Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 (Von Neumann Algebras)

Use complex inner products linear in the first argument throughout. The spectral theorem for compact self-adjoint operators states that, for a compact operator $T=T^{\ast }$ on a Hilbert space, its nonzero spectral values are real eigenvalues, each of finite multiplicity, with no possible accumulation point except zero. Moreover,where $P_{\lambda }$ are the mutually orthogonal eigenspace projections. The operator series converges in operator norm when the nonzero eigenvalues are listed in decreasing absolute value. The sum is at most countable, even if the kernel is nonseparable. Zero need not be an eigenvalue.

First prove that a nonzero compact self-adjoint operator has an eigenvalue of magnitude its norm, without assuming a spectral decomposition. Put $r=\Vert T\Vert >0$ and choose unit vectors ${\xi }_n$ with $\Vert T{\xi }_n\Vert \to r$. Since $\Vert T^2\xi \Vert \le r\Vert T\xi \Vert$, expansion givesCompactness of $T^2$ supplies a subsequence for which $T^2{\xi }_n$ converges; the displayed estimate then forces ${\xi }_n$ itself to converge to a unit vector $\xi$ satisfying $T^2\xi =r^2\xi$. The vectors $\xi +T\xi /r$ and $\xi -T\xi /r$ are eigenvectors for $r$ and $-r$, respectively, whenever nonzero, and they cannot both vanish. This proves the claim.

Self-adjointness gives reality of every eigenvalue and orthogonality of eigenspaces for distinct eigenvalues. An eigenspace for $\lambda \ne 0$ cannot contain an infinite orthonormal sequence: its images under $T$ would have pairwise distance $\sqrt{2}|\lambda |$, contradicting compactness. The same argument proves that there are only finitely many mutually orthogonal eigenvectors with eigenvalues of magnitude at least any given $\epsilon >0$. Hence all nonzero eigenspaces together have a countable orthonormal basis, and their eigenvalues tend to zero if there are infinitely many.

Let $K$ be the orthogonal complement of the closed span of all these eigenspaces. It is invariant under $T$. If $T{|}_K$ were nonzero, the norm-attaining eigenvalue argument applied to that compact self-adjoint restriction would produce another nonzero eigenvector, a contradiction. Thus $K\subseteq \ker T$, and the converse inclusion follows from orthogonality. This establishes the asserted decomposition. On its orthogonal summands the tail of the operator series has norm equal to the supremum of the omitted eigenvalue magnitudes, proving norm convergence.

Finally, a nonzero complex number not among these eigenvalues has positive distance from their set and from zero. Inverting $T-zI$ separately on the eigenspaces and kernel therefore gives a bounded inverse. This proves that no other nonzero spectral values occur and completes the theorem.

For a finite-rank operator $T$, define $|T|=(T^{\ast }T{)}^{1/2}$ and its trace norm bywhere $s_j>0$ are its nonzero singular values, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T={\sum }_j{s}_j{R}_{u_j,v_j}$, where $R_{x,y}z=⟨z,y⟩x$.

For every bounded operator $B$,Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their orthogonal complement. It attains equality, soThis formula proves the triangle inequality and absolute homogeneity. Also $\Vert T\Vert \le \Vert T{\Vert }_1$, so vanishing of the trace norm forces $T=0$. Thus it is a norm on the vector space of finite-rank operators.

The map $B\mapsto F_B$, where $F_B(T)=\mathrm{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its norm is at most $\Vert B\Vert$. Rank-one tests give the reverse inequality: $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, so taking unit vectors $x,y$ gives $\Vert F_B\Vert =\Vert B\Vert$.

Conversely, let $F$ be a bounded linear functional for the trace norm. The sesquilinear form $\beta (x,y)=F(R_{x,y})$ satisfiesThe Riesz representation theorem gives a unique bounded operator $B$ with $\beta (x,y)=⟨Bx,y⟩$, and $\Vert B\Vert \le \Vert F\Vert$. Since every finite-rank operator is a finite sum of rank-one operators, linearity implies $F(T)=\mathrm{Tr}(BT)$ for every $T$. HenceThe same identification extends to the completion, the trace-class operators. This is trace-class duality; completeness of the original finite-rank space is not required to identify its bounded dual.

