PROCEEDINGS OF THE UNIVERSITY OF VAASA
WORKING PAPERS 10
MATHEMATICS 6
Completion and extension of 
operators in Krein spaces
D. BAIDIUK
 II  
 
 
Publisher  Date of publication  
Vaasan yliopisto  February 2016 
Author Type of publication  
D. Baidiuk Working papers 
Name and number of series  
Proceedings of the University of Vaasa. 
Working Papers, 10 
Proceedings of the University of Vaasa. 
Working papers, Mathematics, 6 
Contact information ISBN  
University of Vaasa 
Faculty of Technology 
Department of Mathematics and 
Statistics 
P.O. Box 700 
FI-65101 Vaasa 
Finland 
978-952-476-668-5 (online) 
ISSN  
1799-7658 (Proceedings of the Universi-
ty of Vaasa. Working Papers 10, online) 
Number of pages Language 
23 English 
Title of publication  
Completion and extension of operators in Kreǐn spaces 
Abstract 
   A generalization of the well-known results of M.G. Krein about the description 
of selfadjoint contractive extension of a hermitian contraction is obtained. This 
generalization concerns the situation, where the selfadjoint operator $A$ and 
extensions $\widetilde A$ belong to a Krein space or a Pontryagin space  and 
their defect operators are allowed to have a fixed number of negative eigenval-
ues.   
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 Krein space. 
   This paper is a natural continuation of S. Hassi's and author's paper [5]. 
Keywords 
Completion, extension of operators, Krein and Pontryagin spaces 
MSC 2010 
Primary 46C20, 47A20, 47A63; Secondary 47B25 
 III
ii D. Baidiuk
Contents
1. Introduction 1
2. A completion problem for block operators in Kre˘ın spaces 3
3. Some inertia formulas 5
4. A pair of completion problems in a Kre˘ın space 7
5. Completion problem in a Pontryagin space 10
5.1. Defect operators and link operators 10
5.2. Lemmas on negative indices of certain block operators 11
5.3. Contractive extensions of contractions with minimal negative
indices 15
References 19
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Completion and extension of operators in Kre˘ın spaces. 1
1. Introduction
In 1947 M.G. Kre˘ın published one of his famous papers [17] on a description of
a nonnegative selfadjoint extensions of a densely defined nonnegative operator
A in a Hilbert space. Namely, all nonnegative selfadjoint extensions A˜ of A
can be characterized by the following two inequalities:
(AF + a)
−1 ≤ (A˜+ a)−1 ≤ (AK + a)−1, a > 0,
where the Friedrichs (hard) extension AF and the Kre˘ın-von Neumann (soft)
extension AK of A. He proved these results by transforming the problems the
study of contractive operators.
The first result of the present paper is a generalization of a result due
to Shmul’yan [19] on completions of nonnegative block operators where the
result was applied for introducing so-called Hellinger operator integrals. This
result was extended in [5] for block operators in a Hilbert space by allowing
a fixed number of negative eigenvalues. In Section 2 this result is further
extended to block operators which act in a Kre˘ın space.
In paper [5] we studied classes of “quasi-contractive” symmetric opera-
tors T1 allowing a finite number of negative eigenvalues for the associated de-
fect operator I−T ∗1 T1, i.e., ν−(I−T ∗1 T1) <∞ as well as “quasi-nonnegative”
operators A with ν−(A) <∞ and the existence and description of all possi-
ble selfadjoint extensions T and A˜ of them which preserve the given negative
indices ν−(I − T 2) = ν−(I − T ∗1 T1) and ν−(A˜) = ν−(A), and proved precise
analogs of the above mentioned results of M.G. Kre˘ın under a minimality
condition on the negative indices ν−(I − T ∗1 T1) and ν−(A), respectively. It
was an unexpected fact that when there is a solution then the solution set
still contains a minimal solution and a maximal solution which then describe
the whole solution set via two operator inequalities, just as in the origi-
nal paper of M.G. Kre˘ın. In this paper analogous results are established for
”quasi-contractive” operators acting in a Kre˘ın space; see Theorems 4.2, 5.7.
