# Direct observation of magnon BEC in an out-of-plane magnetized yttrium iron
garnet film

G. A. Knyazev  M-granat, Russian Quantum Center, Skolkovo, Bolshoy Bulvar, 30,
121205 Moscow, Russia  Photonic and Quantum technologies school, Lomonosov
Moscow State University, Leninskie gori, 119991 Moscow, Russia  A. N.
Kuzmichev  M-granat, Russian Quantum Center, Skolkovo, Bolshoy Bulvar, 30,
121205 Moscow, Russia  P. E. Petrov  M-granat, Russian Quantum Center,
Skolkovo, Bolshoy Bulvar, 30, 121205 Moscow, Russia  Photonic and Quantum
technologies school, Lomonosov Moscow State University, Leninskie gori, 119991
Moscow, Russia  I. V. Savochkin  M-granat, Russian Quantum Center, Skolkovo,
Bolshoy Bulvar, 30, 121205 Moscow, Russia  P. M. Vetoshko  M-granat, Russian
Quantum Center, Skolkovo, Bolshoy Bulvar, 30, 121205 Moscow, Russia
Kotelnikov Institute of Radioengineering and Electronics, Mokhovaya 11-7,
125009 Moscow, Russia  V. I. Belotelov  M-granat, Russian Quantum Center,
Skolkovo, Bolshoy Bulvar, 30, 121205 Moscow, Russia  Photonic and Quantum
technologies school, Lomonosov Moscow State University, Leninskie gori, 119991
Moscow, Russia  Yu. M. Bunkov<EMAIL_ADDRESS>M-granat, Russian Quantum
Center, Skolkovo, Bolshoy Bulvar, 30, 121205 Moscow, Russia

###### Abstract

Bose-Einstain condensation occurs at an appropriate density of bosonic
particles, depending on their mass and temperature. We were able to
experimentally observe the transition from the spin wave regime to the magnon
Bose-Einstein condensed state (mBEC) with increasing magnon density by a
microwave pumping. We used optical methods to register the spatial
distribution of the magnon density and phase. For the first time, a coherent
state of stationary magnons was demonstrated far from the region of their
excitation.

Magnetism is, in principle, a quantum phenomenon, which is usually described
in the semiclassical approximation. However, there are a number of phenomena
to which the semiclassical consideration is not applicable. And first of all,
this is Bose - Einstein condensation of magnons — elementary excitations of
the ground magnets state. It follows from quantum statistics that magnons
should form a coherent quantum state (Bose-Einstein condensed state, mBEC) at
a concentration above the critical value $N_{c}$. Magnons density is
determined by the temperature and under stationary conditions it is always
below $N_{c}$. However, the density of magnons can be significantly increased
by exciting them by radio-frequency (RF) photons. This process corresponds to
the magnetic resonance — deflection and precession of magnetization in the
quasi classical model of magnetism. In this paper, we study the properties of
magnons with $\vec{k}=0$ in an out-of-plane magnetized YIG film at room
temperature. Under these conditions, the homogeneous precession corresponds to
the energy minimum and does not decay into spin waves [1]. The critical magnon
concentration $N_{c}$ for this case was calculated in [2] and corresponds to a
deviation of the precessing magnetization by about 3∘.

In this article, we restrict ourselves to considering only mBEC for stationary
magnons excited by resonant microwave pumping, whose properties are similar to
those of an atomic Bose condensate. The coherent state of traveling magnons
observed in in-plane magnetized YIG films has completely different properties
and is not considered in this article.

Bose condensation of stationary magnons was previously discovered in
antiferromagnetic superfluid 3He-B [3]. It leads to the formation of a long-
lived induction signal, which decayed orders of magnitude slower than it
should be due to the inhomogeneity of the magnetic field. Spontaneous recovery
of coherence after the decay of homogeneous precession [4], as well as a
thousandfold narrowing of the resonance line [5], clearly indicated the
formation of mBEC state. The discovered phenomenon of magnon supercurrent [6]
is also a consequence of Bose condensation.

Despite the fact that antiferromagnetic 3He is a superfluid liquid, its
superfluidity does not play any role in the formation of mBEC and magnon
supercurrent. Superfluid properties are not included in any of the mBEC
parameters. Thus, mBEC similar to that obtained in 3He can also be observed in
solid magnets. In particular, the properties of magnons in 3He have many
analogies with magnons in a YIG film [7]. Magnons in an out-of-plane
magnetized YIG film are characterized by repulsion, as in 3He-B, which leads
to an upward frequency shift when the magnetization deviates. Therefore, mBEC
can be similar to that in 3He-B.

The main advantage of antiferromagnetic superfluid 3He is the extremely long
lifetime of magnons. The Gilbert damping constant is of the order of 10 -8,
which makes it possible to observe Bose magnon condensation after turning off
the RF excitation. For YIG the Gilbert constant is by 3 orders of magnitude
larger, which makes it problematic to observe the formation of mBEC after
turning off the RF excitation.

In this article, we used another method for studying magnons, which also
categorically confirms the formation of mBEC in a YIG film. Using an optical
setup, we observed the spatial distribution of the magnon state outside the
region of their excitation. These experiments showed the formation of mBEC at
magnon concentrations above the critical value in the region where there is no
RF excitation.

##  I Experimental setup

The experiments were carried out on an elliptical YIG film 6 $\mu$m thick and
4.5 × 1.5 mm in size. The geometry of the experiment is schematically shown in
Fig. 1. The sample was grown by epitaxy on a gallium gadolinium garnet
substrate. The elliptical shape of the sample was chosen to reduce the
possibility of the secondary resonance modes formation. Magnetic resonance was
excited by a narrow strip line 0.2 mm wide, oriented perpendicular to the main
axis of the sample. It was located at a distance of 1 mm from one of the sides
of the sample.

Figure 1: The scheme of the experiment geometry. A beam of linearly polarized
light illuminates the sample through a prism. The reflected beam is directed
on the detectors. The polarization of the optical beam changes due to the
Faraday rotation on the magnetization component M along the light beam. The
received signal contains information about the angle of magnetization
deflection $\Theta$ and phase of precession $\phi$.

The states of magnons along the main axis of the sample were studied under
local excitation of magnons by the radio-frequency pumping. A beam of linearly
polarized light was sent to the sample through a prism. The position of the
laser beam was scanned along the sample. After the reflection of the beam, its
optical polarization changes due to the Faraday rotation when interacting with
the component of the magnetization M along light beam. Therefore, the Faraday
angle is sensitive to the magnetization dynamics, in particular to its
deflection angle $\Theta$ and phase $\phi$. The reflected beam of light was
directed to a balance photodetector through a Wollaston prism. Therefore, the
signals from detectors carried information about the amplitude and phase of
the precessing magnetization. To translate studies to lower frequencies, we
used light modulation at a frequency shifted from the resonant frequency by
about 12 kHz. This made it possible to record the parameters of the reflected
signal using low-frequency detectors. After detecting the signals, we obtained
the amplitude and phase of the magnon precession at a given point of the
sample and for a given magnetic field. By scanning the magnetic field and the
illumination point of the sample, we obtained the amplitude and phase
distribution of magnons in the sample depending on the position and field. A
detailed description of our optical setup was presented in [8] where the state
of magnons in the region of excitation was investigated.

##  II Experimental results

In this article, we demonstrate the experimental observation of magnons state
outside the excitation region. Experimental records of spatial destribution of
amplitude and phase of the magnetization precession as a function of the
magnetic field at different excitation energies are shown in the figures 2.

