Regular and singular maximal abelian algebras

Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. Self-checked by the writing AI. Original text: CC0 1.0.

Maximal abelianness says that every operator commuting with an abelian algebra already belongs to it. Regularity asks a different question: do the unitaries carrying the algebra onto itself generate the whole ambient algebra? We prove regularity for the purely atomic and the diffuse maximal abelian algebras in $B(H)$, with $H$ separable. Then we prove singularity of the algebra generated by one free-group generator inside $L(\mathbb F_2)$.

The singularity proof has a concrete mechanism. Reduced words make an infinite cyclic subgroup malnormal: its conjugates intersect it trivially unless the conjugating word lies in that subgroup. This forces mixed coefficient expectations to vanish along sufficiently large powers of the generator. A unitary normalizing the cyclic algebra must consequently have all its coefficients in that algebra.

We use the normal group trace and coefficient uniqueness from Section 4 of Crossed-product coefficients and factor tests. The trace-preserving expectation theorem is Theorem 9.1 of Traces, part B. Spatial isomorphism and diffuse classification are Theorems 7.4 and 8.4 of Abelian operator algebras, together with its countable-generation Theorem 8.2. We use those existing classification proofs without repeating them. The double commutant theorem is in The double commutation theorem. Rational-translation ergodicity and the expected-MASA trace restriction are Lemma 4.1 and Theorem 1.2 of Expected maximal abelian algebras and factor types. Elementary tools are reduced words, Hilbert-space orthogonal projection and the trace Cauchy–Schwarz inequality. The linked programme lessons contain the complete prerequisite proofs. Anantharaman and Popa’s freely readable draft treats finite tracial normalizers; Sinclair and Smith’s freely readable paper develops the asymptotic-homomorphism method and its implication for normalizers; Takesaki’s book provides further context. Here the reduced-word lemma proves the required group property directly, and the full two-step polynomial estimate and expectation argument control every normalizing unitary. The conclusions concern regularity and singularity, with no quantitative strong-singularity bound, Popa invariant or hyperbolic-group theorem asserted.

1. Three normalizer conditions

For a unital abelian von Neumann algebra $D\subseteq M$, set $$\mathcal N_M(D)= \{v\in\mathcal U(M):vDv^*=D\},\qquad N_M(D)=\mathcal N_M(D)''. \tag{1.1}$$ Every unitary of $D$ belongs to $\mathcal N_M(D)$, and those unitaries generate $D$. Indeed, continuous functional calculus writes each self-adjoint contraction $d$ as the real part of the unitary $d+i(1-d^2)^{1/2}$; scaling and linear decomposition give all of $D$.

For a maximal abelian $D$ in a factor $M$, the source terminology is:

• regular: $N_M(D)=M$;
• semiregular: $N_M(D)$ is a factor;
• singular: $\mathcal N_M(D)=\mathcal U(D)$, equivalently $N_M(D)=D$.

We will also use the regular and singular equalities in general ambient algebras. A diffuse singular maximal abelian algebra in a factor is not semiregular: its normalizer generates a diffuse abelian algebra, whose centre is itself.

2. Maximal abelian algebras in $B(H)$

Assume throughout this section that $H$ is a nonzero separable Hilbert space and $D\subseteq B(H)$ is maximal abelian.

Proposition 2.1. If $D$ is atomic, then $D$ is regular.

Proof. Let $(p_i)_{i\in I}$ be its minimal projections, with sum $1$. The index set is finite or countable, since their nonzero ranges are pairwise orthogonal in a separable Hilbert space.

Each $p_i$ has rank one. For $T\in B(p_iH)$, regard $T$ as zero on $(1-p_i)H$. Every $d\in D$ is scalar on $p_iH$, because $Dp_i=\mathbb Cp_i$, so $T$ commutes with $D$. Maximal abelianness gives $T\in D$, and then $T\in Dp_i=\mathbb Cp_i$. Thus $B(p_iH)=\mathbb Cp_i$, which forces $\dim p_iH=1$.

