On the Unbounded Picture of $KK$-Theory

On the Unbounded Picture of $\boldsymbol{KK}$-Theory††This paper is a
contribution to the Special Issue on Noncommutative Manifolds and their
Symmetries in honour of Giovanni Landi. The full collection is available at
https://www.emis.de/journals/SIGMA/Landi.html

Jens KAAD

J. Kaad

Department of Mathematics and Computer Science, The University of Southern
Denmark,
Campusvej 55, DK-5230 Odense M, Denmark<EMAIL_ADDRESS>https://portal.findresearcher.sdu.dk/en/persons/kaad

Received October 22, 2019, in final form August 05, 2020; Published online
August 22, 2020

In the founding paper on unbounded $KK$-theory it was established by Baaj and
Julg that the bounded transform, which associates a class in $KK$-theory to
any unbounded Kasparov module, is a surjective homomorphism (under a
separability assumption). In this paper, we provide an equivalence relation on
unbounded Kasparov modules and we thereby describe the kernel of the bounded
transform. This allows us to introduce a notion of topological unbounded
$KK$-theory, which becomes isomorphic to $KK$-theory via the bounded
transform. The equivalence relation is formulated entirely at the level of
unbounded Kasparov modules and consists of homotopies together with an extra
degeneracy condition. Our degenerate unbounded Kasparov modules are called
spectrally decomposable since they admit a decomposition into a part with
positive spectrum and a part with negative spectrum.

$KK$-theory; unbounded $KK$-theory; equivalence relations; bounded transform

19K35; 58B34

Dedicated to Gianni Landi on the occasion
of his 60th birthday

##  1 Introduction

$KK$-theory, as introduced by Kasparov in [16, 18], has its roots in the
Brown–Douglas–Fillmore extension theory of commutative $C^{*}$-algebras [5],
and in Atiyah’s axiomatization of properties of elliptic operators on
manifolds, [1]. But $KK$-theory extends far beyond the context of commutative
$C^{*}$-algebras and has become an important tool for accessing the algebraic
topology of $C^{*}$-algebras with applications ranging from Elliott’s
classification program to key aspects of index theory.

In practice, explicit classes in $KK$-theory often come from unbounded
operators acting on Hilbert $C^{*}$-modules and these unbounded operators
constitute the main ingredient in the unbounded picture of $KK$-theory. The
cycles in the unbounded picture are called unbounded Kasparov modules and are
often of a differential geometric origin with prototypical examples being
Dirac operators (in the case of $K$-homology) or multiplication operators by
symbols of Dirac operator (in the case of $K$-theory). In the unbounded
picture, the relationship between $KK$-theory and the program of Connes on
noncommutative geometry is in fact immediate: spectral triples are, without
any further modifications, examples of unbounded Kasparov modules [6, 7].

The passage from the unbounded picture to the more commonly encountered
bounded picture of $KK$-theory is furnished by the bounded transform which
turns an unbounded Kasparov module into a class in $KK$-theory via the smooth
approximation of the sign function given by $t\mapsto
t\big{(}1+t^{2}\big{)}^{-1/2}$ (and the functional calculus on Hilbert
$C^{*}$-modules). The richness of the unbounded picture is witnessed by a
fundamental theorem of Baaj and Julg stating that any class in $KK$-theory
comes from an unbounded Kasparov module so that the bounded transform is in
fact a surjection (under a mild separability condition) [2]. See also [20, 24]
for other interesting and related lifting results.

In this paper, we construct an equivalence relation on unbounded Kasparov
modules which captures the kernel of the bounded transform and this
equivalence relation is “geometric” in the sense that it can be formulated
without any reference to the bounded picture of $KK$-theory. Our equivalence
relation relies on an extra degeneracy condition on unbounded Kasparov modules
together with a notion of homotopies using families of unbounded Kasparov
modules parametrized by the unit interval. Our degeneracy condition on an
unbounded Kasparov module is formulated in terms of a spectral decomposition
of the unbounded operator in question building on the simple observation that
the phase of an unbounded selfadjoint and regular operator with strictly
positive spectrum is equal to the identity operator. It was pointed out to us
by the referee that spectral decomposability is related to the concept of weak
degeneracy from [8, Definition 3.1] and that it can be formulated
alternatively as a condition on the bounded transform. We are of course
grateful for these comments.

In summary, we introduce a notion of topological unbounded $KK$-theory and
show that topological unbounded $KK$-theory is isomorphic to $KK$-theory via
the bounded transform. We hereby give an affirmative answer to the question
raised by Deeley, Goffeng, and Mesland on page $3$ in [8]. In fact, it turns
out that our topological unbounded $KK$-theory is independent of the choice of
a dense $*$-subalgebra of a $C^{*}$-algebra as long as this $*$-subalgebra is
countably generated. It would thus be interesting to investigate the
relationship between topological unbounded $KK$-theory and the bordism group
introduced in [8], where the equivalence relation comes from Hilsum’s notion
of bordisms of unbounded Kasparov modules [9].

The idea for proving the injectivity of the bounded transform is to apply the
lifting procedure introduced by Baaj and Julg to a homotopy at the bounded
level, where the homotopic elements are bounded transforms of some given
unbounded Kasparov modules. The problem with this idea is that it might very
well happen that the unbounded homotopy achieved from this process connects
two unbounded Kasparov module that are very different from the original ones.
To see what might happen, notice that an unbounded Kasparov module could
satisfy that commutators extend to _bounded_ operators for all elements in a
dense subalgebra whereas a Baaj–Julg lift can always be chosen such that
commutators extend to _compact_ operators for all elements in a (perhaps
different) dense subalgebra. In this paper, we resolve this problem by
studying the concept of a spectrally decomposable unbounded Kasparov module
which provides a type of degenerate elements at the unbounded level, related
to the idea that a strictly positive unbounded operator should not contain any
topological information.

After this paper was written and made available on the arXiv a strengthening
of our results was obtained by van den Dungen and Mesland [26]. Among other
things these authors were able to prove that a spectrally decomposable
unbounded Kasparov module is in fact null-homotopic at the unbounded level,
see [26, Corollary 4.9]. Notably, using ideas related to [2, 20, 24], van den
Dungen and Mesland also establish a lifting result regarding homotopies of
Kasparov modules which is stronger than the lifting results applied in the
present text, see [26, Theorem 2.9].

We emphasize that the word topological is a keyword in connection with our
definition of unbounded $KK$-theory. In other approaches to unbounded
$KK$-theory, the aim is to find an interesting equivalence relation which
captures geometric content at the level of unbounded Kasparov modules while
still admitting explicit formulae for the interior Kasparov product. The
geometric content which could be valuable in this respect relates to the
asymptotic behaviour of eigenvalues and the spectral metric aspects of
noncommutative geometry. Certainly, this geometric content is invisible from a
topological point of view and thus in particular from the point of view of
topological unbounded $KK$-theory. The delicate questions on the geometric
nature of unbounded $KK$-theory are part of ongoing research on the unbounded
Kasparov product and the interested reader can consult the following
(incomplete list of) references: [4, 11, 14, 15, 19, 20, 23, 24].

###  1.1 Standing assumptions

Throughout this text $A=A_{0}\oplus A_{1}$ and $B=B_{0}\oplus B_{1}$ will be
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-algebras with $A$ separable and $B$
$\sigma$-unital (meaning that $B$ has a countable approximate identity). We
moreover fix a norm-dense $\mathbb{Z}/2\mathbb{Z}$-graded $*$-subalgebra
$\mathscr{A}\subseteq A$, which we require to be generated as a $*$-algebra by
some countable subset $\\{x_{j}\,|\,j\in\mathbb{N}\\}\subseteq\mathscr{A}$.
Remark that the grading on $\mathscr{A}$ is compatible with the grading on $A$
meaning that the $\mathbb{Z}/2\mathbb{Z}$-grading operator $\gamma_{A}\colon
A\to A$ (being $1$ on $A_{0}$ and $-1$ on $A_{1}$) induces the
$\mathbb{Z}/2\mathbb{Z}$-grading operator
$\gamma_{\mathscr{A}}\colon\mathscr{A}\to\mathscr{A}$.

##  2 Kasparov modules and $\boldsymbol{KK}$-theory

In this section we give a brief summary of the main definitions concerning
Kasparov modules and $KK$-theory. For more details the reader can consult the
following references: [3, 10, 18]. The commutators appearing in this section
are all graded commutators. For a $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondence $X$ from $A$ to $B$ we usually suppress the even
$*$-homomorphism $\pi_{X}\colon A\to\mathbb{L}(X)$, which determines the left
action of $A$ on $X$ (where $\mathbb{L}(X)$ denotes the
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-algebra of bounded adjointable
operators on $X$).

######  Definition 2.1.

A _Kasparov module_ from $A$ to $B$ is a pair $(X,F)$ where $X$ is a countably
generated $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence from $A$ to
$B$ and $F\colon X\to X$ is an odd bounded adjointable operator such that

$a(F-F^{*}),\ a\big{(}F^{2}-1\big{)},\ [F,a]\colon\ X\to X$

are compact operators for all $a\in A$.

A Kasparov module $(X,F)$ from $A$ to $B$ is _degenerate_ when

$a(F-F^{*})=a\big{(}F^{2}-1\big{)}=[F,a]=0\qquad\mbox{for all}\quad a\in A.$

######  Definition 2.2.

Two Kasparov modules $(X_{0},F_{0})$ and $(X_{1},F_{1})$ from $A$ to $B$ are
_unitarily equivalent_ when there exists an even unitary isomorphism of
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondences $U\colon X_{0}\to
X_{1}$ such that $UF_{0}U^{*}=F_{1}$. In this case we write
$(X_{0},F_{0})\sim_{u}(X_{1},F_{1})$. Remark that $U$ (by definition) has to
intertwine the left actions as well so that
$U\pi_{X_{0}}(a)U^{*}=\pi_{X_{1}}(a)$ for all $a\in A$.

When given a $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-algebra $C$ and an even
$*$-homomorphism $\beta\colon B\to C$ we may “change the base” of a
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence $X$ from $A$ to $B$.
Indeed, we may consider $C$ as a $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondence from $B$ to $C$ and form the interior tensor product
$X\widehat{\otimes}_{\beta}C$ which is a $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondence from $A$ to $C$. Any bounded adjointable operator
$T\colon X\to X$ then induces a bounded adjointable operator
$T\widehat{\otimes}_{\beta}1\colon X\widehat{\otimes}_{\beta}C\to
X\widehat{\otimes}_{\beta}C$ and this operation yields an even
$*$-homomorphism
$\mathbb{L}(X)\to\mathbb{L}\big{(}X\widehat{\otimes}_{\beta}C\big{)}$, see for
example [21, Chapter 4].

######  Definition 2.3.

Two Kasparov modules $(X_{0},F_{0})$ and $(X_{1},F_{1})$ both from $A$ to $B$
are _homotopic_ when there exists a Kasparov module $(X,F)$ from $A$ to
$C([0,1],B)$ such that

$\big{(}X\widehat{\otimes}_{{\rm ev}_{0}}B,F\widehat{\otimes}_{{\rm
ev}_{0}}1\big{)}\sim_{u}(X_{0},F_{0})\qquad\mbox{and}\qquad\big{(}X\widehat{\otimes}_{{\rm
ev}_{1}}B,F\widehat{\otimes}_{{\rm ev}_{1}}1\big{)}\sim_{u}(X_{1},F_{1}),$

where ${\rm ev}_{t}\colon C([0,1],B)\to B$ denotes the even $*$-homomorphism
given by evaluation at $t\in[0,1]$. In this case we write
$(X_{0},F_{0})\sim_{h}(X_{1},F_{1})$.

It is a non-trivial fact that the above homotopy relation is an equivalence
relation and it can be difficult to find a record of this result in the
standard literature on $KK$-theory. We state the result as a proposition here
and notice that the proof is very similar to the proof given in the unbounded
setting later on, see Proposition 4.7 and in particular Lemma 4.6 which can be
applied to prove the transitivity of the relation $\sim_{h}$.

######  Proposition 2.4.

Homotopy of Kasparov modules is an equivalence relation.

######  Definition 2.5.

_$KK$ -theory_ from $A$ to $B$ consists of Kasparov modules from $A$ to $B$
modulo homotopies. $KK$-theory from $A$ to $B$ is denoted by $KK(A,B)$.

We may form the _direct sum_ of two Kasparov modules $(X,F)$ and
$(X^{\prime},F^{\prime})$ from $A$ to $B$ and this is the Kasparov module from
$A$ to $B$ given by

$(X,F)+(X^{\prime},F^{\prime}):=(X\oplus X^{\prime},F\oplus F^{\prime}).$

The _zero module_ is the Kasparov module $(0,0)$ from $A$ to $B$.

We quote the following two results from [3, Chapter 17]:

######  Proposition 2.6.

Any degenerate Kasparov module from $A$ to $B$ is homotopic to the zero
module.

######  Theorem 2.7.

The direct sum operation and the zero module provide $KK$-theory from $A$ to
$B$ with the structure of an abelian group.

##  3 Unbounded Kasparov modules

In this section we review the main results of the paper [2], which can be
regarded as the founding paper on unbounded $KK$-theory. We recall that a
symmetric unbounded operator $D\colon\operatorname{Dom}(D)\to X$ acting on a
Hilbert $C^{*}$-module $X$ over $B$ is selfadjoint and regular when the
operators $D\pm i\colon\operatorname{Dom}(D)\to X$ are surjective, see [21,
Lemmas 9.7 and 9.8]. Unbounded selfadjoint and regular operators admit a
continuous functional calculus as developed in [27, 28], see also [21, Theorem
10.9]. Notice that in our convention all unbounded operators are densely
defined.

######  Definition 3.1.

An _unbounded Kasparov module_ from $\mathscr{A}$ to $B$ is a pair $(X,D)$,
where $X$ is a countably generated $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondence from $A$ to $B$ and $D\colon\operatorname{Dom}(D)\to X$
is an odd unbounded selfadjoint and regular operator such that

  1. 1)

each $a\in\mathscr{A}$ preserves the domain of $D$ and the graded commutator
$[D,a]\colon\operatorname{Dom}(D)\to X$ extends to a bounded operator
$d(a)\colon X\to X$;

  2. 2)

the operator $a\cdot(i+D)^{-1}\colon X\to X$ is compact for all
$a\in\mathscr{A}$.

An unbounded Kasparov module $(X,D)$ from $\mathscr{A}$ to $B$ is _Lipschitz
regular_ when the graded commutator

$[|D|,a]\colon\ \operatorname{Dom}(D)\to X$

extends to a bounded operator on $X$ for all $a\in\mathscr{A}$.

