Regular group operator foundations

Original text: public domain (CC0 1.0).

This lesson constructs the bounded operators and group traces needed for the subgroup-orbit and affine MASA arguments. Compression to a subgroup supplies its expectation. Positive inverses recover kernel projections, while explicit matrix entries compute the regular commutants. The final sections prove normal group-trace uniqueness and the intrinsic finite or countably infinite matrix size used for MASA multiplicity.

H00–H04 begin with square-summable coordinates. Their C*-algebra calculus input is the complete F03–F08 proof chain in Infinite tensor products. G00–G05 then construct the group algebra and subgroup maps. T04a–T04c prove polar decomposition, trace bounds for projection joins and continuity of an order-normal functional along bounded trace-square-null sequences. T04 applies these proofs to preserve the group trace under an abstract isomorphism. The source comparisons at the end of G05 identify the freely readable group-operator arguments.

H00. Square-summable coordinates and bounded operators

Our scalar boundary is the complete real ordered field and its complex field. For any set $I$, let $\ell^2(I)$ consist of the functions $\xi$ with $\sum_{i\in I}|\xi(i)|^2<\infty$, where a sum of nonnegative terms is the supremum of its finite partial sums. Every such vector has at most countable support: for each positive integer $m$, only finitely many coordinates can have modulus at least $1/m$, and the union of those finite sets contains every nonzero coordinate. The inner product is $\langle\xi,\eta\rangle=\sum_i\overline{\xi(i)}\eta(i)$, linear in the second variable. Finite-dimensional Cauchy–Schwarz follows by minimizing $\|\xi-z\eta\|^2$ over $z\in\mathbb C$; if $\eta=0$ its assertion is immediate. Taking finite partial sums proves absolute convergence of the displayed inner product and Cauchy–Schwarz for $\ell^2(I)$.

This space is complete. If $(\xi_j)$ is norm Cauchy, each coordinate is Cauchy and has a limit $\xi(i)$. A uniform bound on $\|\xi_j\|$ bounds every finite sum of squares of these limits, so $\xi\in\ell^2(I)$. For a chosen $\varepsilon>0$, sufficiently late $j,k$ have $\|\xi_j-\xi_k\|<\varepsilon$. Taking $k$ to infinity in each finite-coordinate sum and then taking the supremum gives $\|\xi_j-\xi\|\le\varepsilon$. Finite-coordinate truncations converge in norm; the coordinate vectors are therefore an orthonormal basis. A bijection between two coordinate sets induces a unitary by carrying one basis onto the other. The Hilbert tensor product of two such spaces is $\ell^2(I\times J)$: finite elementary tensors have the product inner product, and their coordinate tensors span a dense subspace of that complete space.

A bounded linear functional $f$ on $\ell^2(I)$ has the form $f(\eta)=\langle\xi,\eta\rangle$. Indeed set $\overline{\xi(i)}=f(\delta_i)$. Testing finite linear combinations with coefficients $\xi(i)$ gives $\sum_{i\in F}|\xi(i)|^2\le\|f\|^2$ for every finite $F$, by dividing by the square root of that sum when it is nonzero. Thus $\xi\in\ell^2(I)$, and agreement on finite-coordinate vectors extends by continuity. This also gives $\|f\|=\|\xi\|$. Applied to $\eta\mapsto\langle\zeta,T\eta\rangle$, it constructs the adjoint $T^*$ of every bounded operator, with $\|T^*\|=\|T\|$. The operator norm is complete: a norm-Cauchy sequence of operators converges on each vector, and its uniform operator-norm bound makes the limit bounded and the convergence uniform on the unit ball. Furthermore

give $\|T^*T\|=\|T\|^2$. Thus bounded operators form a unital C*-algebra. The continuous calculus proved in Infinite tensor products, F03–F08 applies to it and its norm-closed self-adjoint subalgebras.

H01. Orthogonal projections and strong limits in commutants

Every closed linear subspace $V$ of a Hilbert space has an orthogonal projection. For $\xi$, let $d=\inf_{v\in V}\|\xi-v\|$, and take a sequence $v_j$ approaching this infimum. The parallelogram identity gives

The limit $v\in V$ attains $d$. Minimality of $\|\xi-v-tv'\|^2$ for real and purely imaginary $t$, for every $v'\in V$, gives $\xi-v\perp V$. This decomposition is unique, because $V\cap V^\perp=\{0\}$. It gives a linear projection $P_V$, with $\|P_V\|\le1$, $P_V^*=P_V$, and range $V$. If every operator in a self-adjoint family preserves $V$, it preserves $V^\perp$ as well, so $P_V$ commutes with that family.

