Abelian pinching and relative commutants

Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. Self-checked by the writing AI. New original text: public domain (CC0).

A finite partition of the identity removes the off-diagonal corners of an operator. If the partition lies in a centralizer, this operation is an orthogonal projection in the state GNS space. Refining the partition can make its surviving diagonal part nearly scalar on each atom. These two projections give a useful test for maximal abelianness and the preparation for constructing MASAs inside expected subfactors.

Throughout, $M\ne0$ is a von Neumann algebra with a faithful normal state $\varphi$. Put

No factor or separability hypothesis is imposed in this lesson. We use the modular expectation theorem in its state specialization: a modular invariant unital subalgebra $B$ has a unique normal $\varphi$-preserving expectation $E_B$, and

This is the exact general expectation prerequisite, proved in the modular course; we use it rather than develop general weight theory here. Subalgebras of $M_\varphi$, their relative commutants, and the algebras generated by those two commuting algebras are modular invariant. The restriction of $\varphi$ to $M_\varphi$ is a faithful normal trace.

1. The two projections of a finite partition

Let $e_1,\ldots,e_r\in M_\varphi$ be nonzero mutually orthogonal projections with sum $1$, and let $A=\sum_i\mathbb Ce_i$. Define

Lemma 1.1. $T_A=E_{A'\cap M}$ and $S_A=E_A$.

Proof. $A'\cap M=\sum_i e_iMe_i$, so $T_A$ is a normal unital positive retraction onto that algebra and is bimodular there. Centralizer membership gives

Thus it is the expectation in (2). The coefficients in $S_A$ are normal states of the corners; its formula gives a normal positive retraction fixing every $e_i$, preserving $\varphi$, and bimodular over $A$. Uniqueness gives its identification. Faithfulness ensures that every displayed denominator is positive. $\square$

The corner vectors are pairwise orthogonal:

Indeed, unequal left indices give zero by projection multiplication. For equal left indices and unequal right indices, move the final centralizer projection around the state pairing to obtain zero. This explains both identities.

The GNS projections satisfy $S_A\le T_A$, so $T_A-S_A$ is another orthogonal projection. Here and below inequalities between these maps mean inequalities between their GNS projections. If $A\subset D\subset M_\varphi$ are finite-dimensional abelian algebras, their partitions refine one another and

For the last assertion, the range of $T_D-S_D$ lies in the range of $T_A$ and is orthogonal to the range of $S_A$. Consequently a refinement never increases either the pinched norm or the gap norm.

2. A maximal-abelian test for increasing partitions

Theorem 2.1. Let $(A_i)$ be an increasing directed net of finite-dimensional abelian subalgebras of $M_\varphi$, and put $A=(\bigcup_i A_i)''$. Then $A$ is maximal abelian in $M$ if and only if

Proof. The subspaces $\overline{A_i\Omega}$ increase to $\overline{A\Omega}$, by bounded strong density of the increasing union. Thus their projections converge strongly to $P_A$, and $S_{A_i}(x)\to E_A(x)$ in $\varphi$-norm.

The projections onto $\overline{(A_i'\cap M)\Omega}$ decrease. Call their strong limit $P$. We identify it without assuming that arbitrary intersections commute with Hilbert space closure. Fix $x\in M$. The elements $T_{A_i}(x)$ have norm at most $\|x\|$, and their GNS vectors converge to $P(x\Omega)$. An ultraweakly convergent subnet has a limit $b\in M$. For every $z\in M$, normality of $\varphi(z^*\,\cdot)$ identifies

The subnet is eventually in $A_j'\cap M$ for each fixed $j$; these algebras are ultraweakly closed. Hence $b\in A'\cap M$. Conversely $\overline{(A'\cap M)\Omega}$ lies in every one of the decreasing subspaces. Since $M\Omega$ is dense, (7) proves that $P=P_{A'\cap M}$.

Therefore $T_{A_i}(x)\to E_{A'\cap M}(x)$ in $\varphi$-norm. Condition (6) is equivalent to $E_{A'\cap M}=E_A$, by faithfulness, and hence to $A'\cap M=A$. $\square$

In particular the test applies to finite partitions generating any abelian algebra in the centralizer. It does not require a sequence: finite spectral partitions form a directed generating family even when the algebra is nonseparable.

3. Relative commutants survive a corner

For this section let $N\subset M$ be any unital von Neumann subalgebra, and let $e\in N$ be a nonzero projection. Write $N_e=eNe$, $M_e=eMe$, and $N^c=N'\cap M$.

Lemma 3.1. With $z=c_N(e)$, compression gives an isomorphism

In particular

Proof. Choose partial isometries $v_i\in N$ with $v_0=e$, $v_i^*v_i\le e$, and mutually orthogonal final projections summing strongly to $z$. Such a family exists by maximal exhaustion: any nonzero remaining projection under $z$ has a nonzero corner with $Ne$, since $z$ is the central support of $e$; its polar part supplies another member.

