Double cosets and intrinsic MASA multiplicity

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

A MASA acts on the trace Hilbert space from the left and from the right. For a malnormal abelian subgroup, every nontrivial double coset carries the same regular representation of a product group. The number of those copies becomes a type I multiplicity. Separating the copy coming from the subgroup itself requires an additional argument: a reducing subspace alone need not give a central summand.

We use the group trace and subgroup criterion and its regular-group operator foundations. G00 proves the regular commutants; H03–H04 supply separable predual and the type $\mathrm{II}_1$ conclusion. T03 proves uniqueness of the normal normalized trace for an ICC group algebra. T04 proves trace preservation for an abstract isomorphism of the group factors, using positive-supremum preservation and the bounded trace-square continuity proved in T04c; T05–T06 prove the intrinsic finite or countably infinite size of the particular matrix blocks below. These give the precise trace and multiplicity inputs without a general projection-comparison or type-classification argument.

The freely readable comparison is Sinclair–Smith, The Pukánszky invariant for masas in group von Neumann factors, §2, manuscript p. 4, for the intrinsic complement invariant; Theorem 3.2, pp. 8–10, for double-coset intertwiners; and Theorem 4.1 with its proof, p. 11, for the stabilizer multiplicity calculation. Under (1), every off-subgroup stabilizer is trivial, so all nontrivial double cosets belong to a single class and the Pukánszky invariant is the singleton $\{n\}$. The explicit regular-product model and full matrix commutant are proved below. Anantharaman–Popa, An introduction to II1 factors, Theorem 1.3.6, Proposition 4.1.3 and Lemma 12.1.9, supply freely readable proofs of regular commutation, factor-trace uniqueness and the central subgroup projection.

1. The left/right algebra and its canonical projection

Let $G$ be a countable discrete ICC group and $H\subset G$ an infinite abelian subgroup such that

Put $M=L(G)$, $A=L(H)$. Condition (1) makes the $H$-conjugacy orbit of every $g\notin H$ a free $H$-orbit, hence infinite. The subgroup criterion makes $A$ a MASA in the separable $\mathrm{II}_1$ factor $M$.

On $\ell^2(G)=L^2(M,\tau)$, use

The conjugation $J$ is antilinear. Since $H$ is abelian and left and right translations commute, $\mathcal B$ is abelian. Let $e_A$ denote the orthogonal projection onto $\ell^2(H)=L^2(A)$.

Lemma 1.1. The projection $e_A$ belongs to $\mathcal B$, and is therefore central in $\mathcal C$.

Proof. Enumerate $H$ as $(h_j)$, put $U_j=\lambda_{h_j}\rho_{h_j}$, and form the norm-convergent positive sum

Its kernel consists exactly of the vectors fixed by all $U_j$: the quadratic form is the sum of the nonnegative numbers $2^{-j}\|(U_j-1)\xi\|^2$. These unitaries act by $H$-conjugation on basis vectors. All elements of $H$ are fixed; every orbit outside $H$ is infinite by (1). A square-summable coefficient function constant on an infinite orbit vanishes there. Thus $\ker S=\ell^2(H)$.

For each positive integer $m$, the inverse $(1+mS)^{-1}$ belongs to $\mathcal B$ and has norm at most $1$, by H02. Those inverses converge strongly to the projection onto $\ker S$: they fix that kernel, while $(1+mS)^{-1}S=(1-(1+mS)^{-1})/m$ tends to zero in norm, and $\overline{\operatorname{ran}S}=(\ker S)^\perp$. Strong closedness of $\mathcal B$ therefore puts this projection, $e_A$, in $\mathcal B$. Every element of $\mathcal C=\mathcal B'$ commutes with it. Since $\mathcal B$ is abelian, $e_A$ also belongs to $\mathcal C$, proving centrality. $\square$

The role of infinitude is precise. If $H$ is finite, conjugacy orbits outside $H$ can support nonzero constant square-summable vectors. For example, with $G=\mathbb F_2$, $H=\{e\}$, condition (1) holds, but $A=\mathbb C$ is not a MASA, $\mathcal B=\mathbb C1$, and $\mathcal C=B(\ell^2(G))$. The nonzero projection onto $\delta_e$ is not central in $\mathcal C$. Thus the nonzero abelian central summand associated with $L^2(A)$ cannot be inferred from malnormality alone. The infinite-subgroup hypothesis supplies the missing separation in the exercise's intended MASA setting.

2. Every nontrivial double coset is a regular product orbit

Let $\Gamma=H\backslash G/H$, $\Gamma_0=\Gamma\setminus\{H\}$, and select a representative $g_\gamma$ for each $\gamma\in\Gamma_0$. The disjoint basis decomposition is

Every summand reduces both $\lambda(H)$ and $\rho(H)$: left and right subgroup multiplication preserve each double coset, as do their inverses.