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

A Hadamard space is a complete CAT(0) space. A useful equivalent definition, which makes the remaining assertions consequences rather than assumptions, is a complete metric space in which every pair $x,y$ has a midpoint $m$, with $d(x,m)=d(y,m)=d(x,y)/2$, satisfyingThis is the nonpositive-curvature midpoint inequality. Euclidean Hilbert spaces satisfy it with equality in the parallelogram formula. The comparison argument in (c) shows that this definition gives the usual CAT(0) triangle condition.

Write $L=d(x,y)$. If $m^{\prime }$ is any other metric midpoint, apply the inequality from (a) to the chosen midpoint $m$ with $z=m^{\prime }$. The right side is $\frac{1}{2}(L/2{)}^2+\frac{1}{2}(L/2{)}^2-L^2/4=0$, so $m=m^{\prime }$. Every pair has a unique midpoint.

Choose successive midpoints to define $\gamma (k/2^n)$ on all dyadic parameters. Consecutive points at level $n$ have distance $L/2^n$. The triangle inequality gives $d(\gamma (s),\gamma (t))\le (t-s)L$ for dyadic $s<t$. Conversely,which gives the reverse bound. Thus equality holds. Completeness extends this map uniquely and continuously to $[0,1]$, retaining $d(\gamma (s),\gamma (t))=|s-t|L$. It is a constant-speed geodesic.

Every other geodesic has the same midpoint, and then the same successive midpoints, so agrees on all dyadic parameters. Continuity gives agreement everywhere. Hence $\boxed{\text{there is exactly one geodesic segment between any two points}.}$

The comparison triangle must have third side $d(x,y)$: the printed $d(y,z)$ contains an undefined $z$. Use $A=d(a,x)$, $B=d(a,y)$ and $C=d(x,y)$.

First extend the midpoint inequality along any constant-speed geodesic $\gamma$ from $p$ to $q$. If $L=d(p,q)$, the function $d(z,\gamma (t){)}^2-L^2{t}^2$ is midpoint-convex by (a). Iteration proves its convexity bound for dyadic $t$, and continuity then gives the Hadamard squared-distance inequalityApply this first to $[a,x]$ with $z=y_t$, and then to $[a,y]$ with $z=x$:Simplification givesIn a Euclidean comparison triangle, the cosine rule gives the scalar product of the two vectors from the common vertex as $(A^2+B^2-C^2)/2$. The squared distance between their multiples $s$ and $t$ is exactly the right side displayed above. Taking nonnegative square roots proves the required triangle comparison, including degenerate Euclidean triangles.

Let $p,q$ lie in the closed ball of center $z$ and radius $R$. On their unique geodesic, the Hadamard squared-distance inequality givesThus every point of that geodesic lies in the ball. By uniqueness there are no other geodesics to consider, so every closed ball is geodesically convex.

The Stone theorem for the circle group says that a strongly continuous unitary representation $U$ of $\mathbb{T}=\{z:|z|=1\}$ has an orthogonal decompositionEquivalently, $U_z={\sum }_n{z}^n{P}_n$ strongly, with mutually orthogonal projections $P_n$ summing strongly to the identity. Define these projections using normalized Haar measure:Haar invariance gives $U_w{P}_n=w^n{P}_n$. Changing $z$ to $z^{-1}$ in the adjoint integral proves $P_n^{\ast }=P_n$. Applying the defining integral to $P_m\xi$ gives $P_n{P}_m={\delta }_{nm}{P}_m$ by orthogonality of the characters $z^k$. Thus $P_n$ are orthogonal projections, and their ranges are exactly the indicated character subspaces.

It remains to prove completeness of these subspaces. The Fejér kernel has the finite expansionIt is nonnegative, has normalized integral one, and its integral outside any fixed neighborhood of the identity tends to zero, since the denominator in the usual geometric-sum formula is bounded away from zero there. Strong continuity therefore givessplit the integral into that neighborhood, where $\Vert U_z\xi -\xi \Vert$ is small, and its complement, where it is bounded by $2\Vert \xi \Vert$. The integral on the left is ${\sum }_{|n|<N}(1-|n|/N)P_n\xi$. Hence the closed span of all $H_n$ is $H$. Orthogonality then gives the unweighted strong sum and the stated representation formula.

