# Twisted $K$-homology, Geometric cycles and $T$-duality

Bei Liu
Mathematisches Institut, Georg-August-Universit$\ddot{a}$ G$\ddot{o}$ettingen
<EMAIL_ADDRESS>
###### Abstract

Twisted $K$-homology corresponds to $D$-branes in string theory . In this
paper we compare two different models of geometric twisted $K$-homology and
get their equivalence. Moreover, we give another description of geometric
twisted $K$-homology using bundle gerbes. In the last part we construct
$T$-duality transformation for geometric twisted $K$-homology.

##  1 Introduction

String theory, as a candidate for quantum gravity, abstracts interests from
both physicians and mathematicians. Its ultimate goal is to construct quantum
theories of the basic structures of our universe. The starting point of string
theory is that the fundamental units of our universe are 1-dimensional strings
instead of point particles.String theorists find that there are five kinds of
different string theories, i.e type I, type IIA, type IIB, heterotic $E_{8}$
and heterotic $SO(32)$ string theories, all of which are mathematically
consistent (see [18]). While there is only one universe, therefore it becomes
important to study relations between different string theories. These
relations are called duality. There are three kinds of duality in string
theory: $S$-duality, $T$-duality and $U$-duality. $S$-duality is also called
electric-magnetic duality [14]. $T$-duality, which is one of the main topic of
this paper, is a duality exchanging winding number and momentum in the dynamic
equation of $D$-branes. While $U$-duality can be seen as composition of
$S$-duality and $T$-duality. $T$-duality provides an equivalence between type
IIA and type IIB string theory. In physics, this equivalence is reflected by
two important notions in string theory, $D$-branes and Ramond-Ramond fields
living on $D$-branes. Briefly speaking, $D$-branes can be see as a submanifold
of the spacetime manifold on which strings can end. A $Dp$-brane is a
$p$-dimensional submanifold with some other structures on it.

After Witten suggested that the Ramond-Ramond fields should be classified in
(twisted) $K$-theory instead of de Rham cohomology in [21], twisted $K$-theory
has been studied extensively (see [1] and [5]). There are mainly three
different approaches to understand twisted $K$-theory:

  1. 1.

bundle gerbes, which only works when twists are torsion cohomology classes

  2. 2.

homotopy classes of sections of Fredlhom bundles

  3. 3.

K-theory of the $C^{\ast}$-algebra of compact operator bundles determined by
the twisting class

The dual theory of twisted $K$-theory i.e twisted $K$-homology has also been
studied as a mathematical interpretation of $D$-branes. The analytic approach
to twisted $K$-homology is the same as the third for twisted $K$-theory we
list above. However, geometric approaches are more useful for us to understand
$D$-branes in string theory. In [2] and [20] they both give a construction of
geometric twisted $K$-homology groups. We will show that they are equivalent
to each other in this paper. After that, we will construct some properties of
geometric twisted $K$-homology and then use the geometric twisted $K$-cycle in
[20] to construct the $T$-duality transformation for spacetime compactified
over $S^{1}$.

Now we give the structure of this paper. In the next section, we review
different approaches to twisted $K$-homology. In section $3$ we show the
equivalence of that two versions of geometric twisted $K$-homology. In section
$4$ we establish some properties of geometric twisted $K$-homology. In section
$5$ we discuss the charge map in [2] and get a positive answer to the question
on the charge map in the end of [2]. In section $6$ we introduce another
approach to define geometric twisted $K$-cycle using bundle gerbes. In section
$7$ we construct three transformations whose composition give $T$-duality
transformation of geometric twisted $K$-homology. In the last section we will
prove that the $T$-duality transformation is an isomorphism.

Acknowledgement The author thanks the Research Training Group 1493
”Mathematical Structures in Modern Quantum Physics” for the support during his
Ph.D period. The author deeply thanks Thomas Schick for many helpful comments
and discussions. Besides, the author also thanks Bailing Wang’s wonderful
lecture on geometric twisted $K$-homology and discussions.