For $x\in(N_e)'\cap M_e$, define

The sum is strong and bounded by $\|x\|$, because its summands act on orthogonal final subspaces and $x$ commutes with each initial projection. For $a\in Nz$, the coefficients $v_i^*av_j\in eNe$ commute with $x$. Multiplying the two bounded sums therefore gives $ya=ay$. Also $ye=x$: the other final projections are orthogonal to $e$. Thus $y\in N^c z$.

If $y\in N^c z$ and $ye=0$, then $yv_i=v_i ye=0$ for every $i$, so $y=0$. This proves uniqueness and injectivity; compression preserves products because $y$ commutes with $e$. Finally, $N$ and $N^c$ commute, and products $ab$, $a\in N,b\in N^c$, span an ultraweakly dense subalgebra of their join. Their compressions are $(eae)(be)$. Equation (8) now gives (9). $\square$

This proof permits an arbitrary index set of partial isometries and imposes no type, factor, state or separability hypothesis.

4. Moving an operator away from itself

Return to $N\subset M_\varphi$, and put $N^c=N'\cap M$.

Lemma 4.1. If $x\ne0$, $E_{N^c}(x)=0$, and $\eta>0$, there is a unitary $u\in N$ such that

Proof. Let $C$ be the ultraweakly closed convex hull of the unitary conjugates of $x$. It is compact and bounded. Its GNS image is weakly compact, hence norm closed: the map is weakly continuous on a bounded set, first against $z\Omega$ by the normal pairings $\varphi(z^*\,\cdot)$, then against every vector by density and uniform bounds. It has a unique vector of least norm, represented by $y_0\in C$.

Conjugation by $v\in\mathcal U(N)$ is a $\varphi$-norm isometry preserving $C$. Uniqueness and faithfulness give $vy_0v^*=y_0$, so $y_0\in N^c$. For $b\in N^c$ and $u\in\mathcal U(N)$,

Every vector of $C\Omega$ is thus orthogonal to $\overline{N^c\Omega}$; in particular $y_0\Omega$ is orthogonal to itself. Hence $y_0=0$.

If every unitary failed (11), expanding the squared norm and using its isometry would give

The same inequality holds on the convex hull and its ultraweak closure, since this pairing is normal. It cannot hold at $0\in C$. $\square$

Lemma 4.2. If $u\in\mathcal U(M_\varphi)$ and $\|uxu^*-x\|_\varphi>\|x\|_\varphi$, there is a finite spectral partition $e_1,\ldots,e_r$ of $u$ such that

Proof. Approximate $u$ in operator norm by $v=\sum_i\lambda_i e_i$, $|\lambda_i|=1$, retaining $\|vxv^*-x\|_\varphi>\|x\|_\varphi$. Right multiplication by a centralizer element has norm at most its operator norm on $H_\varphi$, so this follows from norm approximation of $u$. The orthogonal blocks in (4) give

At least one quarter of the total energy is off diagonal; subtract it using (4). $\square$

5. Simultaneous small pinches

Put $B=N\vee N^c$.

Theorem 5.1. Given $x_1,\ldots,x_s\in\ker E_B$ and $\varepsilon>0$, there is a finite partition $(e_i)$ of $1$ in $N$ with

If $N^c=\mathbb C$, the same conclusion applies to $x_j-E_N(x_j)$.

Proof. First consider one nonzero $x$. Since $N^c\subset B$, (2) gives $E_{N^c}(x)=0$. Lemmas 4.1 and 4.2 give a partition reducing its squared pinched norm by $3/4$.

Suppose a partition has already been selected. In a nonzero corner $eMe$, use the faithful normal state $\varphi_e=\varphi(e)^{-1}\varphi|_{eMe}$. The corner algebra $eNe$ is in its centralizer. Equation (9) identifies its join with its relative commutant as $eBe$. For $b\in B$,

using centralizer membership of $e$ and $E_B(x)=0$. Thus $exe$ is orthogonal to $eBe$ in the corner GNS space. Apply the first step in every nonzero corner and add their refined partitions. Their energies add by (4), giving another factor $3/4$. After $m$ iterations the squared norm is at most $(3/4)^m\|x\|_\varphi^2$, which proves (14) for one element.

For a finite list, run this procedure for the first element and then for each next element in the current corners. The previous pinched norms cannot increase, by (5). Zero elements need no additional step. This proves (14) simultaneously. If $N^c=\mathbb C$, then $B=N$, and the final assertion follows. $\square$

6. Small gaps and equal atoms

Theorem 6.1. Suppose $N\subset M_\varphi$ and $N'\cap M\subset N$. For a finite $F\subset M$ and $\varepsilon>0$, there is a finite-dimensional abelian $A\subset N$ such that

If $N$ is a $\mathrm{II}_1$ factor, $A$ can be chosen with $2^k$ mutually equivalent minimal projections, with $k$ as large as prescribed. If $N$ is a finite type I factor, it can be chosen as a full diagonal algebra.