Conversely, any orthogonal decomposition indexed by the integers gives this representation; continuity follows first on finite sums and then on all vectors by the uniform unitary norm bound. With $z=e^{it}$ its generator isThis diagonal operator is self-adjoint and $U_{e^{it}}=e^{itA}$. Thus the circle theorem is the integer-spectrum version of the one-parameter theorem.

The Stone theorem for one-parameter unitary groups states that every strongly continuous unitary representation $U_t$ of $\mathbb{R}$ is uniquely $U_t=e^{itA}$ for a possibly unbounded self-adjoint operator $A$. Its domain isConversely every self-adjoint $A$ gives such a group by the Borel functional calculus for a normal operator.

Prove the forward direction by putting $G=iA$, initially defined as the norm derivative at zero. Its domain is dense: for $f\in C_c^{\infty }(\mathbb{R})$, the vector ${\xi }_f=\int f(t)U_t\xi dt$ has derivative $G{\xi }_f=-\int f^{\prime }(t)U_t\xi dt$. Smooth approximate identities make ${\xi }_f\to \xi$. The group law gives $U_{t}D(G)=D(G)$ and $G{U}_t=U_{t}G$ on that domain. Differentiation of preserved inner products shows that $G$ is skew-symmetric, so $A=-iG$ is symmetric.

The derivative operator is closed. Indeed, if ${\xi }_n\to \xi$ and $G{\xi }_n\to \eta$, integrate the orbit derivative to obtainTaking limits gives the same identity with $\xi ,\eta$; dividing by $t$ at zero proves $\xi \in D(G)$ and $G\xi =\eta$.

Now define bounded Laplace integralsShifting the integration variable in $U_h{R}_{\pm }\xi$ and taking the derivative at zero shows that $R_{\pm }\xi \in D(G)$ andHence $I-G$ and $I+G$ are both surjective. Since $G=iA$, this means $A+iI$ and $A-iI$ are surjective.

Here is why these range conditions establish self-adjointness rather than just symmetry. If $\eta \in D(A^{\ast })$, solve $(A-iI)\xi =(A^{\ast }-iI)\eta$. Then $\eta -\xi \in \ker (A^{\ast }-iI)=\mathrm{Ran}(A+iI{)}^{\perp }=\{0\}$. Thus $D(A^{\ast })\subseteq D(A)$, and symmetry gives equality with $A^{\ast }=A$.

The permitted functional calculus now gives $V_t=e^{itA}$. It is unitary, strongly continuous by dominated convergence against spectral measures, and differentiable with derivative $iA{V}_t\xi$ on $D(A)$. Differentiate $V_{-t}{U}_t\xi$ for $\xi \in D(A)$; the two derivative terms cancel because both groups preserve the domain and commute with their generator. Its value is consequently the constant $\xi$. Density of the domain implies $U_t=V_t$ on all of $H$.

For completeness, the converse and the exact domain follow from the same calculus. If $\xi \in D(A)$, dominated convergence applied to $|(e^{it\lambda }-1)/t|\le |\lambda |$ gives the derivative $iA\xi$. If that norm derivative exists, Fatou's lemma applied to its squared norm gives $\int {\lambda }^{2}d⟨E_A(\lambda )\xi ,\xi ⟩<\infty$, so $\xi \in D(A)$. The derivative characterization also proves uniqueness. Thus

For a discrete group, one standard definition of amenability is the fixed-point property: every continuous affine action on a nonempty compact convex set in a locally convex space has a fixed point. The equivalent state formulation, which may be stated without proof here, is that there is a left-invariant invariant mean on ${\ell }^{\infty }(\Gamma )$. Equivalently, every unital left-translation-invariant star-subalgebra of ${\ell }^{\infty }(\Gamma )$ admits a left-invariant state, with continuity for the supremum norm; one may equally pass to its norm closure. A state means a positive linear functional of norm one with value one on the constant function one. Restriction proves the subalgebra formulation from the full-algebra version, and taking the full algebra proves the converse. Such a mean is finitely additive on indicator functions, not necessarily countably additive.

Write $(\lambda (g)\mu )(h)=\mu (g^{-1}h)$ on ${\ell }^1(\Gamma )$ and $(L_{g}f)(h)=f(g^{-1}h)$ on bounded functions. We prove equivalence of this mean condition, the Reiter condition, and existence of a Følner sequence.

