Complete prerequisite proofs: unbounded generators, scalar bounds and the full CAR/factor route.

Bogoliubov automorphisms and the Hilbert–Schmidt criterion

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

A unitary change of one-particle coordinates acts on the CAR algebra. In its tracial representation this action extends normally to the hyperfinite factor. The innerness criterion below answers when its implementer belongs to that factor. Sufficiency follows from the eigenvectors of the compact difference from the identity; necessity uses a controlled diagonalizing perturbation. The eigenmode calculation constructs the local implementing matrices and their convergent product. The normal-extension and continuity theorem then describes the action as a map of topological groups. The complete analytic arguments and twelve worked problems supply the details used throughout.

The concrete CAR, its half-state trace and its completed pointed matrix product are supplied by Fermions, Fock space and quasi-free factors, Sections 10 and 15–16. The diagonal step uses the proved product-innerness theorem. Section 6 below supplies the whole analytic passage to arbitrary unitaries: the Hilbert–Schmidt ideal, compact diagonalization, interval projections, a bounded logarithm, and a dyadic diagonal perturbation with an explicit square-sum bound. It also constructs the unitary groups and the resolvent criterion needed for the possibly unbounded statement in Lemma 3.1. Its continuous calculus is the complete local ITP F03–F04, F06 and F08 argument and its scalar proofs; the underlying Hilbert estimates are ITP Sections 10.1–10.2. Normal continuity is proved directly on square-summable vector coefficients. This route uses neither a factor uniqueness theorem nor a trace-density theorem.

Throughout, $K$ is a separable infinite-dimensional complex Hilbert space, $A=\operatorname{CAR}(K)$, and

The preceding CAR lesson proves that $R$ is an AFD factor of type $\mathrm{II}_1$. We identify $A$ with its faithful represented image. Write $\|x\|_{2,\tau}=\tau(x^*x)^{1/2}$; the one-particle Hilbert–Schmidt norm is written $\|\cdot\|_{\mathrm{HS}}$.

Our annihilator $a(f)$ is conjugate-linear in $f$. Thus

In particular an eigenvalue $e^{is}$ multiplies the annihilator by $e^{-is}$. The one-mode implementers below reflect this convention. Changing to a linear creation generator changes that sign and the placement of the occupation projections together; the trace criterion is unchanged.

4. The general innerness theorem

Theorem 4.1. For every unitary $u$ on the separable infinite-dimensional space $K$,

Proof. Suppose first that $u-1\in\mathcal L^2(K)$. This difference is compact by Section 6.1. Lemma 3.2 therefore gives an orthonormal eigenbasis for $u$. If its eigenvalues are $e^{is_j}$, basis independence gives

The diagonal criterion of Theorem 2.1 makes $\alpha_u$ inner. Its phase-normalized matrix products give an implementing unitary in $R$, with any finite zero-overlap prefix retained. Thus the sufficient implication is already established for every one-particle unitary with Hilbert–Schmidt difference.

Now suppose that $\alpha_u$ is inner. Apply Lemma 3.3 to obtain $v,w$, with $v-1$ Hilbert–Schmidt and $w=uv$ diagonalizable. The sufficient implication just proved applies to $v$; consequently $\alpha_v$ is inner. Since $\alpha_w=\alpha_u\alpha_v$, composition with this inner automorphism gives

In particular $\alpha_w$ is inner, and its eigenbasis lets Theorem 2.1 conclude that $w-1$ is Hilbert–Schmidt. The two exact difference identities are

The second identity and the linear two-sided ideal estimate of Section 6.1 put $u-1$ in the same class. This proves necessity and completes both implications. $\square$

Example 4.2. A nontrivial scalar gauge $u=e^{is}1$, $s\notin2\pi\mathbb Z$, produces an outer automorphism of $R$: $u-1$ has the same nonzero norm on infinitely many orthonormal vectors and is not Hilbert–Schmidt. Every finite-mode restriction is nevertheless inner by (6). Conversely every unitary equal to the identity off a finite-dimensional subspace gives an inner automorphism.