For a family $\mathcal S$ of bounded operators, its commutant is $\mathcal S'=\{T:Ts=sT\text{ for all }s\in\mathcal S\}$. If $\mathcal S$ is self-adjoint, its commutant is a unital self-adjoint algebra. It is norm closed and strongly closed: if $T_j\xi\to T\xi$ for every $\xi$, then $T_js\xi=sT_j\xi$ passes to the limit. The relation $\mathcal S'''=\mathcal S'$ follows directly from the inclusions $\mathcal S\subset\mathcal S''$ and the fact that $\mathcal S'$ commutes with $\mathcal S''$. In particular $M=\mathcal S''$ satisfies $M''=M$. An operator in $M$ invertible on the Hilbert space has its inverse in $M$, since its inverse commutes with every operator commuting with it. These facts use the definition of a commutant, rather than a density theorem for operator polynomials.

H02. Kernel projections from positive inverses

An operator $S=S^*$ is positive if $\langle\xi,S\xi\rangle\ge0$ for all $\xi$. For $t>0$, the operator $1+tS$ is invertible and its inverse $Q_t$ has norm at most $1$. Indeed

proves the lower bound $\|(1+tS)\xi\|\ge\|\xi\|$. It has zero kernel and closed range: a Cauchy sequence of images has Cauchy preimages by this bound. The orthogonal complement of its range is the kernel of its adjoint, hence zero. The range is therefore both dense and closed, and is the whole space. The same bound proves the inverse norm assertion.

The operators $Q_t$ fix $\ker S$. They commute with $S$, and

Thus they converge to zero on $\operatorname{ran}S$, and, by their uniform norm bound, on its closure. Since $S=S^*$, $(\operatorname{ran}S)^\perp=\ker S$. H01 decomposes the space as $\ker S\oplus\overline{\operatorname{ran}S}$; hence $Q_t$ converges strongly to the kernel projection as $t\to\infty$. If $S\in M$, each $Q_t\in M$ by H01, and the projection belongs to $M$ by strong closedness.

The positive quadratic form also obeys Cauchy–Schwarz, by expanding $\langle\xi+z\eta,S(\xi+z\eta)\rangle\ge0$ and minimizing over $z$. If one diagonal value is zero, varying the magnitude and phase of $z$ first shows the corresponding mixed value is zero. Consequently $\langle\xi,S\xi\rangle=0$ implies $\langle\eta,S\xi\rangle=0$ for every $\eta$, and then $S\xi=0$. This supplies a direct kernel argument without a square-root assumption.

The square-root formulation is also available. For a positive $S$, a real $\lambda<0$ gives an invertible $S-\lambda$ by the same range argument. For nonreal $\lambda$, the imaginary part of $\langle\xi,(S-\lambda)\xi\rangle$ gives the lower bound $|\operatorname{Im}\lambda|\|\xi\|$, and its adjoint has the same bound, so the range argument again proves invertibility. The spectrum of $S$ is therefore nonnegative. The continuous calculus F06–F08 in Infinite tensor products supplies $S^{1/2}$, and $\langle\xi,S\xi\rangle=\|S^{1/2}\xi\|^2$. Conversely a self-adjoint operator with nonnegative spectrum has this self-adjoint square root, so its quadratic form is nonnegative. Thus spectral positivity and quadratic-form positivity agree. This justifies either version of the kernel calculation. Its support projection is $1-P_{\ker S}$, the projection onto $\overline{\operatorname{ran}S}$. If $S\in pMp$, this range lies in $p\mathcal H$, so the support is at most $p$ and belongs to $pMp$. A nonzero $S$ has nonzero support.