In Fig. 2 (a,b) the amplitude of signal in units of magnetization deviation
$\Theta$ at a low excitation level of 0.05 mW (a) and at a high excitation
level of 6 mW (b) is shown. The dependences of the signal amplitude on the
magnetic field shift and the position of testing laser beam are shown. The
deviation angle $\Theta$ was calibrated from the field shift from the
resonance. The excitation strip line was located between the positions of 3.5
and 3.7 mm. At the low excitation, the resonance is clearly visible with a
maximum in the region of the strip line and a resonance field of about 2621 Oe
and slightly below.

Figure 2: (a)–(b) The spatial distribution of magnon density in units of
magnetization deflection angle as a function of sweeping down magnetic field
at 0.05 mW (a), and at 6 mW (b) of pumping energy. (c)–(d) The spatial
distribution of the magnon phase as function of sweeping down magnetic field
at 0.05 mW (c) and at 6 mW (d) pumping energy.

This shift from resonanse takes place due to effect of “Foldover” rsonanse,
when the resonanse field shifts due to decrease of demagnetization field at
magnetization deflection. See [8, 9] for details. The spatial distribution of
magnons changes dramatically at a higher level of excitation. Magnons filled
the entire sample in the range of fields from 2621 to 2598 Oe. At a lower
field, magnons do not fill the part of the sample below the strip line, but
continue to fill the upper part of the sample up to 2562 Oe. This field shift
is determined by magnon relaxation, which is proportional to the square of the
deflection angle and the distance from the excitation region to the edge of
the sample. The filling with magnons disappears as soon as their flow from the
excitation region ceases to compensate for the losses.

Of great interest is the spatial distribution of the precession phase in these
experiments, shown in Fig. 2 (c,d). We see that at low excitation power
magnons propagate from the excitation region in the form of spin waves. The
length of these waves changes with the shift of the magnetic field. Naturally,
the magnetization precession frequency should coincide with the RF pump
frequency. In the absence of an additional radio frequency field, the
frequency shift is provided by the gradient energy of spin waves. Therefore,
as the magnetic field decreases, the length of the spin waves also decreases.
This effect is clearly seen in computer micromagnetic simulation of
experimental conditions in the framework of the semiclassical Landau-Lifshitz
theory.

This state changes completely with increasing RF pump energy, as shown in Fig.
2 (d). We see that a spatially homogeneous state with a coherent phase
distribution arises outside the pump region. In this case, it should be
emphasized that coherence is a property of magnons, since it is not supported
by an external RF field. This state directly demonstrates magnon Bose
condensation at a high concentration of magnons. Indeed, Fig. 2(d) clearly
shows that magnons fill the entire sample at a magnon concentration
corresponding to the magnetization deviation angle of more than 3∘ , as
predicted in the work [2].

Let’s consider the spatial distribution of magnon amplitude and phase at a
fixed field in more detail. In Fig. 3 (a) the spatial distribution of the
signal amplitude is shown for an external magnetic field of 2607 Oe and 2615
Oe, marked by a dashed lines in Fig. 2 (a), and at an RF pump power of 0.05
mW. The critical magnon concentration is reached only in the RF pump region,
while outside this region the magnon concentration is much lower and a magnon
gas can be described in the semiclassical approximation. The very different
destribution is shown in Fig. 3 (b) at an excitation power of 6 mW. In this
case, the magnons filled the entire sample in approximately the same
concentration, excluding the edge regions.

The spatial distribution of the magnon precession phase is shown in Fig. 3
(c,d). With a small excitation of 0.05 mW, a phase rotation is observed, which
depends on the distance from the excitation region, which reaches 7$\pi$ per
mm in a field of 2615 Oe and 17$\pi$ per mm in a field of 2607 Oe. We
calculated the spatial destribution of phase precessing for the experimental
conditions by micromagnetic modeling using the MuMax3 [10] program and
obtained excellent agreement with the experimental results.

Figure 3: (a)–(b) The spatial distribution of the magnon density in units of
the magnetization deflection angle in the fields corresponding to the dashed
lines in Fig.2 (a,b) at a pump energy of 0.05 mW (a) and 6 mW (b). (c)–(d) The
spatial distribution of the magnon phase at fields corresponding to the dashed
lines in Fig.2 (c,d) at a pump energy of 0.05 mW (c) and 6 mW (d).

The phase distribution changes drastically at 6 mW excitation as shown in Fig.
3 (d). In this case, a sharp turn of the precession phase near the excitation
region corresponds to the flow of magnons from the excitation region. For
small magnon relaxation, this turn corresponds to about 180∘. It decreases
with increasing relaxation. This experimental result requires further
theoretical study. It is important to note that micromagnetic modeling by
MuMax3 program leads to the formation of waves at any excitation amplitude. We
can conclude that the formation of a spatially homogeneous coherent precession
at high excitation is a consequence of magnons Bose condensation, which,
naturally, lies beyond the scope of the semiclassical theory. The MuMax3
program is based on the classical theory of Landau-Lifshitz magnetization
precession and, of course, cannot simulate Bose condensation, which has a
quantum nature.

##  III Conclusion

Magnon Bose condensation and the associated magnon supercurrent in
antiferromagnetic superfluid 3He are well known quantum phenomena that have
received worldwide recognition [11] and are awarded by F. London Memory Prize
for this discovery. The formation of a Bose condensate of stationary magnons
in solid-state magnets has caused much controversy. Several experimental
results obtained earlier in the out of plane magnetized YIG film were
considered as indirect evidence of magnons Bose condensation [9, 12]. In this
article, magnon Bose condensation is experimentally demonstrated by direct
optical observation of the coherent precession of magnetization far beyond the
excitation region. This result contradicts the semiclassical Landau-Lifshitz
theory, which directly indicates the quantum nature of this effect. It opens
up new perspectives for research in quantum physics, as well as for some
modern technological applications of magnonics, quantum communications, and
quantum computing.

This work was supported by Rosatom in the framework of the Roadmap for Quantum
computing (Contract No. 868-1.3-15/15-2021 dated October 5).

## References

  * [1] Tupitsyn, I. S., Stamp, P. C. E., & Burin A.L. “Stability of Bose-Einstein Condensates of Hot Magnons in Yttrium Iron Garnet Films”. Phys. Rev. Lett. 100, 257202 (2008)
  * [2] Yu. M. Bunkov and V. L. Safonov, “Magnon condensation and spin superfluidity”. J. Mag. and Mag. Mat. 452, 30 (2018).
  * [3] A. S. Borovik-Romanov, Yu. M. Bunkov, V. V. Dmitriev and Yu. M. Mukharskii, “Long-lived induction signal in superfluid 3He-B”. JETP Lett. 40, 1033 (1984).
  * [4] Yu.M. Bunkov and G. E. Volovik “Magnon Bose Einstein Condensation and Spin Superfluidity J. Phys.: Condens. Matter 22 164210 (2010).
  * [5] Yu. M. Bunkov, “Helium-3; Cosmological and atomic physics experiments” Phil. Trans. R. Soc. A, 366, 2821 (2008).
  * [6] A. S. Borovik-Romanov, Yu. M. Bunkov, V. V. Dmitriev, Yu. M. Mukharskiy, D. A. Sergatskov, ”Investigation of Spin Supercurrent in 3He-B”, Phys. Rev. Lett., 62, 1631 (1989).
  * [7] G.E. Volovik, ”Twenty Years of Magnon Bose Condensation and Spin Current Superfluidity in 3He-B”, J. of Low Temp. Phys., 153, 135 (2008).
  * [8] P. E. Petrov, P. O. Kapralov, G. A. Knyazev, A. N. Kuzmichev, P. M. Vetoshko, Yu. M. Bunkov, V. I. Belotelov, ”Magneto-optical imaging of coherent spin dynamics in ferrites” Optics Express 30, 1737 (2022).
  * [9] Yu. M. Bunkov, A. N. Kuzmichev, T. R. Safin, P. M. Vetoshko, V. I. Belotelov, and M. S. Tagirov, “Quantum paradigm of the foldover magnetic resonance” Scientific Reports, 11, 7673 (2021).
  * [10] A. Vansteenkiste et al., ”The design and verification of MuMax3” AIP Advances, 4, 107133 (2014).
  * [11] Yu. M. Bunkov and G. E. Volovik, “Spin Superfluidity and Magnon BEC” in Novel Superfluids Ch.4, (eds. Bennemann, K. H. & Ketterson, J. B. Oxford Univ. Press, Oxford, (2013).
  * [12] P. M. Vetoshko, G. A. Knyazev A. N. Kuzmichev, A. A. Holin, V. I. Belotelov, and Yu. M. Bunkov “Bose condensation and spin superfluidity of magnons in a perpendicularly magnetized film of yttrium iron garnet”, JETP Lett., 112, 299 (2020).