Choose unit vectors $\delta_i\in p_iH$. They form an orthonormal basis, and $D$ is exactly the diagonal algebra in that basis. One inclusion follows from commutativity with the $p_i$; the other follows because every bounded diagonal is a strong limit of its finite diagonal compressions.

The unitary swapping two basis vectors and fixing all the others normalizes $D$. Denote it by $U_{ij}$. Then, for $i\ne j$, $$p_iU_{ij}p_j=|\delta_i\rangle\langle\delta_j|. \tag{2.1}$$ Since $p_i,p_j\in D\subseteq N_{B(H)}(D)$, all rank-one matrix units lie in the normalizer algebra. Finite matrix compressions of any $T\in B(H)$ are finite combinations of these matrix units and converge strongly to $T$. Hence $N_{B(H)}(D)=B(H)$. The one-dimensional case is immediate. $\square$

Proposition 2.2. If $D$ is nonatomic, then $D$ is regular.

Proof. An algebra represented on a separable Hilbert space is countably generated in the strong topology, and every orthogonal family of nonzero projections is countable. The abelian classification prerequisites therefore identify $D$ abstractly with $L^\infty[0,1]$. Theorem 7.4 of Abelian operator algebras implements this isomorphism by a unitary because both represented algebras are maximal abelian. Removing the null endpoint identifies the multiplication model with $$D_0=\{M_f:f\in L^\infty(\mathbb T)\} \quad\text{on }L^2(\mathbb T).$$ Regularity is preserved by unitary conjugation, so it suffices to prove it for $D_0$.

For $q\in\mathbb Q/\mathbb Z$, let $$(V_q\xi)(x)=\xi(x-q).$$ Lebesgue invariance makes $V_q$ unitary, and $$V_qM_fV_q^*=M_{f(\,\cdot-q)}. \tag{2.2}$$ Thus all $V_q$ normalize $D_0$. Let $B$ be the von Neumann algebra generated by $D_0$ and the $V_q$. Since $D_0'=D_0$, every $T\in B'$ has the form $M_f$. Commutation with every $V_q$ means that $f$ is fixed by all rational translations. The previously proved translation lemma makes $f$ constant almost everywhere. Hence $B'=\mathbb C1$, and the double commutant theorem gives $$B=B''=(\mathbb C1)'=B(L^2(\mathbb T)).$$ The normalizer contains all these generators, proving regularity. $\square$

Corollary 2.3. A nonatomic maximal abelian algebra in $B(H)$ is regular but has no normal norm-one retraction from $B(H)$ onto it.

Proof. Regularity is Proposition 2.2. If there were such a retraction, the expected-MASA trace theorem would make the canonical operator trace semifinite on $D$. There would be a nonzero $a\in D_+$ with finite operator trace. A nonzero spectral projection $p=1_{[\varepsilon,\infty)}(a)$ would then have finite rank. The nonzero finite-dimensional algebra $Dp$ contains a minimal projection, which is also minimal in $D$. This contradicts nonatomicity. $\square$

Thus regularity does not imply existence of a normal expectation.

Proposition 2.4 (the mixed case). Let $z$ be the sum of the minimal projections of $D$. If $z\ne0,1$, then $$N_{B(H)}(D)=B(zH)\oplus B((1-z)H). \tag{2.3}$$ In particular $D$ is neither regular nor semiregular.

Proof. A unitary normalizing $D$ permutes its minimal projections, and hence fixes their sum $z$. It is block diagonal for $H=zH\oplus(1-z)H$. This proves the inclusion $\subseteq$ in (2.3).

