Completion and extension of operators in Krein spaces
A generalization of the well-known results of M.G. Kreiın on the description of the self-adjoint contractive extension of a Hermitian contraction is obtained. This generalization concerns the situation where the self-adjoint operator A and extensions A belong to a Kreiın space or a Pontryagin space,...
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irk-123456789-1450842019-01-15T01:23:47Z Completion and extension of operators in Krein spaces Baidiuk, D. A generalization of the well-known results of M.G. Kreiın on the description of the self-adjoint contractive extension of a Hermitian contraction is obtained. This generalization concerns the situation where the self-adjoint operator A and extensions A belong to a Kreiın space or a Pontryagin space, and their defect operators are allowed to have a fixed number of negative eigenvalues. A result of Yu. L. Shmul’yan on completions of nonnegative block operators is generalized for block operators with a fixed number of negative eigenvalues in a Kreiın space. 2016 Article Completion and extension of operators in Krein spaces / D. Baidiuk // Український математичний вісник. — 2016. — Т. 13, № 4. — С. 452-472. — Бібліогр.: 27 назв. — англ. 1810-3200 http://dspace.nbuv.gov.ua/handle/123456789/145084 2010 MSC. 46C20, 47A20, 47A63 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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A generalization of the well-known results of M.G. Kreiın on the description of the self-adjoint contractive extension of a Hermitian contraction is obtained. This generalization concerns the situation where the self-adjoint operator A and extensions A belong to a Kreiın space or a Pontryagin space, and their defect operators are allowed to have a fixed number of negative eigenvalues. A result of Yu. L. Shmul’yan on completions of nonnegative block operators is generalized for block operators with a fixed number of negative eigenvalues in a Kreiın space. |
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Baidiuk, D. Completion and extension of operators in Krein spaces Український математичний вісник |
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Baidiuk, D. |
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Baidiuk, D. |
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Completion and extension of operators in Krein spaces |
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Completion and extension of operators in Krein spaces |
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Completion and extension of operators in Krein spaces |
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Completion and extension of operators in Krein spaces |
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Completion and extension of operators in Krein spaces |
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completion and extension of operators in krein spaces |
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Інститут прикладної математики і механіки НАН України |
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Completion and extension of operators in Krein spaces / D. Baidiuk // Український математичний вісник. — 2016. — Т. 13, № 4. — С. 452-472. — Бібліогр.: 27 назв. — англ. |
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Український математичний вісник |
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AT baidiukd completionandextensionofoperatorsinkreinspaces |
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2025-07-10T20:48:55Z |
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1837294459795013632 |
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Український математичний вiсник
Том 13 (2016), № 4, 452 – 472
Completion and extension of operators
in Krĕın spaces
Dmytro Baidiuk
(Presented by M.M. Malamud)
Abstract. A generalization of the well-known results of M.G. Krĕın
about the description of selfadjoint contractive extension of a hermi-
tian contraction is obtained. This generalization concerns the situation,
where the selfadjoint operator A and extensions à belong to a Krĕın
space or a Pontryagin space and their defect operators are allowed to have
a fixed number of negative eigenvalues. Also a result of Yu. L. Shmul’yan
on completions of nonnegative block operators is generalized for block
operators with a fixed number of negative eigenvalues in a Krĕın space.
This paper is a natural continuation of S. Hassi’s and author’s recent
paper [7].
2010 MSC. 46C20, 47A20, 47A63.
Key words and phrases. Completion, extension of operators, Krĕın
and Pontryagin spaces.
1. Introduction
Let A be a densely defined lower semi-bounded operator in a sepa-
rable Hilbert space H, A ≥ mAI. A problem of existing of selfadjoint
extensions preserving the lower bound mA of A was formulated by J. von
Neumann [4]. He solved it for the case of an operator with finite deficiency
indices. A solution to this problem for operators with arbitrary deficiency
indices was obtained by M. Stone, H. Freudental, and K. Friedrichs [4].
M. G. Krĕın in his seminal paper [19] (see also [1]) described the set
ExtA(0,∞) of all nonnegative selfadjoint extensions à of A ≥ 0 as fol-
lows
(AF + a)−1 ≤ (Ã+ a)−1 ≤ (AK + a)−1, a > 0, Ã ∈ ExtA(0,∞).
Received 27.12.2016
The author thanks his supervisor Seppo Hassi for several detailed discussions on
the results of this paper and also Mark Malamud for comments.
ISSN 1810 – 3200. c⃝ Iнститут математики НАН України
D. Baidiuk 453
Here AF and AK are the Friedrichs (hard) and Krĕın (soft) extensions of
A, respectively.
To obtain such a description he used a special form of the Cayley
transform
T1 = (I −A)(I +A)−1, T = (I − Ã)(I + Ã)−1,
to reduce the study of unbounded operators to the study of contractive
selfadjoint extensions T of a Hermitian nondensely defined contraction
T1 ∈ [H1,H], where H1 = ran (I+A). The set of all selfadjoint contractive
extensions of T1 is denoted by Ext T1(−1, 1). M.G. Krĕın proved that the
set Ext T1(−1, 1) forms an operator interval with minimal and maximal
entries Tm and TM , respectively,
Tm ≤ T ≤ TM , T ∈ Ext T1(−1, 1).
T. Ando and K. Nishio [2] extended main results of the Krĕın theory
to the case of nondensely defined symmetric operators A. For the case
of linear relations (multivalued linear operators) A ≥ 0 it was done by
E.A. Coddington and H.S.V. de Snoo [9].
With respect to the orthogonal decomposition H = H1 ⊕ H2 a con-
traction T1 ∈ [H1,H] admits a block-matrix representation T1 =
(
T11
T21
)
.