H03. A concrete separable predual

For a separable Hilbert space $\mathcal H$, a countable dense sequence gives a countable orthonormal basis: successively subtract its projections on the finite span of earlier chosen vectors, and normalize each nonzero remainder. Every member of the dense sequence lies in the resulting closed span, so that span is the whole space. Finite orthogonal truncations consequently converge to every vector. The coordinate map is an isometry onto $\ell^2(I)$, because finite-coordinate vectors are in its image and completeness makes that image closed. This also includes finite-dimensional and zero spaces and transfers the representation argument of H00. Form the algebraic tensor product $\overline{\mathcal H}\otimes\mathcal H$, and give it the projective norm

This is a norm. A nonzero algebraic tensor lies in the tensor product of two finite-dimensional spans; coordinate functionals in bases of those spans detect a nonzero coefficient and extend to bounded functionals using H00 and H01. Their product bounds its value by a constant times every displayed sum, so $\|u\|_\pi>0$. The triangle inequality follows by concatenating representations. Let $E$ be the completion. Finite rational complex combinations of tensors of a countable orthonormal basis are dense, because finite-coordinate truncations approximate each factor in the projective norm. Thus $E$ is separable.

The dual of $E$ is isometrically $B(\mathcal H)$. A bounded operator $T$ acts by $\overline\xi\otimes\eta\mapsto\langle\xi,T\eta\rangle$, with dual norm $\|T\|$. Conversely a bounded functional on $E$ gives a bounded form, conjugate linear in $\xi$ and linear in $\eta$. H00 represents this form as $\langle\xi,T\eta\rangle$, and its bound makes $T$ bounded. Every $u\in E$ can be represented by a series $\sum_j\overline{\xi_j}\otimes\eta_j$ with $\sum_j\|\xi_j\|\|\eta_j\|<\infty$: approximate $u$ by algebraic tensors with geometrically decreasing errors, represent each successive difference within a geometrically decreasing error of its projective norm, and concatenate the finite representations. Conversely every such series converges in $E$. The weak* topology from $E$ is precisely the ultraweak topology defined by the functionals

If $M=\mathcal S''$, its commutation constraints with $\mathcal S'$ are ultraweak continuous: $\langle\xi,(Ta-aT)\eta\rangle$ is the difference of two one-term functionals. Let $N\subset E$ be the norm-closed span of the corresponding tensors, for all $a\in\mathcal S'$ and $\xi,\eta\in\mathcal H$. Then $M=N^\perp\subset E^*$, exactly by these constraints. The dual of $E/N$ is $N^\perp$: a bounded functional on the quotient pulls back to a functional killing $N$, and such a functional descends with the same norm, by the quotient norm definition. The quotient is complete: from a quotient-Cauchy sequence choose a subsequence whose successive quotient distances are below $2^{-j}$, lift each difference to a vector of norm below $2^{1-j}$, and sum those lifts in $E$. Its image is the limit of the subsequence, and Cauchyness gives the same limit for the full sequence. The image of a countable dense set in $E$ is dense in the quotient. Therefore $E/N$ is a separable predual for $M$. In particular every countable group algebra in its regular representation has separable predual. This proof does not need a compact-operator spectral decomposition.

On a norm-bounded set, weak operator convergence implies ultraweak convergence. Given a series-vector functional, choose a finite initial segment whose remaining sum of products of vector norms is small. The finite segment converges by weak operator convergence; the remaining segment is bounded uniformly by the common operator-norm bound times that remaining sum. Applying this also to the limit operator proves convergence of the whole series. This argument works for nets, since the finite initial segment imposes only finitely many convergence conditions.

For an arbitrary coordinate Hilbert space $\ell^2(I)$, the same projective-tensor, dual and commutation-constraint arguments construct a predual and give this series-vector definition of the ultraweak topology. The separability conclusion is asserted only when $I$ is countable. H00 supplies the representation and completeness facts for those arbitrary coordinate spaces as well.

H04. Why an infinite-dimensional finite factor has no minimal projection

Let $M$ be a factor with a faithful normalized trace $\tau$. If it has a minimal nonzero projection $p$, then $pMp=\mathbb Cp$. To prove this, suppose a self-adjoint element $a\in pMp$ were nonscalar. Its spectrum has at least two points: F04 of Infinite tensor products makes a self-adjoint element with singleton spectrum a scalar. Choose a real $c$ between two spectral points. The continuous positive and negative parts of $a-cp$, supplied by F06–F08, are nonzero and have zero product. H02 gives their nonzero support projections in $pMp$; their ranges are orthogonal because their product is zero. This contradicts minimality of $p$. Thus every self-adjoint element of the corner is scalar, and its real and imaginary parts show the same for every element.