For an unbounded Kasparov module $(X,D)$ it follows automatically that
$d(a)\colon X\to X$ is adjointable for all $a\in\mathscr{A}$ and we have the
formulae

$d(a)^{*}=\begin{cases}-d(a^{*})&\text{for}\ a\text{ even},\\\
d(a^{*})&\text{for}\ a\text{ odd}.\end{cases}$

######  Definition 3.2.

The _direct sum_ of two unbounded Kasparov modules $(X,D)$ and
$(X^{\prime},D^{\prime})$ from $\mathscr{A}$ to $B$ is the unbounded Kasparov
module

$(X,D)+(X^{\prime},D^{\prime}):=(X\oplus X^{\prime},D\oplus D^{\prime})$

from $\mathscr{A}$ to $B$. The _zero module_ from $\mathscr{A}$ to $B$ is the
unbounded Kasparov module $(0,0)$.

It was proved in [2] that every unbounded Kasparov module represents a class
in $KK$-theory:

######  Theorem 3.3.

Suppose that $(X,D)$ is an unbounded Kasparov module from $\mathscr{A}$ to
$B$. Then the pair $\big{(}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{)}$ is a
Kasparov module from $A$ to $B$. In particular, we have an associated class
$\big{[}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{]}\in KK(A,B)$ in $KK$-theory.

We refer to the assignment
$(X,D)\mapsto\big{[}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{]}$ which sends an
unbounded Kasparov module from $\mathscr{A}$ to $B$ to its associated class in
$KK(A,B)$ as the _Baaj–Julg bounded transform_.

It turns out that every class in $KK$-theory can be represented by an
unbounded Kasparov module. This result is also due to Baaj and Julg [2]. The
standing hypothesis that $\mathscr{A}\subseteq A$ is countably generated as a
$*$-algebra plays a crucial role in the proof.

Notice that a bounded positive operator $\Delta\colon X\to X$ is strictly
positive precisely when the image of $\Delta\colon X\to X$ is dense in $X$ and
in this case $\Delta^{-1}\colon\operatorname{Im}(\Delta)\to X$ is an unbounded
positive and regular operator, see [21, Lemma 10.1]. In particular,
$\operatorname{Dom}\big{(}\Delta^{-1}\big{)}:=\operatorname{Im}(\Delta)$.

######  Theorem 3.4.

Suppose that $\mathscr{A}\subseteq A$ is a norm-dense and countably generated
$\mathbb{Z}/2\mathbb{Z}$-graded $*$-subalgebra. Suppose moreover that $(X,F)$
is a Kasparov module from $A$ to $B$ with $F=F^{*}$ and $F^{2}=1$. Then there
exists an even strictly positive compact operator $\Delta\colon X\to X$ such
that

  1. $1)$

the operator $F$ preserves the domain of $\Delta^{-1}$ and
$\big{[}F,\Delta^{-1}\big{]}=0$ on
$\operatorname{Dom}\big{(}\Delta^{-1}\big{)}$;

  2. $2)$

each $a\in\mathscr{A}$ preserves the domain of $\Delta^{-1}$ and
$\big{[}a,\Delta^{-1}\big{]}\colon\operatorname{Dom}\big{(}\Delta^{-1}\big{)}\to
X$ extends to a compact operator on $X$;

  3. $3)$

for each $a\in\mathscr{A}$, the image of the graded commutator $[F,a]$ is
contained in $\operatorname{Dom}\big{(}\Delta^{-1}\big{)}$ and
$\Delta^{-1}[F,a]\colon X\to X$ is a compact operator.

Moreover, with
$D:=\Delta^{-1}F\colon\operatorname{Dom}\big{(}\Delta^{-1}\big{)}\to X$ we
have that $(X,D)$ is an unbounded Kasparov module from $\mathscr{A}$ to $B$
satisfying that

$\big{[}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{]}=[X,F]$

in $KK(A,B)$. In particular, it holds that the Baaj–Julg bounded transform is
surjective.

In the context of the above theorem, it is worthwhile to notice that the
graded commutator
$\big{[}\Delta^{-1}F,a\big{]}\colon\operatorname{Dom}\big{(}\Delta^{-1}\big{)}\to
X$ does in fact extend to a _compact_ operator for all $a\in\mathscr{A}$ and
that $\big{(}i+\Delta^{-1}F\big{)}^{-1}\colon X\to X$ is compact even though
the separable $C^{*}$-algebra $A$ need not be unital.

##  4 Equivalence relations on unbounded Kasparov modules

In this section we introduce an equivalence relation on unbounded Kasparov
modules and use this equivalence relation to construct the topological
unbounded $KK$-theory. A key ingredient in our approach is the following
notion of a degenerate cycle:

######  Definition 4.1.

An unbounded Kasparov module $(X,D)$ from $\mathscr{A}$ to $B$ is _spectrally
decomposable_ when there exists an orthogonal projection $P\colon X\to X$ such
that

  1. 1)

$P$ preserves the domain of $D$ and $DP-PD=0$ on $\operatorname{Dom}(D)$;

  2. 2)

$DP$ and $D(P-1)$ are unbounded positive operators;

  3. 3)

$aP=Pa$ for all even elements $a\in A$ and $aP=(1-P)a$ for all odd elements
$a\in A$;

  4. 4)

$\gamma P=(1-P)\gamma$, where $\gamma\colon X\to X$ is the
$\mathbb{Z}/2\mathbb{Z}$-grading operator on $X$.

Notice that it follows from Definition 4.1$(1)$ and the regularity and
selfadjointness of $D$ that $DP$ and $D(P-1)$ are selfadjoint and regular as
well.

For a spectrally decomposable unbounded Kasparov module $(X,D)$ with spectral
decomposition given by an orthogonal projection $P\colon X\to X$, we apply the
notation

$D_{+}:=DP\colon\ \operatorname{Dom}(DP)\to X\qquad\text{and}\qquad
D_{-}:=D(P-1)\colon\ \operatorname{Dom}(D(P-1))\to X$

for the associated unbounded positive and regular operators.

We remark that $\operatorname{Dom}(D_{+})=\\{\xi\in
X\,|\,P\xi\in\operatorname{Dom}(D)\\}$ and
$\operatorname{Dom}(D_{-})=\\{\xi\in
X\,|\,(P-1)\xi\in\operatorname{Dom}(D)\\}$ so that we obtain the
decomposition:

$\operatorname{Dom}(D)=\operatorname{Dom}(D_{+})\cap\operatorname{Dom}(D_{-}),\qquad
D=D_{+}-D_{-}.$

We refer to [12, Section 6] for more information on products of unbounded
selfadjoint and regular operators with bounded adjointable operators.

######  Lemma 4.2.

Suppose that an unbounded Kasparov module $(X,D)$ from $\mathscr{A}$ to $B$ is
spectrally decomposable. Then the class
$\big{[}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{]}\in KK(A,B)$ is equal to zero.

###### Proof.

By $(3)$ and $(4)$ in Definition 4.1, the pair $(X,2P-1)$ is a degenerate
Kasparov module from $A$ to $B$ and by Proposition 2.6 it therefore suffices
to show that $\big{(}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{)}$ is homotopic to
$(X,2P-1)$. We let $a\in A$ and show that $D\big{(}1+D^{2}\big{)}^{-1/2}Pa-Pa$
and $D\big{(}1+D^{2}\big{)}^{-1/2}(1-P)a-(P-1)a$ are compact operators on $X$.
By $(1)$ in Definition 4.1, it holds that

$1+D_{+}^{2}=\big{(}1+D^{2}\big{)}P+1-P,\qquad
1+D_{-}^{2}=\big{(}1+D^{2}\big{)}(1-P)+P,$

implying the identities

$\big{(}1+D_{+}^{2}\big{)}^{-1/2}P=\big{(}1+D^{2}\big{)}^{-1/2}P,\qquad\big{(}1+D_{-}^{2}\big{)}^{-1/2}(1-P)=\big{(}1+D^{2}\big{)}^{-1/2}(1-P).$

We thus conclude that

$\displaystyle
D\big{(}1+D^{2}\big{)}^{-1/2}P=D_{+}\big{(}1+D_{+}^{2}\big{)}^{-1/2},$
$\displaystyle
D\big{(}1+D^{2}\big{)}^{-1/2}(1-P)=-D_{-}\big{(}1+D_{-}^{2}\big{)}^{-1/2}.$

Using $(2)$ in Definition 4.1, we see that
$D_{+}-\big{(}1+D_{+}^{2}\big{)}^{1/2}\colon\operatorname{Dom}(D_{+})\to X$
and $D_{-}-\big{(}1+D_{-}^{2}\big{)}^{1/2}\colon\operatorname{Dom}(D_{-})\to
X$ extend to bounded adjointable operators on $X$. But this implies that

$\displaystyle D\big{(}1+D^{2}\big{)}^{-1/2}Pa-
Pa=\big{(}D_{+}-\big{(}1+D_{+}^{2}\big{)}^{1/2}\big{)}\big{(}1+D_{+}^{2}\big{)}^{-1/2}Pa$
$\displaystyle\hphantom{D\big{(}1+D^{2}\big{)}^{-1/2}Pa-
Pa}{}=\big{(}D_{+}-\big{(}1+D_{+}^{2}\big{)}^{1/2}\big{)}\big{(}1+D^{2}\big{)}^{-1/2}Pa$

and

$\displaystyle
D\big{(}1+D^{2}\big{)}^{-1/2}(1-P)a-(P-1)a=-\big{(}D_{-}-\big{(}1+D_{-}^{2})^{1/2}\big{)}\big{(}1+D_{-}^{2}\big{)}^{-1/2}(1-P)a$
$\displaystyle\hphantom{D\big{(}1+D^{2}\big{)}^{-1/2}(1-P)a-(P-1)a}{}=-\big{(}D_{-}-\big{(}1+D_{-}^{2}\big{)}^{1/2}\big{)}\big{(}1+D^{2}\big{)}^{-1/2}(1-P)a$

are compact operators on $X$. ∎

######  Remark 4.3.

As pointed out to us by the referee, an alternative proof of Lemma 4.2 can be
given using a result of Skandalis [25, Lemma 11]. Indeed, with notation as in
Lemma 4.2 we obtain that

$(2P-1)D\big{(}1+D^{2}\big{)}^{-1/2}+D\big{(}1+D^{2}\big{)}^{-1/2}(2P-1)=2(D_{+}+D_{-})\big{(}1+D^{2}\big{)}^{-1/2}\geq
0.$

It then follows from [25, Lemma 11] that the Kasparov modules
$\big{(}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{)}$ and $(X,2P-1)$ are operator
homotopic and hence also homotopic.

In fact, as was also remarked by the referee, even more is true: an unbounded
Kasparov module $(X,D)$ from $\mathscr{A}$ to $B$ is spectrally decomposable
if and only if there exists a selfadjoint unitary operator $F\colon X\to X$
satisfying that $(X,F)$ is a degenerate Kasparov module together with the
conditions $FD\big{(}1+D^{2}\big{)}^{-1/2}-D\big{(}1+D^{2}\big{)}^{-1/2}F=0$
and $FD\big{(}1+D^{2}\big{)}^{-1/2}+D\big{(}1+D^{2}\big{)}^{-1/2}F\geq 0$. As
such, spectral decomposability can be viewed as a condition on the bounded
transform of $(X,D)$.

######  Definition 4.4.

Two unbounded Kasparov modules $(X,D)$ and $(X^{\prime},D^{\prime})$ are
_unitarily equivalent_ when there exists an even unitary isomorphism of
$C^{*}$-correspondences $U\colon X\to X^{\prime}$ such that
$UDU^{*}=D^{\prime}$.

Unitary equivalence of unbounded Kasparov modules is indeed an equivalence
relation and we denote it by $\sim_{u}$.

Suppose now that $C$ is an extra $\mathbb{Z}/2\mathbb{Z}$-graded
$\sigma$-unital $C^{*}$-algebra and that $\beta\colon B\to C$ is an even
$*$-homomorphism. As in Section 2 we have the change of base operation given
by the interior tensor product of $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondences: $X\widehat{\otimes}_{\beta}C$. Moreover, any
unbounded selfadjoint and regular operator $D\colon\operatorname{Dom}(D)\to X$
induces an unbounded selfadjoint and regular operator
$D\widehat{\otimes}_{\beta}1\colon\operatorname{Dom}\big{(}D\widehat{\otimes}_{\beta}1\big{)}\to
X\widehat{\otimes}_{\beta}C$, which has resolvents given by

$\big{(}D\widehat{\otimes}_{\beta}1+i\lambda\big{)}^{-1}=(D+i\lambda)^{-1}\widehat{\otimes}_{\beta}1\qquad\text{for
all}\quad\lambda\in\mathbb{R}\setminus\\{0\\}.$

In particular, if $a(i+D)^{-1}\colon X\to X$ is a compact operator for some
$a\in A$, then

$a\big{(}D\widehat{\otimes}_{\beta}1+i\big{)}^{-1}=a(D+i)^{-1}\widehat{\otimes}_{\beta}1\colon\
X\widehat{\otimes}_{\beta}C\to X\widehat{\otimes}_{\beta}C$

is a compact operator as well, see [21, Proposition 4.7]. These observations
allow us to formulate our notion of homotopy at the level of unbounded
Kasparov modules:

######  Definition 4.5.

Two unbounded Kasparov modules $(X_{0},D_{0})$ and $(X_{1},D_{1})$ both from
$\mathscr{A}$ to $B$ are _homotopic_ when there exists an unbounded Kasparov
module $(X,D)$ from $\mathscr{A}$ to $C([0,1],B)$ such that

$\big{(}X\widehat{\otimes}_{{\rm ev}_{0}}B,D\widehat{\otimes}_{{\rm
ev}_{0}}1\big{)}\sim_{u}(X_{0},D_{0})\qquad\mbox{and}\qquad\big{(}X\widehat{\otimes}_{{\rm
ev}_{1}}B,D\widehat{\otimes}_{{\rm ev}_{1}}1\big{)}\sim_{u}(X_{1},D_{1}),$

where ${\rm ev}_{t}\colon C([0,1],B)\to B$ denotes the even $*$-homomorphism
given by evaluation at $t\in[0,1]$. In this case we write
$(X_{0},D_{0})\sim_{h}(X_{1},D_{1})$.