# Hilbert $C^{*}$-modules with Hilbert dual and $C^{*}$-Fredholm operators

Vladimir Manuilov  and  Evgenij Troitsky  Moscow Center for Fundamental and
Applied Mathematics, Lomonosov Moscow State University, 119991 Moscow, Russia
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We study Hilbert $C^{*}$-modules over a $C^{*}$-algebra $A$ for which the
Banach $A$-dual module carries a natural structure of Hilbert $A$-module. In
this direction we prove that if $A$ is monotone complete, $M$ and $N$ are
Hilbert $A$-modules, $M$ is self-dual, and both $T:M\to N$ and its Banach
$A$-dual $T^{\prime}:N^{\prime}\to M^{\prime}$ have trivial kernels and
cokernels then $M\cong N^{\prime}$. With the help of this result, for a
monotone complete $C^{*}$-algebra $A$, we prove that the index of any
$A$-Fredholm operator can be calculated as the difference of its kernel and
cokernel as in the Hilbert space case.

###### Key words and phrases:

Hilbert $C^{*}$-module, monotone complete $C^{*}$-algebra, $A$-Fredholm
operator, dual module, self-dual module, orthogonal complement, polar
decomposition

###### 2000 Mathematics Subject Classification:

46L08; 58B34

The work is supported by the Russian Science Foundation under grant
21-11-00080

##  1\. Introduction

Hilbert $C^{*}$-modules were introduced by W. Paschke in [14] as a
generalization of Hilbert spaces to modules over a $C^{*}$-algebra $A$ with an
inner product taking values also in $A$.

In this paper we deal with right Hilbert $C^{*}$-modules, i.e. $A$ acts on a
module $M$ from the right, and the inner product satisfies $\langle
m,na\rangle=\langle m,n\rangle a$, $\langle ma,n\rangle=a^{*}\langle
m,n\rangle$, $m,n\in M$, $a\in A$. The dual module $M^{\prime}$ consists of
all bounded anti-$A$-linear maps $f:M\to A$ (called functionals), i.e., maps
satisfying $f(ma)=a^{*}f(m)$. The dual module is a Banach $A$-module with the
multiplication defined by $(fa)(m)=f(m)a$, but not necessarily a Hilbert
$C^{*}$-module, and the map $m\mapsto\widehat{m}=\langle\cdot,m\rangle$ is an
isometric inclusion $M\subset M^{\prime}$. A module $M$ is _self-dual_ if this
inclusion is an isomorphism. Self-dual modules have a lot of nice properties,
e.g. [14, 1] (and also [12]), but they are rare. Unlike the first dual module,
the second dual one, $M^{\prime\prime}$, has a natural structure of a Hilbert
$C^{*}$-module, and there is an isometric inclusion $M\subset
M^{\prime\prime}\subset M^{\prime}$. This is the end of the story, as
$M^{\prime\prime\prime}$ is canonically isomorphic to $M^{\prime}$ [15,
Theorem 2.4] (see Theorem 2 below). A module $M$ is _reflexive_ if
$M=M^{\prime\prime}$. Reflexive modules are not as nice as self-dual ones, but
still share some nice properties of the latter, see e.g. [1, 5, 4] (and also
[12]). The last remaining equality, $M^{\prime}=M^{\prime\prime}$, was
discussed in [10], where such modules were called _modules with Hilbert dual_.

Hilbert $C^{*}$-modules over $W^{*}$-algebras have some special properties
that make them closer to Hilbert spaces. In particular, although they need not
be self-dual, the inner product extends to the dual modules, making them
Hilbert $W^{*}$-modules as well. The same is true for monotone complete
$C^{*}$-algebras [3], so any Hilbert $C^{*}$-module over such $C^{*}$-algebra
has Hilbert dual.

It is well known that if there is a bounded operator $T:H_{1}\to H_{2}$
between two Hilbert spaces with trivial kernel and cokernel then $H_{1}$ and
$H_{2}$ are isomorphic. Although Hilbert spaces differ only by cardinality,
the construction of an isomorphism uses a polar decomposition for $T$, which
need not to be invertible.

When we replace Hilbert spaces by Hilbert $C^{*}$-modules, the result becomes
more difficult, and definitely not true in full generality. Our aim is to
clarify the conditions that allow to prove that two Hilbert $C^{*}$-modules,
$M$ and $N$, over a $C^{*}$-algebra $A$ are isomorphic when connected by a
bounded operator $T:M\to N$. Our main result claims that if $A$ is monotone
complete, $M$ self-dual, and both $T$ and its dual $T^{\prime}:N^{\prime}\to
M^{\prime}$ have trivial kernels and cokernels then $M\cong N^{\prime}$
(Theorem 9). As an application, we generalize a result on decomposition of
$A$-Fredholm operators from [2] to the case when $A$ is monotone complete
(Theorem 15) and correct the proof in the $W^{*}$ case.

Acknowledgment: The authors are indebted to Michael Frank and Stefan Ivkovic
for valuable discussions and to an anonymous reviewer for careful reading of
the first version of the paper and very helpful and friendly criticism.

##  2\. Preliminaries

Definitions of Hilbert $C^{*}$-modules, adjointable and non-adjointable
operators and their basic properties can be found in textbooks, e.g. [14, 12,
7, 11].

Recall that the canonical inclusions for dual and double dual modules [15].
For $m\in M$, define $\dot{m}\in M^{\prime\prime}$ by

(1) $\dot{m}(f)=(f(m))^{*},\quad f\in M^{\prime}.$

The map $m\mapsto\dot{m}$ is an isometric module map of $M$ to
$M^{\prime\prime}$. For $X\in M^{\prime\prime}$ define $\widetilde{X}\in
M^{\prime}$ by $\widetilde{X}(m)=X(\widehat{m})$, where $m\mapsto\widehat{m}$,
$M\to M^{\prime}$ was defined above. So, if we identify $M$ with its image
under $m\mapsto\widehat{m}$, then $\widetilde{X}$ is the restriction of $X$ on
$M$. Notice that

(2) $(\dot{m})\\!{\widetilde{\phantom{m}}}=\widehat{m}\quad\mbox{ for all
}m\in M.$

It is clear that the map $X\mapsto\widetilde{X}$ is a module map of
$M^{\prime\prime}$ into $M^{\prime}$ and that
$\|\widetilde{X}\|_{M^{\prime}}\leq\|X\|_{M^{\prime\prime}}$ for $X\in
M^{\prime\prime}$. Moreover, this map is an isometry as it follows from the
following proposition.