Lemma 2.1. The map

is unitary, and intertwines the pair representation with

Proof. The underlying map $(a,b)\mapsto ag_\gamma b$ is onto. If $ag_\gamma b=a'g_\gamma b'$, then $(a'^{-1}a)g_\gamma(bb'^{-1})=g_\gamma$. Condition (1) forces both subgroup factors to be the identity, hence $a=a'$, $b=b'$. The bijection of orthonormal bases proves unitarity. Finally,

which is the image under (5) of $\lambda_H(h)\delta_a\otimes\rho_H(k)\delta_b$. $\square$

Write $\mathcal H_H=\ell^2(H)$, $\mathcal K=\ell^2(\Gamma_0)$. Combining the maps (5) identifies the off-subgroup space with

and the pair representation there is the identical copy (6) on every multiplicity coordinate.

3. Computing the commutant, including inter-copy operators

Let $A_H=\lambda_H(H)''$ on $\mathcal H_H$. The direct regular-group commutant proof, G00, applied to $H$, gives

where the last equality uses $\rho_H(h)=\lambda_H(h^{-1})$ for abelian $H$. Thus $A_H$ is maximal abelian in its regular Hilbert-space representation. Restricting $A$ to $\ell^2(H)$ identifies it faithfully and normally with $A_H$: its trace vector is separating, so the restriction has zero kernel.

On $\mathcal H_H\otimes\mathcal H_H$, put

Here the spatial von Neumann tensor product means the bicommutant generated by its coordinate algebraic tensor operators, so it is strongly closed by H01. The unitaries in (6) generate $D$. To verify that their bicommutant contains the full coordinate algebras, write an operator commuting with $\lambda_H(H)\otimes1$ in matrix entries over the second coordinate. Every entry commutes with $\lambda_H(H)$, hence with $A_H=\lambda_H(H)''$. Thus $A_H\otimes1$ belongs to the indicated bicommutant; the other coordinate follows in the same way. The reverse inclusion holds because the generating unitaries already belong to the spatial tensor product. This algebra is itself maximal abelian: the representation is the regular representation of the abelian group $H\times H$, with inversion in the second group coordinate. The same trace commutation argument proves $D'=D$.

Let $n=|\Gamma_0|$, a positive integer or countable infinity. Lemma 1.1 separates the two blocks centrally, so no intertwiner in $\mathcal C$ connects $\ell^2(H)$ with its orthogonal complement. On the first block its commutant is $A_H$. On the second, its represented algebra is $D\otimes1_{\mathcal K}$.

Theorem 3.1. With the hypotheses of Section 1,

The second summand is homogeneous of type $\mathrm I_n$.

Proof. Only the second-block commutant remains to compute. Write an operator on $(\mathcal H_H\otimes\mathcal H_H)\otimes\mathcal K$ in matrix entries $T_{\gamma\eta}$. It commutes with every $d\otimes1$, $d\in D$, exactly when every entry commutes with $D$, hence belongs to $D$.

Conversely, operators of $D\bar\otimes B(\mathcal K)$ commute with $D\otimes1$. If a bounded matrix has every entry in $D$, each finite-coordinate compression belongs to $D\otimes M_m$. These compressions converge strongly to the operator, so it belongs to the spatial von Neumann tensor product. This proves (11), including all operators that interchange equivalent copies.

Commutation with every constant multiplicity matrix unit forces a central operator to have zero off-diagonal entries and one identical diagonal entry $d\in D$. Conversely $d\otimes1$ is central because $D$ is abelian. Thus the center is $D\otimes1$, in the finite and countably infinite cases alike. The diagonal rank-one projections in $B(\mathcal K)$ give mutually equivalent abelian projections $1_D\otimes e_{\gamma\gamma}$: each corner is $D$. Their central support is the second summand's identity, since a central projection $z\otimes1$ dominating one of them has $z=1$. Their matrix units identify that summand as matrices of size $n$ over the abelian center $D$, or countably infinite matrices if $n=\infty$. This is precisely homogeneous type $\mathrm I_n$. $\square$

The individual double-coset projections are generally not central in $\mathcal C$. All nontrivial double-coset representations are equivalent, so their commutant contains off-diagonal matrix units. The central separation is between the subgroup block and the entire multiplicity block.