Before proving that homotopies of unbounded Kasparov modules yields an
equivalence relation it is worthwhile to spend a little time on a glueing
construction for $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondences.
Consider two countably generated $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondences $X$ and $X^{\prime}$ both from $A$ to $C([0,1],B)$ and
suppose that

$U\colon\ X\widehat{\otimes}_{{\rm ev}_{1}}B\to
X^{\prime}\widehat{\otimes}_{{\rm ev}_{0}}B$

is an even unitary isomorphism of $C^{*}$-correspondences. This data gives
rise to a $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence
$X\times_{U}X^{\prime}$ from $A$ to $C([0,1],B)$ obtained by glueing $X$ and
$X^{\prime}$ using the unitary $U$ to identify the fibres sitting at $1$ and
$0$, respectively. Indeed, we may define

$X\times_{U}X^{\prime}:=\big{\\{}(\xi,\xi^{\prime})\in X\oplus
X^{\prime}\,|\,U\big{(}\xi\widehat{\otimes}_{{\rm
ev}_{1}}1\big{)}=\xi^{\prime}\widehat{\otimes}_{{\rm ev}_{0}}1\big{\\}}$

and endow this set with the vector space structure, left action of $A$ and
$\mathbb{Z}/2\mathbb{Z}$-grading inherited from the direct sum $X\oplus
X^{\prime}$. To construct the right action of $C([0,1],B)$ and the inner
product on $X\times_{U}X^{\prime}$, we introduce the even $*$-endomorphisms
$\beta_{0},\beta_{1}\colon C([0,1],B)\to C([0,1],B)$ by
$\beta_{0}(f)(t)=f(t/2)$ and $\beta_{1}(f)(t)=f((t+1)/2)$ for all $f\in
C([0,1],B)$ and $t\in[0,1]$. We put

$\displaystyle\langle(\xi,\xi^{\prime}),(\eta,\eta^{\prime})\rangle(t):=\begin{cases}\langle\xi,\eta\rangle(2t)&\text{for}\
t\in[0,1/2],\\\ \langle\xi^{\prime},\eta^{\prime}\rangle(2t-1)&\text{for}\
t\in[1/2,1],\end{cases}\qquad\text{and}$ $\displaystyle(\xi,\xi^{\prime})\cdot
f:=(\xi\cdot\beta_{0}(f),\xi^{\prime}\cdot\beta_{1}(f)),$

for all $(\xi,\xi^{\prime}),(\eta,\eta^{\prime})\in X\times_{U}X^{\prime}$,
$f\in C([0,1],B)$ and $t\in[0,1]$.

######  Lemma 4.6.

The $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence
$X\times_{U}X^{\prime}$ from $A$ to $C([0,1],B)$ is countably generated.
Moreover, if $K\colon X\to X$ and $K^{\prime}\colon X^{\prime}\to X^{\prime}$
are compact operators such that $U(K\widehat{\otimes}_{{\rm
ev}_{1}}1)U^{*}=K^{\prime}\widehat{\otimes}_{{\rm ev}_{0}}1$, then the direct
sum $K\oplus K^{\prime}\colon X\oplus X^{\prime}\to X\oplus X^{\prime}$
restricts to a compact operator $K\oplus K^{\prime}\colon
X\times_{U}X^{\prime}\to X\times_{U}X^{\prime}$.

###### Proof.

We start by constructing an adjointable isometry $S\colon
X\times_{U}X^{\prime}\to\ell^{2}(\mathbb{N},C([0,1],B))$, where
$\ell^{2}(\mathbb{N},C([0,1],B))$ denotes the standard module over
$C([0,1],B)$. Since $B$ is $\sigma$-unital by our standing assumptions this
will imply that $X\times_{U}X^{\prime}$ is countably generated.

Since $X$ and $X^{\prime}$ are countably generated over $C([0,1],B)$ it
follows by Kasparov’s stabilization theorem, [17, Theorem 2], that we may find
unitary isomorphisms of Hilbert $C^{*}$-modules

$\displaystyle W\colon\
X\oplus\ell^{2}(\mathbb{N},C([0,1],B))\to\ell^{2}(\mathbb{N},C([0,1],B))\qquad\text{and}$
$\displaystyle W^{\prime}\colon\
X^{\prime}\oplus\ell^{2}(\mathbb{N},C([0,1],B))\to\ell^{2}(\mathbb{N},C([0,1],B)).$

We let $V\colon\ell^{2}(\mathbb{N},B)\to\ell^{2}(\mathbb{N},B)$ denote the
unique unitary isomorphism of Hilbert $C^{*}$-modules, which makes the diagram
here below commute:

$\begin{CD}\big{(}X\widehat{\otimes}_{{\rm
ev}_{1}}B\big{)}\oplus\ell^{2}(\mathbb{N},B)@>{U\oplus
1}>{}>\big{(}X^{\prime}\widehat{\otimes}_{{\rm
ev}_{0}}B\big{)}\oplus\ell^{2}(\mathbb{N},B)\\\ @V{\cong}V{}V@V{}V{\cong}V\\\
\big{(}X\oplus\ell^{2}(\mathbb{N},C([0,1],B))\big{)}\widehat{\otimes}_{{\rm
ev}_{1}}B&&\big{(}X^{\prime}\oplus\ell^{2}(\mathbb{N},C([0,1],B))\big{)}\widehat{\otimes}_{{\rm
ev}_{0}}B\\\ @V{W\widehat{\otimes}_{{\rm
ev}_{1}}1}V{}V@V{}V{W^{\prime}\widehat{\otimes}_{{\rm ev}_{0}}1}V\\\
\ell^{2}(\mathbb{N},C([0,1],B))\widehat{\otimes}_{{\rm
ev}_{1}}B&&\ell^{2}(\mathbb{N},C([0,1],B))\widehat{\otimes}_{{\rm ev}_{0}}B\\\
@V{\cong}V{}V@V{}V{\cong}V\\\
\ell^{2}(\mathbb{N},B)@>{}>{V}>\ell^{2}(\mathbb{N},B).\end{CD}$

We specify that the lower vertical isomorphisms are induced by
$f\widehat{\otimes}_{{\rm ev}_{i}}b\mapsto f(i)\cdot b$, for $i\in\\{0,1\\}$,
and the top vertical unitary isomorphisms come from the distributivity of the
interior tensor product together with the lower vertical isomorphisms.

The notation $\iota\colon X\to X\oplus\ell^{2}(\mathbb{N},C([0,1],B))$ and
$\iota^{\prime}\colon X^{\prime}\to
X^{\prime}\oplus\ell^{2}(\mathbb{N},C([0,1],B))$ refers to the standard
inclusions given on matrix form as $\left(\begin{smallmatrix}1\\\
0\end{smallmatrix}\right)$. We define our adjointable isometry $S\colon
X\times_{U}X^{\prime}\to\ell^{2}(\mathbb{N},C([0,1],B))$ by the formula

$S(\xi,\xi^{\prime})(t):=\begin{cases}V(W\iota\xi)(2t)&\text{for}\
t\in[0,1/2],\\\ (W^{\prime}\iota^{\prime}\xi^{\prime})(2t-1)&\text{for}\
t\in[1/2,1].\end{cases}$

for all $(\xi,\xi^{\prime})\in X\times_{U}X^{\prime}$ and $t\in[0,1]$. We
leave it to the reader to verify that $S$ is well-defined (in particular that
$V(W\iota\xi)(1)=(W^{\prime}\iota^{\prime}\xi^{\prime})(0)$). The adjoint of
$S$ is given explicitly by

$S^{*}(f)=(\iota^{*}W^{*}\sigma_{0}(f),(\iota^{\prime})^{*}(W^{\prime})^{*}\beta_{1}(f)),$

where $\sigma_{0}(f)(t)=V^{*}f(t/2)$ and $\beta_{1}(f)=f((t+1)/2)$ for all
$f\in\ell^{2}(C([0,1],B))$ and $t\in[0,1]$.

The compactness of $K\oplus K^{\prime}\colon X\times_{U}X^{\prime}\to
X\times_{U}X^{\prime}$ is equivalent to the compactness of $S(K\oplus
K^{\prime})S^{*}\colon\ell^{2}(\mathbb{N},C([0,1],B))\to\ell^{2}(\mathbb{N},C([0,1],B))$.
The compact operators on the standard module $\ell^{2}(\mathbb{N},C([0,1],B))$
can be identified with the operator norm continuous maps from $[0,1]$ to the
compact operators on $\ell^{2}(\mathbb{N},B)$. Using this identification, we
compute that

$S(K\oplus K^{\prime})S^{*}(t)=\begin{cases}VL(2t)V^{*}&\text{for}\
t\in[0,1/2],\\\ L^{\prime}(2t-1)&\text{for}\ t\in[1/2,1],\end{cases}$

where $L:=W\iota K\iota^{*}W^{*}$ and
$L^{\prime}:=W^{\prime}\iota^{\prime}K^{\prime}(\iota^{\prime})^{*}(W^{\prime})^{*}\colon\ell^{2}(\mathbb{N},C([0,1],B))\to\ell^{2}(\mathbb{N},C([0,1],B))$.
Since $L$ and $L^{\prime}$ are compact by assumption this proves the
compactness of $K\oplus K^{\prime}\colon X\times_{U}X^{\prime}\to
X\times_{U}X^{\prime}$. ∎

######  Proposition 4.7.

Homotopy of unbounded Kasparov modules is an equivalence relation.

###### Proof.

Reflexivity: For an unbounded Kasparov module $(X,D)$ from $\mathscr{A}$ to
$B$, we have that $(X,D)\sim_{h}(X,D)$ via the unbounded Kasparov module
$(C([0,1],X),E)$ from $\mathscr{A}$ to $C([0,1],B)$, where
$E\colon\operatorname{Dom}(E)\to C([0,1],X)$ is the unbounded selfadjoint and
regular operator defined by

$\displaystyle\operatorname{Dom}(E):=\\{\xi\in
C([0,1],X)\,|\,\xi(t)\in\operatorname{Dom}(D)\ \text{for all}\ t\in[0,1]$
$\displaystyle\hphantom{\operatorname{Dom}(E):=\\{}{}\text{and}\ t\mapsto
D\xi(t)\ \text{is continuous}\\},$ $\displaystyle
E(\xi)(t):=D\xi(t)\qquad\text{for all }t\in[0,1].$

Symmetry: Suppose that two unbounded Kasparov modules $(X_{0},D_{0})$ and
$(X_{1},D_{1})$ from $\mathscr{A}$ to $B$ are homotopic via the unbounded
Kasparov module $(X,D)$ from $\mathscr{A}$ to $C([0,1],B)$. Define the even
$*$-automorphism $\beta\colon C([0,1],B)\to C([0,1],B)$ by
$\beta(f)(t)=f(1-t)$ for all $f\in C([0,1],B)$ and $t\in[0,1]$. Then it holds
that $(X_{1},D_{1})$ and $(X_{0},D_{0})$ are homotopic via the unbounded
Kasparov module
$(X\widehat{\otimes}_{\beta}C([0,1],B),D\widehat{\otimes}_{\beta}1)$ from
$\mathscr{A}$ to $C([0,1],B)$.

Transitivity: Suppose that $(X_{0},D_{0})$, $(X_{1},D_{1})$ and
$(X_{1},D_{1}^{\prime})$ are unbounded Kasparov modules from $\mathscr{A}$ to
$B$ such that $(X_{0},D_{0})\sim_{h}(X_{1},D_{1})$ and
$(X_{1},D_{1})\sim_{h}(X_{1}^{\prime},D_{1}^{\prime})$ via the unbounded
Kasparov modules $(X,D)$ and $(X^{\prime},D^{\prime})$, respectively. Let us
choose an even unitary isomorphism of $C^{*}$-correspondences

$U\colon\ X\widehat{\otimes}_{{\rm ev}_{1}}B\to
X^{\prime}\widehat{\otimes}_{{\rm ev}_{0}}B$

such that $U(D\widehat{\otimes}_{{\rm
ev}_{1}}1)U^{*}=D^{\prime}\widehat{\otimes}_{{\rm ev}_{0}}1$. We study the
associated $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence
$X\times_{U}X^{\prime}$ from $A$ to $C([0,1],B)$ and notice that
$X\times_{U}X^{\prime}$ is countably generated by Lemma 4.6. We define the odd
unbounded selfadjoint and regular operator
$D^{\prime\prime}\colon\operatorname{Dom}(D^{\prime\prime})\to
X\times_{U}X^{\prime}$ by

$\displaystyle\operatorname{Dom}(D^{\prime\prime}):=\\{(\xi,\xi^{\prime})\in
X\times_{U}X^{\prime}\,|\,\xi\in\operatorname{Dom}(D),\,\xi^{\prime}\in\operatorname{Dom}(D^{\prime})\\},$
$\displaystyle
D^{\prime\prime}(\xi,\xi^{\prime}):=(D\xi,D^{\prime}\xi^{\prime}).$

To see that $D^{\prime\prime}$ is indeed selfadjoint and regular, we notice
that $D^{\prime\prime}$ is symmetric and that the resolvents are given by

$(D^{\prime\prime}+\lambda i)^{-1}(\xi,\xi^{\prime})=\big{(}(D+\lambda
i)^{-1}(\xi),(D^{\prime}+\lambda i)^{-1}(\xi^{\prime})\big{)}$

for all $\lambda\in\mathbb{R}\setminus\\{0\\}$ and all $(\xi,\xi^{\prime})\in
X\times_{U}X^{\prime}$.

We claim that $(X\times_{U}X^{\prime},D^{\prime\prime})$ is an unbounded
Kasparov module from $\mathscr{A}$ to $C([0,1],B)$. It is indeed clear that
each $a\in\mathscr{A}$ preserves the domain of $D^{\prime\prime}$ and that the
graded commutator
$[D^{\prime\prime},a]\colon\operatorname{Dom}(D^{\prime\prime})\to
X\times_{U}X^{\prime}$ extends to a bounded adjointable operator on
$X\times_{U}X^{\prime}$. Moreover, it follows from Lemma 4.6 that
$a\cdot(i+D^{\prime\prime})^{-1}\colon X\times_{U}X^{\prime}\to
X\times_{U}X^{\prime}$ is a compact operator for all $a\in A$.

The unbounded Kasparov module $(X\times_{U}X^{\prime},D^{\prime\prime})$ from
$\mathscr{A}$ to $C([0,1],B)$ implements the homotopy from $(X_{0},D_{0})$ to
$(X_{1}^{\prime},D_{1}^{\prime})$ and this ends the proof of the proposition.
∎

Let us shortly discuss the notion of bounded perturbations of unbounded
Kasparov modules:

######  Definition and Proposition 4.8.

Let $(X,D)$ be an unbounded Kasparov module from $\mathscr{A}$ to $B$ and let
$T\colon X\to X$ be an odd bounded selfadjoint operator. Then the pair
$(X,D+T)$ is an unbounded Kasparov module from $\mathscr{A}$ to $B$ and
$(X,D)$ and $(X,D+T)$ are homotopic. We say that $(X,D+T)$ is a _bounded
perturbation_ of $(X,D)$.