The corner algebras $Dz$ and $D(1-z)$ are maximal abelian on their respective Hilbert spaces. For example an operator on $zH$ commuting with $Dz$, extended by zero on the other block, commutes with $D$ and so belongs to $Dz$. The first corner is atomic, the second nonatomic. Propositions 2.1 and 2.2 make their normalizers generate their respective full operator algebras. Extend each corner normalizer by the identity on the other block; the resulting unitary normalizes $D$. Since $z,1-z\in D$, compression of the generated algebra yields both whole blocks in (2.3). The equality follows. Its centre is $\mathbb Cz+\mathbb C(1-z)$, so it is not a factor and is properly smaller than $B(H)$. $\square$

3. Subgroup coefficients and the trace Hilbert space

For a countable discrete group $G$, let $u_g$ be left translation on $\ell^2(G)$ and $$M=L(G),\qquad \tau(x)=\langle x\delta_e,\delta_e\rangle,\qquad \|x\|_2=\tau(x^*x)^{1/2}. \tag{3.1}$$ The earlier group-trace proof gives a faithful normal tracial state. Its coefficients are $$x_g=\tau(xu_g^*),\qquad \|x\|_2^2=\sum_g|x_g|^2. \tag{3.2}$$ The map $x\mapsto x\delta_e$ extends to a unitary from the completion $L^2(M,\tau)$ onto $\ell^2(G)$: it is isometric by (3.1), and its range contains the basis $u_g\delta_e=\delta_g$. Thus finite group polynomials approximate every $x\in M$ in the $2$-norm. This is a Hilbert-space assertion, and does not assert convergence of unordered operator Fourier sums.

Let $K\leq G$, and put $A_K=\{u_k:k\in K\}''\subseteq M$.

Lemma 3.1. There is a faithful normal trace-preserving expectation $E_K:M\to A_K$, and its coefficients satisfy $$(E_K(x))_g= \begin{cases} x_g,&g\in K,\\ 0,&g\notin K. \end{cases} \tag{3.3}$$ It is the $L^2$ orthogonal projection onto $\ell^2(K)$ under the preceding identification. In particular $x\in A_K$ exactly when $x_g=0$ for $g\notin K$.

Proof. The trace restriction to $A_K$ is faithful, finite and normal, so Theorem 9.1 of the trace prerequisite applies. It gives $E_K$, bimodularity and $$\tau(E_K(x)b)=\tau(xb)\qquad(b\in A_K). \tag{3.4}$$ For $g\in K$, take $b=u_g^*$ to get the first part of (3.3). Every element of $A_K$ has zero coefficients off $K$: this holds for polynomials in $u_k$, and then holds on their ultraweak closure because each coefficient functional is normal. This gives the second part.

Normal coefficient uniqueness now shows that an $x$ with all coefficients off $K$ zero equals $E_K(x)$. The $L^2$ description follows from (3.3). Alternatively (3.4) says directly that $x-E_K(x)$ is orthogonal to $A_K$. Its Hilbert closure is $\ell^2(K)$, since it contains the basis $(\delta_k)_{k\in K}$. $\square$

We will repeatedly use $$\|E_K(x)\|_2\leq\|x\|_2,\quad \|xy\|_2\leq\|x\|\|y\|_2,\quad \|xy\|_2\leq\|x\|_2\|y\|. \tag{3.5}$$ The first inequality is orthogonal projection. The second follows from $x^*x\leq\|x\|^2 1$; the third follows by taking adjoints and using the trace identity.

4. Reduced words in the free group

Let $G=\mathbb F_2=\langle a,b\rangle$, the group of reduced words in $a^{\pm1},b^{\pm1}$, and let $K=\langle a\rangle$. Every word $g\notin K$ can be written $$g=a^r v a^s,\qquad r,s\in\mathbb Z, \tag{4.1}$$ where $v$ is a nonempty reduced word whose first and last letters are $b$ or $b^{-1}$. Obtain this form by removing the maximal initial and terminal strings of $a$-letters. There is no cancellation at either displayed boundary.

Lemma 4.1. The subgroup $K$ is malnormal: $$gKg^{-1}\cap K=\{e\}\qquad(g\notin K). \tag{4.2}$$ Moreover, the conjugation orbit $\{a^n g a^{-n}:n\in\mathbb Z\}$ is infinite for every $g\notin K$.