######  Proposition 1 (Prop. 2.1 in [15]).

Let $N$ be a submodule of $M^{\prime}$ containing $\widehat{M}$. For any $f\in
N^{\prime}$, we have $\|f\|_{N^{\prime}}=\|f|_{\widehat{M}}\|$, where

(3) $\|f\|_{N^{\prime}}=\sup_{m\in N,\|m\|\leq 1}\|f(m)\|.$

After making all permissible identifications, we can then array the modules
$M$, $M^{\prime}$, and $M^{\prime\prime}$ as

(4)
$M\overset{D}{\subseteq}M^{\prime\prime}\overset{I}{\subseteq}M^{\prime},\qquad
D(m)=\dot{m},\quad I(X)=\widetilde{X},\qquad I\circ D(m)=\widehat{m}$

by (2).

In [15] the inner product

(5) $\langle\cdot,\cdot\rangle^{\prime\prime}:M^{\prime\prime}\times
M^{\prime\prime}\to A,\qquad\langle X,Y\rangle^{\prime\prime}=Y(I(X)),$

is introduced and the following statement is proved.

######  Theorem 2 (Theorem 2.4 in [15]).

The map _( 5)_ is an $A$-valued inner product on $M^{\prime\prime}$. The norm
obtained from this product coincides with the initial (operator) norm on
$M^{\prime\prime}$. The map $I$ from _( 4)_ is an isometry of
$M^{\prime\prime}$ into $M^{\prime}$.

##  3\. Hilbert $C^{*}$-modules with Hilbert dual

######  Definition 3 ([10]).

A Hilbert $C^{*}$-module $M$ over a $C^{*}$-algebra $A$ is called a module
with a Hilbert dual if the inner product $\langle\cdot,\cdot\rangle$ on $M$
extends to an inner product $\langle\cdot,\cdot\rangle^{\prime}$ on
$M^{\prime}$ such that

  * •

$f(m)=\langle\widehat{m},f\rangle^{\prime}$ for any $m\in M$ and any $f\in
M^{\prime}$ (in particular,
$\langle\widehat{m},\widehat{n}\rangle^{\prime}=\widehat{n}(m)=\langle
m,n\rangle$ for any $m,n\in M$);

  * •

$M^{\prime}$ is complete with respect to the norm determined by the inner
product $\langle\cdot,\cdot\rangle^{\prime}$.

######  Proposition 4.

$M$ is a Hilbert $C^{*}$-module with a Hilbert dual if and only if
$I(M^{\prime\prime})=M^{\prime}$.

###### Proof.

If $I(M^{\prime\prime})=M^{\prime}$, we can consider the induced inner product
on $M^{\prime}$: $\langle f,g\rangle^{\prime}:=\langle
I^{-1}(f),I^{-1}(g)\rangle^{\prime\prime}$, where $f,g\in M^{\prime}$. Let
$X=I^{-1}(f)$. Then, by (2) and (5),

$\displaystyle\langle\widehat{m},f\rangle^{\prime}$ $\displaystyle=$
$\displaystyle\langle I^{-1}(\widehat{m}),X\rangle^{\prime\prime}=\langle
I^{-1}I(\dot{m}),X\rangle^{\prime\prime}$ $\displaystyle=$
$\displaystyle(\langle
X,\dot{m}\rangle^{\prime\prime})^{*}=(\dot{m}(I(X)))^{*}=(\dot{m}(f))^{*}$
$\displaystyle=$ $\displaystyle f(m).$

Conversely let $M$ be a Hilbert $C^{*}$-module with a Hilbert dual, and let
$\langle\cdot,\cdot\rangle^{\prime}$ be an extension of the inner product on
$M$ to $M^{\prime}$. Then, besides the standard (operator) norm
$\|f\|_{M^{\prime}}$ on $M^{\prime}$ (see (3)), there is also the norm
$\|f\|^{\prime}=\|\langle f,f\rangle^{\prime}\|^{1/2}$, $f\in M^{\prime}$. As
$M^{\prime}$ is complete with respect to both norms, and we have the following
estimation: by Definition 3 $\langle
f,\widehat{m}\rangle^{\prime}\langle\widehat{m},f\rangle^{\prime}\leq(\|\widehat{m}\|^{\prime})^{2}\langle
f,f\rangle^{\prime}=(\|m\|)^{2}\langle f,f\rangle^{\prime}$ (see e.g. [12,
Prop. 1.2.4]) and

$\displaystyle(\|f\|^{\prime})^{2}$ $\displaystyle=$ $\displaystyle\|\langle
f,f\rangle^{\prime}\|\geq\Bigl{(}\sup_{m\in M,\>\|m\|\leq 1}\|\langle
f,m\rangle^{\prime}\|\Bigr{)}^{2}$ $\displaystyle=$
$\displaystyle\Bigl{(}\sup_{m\in M,\>\|m\|\leq
1}\|f(m)\|\Bigr{)}^{2}=\left(\|f\|_{M^{\prime}}\right)^{2}.$

Thus
$\mathrm{Id}:(M^{\prime},\|.\|^{\prime})\to(M^{\prime},\|.\|_{M^{\prime}})$ is
an injective bounded epimorphism and the norms are equivalent, i.e. there
exists $c>0$ such that $\|f\|_{M^{\prime}}\leq\|f\|^{\prime}\leq
c\|f\|_{M^{\prime}}$ for any $f\in M^{\prime}$. Given $f\in M^{\prime}$,
consider the map $g\mapsto\langle g,f\rangle^{\prime}$, $M^{\prime}\to A$. It
is obviously anti-$A$-linear. As the two norms are equivalent, this map is
bounded by $c\|f\|_{M^{\prime}}$. Hence this map is a functional on
$M^{\prime}$, i.e. is an element of $M^{\prime\prime}$. This gives a bounded
map $\Phi:M^{\prime}\to M^{\prime\prime}$, $\|\Phi\|\leq c^{2}$, because

$\|\Phi(f)(g)\|=\|\langle
g,f\rangle^{\prime}\|\leq\|g\|^{\prime}\|f\|^{\prime}\leq
c^{2}\|g\|_{M^{\prime}}\|f\|_{M^{\prime}}.$

Consider the composition $I\circ\Phi:M^{\prime}\to M^{\prime}$. Then, for any
$f\in M^{\prime}$, $m\in M$,

$I\circ\Phi(f)(m)=\Phi(f)(\widehat{m})=\langle\widehat{m},f\rangle^{\prime}=f(m).$

Thus, the map $I\circ\Phi$ is the identity map. Therefore, $I$ is an
epimorphism and $I(M^{\prime\prime})=M^{\prime}$. Also, since $I$ is an
isometry, $\Phi$ as an isometric isomorphism and one can _aposteriori_ take
$c=1$. ∎

The end of the proof shows that all inner products on $M^{\prime}$ for a
Hilbert module with a Hilbert dual generate the same norm, hence coincide with
the operator norm on $M^{\prime}$ (see also [10, Propostion 3]).

It also follows that if $M$ has Hilbert dual then $M^{\prime}$ (which is a
Hilbert $C^{*}$-module in this case) is self-dual.