Block matrix representations of the operators Tm and TM were obtained
in [6, 18], and [16], (see also [4, 12,13,27] Namely, it is shown that
Tm =
(
T11 DT11V
∗
V DT11 −I + V (I − T11)V ∗
)
,
TM =
(
T11 DT11V
∗
V DT11 I − V (I + T11)V
∗
)
,
(1.1)
where DT11 := (I−T 2
11)
1/2 and V is given by V := clos (T21D
[−1]
T11
). Based
on these formulas a complete parametrization of the set Ext T1(−1, 1) as
well as main results of the Krĕın theory have also been obtained there.
In turn, the proof of formulas for Tm and TM was based on a result of
Yu. L. Shmul’yan [26] (see also [27]) of nonnegative completions of a
nonnegative block operator.
Recently, S. Hassi and the author [7] extended the main result of [16]
to the case of “quasi-contractive” symmetric operators T1. Recall, that
the “quasi-contractivity” means that ν−(I − T ∗T ) <∞, where
ν−(K) = dim (EK(−∞, 0)H).
For this purpose the above mentioned result of Shmul’yan was generalized
there. Also an analog of block matrix formulas for the operators Tm and
454 Completion of operators in Krĕın spaces
TM was established. Formulas Tm and TM in this case look similar to
(1.1) but the entries V (I±T11)V ∗ are replaced by V (I±T11)JV ∗, where
J = sign (I − T 2
11) and DT11 := |I − T 2
11|1/2.
The first result of the present paper is a further generalization of
Shmul’yan’s result [26] to the case of block operators acting in a Krĕın
space and having a fixed number of negative eigenvalues.
In Section 4 a first Krĕın space analog of completion problem is for-
mulated and a description of its solutions is found. Namely, we consider
classes of “quasi-contractive” symmetric operators T1 in a Krĕın space
with ν−(I−T ∗
1 T1) <∞ and describe all possible selfadjoint (in the Krĕın
space sense) extensions T of T1 which preserve the given negative index
ν−(I − T ∗T ) = ν−(I − T ∗
1 T1). This problem is close to the completion
problem studied in [7] and has a similar description for its solutions. For
related problems see also [3–5,10–16,18,20,22–25,27].
The main result of the present paper is Theorem 5.7. Namely, we
consider classes of “quasi-contractive” symmetric operators T1 in a Pon-
tryagin space (H, J) with
ν−[I − T [∗]
1 T1] := ν−(J(I − T [∗]
1 T1)) <∞ (1.2)
and we establish a solvability criterion and a description of all possi-
ble selfadjoint extensions T of T1 (in the Pontryagin space sense) which
preserve the given negative index ν−[I − T [∗]T ] = ν−[I − T [∗]
1 T1]. The
formulas for Tm and TM are also extended in an appropriate manner (see
(5.16)). It should be emphasized that in this more general setting for-
mulas (5.16) involve so-called link operator LT which was introduced by
Arsene, Constantintscu and Gheondea in [5] (see also [4, 10, 11,21]).
2. A completion problem for block operators in Krĕın
spaces
By definition the modulus |C| of a closed operator C is the nonneg-
ative selfadjoint operator |C| = (C∗C)1/2. Every closed operator admits
a polar decomposition C = U |C|, where U is a (unique) partial isometry
with the initial space ran |C| and the final space ranC, cf. [17]. For a
selfadjoint operator H =
∫
R t dEt in a Hilbert space H the partial isome-
try U can be identified with the signature operator, which can be taken
to be unitary: J = sign (H) =
∫
R sign (t) dEt, in which case one should
define sign (t) = 1 if t ≥ 0 and otherwise sign (t) = −1.
Let H be a Hilbert space, and let JH be a signature operator in it,
i.e., JH = J∗
H = J−1
H . We interpret the space H as a Krĕın space (H, JH)
D. Baidiuk 455
(see [6,8]) in which the indefinite scalar product is defined by the equality
[φ,ψ]H = (JHφ,ψ)H.
Let us introduce a partial ordering for selfadjoint Krĕın space operators.
For selfadjoint operators A and B with the same domains A ≥J B if and
only if [(A−B)f, f ] ≥ 0 for all f ∈ domA. If not otherwise indicated the
word "smallest" means the smallest operator in the sense of this partial
ordering.
Consider a bounded incomplete block operator
A0 =
(
A11 A12
A21 ∗
)(
(H1, J1)
(H2, J2)
)
→
(
(H1, J1)
(H2, J2)
)
(2.1)
in the Krĕın space H = (H1⊕H2, J), where (H1, J1) and (H2, J2) are Krĕın
spaces with fundamental symmetries J1 and J2, and J =
(
J1 0
0 J2
)
.
Theorem 2.1. Let H = (H1 ⊕ H2, J) be an orthogonal decomposition of
the Krĕın space H and let A0 be an incomplete block operator of the form
(2.1). Assume that A11 = A
[∗]
11 and A21 = A
[∗]
12 are bounded, the num-
bers of negative squares of the quadratic form [A11f, f ] (f ∈ domA11)
ν−[A11] := ν−(J1A11) = κ < ∞, where κ ∈ Z+, and let us introduce
J11 := sign (J1A11) the (unitary) signature operator of J1A11. Then:
(i) There exists a completion A ∈ [(H, J)] of A0 with some operator
A22 = A
[∗]
22 ∈ [(H2, J2)] such that ν−[A] = ν−[A11] = κ if and only
if
ranJ1A12 ⊂ ran |A11|1/2.
(ii) In this case the operator S = |A11|[−1/2]J1A12, where |A11|[−1/2]
denotes the (generalized) Moore–Penrose inverse of |A11|1/2, is well
defined and S ∈ [(H2, J2), (H1, J1)]. Moreover, S[∗]J1J11S is the
“smallest” operator in the solution set
A :=
{
A22 = A
[∗]
22 ∈ [(H2, J2)] : A = (Aij)
2
i,j=1 : ν−[A] = κ
}
and this solution set admits a description
A =
{
A22 ∈ [(H2, J2)] : A22 = J2(S
∗J11S + Y ) = S[∗]J1J11S + J2Y,
where Y = Y ∗ ≥ 0
}
.