Assume first that an invariant mean $m$ exists. Fix a finite set $K\subseteq \Gamma$. In the real Banach space ${⨁}_{g\in K}{\ell }^1(\Gamma )$ with sum norm, consider the convex setIf zero were outside its norm closure, the Hahn-Banach separation theorem would give bounded real functions $f_g$ and $\delta >0$ such thatfor every such $\mu$. Testing point masses at $h$ showsApplying the positive invariant mean makes the left side zero and the right side at least $\delta$, a contradiction. Therefore zero is in the closure: for any $\epsilon >0$ a finitely supported probability measure makes the sum of the translation defects over $K$ less than $\epsilon$. Enumerate the countable group, take the first $n$ elements for $K$ and $\epsilon =1/n$, and obtain a sequence satisfying the Reiter condition. This proves (1) implies (2).

Next assume (2). An ${\ell }^1$ probability measure can be approximated in norm by finitely supported probability measures, by restricting to a large finite set and renormalizing. Replacing $\mu$ by $\nu$ changes each defect by at most $2\Vert \mu -\nu {\Vert }_1$. Thus simultaneous small defects for finitely many translations can be achieved with finite support.

For such $\nu$, set $F_t=\{h:\nu (h)>t\}$. The elementary identity $|r-s|={\int }_0^{\infty }|1_{r>t}-1_{s>t}|dt$ for $r,s\ge 0$ givesIf the sum of defects over $K$ is less than $\epsilon$, some $t$ with $F_t\ne \varnothing$ must satisfy ${\sum }_{g\in K}|F_t△g{F}_t|<\epsilon |F_t|$; otherwise integration would contradict that strict inequality. Applying this layer-cake extraction of Følner sets to successive finite sets of translations gives nonempty finite $F_n$ with each relative boundary tending to zero. Hence (2) implies (3).

Conversely, normalized counting measures ${\mu }_n=1_{F_n}/|F_n|$ satisfyso (3) implies (2). Finally, from (2) define states $m_n(f)={\sum }_h{\mu }_n(h)f(h)$. The dual unit ball of ${\ell }^{\infty }(\Gamma )$ is weak-star compact by the Banach-Alaoglu theorem; take a convergent subnet, since this compactness need not be metrizable. Its limit $m$ is positive with $m(1)=1$, andThus $m$ is invariant, proving (2) implies (1). Together these implications establish all three equivalences. Countability is used to obtain sequences from finite tests; it is not a justification for replacing the compactness subnet by a subsequence.

Assume the invariant set $C$ is nonempty, as required for any fixed-point assertion. For $x\in C$, put $r(x)={\sup }_{c\in C}\Vert x-c\Vert$ and $r_{\ast }={\inf }_{x\in C}r(x)$. Boundedness makes these finite. The parallelogram identity gives, for $x,y\in C$,The midpoint lies in $C$, so a minimizing sequence $x_n$ satisfiesSince $C$ is closed in a Hilbert space, the sequence converges to $x_{\ast }\in C$. The function $r$ is 1-Lipschitz, hence $r(x_{\ast })=r_{\ast }$. The same midpoint inequality proves uniqueness of this minimizing center.

An affine isometry from the group maps $C$ onto itself, since its inverse also preserves $C$. Therefore it preserves $r$ and sends its unique minimizer to another minimizer. Uniqueness givesThis proof applies to a closed bounded convex subset, the substantive meaning of the printed “convex subspace”; if it is literally a bounded vector subspace, it is just the zero subspace.

The topology needed here is the weak-star topology $\sigma (E^{\ast },E)$, meaning pointwise convergence on $E$. With the literal weak topology $\sigma (E^{\ast },E^{\ast \ast })$, the claimed compactness is false. For example, $E=c_0$ is separable and $E^{\ast }={\ell }^1$. Its unit vectors $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto {\sum }_j{x}_j$ has value one throughout. Thus the dual unit ball is not weakly compact. We prove the intended weak-star assertion in full.

Choose a countable dense set $(x_k)$ in the unit ball of $E$. On $E_1^{\ast }$ defineEach term is at most $2\cdot 2^{-k}$, so the series converges. The triangle inequality is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\Vert \varphi \Vert ,\Vert \psi \Vert \le 1$, approximation of any unit vector by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the weak-star topology, not just agreement of its convergent sequences.