The closed span of the ranges of $upu^*$, over the unitaries $u\in M$, has a projection $z\in M$. Indeed every such range reduces $M'$; H01 gives its projection in $M''=M$. Conjugation by any unitary of $M$ preserves this span, so $z$ commutes with those unitaries. They linearly span $M$: for a self-adjoint contraction $h$, the continuous calculus gives the unitary $h+i(1-h^2)^{1/2}$, whose real part is $h$; scaling and taking real and imaginary parts treats any element. Thus $z$ is central. Since $p\ne0$, the factor property gives $z=1$.

For any nonzero projection $q\in M$, it follows that $qMp\ne0$. Otherwise $qup=0$ for every unitary $u$, making $q$ vanish on the dense span just described. Choose $x=qap\ne0$. Since $x^*x\in pMp$, it equals $\alpha p$ for $\alpha>0$. Then $v=\alpha^{-1/2}x$ satisfies $v^*v=p$, and $vv^*\le q$ is a projection equivalent to $p$.

Starting with $p$, repeat this in the complement of the sum of the projections already chosen. Each new projection has trace $\tau(p)>0$. Since the sum has trace at most $1$, after finitely many steps there can be no nonzero complement. We obtain $p_1+\cdots+p_m=1$, with each $p_i$ equivalent to $p$. Choose $v_i^*v_i=p$, $v_iv_i^*=p_i$. The operators $v_iv_j^*$ are matrix units, and for $x\in M$, every $v_i^*xv_j$ lies in $pMp=\mathbb Cp$. Expanding $x=\sum_{i,j}p_ixp_j$ therefore proves $M\cong M_m(\mathbb C)$. An infinite-dimensional factor with a faithful normalized trace consequently has no minimal projections; it is a type $\mathrm{II}_1$ factor. An infinite discrete group's unitaries are linearly independent on $\delta_e$, so its ICC group factor is infinite dimensional and this conclusion applies.

This also gives the formulation in terms of abelian projections. The central-support argument above applies to every nonzero projection $r$, not just a minimal one; hence $qMr\ne0$ for any nonzero projections $q,r$ in a factor. If a nonzero projection $e$ has commutative corner $eMe$ and $0<q<e$, put $r=e-q$. Commutativity gives $qMr=q(eMe)r=0$, a contradiction. Thus an abelian projection in a factor is minimal. The infinite-dimensional case has no nonzero abelian projections. Finally the trace makes the identity finite: if $v^*v=1$, then $\tau(vv^*)=1$, so faithfulness applied to $1-vv^*$ gives $vv^*=1$. This proves both the type-II and finite parts of the type $\mathrm{II}_1$ designation.

Regular representations and conventions

Let $G$ be a discrete group with identity $e$. The space

has inner product $\langle\xi,\eta\rangle=\sum_s\overline{\xi(s)}\eta(s)$, antilinear in its first variable. Its usual orthonormal basis is $(\delta_s)_{s\in G}$. All sums over $G$ below use finite partial sums and are independent of ordering. H00 shows that square-summable vectors have countable support; whenever a scalar product is rearranged, its absolute convergence is proved first. Thus the commutant and subgroup arguments also apply to arbitrary discrete groups.

H00 proves the bounded-operator, adjoint and Hilbert-space representation facts used below, including density of finitely supported vectors. A bounded operator $T$ is positive when $\langle\xi,T\xi\rangle\geq0$ for every $\xi$. Order is the order defined by positive differences. We will prove both preservation of bounded increasing positive suprema and ultraweak continuity for the trace and subgroup expectation. The elementary existence of these suprema is proved below.

For a set $S$ of bounded operators write

The left and right regular representations are

These operators permute an orthonormal basis, so they are unitary. Their products satisfy $\lambda(g)\lambda(h)=\lambda(gh)$, $\rho(g)\rho(h)=\rho(gh)$, and $\lambda(g)\rho(h)=\rho(h)\lambda(g)$. We define the group von Neumann algebra by

This definition lets us establish the particular commutant identities needed here directly. No operator approximation by finite Fourier sums is needed.

G00. Matrix products and the regular commutants

For $T\in B(\ell^2(G))$, let $T_{s,t}=\langle\delta_s,T\delta_t\rangle$. Its $t$-th column lies in $\ell^2(G)$, with norm $\|T\delta_t\|\leq\|T\|$. Its $s$-th row lies in $\ell^2(G)$, with norm $\|T^*\delta_s\|\leq\|T\|$. Indeed, $(T^*)_{t,s}=\overline{T_{s,t}}$.