456 Completion of operators in Krĕın spaces
Proof. Let us introduce a block operator
Ã0 =
(
Ã11 Ã12
Ã21 ∗
)
=
(
J1A11 J1A12
J2A21 ∗
)
.
The blocks of this operator satisfy the identities Ã11 = Ã∗
11, Ã
∗
21 = Ã12
and
ranJ1A11 = ran Ã11 ⊂ ran |Ã11|1/2 = ran (Ã∗
11Ã11)
1/4
= ran (A∗
11A11)
1/4 = ran |A11|1/2.
Then due to [7, Theorem 1] a description of all selfadjoint operator
completions of Ã0 admits representation à =
(
Ã11 Ã12
Ã21 Ã22
)
with Ã22 =
S̃∗J11S̃ + Y , where S̃ = |Ã11|[−1/2]Ã12 and Y = Y ∗ ≥ 0.
This yields description for the solutions of the completion problem.
The set of completions has the form A =
(
A11 A12
A21 A22
)
, where
A22 = J2Ã22 = J2A21J1|A11|[−1/2]J11|A11|[−1/2]J1A12 + J2Y
= J2S
∗J11S + J2Y = S[∗]J1J11S + J2Y.
3. Some inertia formulas
Some simple inertia formulas are now recalled. The factorization H =
B[∗]EB clearly implies that ν±[H] ≤ ν±[E], cf. (1.2). If H1 and H2 are
selfadjoint operators in a Krĕın space, then
H1 +H2 =
(
I
I
)[∗](
H1 0
0 H2
)(
I
I
)
shows that ν±[H1 + H2] ≤ ν±[H1] + ν±[H2]. Consider the selfadjoint
block operator H ∈ [(H1, J1)⊕ (H2, J2)], where Ji = J∗
i = J−1
i , (i = 1, 2)
of the form
H = H [∗] =
(
A B[∗]
B I
)
,
By applying the above mentioned inequalities shows that
ν±[A] ≤ ν±[A−B[∗]B] + ν±(J2). (3.1)
Assuming that ν−[A−B∗J2B] and ν−(J2) are finite, the question when
ν−[A] attains its maximum in (3.1), or equivalently, ν−[A − B∗J2B] ≥
D. Baidiuk 457
ν−[A] − ν−(J2) attains its minimum, turns out to be of particular in-
terest. The next result characterizes this situation as an application of
Theorem 2.1. Recall that if J1A = JA|A| is the polar decomposition of
J1A, then one can interpret HA = (ranJ1A, JA) as a Krĕın space gener-
ated on ranJ1A by the fundamental symmetry JA = sign (J1A).
Theorem 3.1. Let A ∈ [(H1, J1)] be selfadjoint, B ∈ [(H1, J1), (H2, J2)],
Ji = J∗
i = J−1
i ∈ [Hi], (i = 1, 2), and assume that ν−[A], ν−(J2) <∞. If
the equality
ν−[A] = ν−[A−B[∗]B] + ν−(J2)
holds, then ranJ1B
[∗] ⊂ ran |A|1/2 and J1B
[∗] = |A|1/2K for a unique
operator K ∈ [(H2, J2),HA] which is J-contractive: J2 −K∗JAK ≥ 0.
Conversely, if B[∗] = |A|1/2K for some J-contractive operator K ∈
[(H2, J2),HA], then the equality (3.1) is satisfied.
Proof. Assume that (3.1) is satisfied. The factorization
H =
(
A B[∗]
B I
)
=
(
I B[∗]
0 I
)(
A−B[∗]B 0
0 I
)(
I 0
B I
)
shows that ν−[H] = ν−[A − B[∗]B] + ν−(J2), which combined with the
equality (3.1) gives ν−[H] = ν−[A]. Therefore, by Theorem 2.1 one
has ranJ1B
[∗] ⊂ ran |A|1/2 and this is equivalent to the existence of
a unique operator K ∈ [(H2, J2),HA] such that J1B
[∗] = |A|1/2K; i.e.
K = |A|[−1/2]J1B
[∗]. Furthermore, K [∗]J1JAK ≤J2 I by the minimal-
ity property of K [∗]J1JAK in Theorem 2.1, in other words K is a J-
contraction.
Converse, if J1B
[∗] = |A|1/2K for some J-contractive operator K ∈
[(H2, J2),HA], then clearly ranJ1B
[∗] ⊂ ran |A|1/2. By Theorem 2.1
the completion problem for H0 has solutions with the minimal solution
S[∗]J1JAS, where
S = |A|[−1/2]J1B
[∗] = |A|[−1/2]|A|1/2K = K.
Furthermore, by J-contractivity of K one has K [∗]J1JAK ≤J2 I, i.e. I
is also a solution and thus ν−[H] = ν−[A] or, equivalently, the equality
(3.1) is satisfied.
4. A pair of completion problems in a Krĕın space
In this section we introduce and describe the solutions of a Krĕın
space version of a completion problem that was treated in [7].
458 Completion of operators in Krĕın spaces
Let (Hi, (Ji·, ·)) and (H, (J ·, ·)) be Krĕın spaces, where H = H1 ⊕
H2,J =
(
J1 0
0 J2
)
, and Ji are fundamental symmetries (i = 1, 2), let
T11 = T
[∗]
11 ∈ [(H1, J1)] be an operator such that ν−(I−T ∗
11T11) = κ <∞.