##  2 Reviews on twisted K-homology

Let $X$ be a locally finite $CW$-complex and $\alpha:X\rightarrow
K(\mathbb{Z},3)$ be a twist over $X$. Denote the projective unitary group over
a separable Hilbert space $\mathbb{H}$ by $PU(\mathbb{H})$ and denote the
$C^{\ast}$-algebra of compact operators by $\mathcal{K}$. Since the
classifying space of $PU(\mathbb{H})$ is a model of $K(\mathbb{Z},3)$ (see
[1]), therefore $\alpha$ determines a principal $PU(\mathbb{H})$-bundle
$\mathfrak{P}$ over $X$. We denote the associated $\mathcal{K}$-bundle of
$\mathfrak{P}$ by $\mathcal{A}$. Then all of the continuous sections with
compact support of $\mathcal{A}$ give rise to a $C^{\ast}$-algebra which we
denote by $C^{\ast}(X,\alpha)$. A direct way to construct a twisted
$K$-homology theory for $(X,\alpha)$ is to use the $K$-homology of the
continuous trace $C^{\ast}$-algebra of $C^{\ast}(X,\alpha)$. We denote this
twisted $K$-homology group by $K^{a}_{\ast}(X,\alpha)$. However, it is very
difficult to see the geometric meaning $K$-cycles through this approach. In
[2] and [20], more topological and geometric models are constructed. Let
$\mathfrak{K}$ be the complex $K$-theory spectrum and
$\mathcal{P}_{\alpha}(\mathfrak{K})$the corresponding bundle of based spectra
over $X$. In [20], the topological twisted $K$-homology group is defined to be

$K^{t}_{n}(X,\alpha):=\lim_{k\rightarrow\infty}[S^{n+2k},\mathcal{P}_{\alpha}(\Omega^{2k}\mathfrak{K})/X]$
(2.1)

This definition comes from the classical definition of homology theory by
spectra, which is automatically a homology theory. In [20], B.L. Wang gave a
geometric twisted $K$-homology. Before giving his constructions, we first
review the definition of geometric cycles of $K$-homology in [4]. A geometry
cycle on a pair of space $(X,Y)(Y\subset X)$ is a triple $(M,f,[E])$, such
that

  * •

$M$ is a compact spinc-manifold(probably with boundary);

  * •

$f$ is a continuous map from $M$ to $X$ such that $f(\partial M)\subset Y$;

  * •

$[E]$ is a $K^{0}$-class of $M$

The definition of twisted cases is given in [20] as follows.

######  Definition 2.1.

A geometric cycle for $(X,Y,\alpha)$ is a quintuple
$(M,\iota,\upsilon,\eta,[E])$ such that

  * •

$M$ is a $\alpha$-twisted spinc-manifold,i.e $M$ is a compact oriented
manifold and the following diagram exists

${M}$BSO${X}$${\
K(\mathbb{Z},3)}$${\iota}$${\upsilon}$${\alpha}$${W_{3}}$${\eta}$ (2.2)

Here $\upsilon$ and $W_{3}$ are classifying maps of the stable normal bundle
of $M$ and the third integral Stiefel-Whitney class respectively, $\eta$ is a
homotopy between $\alpha\circ\iota$ and $W_{3}\circ\upsilon$. Moreover we
require $\iota(\partial M)\subset Y$.

  * •

$[E]$ is an element class of $K^{0}(X)$ which is represented by a
$\mathbb{Z}_{2}$-graded vector bundle $E$.