Spatial implementation by $W_u$ is always available. Innerness asks the stronger question whether an implementer belongs to $R$.

2. Eigenvectors and tensor coordinates

Suppose $uf_j=e^{is_j}f_j$ for an orthonormal basis $(f_j)$. Put

The CAR lesson proves that the matrix algebras $N_j$ generated by $b_j$ commute, generate $R$, and identify it with the tracial product $\bar\bigotimes_j(M_2,\operatorname{tr}_2)$. Their matrix units have $e_{11}^{(j)}=1-n_j$, $e_{22}^{(j)}=n_j$, and $e_{12}^{(j)}=b_j$.

Since $\alpha_u(n_j)=n_j$, every parity factor in (5) is fixed. Hence

Define the even one-mode unitary

Matrix-unit multiplication gives

The diagonal projections and the adjoint generator transform as required. Thus $N_j$ is invariant, its restriction is $\operatorname{Ad}w_j$, and

The identity first holds on every finite tensor prefix and then holds normally on $R$.

Theorem 2.1. Under the eigenbasis assumption,

Proof. Equations (7)–(8) and the already proved product-innerness theorem give the first equivalence, including finitely many zero overlaps. Put $t_j=|\cos(s_j/2)|$. Then

so

Summing and using the basis definition of Hilbert–Schmidt class proves the second equivalence. $\square$

For sufficiency one can also see the implementer: adjust the scalar phase of each $w_j$ so that its trace is nonnegative. The partial products converge strongly, together with their adjoints, by the product theorem's tail estimate. Their limit is a unitary in $R$. Finitely many vanishing traces affect a finite prefix only.

3. Perturbing self-adjoint generators

We first justify the analytic estimate used to pass from an arbitrary unitary to a diagonal one. Section 6 proves its Hilbert–Schmidt facts and its unbounded-operator inputs.

Lemma 3.1. Let $h$ be any self-adjoint operator on $K$, possibly unbounded, and let $k=k^*\in\mathcal L^2(K)$. On $\operatorname{Dom}h$, $h+k$ is self-adjoint. For every real $t$,

where the integral converges in Hilbert–Schmidt norm. Moreover

Proof. Hilbert–Schmidt operators are bounded. For sufficiently large $r>0$,

has a bounded inverse, since $\|k(h\mp ir)^{-1}\|\le\|k\|/r<1$. Bounded addition preserves closedness on $\operatorname{Dom}h$, and the sum is symmetric. Surjectivity at both imaginary points is the self-adjoint resolvent criterion proved in Section 6.6, so $h+k$ is self-adjoint on that domain.

The groups constructed in Section 6.6 preserve their generator domains and have the stated strong derivatives there. Thus the unitary group of $h+k$ preserves $\operatorname{Dom}(h+k)=\operatorname{Dom}h$. For $\xi$ in this common domain the strong product derivative is legitimate and equals

Indeed the two terms containing $h$, evaluated on $e^{is(h+k)}\xi$, cancel. Integrating gives (11) on the dense common domain.

The needed graph continuity is explicit: $h e^{is(h+k)}\xi=e^{is(h+k)}(h+k)\xi-k e^{is(h+k)}\xi$ is continuous in $s$. Thus the domain differentiation uses a graph-continuous vector path, not an operator-norm derivative of the unbounded generator's group.