If $A,B$ are bounded, then

The inequality is Cauchy–Schwarz. To verify the equality, expand $B\delta_t$ in the orthonormal basis. Its finite partial sums converge in Hilbert space norm. Applying $A$, then taking the $s$-th coordinate, gives (3). Operators with the same matrix entries agree: they agree on each basis vector and hence, by boundedness and density, on every vector.

Lemma 1. A bounded operator $x$ commutes with $\rho(G)$ if and only if there is a function $a\in\ell^2(G)$ such that

A bounded operator $y$ commutes with $\lambda(G)$ if and only if there is a function $b\in\ell^2(G)$ such that

Proof. If $x\rho(g)=\rho(g)x$, put $a=x\delta_e$. Since $\delta_t=\rho(t^{-1})\delta_e$,

whose $s$-th coordinate is $a(st^{-1})$. Conversely, if (4) holds, the $s$-th coordinates of $x\rho(g)\delta_t$ and $\rho(g)x\delta_t$ are both $a(sgt^{-1})$. Equality on the basis proves commutation.

For the second assertion put $b=y\delta_e$, and use $\delta_t=\lambda(t)\delta_e$. This gives $y\delta_t=\lambda(t)b$ and (5). Conversely, the $s$-th coordinates of $y\lambda(g)\delta_t$ and $\lambda(g)y\delta_t$ are both $b(t^{-1}g^{-1}s)$. Again equality on the basis suffices. $\square$

Theorem 2 (regular commutants). For every discrete $G$,

Proof. Since every $\rho(g)$ commutes with $\lambda(G)$, we have $\rho(G)\subseteq\lambda(G)'$; taking commutants gives $\lambda(G)''\subseteq\rho(G)'$.

For the reverse inclusion take $x\in\rho(G)'$ and an arbitrary $y\in\lambda(G)'$. Let $a,b$ be their functions in (4) and (5). Formula (3) gives

and

Both sums converge absolutely. For example the maps $r\mapsto sr^{-1}$ and $r\mapsto t^{-1}r$ are bijections of $G$, so the sum of absolute values in (7) is at most $\|a\|_2\|b\|_2$; the same argument applies to (8). In (8) substitute

This is a bijective change of index, and it gives $u^{-1}s=t^{-1}r$, $ut^{-1}=sr^{-1}$. Thus (7) and (8) are equal. All matrix entries of $xy$ and $yx$ agree, hence $xy=yx$. Since $y$ was arbitrary in $\lambda(G)'$, $x\in\lambda(G)''$.

Finally, for every set $S$, $S'''=S'$: the inclusion $S\subseteq S''$ yields $S'''\subseteq S'$, while each element of $S'$ commutes with each element of $S''$ by the definition of $S''$. Taking commutants of the first identity in (6) therefore gives the second. $\square$

In particular $L(G)=\rho(G)'$ is a unital algebra closed under adjoints: if $x$ commutes with all $\rho(g)$, then $x^*$ does too, by taking adjoints and using $\rho(g)^*=\rho(g^{-1})$. It is closed in both the weak and strong operator topologies. For example, if $x_i\to x$ strongly and each $x_i$ commutes with a fixed $\rho(g)$, then for every vector $\xi$,

For weak convergence take the matrix coefficient against an arbitrary vector on both sides of the same equality. This proves the closure assertions directly.

G01. Increasing positive operators and the canonical trace

We first record an order fact so that normality will not hide a density or continuity theorem.

Lemma 3 (monotone convergence of bounded operators). Let $(T_i)$ be an increasing net of positive bounded operators on a Hilbert space, with $\sup_i\|T_i\|\leq C<\infty$. There is a positive operator $T$ such that $T_i\to T$ strongly, and $T$ is their least upper bound. If every $T_i$ belongs to a strongly closed algebra, $T$ belongs to that algebra.

Proof. A positive operator $A$ has the positive-form Cauchy–Schwarz inequality

To prove it, expand $\langle\eta+z\xi,A(\eta+z\xi)\rangle\geq0$ for $z\in\mathbb C$ and minimize the resulting quadratic polynomial. If $\langle\xi,A\xi\rangle=0$, the linear term must vanish for every $z$, which proves the zero case too. The positivity condition also implies self-adjointness, by polarization of this quadratic form.