Let $\Gamma(X,\alpha)$ be the collections of all geometric cycles for
$(X,\alpha)$. To get geometric twisted $K$-homology, we still need an
equivalence relation on $\Gamma(X,\alpha)$, which is generated by the
following basic relations:

  * •

Direct sum - disjoint union If $(M,\iota,\upsilon,\eta,[E_{1}])$ and
$(M,\iota,\upsilon,\eta,[E_{2}])$ are geometric cycles over $(X,\alpha)$, then

$(M,\iota,\upsilon,\eta,[E_{1}])\cup(M,\iota,\upsilon,\eta,[E_{2}])\sim(M,\iota,\upsilon,\eta,[E_{1}]+[E_{2}])$
(2.3)

  * •

Bordism Given two geometric cycle $(M,\iota,\upsilon,\eta,[E_{1}])$ and
$(M,\iota,\upsilon,\eta,[E_{2}])$ , if there exists a $\alpha$-twisted spinc-
manifold $(W,\iota,\upsilon,\eta)$ and $[E]\in K^{0}(W)$ such that

$\delta(W,\iota,\upsilon,\eta)=-(M_{1},\iota_{1},\upsilon_{1},\eta_{1})\cup(M_{2},\iota_{2},\upsilon_{2},\eta_{2})$
(2.4)

and $\delta([E])=[E_{1}]\cup[E_{2}]$. Here
$-(M_{1},\iota_{1},\upsilon_{1},\eta_{1})$ means the manifold $M_{1}$ with the
opposite $\alpha$-twisted $Spin^{c}$ structure.

  * •

Spinc vector bundle modification Given a geometric cycle
$(M,\iota,\upsilon,\eta,E)$ and a spinc vector bundle $V$ over $M$ with even
dimensional fibers, we can choose a Riemannian metric on $V\oplus\mathbb{R}$
and get the sphere bundle $\hat{M}=S(V\oplus\mathbb{R})$. Then the vertical
tangent bundle $T^{v}(\hat{M})$ admits a natural spinc structure. Let
$S^{+}_{V}$ be the associated positive spinor bundle and
$\rho:\hat{M}\rightarrow M$ be the projection. Then

$(M,\iota,\upsilon,\eta,E)\sim(\hat{M},\iota\circ\rho,\upsilon\circ\rho,\eta\circ\rho,\rho^{\ast}E\otimes
S^{+}_{V}).$ (2.5)

Here $\upsilon^{\prime}$ is a classifying map of the stable normal bundle of
$\hat{M}$ and $\eta^{\prime}$ is a chosen homotopy between
$W_{3}\circ\upsilon^{\prime}$ and $\alpha\circ\iota\circ\rho$.

######  Definition 2.2.

$K^{g}_{\ast}:=\Gamma(X,\alpha)/\sim$. Addition is given by disjoint union
relation above. Let $K^{g}_{0}(X,\alpha)$(respectively $K^{g}_{1}(X,\alpha)$)
be the subgroup of $K^{g}_{\ast}(X,\alpha)$ determined by all geometric cycles
with even (respectively odd) dimensional $\alpha$-twisted spinc-manifolds.

There is an natural isomorphism $\mu$ between $K^{g}_{0/1}(X,\alpha)$ and
$K^{a}_{0/1}(X,\alpha)$:

$\mu(M,\iota,\upsilon,\eta,[E])=\iota_{\ast}\circ\eta_{\ast}\circ
I^{\ast}\circ PD([E])$ (2.6)

Here $PD:K^{i}(M)\rightarrow K_{n+i}(M,W_{3}\circ\tau)$ is the
Poincar$\check{e}$ duality map between $K$-group and $K$-homology group,
$\iota_{\ast}$ is the push-forward map induced by $\iota$, $\eta_{\ast}$ is
the canonical isomorphism induced by $\eta$

$\eta_{\ast}:K^{a}_{0/1}(M,W_{3}\circ\upsilon)\rightarrow
K^{a}_{1/0}(M,\alpha\circ\iota)$ (2.7)

and $I:K^{a}_{0/1}(M,W_{3}\circ\tau)\rightarrow
K^{a}_{0/1}(M,W_{3}\circ\upsilon)$ is a natural isomorphism induced by the
anti-isomorphism of the associated $\mathcal{K}(\mathbb{H})$-bundles. The
following theorem in [20] states that $\mu$ is an isomorphism.