For a finite-rank middle operator, the integrand in (11) is Hilbert–Schmidt norm continuous: write it as a finite sum of rank-one operators and use strong continuity of both unitary groups and their adjoints. Approximate a general $k$ in Hilbert–Schmidt norm by finite-rank operators, keeping both unitary groups fixed and replacing only their middle factor. The ideal bound and unitarity make that approximation uniform in $s$. The integrand is therefore Hilbert–Schmidt norm continuous, with norm $\|k\|_{\mathrm{HS}}$. Its Banach-space integral has norm at most $|t|\|k\|_{\mathrm{HS}}$, and is a bounded operator. Density extends the integrated domain identity to all of $K$. This proves (11)–(12), with the usual reversed orientation for $t<0$. $\square$

Lemma 3.2. If $v$ is unitary and $v-1$ is compact, then $v$ has an orthonormal basis of eigenvectors.

Proof. The compact operator $T=v-1$ is normal, because both $T^*T$ and $TT^*$ equal $2-v-v^*$. Its real and imaginary parts are commuting compact self-adjoint operators. Decompose by the compact self-adjoint diagonalization proved in Section 6.2 for $\operatorname{Re}T$. Every nonzero eigenspace is finite dimensional and is preserved by $\operatorname{Im}T$, so diagonalize the latter on each. On $\ker\operatorname{Re}T$, diagonalize its compact self-adjoint restriction $\operatorname{Im}T$, including a basis of the joint kernel. The resulting orthonormal basis diagonalizes both parts, hence $T$ and $v$. $\square$

Lemma 3.3. For every unitary $u$ on $K$ and every $\varepsilon>0$, there is a unitary $v$ with

such that $w=uv$ has an orthonormal basis of eigenvectors.

Proof. Section 6.4 constructs a self-adjoint logarithm $h$, with $\|h\|\le\pi$, such that $u=e^{ih}$, from continuous functions of its real part and the positive and negative subspaces of its imaginary part. Section 6.5 proves the required bounded Weyl–von Neumann statement at separable Hilbert-space generality: it supplies a self-adjoint $k\in\mathcal L^2(K)$, with $\|k\|_{\mathrm{HS}}<\varepsilon$, for which $h+k$ has an orthonormal eigenbasis. Set

These operators are unitary. Lemma 3.1 at $t=1$ gives the required Hilbert–Schmidt estimate, and $w$ is diagonal on the eigenbasis of $h+k$. $\square$

Separability enters through this diagonal perturbation input. The argument does not assert the same diagonal perturbation theorem on an arbitrary nonseparable space.

1. Normal extension and continuity

Every tracial state on $A$ restricts to the normalized trace on each finite-mode full matrix algebra. Their union is norm dense, so $\tau$ is the unique tracial state. Hence $\tau\alpha_u=\tau$.

Theorem 1.1. Each $\alpha_u$ has a canonical normal extension to $R$, still denoted by $\alpha_u$. The map

is an injective continuous homomorphism, for the strong operator topology on $\mathcal U(K)$ and the point-predual norm topology on $\operatorname{Aut}(R)$.

Proof. On the dense tracial GNS subspace put

Trace preservation gives $\|W_u(x\Omega)\|=\|x\Omega\|$, and $W_{u^{-1}}$ gives its inverse. Thus $W_u$ is unitary. On $A$, $W_u xW_u^*=\alpha_u(x)$, so conjugation maps $R=A''$ onto itself and gives a normal extension. Two normal extensions agreeing on an ultraweakly dense algebra agree everywhere. Formula (3) also gives the homomorphism law.

If $u_j\to u$ strongly, the CAR norm identity gives

Finite product telescoping and norm density give convergence on every $x\in A$. Consequently $W_{u_j}\to W_u$ strongly. Also $u_j^*\to u^*$ strongly, since $\|(u_j^*-u^*)f\|=\|f-u_j u^*f\|$.

Here is the entire normal-functional argument, without a trace density. On the concrete $R\subset B(H_\tau)$, a normal functional is the restriction of

The exact coefficient topology and its continuous dual are explained in Section 6.7. Conjugation by $W_{u_j}$ pulls this functional back to the same series with $\xi_n,\eta_n$ replaced by $W_{u_j}^*\xi_n,W_{u_j}^*\eta_n$. Strong convergence of these adjoints follows from the already proved convergence for $u_j^{-1}$. It implies convergence on each square-summable list in its square-sum norm: first retain finitely many vectors, and then bound each omitted difference by twice the vector's norm. Cauchy–Schwarz bounds the functional-norm difference by

This tends to zero and proves convergence in the specified point-predual norm topology. The finite-list and tail argument works for nets as well as sequences.