The decomposition proof also works for arbitrary discrete $G$, infinite abelian $H$ satisfying (1), and an arbitrary double-coset index set. In Lemma 1.1 one may choose a countably infinite subgroup $H_0\subset H$: select countably many distinct elements of $H$ and take the subgroup they generate; the finite words in these generators and their inverses form a countable set. Then its conjugation action is still free outside $H$, and its fixed vectors are exactly $\ell^2(H)$. Finite-coordinate compressions in Theorem 3.1 then form a net rather than a sequence. Formula (11) remains valid with $\mathcal K=\ell^2(\Gamma_0)$; countability is used here only for the countable multiplicity and separable-factor formulation of the source exercise.

4. Why the multiplicity is intrinsic to the MASA

For a MASA in a finite factor, define intrinsically

in its canonical trace representation. The distinguished projection $e_A$ comes from $L^2(A)$, not from a chosen group presentation. For the present group MASAs it is central, and Theorem 3.1 says that $\mathcal C_A^0$ is homogeneous of type $\mathrm I_n$.

Theorem 4.1. If an isomorphism between finite factors carries one of these group MASAs onto another, their numbers $n$ are equal.

Proof. Let $\theta:M\to N$ carry $A$ onto $A_1$. The abstract $*$-isomorphism $\theta$ preserves increasing positive suprema. Therefore $\tau_N\theta$ is a normalized order-normal trace on the ICC group algebra $M$. T04 proves its equality with $\tau_M$: T04c supplies continuity along the bounded trace-square-convergent conjugation averages of T03. Thus $\tau_N\theta=\tau_M$, under the stated abstract-isomorphism hypothesis. Hence

extends to a unitary of the trace Hilbert spaces. It intertwines left multiplication, takes $\overline{A\Omega_M}$ onto $\overline{A_1\Omega_N}$, and satisfies

because both sides send $x\Omega_M$ to $\theta(x)^*\Omega_N$. It therefore carries $e_A$ to $e_{A_1}$ and unitarily conjugates the two algebras in (12). The two complementary blocks have the explicit matrix models $D\bar\otimes B(\ell^2(n))$ and $D_1\bar\otimes B(\ell^2(m))$. T05–T06 prove their sizes intrinsic: a finite block has exactly its matrix size in any orthogonal full-support abelian projection decomposition of the unit, while a countably infinite block has a proper shift isometry and a finite block has none. The induced isomorphism therefore forces $n=m$. $\square$

This distinguishes MASAs as embedded pairs $(M,A)$. It does not prove that their ambient factors are nonisomorphic. It also distinguishes the $n=1$ case correctly: even when both summands of (11) are abelian, the canonical projection $e_A$ still singles out the complement in (12).

5. Exercises with complete solutions

Exercise 1. Verify the identity $J\lambda_hJ=\rho_h$, including an arbitrary complex coefficient.

Solution. On $c\delta_t$, the first $J$ gives $\overline c\delta_{t^{-1}}$, left translation gives $\overline c\delta_{ht^{-1}}$, and the second $J$ gives $c\delta_{th^{-1}}$. Thus the composite is linear and equals $\rho_h$. Omitting either complex conjugation would make $J$ fail to be the trace conjugation.

Exercise 2. Show why reducing each summand in (4) does not itself imply that their projections are central in $\mathcal C$.

Solution. A reducing projection commutes with $\mathcal B$, so it belongs to $\mathcal C$. Centrality additionally requires commutation with every element of $\mathcal C$. Two equivalent nontrivial orbit representations have an intertwiner between them, which belongs to $\mathcal C$ and fails to commute with their separate projections. Lemma 1.1 proves the stronger fact $e_A\in\mathcal B$, which gives centrality for the subgroup projection.

Exercise 3. Prove directly that the kernel of (3) is the common fixed-vector space.

Solution. If $S\xi=0$, positivity gives $0=\langle\xi,S\xi\rangle=\sum_j2^{-j}\|(U_j-1)\xi\|^2$. Every summand is nonnegative, so every term vanishes. Conversely a common fixed vector is killed by every summand and hence by the norm-convergent sum. A vector with zero positive quadratic form belongs to the kernel, since $\langle\xi,S\xi\rangle=\|S^{1/2}\xi\|^2$.

Exercise 4. For $H=\{e\}\subset\mathbb F_2$, compute $\mathcal B$, $\mathcal C$ and the centrality of $e_A$.

Solution. Both subgroup representations are the identity, so $\mathcal B=\mathbb C1$ and $\mathcal C=B(\ell^2(\mathbb F_2))$. The projection $e_A$ has rank one, onto $\delta_e$. An operator carrying $\delta_e$ to $\delta_a$, for a free generator $a$, does not commute with it. Thus it is not central. This is why malnormality with a finite subgroup is insufficient for the intended nonzero central abelian summand.