Denote T̃11 = J1T11, then T̃11 = T̃ ∗
11 in the Hilbert space H1. Rewrite
ν−(I − T ∗
11T11) = ν−(I − T̃ 2
11). Denote
J+ = sign (I − T̃11), J− = sign (I + T̃11), and J11 = sign (I − T̃ 2
11),
and let κ+ = ν−(J+) and κ− = ν−(J−). It is easy to get that J11 =
J−J+ = J+J−. Moreover, there is an equality κ = κ−+κ+ (see [7, Lemma
5.1]). We recall the results for the operator T̃11 from the paper [7] and
after that reformulate them for the operator T11. We recall completion
problem and its solutions that was investigated in a Hilbert space setting
in [7]. The problem concerns the existence and a description of selfadjoint
operators T̃ such that Ã+ = I+T̃ and Ã− = I−T̃ solve the corresponding
completion problems
Ã0
± =
(
I ± T̃11 ±T̃ ∗
21
±T̃21 ∗
)
, (4.1)
under minimal index conditions ν−(I + T̃ ) = ν−(I + T̃11), ν−(I − T̃ ) =
ν−(I − T̃11), respectively. The solution set is denoted by Ext
T̃1,κ
(−1, 1).
The next theorem gives a general solvability criterion for the comple-
tion problem (4.1) and describes all solutions to this problem.
Theorem 4.1. ([7, Theorem 5]) Let T̃1 =
(
T̃11
T̃21
)
: H1 →
(
H1
H2
)
be a
symmetric operator with T̃11 = T̃ ∗
11 ∈ [H1] and ν−(I− T̃ 2
11) = κ <∞, and
let J11 = sign (I − T̃ 2
11). Then the completion problem for Ã0
± in (4.1)
has a solution I ± T̃ for some T̃ = T̃ ∗ with ν−(I − T̃ 2) = κ if and only
if the following condition is satisfied:
ν−(I − T̃ 2
11) = ν−(I − T̃ ∗
1 T̃1). (4.2)
If this condition is satisfied then the following facts hold:
(i) The completion problems for Ã0
± in (4.1) have minimal solutions
ñ.
(ii) The operators T̃m := Ã+ − I and T̃M := I − Ã− ∈ Ext
T̃1,κ
(−1, 1).
D. Baidiuk 459
(iii) The operators T̃m and T̃M have the block form
T̃m =
(
T̃11 D
T̃11
V ∗
V D
T̃11
−I + V (I − T̃11)J11V ∗
)
,
T̃M =
(
T̃11 D
T̃11
V ∗
V D
T̃11
I − V (I + T̃11)J11V
∗
)
,
(4.3)
where D
T̃11
:= |I − T̃ 2
11|1/2 and V is given by V := clos (T̃21D
[−1]
T̃11
).
(iv) The operators T̃m and T̃M are extremal extensions of T̃1:
T̃ ∈ Ext
T̃1,κ
(−1, 1) iff T̃ = T̃ ∗ ∈ [H], T̃m ≤ T̃ ≤ T̃M .
(v) The operators T̃m and T̃M are connected via
(−T̃ )m = −T̃M , (−T̃ )M = −T̃m.
For what follows it is convenient to reformulate the above theorem in
a Krĕın space setting. Consider the Krĕın space (H, J) and a selfadjoint
operator T in this space. Now the problem concerns selfadjoint operators
A+ = I + T and A− = I − T in the Krĕın space (H, J) that solve the
completion problems
A0
± =
(
I ± T11 ±T [∗]
21
±T21 ∗
)
, (4.4)
under minimal index conditions ν−(I +JT ) = ν−(I +J1T11) and ν−(I −
JT ) = ν−(I−J1T11), respectively. The set of solutions T to the problem
(4.4) will be denoted by Ext J2T1,κ(−1, 1).
Denote
T1 =
(
T11
T21
)
: (H1, J1)→
(
(H1, J1)
(H2, J2)
)
, (4.5)
so that T1 is symmetric (nondensely defined) operator in the Krĕın space
[(H1, J1)], i.e. T11 = T
[∗]
11 .
Theorem 4.2. Let T1 be a symmetric operator in a Krĕın space sense
as in (4.5) with T11 = T
[∗]
11 ∈ [(H1, J1)] and ν−(I − T ∗
11T11) = κ < ∞,
and let J = sign (I − T ∗
11T11). Then the completion problems for A0
± in
(4.4) have a solution I ± T for some T = T [∗] with ν−(I − T ∗T ) = κ if
and only if the following condition is satisfied:
ν−(I − T ∗
11T11) = ν−(I − T ∗
1 T1). (4.6)
If this condition is satisfied then the following facts hold:
460 Completion of operators in Krĕın spaces
(i) The completion problems for A0
± in (4.4) have “minimal” (J2-mi-
nimal) solutions A±.
(ii) The operators Tm := A+−J and TM := J−A− ∈ Ext J2T1,κ(−1, 1).
(iii) The operators Tm and TM have the block form
Tm =
(
T11 J1DT11V
∗
J2V DT11 −J2 + J2V (I − J1T11)J11V ∗
)
,
TM =
(
T11 J1DT11V
∗
J2V DT11 J2 − J2V (I + J1T11)J11V
∗
)
,
(4.7)
where DT11 := |I−T ∗
11T11|1/2 and V is given by V := clos (J2T21D
[−1]
T11
).
(iv) The operators Tm and TM are J2-extremal extensions of T1:
T ∈ Ext J2T1,κ(−1, 1) iff T = T [∗] ∈ [(H, J)], Tm ≤J2 T ≤J2 TM .
(v) The operators Tm and TM are connected via
(−T )m = −TM , (−T )M = −Tm.
Proof. The proof is obtained by systematic use of the equivalence that T
is a selfadjoint operator in a Krĕın space if and only if T̃ is a selfadjoint
in a Hilbert space. In particular, T gives solutions to the completion
problems (4.4) if and only if T̃ solves the completion problems (4.4). In
view of
I − T ∗
11T11 = I − T ∗
11JJT11 = I − T̃ 2
11,
we are getting formula (4.6) from (4.2). Then formula (4.7) follows by
multiplying the operators in (4.3) by the fundamental symmetry.