###### Proof.

The domain of $D+T$ agrees with the domain of $D$ and
$D+T\colon\operatorname{Dom}(D)\to X$ is selfadjoint and regular by [27,
Example 1]. The commutator condition in Definition 3.1 is immediately verified
and the resolvent condition in Definition 3.1 follows from the resolvent
identity:

$a\cdot(i+D+T)^{-1}=a\cdot(i+D)^{-1}-a\cdot(i+D)^{-1}T(i+D+T)^{-1}.$

We thus have that $(X,D+T)$ is an unbounded Kasparov module from $\mathscr{A}$
to $B$. To see that $(X,D)$ and $(X,D+T)$ are homotopic we apply the unbounded
Kasparov module from $\mathscr{A}$ to $C([0,1],B)$ given by $(C([0,1],X),E)$
where the corresponding odd unbounded selfadjoint and regular operator is
defined by

$\displaystyle\operatorname{Dom}(E):=\\{\xi\in
C([0,1],X)\,|\,\xi(t)\in\operatorname{Dom}(D)\ \text{for all}\ t\in[0,1]$
$\displaystyle\hphantom{\operatorname{Dom}(E):=\\{}{}\text{and}\ t\mapsto
D\xi(t)\ \text{is continuous}\\},$ $\displaystyle E(\xi)(t):=(D+t\cdot
T)\xi(t)\qquad\text{for all }t\in[0,1].$ ∎

We remark that the above notion of bounded perturbations yields an equivalence
relation on unbounded Kasparov modules. We will not discuss this equivalence
relation any further at this point but refer the reader to [11] for more
details.

In this paper the relevant equivalence relation on unbounded Kasparov modules
is a stabilized version of homotopies of unbounded Kasparov modules, where we
are using spectrally decomposable unbounded Kasparov modules in the
stabilization procedure:

######  Definition 4.9.

Two unbounded Kasparov modules $(X_{0},D_{0})$ and $(X_{1},D_{1})$ from
$\mathscr{A}$ to $B$ are _stably homotopic_ when there exist two spectrally
decomposable unbounded Kasparov modules $(Y_{0},D_{0}^{\prime})$ and
$(Y_{1},D_{1}^{\prime})$ from $\mathscr{A}$ to $B$ such that

$(X_{0}\oplus Y_{0},D_{0}\oplus D_{0}^{\prime})\sim_{h}(X_{1}\oplus
Y_{1},D_{1}\oplus D_{1}^{\prime}).$

In this case we write $(X_{0},D_{0})\sim_{sh}(X_{1},D_{1})$.

We remark that the relation $\sim_{sh}$ is indeed an equivalence relation and
that this can be verified using Proposition 4.7 together with the fact that
the direct sum of two spectrally decomposable unbounded Kasparov modules is
again spectrally decomposable.

######  Definition 4.10.

The _topological unbounded $KK$-theory_ from $\mathscr{A}$ to $B$ consists of
the unbounded Kasparov modules from $\mathscr{A}$ to $B$ modulo stable
homotopies, thus modulo the equivalence relation $\sim_{sh}$. The topological
unbounded $KK$-theory from $\mathscr{A}$ to $B$ is denoted by $UK^{{\rm
top}}(\mathscr{A},B)$.

We may equip the topological unbounded $KK$-theory from $\mathscr{A}$ to $B$
with the structure of a commutative monoid, where the addition is induced by
the direct sum operation from Definition 3.2 and the neutral element is the
class of the zero module $(0,0)$. We shall see in Section 6 that $UK^{{\rm
top}}(\mathscr{A},B)$ is in fact an abelian group.

The next result is a combination of Theorem 3.4 and Lemma 4.2 together with
the observation that the Baaj–Julg bounded transform is compatible with direct
sums and homotopies of unbounded Kasparov modules.

######  Theorem 4.11.

The Baaj–Julg bounded transform
$(X,D)\mapsto\big{[}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{]}$ induces a well-
defined surjective homomorphism

$\mathcal{F}\colon\ UK^{{\rm top}}(\mathscr{A},B)\to KK(A,B),$

which we also refer to as the Baaj–Julg bounded transform.

The main result of this paper is that the surjective homomorphism
$\mathcal{F}\colon UK^{{\rm top}}(\mathscr{A},B)\to KK(A,B)$ is in fact an
isomorphism. In particular, it holds that $UK^{{\rm top}}(\mathscr{A},B)$ is
independent of the norm-dense $\mathbb{Z}/2\mathbb{Z}$-graded $*$-subalgebra
$\mathscr{A}\subseteq A$ as long as $\mathscr{A}$ is countably generated. This
will be proved in Section 7.

##  5 Lipschitz regularity and invertibility

We shall now see that given a class in topological unbounded $KK$-theory one
may always choose a Lipschitz regular representative $(X,D)$ with the extra
property that the unbounded selfadjoint and regular operator
$D\colon\operatorname{Dom}(D)\to X$ is invertible (so that $D^{-1}\colon X\to
X$ is a bounded adjointable operator with image equal to the domain of $D$).

We start with Lipschitz regularity:

######  Proposition 5.1.

Let $r\in(0,1/2)$ and suppose that $(X,D)$ is an unbounded Kasparov module
from $\mathscr{A}$ to $B$. Then $\big{(}X,D\big{(}1+D^{2}\big{)}^{-r}\big{)}$
is a Lipschitz regular unbounded Kasparov module from $\mathscr{A}$ to $B$ and
it holds that

$\big{(}X,D\big{(}1+D^{2}\big{)}^{-r}\big{)}\sim_{h}(X,D).$

###### Proof.

For each $t\in[0,1]$, define the functions

$f_{t}^{\pm}\colon\ \mathbb{R}\to\mathbb{C},\qquad f_{t}^{\pm}(x):=\big{(}i\pm
x\big{(}1+x^{2}\big{)}^{-tr}\big{)}^{-1}.$

We remark that $f_{t}^{\pm}\in C_{0}(\mathbb{R})$ and that the maps $[0,1]\to
C_{0}(\mathbb{R})$ given by $t\mapsto f_{t}^{\pm}$ are continuous with respect
to the supremum norm on $C_{0}(\mathbb{R})$. Notice in this respect that we
have the estimates $|f_{t}^{\pm}(x)|\leq 2^{r}|x|^{2r-1}$ whenever $|x|\geq 1$
and $t\in[0,1]$.

For each $a\in A$, we thus have that

$a\cdot\big{(}i\pm D\big{(}1+D^{2}\big{)}^{-tr}\big{)}^{-1}=a\cdot
f_{t}^{\pm}(D)\colon\ X\to X$

are compact operators for all $t\in[0,1]$ and that the maps
$[0,1]\to\mathbb{L}(X)$ given by $t\mapsto\big{(}i\pm
D\big{(}1+D^{2}\big{)}^{-tr}\big{)}^{-1}$ are continuous in operator norm.

For $p\in(0,1/2]$, we are going to apply the integral formula

$\displaystyle\big{(}1+D^{2}\big{)}^{-p}=\frac{\sin(p\pi)}{\pi}\int_{0}^{\infty}\lambda^{-p}\big{(}1+\lambda+D^{2}\big{)}^{-1}d\lambda,$
(5.1)

where the integrand is continuous in operator norm and the integral converges
absolutely in operator norm.

Let $a\in\mathscr{A}$ be homogeneous. For each $t\in[0,1]$, the domain of $D$
is a core for $D\big{(}1+D^{2}\big{)}^{-tr}$ and for each
$\xi\in\operatorname{Dom}(D)$, we compute the graded commutator

$\displaystyle\big{[}D\big{(}1+D^{2}\big{)}^{-tr},a\big{]}(\xi)=\big{(}1+D^{2}\big{)}^{-tr}[D,a](\xi)+(-1)^{\deg(a)}\big{[}\big{(}1+D^{2}\big{)}^{-tr},a\big{]}D(\xi).$
(5.2)

The first term extends to the bounded adjointable operator
$\big{(}1+D^{2}\big{)}^{-tr}d(a)\colon X\to X$ and we remark that the map
$[0,1]\to\mathbb{L}(X)$ defined by $t\mapsto\big{(}1+D^{2}\big{)}^{-tr}d(a)$
is continuous with respect to the strict operator topology on $\mathbb{L}(X)$.
The second term in equation (5.2) is more complicated and, using the integral
formula in equation (5.1), we obtain the expression

$\displaystyle\big{[}\big{(}1+D^{2}\big{)}^{-tr},a\big{]}D$
$\displaystyle\qquad{}=(-1)^{\deg(a)+1}\frac{\sin(tr\pi)}{\pi}\int_{0}^{\infty}\lambda^{-tr}\big{(}1+\lambda+D^{2}\big{)}^{-1}d(a)D^{2}\big{(}1+\lambda+D^{2}\big{)}^{-1}d\lambda$
$\displaystyle\qquad\quad{}-\frac{\sin(tr\pi)}{\pi}\int_{0}^{\infty}\lambda^{-tr}D\big{(}1+\lambda+D^{2}\big{)}^{-1}d(a)D\big{(}1+\lambda+D^{2}\big{)}^{-1}d\lambda$
(5.3)

for all $t\in(0,1]$, where the left hand side only makes sense on
$\operatorname{Dom}(D)$. The right hand side does however make sense as a
bounded adjointable operator on $X$ and the operator norm is dominated by

$\frac{2\sin(tr\pi)\|d(a)\|}{\pi}\int_{0}^{\infty}\lambda^{-tr}(1+\lambda)^{-1}d\lambda=2\|d(a)\|.$

For each $t\in[0,1]$ we denote the bounded adjointable extension of
$\big{[}\big{(}1+D^{2}\big{)}^{-tr},a\big{]}D\colon\operatorname{Dom}(D)\allowbreak\to
X$ by $G_{t}(a)\colon X\to X$. Notice that $G_{0}(a)=0$ and that $G_{t}(a)$ is
given explicitly by the right hand side of equation (5.3) for $t\in(0,1]$. In
particular, it holds that $\|G_{t}(a)\|\leq 2\|d(a)\|$ for all $t\in[0,1]$.
Using the identity in equation (5.2) one may also verify that

$G_{t}(a)^{*}+(-1)^{\deg(a)}G_{t}(a^{*})=\big{[}d(a^{*}),\big{(}1+D^{2}\big{)}^{-tr}\big{]}$
(5.4)

for all $t\in[0,1]$. We claim that the map $[0,1]\to\mathbb{L}(X)$ given by
$t\mapsto G_{t}(a)$ is strictly continuous. Since $\mathscr{A}$ is a
$*$-algebra and since the right hand side of the identity in equation (5.4)
defines a strictly continuous map on $[0,1]$ we only need to show that the map
$t\mapsto G_{t}(a)\xi$ is norm continuous for every $\xi\in X$. In fact,
because of the uniform bound on the operator norm of $G_{t}(a)$ for
$t\in[0,1]$ and the density of $\operatorname{Dom}(D)$ in $X$, we may restrict
our attention to elements $\xi\in\operatorname{Dom}(D)$. But for
$\xi\in\operatorname{Dom}(D)$ the norm continuity of the map $t\mapsto
G_{t}(a)\xi$ follows since
$t\mapsto\big{[}\big{(}1+D^{2}\big{)}^{-tr},a\big{]}$ is strictly continuous
and since $G_{t}(a)\xi=\big{[}\big{(}1+D^{2}\big{)}^{-tr},a\big{]}D\xi$ for
all $t\in[0,1]$.

Our efforts so far can be summarized as follows: we have an unbounded Kasparov
module $(C([0,1],X),E)$ from $\mathscr{A}$ to $C([0,1],B)$, where the
unbounded selfadjoint and regular operator $E\colon\operatorname{Dom}(E)\to
C([0,1],X)$ is defined by

$\displaystyle\operatorname{Dom}(E):=\big{\\{}\xi\in
C([0,1],X)\,|\,\xi(t)\in\operatorname{Dom}\big{(}D(1+D^{2})^{-tr}\big{)}\
\text{for all}\ t\in[0,1]$
$\displaystyle\hphantom{\operatorname{Dom}(E):=\big{\\{}}{}\text{and}\
t\mapsto D\big{(}1+D^{2}\big{)}^{-tr}\xi(t)\ \text{is continuous}\big{\\}},$
$\displaystyle E(\xi)(t):=D\big{(}1+D^{2}\big{)}^{-tr}\xi(t)\qquad\text{for
all }t\in[0,1].$

In particular, we have that the unbounded Kasparov modules $(X,D)$ and
$\big{(}X,D\big{(}1+D^{2}\big{)}^{-r}\big{)}$ are homotopic.

To finish the proof of the proposition we only need to argue that the
unbounded Kasparov module $\big{(}X,D\big{(}1+D^{2}\big{)}^{-r}\big{)}$ is
Lipschitz regular. Let $a\in\mathscr{A}$ be homogeneous. Since the function
$x\mapsto|x|-x^{2}\big{(}1+x^{2}\big{)}^{-1/2}$ is bounded on $\mathbb{R}$, we
just have to prove that the graded commutator

$\big{[}D\big{(}1+D^{2}\big{)}^{-1/2}D\big{(}1+D^{2}\big{)}^{-r},a\big{]}\colon\
\operatorname{Dom}(D)\to X$

extends to a bounded operator on $X$. Notice in this respect that

$D\big{(}1+D^{2}\big{)}^{-1/2}D\big{(}1+D^{2}\big{)}^{-r}\colon\
\operatorname{Dom}\big{(}|D|^{1-2r}\big{)}\to X$

agrees with

$\big{|}D\big{(}1+D^{2}\big{)}^{-r}\big{|}=|D|\big{(}1+D^{2}\big{)}^{-r}\colon\
\operatorname{Dom}\big{(}|D|^{1-2r}\big{)}\to X$

up to a bounded selfadjoint operator and moreover that
$\operatorname{Dom}(D)\subseteq X$ is a core for
$\big{|}D\big{(}1+D^{2}\big{)}^{-r}\big{|}$. Since we already know that

$[D,a]D\big{(}1+D^{2}\big{)}^{-r-1/2}\ \text{and}\
D\big{(}1+D^{2}\big{)}^{-1/2}\big{[}D\big{(}1+D^{2}\big{)}^{-r},a\big{]}\colon\
\operatorname{Dom}(D)\to X$

extend to bounded operators on $X$ we are left with proving that

$D\big{[}\big{(}1+D^{2}\big{)}^{-1/2},a\big{]}D\big{(}1+D^{2}\big{)}^{-r}\colon\
\operatorname{Dom}(D)\to X$

extends to a bounded operator on $X$. But this follows since the integral
formula in equation (5.1) implies that

$\displaystyle
D\big{[}\big{(}1+D^{2}\big{)}^{-1/2},a\big{]}D\big{(}1+D^{2}\big{)}^{-r}$
$\displaystyle\qquad{}=(-1)^{\deg(a)+1}\frac{1}{\pi}\int_{0}^{\infty}\lambda^{-1/2}D\big{(}1+\lambda+D^{2}\big{)}^{-1}d(a)D^{2}\big{(}1+\lambda+D^{2}\big{)}^{-1}\big{(}1+D^{2}\big{)}^{-r}d\lambda$
$\displaystyle\qquad\quad{}-\frac{1}{\pi}\int_{0}^{\infty}\lambda^{-1/2}D^{2}\big{(}1+\lambda+D^{2}\big{)}^{-1}d(a)D\big{(}1+\lambda+D^{2}\big{)}^{-1}\big{(}1+D^{2}\big{)}^{-r}d\lambda,$
(5.5)

where the left hand side only makes sense on
$\operatorname{Dom}\big{(}|D|^{1-2r}\big{)}$, but the right hand side makes
sense as a bounded operator on $X$. Indeed, both of the integrals in equation
(5.5) have operator norm continuous integrands and converge absolutely because
of the operator norm estimates

$\displaystyle\big{\|}\lambda^{-1/2}D\big{(}1+\lambda+D^{2}\big{)}^{-1}d(a)D^{2}\big{(}1+\lambda+D^{2}\big{)}^{-1}\big{(}1+D^{2}\big{)}^{-r}\big{\|},$
$\displaystyle\big{\|}\lambda^{-1/2}D^{2}\big{(}1+\lambda+D^{2}\big{)}^{-1}d(a)D\big{(}1+\lambda+D^{2}\big{)}^{-1}\big{(}1+D^{2}\big{)}^{-r}\big{\|}$
$\displaystyle\qquad{}\leq\|d(a)\|\cdot\lambda^{-1/2}(1+\lambda)^{-1/2-r},$

which are valid for all $\lambda\in(0,\infty)$. ∎

We continue with invertibility:

######  Proposition 5.2.

Suppose that $(X,D)$ is an unbounded Kasparov module from $\mathscr{A}$ to
$B$. Then $(X,D)$ is homotopic to a Lipschitz regular unbounded Kasparov
module $(X^{\prime},D^{\prime})$ with $D^{\prime}\colon$
$\operatorname{Dom}(D^{\prime})\to X^{\prime}$ invertible.

