MATHEMATICS

V. V. PASHENKOV

ON THE STRUCTURE OF SUBSPACES OF A HILBERT SPACE

(Presented by Academician A. I. Mal'tsev, 7 II 1961)

It is known ((^6)) that the completion by intersections (R) of a distributive structure (L) need not be a Dedekind structure. On the other hand, the completion by intersections of a Boolean algebra is a Boolean algebra. The question remained open whether the completion of a Dedekind structure with complements would necessarily be Dedekind (((^2),) Problem 57). In the present note it is shown that even for a Dedekind structure with orthocomplements the completion by intersections need not be Dedekind.

Let (H) be a Hilbert space. Denote the set of all subspaces of (H) by (L(H)), the set of all closed subspaces by (T(H)), and the set of all finite-dimensional subspaces and their orthogonal complements by (S(H)). Since a finite-dimensional subspace is closed (((^1),) p. 60) and the orthogonal complement of any set is a closed subspace (((^1),) p. 83), we have (L(H) \supseteq T(H) \supseteq S(H)). We shall assume that the sets (L(H)), (T(H)), and (S(H)) are ordered by inclusion.

Proposition 1. Let (H_1) and (A) be subspaces in (H), with (H_1) closed and (A) finite-dimensional. Then (H_1 + A) is a closed subspace (((^3),) p. 50).

Proposition 2. Every closed subspace is the orthogonal complement of its orthogonal complement (((^4),) p. 102).

We shall agree to denote by (H_1^0) the orthogonal complement of the subspace (H_1).

Proposition 3. If (H_1) is a closed subspace, then every element (x \in H) is uniquely representable in the form (x = x_1 + x_2), where (x_1 \in H_1), (x_2 \in H_1^0) (((^4),) p. 101).

If it is necessary to emphasize that the subspace (H_1) is finite-dimensional (is the orthogonal complement of a finite-dimensional subspace), then we shall place the symbol (\vee) ((\wedge)) over the letter (H_1).

Let (H_1) and (H_2) be subspaces, with (H_1) closed. On the basis of Proposition 3, every element (x_\alpha \in H_2) can be represented in the form (x_\alpha = x'\alpha + x''\alpha), where (x'\alpha \in H_1), (x''\alpha \in H_1^0). The subspace consisting of the elements (x''\alpha \in H_1^0) for all (x\alpha \in H_2) will be denoted by (H_2^{H_1}).

Lemma 1. Let the closed subspace (H_1) contain the subspace (\widehat{H}_2). Then (H_1) is the orthogonal complement of a finite-dimensional subspace.

Proof. Since (H_1 \supseteq H_2), it follows that (H_1^0 \subseteq H_2^0). But (H_2^0) is finite-dimensional by assumption; hence (H_1^0) is also finite-dimensional. It remains to use Proposition 2.

Lemma 2. The subspace (\widehat{H}_1 + H_2) is closed, whatever the subspaces (\widehat{H}_1) and (H_2) may be.

Proof. Let us verify the validity of the relation

[
\hat H_1+H_2=\hat H_1+H_2^{H_1}.
]

Let (x\in H_1+H_2), i.e., (x=h_1+h_2) ((h_1\in H_1,\ h_2\in H_2)). Take a representation of the element (h_2=h_2'+h_2'') ((h_2'\in H_1,\ h_2''\in H_1^0)). Then
[
x=h_1+h_2=(h_1+h_2')+h_2''.
]
But ((h_1+h_2')\in H_1,\ h_2''\in H_2^{H_1}), whence (x\in H_1+H_2^{H_1}). Conversely, let (y\in H_1+H_2^{H_1}), i.e., (y=h_1+h_2'') ((h_1\in H_1,\ h_2''\in H_2^{H_1})). Then for some (h_2\in H_2) we have (h_2=h_2'+h_2''), where (h_2'\in H_1), whence
[
y=h_1+h_2''=h_1+h_2-h_2'=(h_1-h_2')+h_2\in H_1+H_2.
]

Since (H_2^{H_1}\leq H_1^0) and (H_1^0) is finite-dimensional by assumption, (H_2^{H_1}) is also finite-dimensional. It remains to use Proposition 1.

If in a partially ordered (p.o.) set there exist greatest or least elements, we shall denote them by 1 and 0, respectively.

Lemma 3. Let the p.o. set (A) lie in the lattice (B), which preserves the order relation in (A). Then, if
[
x=\sup_B{x_\alpha}\quad \left(x=\inf_B{x_\alpha}\right)^*,
]
(x\in A) and ({x_\alpha}\subseteq A), then
[
x=\sup_A{x_\alpha}\quad \left(x=\inf_A{x_\alpha}\right).
]

Proof is trivial.