Any self-dual Hilbert $C^{*}$-module is obviously a module with a Hilbert
dual. It was shown in [14] that any Hilbert $C^{*}$-module over a
$W^{*}$-algebra is a module with a Hilbert dual, and it was shown in [6, 9, 3]
that the same holds for a slightly bigger class of $C^{*}$-algebras, namely
for monotone complete ones. Moreover, Theorem 4.7 of [3] (an immediate
consequence of [6]) claims that any Hilbert $C^{*}$-module over a
$C^{*}$-algebra $A$ has a Hilbert dual if and only if $A$ is monotone
complete.

There are also examples of Hilbert $C^{*}$-modules with Hilbert dual over
$C^{*}$-algebras beyond the class of monotone complete ones.

######  Example 5.

Let $X$ be a locally compact Hausdorff space, $M=C_{0}(X)$ the algebra of
continuous functions on $X$ vanishing at infinity, and let $A=C_{b}(X)$ be its
multiplier algebra, i.e. the algebra of all bounded continuous functions on
$X$. Then we can consider $M$ as a Hilbert $C^{*}$-module over $A$.

Let us verify that $M^{\prime}=C_{b}(X)$. The inclusion $C_{b}(X)\subseteq
M^{\prime}$ is evident. Let now ${\varphi}\in M^{\prime}$. Consider a
countable partition of unity $\\{f_{i}\\}$ on $X$ consisting of real-valued
functions with compact supports. Then, for any $a\in M$, one has
$a=\sum_{i}af_{i}$ uniformly. Then
${\varphi}(a)=\sum_{i}{\varphi}(af_{i})=\sum_{i}{\varphi}(f_{i}a)=\sum_{i}a^{*}{\varphi}(f_{i})$
uniformly. So, $\sum_{i}{\varphi}(f_{i})$ is a multiplier of $C_{0}(X)$
corresponding to ${\varphi}$.

So, $M^{\prime}=A$ is a Hilbert module over $A$. Then it is clear that
$M^{\prime\prime}=M^{\prime}$, because each $A$-functional on a unital $A$ is
determined by its value at $1_{A}$.

Recall that a commutative algebra is monotone complete iff it is an
AW*-algebra iff its spectrum is extremally disconnected. So if $X$ is not
extremely disconnected (e.g. $X=(0,1)$) then $C_{0}(X)$ provides an example of
a Hilbert $C^{*}$-module with Hilbert dual, although the underlying
$C^{*}$-algebra $C_{b}(X)$ is not monotone complete.

##  4\. Around polar decomposition in Hilbert $C^{*}$-modules

Proposition 2 of [2] claims that if $T:M\to N$ is an injective bounded module
homomorphism of Hilbert $C^{*}$-modules over a $W^{*}$-algebra,
$T(M)^{\perp\perp}=N$ and if $M$ is self-dual then $T(T^{*}T)^{-1/2}$ is a
bounded module isomorphism $M\cong N$ (although $(T^{*}T)^{-1/2}$ may be
unbounded). S. Ivkovic has informed us that there is a problem in the proof,
thus urging us to revise this claim. Here is a counterexample.

######  Example 6.

Let $A=l_{\infty}$ be the $W^{*}$-algebra of bounded sequences, $M=A$. We
shall use the notation $\delta_{n}\in A$ for the sequence of zeroes with the
only unit at the $n$-th place, $\delta_{n}=(0,...,0,1,0,...)$. Consider the
following closed submodule $N\subset l_{2}(A)$ of all sequences of the form
$x=(a_{1},a_{2},...)$ where each $a_{n}\in A$, $n\in\mathbb{N}$, satisfies
$a_{n}=\delta_{n}a_{n}$. If this set is not closed then take the closure.

Fix $\lambda_{n}=n^{-1}$, and set
$T(a)=(\lambda_{1}\delta_{1}a,\lambda_{2}\delta_{2}a,...)\in N$, $a\in A$.
Then $T:M\to N$ is injective, and $T(M)^{\perp}=0$ in $N$. Set
$f=T(T^{*}T)^{-1/2}$, then $f(a)=(\delta_{1}a,\delta_{2}a,...)$. The range of
$f$ is greater than $N$ (e.g. it contains $f(1)=(\delta_{1},\delta_{2},...)$,
which is not in $N$). Moreover, $M=A$ is not isometrically isomorphic to $N$.

The reason for the error in the proof of Proposition 2 of [2] is that $f$ maps
$M$ not to $N$, but to the dual module $N^{\prime}$. The problem with the
Lin’s proof of the polar decomposition [8] is similar: he uses the polar
decomposition of an operator over the enveloping $W^{*}$-algebra, but the
unitary in this decomposition maps into the dual module.

The easiest way to fix Proposition 2 of [2] is to add the requirement of self-
duality for $N$ (see Corollary 10 below).

We want to pursue this issue further and to remove the requirement of self-
duality for $N$.

Let $T:K\to M$ be an operator between two Hilbert $C^{*}$-modules. Define
$T^{\\#}:M\to K^{\prime}$ by $(T^{\\#}m)(k)=\langle Tk,m\rangle$, $m\in M$,
$k\in K$. If $K\subset M$ then we denote by $J$ the corresponding inclusion,
$J:K\to M$.

######  Lemma 7.

Suppose that $K\subset M$, $K^{\perp}=0$, $M$ is self-dual and $K$ has a
Hilbert dual. Then $\|\langle J^{\\#}m,J^{\\#}m\rangle^{\prime}\|=\|\langle
m,m\rangle\|$ for any $m\in M$. Also $J^{\\#}:M\to K^{\prime}$ is an
isomorphism.

###### Proof.

The map $J^{\\#}$ is given by $(J^{\\#}m)(k)=\langle J(k),m\rangle$. Note that
$J^{\\#}$ is a contraction. Indeed,

$\displaystyle\|J^{\\#}(m)\|$ $\displaystyle=$ $\displaystyle\sup_{k\in
K,\|k\|\leq 1}\|J^{\\#}(m)(k)\|$ $\displaystyle=$ $\displaystyle\sup_{k\in
K,\|k\|\leq 1}\|\langle J(k),m\rangle\|\leq\|m\|.$

Thus,

(6) $\|\langle J^{\\#}m,J^{\\#}m\rangle^{\prime}\|\leq\|\langle m,m\rangle\|$

for any $m\in M$.

For $k,k^{\prime}\in K$, we have $J^{\\#}(J(k))(k^{\prime})=\langle
J(k^{\prime}),J(k)\rangle=\langle k^{\prime},k\rangle$, i.e.,
$J^{\\#}(J(k))=\widehat{k}$.

Let us define a map $Q:K^{\prime}\to M$. Let $\tau\in K^{\prime}$. Then the
map

$m\mapsto\bar{\tau}(m)=\langle J^{\\#}m,\tau\rangle^{\prime}$

is a functional on $M$ (algebraic properties are obvious, and boundedness
follows from that of $J^{\\#}$). Then self-duality of $M$ implies existence of
a unique $m_{\tau}\in M$ such that $\bar{\tau}(m)=\langle m,m_{\tau}\rangle$
for any $m\in M$. Set $Q(\tau)=m_{\tau}$.

Then $\bar{\tau}(m_{\tau})=\langle m_{\tau},m_{\tau}\rangle$, and by (6) one
has

$\displaystyle\|m_{\tau}\|^{2}$ $\displaystyle=$ $\displaystyle\|\langle
m_{\tau},m_{\tau}\rangle\|=\|\langle\tau,J^{\\#}(m_{\tau})\rangle^{\prime}\|$
$\displaystyle\leq$ $\displaystyle\|\tau\|\cdot\|\langle
J^{\\#}(m_{\tau}),J^{\\#}(m_{\tau})\rangle^{\prime}\|^{1/2}$
$\displaystyle\leq$ $\displaystyle\|\tau\|\cdot\|\langle
m_{\tau},m_{\tau}\rangle\|^{1/2}$ $\displaystyle=$
$\displaystyle\|\tau\|\cdot\|m_{\tau}\|,$

whence $\|m_{\tau}\|\leq\|\tau\|$, and $\|Q\|\leq 1$.