Finally, if $\alpha_u=\operatorname{id}$, then $a(uf-f)=0$. The identity $\|a(g)\|=\|g\|$ makes $uf=f$ for every $f$, so $u=1$. $\square$

5. Exercises with complete solutions

Exercise 1. Show directly that trace preservation in Section 1 follows from finite-mode matrix traces even if $u$ does not preserve a chosen increasing sequence of finite-mode algebras.

Solution. Fix a finite-dimensional $E\subset K$, and set $d=\dim E$. Matching the fields sends $\operatorname{CAR}(E)$ onto $\operatorname{CAR}(uE)$; the inverse matches them using $u^*$. Both are full matrices of size $2^d$.

The normalized trace can be recovered from matrix-unit multiplication alone. A tracial state $t$ vanishes on $e_{ab}$ for $a\ne b$, since $e_{aa}e_{ab}=e_{ab}$ but $e_{ab}e_{aa}=0$. Also

Their sum is $t(1)=1$, so each diagonal value is $2^{-d}$. Consequently the pullback of the trace on $\operatorname{CAR}(uE)$ is the trace on $\operatorname{CAR}(E)$, and $\tau(\alpha_u(x))=\tau(x)$ for every $x$ in that algebra.

Every CAR polynomial lies in some finite $E$. For $x\in A$, choose such a polynomial $x_0$ with $\|x-x_0\|<\varepsilon$. The automorphism is isometric and the state has norm one, giving

Letting $\varepsilon\to0$ proves trace preservation on $A$. This uses the directed collection of all finite subspaces, including $uE$; a fixed exhaustion need not be invariant.

Exercise 2. Verify the one-mode phase in (7), including its action on the occupation projection.

Solution. Write $q=1-n$. The one-mode relations give $qb=b$, $nb=0$, $bq=0$, and $bn=b$. For $w=e^{-is/2}q+e^{is/2}n$,

Taking adjoints gives $wb^*w^*=e^{is}b^*$. Orthogonality of $q,n$ proves $w^*w=ww^*=q+n=1$, and $w$ commutes with each projection, so $wnw^*=n$ and $wqw^*=q$. The tracial one-mode values are $\tau(q)=\tau(n)=1/2$; hence

The annihilator convention is essential: $a(e^{is}f)=e^{-is}a(f)$, exactly the phase just computed.

Exercise 3. Explain why diagonalizing a one-particle unitary does not make the raw one-mode CAR algebras commute, and why the tensor calculation still works.

Solution. A coordinate change preserves the CAR. For $i\ne j$, $a_i a_j=-a_j a_i$, and this product is nonzero: it removes two occupied modes from their common finite occupation vector. Thus the raw one-mode algebras still do not commute.

Let $S_j=Z_1\cdots Z_{j-1}$, so $b_j=S_j a_j$. For $i<j$, the string $S_j$ contains $Z_i$, which anticommutes with $a_i$, while the other string factors commute with it. The earlier string commutes with $a_j$. Therefore

Replacing either field by its adjoint gives the same two canceling signs. These are the commuting full matrix legs $N_j$.

Furthermore $b_j^*b_j=n_j$. A diagonal $\alpha_u$ fixes $n_j$, hence $Z_j$ and every $S_j$, and sends $b_j$ to $e^{-is_j}b_j$. Exercise 2 supplies the even implementer $w_j$ on each $N_j$. The resulting product calculation applies to these commuting legs and their normal closure $R$, with overlaps $\cos(s_j/2)$; it does not assert commutation of the raw fields.