###### Proof.

By Proposition 5.1 we may assume that $(X,D)$ is already Lipschitz regular.
Let us denote the $\mathbb{Z}/2\mathbb{Z}$-grading operator on $X$ by
$\gamma\colon X\to X$. Define the $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-correspondence $\widetilde{X}$ from $A$ to $B$ which agrees with $X$
as a Hilbert $C^{*}$-module over $B$, but $\widetilde{X}$ has grading operator
$-\gamma\colon\widetilde{X}\to\widetilde{X}$ and the left action of $A$ is
trivial. Then the unbounded Kasparov module $\big{(}\widetilde{X},-D\big{)}$
is homotopic to the zero module $(0,0)$. Indeed, we may consider the
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence
$C_{0}\big{(}(0,1],\widetilde{X}\big{)}$ from $A$ to $C([0,1],B)$ equipped
with the odd unbounded selfadjoint and regular operator
$E\colon\operatorname{Dom}(E)\to C_{0}\big{(}(0,1],\widetilde{X}\big{)}$
defined by

$\displaystyle\begin{split}&\operatorname{Dom}(E):=\big{\\{}\xi\in
C_{0}\big{(}(0,1],\widetilde{X}\big{)}\,|\,\xi(t)\in\operatorname{Dom}(D)\
\text{for all}\ t\in(0,1]\\\
&\hphantom{\operatorname{Dom}(E):=\big{\\{}}{}\text{and}\ t\mapsto-D\xi(t)\
\text{is continuous and vanishes at zero}\big{\\}},\\\
&E(\xi)(t):=-D\xi(t)\qquad\text{for all }t\in(0,1].\end{split}$

Since the left action of $A$ on $C_{0}\big{(}(0,1],\widetilde{X}\big{)}$ is
trivial we have that $\big{(}C_{0}\big{(}(0,1],\widetilde{X}\big{)},E\big{)}$
is an unbounded Kasparov module from $\mathscr{A}$ to $C([0,1],B)$ thus
realizing the homotopy from $\big{(}\widetilde{X},-D\big{)}$ to $(0,0)$. The
result of the proposition now follows from Proposition 4.8 by noting that
$(X,D)+\big{(}\widetilde{X},-D\big{)}=\big{(}X\oplus\widetilde{X},D\oplus(-D)\big{)}$
is a bounded perturbation of the Lipschitz regular unbounded Kasparov module

$\left(X\oplus\widetilde{X},\left(\begin{matrix}D&1\\\
1&-D\end{matrix}\right)\right)$

from $\mathscr{A}$ to $B$. Remark in this respect that the square of
$\left(\begin{smallmatrix}D&1\\\
1&-D\end{smallmatrix}\right)\colon\operatorname{Dom}(D)\oplus\operatorname{Dom}(D)\to
X\oplus\widetilde{X}$ is given by

$\left(\begin{matrix}D&1\\\
1&-D\end{matrix}\right)^{2}=\left(\begin{matrix}1+D^{2}&0\\\
0&1+D^{2}\end{matrix}\right)\colon\
\operatorname{Dom}\big{(}D^{2}\big{)}\oplus\operatorname{Dom}\big{(}D^{2}\big{)}\to
X\oplus\widetilde{X},$

which is indeed an invertible operator. ∎

##  6 Group structure

We show in this section that the commutative monoid $UK^{{\rm
top}}(\mathscr{A},B)$ is in fact an abelian group. This result relies on a
more general proposition stating (at least roughly speaking) that two
unbounded Kasparov modules $(X,D)$ and $(X,D^{\prime})$ from $\mathscr{A}$ to
$B$ are stably homotopic when the odd unbounded selfadjoint and regular
operators $D$ and $D^{\prime}$ have the same phase. This proposition will also
be of key importance later on when we prove the injectivity of the Baaj–Julg
bounded transform.

######  Definition 6.1.

The _inverse_ of an unbounded Kasparov modules $(X,D)$ from $\mathscr{A}$ to
$B$ is the unbounded Kasparov module from $\mathscr{A}$ to $B$ given by

$-(X,D):=\big{(}X^{{\rm op}},-D\big{)},$

where $X^{{\rm op}}$ agrees with $X$ as a Hilbert $C^{*}$-module over $B$, but
$X^{{\rm op}}$ is equipped with the opposite $\mathbb{Z}/2\mathbb{Z}$-grading
and with left action $\pi_{X^{{\rm op}}}\colon A\to\mathbb{L}\big{(}X^{{\rm
op}}\big{)}$ defined by

$\pi_{X^{{\rm op}}}(a)=\begin{cases}\pi_{X}(a)&\mbox{for}\ a\mbox{ even},\\\
-\pi_{X}(a)&\mbox{for}\ a\mbox{ odd}.\end{cases}$

######  Proposition 6.2.

Let $(X,D)$ and $(X,D^{\prime})$ be two unbounded Kasparov modules from
$\mathscr{A}$ to $B$ and suppose there exists an odd selfadjoint unitary
operator $F\colon X\to X$ such that

  1. $1)$

the operator $F$ preserves the domain of $D$ and of $D^{\prime}$ and the
commutators $FD-DF\colon$ $\operatorname{Dom}(D)\to X$ and
$FD^{\prime}-D^{\prime}F\colon\operatorname{Dom}(D^{\prime})\to X$ have
bounded extensions to $X$;

  2. $2)$

the unbounded operators $DF$ and $D^{\prime}F$ are bounded perturbations of
even unbounded positive and regular operators
$\Delta\colon\operatorname{Dom}(DF)\to X$ and
$\Delta^{\prime}\colon\operatorname{Dom}(D^{\prime}F)\to X$;

  3. $3)$

for each $a\in\mathscr{A}$, the image of the graded commutator $[F,a]$ is
contained in $\operatorname{Dom}(D)\cap\operatorname{Dom}(D^{\prime})$ and the
operators $D[F,a],D^{\prime}[F,a]\colon X\to X$ are bounded.

Then $(X,D)-(X,D^{\prime})$ is stably homotopic to the zero module $(0,0)$.

###### Proof.

We are going to show that

$(X,D)-(X,D^{\prime})=\big{(}X\oplus X^{{\rm op}},D\oplus(-D^{\prime})\big{)}$

is homotopic to a spectrally decomposable unbounded Kasparov module. We denote
the grading operator on $X$ by $\gamma\colon X\to X$ so that the grading
operator on $X^{{\rm op}}$ is given by $-\gamma\colon X\to X$.

We remark that the graded commutator $[F,a]\colon X\to X$ is compact for all
$a\in A$. Indeed, for any $a\in\mathscr{A}$ and $a^{\prime}\in A$, it follows
from assumption $(3)$ and the fact that $(X,D)$ is an unbounded Kasparov
module that

$a^{\prime}[F,a]=a^{\prime}(i+D)^{-1}(i+D)[F,a]\colon\ X\to X$

is a compact operator.

Define the orthogonal projections

$P_{+}:=\frac{1+F}{2},\ P_{-}:=\frac{1-F}{2}=1-P_{+}\colon\ X\to X$

and for each $t\in[0,1]$, define the unitary automorphisms of the Hilbert
$C^{*}$-module $X\oplus X$:

$\displaystyle U_{t}:=\left(\begin{matrix}\cos(t\pi/2)\cdot
P_{-}+P_{+}&\sin(t\pi/2)\cdot P_{-}\\\ -\sin(t\pi/2)\cdot
P_{-}&\cos(t\pi/2)\cdot P_{-}+P_{+}\end{matrix}\right)\qquad\text{and}$
$\displaystyle V_{t}:=\left(\begin{matrix}\cos(t\pi/2)&\sin(t\pi/2)\\\
-\sin(t\pi/2)&\cos(t\pi/2)\end{matrix}\right).$ (6.1)

We study the $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence $Y$ from
$A$ to $C([0,1],B)$, which as a Hilbert $C^{*}$-module over $C([0,1],B)$
agrees with $C([0,1],X\oplus X)$, but with left action $\pi\colon
A\to\mathbb{L}(Y)$ defined by putting

$\pi_{t}(a):=\begin{cases}U_{t}\left(\begin{matrix}\pi_{X}(a)&0\\\
0&\pi_{X}(a)\end{matrix}\right)U_{t}^{*}&\text{for}\ a\text{
even},\vspace{1mm}\\\ U_{t}\left(\begin{matrix}0&-\pi_{X}(a)\\\
-\pi_{X}(a)&0\end{matrix}\right)V_{t}U_{t}^{*}&\text{for}\ a\text{
odd},\end{cases}$

for all $t\in[0,1]$, and with grading operator $\sigma\colon Y\to Y$ defined
by

$\sigma_{t}:=\left(\begin{matrix}0&-\gamma\\\
-\gamma&0\end{matrix}\right)\colon\ X\oplus X\to X\oplus X\qquad\text{for all
}t\in[0,1].$

It is useful to notice that

$\sigma_{t}=U_{t}\left(\begin{matrix}0&-\gamma\\\
-\gamma&0\end{matrix}\right)V_{t}U_{t}^{*}\qquad\text{for all }t\in[0,1].$

Using assumption $(1)$, we define the unbounded selfadjoint and regular
operators

$\displaystyle D_{+}:=P_{+}DP_{+}\colon\ \operatorname{Dom}(DP_{+})\to
X,\qquad D_{-}:=P_{-}DP_{-}\colon\ \operatorname{Dom}(DP_{-})\to
X\qquad\text{and}$ $\displaystyle D^{\prime}_{+}:=P_{+}D^{\prime}P_{+}\colon\
\operatorname{Dom}(D^{\prime}P_{+})\to X,\qquad
D^{\prime}_{-}:=P_{-}D^{\prime}P_{-}\colon\ \operatorname{Dom}(DP_{-})\to X.$

One may then verify directly that

$\displaystyle D_{+}-D_{-}^{\prime}\colon\
\operatorname{Dom}(DP_{+})\cap\operatorname{Dom}(D^{\prime}P_{-})\to
X\qquad\text{and}$ $\displaystyle D_{-}-D_{+}^{\prime}\colon\
\operatorname{Dom}(DP_{-})\cap\operatorname{Dom}(D^{\prime}P_{+})\to X$

are unbounded selfadjoint and regular operators. The resolvent of
$D_{+}-D_{-}^{\prime}$ is for example given by

$(i\lambda+D_{+}-D_{-}^{\prime})^{-1}=P_{+}(i\lambda+D_{+})^{-1}+P_{-}(i\lambda-
D_{-}^{\prime})^{-1}$

for all $\lambda\in\mathbb{R}\setminus\\{0\\}$. Alternatively, we refer to
[13, Section 7] or [22, Theorem 4.5] for much more general results on sums of
unbounded selfadjoint and regular operators.

We claim that the pair
$\left(Y,\left(\begin{smallmatrix}D_{+}-D^{\prime}_{-}&0\\\
0&D_{-}-D_{+}^{\prime}\end{smallmatrix}\right)\right)$ is an unbounded
Kasparov module from $\mathscr{A}$ to $C([0,1],B)$ (where it is understood
that the unbounded selfadjoint and regular operator in question acts as
$(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime})$ in each fibre). For each
$t\in[0,1]$ we thus have the fibre
$(Y_{t},(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$ where $Y_{t}$ is
the countably generated $\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence
from $A$ to $B$ which agrees with $X\oplus X$ as a Hilbert $C^{*}$-module but
with grading operator $\sigma_{t}=\left(\begin{smallmatrix}0&-\gamma\\\
-\gamma&0\end{smallmatrix}\right)$ and with left action given by the even
$*$-homomorphism $\pi_{t}\colon A\to\mathbb{L}(X\oplus X)$.