Given a sequence $({\varphi }_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform norm bounds then make it pointwise Cauchy at every $x\in E$. Define $\varphi (x)$ as the limit. Linearity passes to the limit, and $|\varphi (x)|\le \Vert x\Vert$, so $\varphi \in E_1^{\ast }$. The subsequence converges to it in the metric. Thus this metric space is sequentially compact, hence compact, proving

Use the same intended weak-star topology as in (b), and assume $C\ne \varnothing$. The induced operator is $T=S^{\ast }$, so $T\varphi (x)=\varphi (Sx)$; it is weak-star continuous because evaluation after applying $T$ is still evaluation at a vector of $E$.

Choose $\varphi \in C$ and form the averagesInvariance and convexity put every ${\varphi }_n$ in $C$. Weak-star closedness and (b) make $C$ compact, so some subsequence converges to $\psi \in C$. Telescoping givesbecause both orbit points lie in the dual unit ball. Weak-star continuity now gives $T\psi -\psi =0$. Hence $\boxed{T\text{ has a fixed point in }C.}$ This is the weak-star fixed point theorem for an adjoint operator; it does not require $T$ to be a contraction on the entire dual space.

If “weakly closed” is interpreted literally as closed for $\sigma (E^{\ast },E^{\ast \ast })$, the result can fail. Take $E=c_0$, let $S$ be the left shift and $T=S^{\ast }$ the right shift on ${\ell }^1$. The setis a nonempty weakly closed convex subset of the unit ball: its defining coordinate inequalities and sum equality are weakly closed. It is invariant under the right shift. But a fixed vector of that shift has first coordinate zero and then every coordinate zero, so none belongs to $C$. This supplies a counterexample to the literal topology and explains why the weak-star qualification is essential.

Put $F=\ker (I-U)$. For a unitary operator, $\ker (I-U^{\ast })=F$, soThis follows from the general identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$. The averaging operators have $\Vert T_n\Vert \le 1$ and equal the identity on $F$. On a vector $(I-U)\eta$, telescoping givesThe uniform norm bound extends this convergence to the closure of the range. Thus $T_n$ tends to identity on $F$ and to zero on its orthogonal complement. Consequentlywhich is exactly convergence in the strong operator topology. This proves the Von Neumann mean ergodic theorem in the requested unitary case.

For a unital star-subalgebra $A\subseteq B(H)$, the Von Neumann double commutant theorem statesHere $A^{\prime }$ is the commutant of an operator algebra, consisting of operators commuting with every element of $A$. The double commutant is closed in either topology: multiplication by a fixed bounded operator is continuous in both. Therefore both closures of $A$ lie in $A^{\prime \prime }$.

For the converse, let $T\in A^{\prime \prime }$ and choose finitely many vectors ${\xi }_1,\dots ,{\xi }_n$. In $H^n$ putIt is invariant under the diagonal action of $A$ and its adjoints, so its orthogonal projection $Q$ commutes with that diagonal action. Consequently each matrix entry $Q_{ij}$ belongs to $A^{\prime }$. The diagonal operator with entry $T$ therefore commutes with $Q$. Since $I\in A$, the tuple $\xi =({\xi }_1,\dots ,{\xi }_n)$ belongs to $K$, and hence so does $(T{\xi }_1,\dots ,T{\xi }_n)$. By the definition of $K$, one element $a\in A$ approximates this entire tuple arbitrarily closely. This is exactly approximation in every strong-operator neighborhood of $T$. It proves $A^{\prime \prime }\subseteq {\overline{A}}^{\mathrm{SOT}}$ and thus both asserted equalities. In particular, $M=A^{\prime \prime }$ is a Von Neumann algebra.

The Kaplansky density theorem strengthens the result by preserving norm bounds:It also holds for the self-adjoint and positive unit balls; the full ball can in fact be approximated in the strong-star topology, meaning strong convergence of both operators and adjoints. We prove the bound rather than merely scaling the unbounded approximants furnished by the double commutant theorem.