Exercise 4. Let $u f_j=e^{i/j}f_j$. Is $\alpha_u$ inner? Is $u-1$ trace class?

Solution. Put $T=u-1$. Its diagonal entries are $d_j=e^{i/j}-1$. Integrating the scalar derivative of $e^{it}$ gives $|e^{ix}-1|\le|x|$, and hence

For example, the block $2^n\le j<2^{n+1}$ contributes at most $2^{-n}$, so the last sum is at most $2$.

Innerness also has a concrete limit here. Take the $w_j$ of (6), whose traces $r_j=\cos(1/(2j))$ are positive, and let $W_m=\prod_{j\le m}w_j$. For $n>m$,

The same estimate on every finite-support vector and for adjoints gives strong* convergence to a unitary $W\in R$. Its conjugation on each finite leg is eventually $\alpha_u$, so $W$ implements the normal automorphism.

For trace class, $T^*T$ is diagonal with entries $|d_j|^2$, so its positive square root has eigenvalues

The inequality follows from the chord bound $\sin t\ge2t/\pi$ on $[0,\pi/2]$. The harmonic sum diverges: every block $2^n\le j<2^{n+1}$ contributes at least $1/2$. Thus these singular values are not summable and $T$ is not trace class.

Exercise 5. Repeat Exercise 4 with $u f_j=e^{i/\sqrt j}f_j$. Show that compactness alone cannot replace Hilbert–Schmidt class.

Solution. Here the diagonal entries are $d_j=e^{i/\sqrt j}-1$. If $P_m$ projects onto the first $m$ basis vectors, then $TP_m$ is finite rank and

Thus $T=u-1$ is compact. The same scalar chord estimate as above gives

Its Hilbert–Schmidt square sum diverges. Its singular-value sum diverges as well, so the answer to the trace-class question repeated from Exercise 4 is also negative.

To see the obstruction in tensor coordinates, set $r_j=\cos(1/(2\sqrt j))$. Since $|d_j|^2=4(1-r_j)(1+r_j)\le8(1-r_j)$, the overlap defects have divergent sum. If a unitary $U\in R$ implemented the product action, the expectation argument of Theorem 7.2 in the tensor-product lesson would give

Choosing $m$ with $|a_m|>0$ would bound every subsequent defect sum by $-\log|a_m|$, a contradiction. Hence $\alpha_u$ is outer. The explicit norm approximation above shows why compactness does not suffice.

Exercise 6. Can an inner diagonal Bogoliubov automorphism have a one-mode implementer of trace zero? Can it have infinitely many such modes?

Solution. The overlap vanishes exactly when $\cos(s_j/2)=0$, that is, when $e^{is_j}=-1$. Suppose first that this occurs on a finite set $F$, and the remaining eigenvalues are $1$. Then

is an even finite CAR unitary implementing the required sign changes. For a single such mode, $u-1$ has rank one. Thus one trace-zero mode, or any finite number, is compatible with innerness.

If there are infinitely many such modes, each contributes $4$ to that square sum and $1$ to the overlap-defect sum in (9). The diagonal necessity criterion rules out innerness.

A finite zero-overlap prefix also causes no problem when the later modulus defects are summable. Past that prefix, replace each $w_j$ by a scalar multiple with trace $r_j=|\cos(s_j/2)|>0$; scalar multiplication leaves its conjugation unchanged. For these normalized partial products, the squared difference on a local vector is $2\|\xi\|^2(1-\prod_{\text{tail}}r_j)$; multiplying by the fixed prefix leaves this norm unchanged. The adjoint estimate is identical. Thus tail convergence can produce a unitary even though the trace of the whole prefix is zero.

Exercise 7. Prove the continuity of the Hilbert–Schmidt integrand in (11) for a rank-one operator, without operator-norm continuity of an unbounded generator's unitary group.