Exercise 5. Derive the injectivity of the basis map (5), and identify exactly where malnormality enters.

Solution. Equality $ag_\gamma b=a'g_\gamma b'$ gives $(a'^{-1}a)g_\gamma(bb'^{-1})=g_\gamma$. Both outer factors belong to $H$, and $g_\gamma\notin H$. Apply (1), with $h=a'^{-1}a$ and $k^{-1}=bb'^{-1}$, to obtain $a=a'$, $b=b'$. Without this conclusion the map would quotient a nontrivial stabilizer and need not be a regular product representation.

Exercise 6. Why does (6) generate $A_H\bar\otimes A_H$, even though it uses the right regular representation in its second coordinate?

Solution. Set $k=e$ to obtain $\lambda_H(h)\otimes1$, and set $h=e$ to obtain $1\otimes\rho_H(k)$. For abelian $H$, $\rho_H(k)=\lambda_H(k^{-1})$, so the two generated coordinate algebras are both $A_H$. Their joint von Neumann algebra is the spatial tensor product in (10).

Exercise 7. In the $n=2$ case, exhibit an operator in $\mathcal C$ that interchanges the two nontrivial double-coset copies.

Solution. Under (8), take $1_D\otimes(e_{12}+e_{21})$ on the second block and zero on the subgroup block. It commutes with $D\otimes1$, so belongs to $\mathcal C$. It swaps the two multiplicity coordinates and does not commute with the projection $1_D\otimes e_{11}$. The latter projection is therefore not central.

Exercise 8. Prove the strong convergence of finite-coordinate compressions used in Theorem 3.1.

Solution. Let $P_F$ project onto the coordinates of a finite subset $F\subset\Gamma_0$. These projections tend strongly to $1$ along inclusion. For a bounded operator $T$, $P_FTP_F\xi-T\xi=P_FT(P_F\xi-\xi)+(P_F-1)T\xi$, whose norm tends to zero. Thus the finite matrices recover every bounded operator with entries in $D$. This works as a net for an uncountable index set as well.

Exercise 9. What happens to the invariant when there are exactly two total double cosets?

Solution. There is one nontrivial double coset, so $n=1$ and $\mathcal C_A^0\cong D$ is abelian, of type $\mathrm I_1$. The full algebra is $A_H\oplus D$, also abelian. The canonical subgroup projection $e_A$ is nevertheless preserved under pair isomorphisms, so the complement still has a well-defined multiplicity $1$; it cannot be confused with an absent complement or an arbitrary abelian summand.

Exercise 10. Explain why the unitary (13) preserves the right action and not just the left action.

Solution. The trace conjugation sends $x\Omega$ to $x^*\Omega$, and (14) follows on a dense set. Right multiplication by $a$ is $J a^*J$ when $a$ denotes left multiplication. Therefore intertwining the left action and $J$ also intertwines the right action. Since $\theta(A)=A_1$, the whole left/right generated algebra and its commutant are conjugated, along with $e_A$.

References

Lajos Pukánszky, On Maximal Abelian Subrings of Factors of Type II₁, Canadian Journal of Mathematics 12 (1960), 289–296. Lemma 1, p. 290, transports factor isomorphisms to the trace completion and both multiplication actions; Lemma 4, pp. 293–295, identifies the off-subgroup double-coset spaces with equivalent regular product representations and separates the subgroup trace-space projection.

Allan M. Sinclair and Roger R. Smith, The Pukánszky invariant for masas in group von Neumann factors, author-hosted manuscript of the 2005 Illinois Journal of Mathematics article, 49, 325–343. §2, manuscript p. 4, defines the complement invariant and proves centrality of the MASA trace-space projection; Theorem 3.2 and its proof, pp. 8–10, describe double-coset intertwiners; Theorem 4.1 and its proof, p. 11, turn equal stabilizers into homogeneous type I multiplicity. The present malnormal case has trivial stabilizers throughout and gives the singleton invariant $\{n\}$.

Claire Anantharaman and Sorin Popa, An introduction to II1 factors, author-hosted draft. Theorem 1.3.6 and its supporting Lemma 1.3.4 and Proposition 1.3.5, printed pp. 7–8 (PDF pp. 13–14), prove regular commutation, applied here to $H$ and $H\times H$. Proposition 4.1.3 with Lemma 4.1.1 and Corollary 4.1.2, printed pp. 59–60 (PDF pp. 65–66), proves uniqueness of the factor trace. Lemma 12.1.9, printed pp. 195–196 (PDF pp. 201–202), proves that the MASA trace-space projection belongs to the left/right generated algebra; Lemma 1.1 above gives a complete fixed-vector proof in the present group setting.