First put $C={\overline{A}}^{\Vert \cdot \Vert }$, a unital C-star algebra. Its self-adjoint part is weak-operator dense in $M_{\mathrm{sa}}$: symmetrize any weak approximating net. For a convex subset, strong and weak operator closures agree. To see this explicitly, separation from a strong closure supplies a continuous real functional on finitely many image vectors, hence one of the form $\mathrm{Re}{\sum }_j⟨a{\xi }_j,{\eta }_j⟩$, which is also weak-operator continuous. The Hahn-Banach separation theorem would then separate the point from the weak closure as well. Thus every self-adjoint $x\in M$ is a strong limit of self-adjoint $a_i\in C$, initially without a common norm bound.

The needed functional-calculus continuity follows directly from resolvents. The resolvent identity givesThe first factor has norm at most one, so the right side tends strongly to zero. The same holds at $-i$. Both resolvent nets are uniformly bounded, so their products converge strongly. Their scalar functions generate a self-adjoint algebra separating points of $\mathbb{R}$ and vanishing at infinity. The Stone-Weierstrass theorem, applied to the one-point compactification, makes this algebra uniformly dense in $C_0(\mathbb{R})$. Uniform approximation and the functional-calculus norm bound therefore showThis proves strong continuity of functional calculus through resolvents without an unjustified uniform bound on the original $a_i$.

If $x=x^{\ast }$ and $\Vert x\Vert \le 1$, choose a real compactly supported continuous $f$ equal to the identity on $[-1,1]$ and bounded in absolute value by one. Then $f(a_i)\in C$ are self-adjoint contractions and converge strongly to $f(x)=x$. For a positive contraction $x$, choose instead $0\le f\le 1$ equal to the identity on $[0,1]$; this gives positive contractions. These prove the two restricted versions for $C$.

For an arbitrary contraction $x\in M$, apply the self-adjoint result in the matrix algebra $M_2(C)$ toThe strong closure of $M_2(C)$ is $M_2(M)$, since finitely many entries may be approximated simultaneously. The upper-right entries $c_i$ of the approximating self-adjoint contraction matrices belong to $C$, satisfy $\Vert c_i\Vert \le 1$, and converge strongly to $x$. The lower-left entries are $c_i^{\ast }$ and converge strongly to $x^{\ast }$. This proves the full strong-star assertion for $C$.

Finally transfer the bounds from $C$ to the possibly nonclosed algebra $A$. A contraction $c\in C$ has an approximant $b\in A$ with $\Vert b-c\Vert <\epsilon$, so $b/(1+\epsilon )$ is an $A$-contraction within $2\epsilon$ of $c$. Self-adjoint approximants are obtained by symmetrizing first. For a positive contraction, approximate $c^{1/2}$ by $b\in A$ and use the positive contraction $b^{\ast }b/(1+\epsilon {)}^2$, which tends in norm to $c$ as $\epsilon \downarrow 0$. Combining these norm approximations with each finite-vector strong approximation proves all versions for $A$. The reverse inclusions follow because the relevant norm bounds and the inequalities defining a positive operator are preserved by strong limits.

Interpret the involution as the adjoint for the Hilbert-space structure on $V$. The hypotheses make $\Omega$ a cyclic vector for an operator algebra for both $A$ and $A^{\prime }$. They also make it a separating vector for each: if $a\Omega =0$, then $a$ vanishes on $A^{\prime }\Omega =V$, and similarly for $A^{\prime }$. Thus the Tomita operatoris well defined, antilinear, invertible and satisfies $S^2=I$. Define the modular operator and modular conjugation byWe calculate them explicitly to prove the commutant assertion.

The finite-dimensional structure of a unital star-algebra gives an orthogonal decompositionThis is the usual matrix-block structure: minimal central projections separate the simple summands, and each simple matrix algebra acts as its defining representation with a multiplicity space. Identify each vector block with an $n_j$-by-$m_j$ matrix $D_j$ with its Hilbert-Schmidt inner product. The two algebras act by left and right multiplication. If $D_j$ has rank $r_j$, the left orbit has dimension $n_j{r}_j$ and the right orbit dimension $m_j{r}_j$. Cyclicity on both sides therefore forces $r_j=m_j=n_j$. Every $D_j$ is square and invertible.