Proof. With (4.1), for $k\ne0$, $$g a^k g^{-1}=a^r v a^k v^{-1}a^{-r}. \tag{4.3}$$ This word is reduced at all boundaries involving a displayed power of $a$: the bordering letters of $v$ are $b$-letters. It retains $b$-letters and cannot belong to $K$. This proves (4.2).

If $a^n g a^{-n}=a^m g a^{-m}$ with $n\ne m$, then $g$ commutes with $a^{n-m}$. Consequently $g a^{n-m}g^{-1}=a^{n-m}$ is a nonidentity element of $gKg^{-1}\cap K$, contradicting (4.2). All conjugates are distinct. $\square$

Proposition 4.2. The algebra $$A=A_K=\{u_a\}''$$ is a diffuse maximal abelian subalgebra of $L(\mathbb F_2)$.

Proof. It is abelian because $K$ is cyclic. Suppose $x\in M$ commutes with $u_a$. It is fixed by conjugation with all $u_a^n$. The coefficient conjugation formula makes $x_g$ constant along $\{a^n g a^{-n}\}$. For $g\notin K$, this orbit is infinite by Lemma 4.1. Square summability in (3.2) therefore forces $x_g=0$. Lemma 3.1 puts $x$ in $A$. Thus $A'\cap M=A$.

To see diffuseness, decompose $\ell^2(G)$ over right cosets $Kt$. Left translations by $K$ preserve each $\ell^2(Kt)$, and the unitary $\delta_k\mapsto\delta_{kt}$ identifies this action with the regular action on $\ell^2(K)$. Hence the represented algebra $A$ is a faithful normal amplification of $L(K)$. Since $K\cong\mathbb Z$, the trigonometric orthonormal basis identifies $L(K)$ with the multiplication algebra $L^\infty(\mathbb T)$, as in the earlier circle coefficient example. Every positive-measure set on the circle splits into two positive-measure sets. Thus $L^\infty(\mathbb T)$, and hence $A$, has no minimal projection. $\square$

5. The mixing estimate and singularity

Write $E_A=E_K$, and set $w_n=u_a^n$ for $n\geq1$.

Lemma 5.1. If $x,y\in M$ satisfy $E_A(x)=E_A(y)=0$, then $$\|E_A(xw_n y)\|_2\longrightarrow0. \tag{5.1}$$ For arbitrary $x,y\in M$, this gives $$\|E_A(xw_n y)-E_A(x)w_nE_A(y)\|_2\longrightarrow0. \tag{5.2}$$

Proof. First let $x=u_g$, $y=u_h$ with $g,h\notin K$. Their mixed product is $u_{g a^n h}$. If $g a^n h$ and $g a^m h$ both belong to $K$, then $$(g a^n h)(g a^m h)^{-1} =g a^{n-m}g^{-1}\in K.$$ Malnormality forces $n=m$. Thus $E_A(u_g w_nu_h)$ can be nonzero for at most one integer $n$. For finite polynomials $p,q$ supported off $K$, only finitely many such pairs of words occur. It follows that $E_A(pw_nq)=0$ for all sufficiently large $n$.

For general $x,y$ as in (5.1), their coefficient vectors are supported off $K$. Given $\varepsilon>0$, choose a finite polynomial $p$ supported off $K$ with $$\|x-p\|_2<\frac{\varepsilon}{2(1+\|y\|)}.$$ After $p$ has been chosen, choose a finite polynomial $q$ supported off $K$ with $$\|y-q\|_2<\frac{\varepsilon}{2(1+\|p\|)}.$$ For all sufficiently large $n$, $E_A(pw_nq)=0$, and (3.5) gives $$\begin{aligned} \|E_A(xw_n y)\|_2 &\leq \|(x-p)w_n y\|_2+\|pw_n(y-q)\|_2\\ &\leq\|x-p\|_2\|y\|+\|p\|\|y-q\|_2 <\varepsilon. \end{aligned}$$ No uniform operator-norm bound on the approximating polynomial was assumed.

