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 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

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

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 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,

$\rho _{1}\simeq \rho _{1}'\oplus \sigma _{1}'\oplus \rho _{2}'\oplus \sigma _{2}'\cdots \simeq (\oplus _{i\geq 1}\rho _{i}')\oplus (\oplus _{i\geq 1}\sigma _{i}'),$

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.

See also

• Schröder–Bernstein theorem for measurable spaces
• Schröder–Bernstein property

References

• B. Blackadar, Operator Algebras, Springer, 2006.