Corollary 1. Let the lattice with respect to the order of inclusion (U) consist of subspaces (closed subspaces) of the space (H). Since the set-theoretic intersection of subspaces (closed subspaces) is a subspace (closed subspace), the lattice product of subspaces in (U) will coincide with the set-theoretic intersection, if this intersection belongs to (U).

Definition 1. A lattice (L) with 1 and 0 will be called a lattice with orthocomplementations if there exists a mapping
[
\varphi:\ x\mapsto x'
]
of the lattice (L) into (L), such that: 1) from (x\geq y) it follows that (x'\leq y'); 2) (x+x'=1); 3) (x\cdot x'=0); 4) (x''=x) for any (x,y\in L).

Lemma 4. Let (L) be a lattice with orthocomplementations. Then from
[
x=\sup{x_\alpha}\quad \left(x=\inf{x_\alpha}\right)
]
it follows that
[
x'=\inf{x_\alpha'}\quad \left(x'=\sup{x_\alpha'}\right),
]
where (x') and (x_\alpha') are the orthocomplements of the elements (x) and (x_\alpha).

Proof. Let (x=\sup{x_\alpha}). Since (x\geq x_\alpha), we have (x'\leq x_\alpha') for all (\alpha). Hence (x') is a lower bound for ({x_\alpha'}). If (y') is another lower bound, i.e., (y'\leq x_\alpha'), then (y\geq x_\alpha''=x_\alpha) for all (\alpha), whence
[
y\geq x=\sup{x_\alpha}.
]
But then (y'\leq x'), i.e., (x') is the exact lower bound for ({x_\alpha'}).

Lemma 5. The p.o. set (T(H)) is a complete lattice with orthocomplementations.

Proof. Note that (T(H)\ni H). Since (T(H)) consists of all closed subspaces of (H), by Corollary 1, together with any set of subspaces ({H_\alpha}) in (T(H)), the element
[
\inf_T{H_\alpha}
]
also belongs to (T(H)). Using (2), p. 82, we conclude that (T(H)) is a complete lattice.

To each subspace (X\in T(H)) we put in correspondence its orthogonal complement (X^0). The fulfillment of 2) and 4) of Definition 1 follows in an obvious way from Propositions 3 and 2, respectively. The fulfillment of 1) and 3) is easily checked.

Theorem 1. The p.o. set (S(H)) is a sublattice of the lattice (L(H)^{**}).

* (\sup_B) ((\inf_B)) means that the exact upper (lower) bound is taken in the p.o. set (B).

** The p.o. set (L(H)) is a complete Dedekind lattice ((2), p. 188).

Proof. Let (H_1) and (H_2) belong to (S(H)). Denote
[
H_3=\inf^{L}(H_1,H_2)=H_1\cap H_2
]
(the set-theoretic intersection is meant); (H_4=H_1+H_2).

Consider the following cases.

1. (H_1, H_2); then (H_3) and (H_4) are finite-dimensional; hence (H_3,H_4\in S(H)).

2. (\check H_1,\check H_2); then (H_3\leq H_1), and hence (H_3) is finite-dimensional and (H_3\in S(H)). From Proposition 1 it follows that (H_4) is closed, and since (H_4\geq H_2), by Lemma 1 we have (H_4\in S(H)).

3. (\hat H_1,\hat H_2); let
   [
   H_5=H_1^0+H_2^0=H_1^0+H_2^0.
   ]
   Since (H_5\geq H_1^0) and (H_5\geq H_2^0), it follows that
   [
   H_5^0\leq (H_1^0)^0=H_1,\qquad H_5^0\leq (H_2^0)^0=H_2,
   ]
   whence
   [
   H_5^0\leq H_1\cap H_2=H_3.
   ]
   The subspace (H_5) is finite-dimensional as the sum of the finite-dimensional subspaces (H_1^0) and (H_2^0); moreover, (H_3) is closed as the intersection of closed subspaces; consequently from ((\hat H_5)\leq H_3) and Lemma 1 it follows that (H_3\in S(H)). By Lemma 2 the subspace (H_4) is closed, and since, moreover, (H_4\geq \hat H_1), in view of Lemma 1, the subspace (H_4\in S(H)).

We have shown that together with every pair of subspaces (H_1) and (H_2) from (S(H)), their structural sum and product from (L(H)) also belong to (S(H)), i.e. (S(H)) is a substructure in (L(H)).

Corollary 2. The structure (S(H)) is Dedekind.

Corollary 3. The structure (S(H)) is a substructure in (T(H)).

