Schröder–Bernstein theorems for operator algebras

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

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 PM and PN, onto M and N respectively, are elements of M, M « N if PM « PN.

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


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

So



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 ~.

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 a unitary operator U: H2 → H1 such that φ1(a)U = Uφ2(a), for every a.

In this setting, the Schröder–Bernstein theorem reads:


 * If two representations &rho; and &sigma;, 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 '), $$

and



\sigma_1 \simeq \sigma_1 ' \oplus \rho_2' \oplus \sigma_2 ' \cdots \simeq ( \oplus_{i \geq 2} \rho_i ' ) \oplus ( \oplus_{i \geq 1} \sigma_i '). $$

Now each additional summand in the direct sum expression is obtained using one of the two fixed partial isometries, so



\rho_i ' \simeq \rho_j ' \quad \mbox{and} \quad \sigma_i ' \simeq \sigma_j ' \quad \mbox{for all} \quad i,j \;. $$

This proves the theorem.