})$ , surjectivity of $α$ would make it invariant under $π_{F} ({C A R}_{N})$ ; [F1] therefore makes it trivial or all of $F_{\infty}$ . Thus $π_{h}$ is irreducible.

2.1givenF1F2algebra∎

Suppose a unitary $U : F_{\infty} \to F_{\infty}$ intertwines $π_{F}$ with $π_{h}$ . Then for every $j$ , $U c_{j} = π_{h} (a_{j}) U = c_{j}^{†} U$ . Applying this to the vacuum $Ω = e_{\emptyset}$ gives $c_{j}^{†} U Ω = U c_{j} Ω = 0$ for every $j$ . Since $U Ω \neq 0$ , this contradicts the common-creator-kernel statement in [F1]. No such unitary exists.

Remarks

The explicit obstruction is the absence of a nonzero vector in the infinite occupation space annihilated by every creator. In finite mode number such a fully occupied vector exists, so this argument is specifically about countably infinitely many modes.

5 · Examples, counterexamples and false statements

None yet.