Proof. Let (H_1,H_2\in S(H)); then
[
H_3=\inf^{L}(H_1,H_2)
]
and
[
H_4=H_1+H_2
]
also belong to (S(H)), since (S(H)) is a substructure in (L(H)). From (T(H)\supseteq S(H)) we conclude that, together with (H_1,H_2\in S(H)), the structure (T(H)) contains also (H_3,H_4\in S(H)); whence, in view of (T(H)\subseteq L(H)) and Lemma 3, we obtain
[
H_3=\inf^{T}(H_1,H_2)
]
and (H_4=H_1+H_2), i.e. (S(H)) is a substructure in (T(H)).

Corollary 4. The structure (S(H)) is a structure with orthocomplements.

Proof. The structure (T(H)), by Lemma 5, possesses orthocomplements, and in the substructure (S(H)) of the structure (T(H)), together with every subspace (H_1), its orthocomplement in (T(H)) (the orthogonal complement of (H_1)) also belongs. It is easy to verify the fulfillment of 1)—4) of Definition 1.

Lemma 6. Let a partially ordered set (P) be contained in a complete structure (L), and let (L) preserve the order relation in (P). If every element (x\in L) is the exact lower and exact upper bound for subsets of elements of (P), then (L) is isomorphic to the completion (R) of the partially ordered set (P) by cuts.*

Proof. To each (x\in L) put in correspondence the set
[
I(x)={p\in P;\ p\leq x}:\ x\stackrel{\varphi}{\longrightarrow}I(x).
]
Since
[
x=\sup^{L} I(x),
]
we have
[
I^(x)={p\in P,\ p\geq x},
]
and then
[
x=\inf^{L} I^(x),
]
whence
[
I(x)=I^{+}(x).
]
Consequently, (I(x)\in R) for every (x\in L). Conversely, let (I\in R) and
[
x=\sup^{L} I.
]
Then
[
I^=I^(x),
]
whence
[
I=I^{+}=I^{*+}(x)=I(x).
]
Thus (\varphi) maps (L) onto the whole structure (R). From (x\geq y) it follows that (I(x)\geq I(y)), and conversely. Consequently, the mappings (\varphi) and (\varphi^{-1}) are one-to-one and isotone, i.e. (\varphi) is an isomorphism.

Lemma 7. The structure (T(H)) is isomorphic to the completion by cuts of the structure (S(H)).

* In the sense of ({}^{(2)}), p. 93; from the same source, in the proof, the notations (I^*) and (I^+) will be used.

Proof. Let ({X_\alpha}) be the set of all one-dimensional subspaces from the subspace (H_1 \in T(H)). Then

(H_1 = \overset{T}{\sup}{X_\alpha}). Next, let (H_1^0 = H'1 = \overset{T}{\sup}{Y\beta}) be a representation of the subspace (H_1^0) as a sum of one-dimensional subspaces (Y_\beta). Then, taking into account Proposition 2 and Lemmas 4 and 5, we obtain

(H_1 = (H'1)' = \overset{T}{\inf}{Y'\beta} = \overset{T}{\inf}{Y_\beta^0}). Thus, the conditions of Lemma 6 are satisfied.

Lemma 8. Let (H) be an infinite-dimensional space. Then the structure (T(H)) is not Dedekind.

Proof. Let (H_1) and (H_2) be closed subspaces whose sum (H_3) is not closed ((5), Ch. 1). Let (H_4 = H_1 \overset{T}{+} H_2). Then there exists a vector (a \in H_4 \setminus H_3). Consider the subspace (H_5 = H_1 \overset{L}{+} {a}). By Proposition 1 we obtain (H_5 \in T(H)). Moreover, it is clear that

(H_4 = H_1 \overset{T}{+} H_2 = H_5 \overset{T}{+} H_2). Next, let (H_6 = H_1 \cap H_2). We shall show that (H_6 = H_5 \cap H_2). Suppose the contrary; let the vector (b \in (H_5 \cap H_2) \setminus H_6). Then (b \in H_2 \subseteq H_3); moreover (b \in H_5), i.e., (b = h_1 + \alpha a) ((h_1 \in H_1), (\alpha) is a number), and from (b \notin H_6) it follows that (\alpha \ne 0). This, however, is impossible, since from (b \in H_3) and ((-h_1) \in H_1 \subseteq H_3) it follows that ((-h_1 + b) \in H_3), but ((-h_1) + b = (-h_1) + h_1 + \alpha a = \alpha a \notin H_3). Consequently, (H_1 \cap H_2 = H_5 \cap H_2 = H_6).

From Theorem 1, Corollaries 2 and 4, and Lemmas 7 and 8 it follows:

Theorem 2. There exist Dedekind structures with orthocomplements whose completion by sections is not a Dedekind structure.

The author expresses gratitude to Prof. L. A. Skornyakov for a number of useful suggestions.

Moscow State University
named after M. V. Lomonosov

Received
1 II 1961

References

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