5. Completion problem in a Pontryagin space
5.1. Defect operators and link operators
Let (H, (·, ·)) be a Hilbert space and let J be a symmetry in H, i.e.
J = J∗ = J−1, so that (H, (J ·, ·)), becomes a Pontryagin space. Then
associate with T ∈ [H] the corresponding defect and signature operators
DT = |J − T ∗JT |1/2, JT = sign (J − T ∗JT ), DT = ranDT ,
where the so-called defect subspace DT can be considered as a Pontryagin
space with the fundamental symmetry JT . Similar notations are used
with T ∗:
DT ∗ = |J − TJT ∗|1/2, JT ∗ = sign (J − TJT ∗), DT ∗ = ranDT ∗ .
D. Baidiuk 461
By definition JTD
2
T = J − T ∗JT and JTDT = DTJT with analogous
identities for DT ∗ and JT ∗ . In addition,
(J − T ∗JT )JT ∗ = T ∗J(J − TJT ∗), (J − TJT ∗)JT = TJ(J − T ∗JT ).
Recall that T ∈ [H] is said to be a J-contraction if J −T ∗JT ≥ 0, i.e.
ν−(J − T ∗JT ) = 0. If, in addition, T ∗ is a J-contraction, T is termed as
a J-bicontraction.
For the following consideration an indefinite version of the commuta-
tion relation of the form TDT = DT ∗T is needed; these involve so-called
link operators introduced in [5, Section 4] (see also [7]).
Definition 5.1. There exist unique operators LT ∈ [DT ,DT ∗ ] and LT ∗ ∈
[DT ∗ ,DT ] such that
DT ∗LT = TJDT �DT , DTLT ∗ = T ∗JDT ∗�DT ∗ ; (5.1)
in fact, LT = D
[−1]
T ∗ TJDT �DT and LT ∗ = D
[−1]
T T ∗JDT ∗�DT ∗.
The following identities can be obtained with direct calculations; see
[5, Section 4]:
L∗
TJT ∗�DT ∗ = JTLT ∗ ;
(JT −DTJDT )�DT = L∗
TJT ∗LT ;
(JT ∗ −DT ∗JDT ∗)�DT ∗ = L∗
T ∗JTLT ∗ .
(5.2)
Now let T be selfadjoint in Pontryagin space (H, J), i.e. T ∗ = JTJ .
Then connections between DT ∗ and DT , JT ∗ and JT , LT ∗ and LT can be
established.
Lemma 5.1. Assume that T ∗ = JTJ . Then DT = |I − T 2|1/2 and the
following equalities hold:
DT ∗ = JDTJ, (5.3)
in particular,
DT ∗ = JDT and DT = JDT ∗ ;
JT ∗ = JJTJ ; (5.4)
LT ∗ = JLTJ. (5.5)
Proof. The defect operator of T can be calculated by the formula
DT =
((
I − (T ∗)2
)
JJ(I − T 2)
)1/4
=
((
I − (T ∗)2
)
(I − T 2)
)1/4
.
462 Completion of operators in Krĕın spaces
Then
DT ∗ =
(
J
(
I − (T ∗)2
)
(I − T 2)J
)1/4
= J
((
I − (T ∗)2
)
(I − T 2)
)1/4
J
= JDTJ
i.e. (5.3) holds. This implies
JDT ∗ ⊂ DT and JDT ⊂ DT ∗ .
Hence from the last two formulas we get
DT ∗ = J(JDT ∗) ⊂ JDT ⊂ DT ∗
and similarly
DT = J(JDT ) ⊂ JDT ∗ ⊂ DT .
The formula
JTD
2
T = J − T ∗JT = J(J − TJT ∗)J = JJT ∗D2
T ∗J = JJT ∗JD2
TJJ
= JJT ∗JD2
T
yields the equation (5.4).
The relation (5.5) follows from
DTLT ∗ = T ∗JDT ∗�DT ∗ = JTJDTJ�DT ∗ = JDT ∗LTJ = DTJLTJ.
5.2. Lemmas on negative indices of certain block operators
The first two lemmas are of preparatory nature for the last two lem-
mas, which are used for the proof of the main theorem.
Lemma 5.2. Let
(
J T
T J
)
:
(
H
H
)
→
(
H
H
)
be a selfadjoint operator in
the Hilbert space H2 = H⊕ H. Then∣∣∣∣(J T
T J
)∣∣∣∣1/2 = U
(
|J + T |1/2 0
0 |J − T |1/2
)
U∗,
where U = 1√
2
(
I I
I −I
)
is a unitary operator.
D. Baidiuk 463
Proof. It is easy to check that(
J T
T J
)
= U
(
J + T 0
0 J − T
)
U∗. (5.6)
Then by taking the modulus one gets∣∣∣∣(J T
T J
)∣∣∣∣2 = ((J T
T J
)∗(
J T
T J
))
= U
(
|J + T |2 0
0 |J − T |2
)
U∗.
The last step is to extract the square roots (twice) from the both sides
of the equation:∣∣∣∣(J T
T J
)∣∣∣∣1/2 = U
(
|J + T |1/2 0
0 |J − T |1/2
)
U∗.
The right hand side can be written in this form because U is unitary.
Lemma 5.3. Let T = T ∗ ∈ H be a selfadjoint operator in a Hilbert
space H and let J = J∗ = J−1 be a fundamental symmetry in H with
ν−(J) <∞. Then
ν−(J − TJT ) + ν−(J) = ν−(J − T ) + ν−(J + T ). (5.7)
In particular, ν−(J − TJT ) <∞ if and only if ν−(J ± T ) <∞.