We let $t\in[0,1]$ be given and compute for each even $a\in A$ that

$\displaystyle\pi_{t}(a)-\pi_{0}(a)=\frac{1}{2}\left(\begin{matrix}[F,a]&0\\\
0&[F,a]\end{matrix}\right)\left(\begin{matrix}1-\cos(t\pi/2)&-\sin(t\pi/2)\\\
\sin(t\pi/2)&1-\cos(t\pi/2)\end{matrix}\right)U_{t}^{*}$
$\displaystyle\hphantom{\pi_{t}(a)-\pi_{0}(a)=}{}=-\frac{1}{2}U_{t}\left(\begin{matrix}1-\cos(t\pi/2)&\sin(t\pi/2)\\\
-\sin(t\pi/2)&1-\cos(t\pi/2)\end{matrix}\right)\left(\begin{matrix}[F,a]&0\\\
0&[F,a]\end{matrix}\right)$

and for each odd $a\in A$ that

$\displaystyle\pi_{t}(a)-\pi_{0}(a)=-\frac{1}{2}\left(\begin{matrix}[F,a]&0\\\
0&[F,a]\end{matrix}\right)\left(\begin{matrix}-\sin(t\pi/2)&1-\cos(t\pi/2)\\\
1-\cos(t\pi/2)&\sin(t\pi/2)\end{matrix}\right)V_{t}U_{t}^{*}$
$\displaystyle\hphantom{\pi_{t}(a)-\pi_{0}(a)=}{}=-\frac{1}{2}U_{t}V_{t}^{*}\left(\begin{matrix}-\sin(t\pi/2)&1-\cos(t\pi/2)\\\
1-\cos(t\pi/2)&\sin(t\pi/2)\end{matrix}\right)\left(\begin{matrix}[F,a]&0\\\
0&[F,a]\end{matrix}\right).$

Since the graded commutator $[F,a]\colon X\to X$ is compact, this computation
implies that $\pi_{t}(a)-\pi_{0}(a)\colon X\oplus X\to X\oplus X$ is a compact
operator for all $a\in A$ and $t\in[0,1]$ and moreover that the associated map
$[0,1]\to\mathbb{K}(X\oplus X)$ is continuous in operator norm. Using
assumption $(3)$, the above computation also implies that
$\pi_{t}(a)-\pi_{0}(a)$ preserves the domain of
$(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime})$ for all $a\in\mathscr{A}$
and $t\in[0,1]$ (the image of $\pi_{t}(a)-\pi_{0}(a)$ is in fact contained in
this domain) and moreover that the graded commutator

$\displaystyle[(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}),\pi_{t}(a)-\pi_{0}(a)]\colon$
$\displaystyle\qquad{}\operatorname{Dom}(DP_{+}\cap
D^{\prime}P_{-})\oplus\operatorname{Dom}(DP_{-}\oplus D^{\prime}P_{+})\to
X\oplus X$

extends to a bounded operator on $X\oplus X$ (in fact each of the terms have
this property). The associated map $[0,1]\to\mathbb{L}(X\oplus X)$ given by

$t\mapsto\overline{[(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}),\pi_{t}(a)-\pi_{0}(a)]}$

is continuous in operator norm. These observations imply that
$(Y,(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$ is an unbounded
Kasparov module from $\mathscr{A}$ to $C([0,1],B)$ if and only if the fibre at
$t=1$ is an unbounded Kasparov module from $\mathscr{A}$ to $B$.

The fibre at $t=1$ is given by the pair
$(Y_{1},(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$, where $Y_{1}$ is
unitarily isomorphic to $X\oplus X^{{\rm op}}$ as a
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence via the unitary
operator $U_{1}\colon X\oplus X\to X\oplus X$ defined in equation (6.1).
Moreover, we have that

$\displaystyle
U_{1}(D\oplus(-D^{\prime}))U_{1}^{*}=\left(\begin{matrix}D_{+}-D^{\prime}_{-}&-P_{+}DP_{-}-P_{-}D^{\prime}P_{+}\\\
-P_{-}DP_{+}-P_{+}D^{\prime}P_{-}&D_{-}-D^{\prime}_{+}\end{matrix}\right)\colon$
$\displaystyle\qquad{}\big{(}\operatorname{Dom}(DP_{+})\cap\operatorname{Dom}(D^{\prime}P_{-})\big{)}\oplus\big{(}\operatorname{Dom}(D^{\prime}P_{+})\cap\operatorname{Dom}(DP_{-})\big{)}\to
X\oplus X.$

By assumption $(1)$ we know that the off-diagonal entries extend to bounded
operators on $X$ and it follows that the fibre at $t=1$ agrees with the
unbounded Kasparov module $(X\oplus X^{{\rm op}},D\oplus(-D^{\prime}))$ from
$\mathscr{A}$ to $B$ up to unitary equivalence and bounded perturbations. By
Proposition 4.8 this implies in particular that
$(Y_{1},(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$ is an unbounded
Kasparov module from $\mathscr{A}$ to $B$ which is homotopic to $(X\oplus
X^{{\rm op}},D\oplus(-D^{\prime}))=(X,D)-(X,D^{\prime})$.

We may thus conclude that
$(Y,(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$ is an unbounded
Kasparov module from $\mathscr{A}$ to $C([0,1],B)$.

By what has been achieved so far, we have reduced the proof of the proposition
to showing that the fibre of
$(Y,(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$ at $t=0$ is homotopic
to a spectrally decomposable unbounded Kasparov module from $\mathscr{A}$ to
$B$.

The unbounded Kasparov module sitting as the fibre at $t=0$ is given by the
pair $(Y_{0},(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime}))$, where
$Y_{0}$ agrees with $X\oplus X$ as a Hilbert $C^{*}$-module over $B$ but with
grading operator $\sigma_{0}=\left(\begin{smallmatrix}0&-\gamma\\\
-\gamma&0\end{smallmatrix}\right)$ and left action $\pi_{0}\colon
A\to\mathbb{L}(Y_{0})$ given by

$\pi_{0}(a)=\begin{cases}\left(\begin{matrix}\pi_{X}(a)&0\\\
0&\pi_{X}(a)\end{matrix}\right)&\text{for}\ a\text{ even},\vspace{1mm}\\\
\left(\begin{matrix}0&-\pi_{X}(a)\\\
-\pi_{X}(a)&0\end{matrix}\right)&\text{for}\ a\text{ odd}.\end{cases}$

By assumption $(2)$ we know that
$(D_{+}-D_{-}^{\prime})\oplus(D_{-}-D_{+}^{\prime})$ is a bounded perturbation
of the unbounded selfadjoint and regular operator

$\displaystyle(P_{+}\Delta P_{+}+P_{-}\Delta^{\prime}P_{-})\oplus(-P_{-}\Delta
P_{-}-P_{+}\Delta^{\prime}P_{+})\colon$
$\displaystyle\qquad{}\big{(}\operatorname{Dom}(DP_{+})\cap\operatorname{Dom}(D^{\prime}P_{-})\big{)}\oplus\big{(}\operatorname{Dom}(D^{\prime}P_{+})\cap\operatorname{Dom}(DP_{-})\big{)}\to
X\oplus X,$

where the upper diagonal entry and minus the lower diagonal entry are both
unbounded positive and regular operators. But this shows that the fibre at
$t=0$ is a bounded perturbation of a spectrally decomposable unbounded
Kasparov module. Indeed, the unbounded Kasparov module $(Y_{0},(P_{+}\Delta
P_{+}+P_{-}\Delta^{\prime}P_{-})\oplus(-P_{-}\Delta
P_{-}-P_{+}\Delta^{\prime}P_{+}))$ from $\mathscr{A}$ to $B$ is spectrally
decomposable (using the orthogonal projection $\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)\colon X\oplus X\to X\oplus X$ when verifying the
conditions in Definition 4.1). ∎

######  Remark 6.3.

It is worthwhile to understand the result of Proposition 6.2 in the light of
Skandalis’ work in [25]. This relationship was communicated to us by the
referee. Indeed, under the assumptions of Proposition 6.2 we obtain that
$(X,F)$ is a Kasparov module from $A$ to $B$ (see the proof of the
proposition). With some effort it can moreover be established that the
assumptions imply that
$a\big{(}FD\big{(}1+D^{2}\big{)}^{-1/2}+D\big{(}1+D^{2}\big{)}^{-1/2}F\big{)}a^{*}\geq
0$ modulo the compact operators on $X$ for all $a\in A$ and a similar result
applies to $D^{\prime}\big{(}1+(D^{\prime})^{2}\big{)}^{-1/2}$. Therefore, by
[25, Lemma 11] we obtain that the bounded transforms
$\big{(}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{)}$ and
$\big{(}X,D^{\prime}\big{(}1+(D^{\prime})^{2}\big{)}^{-1/2}\big{)}$ are
operator homotopic since they are both operator homotopic to $(X,F)$.

Notice however that this argument does not yield an alternative proof of
Proposition 6.2 since it only provides information on the bounded transforms
of the unbounded Kasparov modules $(X,D)$ and $(X,D^{\prime})$.

######  Theorem 6.4.

The direct sum of unbounded Kasparov modules and the zero module provide the
topological unbounded $KK$-theory, $UK^{{\rm top}}(\mathscr{A},B)$, with the
structure of an abelian group.

###### Proof.

For an unbounded Kasparov module $(X,D)$ from $\mathscr{A}$ to $B$, we need to
prove that $(X,D)-(X,D)=(X\oplus X^{{\rm op}},D\oplus(-D))$ is stably
homotopic to the zero module $(0,0)$. By Proposition 5.2 we may assume that
$(X,D)$ is Lipschitz regular and that $D\colon\operatorname{Dom}(D)\to X$ is
invertible. The phase of $D$ is then a well-defined odd selfadjoint unitary
operator $F:=D|D|^{-1}\colon X\to X$. The result of the present proposition
will now be a consequence of Proposition 6.2 applied to the case where
$(X,D)=(X,D^{\prime})$: The conditions $(1)$ and $(2)$ are clearly satisfied
and condition $(3)$ follows from the Lipschitz regularity of $(X,D)$. Indeed,
for each homogeneous $a\in\mathscr{A}$ and each $\xi\in\operatorname{Dom}(D)$
it holds that

$[F,a](\xi)=|D|^{-1}[D,a](\xi)+(-1)^{\deg(a)+1}|D|^{-1}[|D|,a]F(\xi)$

so that $[F,a]=|D|^{-1}T(a)$ for some bounded adjointable operator $T(a)\colon
X\to X$. ∎

##  7 Injectivity of the bounded transform

We are now ready to prove the main theorem of this paper:

######  Theorem 7.1.

Suppose that $A$ and $B$ are a $\mathbb{Z}/2\mathbb{Z}$-graded
$C^{*}$-algebras with $A$ separable and $B$ $\sigma$-unital. For any norm-
dense countably generated $\mathbb{Z}/2\mathbb{Z}$-graded $*$-subalgebra
$\mathscr{A}\subseteq A$ we have an isomorphism of abelian groups

$\mathcal{F}\colon\ UK^{{\rm top}}(\mathscr{A},B)\to KK(A,B)$

induced by the Baaj–Julg bounded transform
$(X,D)\mapsto\big{[}X,D\big{(}1+D^{2}\big{)}^{-1/2}\big{]}$.

###### Proof.

By Theorems 6.4 and 4.11 we only need to show that $\mathcal{F}\colon UK^{{\rm
top}}(\mathscr{A},B)\to KK(A,B)$ is injective.

Suppose that two unbounded Kasparov modules $(X_{0},D_{0})$ and
$(X_{1},D_{1})$ from $\mathscr{A}$ to $B$ satisfy that their bounded
transforms $\big{(}X_{0},D_{0}\big{(}1+D_{0}^{2}\big{)}^{-1/2}\big{)}$ and
$\big{(}X_{1},D_{1}\big{(}1+D_{1}^{2}\big{)}^{-1/2}\big{)}$ are homotopic. We
thus have a Kasparov module $(X,F)$ from $A$ to $C([0,1],B)$ and two even
unitary isomorphisms of $C^{*}$-correspondences $U_{0}\colon
X\widehat{\otimes}_{{\rm ev}_{0}}B\to X_{0}$ and $U_{1}\colon
X\widehat{\otimes}_{{\rm ev}_{1}}B\to X_{1}$ implementing unitary equivalences

$\displaystyle\big{(}X_{0},D_{0}\big{(}1+D_{0}^{2}\big{)}^{-1/2}\big{)}\sim_{u}\big{(}X\widehat{\otimes}_{{\rm
ev}_{0}}B,F\widehat{\otimes}_{{\rm ev}_{0}}1\big{)}\qquad\text{and}$
$\displaystyle\big{(}X_{1},D_{1}\big{(}1+D_{1}^{2}\big{)}^{-1/2}\big{)}\sim_{u}\big{(}X\widehat{\otimes}_{{\rm
ev}_{1}}B,F\widehat{\otimes}_{{\rm ev}_{1}}1\big{)}.$

By Proposition 5.1, we may assume without loss of generality that
$(X_{0},D_{0})$ and $(X_{1},D_{1})$ are both Lipschitz regular and, using [3,
Proposition 17.4.3], we may moreover assume that $F\colon X\to X$ is a
selfadjoint contraction. We let $X\oplus\widetilde{X}$ denote the
$\mathbb{Z}/2\mathbb{Z}$-graded $C^{*}$-correspondence from $A$ to
$C([0,1],B)$ which agrees with $X\oplus X$ as a Hilbert $C^{*}$-module over
$C([0,1],B)$ but with grading operator $\gamma\oplus(-\gamma)\colon X\oplus
X\to X\oplus X$ and with left action given by the even $*$-homomorphism
$\pi_{X\oplus\widetilde{X}}:=\pi_{X}\oplus 0\colon
A\to\mathbb{L}\big{(}X\oplus\widetilde{X}\big{)}$. Define the odd bounded
adjointable operator

$G:=\left(\begin{matrix}F&\big{(}1-F^{2}\big{)}^{1/2}\\\
\big{(}1-F^{2}\big{)}^{1/2}&-F\end{matrix}\right)\colon\
X\oplus\widetilde{X}\to X\oplus\widetilde{X}$

and notice that $G^{2}=1$ and $G=G^{*}$. Moreover, it holds that
$\big{(}X\oplus\widetilde{X},G\big{)}$ is a Kasparov module from $A$ to
$C([0,1],B)$ which is homotopic to $(X,F)$, see [3, Section 17.6]. A similar
construction applies to the endpoints yielding Kasparov modules
$\big{(}X_{0}\oplus\widetilde{X_{0}},G_{0}\big{)}$ and
$\big{(}X_{1}\oplus\widetilde{X_{1}},G_{1}\big{)}$ from $A$ to $B$. For each
$i\in\\{0,1\\}$, we define the invertible unbounded selfadjoint and regular
operator

$E_{i}:=\left(\begin{matrix}D_{i}&1\\\ 1&-D_{i}\end{matrix}\right)\colon\
\operatorname{Dom}(D_{i})\oplus\operatorname{Dom}(D_{i})\to
X_{i}\oplus\widetilde{X}_{i}$

and remark that $G_{i}=E_{i}|E_{i}|^{-1}$. We recall from Proposition 5.2 that
$\big{(}X_{i}\oplus\widetilde{X_{i}},E_{i}\big{)}$ is a Lipschitz regular
unbounded Kasparov module from $\mathscr{A}$ to $B$ and that

$\big{[}X_{i}\oplus\widetilde{X_{i}},E_{i}\big{]}=[X_{i},D_{i}]$

in the topological unbounded $KK$-theory, $UK^{{\rm top}}(\mathscr{A},B)$.