For each $\xi$ the numbers $q_i(\xi)=\langle\xi,T_i\xi\rangle$ increase to a finite limit $q(\xi)$, bounded by $C\|\xi\|^2$. Polarization shows that each $\langle\eta,T_i\xi\rangle$ has a limit. These limits form a bounded sesquilinear form, of absolute value at most $C\|\eta\|\|\xi\|$, since this bound holds for each $i$. The Hilbert space representation theorem supplies a bounded operator $T$ with those matrix coefficients. It is positive and $0\leq T_i\leq T\leq CI$.

Put $A_i=T-T_i$. By (9), taking the supremum over unit vectors $\eta$,

Thus $T_i\to T$ strongly. If $S$ is any self-adjoint upper bound, then $\langle\xi,(S-T_i)\xi\rangle\geq0$ for every $i$; taking the limit gives $S\geq T$. This proves that $T$ is the supremum. The last assertion is strong closure. $\square$

Define

Theorem 4 (canonical trace). The functional $\tau_G$ is a faithful normal tracial state on $L(G)$.

Proof. It is linear, positive, and $\tau_G(I)=1$, directly from (11). Also $|\tau_G(x)|\leq\|x\|$. It has norm one because equality holds at $I$.

The vector $\delta_e$ is separating for $L(G)$. Indeed, if $x\delta_e=0$, then, because $x\in\rho(G)'$,

for every $t$. Hence $x=0$. It is also cyclic, since $\lambda(t)\delta_e=\delta_t$ and finitely supported vectors are dense.

For $x\in L(G)$,

This proves faithfulness in the form $\tau_G(x^*x)=0\Rightarrow x=0$. It also proves faithfulness on arbitrary positive operators without invoking a square root: if $p\geq0$ and $\tau_G(p)=0$, inequality (9) gives $\langle\eta,p\delta_e\rangle=0$ for every $\eta$, so $p\delta_e=0$, and the separating property gives $p=0$.

For traciality let $a=x\delta_e$, $c=y\delta_e$, where $x,y\in L(G)=\rho(G)'$. Equations (3) and (4) give

Both sums have sum of absolute values at most $\|a\|_2\|c\|_2$. Changing $u$ to $r^{-1}$ in the second proves $\tau_G(xy)=\tau_G(yx)$ for the arbitrary bounded operators under consideration.

If $x_i$ is an increasing bounded net of positive elements of $L(G)$, Lemma 3 and strong closure give its supremum $x\in L(G)$ and $x_i\delta_e\to x\delta_e$. Therefore

This proves order normality. Formula (11) is a one-term series-vector functional in H03, so the trace is also ultraweakly continuous. $\square$

The trace also directly rules out proper isometries in $L(G)$. If $v^*v=I$, then $vv^*$ is a projection and $\tau_G(I-vv^*)=\tau_G(I)-\tau_G(v^*v)=0$. Faithfulness gives $vv^*=I$. This observation uses no classification of finite factors.

G02. Fourier coefficients and their precise convergence

For $x\in L(G)$ define its Fourier coefficient at $g$ by

Lemma 1 gives the complete matrix from these coefficients:

In particular the coefficients determine $x$ uniquely. They satisfy

and

For the first identity in (16), $\lambda(g)^*\delta_e=\delta_{g^{-1}}$, so (15) gives the coefficient $x_{e,g^{-1}}=\widehat x(g)$. The adjoint identity follows by exchanging the matrix indices in (15) and taking complex conjugates. Formula (17) is the $s,e$ case of (3), with Cauchy–Schwarz as indicated.

Put $\|x\|_2=\tau_G(x^*x)^{1/2}$. By (12),

The map $V_\tau:x\mapsto x\delta_e$ is an isometry from $L(G)$, with the inner product $\tau_G(x^*y)$, into $\ell^2(G)$. Its range is dense because it contains every $\delta_g=\lambda(g)\delta_e$. It therefore extends to a unitary from the Hilbert space completion, denoted $L^2(L(G),\tau_G)$, onto $\ell^2(G)$.

If $F\subset G$ is finite and

then

as the finite subsets exhaust $G$. This is exactly convergence in the Hilbert space just defined. Equation (19) makes no assertion about strong or weak operator convergence of the $x_F$, and gives no uniform operator norm bound for these finite sums.