######  Theorem 2.3.

The assignment
$(M,\iota,\upsilon,\eta,[E])\rightarrow\mu(M,\iota,\upsilon,\eta,[E])$, called
the assembly map, defines a natural homomorphism

$\mu:K^{g}_{0/1}(X,\alpha)\rightarrow K^{a}_{0/1}(X,\alpha)$

which is an isomorphism for any smooth manifold $X$ with a twisting
$\alpha:X\rightarrow K(\mathbb{Z},3)$.

Before giving the definitions, we need to point out that there is another
description of a twist over a space here. Let $X$ be a second countable
locally compact Hausdorff topological space. Then a twist on $X$ in [6] is a
locally trivial bundle $\mathcal{A}$ of elementary $C^{\ast}$-algebras on $X$,
i.e each fiber of $\mathcal{A}$ is an elementary $C^{\ast}$-algebra and with
the structure group the automorphism group of $\mathcal{K}(\mathbb{H})$ for
some Hilbert space $\mathbb{H}$. However, these two description can be
transferred from one to the other. In [2] they used ($X$, $\mathcal{A}$) as
the starting point and defined topological twisted $K$-cycles as follows.

######  Definition 2.4.

An $\mathcal{A}$-twisted $K$-cycles on $X$ is a triple $(M,\sigma,\psi)$ where

  * •

$M$ is a compact spinc-manifold without boundary.

  * •

$\phi:M\rightarrow X$ is a continuous map.

  * •

$\sigma\in K_{0}(\Gamma(M,\phi^{\ast}\mathcal{A}^{op}))$

Let $E(X,\alpha)$ be all of the $K$-cycles over $(X,\mathcal{A})$. Then the
topological $\mathcal{A}$-twisted $K$-homology group over $(X,\mathcal{A})$ is
defined by

$K^{top}_{\ast}(X,\alpha):=E(X,\alpha)/\sim$ (2.8)

Here $\sim$ is a similar equivalence generated by disjoint unions, bordisom
and vector bundle modification. Moreover, a more geometric twisted $K$-cycle
is given in [2], which is called $D$-cycle. In order to introduce $D$-cycle,
we still also need the notion of spinor bundles.

######  Definition 2.5.

A spinor bundle for a twisting $\mathcal{A}$ is a vector bundle $S$ of Hilbert
spaces on $X$ together with a given isomorphism of twisting data

$\mathcal{A}\cong\mathcal{K}(S)$ (2.9)

One fact we need to know about spinor bundle is that a spinor bundle for
$\mathcal{A}$ exists if and only if $DD(\mathcal{A})=0$. Now we can give the
definition of $D$-cycle in [2].

######  Definition 2.6.

A $D$-cycle for $(X,\mathcal{A})$ is a $4$-tuple $(M,E,\phi,S)$ such that

  * •

$M$ is a closed oriented $C^{\infty}$-Riemannian manifold.

  * •

$E$ is a complex vector bundle on $M$.

  * •

$\phi$ is a continuous map from $M$ to $X$.

  * •

$S$ is a spinor bundle for
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\phi^{\ast}\mathcal{A}^{op}$.

Two $D$-cycles $(M,E,\phi,S),(M^{\prime},E^{\prime},\phi^{\prime},S^{\prime})$
for $(X,\mathcal{A})$ are isomorphic if there is an orientation preserving
isometric diffeomorphism $f:M\rightarrow M$ such that the diagram

${M}$${M^{\prime}}$${\ X}$${f}$${\phi}$${\phi^{\prime}}$ (2.10)

commutes, and $f^{\ast}E^{\prime}\cong E,f^{\ast}S^{\prime}\cong S$.