In Section 4 a first Kre˘ın 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 Kre˘ın space with
ν−(I − T ∗1 T1) < ∞ and we describe all possible selfadjoint (in the Kre˘ın
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 prob-
lem studied in [5] and has a similar description for its solutions. For further
history behind this problem see also [1, 2, 3, 7, 8, 9, 10, 11, 12, 14, 15, 16, 20].
The main result of the present paper is proved in Section 5. Namely,
we consider classes of ”quasi-contractive” symmetric operators T1 in a Kre˘ın
space (H, J) with
ν−[I − T [∗]1 T1] := ν−(J(I − T [∗]1 T1)) <∞ (1.1)
and we establish a solvability criterion and a description of all possible selfad-
joint extensions T of T1 (in the Kre˘ın space sense) which preserve the given
negative index ν−[I − T [∗]T ] = ν−[I − T [∗]1 T1]. It should be pointed out that
Proceedings of the University of Vaasa. Working Papers      2
2 D. Baidiuk
in this more general setting the descriptions involve so-called link operator
LT which was introduced by Arsene, Constantintscu and Gheondea in [3]
(see also [2, 7, 8, 18]).
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Completion and extension of operators in Kre˘ın spaces. 3
2. A completion problem for block operators in Kre˘ın spaces
By definition the modulus |C| of a closed operator C is the nonnegative
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. [13]. For a selfadjoint op-
erator H =
∫
R t dEt in a Hilbert space H the partial isometry 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 Kre˘ın space (H, JH) (see
[4, 6]) 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 Kre˘ın 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 Kre˘ın space H = (H1 ⊕ H2, J), where (H1, J1) and (H2, J2) are Kre˘ın
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
Kre˘ın 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 numbers of neg-
ative 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
ran J1A12 ⊂ 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)2i,j=1 : ν−[A] = κ
}
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4 D. Baidiuk
and this solution set admits a description
A =
{
A22 ∈ [(H2, J2)] : A22 = J2(S∗J11S + Y ) = S[∗]J1J11S + J2Y,
where Y = Y ∗ ≥ 0
}
.
Proof. Let us introduce a block operator
A˜0 =
(
A˜11 A˜12
A˜21 ∗
)
=
(
J1A11 J1A12
J2A21 ∗
)
.
The blocks of this operator satisfy the identities A˜11 = A˜
∗
11, A˜
∗
21 = A˜12 and
ran J1A11 = ran A˜11 ⊂ ran |A˜11|1/2 = ran (A˜∗11A˜11)1/4
= ran (A∗11A11)
1/4 = ran |A11|1/2.
Then due to [5, Theorem 1] a description of all selfadjoint operator
completions of A˜0 admits representation A˜ =
(
A˜11 A˜12
A˜21 A˜22
)
with A˜22 =
S˜∗J11S˜ + Y , where S˜ = |A˜11|[−1/2]A˜12 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 = J2A˜22 = J2A21J1|A11|[−1/2]J11|A11|[−1/2]J1A12 + J2Y
= J2S
∗J11S + J2Y = S[∗]J1J11S + J2Y. 
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Completion and extension of operators in Kre˘ın spaces. 5
3. Some inertia formulas
Some simple inertia formulas are now recalled. The factorizationH = B[∗]EB
clearly implies that ν±[H] ≤ ν±[E], cf. (1.1). If H1 and H2 are selfadjoint
operators in a Kre˘ın 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−1i , (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] ≥ ν−[A] −
ν−(J2) attains its minimum, turns out to be of particular interest. 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 in-
terpret HA = (ran J1A, JA) as a Kre˘ın space generated on ran J1A 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 ran J1B
[∗] ⊂ ran |A|1/2 and J1B[∗] = |A|1/2K for a unique oper-
ator 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
ran J1B
[∗] ⊂ ran |A|1/2 and this is equivalent to the existence of a unique oper-
ator K ∈ [(H2, J2),HA] such that J1B[∗] = |A|1/2K; i.e. K = |A|[−1/2]J1B[∗].