G03. The subgroup algebra inside the group algebra

Let $H\leq G$ be any subgroup. Write $\lambda_H,\rho_H$ for its regular representations on $\ell^2(H)$. Applying Theorem 2 to $H$ gives

To identify this algebra inside $L(G)$, partition $G$ into right cosets $Ht$. Choose one representative $t$ in each coset, including $e$ for $H$, and let $\mathcal T$ be the representatives. Then

Each $U_t$ is unitary onto its summand and intertwines $\lambda_H(h)$ with $\lambda_G(h)$ on that summand. Define

The direct sum is bounded: the square of its value on $\xi=\bigoplus_t\xi_t$ has norm at most $\|b\|^2\sum_t\|\xi_t\|^2$. Conversely restriction to the $H$ summand has norm $\|b\|$, so $\|\iota_H(b)\|=\|b\|$. Products and adjoints are computed on each summand, making $\iota_H$ an injective unital *-homomorphism. It is independent of representatives: replacing $t$ by $h_0t$ changes $U_t$ to $U_t\rho_H(h_0^{-1})$, and $b$ commutes with that right regular unitary. Its values on group elements are

Theorem 5 (subgroup embedding). The map $\iota_H$ identifies $L(H)$ with

It preserves the canonical trace and suprema of increasing bounded positive nets.

Proof. Let $P_t$ be the projection onto $\ell^2(Ht)$. If $y\in\lambda_G(H)'$, every block

commutes with $\lambda_H(H)$. Explicitly, for $h\in H$ the intertwining relations and commutation of $y$ give

Thus $y_{u,t}\in\lambda_H(H)'$. For $b\in\lambda_H(H)''$,

for each pair of cosets. These are precisely the blocks of $\iota_H(b)y$ and $y\iota_H(b)$. Equality of all blocks implies equality of the operators: for a vector in one summand their projections to all summands agree, and finite sums of such vectors are dense. Hence $\iota_H(b)\in\lambda_G(H)''$.

Conversely let $x\in\lambda_G(H)''$. Each $P_t$ commutes with $\lambda_G(H)$, so $x$ commutes with $P_t$. Thus $x$ is block diagonal. For any $u,t\in\mathcal T$, the bounded partial isometry

also commutes with $\lambda_G(H)$. Commutation with $x$ says that its diagonal blocks, transported to $\ell^2(H)$, are all the same bounded operator $b$. For each $z\in\lambda_H(H)'$, the diagonal operator $\bigoplus_t U_tzU_t^*$ belongs to $\lambda_G(H)'$, so $x$ commutes with it. Restricting to the $H$ summand gives $bz=zb$. Consequently $b\in\lambda_H(H)''=L(H)$ and $x=\iota_H(b)$. This proves equality of the range with (23). Inclusion in $L(G)$ follows from $\lambda_G(H)\subseteq\lambda_G(G)$: taking commutants reverses the inclusion, and taking commutants again restores it.

Because the $H$ summand uses $t=e$,

For the order assertion let $b_i\uparrow b$ be a bounded increasing net in $L(H)_+$. Lemma 3 gives strong convergence on $\ell^2(H)$. The positive operator $\iota_H(b)$ is an upper bound for all $\iota_H(b_i)$. To verify strong convergence on $\ell^2(G)$, fix $\xi=\bigoplus_t\xi_t$ and bound $\|b-b_i\|$ by a common constant $D$. For a finite collection $K\subset\mathcal T$,

First make the tail as small as desired by choosing $K$, then use strong convergence for its finitely many vectors. Thus $\iota_H(b_i)\to\iota_H(b)$ strongly. Lemma 3, or its least-upper-bound argument, proves that $\iota_H(b)$ is their supremum. $\square$

An element $\iota_H(b)$ has Fourier coefficients zero outside $H$, and its coefficient at $h\in H$ is $\widehat b(h)$, because $\iota_H(b)\delta_e=b\delta_e\in\ell^2(H)$.

G04. Compression and the subgroup expectation

Let $P_H$ denote the orthogonal projection of $\ell^2(G)$ onto $\ell^2(H)$, and let $j_H:\ell^2(H)\hookrightarrow\ell^2(G)$ be inclusion. Thus $j_H^*=P_H$, with the range of $P_H$ viewed as $\ell^2(H)$. For $x\in L(G)$ put