Let $\Gamma^{D}(X,\alpha)$ be the collection of all $D$-cycles over
$(X,\mathcal{A})$. Similar, we can impose an equivalence generated by disjoint
union, bordism and vector bundle modification on $\Gamma^{D}(X,\alpha)$ and
get the geometric $\mathcal{A}$-twisted $K$-homology in [2], which we denote
by $K^{geo}_{\ast}(X,\mathcal{A})$.

  * •

Direct sum - disjoint union If $(M,E_{1},\iota,S)$ and $(M,E_{2},\iota,S)$ are
$D$-cycles for $(X,\alpha)$ then

$(M,E_{1},\iota,S)\cup(M,E_{2},\iota,S)\sim(M,E_{1}\oplus E_{2},\iota,S)$
(2.11)

  * •

Bordism Given two $D$ cycle $(M_{0},E_{0},\iota_{0},S_{0})$ and
$(M_{1},E_{1},\iota_{1},S_{1})$ , if there exists a $4$-tuple $(W,E,\Phi,S)$
such that $W$ is a compact oriented Riemannian manifold with boundary, $E$ is
a complex vector bundle over $W$ , $\Phi$ is a continuous map from $W$ to $X$
and

$(\delta W,E|\delta W,\Phi|\delta W,S^{+}|\delta
W)\cong(M_{0},E_{0},\iota_{0},S_{0})\cup(M_{1},E_{1},\iota_{1},S_{1})$ (2.12)

Here $S^{+}|\delta$ is the positive part of $S$. It is $S$ itself when $W$ is
odd dimensional.

  * •

Spinc-vector bundle modification Given a $D$ cycle $(M,E,\iota,S)$ and a
spinc-vector bundle $V$ over $M$ with even dimensional fibers. Let
$S^{\prime}$ be the spinor bundle of
Cliff${}^{+}_{\mathbb{C}}(T\hat{M})\otimes(\phi\circ\rho)^{\ast}(\mathcal{A})^{op}$.
Use the notation in the definition of geometric $K$-cycles. Then

$(M,E,\iota,S)\sim(\hat{M},\beta\otimes\rho^{\ast}E,\iota\circ\rho,\rho^{\ast}S).$
(2.13)

Here $\beta$ is a complex vector bundle over $\hat{M}$ whose restriction to
each fiber of $\rho$ is the Bott generator vector bundle of $\hat{M}$.

In [2] they gave a natural charge map
$h:K^{geo}_{\ast}(X,\mathcal{A})\rightarrow K^{top}_{\ast}(X,\mathcal{A})$ as
follows: Let $(M,E,\phi,S)$ be a $D$-cycle and choose a normal bundle for $M$
with even dimensional fibers, then

$h(M,E,\phi,S):=(S(v\oplus\underline{\mathbb{R})},\phi\circ\pi,\sigma)$ (2.14)

Here $\sigma$ is defined as the image of $E$ under the composition of $s_{!}$
and $\chi$.

  1. 1.

Let $s:M\rightarrow S(\upsilon\oplus\underline{\mathbb{R}})$ be the canonical
section of unite section on the trivial real line bundle. For simplicity, we
denote the total space of sphere bundle
$S(\upsilon\oplus\underline{\mathbb{R}})$ by $M$ and the bundle map by $\rho$.
Then $s_{!}:K^{0}(M)\rightarrow
K_{0}(\Gamma(\hat{M},\rho^{\ast}(Cliff_{\mathbb{C}}(\upsilon))))$ is the Gysin
homomorphism induced by $s$.

  2. 2.

$\chi:K_{0}(\Gamma(\hat{M},\rho^{\ast}(Cliff_{\mathbb{C}}(\upsilon))))\rightarrow
K_{0}(\Gamma(\hat{M},(\phi\circ\rho)^{\ast}\mathcal{A}^{op}))$ is the
isomorphism induced by the trivialisation of $TM\oplus\upsilon$ and the given
spinor bundle $S$.

They propose a question which asked that if $h$ is an isomorphism. We will
prove in the paper that it is true when $X$ is a countable $CW$-complex.