Furthermore, K [∗]J1JAK ≤J2 I by the minimality 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 ran J1B
[∗] ⊂ ran |A|1/2. By Theorem 2.1 the com-
pletion problem for H0 has solutions with the minimal solution S[∗]J1JAS,
Proceedings of the University of Vaasa. Working Papers      6
6 D. Baidiuk
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. 
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Completion and extension of operators in Kre˘ın spaces. 7
4. A pair of completion problems in a Kre˘ın space
In this section we introduce and describe the solutions of a Kre˘ın space version
of a completion problem that was treated in [5].
Let (Hi, (Ji·, ·)) and (H, (J ·, ·)) be Kre˘ın 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 T˜11 =
J1T11, then T˜11 = T˜
∗
11 in the Hilbert space H1. Rewrite ν−(I − T ∗11T11) =
ν−(I − T˜ 211). Denote
J+ = sign (I − T˜11), J− = sign (I + T˜11), and J11 = sign (I − T˜ 211),
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 [5, Lemma 5.1]).
We recall the results for the operator T˜11 from the paper [5] 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 [5]. The problem
concerns the existence and a description of selfadjoint operators T˜ such that
A˜+ = I + T˜ and A˜− = I − T˜ solve the corresponding completion problems
A˜0± =
(
I ± T˜11 ±T˜ ∗21
±T˜21 ∗
)
, (4.1)
under minimal index conditions ν−(I + T˜ ) = ν−(I + T˜11), ν−(I − T˜ ) =
ν−(I − T˜11), respectively. The solution set is denoted by Ext T˜1,κ(−1, 1).
The next theorem gives a general solvability criterion for the completion
problem (4.1) and describes all solutions to this problem.
Theorem 4.1. ([5, Theorem 5]) Let T˜1 =
(
T˜11
T˜21
)
: H1 →
(
H1
H2
)
be a symmetric
operator with T˜11 = T˜
∗
11 ∈ [H1] and ν−(I − T˜ 211) = κ < ∞, and let J11 =
sign (I − T˜ 211). Then the completion problem for A˜0± in (4.1) has a solution
I ± T˜ for some T˜ = T˜ ∗ with ν−(I − T˜ 2) = κ if and only if the following
condition is satisfied:
ν−(I − T˜ 211) = ν−(I − T˜ ∗1 T˜1). (4.2)
If this condition is satisfied then the following facts hold:
(i) The completion problems for A˜0± in (4.1) have minimal solutions A˜±.
(ii) The operators T˜m := A˜+ − I and T˜M := I − A˜− ∈ Ext T˜1,κ(−1, 1).
(iii) The operators T˜m and T˜M have the block form
T˜m =
(
T˜11 DT˜11V
∗
V DT˜11 −I + V (I − T˜11)J11V ∗
)
,
T˜M =
(
T˜11 DT˜11V
∗
V DT˜11 I − V (I + T˜11)J11V ∗
)
,
(4.3)
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8 D. Baidiuk
where DT˜11 := |I − T˜ 211|1/2 and V is given by V := clos (T˜21D
[−1]
T˜11
).
(iv) The operators T˜m and T˜M are extremal extensions of T˜1:
T˜ ∈ Ext T˜1,κ(−1, 1) iff T˜ = T˜ ∗ ∈ [H], T˜m ≤ T˜ ≤ T˜M .
(v) The operators T˜m and T˜M are connected via
(−T˜ )m = −T˜M , (−T˜ )M = −T˜m.
For what follows it is convenient to reformulate the above theorem in
a Kre˘ın space setting. Consider the Kre˘ın 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 Kre˘ın space (H, J) that solve the com-
pletion problems
A0± =
(
I ± T11 ±T [∗]21
±T21 ∗
)
, (4.4)
underminimal 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 Kre˘ın space
[(H1, J1)], i.e. T11 = T
[∗]
11 .