Proof. Consider block operators
(
J T
T J
)
and
(
J + T 0
0 J − T
)
. Equal-
ity (5.6) yields ν−
(
J T
T J
)
= ν−
(
J + T 0
0 J − T
)
. The negative index
of
(
J + T 0
0 J − T
)
equals ν−(J−T )+ν−(J+T ) and the negative index
of
(
J T
T J
)
is easy to find by using the equality
(
J T
T J
)
=
(
I 0
TJ I
)(
J 0
0 J − TJT
)(
I JT
0 I
)
. (5.8)
Then one gets (5.7).
Let (Hi, (Ji·, ·)) (i = 1, 2) and (H, (J ·, ·)) be Pontryagin spaces, where
H = H1 ⊕ H2 and J =
(
J1 0
0 J2
)
. Consider an operator T11 = T
[∗]
11 ∈
464 Completion of operators in Krĕın spaces
[(H1, J1)] such that ν−[I − T 2
11] = κ <∞; see (1.2). Denote T̃11 = J1T11,
then T̃11 = T̃ ∗
11 in the Hilbert space H1. Rewrite
ν−[I − T 2
11] = ν−(J1(I − T 2
11)) = ν−(J1 − T̃11J1T̃11)
= ν−((J1 − T̃11)J1(J1 + T̃11)).
Furthermore, denote
J+ = sign (J1(I − T11)) = sign (J1 − T̃11),
J− = sign (J1(I + T11)) = sign (J1 + T̃11),
J11 = sign (J1(I − T 2
11))
(5.9)
and let κ+ = ν−[I − T11] and κ− = ν−[I + T11]. Notice that |I ∓ T11| =
|J1 ∓ T̃11| and one has polar decompositions
I ∓ T11 = J1J±|I ∓ T11|. (5.10)
Lemma 5.4. Let T11 = T
[∗]
11 ∈ [(H1, J1)] and T =
(
T11 T12
T21 T22
)
∈ [(H, J)]
be a selfadjoint extension of the operator T11 with ν−[I ± T11] < ∞ and
ν−(J) <∞. Then the following statements
(i) ν−[I ± T11] = ν−[I ± T ];
(ii) ν−[I − T 2] = ν−[I − T 2
11]− ν−(J2);
(iii) ranJ1T
[∗]
21 ⊂ ran |I ± T11|1/2
are connected by the implications (i)⇔ (ii)⇒ (iii).
Proof. The Lemma can be formulated in an equivalent way for the Hil-
bert space operators: the block operator T̃ = JT =
(
T̃11 T̃12
T̃21 T̃22
)
is a
selfadjoint extension of T̃11 = T̃ ∗
11 ∈ [H1]. Then the following statements
(i’) ν−(J1 ± T̃11) = ν−(J ± T̃ )
(ii’) ν−(J − T̃ JT̃ ) = ν−(J1 − T̃11J1T̃11)− ν−(J2);
(iii’) ran T̃12 ⊂ ran |J1 ± T̃11|1/2
are connected by the implications (i′)⇔ (ii′)⇒ (iii′).
Hence it’s sufficient to prove this form of the Lemma.
D. Baidiuk 465
Let us prove the equivalence (i′)⇔ (ii′). Condition (ii’) is equivalent
to
ν−
(
J1 T̃11
T̃11 J1
)
= ν−
(
J T̃
T̃ J
)
. (5.11)
Indeed, in view of (5.8)
ν−
(
J1 T̃11
T̃11 J1
)
= ν−(J1) + ν−(J1 − T̃11J1T̃11)
and
ν−
(
J T̃
T̃ J
)
= ν−(J) + ν−(J − T̃ JT̃ )
= ν−(J1) + ν−(J2) + ν−(J − T̃ JT̃ ).
By using Lemma 5.3, equality (5.11) is equivalent to
ν−(J1 − T̃11) + ν−(J1 + T̃11) = ν−(J − T̃ ) + ν−(J + T̃ ). (5.12)
Hence, (i′)⇒ (ii′).
Because ν−(J1±T̃11) ≤ ν−(J±T̃ ), then (5.12) shows that (ii′)⇒ (i′).
Now we prove implication (ii′) ⇒ (iii′);the arguments here will be
useful also for the proof of Lemma 5.5 below. Use a permutation to
transform the matrix in the right hand side of (5.11):
ν−
(
J T̃
T̃ J
)
= ν−
J1 0 T̃11 T̃12
0 J2 T̃21 T̃22
T̃11 T̃12 J1 0
T̃21 T̃22 0 J2
= ν−
J1 T̃11 0 T̃12
T̃11 J1 T̃12 0
0 T̃21 J2 T̃22
T̃21 0 T̃22 J2
.
Then condition (5.11) implies to the condition
ran
(
0 T̃12
T̃12 0
)
⊂ ran
∣∣∣∣∣
(
J1 T̃11
T̃11 J1
)∣∣∣∣∣
1/2
;
(see Theorem 2.1). By Lemma 5.2 the last inclusion can be rewritten as
ran
(
0 T̃12
T̃12 0
)
⊂ ranU
(
|J1 + T̃11|1/2 0
0 |J1 − T̃11|1/2
)
U∗,
466 Completion of operators in Krĕın spaces
where U = 1√
2
(
I I
I −I
)
is a unitary operator. This inclusion is equiva-
lent to
ranU∗
(
0 T̃12
T̃12 0
)
U = ran
(
T̃12 0
0 −T̃12
)
⊂ ran
(
|J1 + T̃11|1/2 0
0 |J1 − T̃11|1/2
)
and clearly this is equivalent to condition (iii’).
Note that if T̃11 has a selfadjoint extension T̃ satisfying (i’). Then by
applying Theorem 2.1 (or [7, Theorem 1]) it yields (iii’).