Consider the composition $Q\circ J^{\\#}:M\to M$. Set $\tau=J^{\\#}(m)$,
$n=m_{\tau}$. Then $Q\circ J^{\\#}(m)=n$, and

$\langle k,n\rangle=\bar{\tau}(k)=\tau(k)=\langle k,m\rangle$

for any $k\in K$. As $K^{\perp}=0$, we have $n=m$, i.e. $Q\circ
J^{\\#}=\operatorname{id}_{M}$. Since both $Q$ and $J^{\\#}$ are contractions,
$J^{\\#}$ is an isometric embedding.

Taking the composition in the opposite order, we have, for any $\tau\in
K^{\prime}$ and $k\in K$,

$\displaystyle(J^{\\#}\circ Q(\tau))(k)$ $\displaystyle=$
$\displaystyle(J^{\\#}(m_{\tau}))(k)=\langle J(k),m_{\tau}\rangle$
$\displaystyle=$ $\displaystyle\overline{\tau}(J(k))=\langle
J^{\\#}J(k),\tau\rangle^{\prime}$ $\displaystyle=$
$\displaystyle\langle\widehat{k},\tau\rangle^{\prime}=\tau(k)$

and $J^{\\#}\circ Q=\operatorname{id}_{K^{\prime}}$. Hence, as above, $Q$ is
an isometric embedding. Thus, $Q$ and $J^{\\#}$ are isometric isomorphisms. ∎

######  Corollary 8.

The map $J^{\\#}$ satisfies $\langle J^{\\#}m,J^{\\#}m\rangle^{\prime}=\langle
m,m\rangle$ for any $m\in M$.

###### Proof.

By Lemma 7 and its proof, $J^{\\#}$ is an isometry with the inverse isometry
$Q$. Hence,

$\displaystyle\langle m,m\rangle$ $\displaystyle=$ $\displaystyle\langle
QJ^{\\#}m,QJ^{\\#}m\rangle\leq\|Q\|^{2}\langle
J^{\\#}m,J^{\\#}m\rangle^{\prime}$ $\displaystyle=$ $\displaystyle\langle
J^{\\#}m,J^{\\#}m\rangle^{\prime}\leq\|J^{\\#}\|^{2}\langle m,m\rangle$
$\displaystyle=$ $\displaystyle\langle m,m\rangle.$

∎

######  Theorem 9.

Let $A$ be a monotone complete $C^{*}$-algebra, let $M$, $N$ be Hilbert
$C^{*}$-modules over $A$, $M$ self-dual, and let $T:M\to N$ be an operator
with $(T(M))^{\perp\perp}=N$, $\operatorname{Ker}(T)=\\{0\\}$ and
$\operatorname{Ker}(T^{\prime})=\\{0\\}$, where $T^{\prime}:N^{\prime}\to
M^{\prime}$ is the adjoint operator. Then $M\cong N^{\prime}$.

###### Proof.

We have that $T^{\prime}:N^{\prime}\to M^{\prime}$ is a morphism of self-dual
modules (see Theorem 2) with $\operatorname{Ker}(T^{\prime})=\\{0\\}$. Since
$M$ is self-dual, $T$ is adjointable [14]. One has $0=\langle
T^{*}(n),m\rangle=\langle n,T(m)\rangle$ and $\operatorname{Ker}(T)=\\{0\\}$
implies $(T^{*}(N))^{\bot}=\\{0\\}$.

Denote $\Lambda_{M}:M\to M^{\prime}$, $m\mapsto\widehat{m}$, and
$\Lambda_{N}:N\to N^{\prime}$, $n\mapsto\widehat{n}$. Then $T^{*}$ and
$T^{\prime}$ are connected by the following commutative diagram:

(7)
$\textstyle{M^{\prime}}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
N^{\prime}}$$\scriptstyle{T^{\prime}}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\Lambda_{M}}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
N\ignorespaces\ignorespaces\ignorespaces\ignorespaces.}$$\scriptstyle{T^{*}}$$\scriptstyle{\Lambda_{N}}$

Thus $(T^{*}(N))^{\bot}=\\{0\\}$ implies
$(T^{\prime}(N^{\prime}))^{\bot}=\\{0\\}$.

Since $N^{\prime}$ is self-dual, $T^{\prime}$ is adjointable [14]. Let
$S=((T^{\prime})^{*}T^{\prime})^{1/2}:N^{\prime}\to N^{\prime}$. Then
$S(N^{\prime})^{\perp}=\\{0\\}$. Indeed, if $f\in
S(N^{\prime})^{\perp}_{N^{\prime}}$, then for $f^{\prime}=S(f)$, one has

$0=\langle
S(f^{\prime}),f\rangle=\langle(T^{\prime})^{*}T^{\prime}f,f\rangle=\langle
T^{\prime}f,T^{\prime}f\rangle$

and $f=0$. Denote the norm closure of $S(N^{\prime})$ by $K$. If $J$ denotes
the inclusion of $K$ in $N^{\prime}$ then $J^{\\#}:N^{\prime}\to K^{\prime}$
is the above isometric isomorphism (since $A$ is monotone complete, all
modules are with a Hilbert dual). We can apply Lemmas 7 and 8 because
$N^{\prime}$ is self-dual and $K^{\bot}\subseteq
S(N^{\prime})^{\perp}=\\{0\\}$.

Since $\langle T^{\prime}f,T^{\prime}f\rangle=\langle Sf,Sf\rangle$, the map
$U_{0}:S(N^{\prime})\to M^{\prime}$ defined by $U_{0}(Sf):=T^{\prime}f$ is an
isometric embedding and extends by continuity to an isometric embedding
$U:K\to M^{\prime}$ (because $M^{\prime}$ is complete). Note that

$U(K)^{\bot}\subseteq(U_{0}(S(N^{\prime})))^{\bot}=(T^{\prime}(N^{\prime}))^{\bot}=\\{0\\}.$

So, since $M^{\prime}$ is self-dual, we can apply Lemmas 7 and 8. We obtain in
this way an isometric isomorphism $U^{\\#}:M^{\prime}\to K^{\prime}$. Thus
$(J^{\\#})^{-1}\circ U^{\\#}:M^{\prime}\cong K^{\prime}\cong N^{\prime}$ is
the desired isometric isomorphism. ∎

######  Corollary 10.

Let $A$ be a monotone complete $C^{*}$-algebra, let $M$, $N$ be self-dual
Hilbert $C^{*}$-modules over $A$, and let $T:M\to N$ be an operator with
$\operatorname{Ker}(T)=\\{0\\}$. Suppose that $(T(M))^{\perp\perp}=N$. Then
$M\cong N$.