Theorem 4.2. Let T1 be a symmetric operator in a Kre˘ın 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:
(i) The completion problems for A0± in (4.4) have ”minimal”(J2-minimal)
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 .
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Completion and extension of operators in Kre˘ın spaces. 9
(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 Kre˘ın space if and only if T˜ is a selfadjoint in a
Hilbert space. In particular, T gives solutions to the completion problems
(4.4) if and only if T˜ solves the completion problems (4.4). In view of
I − T ∗11T11 = I − T ∗11JJT11 = I − T˜ 211,
we are getting formula (4.6) from (4.2). Then formula (4.7) follows by multi-
plying the operators in (4.3) by the fundamental symmetry. 
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10 D. Baidiuk
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∗ .
By definition JTD
2
T = J − T ∗JT and JTDT = DTJT with analogous identi-
ties 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 commutation
relation of the form TDT = DT∗T is needed; these involve so-called link
operators introduced in [3, Section 4] (see also [5]).
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 [3,
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 5.2. 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)
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Completion and extension of operators in Kre˘ın spaces. 11
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 .
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∗D2T∗J = JJT∗JD2TJJ
= JJT∗JD
2
T
yields the equation (5.4).
The relation (5.5) follows from
DTLT∗ = T
∗JDT∗DT∗ = JTJDTJDT∗ = 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 lemmas,
which are used for the proof of the main theorem.
Lemma 5.3. Let
(
J T
T J
)
:
(
H
H
)
→
(
H
H
)
be a selfadjoint operator in the
Hilbert space H2 = H⊕ H. Then∣∣∣∣(J TT J
)∣∣∣∣1/2 = U (|J + T |1/2 00 |J − T |1/2
)
U∗,
where U = 1√
2
(
I I
I −I
)
is a unitary operator.
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 TT J
)∣∣∣∣2 = ((J TT J
)∗(
J T
T J
))
= U
(|J + T |2 0
0 |J − T |2
)
U∗.
Proceedings of the University of Vaasa. Working Papers      12
12 D. Baidiuk
The last step is to extract the square roots (twice) from the both sides of the
equation: ∣∣∣∣(J TT J
)∣∣∣∣1/2 = U (|J + T |1/2 00 |J − T |1/2
)
U∗.
The right hand side can be written in this form because U is unitary. 
Lemma 5.4. 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
)
. Equality
(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 ∈ [(H1, J1)]
such that ν−[I−T 211] = κ <∞; see (1.1). Denote T˜11 = J1T11, then T˜11 = T˜ ∗11
in the Hilbert space H1. Rewrite
ν−[I − T 211] = ν−(J1(I − T 211)) = ν−(J1 − T˜11J1T˜11)
= ν−((J1 − T˜11)J1(J1 + T˜11)).
Furthermore, denote
J+ = sign (J1(I − T11)) = sign (J1 − T˜11),
J− = sign (J1(I + T11)) = sign (J1 + T˜11),
J11 = sign (J1(I − T 211))
(5.9)
and let κ+ = ν−[I−T11] and κ− = ν−[I+T11]. Notice that |I∓T11| = |J1∓T˜11|
and one has polar decompositions
I ∓ T11 = J1J±|I ∓ T11|. (5.10)
Lemma 5.5. 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
Proceedings of the University of Vaasa. Working Papers      13
Completion and extension of operators in Kre˘ın spaces. 13
(i) ν−[I ± T11] = ν−[I ± T ];
(ii) ν−[I − T 2] = ν−[I − T 211]− 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 Hilbert
space operators: the block operator T˜ = JT =
(
T˜11 T˜12
T˜21 T˜22
)
is a selfadjoint
extension of T˜11 = T˜
∗
11 ∈ [H1]. Then the following statements
(i’) ν−(J1 ± T˜11) = ν−(J ± T˜ )
(ii’) ν−(J − T˜ JT˜ ) = ν−(J1 − T˜11J1T˜11)− ν−(J2);
(iii’) ran T˜12 ⊂ ran |J1 ± T˜11|1/2
are connected by the implications (i′)⇔ (ii′)⇒ (iii′).