Using the Kasparov module $\big{(}X\oplus\widetilde{X},G\big{)}$ from $A$ to
$C([0,1],B)$ as input for Theorem 3.4, we may choose an even strictly positive
compact operator $\Delta\colon X\oplus\widetilde{X}\to X\oplus\widetilde{X}$
such that $(1)$, $(2)$, and $(3)$ in Theorem 3.4 hold. In particular, we
obtain an unbounded Kasparov module
$\big{(}X\oplus\widetilde{X},\Delta^{-1}G\big{)}$ from $\mathscr{A}$ to
$C([0,1],B)$ which implements a homotopy between the unbounded Kasparov
modules

$\big{(}X_{0}\oplus\widetilde{X_{0}},\Delta_{0}^{-1}G_{0}\big{)}\qquad\text{and}\qquad\big{(}X_{1}\oplus\widetilde{X_{1}},\Delta_{1}^{-1}G_{1}\big{)},$

where by definition

$\Delta_{i}:=(U_{i}\oplus U_{i})\big{(}\Delta\widehat{\otimes}_{{\rm
ev}_{i}}1\big{)}(U_{i}^{*}\oplus
U_{i}^{*})\in\mathbb{K}\big{(}X_{i}\oplus\widetilde{X}_{i}\big{)},\qquad\text{for
}i\in\\{0,1\\}.$

Let us fix an $i\in\\{0,1\\}$. Summarizing what has been obtained so far, we
see that the proof of the theorem is finished provided that the identity

$\big{[}X_{i}\oplus\widetilde{X_{i}},\Delta_{i}^{-1}G_{i}\big{]}=\big{[}X_{i}\oplus\widetilde{X_{i}},E_{i}\big{]}$

holds in $UK^{{\rm top}}(\mathscr{A},B)$. We are going to apply Proposition
6.2 for our two unbounded Kasparov modules from $\mathscr{A}$ to $B$ and the
odd selfadjoint unitary operator $G_{i}\colon X_{i}\oplus\widetilde{X}_{i}\to
X_{i}\oplus\widetilde{X}_{i}$. So we need to verify the three conditions in
the statement of Proposition 6.2. For condition $(1)$ we have that $G_{i}$
preserves the domains of $\Delta_{i}^{-1}G_{i}$ and $E_{i}$ and that both the
non-graded commutators
$\big{[}G_{i},\Delta_{i}^{-1}G_{i}\big{]}\colon\operatorname{Dom}\big{(}\Delta_{i}^{-1}G_{i}\big{)}\to
X_{i}\oplus\widetilde{X_{i}}$ and
$[G_{i},E_{i}]\colon\operatorname{Dom}(E_{i})\to X_{i}\oplus\widetilde{X_{i}}$
are in fact trivial. For condition $(2)$ we notice that

$\displaystyle\Delta_{i}^{-1}G_{i}^{2}=\Delta_{i}^{-1}\colon\
\operatorname{Dom}\big{(}\Delta_{i}^{-1}\big{)}\to
X_{i}\oplus\widetilde{X_{i}}\qquad\text{and}$ $\displaystyle
E_{i}G_{i}=|E_{i}|\colon\ \operatorname{Dom}(|E_{i}|)\to
X_{i}\oplus\widetilde{X_{i}}$

are already even unbounded positive and regular operators (so no bounded
perturbations are needed). To check the final condition $(3)$, we let
$a\in\mathscr{A}$ be homogeneous. The graded commutator $[G_{i},a]$ has image
contained in
$\operatorname{Dom}\big{(}\Delta_{i}^{-1}\big{)}=\operatorname{Dom}\big{(}\Delta_{i}^{-1}G_{i}\big{)}$
and

$\Delta_{i}^{-1}G_{i}[G_{i},a]=G_{i}\Delta_{i}^{-1}[G_{i},a]\colon\
X_{i}\oplus\widetilde{X_{i}}\to X_{i}\oplus\widetilde{X_{i}}$

is bounded since $[G,a]\colon X\oplus\widetilde{X}\to X\oplus\widetilde{X}$
has image contained in $\operatorname{Dom}\big{(}\Delta^{-1}\big{)}$ and
$\Delta^{-1}[G,a]$: $X\oplus\widetilde{X}\to X\oplus\widetilde{X}$ is bounded
by construction, see Theorem 3.4. The graded commutator $[G_{i},a]$ also has
image contained in $\operatorname{Dom}(E_{i})=\operatorname{Dom}(|E_{i}|)$
since

$[G_{i},a](\xi)=|E_{i}|^{-1}[E_{i},a](\xi)+(-1)^{\deg(a)+1}|E_{i}|^{-1}[|E_{i}|,a]|E_{i}|^{-1}E_{i}(\xi)$

for all $\xi\in\operatorname{Dom}(E_{i})$ and since
$\big{(}X_{i}\oplus\widetilde{X_{i}},E_{i}\big{)}$ is Lipschitz regular.
Letting $e_{i}(a)$ and $|e_{i}|(a)$ denote the bounded extensions of the
graded commutators $[E_{i},a]$ and
$[|E_{i}|,a]\colon\operatorname{Dom}(E_{i})\to X_{i}\oplus\widetilde{X_{i}}$
we moreover see that

$E_{i}[G_{i},a]=G_{i}\cdot e_{i}(a)+(-1)^{\deg(a)+1}G_{i}\cdot|e_{i}|(a)\cdot
G_{i}\colon X_{i}\oplus\widetilde{X_{i}}\to X_{i}\oplus\widetilde{X_{i}}$

is bounded. It thus follows from Proposition 6.2 and Theorem 6.4 that
$\big{[}X_{i}\oplus\widetilde{X_{i}},\Delta_{i}^{-1}G_{i}\big{]}=\big{[}X_{i}\oplus\widetilde{X_{i}},E_{i}\big{]}$
in $UK^{{\rm top}}(\mathscr{A},B)$ and this ends the proof of the theorem. ∎

### Acknowledgements

The starting point for this paper was a couple of conversations with Bram
Mesland during the thematic programme on “Bivariant $K$-theory in Geometry and
Physics” at the Erwin Schrödinger Institute in Vienna in November 2018. I
would like to thank the ESI for their hospitality and support and the
organizers of the thematic programme for this great opportunity to meet and
discuss the unbounded approach to $KK$-theory and its applications in
mathematical physics. As always, I am also grateful to my friends and
collaborators Magnus Goffeng, Bram Mesland, and Adam Rennie for many good
conversations on unbounded $KK$-theory and its relationship to $KK$-theory.
Finally, I would like to thank the anonymous referee for her/his comments
regarding spectral decomposability and its relationship to the work of
Skandalis.

The author was partially supported by the DFF-Research Project 2
“Automorphisms and Invariants of Operator Algebras”, no. 7014-00145B.

## References

  * [1] Atiyah M.F., Global theory of elliptic operators, in Proc. Internat. Conf. on Functional Analysis and Related Topics (Tokyo, 1969), University of Tokyo Press, Tokyo, 1970, 21–30.
  * [2] Baaj S., Julg P., Théorie bivariante de Kasparov et opérateurs non bornés dans les $C^{*}$-modules hilbertiens, C. R. Acad. Sci. Paris Sér. I Math. 296 (1983), 875–878.
  * [3] Blackadar B., $K$-theory for operator algebras, 2nd ed., Mathematical Sciences Research Institute Publications, Vol. 5, Cambridge University Press, Cambridge, 1998.
  * [4] Brain S., Mesland B., van Suijlekom W.D., Gauge theory for spectral triples and the unbounded Kasparov product, J. Noncommut. Geom. 10 (2016), 135–206, arXiv:1306.1951.
  * [5] Brown L.G., Douglas R.G., Fillmore P.A., Extensions of $C^{*}$-algebras and $K$-homology, Ann. of Math. 105 (1977), 265–324.
  * [6] Connes A., Noncommutative geometry, Academic Press, Inc., San Diego, CA, 1994.
  * [7] Connes A., Gravity coupled with matter and the foundation of non-commutative geometry, Comm. Math. Phys. 182 (1996), 155–176, arXiv:hep-th/9603053.
  * [8] Deeley R.J., Goffeng M., Mesland B., The bordism group of unbounded $KK$-cycles, J. Topol. Anal. 10 (2018), 355–400, arXiv:1503.07398.
  * [9] Hilsum M., Bordism invariance in $KK$-theory, Math. Scand. 107 (2010), 73–89.
  * [10] Jensen K.K., Thomsen K., Elements of $KK$-theory, Mathematics: Theory & Applications, Birkhäuser Boston, Inc., Boston, MA, 1991.
  * [11] Kaad J., Morita invariance of unbounded bivariant $K$-theory, arXiv:1612.08405.
  * [12] Kaad J., Differentiable absorption of Hilbert $C^{*}$-modules, connections, and lifts of unbounded operators, J. Noncommut. Geom. 11 (2017), 1037–1068, arXiv:1407.1389.
  * [13] Kaad J., Lesch M., A local global principle for regular operators in Hilbert $C^{*}$-modules, J. Funct. Anal. 262 (2012), 4540–4569, arXiv:1107.2372.
  * [14] Kaad J., Lesch M., Spectral flow and the unbounded Kasparov product, Adv. Math. 248 (2013), 495–530, arXiv:1110.1472.
  * [15] Kaad J., van Suijlekom W.D., Factorization of Dirac operators on toric noncommutative manifolds, J. Geom. Phys. 132 (2018), 282–300, arXiv:1803.08921.
  * [16] Kasparov G.G., Topological invariants of elliptic operators. I. $K$-homology, Math. USSR-Izv. 9 (1975), 751–792.
  * [17] Kasparov G.G., Hilbert $C^{*}$-modules: theorems of Stinespring and Voiculescu, J. Operator Theory 4 (1980), 133–150.
  * [18] Kasparov G.G., The operator $K$-functor and extensions of $C^{*}$-algebras, Math. USSR-Izv. 16 (1980), 513–572.
  * [19] Kucerovsky D., The $KK$-product of unbounded modules, $K$-Theory 11 (1997), 17–34.
  * [20] Kucerovsky D., A lifting theorem giving an isomorphism of $KK$-products in bounded and unbounded $KK$-theory, J. Operator Theory 44 (2000), 255–275.
  * [21] Lance E.C., Hilbert $C^{*}$-modules. A toolkit for operator algebraists, London Mathematical Society Lecture Note Series, Vol. 210, Cambridge University Press, Cambridge, 1995.
  * [22] Lesch M., Mesland B., Sums of regular self-adjoint operators in Hilbert-$C^{*}$-modules, J. Math. Anal. Appl. 472 (2019), 947–980, arXiv:1803.08295.
  * [23] Mesland B., Unbounded bivariant $K$-theory and correspondences in noncommutative geometry, J. Reine Angew. Math. 691 (2014), 101–172, arXiv:0904.4383.
  * [24] Mesland B., Rennie A., Nonunital spectral triples and metric completeness in unbounded $KK$-theory, J. Funct. Anal. 271 (2016), 2460–2538, arXiv:1502.04520.
  * [25] Skandalis G., Some remarks on Kasparov theory, J. Funct. Anal. 56 (1984), 337–347.
  * [26] van den Dungen K., Mesland B., Homotopy equivalence in unbounded $KK$-theory, Ann. K-Theory 5 (2020), 501–537, arXiv:1907.04049.
  * [27] Woronowicz S.L., Unbounded elements affiliated with $C^{*}$-algebras and noncompact quantum groups, Comm. Math. Phys. 136 (1991), 399–432.
  * [28] Woronowicz S.L., Napiórkowski K., Operator theory in the $C^{*}$-algebra framework, Rep. Math. Phys. 31 (1992), 353–371.


# A New Reduction from Search SVP to Optimization SVP

Gengran Hu, Yanbin Pan
Key Laboratory of Mathematics Mechanization
Academy of Mathematics and Systems Science, Chinese Academy of Sciences
Beijing 100190, China
<EMAIL_ADDRESS><EMAIL_ADDRESS>

###### Abstract

It is well known that search SVP is equivalent to optimization SVP. However,
the former reduction from search SVP to optimization SVP by Kannan needs
polynomial times calls to the oracle that solves the optimization SVP. In this
paper, a new rank-preserving reduction is presented with only one call to the
optimization SVP oracle. It is obvious that the new reduction needs the least
calls, and improves Kannan’s classical result. What’s more, the idea also
leads a similar direct reduction from search CVP to optimization CVP with only
one call to the oracle.

Keywords: Search SVP, Optimization SVP, Lattice, Reduction.

##  1 Introduction

Given a matrix $B=(b_{ij})\in\mathbb{R}^{m\times n}$ with rank $n$, the
lattice $L(B)$ spanned by the columns of $B$ is

$L(B)=\\{\sum_{i=1}^{n}x_{i}b_{i}|x_{i}\in\mathbb{Z}\\},$

where $b_{i}$ is the $i$-th column of $B$. Lattice has many important
applications in cryptography. The shortest vector problem (SVP) and the
closest vector problem (CVP) are two of the most famous problems of lattice.

SVP refers to find the shortest non-zero vector in a given lattice. There are
three different variants of SVP:

1\. Search SVP: Given a lattice basis $B\in\mathbb{Z}^{m\times n}$, find
$v\in\mathcal{L}(B)$ such that $\|v\|=\lambda_{1}(\mathcal{L}(B))$, where
$\lambda_{1}(\mathcal{L}(B))$ is the length of the shortest non-zero vector in
$\mathcal{L}(B)$.

2\. Optimization SVP: Given a lattice basis $B\in\mathbb{Z}^{m\times n}$, find
$\lambda_{1}(\mathcal{L}(B))$.

3\. Decisional SVP: Given a lattice basis $B\in\mathbb{Z}^{m\times n}$ and a
rational $r\in\mathbb{Q}$, decide whether $\lambda_{1}(\mathcal{L}(B))\leq r$
or not.

It has been proved that the three problems are equivalent to each other (see
[2]). It is easy to check that the decisional SVP is as hard as the
optimization SVP and the optimization variant can be reduced to the search
variant.

In 1987, Kannan [1] also showed that the search variant can be reduced to the
optimization variant. The basic idea of his reduction is to recover the
integer coefficients of some shortest vector under the given lattice basis by
introducing small errors to the original lattice basis. However, his reduction
is a bit complex. It needs to call polynomial times optimization SVP oracle,
since it could not determine the signs of the shortest vector’s entries at one
time. It also needs oracle to solve optimization SVP for some lattices with
lower rank besides with the same rank as the original lattice.

In this paper, we propose a new rank-preserving reduction which can solve the
search SVP with only one call to the optimization SVP oracle. It is obvious
that there is no reduction with less calls than ours. Instead of recovering
the shortest vector directly as in [1], we first recover the integer
coefficients of some shortest vector under the given lattice basis, then
recover the shortest vector.

A similar direct reduction from search CVP to optimization CVP with only one
call also holds whereas some popular reductions [2, 3] usually takes
decisional CVP to bridge the search CVP and optimization CVP. The former
reduction from decisional CVP to optimization CVP needs one call to the
optimization CVP oracle, but it needs polynomial times calls to the decisional
CVP oracle to reduce search CVP to decisional CVP.