For (5.2), write $x=E_A(x)+x_0$ and $y=E_A(y)+y_0$. Bimodularity and $w_n\in A$ make the two cross terms zero under $E_A$, so $$E_A(xw_n y)-E_A(x)w_nE_A(y)=E_A(x_0w_n y_0).$$ Apply (5.1). $\square$

Theorem 5.2. The diffuse maximal abelian algebra $A=\{u_a\}''$ is singular in the type-II$_1$ factor $L(\mathbb F_2)$: $$\mathcal N_M(A)=\mathcal U(A). \tag{5.3}$$

Proof. The earlier ICC theorem makes $M=L(\mathbb F_2)$ a type-II$_1$ factor. Let $v\in\mathcal U(M)$ normalize $A$, and put $d=E_A(v)\in A$. Contractivity gives $\|d\|\leq1$. For every $n$, $v w_n v^*$ is a unitary in $A$, so $$E_A(vw_n v^*)=vw_n v^*,\qquad \|E_A(vw_n v^*)\|_2=1.$$ Apply Lemma 5.1 to $x=v$, $y=v^*$. Since $E_A(v^*)=d^*$, it yields $$\|vw_n v^*-d w_n d^*\|_2\longrightarrow0. \tag{5.4}$$ The algebra $A$ is abelian, and thus $d w_n d^*=|d|^2w_n$ has $2$-norm $\||d|^2\|_2$, independent of $n$. Taking norms in (5.4) gives $$\tau(|d|^4)=\||d|^2\|_2^2=1.$$ But $0\leq|d|^4\leq1$ and $\tau(1)=1$. Faithfulness makes $|d|^4=1$, and therefore $|d|^2=1$. Lemma 3.1 makes $E_A$ an orthogonal projection in $L^2$, so $$\|v-d\|_2^2=\|v\|_2^2-\|d\|_2^2=1-1=0.$$ Faithfulness gives $v=d\in A$. Its unitarity in $M$ is then unitarity in $A$. Conversely every unitary of $A$ normalizes $A$. This proves (5.3). $\square$

The proof controls arbitrary normalizing unitaries, not only the group unitaries $u_g$. Malnormality of the subgroup alone would not justify that conclusion without the $L^2$ approximation and expectation argument.

6. Graded exercises with complete solutions

Exercise 6.1 (introductory: diagonal normalizers). Let $D$ be the diagonal algebra on $\ell^2(I)$, where $I$ is finite or countably infinite. Prove that its unitary normalizers are exactly the monomial unitaries $$U\delta_i=z_i\delta_{\sigma(i)},\qquad |z_i|=1,$$ with $\sigma$ a permutation of $I$. Show that the diagonal unitaries and finite transpositions already generate $B(\ell^2(I))$ as a von Neumann algebra.

Solution. A normalizer induces an automorphism of $D$, so it permutes the minimal projections $p_i=|\delta_i\rangle\langle\delta_i|$. Thus $Up_iU^*=p_{\sigma(i)}$ for a permutation $\sigma$, which gives the displayed formula with unimodular $z_i$. Conversely such a unitary takes $M_f$ to $M_{f\circ\sigma^{-1}}$, so it normalizes $D$.

The diagonal unitaries generate $D$, hence all $p_i$. A finite transposition $U_{ij}$ gives $p_iU_{ij}p_j=e_{ij}$ for $i\ne j$, and $e_{ii}=p_i$. Thus every matrix unit lies in the generated algebra. If $P_F=\sum_{i\in F}p_i$, then $P_FTP_F$ is a finite combination of these units and tends strongly to $T$ for every $T\in B(\ell^2(I))$. The generated von Neumann algebra is therefore all of $B(\ell^2(I))$.