Hence it’s sufficient to prove this form of the Lemma.
Let us prove the equivalence (i′)⇔ (ii′). Condition (ii’) is equivalent to
ν−
(
J1 T˜11
T˜11 J1
)
= ν−
(
J T˜
T˜ J
)
. (5.11)
Indeed, in view of (5.8)
ν−
(
J1 T˜11
T˜11 J1
)
= ν−(J1) + ν−(J1 − T˜11J1T˜11)
and
ν−
(
J T˜
T˜ J
)
= ν−(J) + ν−(J − T˜ JT˜ ) = ν−(J1) + ν−(J2) + ν−(J − T˜ JT˜ ).
By using Lemma 5.4, equality (5.11) is equivalent to
ν−(J1 − T˜11) + ν−(J1 + T˜11) = ν−(J − T˜ ) + ν−(J + T˜ ). (5.12)
Hence, (i′)⇒ (ii′).
Because ν−(J1 ± T˜11) ≤ ν−(J ± T˜ ), 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 5.6 below. Use a permutation to transform the
matrix in the right hand side of (5.11):
ν−
(
J T˜
T˜ J
)
= ν−

J1 0 T˜11 T˜12
0 J2 T˜21 T˜22
T˜11 T˜12 J1 0
T˜21 T˜22 0 J2
 = ν−

J1 T˜11 0 T˜12
T˜11 J1 T˜12 0
0 T˜21 J2 T˜22
T˜21 0 T˜22 J2
 .
Then condition (5.11) implies to the condition
ran
(
0 T˜12
T˜12 0
)
⊂ ran
∣∣∣∣∣
(
J1 T˜11
T˜11 J1
)∣∣∣∣∣
1/2
;
Proceedings of the University of Vaasa. Working Papers      14
14 D. Baidiuk
(see Theorem 2.1). By Lemma 5.3 the last inclusion can be rewritten as
ran
(
0 T˜12
T˜12 0
)
⊂ ranU
(
|J1 + T˜11|1/2 0
0 |J1 − T˜11|1/2
)
U∗,
where U = 1√
2
(
I I
I −I
)
is a unitary operator. This inclusion is equivalent
to
ranU∗
(
0 T˜12
T˜12 0
)
U = ran
(
T˜12 0
0 −T˜12
)
⊂ ran
(
|J1 + T˜11|1/2 0
0 |J1 − T˜11|1/2
)
and clearly this is equivalent to condition (iii’).
Note that if T˜11 has a selfadjoint extension T˜ satisfying (i’). Then by
applying Theorem 2.1 (or [5, Theorem 1]) it yields (iii’). 
Lemma 5.6. 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 211] <∞, ν−(J1) <∞, and ν−(J2) <∞.
Then for the conditions
(i) ν−[I1 − T 211] = ν−[I1 − T [∗]1 T1] + ν−(J2);
(ii) ranJ1T
[∗]
21 ⊂ ran |I − T 211|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 211 and B = T21.
A proof of (i)⇒(iii) is quite similar to the proof used in Lemma 5.5.
Statement (i) is equivalent the following equation:
ν−
(
J1 T˜11
T˜11 J1
)
= ν−
(
J T˜1
T˜ ∗1 J1
)
.
Indeed,
ν−
(
J1 T˜11
T˜11 J1
)
= ν−
(
J1 0
0 J1 − T˜11J1T˜11
)
= ν−(J1 − T˜11J1T˜11) + ν−(J1) <∞
and
ν−
(
J T˜1
T˜ ∗1 J1
)
= ν−
(
J 0
0 J1 − T˜ ∗1 JT˜1
)
= ν−(J1 − T˜11J1T˜11 − T˜ ∗21J2T˜21) + ν−(J1) + ν−(J2).
Proceedings of the University of Vaasa. Working Papers      15
Completion and extension of operators in Kre˘ın spaces. 15
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 T˜1
T˜ ∗1 J1
)
= ν−
 J1 0 T˜110 J2 T˜21
T˜11 T˜
∗
21 J1
 = ν−
 J1 T˜11 0T˜11 J1 T˜ ∗21
0 T˜21 J2
 .