Solution. Allow arbitrary $\xi,\eta\in K$, with no generator-domain assumption, and put $\theta_{\xi,\eta}z=\xi\langle\eta,z\rangle$. For the fixed groups $U(s)=e^{-ish}$, $V(s)=e^{is(h+k)}$, unitarity and the adjoint identity give

Parseval gives $\|\theta_{\xi,\eta}\|_{\mathrm{HS}}=\|\xi\|\|\eta\|$. Subtract the expressions at $s,t$, adding and subtracting $\theta_{U(t)\xi,V(s)^*\eta}$, to obtain

Both terms vanish as $s\to t$ by strong continuity; $V(s)^*=e^{-is(h+k)}$ is the negative-time group. This proves Hilbert–Schmidt norm continuity even for unbounded $h$. Only the two individual vector orbits are estimated; no operator-norm continuity of either group is used. Finite sums and uniform Hilbert–Schmidt approximation extend the argument to the middle factors in (11).

Exercise 8. For commuting bounded self-adjoint $h,k$, simplify (11) and check its sign.

Solution. Since bounded $h,k$ commute, their exponential series can be multiplied and regrouped, giving

Thus the left side of (11) is $e^{itk}-1$, and its right side is

because the norm derivative of $e^{isk}$ is $ik e^{isk}$. This also holds for $t<0$, with the oriented integral $\int_0^t=-\int_t^0$. The first-order term $itk$ checks the positive sign. When $k$ is Hilbert–Schmidt, as in Lemma 3.1, unitary invariance of that norm gives the bound $|t|\|k\|_{\mathrm{HS}}$; the bounded-operator identity itself does not require this extra assumption. The same bound follows without an integral after diagonalizing the compact self-adjoint $k$: if its eigenvalues, including zeros, are $\mu_j$, then

Exercise 9. In the proof of Theorem 4.1, explain why the fact that $v-1$ is Hilbert–Schmidt proves innerness of $\alpha_v$ without circular reasoning.

Solution. The argument first reduces $v$ to the already proved diagonal case. Section 6.1 approximates $v-1$ in operator norm by finite-rank operators, so it is compact. It is also normal:

Its commuting compact self-adjoint real and imaginary parts can be diagonalized together as in Section 6.2. Nonzero eigenspaces of the real part are finite-dimensional; diagonalize the imaginary part there and on the real part's kernel, retaining a basis of the joint kernel. This gives an eigenbasis for $v$, including its $1$-eigenspace.

Write its eigenvalues as $e^{it_j}$. Basis independence of the Hilbert–Schmidt norm gives

Theorem 2.1, whose proof uses the product implementer and expectation-tail calculation, therefore proves $\alpha_v$ inner. This argument uses only Lemma 3.2 and the diagonal theorem established before Theorem 4.1. It never applies the desired arbitrary-unitary equivalence to $v$. Only now use $w=uv$ and $\alpha_w=\alpha_u\alpha_v$: composition and inversion of inner automorphisms remain inner, so (16) transfers the question from $u$ to the diagonal $w$.

Exercise 10. Show that $\mathcal U_2(K)=\{u\in\mathcal U(K):u-1\in\mathcal L^2(K)\}$ is a subgroup. Show that it is normal in $\mathcal U(K)$.

Solution. The identity belongs to $\mathcal U_2(K)$. For two members $u,v$,

The two-sided ideal and linearity proved in Section 6.1 put this difference in the Hilbert–Schmidt class. For inverses,

Thus the set is a subgroup. Finally, for any unitary $w$, without any Hilbert–Schmidt assumption on $w-1$,

Conjugation preserves membership, which is precisely normality of this subgroup in $\mathcal U(K)$. The proved equivalence (15) also gives the exact identification

the preimage of the inner subgroup under (2).