Proof. Our relative-commutant hypothesis gives $B=N$. Decompose $x=y+z$, with $y=E_N(x)$, $z=x-E_N(x)$. Theorem 5.1 supplies a finite abelian $A_0\subset N$ with

In each minimal corner $eNe$ of $A_0$, choose a MASA $D_e$. It lies in the centralizer of the corner state, since that state is tracial on $eNe$. Finite spectral partitions generate $D_e$. Theorem 2.1, now in $eNe$, supplies a finite abelian refinement whose gap on the finitely many $eye$ is at most $\delta\|eye\|_\varphi$. Combine these refinements into $A\supset A_0$.

For $y\in N$, formula (3) shows that the ambient and the $N$-internal gap agree. Corner orthogonality gives

Also $S_A(z)=0$, since $A\subset N$, and $T_A\le T_{A_0}$. Therefore

The last step uses the orthogonal decomposition in (2). Choose $\delta$ small enough for all $x\in F$. This proves (16).

The sharper bound $\delta\|x\|_\varphi$ also follows by orthogonality: $(T_A-S_A)(y)\in N$, while $T_A(z)$ is orthogonal to $N$, since $T_A$ is self-adjoint on GNS space and preserves $N$. Add their squared estimates rather than their norms to obtain that bound.

Now let $N$ be a $\mathrm{II}_1$ factor. Its restricted state is its normalized trace $\tau$. For the minimal projections $f_1,\ldots,f_r$ of $A$, choose positive dyadic numbers $m_i/2^k$ summing to $1$ and approaching $\tau(f_i)$. Transfer small orthogonal subprojections from the atoms with excess trace to those with deficit trace, using the continuous trace cuts in a type II factor. This gives a partition $f_i^{(k)}$ of those exact traces with

Their differences are self-adjoint centralizer elements, so they tend strong* to zero in $M$. Bounded multiplication is strong* continuous. Formula (3), normality of $\varphi$, and the positive limiting denominators consequently show that both expectations on each fixed $x\in F$ converge in $\varphi$-norm. Preserve the strict margin in (16).

Split each $f_i^{(k)}$ into $m_i$ projections of trace $2^{-k}$. They are equivalent by finite-factor projection comparison. Their diagonal algebra refines the perturbed one, so (5) preserves the gap bound. Increasing $k$ makes its size arbitrarily large.

Finally, if $N$ is a finite type I factor, a full diagonal algebra has equal rank-one atoms and is maximal abelian in $M$. Indeed $N'\cap M\subset N$ gives $N'\cap M=\mathbb C$; the type I matrix decomposition then gives $M=N$. Its gap is zero. $\square$

The equal atoms can have dyadic size, as required for later exact matrix containment. The triangle inequality in (18) has the factor $\sqrt2$; the orthogonal decomposition gives the sharper estimate described immediately after it.

7. Exercises with complete solutions

Exercise 1. Derive both formulas (3) when $M=M_3$, $\varphi(x)=\operatorname{Tr}(\rho x)$, and $\rho=\operatorname{diag}(1/2,1/3,1/6)$, using the rank-one diagonal partition.

Solution. Each $e_i$ commutes with $\rho$, so is in the centralizer. Pinching keeps the three diagonal entries. Its scalar coefficient in the $i$-th corner is $\varphi(e_i x e_i)/\varphi(e_i)=\rho_{ii}x_{ii}/\rho_{ii}=x_{ii}$. Thus $T_A=S_A=\operatorname{diag}(x_{11},x_{22},x_{33})$. The unequal state weights cancel inside each atom.

Exercise 2. Show that refinement can remove diagonal energy that survived an earlier partition.

Solution. In $M_3$ take the coarse partition $e_1=p_1+p_2,e_2=p_3$ and $x=e_{12}$. Then $T_A(x)=x$. Refining to the three rank-one projections makes every $p_i x p_i=0$, so $T_D(x)=0$. This is consistent with $T_D\le T_A$.

Exercise 3. Prove the final inequality of (5) from subspace inclusions.

Solution. A vector in the range of $T_D-S_D$ belongs to $\overline{(D'\cap M)\Omega}\subset\overline{(A'\cap M)\Omega}$ and is orthogonal to $\overline{D\Omega}\supset\overline{A\Omega}$. It therefore lies in the range of $T_A-S_A$. Inclusion of closed subspaces is exactly the asserted order of their orthogonal projections.

Exercise 4. Why is an ultraweak cluster point in Theorem 2.1 sufficient to identify the decreasing GNS projection?