It follows from [5, Theorem 1] that the condition (i) implies the condition
ran
(
0
T˜ ∗21
)
⊂ ran
∣∣∣∣∣
(
J1 T˜11
T˜11 J1
)∣∣∣∣∣
1/2
= ranU
(
|J1 + T˜11|1/2 0
0 |J1 − T˜11|1/2
)
U∗,
where U = 1√
2
(
I I
I −I
)
is a unitary operator (see Lemma 5.3). Then, equiv-
alently,
ran T˜ ∗21 ⊂ ran |J1 ± T˜11|1/2. 
5.3. Contractive extensions of contractions with minimal negative indices
Following to [5, 12, 14] 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 5.5 the two minimal index
conditions above are equivalent to single condition ν−[I−T 2] = ν−[I−T 211]−
ν−(J2).
It is clear from Theorem 2.1 that the conditions ran J1T
[∗]
21 ⊂ ran |I −
T11|1/2 and ran J1T [∗]21 ⊂ ran |I + T11|1/2 are necessary for the existence of
solutions; however as noted already in [5] they are not sufficient even in the
Hilbert space setting.
The next theorem gives a general solvability criterion for the completion
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 211] − ν−(J2) > 0. The
set of solutions T to the problem (5.13) will be denoted by Ext T1,κ(−1, 1)J2 .
Proceedings of the University of Vaasa. Working Papers      16
16 D. Baidiuk
Theorem 5.7. Let T1 be a symmetric operator as in (5.14) with T11 = T
[∗]
11 ∈
[(H1, J1)] and ν−[I−T 211] = κ <∞, and let JT11 = sign (J1(I−T 211)). 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 211] = 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” solutions 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 211|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)
(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 211] ≤ 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 5.6 one gets
the inclusions ranJ1T
[∗]
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 completion
problems for A0± in (5.13).
(ii) & (iii) By Lemma 5.6 the inclusion ranJ1T
[∗]
21 ⊂ ran |I−T 211|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 5.2 can be now rewritten as in (5.16). Indeed, observe that (see
Proceedings of the University of Vaasa. Working Papers      17
Completion and extension of operators in Kre˘ın spaces. 17
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∗T11J1)
[−1]DT11J1DT11V
∗
= J2V (I + L
∗
T11J1)
[−1](J11 − L∗T11JT∗11LT11)V ∗
= J2V (I + L
∗
T11J1)
[−1](J11 − (L∗T11J1)2J11)V ∗
= J2V (I + L
∗
T11J1)
[−1](I + L∗T11J1)(I − L∗T11J1)J11V ∗
= J2V (I − L∗T11J1)J11V ∗,
where the third equality follows from (5.1) and the fourth from (5.2).
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∗T11J1)[−1]DT11J1DT11V ∗
= J2V (I − L∗T11J1)[−1](J11 − L∗T11JT∗11LT11)V ∗
= J2V (I − L∗T11J1)[−1](J11 − (L∗T11J1)2J11)V ∗
= J2V (I − L∗T11J1)[−1](I − L∗T11J1)(I + L∗T11J1)J11V ∗
= J2V (I + L
∗
T11J1)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 211] − 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 A
0
+ 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 5.5 we get
ν−[I − T 2m] = κ− ν−(J2),
ν−[I − T 2M ] = 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).
Proceedings of the University of Vaasa. Working Papers      18
18 D. Baidiuk
(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). 
Acknowledgements. The author thanks his supervisor Seppo Hassi for several
detailed discussions on the results of this paper.
Proceedings of the University of Vaasa. Working Papers      19
Completion and extension of operators in Kre˘ın spaces. 19
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Proceedings of the University of Vaasa. Working Papers      20
20 D. Baidiuk
Dmytro Baidiuk
Department of Mathematics and Statistics
University of Vaasa
P.O. Box 700, 65101 Vaasa
Finland
e-mail: dbaidiuk@uwasa.fi