Exercise 6.2 (intermediate: a mixed maximal abelian algebra). On $$H=\mathbb C\oplus L^2(\mathbb T),$$ consider $$D=\{\lambda\oplus M_f:\lambda\in\mathbb C,\ f\in L^\infty(\mathbb T)\}.$$ Prove maximal abelianness and compute its normalizer algebra. Decide whether it is regular, semiregular or singular.

Solution. The projection $z=1\oplus0$ belongs to $D$. Any operator commuting with $D$ commutes with $z$, so it is block diagonal. Its first block is scalar; its second commutes with all multiplications and is itself a multiplication. Hence $D'=D$.

The projection $z$ is the sole minimal projection of $D$, since the circle multiplication algebra is diffuse. Every normalizer fixes $z$, so its generated algebra is contained in $\mathbb C\oplus B(L^2(\mathbb T))$. The unitaries $1\oplus V_q$, $q\in\mathbb Q/\mathbb Z$, normalize $D$. Together with its multiplication unitaries they generate the whole second block by Proposition 2.2. Compression by $z,1-z$ therefore gives $$N_{B(H)}(D)=\mathbb C\oplus B(L^2(\mathbb T)).$$ This is properly smaller than $B(H)$ and has a two-dimensional centre, so $D$ is neither regular nor semiregular. It is not singular either: a nonidentity rational translation $1\oplus V_q$ is a normalizer outside $D$. For example it does not commute with multiplication by the indicator of an arc moved by that translation.

Exercise 6.3 (advanced: replacing the generator by its square). In $M=L(\mathbb F_2)$, put $$A=\{u_a\}'',\qquad D=\{u_a^2\}''.$$ Compute $D'\cap M$ and $\mathcal N_M(D)$. Explain why $D$ is not a maximal abelian algebra and why the singularity theorem for $A$ cannot be transferred to it.

Solution. If $x$ commutes with $u_a^2$, its coefficients are constant along the conjugation orbits $a^{2n}g a^{-2n}$. For $g\notin\langle a\rangle$, those orbits are infinite by the malnormality argument: equality at distinct exponents would make $g$ commute with a nonzero even power of $a$. Square summability forces all coefficients off $\langle a\rangle$ to vanish, and Lemma 3.1 puts $x$ in $A$. Conversely $A$ commutes with $D$, so $$D'\cap M=A. \tag{6.1}$$ The inclusion $D\subset A$ is proper: every coefficient of an element of $D$ off $\langle a^2\rangle$ vanishes by Lemma 3.1, whereas the coefficient of $u_a$ at $a$ is $1$. Thus $D$ is not maximal abelian.

If $v$ normalizes $D$, conjugation carries its relative commutant onto itself: $$v(D'\cap M)v^*=(vDv^*)'\cap M=D'\cap M.$$ By (6.1), $v$ normalizes $A$. Theorem 5.2 forces $v\in\mathcal U(A)$. Conversely every unitary of the abelian $A$ commutes with $D$ and normalizes it. Hence $$\mathcal N_M(D)=\mathcal U(A),\qquad N_M(D)=A.$$ In particular $u_a$ is a normalizer outside $D$. The subgroup $\langle a^2\rangle$ also fails malnormality: conjugation by $a\notin\langle a^2\rangle$ fixes that whole subgroup. Both the maximality and the mixing hypothesis used for the primitive generator are lost.

References

[Takesaki] M. Takesaki, Theory of Operator Algebras I, Springer-Verlag, 1979.

[Earlier lessons] The group trace, trace-preserving expectation, abelian classification, translation lemma and expected-MASA trace theorem linked above, with their specific proof locators.

[Anantharaman–Popa] Claire Anantharaman and Sorin Popa, An introduction to II1 factors, author-hosted draft IIunV15.

[Sinclair–Smith] Allan M. Sinclair and Roger R. Smith, Strongly Singular Masas in Type II1 Factors, arXiv:math/0107075v1, 10 July 2001.

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