Exercise 11. Let $u f_j=e^{is_j}f_j$, and let $u_m$ agree with $u$ on $f_1,\ldots,f_m$ and with the identity on the remaining basis vectors. Prove that inner automorphisms $\alpha_{u_m}$ can converge to an outer $\alpha_u$.

Solution. For $f=\sum_j c_jf_j$, the truncation has

The same estimate applies to adjoints. Moreover $u_m-1$ has finite rank. A finite implementer is $W_m=\prod_{j\le m}w_j$ in the corresponding finite CAR algebra; its evenness fixes all later fields.

The convergence on one-particle vectors implies convergence in norm on every CAR generator. Telescoping a polynomial and using norm density then proves point-norm convergence $\alpha_{u_m}(x)\to\alpha_u(x)$ on $A$. For the extensions to $R$, let $V_m,V$ be their trace-preserving GNS unitaries from (3). They converge strongly together with their adjoints. If $\psi(x)=\sum_l\langle\xi_l,x\eta_l\rangle$ is any normal coefficient series, the pullback difference has norm at most

Finite-list convergence followed by a uniformly bounded square-summable tail proves that this tends to zero. Hence convergence also holds in the point-predual norm topology of Theorem 1.1.

If $\sum_j|e^{is_j}-1|^2=\infty$, the limit is outer. Taking $s_j=1/\sqrt j$ gives the compact example of Exercise 5; taking $s_j=\pi$ gives the constant gauge $u=-1$, for which $u-1=-2\,1$ is not Hilbert–Schmidt. In each case the approximating automorphisms are inner and the limit is outer.

The full density assertion for $\operatorname{Int}(R)$ can also be verified from these local tools. Given any normal automorphism $\beta$ of $R$, the normal trace $\tau\beta$ restricts on each finite matrix algebra $A_n$ to its unique normalized trace. Ultraweak density therefore gives $\tau\beta=\tau$. If $(e_{ij})_{1\le i,j\le2^n}$ are its matrix units, $e_{11}$ and $\beta(e_{11})$ have equal trace. Projection comparison in the finite factor, with faithfulness killing any equal-trace remainder, supplies $v^*v=e_{11}$, $vv^*=\beta(e_{11})$. Define

Matrix-unit multiplication gives $Z_n^*Z_n=Z_nZ_n^*=1$ and $Z_ne_{ij}Z_n^*=\beta(e_{ij})$. Thus $\gamma_n=\operatorname{Ad}Z_n$ agrees with $\beta$ on $A_n$. For every $x\in R$,

Here both automorphisms preserve $\tau$, and the finite expectations recover every GNS vector. Their trace-preserving GNS unitaries consequently converge strongly, together with their adjoints, to that of $\beta$. The same square-summable coefficient estimate above proves point-predual norm convergence. Hence $\operatorname{Int}(R)$ is dense in $\operatorname{Aut}(R)$, while the outer example proves it is proper.

Exercise 12. A unitary $u$ is the identity off a $d$-dimensional reducing subspace $E$. Locate an implementer of $\alpha_u$ inside the finite CAR algebra and verify that it commutes with all tail annihilators.

Solution. Choose an orthonormal eigenbasis $f_1,\ldots,f_d$ for the finite unitary $u|_E$, with eigenvalues $e^{is_j}$, and put $n_j=a(f_j)^*a(f_j)$. The unitary

belongs to $\operatorname{CAR}(E)$. Each factor is even, and the factors commute because the occupation projections commute. For $i\ne j$, $n_i$ commutes with $a(f_j)$ by two CAR sign changes. Exercise 2 therefore gives $Wa(f_j)W^*=e^{-is_j}a(f_j)=a(uf_j)$.

For $g\in E^\perp$, the CAR gives

Thus every factor of $W$, and $W$ itself, commutes with every tail annihilator and its adjoint. Its conjugation fixes $a(g)=a(ug)$. Splitting an arbitrary $f$ into $E$ and $E^\perp$ proves the desired action on all generators and hence on the completed algebra and its normal closure. For $d=0$, the empty product is $1$.

