NOTES ON INVARIANT SUBSPACES

William J. DeMeo

October 13, 2009

1 Definitions and Notations• X = 〈X; +,−, 0, fα〉

Let F be a field (or division ring) and let X = 〈X; +,−, 0, fα〉 be a vector space (or F-module), where, for eachα ∈ F, the unary operation symbol fα denotes scalar multiplication by α: fα(x) = αx.

Endow X with an inner product and an associated norm topology with respect to which X is complete andseparable. Thus, X is a separable (skew) Hilbert space. We will denote the norm-closure of a subset S ⊆ X byS̄.

• Sub[X] = 〈Sub[X];∧,∨〉Let Sub[X] denote the subsets of X that are closed under the algebraic operations +, − and fα (but not neces-sarily closed in the norm topology). Algebraists call the sets in Sub[X] subuniverses of X, while analysts callthem either linear manifolds or subspaces. (Often, as below, the term subspace is reserved for the norm-closedlinear manifolds.) Each S ∈ Sub[X] can be endowed with the operations of X, by restriction to S, and then Salong with these restricted operations is a subalgebra of X.

If we partially order the sets in Sub[X] by set inclusion ⊆, then the algebra 〈Sub[X];∧,∨〉, with

V1 ∧ V2 , V1 ∩ V2 and V1 ∨ V2 , V1 + V2 = {v1 + v2 : vi ∈ Vi}, (1)

is a complete, complemented, modular lattice.1

Question 1: Is 〈Sub[X];∧,∨〉 distributive?

Answer: No. Consider a 2-dimensional vector space X. For x ∈ X, let (x) denote the subspace generated byx. If x1,x2 ∈ X are two linearly independent vectors, then (xi) ∨ (xj) = X and (xi) ∧ (xj) = (0) wheneveri 6= j. Thus, any three independent vectors will generate an M3 sublattice of X.

Question 2: Is 〈Sub[X];∧,∨〉 algebraic?

Answer: Yes. 〈Sub[X];⊆〉 is an algebraic closure system, so we need only verify that the meet and joinoperations induced by the partial order ⊆ are those defined above.

• CSub[X] = 〈CSub[X];∧,∨〉Let CSub[X] = {V ∈ Sub[X] : V = V̄ }. That is, CSub[X] is the set of norm-closed subuniverses of X.Clearly CSub[X] is a partially ordered set under set inclusion. However, 〈CSub[X];∧,∨〉 is not a sublatticeof 〈Sub[X];∧,∨〉. For, there are closed subspaces V1, V2 such that the span V1 + V2 is not closed. (See, forexample, page 28 of [1]). Therefore, CSub[X] is not closed under the join defined in (1).

1See exercise 5, page 189, of Hungerford[3]. Complemented means for all V1 ∈ Sub[X] there is a V2 ∈ Sub[X] such that V1 + V2 = X andV1 ∩ V2 = 0, so that V1 ⊕ V2 = X .

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REFERENCES REFERENCES

However, if we redefine ∨ to be the norm-closure of the span,

V1 ∨ V2 , V1 + V2,

then CSub[X] = 〈CSub[X];∧,∨〉 is a lattice. In fact, CSub[X] is a modular lattice if and only if X is finitedimensional and is a distributive lattice if and only if X has dimension 0 or 1. (See, e.g., , Halmos [2].)

Question 3: Is CSub[X] complete?

Answer: Yes, since arbitrary intersections of closed sets are closed,∧α Vα =

⋂α Vα exists in CSub[X] for

any collection {Vα} ⊆ CSub[X].

Question 4: Is CSub[X] algebraic?

To answer this, we might try to check whether it is an algebra closure system as we did above to answer question2. If that fails, we could check whether it is a closure system defined by finitary closure rules (which I don’tthink is the case – check it!)

• 〈algS;∧,∨〉For S ⊆P(X), define

AlgS = {σ ∈ Aut[X] : σ(S) ⊆ S, for all S ∈ S}

andalgS = {σ ∈ Aut[X] : σ(S) = S, for all S ∈ S}.

• 〈latA;∧,∨〉For A ⊆ Aut[X], define

LatA = {S ∈ CSub[X] : σ(S) ⊆ S, for all σ ∈ A}

andlatA = {S ∈ CSub[X] : σ(S) = S, for all σ ∈ A}.

References[1] Paul R. Halmos. Introduction to Hilbert Space and the theory of spectral multiplicty. Chelsea, New York, 1951.

[2] Paul R. Halmos. A Hilbert Space Problem Book. Springer-Verlag, New York, 1984.

[3] Thomas W. Hungerford. Algebra. Springer-Verlag, New York, 1974.

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