Lemma 5.5. Let T11 = T
[∗]
11 ∈ [(H1, J1)] be an operator and let
T1 =
(
T11
T21
)
: (H1, J1)→
(
(H1, J1)
(H2, J2)
)
be an extension of T11 with ν−[I−T 2
11] <∞, ν−(J1) <∞, and ν−(J2) <
∞. Then for the conditions
(i) ν−[I1 − T 2
11] = ν−[I1 − T [∗]
1 T1] + ν−(J2);
(ii) ranJ1T
[∗]
21 ⊂ ran |I − T 2
11|1/2;
(iii) ranJ1T
[∗]
21 ⊂ ran |I ± T11|1/2
the implications (i)⇒ (ii) and (i)⇒ (iii) hold.
Proof. First we prove that (i)⇒(ii). In fact, this follows from Theorem
3.1 by taking A = I − T 2
11 and B = T21.
A proof of (i)⇒(iii) is quite similar to the proof used in Lemma 5.4.
Statement (i) is equivalent the following equation:
ν−
(
J1 T̃11
T̃11 J1
)
= ν−
(
J T̃1
T̃ ∗
1 J1
)
.
Indeed,
ν−
(
J1 T̃11
T̃11 J1
)
= ν−
(
J1 0
0 J1 − T̃11J1T̃11
)
= ν−(J1 − T̃11J1T̃11) + ν−(J1) <∞
D. Baidiuk 467
and
ν−
(
J T̃1
T̃ ∗
1 J1
)
= ν−
(
J 0
0 J1 − T̃ ∗
1 JT̃1
)
= ν−(J1 − T̃11J1T̃11 − T̃ ∗
21J2T̃21) + ν−(J1) + ν−(J2).
Due to (i) the right hand sides coincide and then the left hand sides
coincide as well.
Now let us permutate the matrix in the latter equation.
ν−
(
J T̃1
T̃ ∗
1 J1
)
= ν−
J1 0 T̃11
0 J2 T̃21
T̃11 T̃ ∗
21 J1
= ν−
J1 T̃11 0
T̃11 J1 T̃ ∗
21
0 T̃21 J2
.
It follows from [7, Theorem 1] that the condition (i) implies the condition
ran
(
0
T̃ ∗
21
)
⊂ ran
∣∣∣∣∣
(
J1 T̃11
T̃11 J1
)∣∣∣∣∣
1/2
= ranU
(
|J1 + T̃11|1/2 0
0 |J1 − T̃11|1/2
)
U∗,
where U = 1√
2
(
I I
I −I
)
is a unitary operator (see Lemma 5.2). Then,
equivalently,
ran T̃ ∗
21 ⊂ ran |J1 ± T̃11|1/2.
5.3. Contractive extensions of contractions with minimal neg-
ative indices
Following to [7, 16, 18] we consider the problem of existence and a
description of selfadjoint operators T in the Pontryagin space
(
(H1, J1)
(H2, J2)
)
such that A+ = I+T and A− = I−T solve the corresponding completion
problems
A0
± =
(
I ± T11 ±T [∗]
21
±T21 ∗
)
, (5.13)
under minimal index conditions ν−[I + T ] = ν−[I + T11], ν−[I − T ] =
ν−[I − T11], respectively. Observe, that by Lemma 5.4 the two minimal
index conditions above are equivalent to single condition ν−[I − T 2] =
ν−[I − T 2
11]− ν−(J2).
468 Completion of operators in Krĕın spaces
It is clear from Theorem 2.1 that the conditions ranJ1T
[∗]
21 ⊂ ran |I −
T11|1/2 and ranJ1T
[∗]
21 ⊂ ran |I+T11|1/2 are necessary for the existence of
solutions; however as noted already in [7] they are not sufficient even in
the Hilbert space setting.
The next theorem gives a general solvability criterion for the com-
pletion problem (5.13) and describes all solutions to this problem. As
in the definite case, there are minimal solutions A+ and A− which are
connected to two extreme selfadjoint extensions T of
T1 =
(
T11
T21
)
: (H1, J1)→
(
(H1, J1)
(H2, J2)
)
, (5.14)
now with finite negative index ν−[I − T 2] = ν−[I − T 2
11] − ν−(J2) >
0. The set of solutions T to the problem (5.13) will be denoted by
Ext T1,κ(−1, 1)J2 .
Theorem 5.1. Let T1 be a symmetric operator as in (5.14) with T11 =
T
[∗]
11 ∈ [(H1, J1)] and ν−[I − T 2
11] = κ < ∞, and let JT11 = sign (J1(I −
T 2
11)). Then the completion problem for A0
± in (5.13) has a solution I±T
for some T = T [∗] with ν−[I−T 2] = κ−ν−(J2) if and only if the following
condition is satisfied:
ν−[I − T 2
11] = ν−[I − T [∗]
1 T1] + ν−(J2). (5.15)
If this condition is satisfied then the following facts hold:
(i) The completion problems for A0
± in (5.13) have “minimal” solu-
tions A± (for the partial ordering introduced in the first section).
(ii) The operators Tm := A+−I and TM := I−A− ∈ Ext T1,κ(−1, 1)J2.
(iii) The operators Tm and TM have the block form
Tm =
(
T11 J1DT11V
∗
J2V DT11 −I + J2V (I − L∗
TJ1)J11V
∗
)
,
TM =
(
T11 J1DT11V
∗
J2V DT11 I − J2V (I + L∗
TJ1)J11V
∗
)
,
(5.16)
where DT11 := |I−T 2
11|1/2 and V is given by V := clos (J2T21D
[−1]
T11
).
(iv) The operators Tm and TM are “extremal” extensions of T1:
T ∈ Ext T1,κ(−1, 1)J2 iff T = T [∗] ∈ [(H, J)], Tm ≤J2 T ≤J2 TM .
