Schröder–Bernstein theorems for operator algebras

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The Schröder–Bernstein theorem, from set theory, has analogs in the context operator algebras. This article discusses such operator-algebraic results.

[edit] For von Neumann algebras

Suppose M is a von Neumann algebra and E, F are projections in M. Let ~ denote the Murray-von Neumann equivalence relation on M. Define a partial order « on the family of projections by E « F if E ~ F' ≤ F. In other words, E « F if there exists a partial isometry U ∈ M such that U*U = E and UU* ≤ F.

For closed subspaces M and N where projections P_M and P_N, onto M and N respectively, are elements of M, M « N if P_M « P_N.

The Schröder–Bernstein theorem states that if M « N and N « M, then M ~ N.

A proof, one that is similar to a set-theoretic argument, can be sketched as follows. Colloquially, N « M means that N can be isometrically embedded in M. So

$M = M_0 \supset N_0$

where N₀ is an isometric copy of N in M. By assumption, it is also true that, N, therefore N₀, contains an isometric copy M₁ of M. Therefore one can write

$M = M_0 \supset N_0 \supset M_1.$

By induction,

$M = M_0 \supset N_0 \supset M_1 \supset N_1 \supset M_2 \supset N_2 \supset \cdots .$

It is clear that

$R = \cap_{i \geq 0} M_i = \cap_{i \geq 0} N_i.$

Let

$M \ominus N \stackrel{\mathrm{def}}{=} M \cap (N)^{\perp}.$

$M = \oplus_{i \geq 0} ( M_i \ominus N_i ) \quad \oplus \quad \oplus_{j \geq 0} ( N_j \ominus M_{j+1}) \quad \oplus R$

and

$N_0 = \oplus_{i \geq 1} ( M_i \ominus N_i ) \quad \oplus \quad \oplus_{j \geq 0} ( N_j \ominus M_{j+1}) \quad \oplus R.$

Notice

$M_i \ominus N_i \sim M \ominus N \quad \mbox{for all} \quad i.$

The theorem now follows from the countable additivity of ~.

[edit] Representations of C*-algebras

There is also an analog of Schröder–Bernstein for representations of C*-algebras. If A is a C*-algebra, a representation of A is a *-homomorphism φ from A into L(H), the bounded operators on some Hilbert space H.

If there exists a projection P in L(H) where P φ(a) = φ(a) P for every a in A, then a subrepresentation σ of φ can be defined in a natural way: σ(a) is φ(a) restricted to the range of P. So φ then can be expressed as a direct sum of two subrepresentations φ = φ' ⊕ σ.

Two representations φ₁ and φ₂, on H₁ and H₂ respectively, are said to be unitarily equivalent if there exists an unitary operator U: H₂ → H₁ such that φ₁(a)U = U'φ₂(a), for every a.

In this setting, the Schröder–Bernstein theorem reads:

If two representations ρ and σ, on Hilbert spaces H and G respectively, are each unitarily equivalent to a subrepresentation of the other, then they are unitarily equivalent.

A proof that resembles the previous argument can be outlined. The assumption implies that there exist surjective partial isometries from H to G and from G to H. Fix two such partial isometries for the argument. One has

$\rho = \rho_1 \simeq \rho_1 ' \oplus \sigma_1 \quad \mbox{where} \quad \sigma_1 \simeq \sigma.$

In turn,

$\rho_1 \simeq \rho_1 ' \oplus (\sigma_1 ' \oplus \rho_2) \quad \mbox{where} \quad \rho_2 \simeq \rho .$

By induction,

$\rho_1 \simeq \rho_1 ' \oplus \sigma_1 ' \oplus \rho_2' \oplus \sigma_2 ' \cdots \simeq ( \oplus_{i \geq 1} \rho_i ' ) \oplus ( \oplus_{i \geq 1} \sigma_i '),$