FOM: Invariant Subspace Problem

[Harvey Friedman]

Richard Laver pointed out to me on the phone that the invariant subspace
problem for Hilbert space is a very simple example of Boolean relation
theory.

Let V = the set of linear operators on Hilbert space H. Let K be the set of
closed subspaces of H.

The invariant subspace problem asserts that

*for all f in V there exists A in K such that fA containedin A and A not= H.*

This is Boolean relation theory. To write the invariant subspace problem as
equational Boolean relation theory, let K' be the set of proper closed
subspaces of Hilbert space. The invariant subspace problem asserts that

**for all f in V there exists A in K such that fA containedin A.**

**********

SOME UNDERGRADUATE EXERCISES IN BOOLEAN RELATION THEORY

Let us now look at the situation with regard to all functions from a set E
into E, and the nonempty proper subsets of E.

THEOREM 1. The statement "for all f:E into E there exists a nonempty proper
subset A of E such that fA containedin A" holds if and only if E is
infinite.

THEOREM 2. The statement "for all multivariate f from E into E there exists
a nonempty proper subset A of E such that fA containedin A" holds if and
only if E is uncountable. The same claim is true if "multivariate" is
replaced by "binary."

THEOREM 3. The statement "for all f:E into E there exists a proper subset A
of E of the same cardinality as E such that fA containedin A" is true if
and only if E is uncountable.

THEOREM 4. The statement "for all f:E^2 into E there exists a proper subset
A of E of the same cardinality as E such that fA containedin A" is true if
and only if "for all multivariate functions f from E into E there exists a
proper subset A of the same cardinality as E such that fA containedin A" is
true if and only if the cardinaliity of E is a Jonsson cardinal.

NOTE: It is well known that Jonsson cardinals are of cofinality omega or
weakly inaccessible. Ramsey cardinals and measurable cardinals are Jonsson
cardinals. If V = L then there are no Jonsson cardinals. There are many
results and open problems concerning Jonsson cardinals.