Right multiplication by a suitable unitary is a unitary change of this identification commuting with the left algebra. The polar decomposition of each $D_j$ therefore lets us assume $D_j>0$. In one such block, for an arbitrary matrix $X=aD$, the Tomita operator isIts polar factors areIndeed, $J$ is an antiunitary involution and $J{\Delta }^{1/2}(X)=D^{-1}{X}^{\ast }D$. The fact that ${\Delta }^{1/2}$ is a positive operator is transparent in a basis diagonalizing $D$: it multiplies the matrix unit $E_{kl}$ by the positive ratio $d_k/d_l$. These are therefore the unique polar factors of $S$.

Finally, if $L_a$ denotes left multiplication,As $a$ ranges over the full matrix block, these are exactly its right multiplications. Applying this on every block and transporting back by the unitary identifications provesThis also establishes the required modular definitions intrinsically, independent of the chosen matrix coordinates.

Write $M=\lambda (\Gamma {)}^{\prime \prime }$, the group von Neumann algebra, and define right translations by $\rho (g){\delta }_h={\delta }_{h{g}^{-1}}$. Left and right translations commute, so every $T\in M$ commutes with every $\rho (g)$. A Von Neumann factor means that the center of $M$ is exactly $\mathbb{C}I$.

Suppose every nonidentity conjugacy class is infinite and $T$ is central. Put $\eta =T{\delta }_e$. Since $T$ commutes with both regular actions and $\lambda (g)\rho (g){\delta }_e={\delta }_e$,The action on a basis vector is ${\delta }_h\mapsto {\delta }_{gh{g}^{-1}}$, so the coefficients of $\eta \in {\ell }^2(\Gamma )$ are constant on conjugacy classes. A nonzero constant on an infinite class would have infinite squared sum. Thus all coefficients away from the identity vanish and $\eta =c{\delta }_e$.

Commutation with the right action now gives $T{\delta }_h=c{\delta }_h$ for every $h$, because those vectors form the right orbit of ${\delta }_e$. Hence $T=cI$ and $M$ is a factor. This argument also shows that ${\delta }_e$ is separating for $M$.

Conversely, if a nonidentity class $C$ is finite, thencommutes with every $\lambda (g)$, since conjugation permutes $C$. Therefore $Z\in M\cap \lambda (\Gamma {)}^{\prime }=Z(M)$. It is not scalar: $Z{\delta }_e={\sum }_{h\in C}{\delta }_h$ is nonzero and orthogonal to ${\delta }_e$. Consequently $M$ is not a factor. We have provedThe condition is vacuous for the trivial group, whose algebra $\mathbb{C}I$ is indeed a factor.

The dense commutant orbit makes $\Omega$ separating for $M$, so $S_0(a\Omega )=a^{\ast }\Omega$ is well defined on the dense subspace $M\Omega$. Put $\tau (a)=⟨a\Omega ,\Omega ⟩$. The assumed identity says that this positive vector functional is a tracial positive functional. HenceIt extends to an antiunitary involution $J$ on $H$, with $J\Omega =\Omega$. This is the modular conjugation, and the modular operator in this tracial case is $\Delta =I$.

On the dense subspace $M\Omega$,Thus it acts by right multiplication and commutes with all left multiplications; $JMJ\subseteq M^{\prime }$. We prove the reverse inclusion rather than assuming the general Tomita commutation theorem.

Let $T\in M^{\prime }$, $C=\Vert T\Vert$, and $\eta =T\Omega$. The setis weakly compact and convex. Indeed, trace-class duality and the Banach-Alaoglu theorem make the bounded-operator ball compact in a topology stronger than the weak operator topology. The algebra $M$ is weak-operator closed, and $a\mapsto a\Omega$ is weakly continuous, so its image is weakly compact.

For $b\in M$ use the polar decomposition of a bounded operator $b=uh$, $h=|b|$. Both $h$ and $u$ belong to $M$: continuous functional calculus gives $h$, and $b(h+\epsilon I{)}^{-1}$ tends strongly to $u$. The polar decomposition is proved independently in Question 8. Since $T$ commutes with $M$,The trace identity givesTherefore $|⟨\eta ,b\Omega ⟩|\le C\tau (h)$. Meanwhilewith the supremum attained by $a=Cu$. For the upper bound, traciality gives $\tau (b^{\ast }a)=\tau (h^{1/2}{u}^{\ast }a{h}^{1/2})$; evaluating this on $\Omega$ bounds its absolute value by $\Vert a\Vert \tau (h)$.