###### Proof.

In the self-dual case $\Lambda_{N}$ is an isomorphism and
$T^{\prime}=\Lambda_{M}\circ T^{*}\circ(\Lambda_{N})^{-1}$ (see (7)). If
$T(M)^{\bot}=\\{0\\}$, then $\mathop{\rm Ker}\nolimits(T^{*})=\\{0\\}$.
Indeed, suppose that $T^{*}(n)=0$, then $\langle T^{*}n,m\rangle=0$ for any
$m\in M$, so $\langle n,Tm\rangle=0$ for any $m\in M$, i.e. $n\bot T(M)$.
Hence $n=0$ and $\mathop{\rm Ker}\nolimits(T^{*})=\\{0\\}$. Thus, the property
$\operatorname{Ker}(T^{\prime})=\\{0\\}$ follows from the other conditions in
this case. ∎

##  5\. Decompositions into direct sums

Here we explain that Corollary 3 in [2], which was based on an erroneous
statement, is nevertheless true. As a bonus, we extend it from the case of
Hilbert $C^{*}$-modules over $W^{*}$ algebras to that over monotone complete
$C^{*}$-algebras.

The next statement is known for Hilbert $C^{*}$-modules over $W^{*}$-algebras
([2], [12, Lemma 3.6.1]).

######  Lemma 11.

Let $L$ be a self-dual Hilbert $C^{*}$-module, and let $K\subseteq L$ be a
Hilbert $C^{*}$-submodule with a Hilbert dual. Then we have the orthogonal
decomposition: $L=K^{\bot\bot}\oplus K^{\bot}.$

###### Proof.

For the inclusion $i:K\subseteq L$, consider the Banach adjoint map
$R=i^{\prime}:L^{\prime}\to K^{\prime}$ of self-dual Hilbert modules. The map
$R$ is given by $R(f)(k)=f(i(k))$, $f\in L^{\prime}$, $k\in K$, and satisfies
the formula

(8) $R(\widehat{l}\,)(k)=\widehat{l}(i(k))=\langle
i(k),l\rangle_{L}=\langle\widehat{i(k)},\widehat{l}\rangle_{L^{\prime}},\quad
l\in L,k\in K.$

As both $K^{\prime}$ and $L$ are self-dual, $R$ admits the Hilbert
$C^{*}$-module adjoint morphism $R^{*}:K^{\prime}\to L^{\prime}\cong L$.

We claim that $R^{*}R$ is a projection, and that $\mathop{\rm
Ker}\nolimits(R)=\mathop{\rm Ker}\nolimits(R^{*}R)$.

The first claim would follow if we show that
$RR^{*}=\operatorname{id}_{K^{\prime}}$. Note that $R^{*}$ satisfies $\langle
f,R(\widehat{l}\,)\rangle_{K^{\prime}}=\langle
R^{*}(f),\widehat{l}\,\rangle_{L^{\prime}}$, $f\in K^{\prime}$, $l\in L$.
Taking here $f=\widehat{k}$, we get
$\langle\widehat{k},R(\widehat{l}\,)\rangle_{K^{\prime}}=\langle
R^{*}(\widehat{k}),\widehat{l}\,\rangle_{L^{\prime}}$. Compairing this with
(8), we see that $R^{*}(\widehat{k})=\widehat{i(k)}$ for any $k\in K$. Next,

$\displaystyle\langle RR^{*}(\widehat{k}),\widehat{r}\rangle_{K^{\prime}}$
$\displaystyle=$ $\displaystyle\langle
R^{*}(\widehat{k}),R^{*}(\widehat{r})\rangle_{L^{\prime}}$ $\displaystyle=$
$\displaystyle\langle\widehat{i(k)},\widehat{i(r)}\rangle_{L^{\prime}}=\langle
i(k),i(r)\rangle_{L}$ $\displaystyle=$ $\displaystyle\langle
k,r\rangle_{K}=\langle\widehat{k},\widehat{r}\rangle_{K^{\prime}}$

for any $k,r\in K$, hence $\langle
RR^{*}(\widehat{k})-\widehat{k},\widehat{r}\rangle_{K^{\prime}}=0$, which
implies that $RR^{*}(\widehat{k})=\widehat{k}$ for any $k\in K$. Now let $f\in
K^{\prime}$. Then

$\langle RR^{*}(f),\widehat{k}\rangle_{K^{\prime}}=\langle
f,RR^{*}(\widehat{k})\rangle_{K^{\prime}}=\langle
f,\widehat{k}\rangle_{K^{\prime}},$

or, alternatively, $RR^{*}(f)(k)=f(k)$ for any $k\in K$, hence
$RR^{*}=\operatorname{id}_{K^{\prime}}$.

For the second claim, it is clear that $\mathop{\rm Ker}\nolimits
R\subset\mathop{\rm Ker}\nolimits R^{*}R$. To prove the opposite inclusion
suppose that $R(x)\neq 0$ for some $x\in L^{\prime}=L$. Then $\langle
R(x),R(x)\rangle=\langle R^{*}R(x),x\rangle\neq 0$, hence, $R^{*}Rx\neq 0$.

It remains to observe that $K^{\bot}=\mathop{\rm Ker}\nolimits(R)$ and thus is
orthogonally complemented. ∎

######  Corollary 12.

Suppose that $M$ and $L$ are self-dual modules over a monotone complete
algebra, $T:M\to L$ is an (adjointable) morphism with $\mathop{\rm
Ker}\nolimits(T)=\\{0\\}$. Then $M\cong(T(M))^{\bot\bot}$.

###### Proof.

Recall that all Hilbert $C^{*}$-modules over a monotone complete
$C^{*}$-algebra have Hilbert dual. By Lemma 11, we have
$L=(T(M))^{\bot\bot}\oplus(T(M))^{\bot}$. Set $N:=(T(M))^{\bot\bot}\supseteq
T(M)$. Then $N=(T(M))^{\bot\bot}$ is a direct orthogonal summand of a self-
dual module and thus a self-dual module itself. Consider $T$ as a map from $M$
to $N$ (we keep the same notation for this morphism). Then $T:M\to N$
satisfies the assumptions of Corollary 10, whence $M\cong N^{\prime}$. Self-
duality of $N$ finishes the proof. ∎

Note that, for an adjointable operator $F:M\to N$, we have

$\displaystyle(\mathop{\rm Im}\nolimits F)^{\bot}$ $\displaystyle=$
$\displaystyle\\{y\in N\colon\langle Fx,y\rangle=0\mbox{ for all }x\in M\\}$
$\displaystyle=$ $\displaystyle\\{y\in N\colon\langle x,F^{*}y\rangle=0\mbox{
for all }x\in M\\}$ $\displaystyle=$ $\displaystyle\\{y\in N\colon
F^{*}y=0\\}$ $\displaystyle=$ $\displaystyle\mathop{\rm Ker}\nolimits F^{*}.$

Hence,

(9) $(\overline{\mathop{\rm Im}\nolimits F})^{\bot}=(\mathop{\rm Im}\nolimits
F)^{\bot}=\mathop{\rm Ker}\nolimits F^{*}\qquad\mbox{for an adjointable }F.$

From this observation and Lemma 11 we obtain the following statement.

######  Corollary 13.

Suppose that $L$ is a self-dual Hilbert $C^{*}$-module and let $K\subseteq L$
be a Hilbert $C^{*}$-submodule with a Hilbert dual that is the kernel of a
morphism $F:L\to M$, $K=\mathop{\rm Ker}\nolimits F$. Since $L$ is self-dual,
$F$ is adjointable and we require $R:=\overline{\mathop{\rm Im}\nolimits
F^{*}}$ to be a module with Hilbert dual. Then $K=K^{\bot\bot}$ and we have
the orthogonal decomposition: $L=K^{\bot\bot}\oplus K^{\bot}=K\oplus
K^{\bot}.$

###### Proof.

By (9), $K=\mathop{\rm Ker}\nolimits F=(\overline{\mathop{\rm Im}\nolimits
F^{*}})^{\bot}=R^{\bot}$. By Lemma 11 we have orthogonal decompositions
$L=R^{\bot\bot}\oplus R^{\bot}$ and $L=K^{\bot\bot}\oplus K^{\bot}$. From the
first decomposition we obtain $L=K\oplus K^{\bot}$. Then from the second one
we obtain $K=K^{\bot\bot}$. ∎

##  6\. Application to Fredholm operators

Until now, all direct sums of submodules were orthogonal, but now we have to
pay atention to the difference between Banach direct sums and Hilbert ones.
Our notation $\oplus$ means the Banach direct sum, and it should be clear from
the context, whether the summands are orthogonaal or not.