##  2 The New Reduction

For simplicity, we just give the new reduction for the full rank lattice, i.e.
$n=m$, as in [1]. It is easy to general the new reduction for the lattices
with rank $n<m$.

###  2.1 Some Notations

Given a lattice basis $B=(b_{ij})\in\mathbb{R}^{n\times n}$, let
$M(B)=\max|b_{ij}|$. For lattice $L(B)$, we define its SVP solution set
$S_{B}$ as:

$S_{B}=\\{x\in\mathbb{Z}^{n}|\|Bx\|=\lambda_{1}(\mathcal{L}(B))\\}$

Denote by $poly(n)$ the polynomial in $n$.

###  2.2 Some Lemmas

We need some lemmas to prove our main theorem.

######  Lemma 1.

For every positive integer $n$, there exist $n$ positive integers
$a_{1}<a_{2}<\ldots<a_{n}$ s.t. all the $a_{i}+a_{j}(i\leq j)$’s are distinct
and $a_{n}$ is bounded by $poly(n)$.

###### Proof.

We can take $a_{k}=(n^{2}+k-1)^{2}$ for $k=1,2,\cdots,n$. Suppose
$a_{i_{1}}+a_{j_{1}}=a_{i_{2}}+a_{j_{2}}$ for some $i_{1},j_{1},i_{2},j_{2}$,
we get
$(i_{1}-1)^{2}+(j_{1}-1)^{2}+2n^{2}((i_{1}-1)+(j_{1}-1))=(i_{2}-1)^{2}+(j_{2}-1)^{2}+2n^{2}((i_{2}-1)+(j_{2}-1))$.
Since $(i_{1}-1)^{2}+(j_{1}-1)^{2},(i_{2}-1)^{2}+(j_{2}-1)^{2}<2n^{2}$, we
have $(i_{1}-1)^{2}+(j_{1}-1)^{2}=(i_{2}-1)^{2}+(j_{2}-1)^{2}$ and
$i_{1}+j_{1}=i_{2}+j_{2}$, which leads $\\{i_{1},j_{1}\\}=\\{i_{2},j_{2}\\}$.
Hence all the $a_{i}+a_{j}(i\leq j)$’s are distinct. It is obvious that
$a_{n}\leq(n^{2}+n-1)^{2}$. ∎

######  Lemma 2.

Given positive odd integer $p>2$, and any positive integer $n$, which
satisfies $n=\sum_{i=0}^{k}n_{i}p^{i}$ where $|n_{i}|\leq\lfloor p/2\rfloor$,
then we can recover the coefficients $n_{i}$’s in polynomial time.

###### Proof.

We can recover $n_{0}$ by computing $a\equiv n\mbox{ mod }p$ and choose $a$ in
the interval from $-\lfloor p/2\rfloor$ to $\lfloor p/2\rfloor$. After
obtaining $n_{0}$, we get another integer $(n-n_{0}*p^{0})/p$. Recursively, we
can recover all the coefficients. This can be done in polynomial time
obviously. ∎

######  Lemma 3.

For bivariate polynomial $f(x,y)=xy$, given any lattice basis matrix
$B\in\mathbb{Z}^{n\times n}$, $\lambda_{1}(L(B))$ has an upper bound $f(M,n)$,
where $M=M(B)$. What’s more, for every $x\in S_{B}$, $|x_{i}|$
$(i=1,\cdots,n)$ has an upper bound $f(M^{n},n^{n})$.

###### Proof.

The length of any column of $B$ is an upper bound of $\lambda_{1}(L(B))$, so
$\lambda_{1}(L(B))\leq n^{1/2}M\leq nM$.

For $x\in S_{B}$, we let $y=Bx$, then $\|y\|=\lambda_{1}(L(B))\leq\sqrt{n}M$.
By Cramer’s rule, we know that

$x_{i}=\dfrac{\det(B^{(i)})}{\det(B)},$

where $B^{(i)}$ is formed by replacing the $i$-th column of $B$ by $y$. By
Hadamard’s inequality, $|\det(B^{(i)})|\leq n^{n/2}M^{n}\leq n^{n}M^{n}$. We
know $|\det(B)|\geq 1$ since $\det(B)$ is a non-zero integer. Hence
$|x_{i}|\leq n^{n}M^{n}$. ∎

###  2.3 The Main Theorem

######  Theorem 1.

Assume there exists an oracle $\mathcal{O}$ that can solve the optimization
SVP for any lattice $L(B^{\prime})$ with basis
$B^{\prime}\in\mathbb{Z}^{n\times n}$, then there is an algorithm that can
solve the search SVP for any lattice $L(B)$ with basis
$B\in\mathbb{Z}^{n\times n}$ with only one call to $\mathcal{O}$ in
$poly(\log_{2}{M},n,\log_{2}{n})$ time, where $M=M(B)$.

###### Proof.

The main steps of the algorithm are as below:
(1) Constructing a new lattice basis $B_{\epsilon}\in\mathbb{Z}^{n\times n}$.

We construct $B_{\epsilon}$ from the original lattice $B$:

$B_{\epsilon}=\epsilon_{n+1}B+\left(\begin{array}[]{cccc}\epsilon_{1}&\epsilon_{2}&\dots&\epsilon_{n}\\\
0&0&\dots&0\\\ \vdots&\vdots&&\vdots\\\ 0&0&\dots&0\\\ \end{array}\right)$

where the $\epsilon_{i}$ will be determined as below.

For any $x\in\mathbb{Z}^{n}$, we difine $c(x)=\sum_{i=1}^{n}b_{1i}x_{i}$. For
$x\in S_{B}$, by Lemma 3, $|x_{i}|$ has an upper bound $f(M^{n},n^{n})$. Let
$M_{1}=2f((M+1)^{n},n^{n})$. In addition, $\|Bx\|=\lambda_{1}(L(B))$ is
bounded by $f(M,n)$. Let $M_{2}=f(M+1,n)$. $|c(x)|$ is also bounded by $M_{2}$
since $|c(x)|\leq\|Bx\|$. We let

$p=2*\max{\\{M_{2}^{2},2M_{1}M_{2},2M_{1}^{2}\\}}+1.$

By Lemma 1, we can choose $n+1$ positive integers
$a_{1}<a_{2}<\ldots<a_{n+1}$, such that all the $a_{i}+a_{j}(i\leq j)$’s are
distinct where $a_{n+1}$ is bounded by $poly(n)$. Let

$\epsilon_{i}=p^{a_{i}}.$

We first show that
$|\det{(\frac{1}{\epsilon_{n+1}}B_{\epsilon})|\geq\frac{1}{2}}$, so
$B_{\epsilon}$ is indeed a lattice basis. Notice that

$\det{(\frac{1}{\epsilon_{n+1}}B_{\epsilon})}=\det(B)+\sum_{i=1}^{n}\alpha_{i}\frac{\epsilon_{i}}{\epsilon_{n+1}},$

where $\alpha_{i}$ is the cofactor of $B_{1i}$ in $B$. Since
$\frac{\epsilon_{i}}{\epsilon_{n+1}}\leq\frac{1}{p^{2}}$ and $|\alpha_{i}|\leq
M^{n-1}(n-1)^{n-1}$,
$|\sum_{i=1}^{n}\alpha_{i}\frac{\epsilon_{i}}{\epsilon_{n+1}}|\leq\frac{1}{p^{2}}M^{n-1}n^{n}<\frac{1}{2}$.
By the fact $\det(B)$ is a non-zero integer, we get

$|\det{(\frac{1}{\epsilon_{n+1}}B_{\epsilon})|\geq\frac{1}{2}}.$ (1)

We claim that $S_{B_{\epsilon}}\subseteq S_{B}$. Since
$S_{B_{\epsilon}}=S_{\frac{1}{\epsilon_{n+1}}B_{\epsilon}}$, it is enough to
prove $S_{\frac{1}{\epsilon_{n+1}}B_{\epsilon}}\subseteq S_{B}$.

For any $x\in S_{\frac{1}{\epsilon_{n+1}}B_{\epsilon}}$, by (1) and the proof
of Lemma 3, we know that $|x_{i}|\leq M_{1}$, $|c(x)|\leq M_{2}$. By the
choice of $p$, $x_{i}^{2},2c(x)x_{i},2x_{i}x_{j}$ are in the interval
$[-\lfloor p/2\rfloor,\lfloor p/2\rfloor]$. Together with the fact that
$\frac{\epsilon_{i}\epsilon_{j}}{\epsilon_{n+1}^{2}}(i\leq j)$’s are different
powers of $p$, we have

$\begin{array}[]{rcl}\lambda_{1}(L(\frac{1}{\epsilon_{n+1}}B_{\epsilon}))^{2}&=&\|\frac{1}{\epsilon_{n+1}}B_{\epsilon}x\|^{2}\\\
&=&\|Bx\|^{2}+\sum_{i=1}^{n}x_{i}^{2}(\frac{\epsilon_{i}}{\epsilon_{n+1}})^{2}+\sum_{i=1}^{n}2c(x)x_{i}\frac{\epsilon_{i}}{\epsilon_{n+1}}+\sum_{i<j}2x_{i}x_{j}\frac{\epsilon_{i}\epsilon_{j}}{\epsilon_{n+1}^{2}}\\\
&>&\|Bx\|^{2}-(\lfloor
p/2\rfloor+1)\frac{\epsilon_{n}}{\epsilon_{n+1}}.\end{array}$ (2)

Similarly, for any $y\in S_{B}$, we have

$\begin{array}[]{rcl}\|\frac{1}{\epsilon_{n+1}}B_{\epsilon}y\|^{2}&=&\|By\|^{2}+\sum_{i=1}^{n}y_{i}^{2}(\frac{\epsilon_{i}}{\epsilon_{n+1}})^{2}+\sum_{i=1}^{n}2c(y)y_{i}\frac{\epsilon_{i}}{\epsilon_{n+1}}+\sum_{i<j}2y_{i}y_{j}\frac{\epsilon_{i}\epsilon_{j}}{\epsilon_{n+1}^{2}}\\\
&<&\lambda_{1}(L(B))^{2}+(\lfloor
p/2\rfloor+1)\frac{\epsilon_{n}}{\epsilon_{n+1}}\end{array}$ (3)

Next, we prove $S_{\frac{1}{\epsilon_{n+1}}B_{\epsilon}}\subseteq S_{B}$.
Suppose there exists $x\in S_{\frac{1}{\epsilon_{n+1}}B_{\epsilon}}$ but
$x\not\in S_{B}$, then

$\|Bx\|^{2}\geq\lambda_{1}(L(B))^{2}+1.$ (4)

Notice that $\frac{\epsilon_{n}}{\epsilon_{n+1}}<\frac{1}{p^{2}}$, we have
$0<(\lfloor p/2\rfloor+1)\frac{\epsilon_{n}}{\epsilon_{n+1}}<\frac{1}{2}$.
Together with (2), (3) and (4), we have

$\begin{array}[]{rcl}\lambda_{1}(L(\frac{1}{\epsilon_{n+1}}B_{\epsilon}))^{2}&>&\|Bx\|^{2}-(\lfloor
p/2\rfloor+1)\frac{\epsilon_{n}}{\epsilon_{n+1}}\\\
&\geq&\lambda_{1}(L(B))^{2}+1-(\lfloor
p/2\rfloor+1)\frac{\epsilon_{n}}{\epsilon_{n+1}}\\\
&>&\lambda_{1}(L(B))^{2}+(\lfloor
p/2\rfloor+1)\frac{\epsilon_{n}}{\epsilon_{n+1}}\\\
&>&\|\frac{1}{\epsilon_{n+1}}B_{\epsilon}y\|^{2},\end{array}$

which is an contradiction, since $\frac{1}{\epsilon_{n+1}}B_{\epsilon}y\in
L(\frac{1}{\epsilon_{n+1}}B_{\epsilon})$. Hence $S_{B_{\epsilon}}\subseteq
S_{B}$.
(2) Querying the oracle $\mathcal{O}$ with $B_{\epsilon}$ once, we get
$\lambda_{1}(\mathcal{L}(B_{\epsilon}))$.

So there exists $x=(x_{1},\ldots,x_{n})^{\mathrm{T}}\in
S_{B_{\epsilon}}\subseteq S_{B}$, such that

$\|Bx\|^{2}\epsilon_{n+1}^{2}+\sum_{i=1}^{n}x_{i}^{2}\epsilon_{i}^{2}+\sum_{i=1}^{n}2c(x)x_{i}\epsilon_{n+1}\epsilon_{i}+\sum_{i<j}2x_{i}x_{j}\epsilon_{i}\epsilon_{j}=\lambda_{1}(\mathcal{L}(B_{\epsilon}))^{2}$

(3) Recovering all the $x_{i}$’s and output $Bx$.

Since $x\in S_{B}$, every coefficient
$\|Bx\|^{2},x_{i}^{2},2c(x)x_{i},2x_{i}x_{j}$ is in the interval $[-\lfloor
p/2\rfloor,\lfloor p/2\rfloor]$ and $\epsilon_{i}\epsilon_{j}$ $(i\leq j)$’s
are different powers of $p$. Hence,
$\log_{2}{(\lambda_{1}(\mathcal{L}(B_{\epsilon})))}$ is bounded by
$poly(\log_{2}M,n,\log_{2}n)$. Furthermore, by Lemma 2, we can recover all the
coefficients in $poly(\log_{2}M,n,\log_{2}n)$ time. Especially, we can recover
all $x_{i}^{2}$ and $x_{i}x_{j}(i\neq j)$. Let $k=\min\\{i|x_{i}\neq 0\\}$. We
fix $x_{k}=\sqrt{x_{k}^{2}}>0$, and can recover all the remaining
$x_{j}=sign(x_{k}x_{j})\sqrt{x_{j}^{2}}$ according to $x_{j}^{2}$ and
$x_{k}x_{j}(k\neq j)$.

It is easy to check that the complexity of every step is bounded by
$poly(\log_{2}{M},n,\log_{2}n)$. ∎

######  Remark 1.

For any search CVP instant $(B,t)$, given an oracle which can solve the
optimization CVP, we can call the oracle with $(B_{\epsilon},\epsilon_{n+1}t)$
only once to solve the search CVP similarly.

##  3 Conclusions

In this paper, we give a new reduction from search SVP to optimization SVP
with only one call, which is the least, to the optimization SVP oracle. A
similar result for CVP also holds. However, it seems hard to apply the idea
for GapSVP or GapCVP, since the new reduction is also sensitive to the error.

## References

  * [1] R. Kannan, Minkowski’s convex body theorem and integer programming, Mathematics of Operation Research, 12(3): 415-440, 1987.
  * [2] D. Micciancio, S. Goldwasser, Complexity of Lattice Problems: A Cryptography Perspective, Kluwer Academic Publishes, 2002.
  * [3] O. Regev, Lattices in computer science, Lecture notes of a course given in Tel Aviv University, 2004.