(5.17)
D. Baidiuk 469
(v) The operators Tm and TM are connected via
(−T )m = −TM , (−T )M = −Tm. (5.18)
Proof. It is easy to see by (3.1) that κ = ν−[I − T 2
11] ≤ ν−[I − T [∗]
1 T1] +
ν−(J2) ≤ ν−[I−T 2]+ν−(J2). Hence the condition ν−[I−T 2] = κ−ν−(J2)
implies (5.15). The sufficiency of this condition is obtained when proving
the assertions (i)–(iii) below.
(i) If the condition (5.15) is satisfied then by using Lemma 5.5 one
gets the inclusions ran J1T
[∗]
21 ⊂ ran |I ± T11|1/2, which by Theorem 2.1
means that each of the completion problems, A0
± in (5.13), is solvable.
It follows that the operators
S− = |I + T11|[−1/2]J1T
[∗]
21 , S+ = |I − T11|[−1/2]J1T
[∗]
21 (5.19)
are well defined and they provide the minimal solutions A± to the com-
pletion problems for A0
± in (5.13).
(ii) & (iii) By Lemma 5.5 the inclusion ranJ1T
[∗]
21 ⊂ ran |I − T 2
11|1/2
holds. This inclusion is equivalent to the existence of a (unique) bounded
operator V ∗ = D
[−1]
T11
J1T
[∗]
21 with ker V ⊃ ker DT11 , such that J1T
[∗]
21 =
DT11V
∗. The operators Tm := A+ − I and TM := I − A− (see proof
of (i)) by using (5.1), (5.2), and 5.1 can be now rewritten as in (5.16).
Indeed, observe that (see Theorem 2.1, (5.9), and (5.10))
J2S
∗
−J−S− = J2V DT11 |I + T11|[−1/2]J−|I + T11|[−1/2]DT11V
∗
= J2V DT11(J1(I + T11))
[−1]DT11V
∗
= J2V DT11D
[−1]
T11
(I + L∗
T11
J1)
[−1]DT11J1DT11V
∗
= J2V (I + L∗
T11
J1)
[−1](J11 − L∗
T11
JT ∗
11
LT11)V
∗
= J2V (I + L∗
T11
J1)
[−1](J11 − (L∗
T11
J1)
2J11)V
∗
= J2V (I + L∗
T11
J1)
[−1](I + L∗
T11
J1)(I − L∗
T11
J1)J11V
∗
= J2V (I − L∗
T11
J1)J11V
∗,
where the third equality follows from (5.1) and the fourth from (5.2).
470 Completion of operators in Krĕın spaces
And similarly for
J2S
∗
+J+S+ = J2V DT11 |I − T11|[−1/2]J+|I − T11|[−1/2]DT11V
∗
= J2V DT11(J1(I − T11))[−1]DT11V
∗
= J2V DT11D
[−1]
T11
(I − L∗
T11
J1)
[−1]DT11J1DT11V
∗
= J2V (I − L∗
T11
J1)
[−1](J11 − L∗
T11
JT ∗
11
LT11)V
∗
= J2V (I − L∗
T11
J1)
[−1](J11 − (L∗
T11
J1)
2J11)V
∗
= J2V (I − L∗
T11
J1)
[−1](I − L∗
T11
J1)(I + L∗
T11
J1)J11V
∗
= J2V (I + L∗
T11
J1)J11V
∗,
which implies the representations for Tm and TM in (5.16). Clearly, Tm
and TM are selfadjoint extensions of T1, which satisfy the equalities
ν−[I + Tm] = κ−, ν−[I − TM ] = κ+.
Moreover, it follows from (5.16) that
TM − Tm =
(
0 0
0 2(I − J2V J11V ∗)
)
. (5.20)
Now the assumption (5.15) will be used again. Since ν−[I−T [∗]
1 T1] =
ν−[I−T 2
11]−ν−(J2) and T21 = J2V DT11 it follows from Theorem 3.1 that
V ∗ ∈ [H2,DT11 ] is J-contractive: J2 − V J11V ∗ ≥ 0. Therefore, (5.20)
shows that TM ≥J2 Tm and I+TM ≥J2 I+Tm and hence, in addition to
I + Tm, also I + TM is a solution to the problem A0
+ and, in particular,
ν−[I + TM ] = κ− = ν−[I + Tm]. Similarly, I − TM ≤J2 I − Tm which
implies that I − Tm is also a solution to the problem A0
−, in particular,
ν−[I − Tm] = κ+ = ν−[I − TM ]. Now by applying Lemma 5.4 we get
ν−[I − T 2
m] = κ− ν−(J2),
ν−[I − T 2
M ] = κ− ν−(J2).
Therefore, Tm, TM ∈ Ext T1,κ(−1, 1)J2 which in particular proves that
the condition (5.15) is sufficient for solvability of the completion problem
(5.13).
(iv) Observe, that T ∈ Ext T1,κ(−1, 1)J2 if and only if T = T [∗] ⊃ T1
and ν−[I ± T ] = κ∓. By Theorem 2.1 this is equivalent to
J2S
∗
−J−S− − I ≤J2 T22 ≤J2 I − J2S∗
+J+S+. (5.21)
The inequalities (5.21) are equivalent to (5.17).
(v) The relations (5.18) follow from (5.19) and (5.16).
D. Baidiuk 471
Remark 5.1. In case of a contraction operator T1 this result coincides
with the main result of [16] and in case of a “quasi-contraction” operator
T1 with finite negative index it coincides with the result of [7, Theorem 5].
References
[1] N. I. Akhiezer, I. M. Glazman,Theory of Linear Operators in Hilbert Space, Nauka,
Moscow, 1966.
[2] T. Ando, K. Nishio, Positive self-adjoint Extensions of Positive Symmetric Op-
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Contact information
Dmytro Baidiuk Department of Mathematics and Statistics
University of Vaasa
Vaasa, Finland
E-Mail: dbaidiuk@uwasa.fi
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