The following definition of Mishchenko-Fomenko $C^{*}$-Fredholm operator [13]
plays an important role in noncommutative geometry.

######  Definition 14.

An adjointable operator $F:M\to N$ between countably generated Hilbert
$C^{*}$-modules over a unital $C^{*}$-algebra $A$ is called $A$-_Fredholm_ if
there exist direct sum decompositions ${M}={M}_{0}\oplus{M}_{1}$ and
${N}={N}_{0}\oplus{N}_{1}$ such that

(10) $F=\left(\begin{matrix}F_{0}&0\\\
0&F_{1}\end{matrix}\right):{M}_{0}\oplus{M}_{1}\to{N}_{0}\oplus{N}_{1},$

$F_{0}:{M}_{0}\cong{N}_{0}$, $F_{1}:{M}_{1}\to{N}_{1}$, where ${M}_{1}$ and
${N}_{1}$ are projective finitely generated modules and $M_{0}$ is orthogonal
to $M_{1}$. Its _index_ is $\mathrm{index}(F)=[M_{1}]-[N_{1}]\in K_{0}(A)$,
where $[L]$ denotes the $K$-theory class of the projective finitely generated
module $L$ over $A$.

Index is stable under $A$-compact perturbations ([13], see also [12]), in
particular,

$\mathrm{index}\left(\begin{matrix}F_{0}&0\\\
0&F_{1}\end{matrix}\right)=\mathrm{index}\left(\begin{matrix}F_{0}&0\\\
0&0\end{matrix}\right).$

For the latter operator the index can be calculated like in the classical
(Hilbert space) case: $[\mathop{\rm Ker}\nolimits]-[\mathrm{Coker}]$. But for
developing of elements of Hodge theory in this setting (see e.g. [2]) one
needs to have something like this honestly (i.e. not after an $A$-compact
perturbation). The next statement solves this problem.

######  Theorem 15.

Suppose that $M$ and $N$ are countably generated $A$-modules and $F:M\to N$ is
an $A$-Fredholm operator, where $A$ is a monotone complete $C^{*}$-algebra.
Then $\mathop{\rm Ker}\nolimits F$ and $(\mathop{\rm Im}\nolimits F)^{\bot}$
are finitely generated projective $A$-modules and

$\mathrm{index}(F)=[\mathop{\rm Ker}\nolimits F]-[(\mathop{\rm Im}\nolimits
F)^{\bot}]\in K_{0}(A).$

###### Proof.

For the decomposition from the above definition, we can assume that the sum on
the right-hand side of (10) is also orthogonal. Indeed, since
$F_{0}:{M}_{0}\cong{N}_{0}$ is an adjointable isomorphism, ${N}_{0}$ has an
orthogonal complement: $N={N}_{0}\oplus({N}_{0})^{\bot}$ (see e.g. [12,
Theorem 2.3.3], one can write down the corresponding orthogonal projection
explicitly: $P=F(F^{*}F)^{-1}F^{*}$). Then, for

$\widetilde{F}=\left(\begin{matrix}F_{0}&0\\\
0&(1-P)F_{1}\end{matrix}\right):{M}_{0}\oplus{M}_{1}\to{N}_{0}\oplus({N}_{0})^{\bot},$

we have $({N}_{0})^{\bot}\cong{N}_{1}$, so
$\mathrm{index}(F)=[M_{1}]-[N_{1}]=[M_{1}]-[({N}_{0})^{\bot}]=\mathrm{index}(\widetilde{F})$,
$\mathop{\rm Ker}\nolimits\widetilde{F}=\mathop{\rm Ker}\nolimits F$,
$\mathop{\rm Im}\nolimits\widetilde{F}=\mathop{\rm Im}\nolimits F$. So, we
assume that ${N}={N}_{0}\oplus{N}_{1}$ with $N_{0}\perp N_{1}$.

Let $x=x_{0}+x_{1}$, $x_{0}\in{M}_{0}$, $x_{1}\in{M}_{1}$ and $F(x)=0$, so
$0=F_{0}(x_{0})+F_{1}(x_{1})\in{N}_{0}\oplus{N}_{1}$. Thus $F_{0}(x_{0})=0$,
$F_{1}(x_{1})=0$, so $x_{0}=0$ and $x\in{M}_{1}$. Thus $\mathop{\rm
Ker}\nolimits F=\mathop{\rm Ker}\nolimits F_{1}\subseteq{M}_{1}$.

Consider $F_{1}:M_{1}\to N_{1}$. Then $M_{1}$ is self-dual, because it is
finitely generated projective module. Since $A$ is monotone complete, all
modules have Hilbert dual, in particular, so is $\mathop{\rm Ker}\nolimits
F_{1}$. Therefore Corollary 13 gives $M_{1}=\mathop{\rm Ker}\nolimits
F_{1}\oplus(\mathop{\rm Ker}\nolimits F_{1})^{\bot}$. Denote
$M_{2}=(\mathop{\rm Ker}\nolimits F_{1})^{\bot}$. Also we have $\mathop{\rm
Im}\nolimits F=N_{0}\oplus F(M_{2})$.

Note that $(\mathop{\rm Im}\nolimits F)^{\bot}\subseteq N_{1}$ is the kernel
of the adjoint operator: $(\mathop{\rm Im}\nolimits F)^{\bot}=\mathop{\rm
Ker}\nolimits F^{*}$ (see (9)). Hence $(\mathop{\rm Im}\nolimits F)^{\bot}$ is
an orthogonal complementable finitely generated projective submodule of
$N_{1}$. Evidently, $F(M_{2})\subseteq((\mathop{\rm Im}\nolimits
F)^{\bot})^{\bot}_{N_{1}}$ and we obtain the following form for $F$:

$\left(\begin{smallmatrix}F_{0}&0&0\\\ 0&F_{2}&0\\\
0&0&0\end{smallmatrix}\right):M_{0}\oplus(M_{2}\oplus\mathop{\rm Ker}\nolimits
F)\to N_{0}\oplus((\mathop{\rm Im}\nolimits
F)^{\bot})^{\bot}_{N_{1}}\oplus(\mathop{\rm Im}\nolimits F)^{\bot},$

where $F_{2}=F_{1}|_{M_{2}}$, and all summands, on both sides, are orthogonal.
By Corollary 12, $M_{2}\cong((\mathop{\rm Im}\nolimits
F)^{\bot})^{\bot}_{N_{1}}$ and

$\displaystyle\mathrm{index}(F)$ $\displaystyle=$
$\displaystyle[M_{1}]-[N_{1}]$ $\displaystyle=$
$\displaystyle[M_{2}]+[\mathop{\rm Ker}\nolimits F]-[((\mathop{\rm
Im}\nolimits F)^{\bot})^{\bot}_{N_{1}}]-[(\mathop{\rm Im}\nolimits F)^{\bot}]$
$\displaystyle=$ $\displaystyle[\mathop{\rm Ker}\nolimits F]-[(\mathop{\rm
Im}\nolimits F)^{\bot}]$

as desired. ∎

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