Solution. The pinched operators are bounded. Against every vector $z\Omega$, their GNS pairings are the normal functionals $\varphi(z^*\,\cdot)$, so the cluster point $b$ has GNS vector equal to the already existing projection limit. Eventual membership in each $A_j'\cap M$ puts $b$ in their intersection. Density of $M\Omega$ identifies the entire projection, not merely its value on one test vector.

Exercise 5. Check (8) in $N=M_2\otimes1\subset M=M_2\bar\otimes P$, with $e=e_{11}\otimes1$.

Solution. $N'\cap M=1\otimes P$. The corner $eMe=e_{11}\otimes P$, while $eNe=\mathbb Ce$, so its relative commutant is the entire corner. Compression sends $1\otimes y$ to $e_{11}\otimes y$, an isomorphism. Equation (9) is then $e_{11}\otimes P=eMe$.

Exercise 6. Explain where central support enters the uniqueness in (8).

Solution. If $y\in N^c z$ and $ye=0$, commutation gives $yv_i=v_i ye=0$. The final projections $v_i v_i^*$ sum to $z$, so $yz=0$, and its support condition gives $y=0$. Without that support restriction an element on $1-z$ would also compress to zero.

Exercise 7. Verify the coefficient $4$ in (13) for $u=\operatorname{diag}(1,-1)$ and $x=e_{12}$ with normalized trace on $M_2$.

Solution. $uxu^*=-x$, so the squared displacement is $4\|x\|_2^2=2$. The sole nonzero block is off diagonal, with coefficient $|1(-1)-1|^2=4$. Thus (13) is an equality. The pinched operator is zero.

Exercise 8. How many single-element refinement rounds suffice to make a pinched norm at most $10^{-2}\|x\|_\varphi$?

Solution. It suffices that $(3/4)^m\le10^{-4}$, or $m\ge\log(10^{-4})/\log(3/4)$. This quotient is between $32$ and $33$, so $33$ rounds suffice. Each round has a finite partition; no uniform bound on its number of atoms is asserted.

Exercise 9. Prove (15) directly using the GNS orthogonality of $x$ to $B$.

Solution. For $b\in B$, centralizer membership of $e$ gives $\varphi(b^*exe)=\varphi(eb^*ex)$. Since $ebe\in B$, this is the inner product of $x\Omega$ with $(ebe)\Omega$, hence zero. Dividing by $\varphi(e)$ gives the identical corner-state orthogonality.

Exercise 10. Give a finite example showing why the last bound in (18) needs $\sqrt2$.

Solution. With normalized trace on $M_2$ and $N$ its diagonal algebra, take $x=e_{11}+e_{12}$. Then $E_N(x)=e_{11}$, $x-E_N(x)=e_{12}$, and both component norms are $1/\sqrt2$. Their sum is $\sqrt2$, whereas $\|x\|_2=1$. Orthogonality gives the corrected factor exactly.

Exercise 11. Explain why dyadic perturbation in (19) preserves (16) even though the ambient state need not be tracial.

Solution. The projection differences lie in $N$, where $\varphi$ is tracial, and are self-adjoint. Their small $2$-norm therefore gives small symmetric state seminorm, hence bounded strong* convergence in $M$. Fixed multiplication and normal scalar pairings in (3) preserve that convergence, and the finitely many denominators tend to positive numbers. Both expectation values on each prescribed $x$ converge in $\varphi$-norm. A strict finite margin remains for sufficiently large $k$.

Exercise 12. Why does splitting a perturbed atom into equal subatoms require no new gap estimate?

Solution. The new diagonal algebra contains the old one. Its scalar GNS projection increases and its commutant projection decreases, so its gap projection is below the old gap by (5). Applying the two projections to any $x\Omega$ proves the required norm inequality immediately.

Reading and prerequisites

Sorin Popa, On a problem of R. V. Kadison on maximal abelian *-subalgebras in factors, INCREST preprint 41/1981, May 1981, second version; the journal article appeared in Inventiones Mathematicae 65 (1981), 269–281, DOI 10.1007/BF01389015. Lemmas 1.1–1.5 and Remark 1.4, printed pp.4–10 of this preprint, give the finite tracial MASA test, corner lifting, displacement and spectral pinching constructions. The present proofs extend their mechanism to arbitrary faithful normal states with the pinching algebra in the centralizer. The GNS projection argument handles directed nets, the corner lifting permits arbitrary index sets and all types, and the dyadic perturbation is checked in the ambient state topology. Popa’s Lemma 1.5 uses half the desired error for each component; its triangle estimate needs no correction.

General modular expectations and their GNS projection identity remain the selected modular-course prerequisite. Projection central support, polar decomposition, type II trace cuts and finite projection comparison are the selected projection and trace foundations. All broader hypotheses are stated in the corresponding results and proved above; transitive source verification of those foundations remains pending.