6. Complete analytic support for the proof route

These supporting arguments do not add numbered results to the original six statements. They prove the exact analytic inputs used there. We work with complete complex Hilbert spaces, bounded adjoints, and the continuous calculus proved in ITP F03–F04, F06 and F08, whose scalar completeness, polynomial approximation, exponential series and Banach-valued integral proofs are in Elementary calculus for infinite products. These are local proof providers, rather than a citation to a projection-valued spectral theorem. The Hilbert-space completion, orthogonal projection, Riesz and adjoint arguments are the full ITP Sections 10.1–10.2 constructions.

6.1. The Hilbert–Schmidt space and its whole ideal estimate

A countable orthonormal basis of a separable Hilbert space is obtained by applying Gram–Schmidt to a countable dense sequence and discarding zero residuals. Each original vector belongs to the span of the residuals through its stage. The completed span is therefore the whole space. Orthogonal finite partial sums give Pythagoras; their distance to a vector tends to zero by this density. This proves Parseval. Bessel's inequality is the same finite Pythagoras estimate without totality. The assertions also cover finite-dimensional and zero subspaces.

For a bounded $T$, define

on an orthonormal basis $(e_j)$. Nonnegative double sums can be interchanged: both iterated sums are the supremum of all finite rectangle sums. Thus Parseval on a second basis $(f_l)$ gives

Keep the second basis fixed and change the first; this proves basis independence. It also proves $\|T^*\|_{\mathrm{HS}}=\|T\|_{\mathrm{HS}}$, allowing either side to be infinite.

For a finite sum $x=\sum_j c_je_j$, Cauchy–Schwarz gives

Density proves $\|T\|\le\|T\|_{\mathrm{HS}}$. Conversely any square-summable matrix $(t_{lj})$ defines an operator on finite coordinate vectors with exactly this bound; it extends boundedly to the Hilbert completion and has the prescribed columns. Consequently the Hilbert–Schmidt space is the complete Hilbert space of these matrices, with its square-sum inner product. Its norm triangle inequality and its linearity are the Hilbert-space ones, with no spectral or singular-value premise.

For $S,T$ Hilbert–Schmidt or bounded as appropriate,

This proves the linear two-sided ideal property. If $B,C$ are unitary, apply the bounds both to $BTC$ and to $B^*(BTC)C^*$; hence

Writing $P_m$ for the first $m$ basis vectors,

Thus finite-rank operators are dense in this norm and every Hilbert–Schmidt operator is compact. To justify that last word explicitly, a finite-rank operator maps the unit ball into a bounded finite-dimensional set, which has finite nets of every positive radius by coordinate compactness. A norm limit of such operators has the same finite-net property, by approximation in operator norm. Any sequence in its image then has a Cauchy subsequence, by successive finite-net extraction; Hilbert completeness gives its limit. This is compactness.

Conversely, let $T$ be any compact bounded operator. Its unit-ball image has a finite $\delta$-net for every $\delta>0$: otherwise successive choices at pairwise distance at least $\delta$ would give a sequence with no convergent subsequence, contrary to compactness. Take net points $y_1,\ldots,y_m$ and the orthogonal projection $P$ onto their finite-dimensional span. Since that projection gives the nearest vector in its range,

Thus every compact operator is a norm limit of finite-rank operators. Adjoint isometry gives $T^*=\lim(PT)^*$ in norm, and $(PT)^*=T^*P$ has finite-dimensional range. The first direction above makes $T^*$ compact. Finite sums and scalar multiples are compact by adding their finite-rank approximations. This proves, in particular, compactness of both real and imaginary parts for the arbitrary compact $T=v-1$ in Lemma 3.2; it does not assume that $T$ is Hilbert–Schmidt.

For the rank-one operator $\theta_{\xi,\eta}z=\xi\langle\eta,z\rangle$, Parseval gives