######  Remark 2.7.

We should note that all of these definitions of geometric twisted $K$-homology
group cannot work for general twists. The twists should satisfy the Freed-
Witten condition (see [20]), i.e, there exists a manifold $M$ and a continuous
map $l:M\rightarrow X$ s.t

$l^{\ast}(H)+W_{3}(M)=0$ (2.15)

##  3 Equivalence between two versions of geometric twisted K-homology

We list the following theorem from [17] first.

######  Theorem 3.1.

Let $E$ be a vector bundle over a space $X$ and $\tilde{E}$ be the Clifford
bundle of $E$. Let $W_{3}(E)$ be the third integral Stiefel-Whitney class of
$E$. Then we have:

$\displaystyle W_{3}(E)=\begin{cases}DD(\tilde{E})&if\ $E$\ has\ even\
dimension\\\ DD(\tilde{E}^{+})&if\ $E$\ has\ odd\ dimension.\end{cases}$ (3.1)

######  Remark 3.2.

Through the above theorem we can get a bit of idea why the two versions of
geometric cycle are equivalent. The existence of spinor bundle implies the
triviality of Dixmier-Douady class of
$DD(Cliff^{+}(TM)\otimes\phi^{\ast}(\mathcal{A}^{op})$, so the choice of the
spinor bundle $S$ for $Cliff^{+}(TM)\otimes\phi^{\ast}\mathcal{A}^{op}$ in the
definition of $D$-cycle corresponds to the choice of the homotopy $\eta$ in
the definition of geometric $K$-cycle in [20].

Let $X$ be a locally finite $CW$-complex and $\alpha:X\rightarrow
K(\mathbb{Z},3)$ be a twist in the sense of [20]. From the discussion in the
beginning of section $2$ we know that $\alpha$ determines an associated
$\mathcal{K}$-bundle $\mathcal{A}$ over $X$. While $\mathcal{A}$ is exactly a
twist in the sense of [2]. Therefore, we get the basic set up data of the two
definitions of geometric twisted $K$-cycles are equivalent.

Moreover, we can consider twisting datas in different categories. Define
$Twist_{1}(X)$ to be a category with continuous maps $\alpha:X\rightarrow
K(\mathbb{Z},3)$ as objects. The morphisms of $Twist_{1}(X)$ are homotopies
between objects. Define $Twist_{2}(X)$ to be a category with
$\mathcal{H}$-bundles over $X$ as objects. The morphisms of $Twist_{2}(X)$ are
isomorphisms of $\mathcal{K}$-bundles. Obviously, we can see that
$Twist_{1}(X)$ and $Twist_{2}(X)$ are groupoids and the sets of equivalence
classes for them are both $H^{3}(X,\mathbb{Z})$. Now we give two lemmas which
is used later, whose proof are probably already known and so we don’t claim
that they are original.

######  Lemma 3.3.

Denote the projection from $X\times I$ to $X$ by $p$. For every
$\mathcal{K}$-bundle $\mathcal{A}$ over $X\times I$, there exists a
$\mathcal{K}$-bundle $\mathcal{A}^{\prime}$ over $X$ such that
$\mathcal{A}\cong p^{\ast}(\mathcal{A}^{\prime})$.

The proof is obvious.

######  Lemma 3.4.

Let $\mathcal{A}$ be a $\mathcal{K}$-bundle over $X$, then there is a
canonical spinor bundle for $\mathcal{A}\otimes\mathcal{A}^{op}$.

The proof can be found in [2].

Now we give a construction which is useful in the proof of the main theorem in
this section. Assume $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ are two
$\mathcal{K}$-bundles over $X$ and
$\lambda:\mathcal{K}_{1}\rightarrow\mathcal{K}_{2}$ is an isomorphism. Then we
can get a $\mathcal{K}$-bundle over $X\times[0,1]$ as follows. First we can
see that $\mathcal{K}_{1}\times[0,1/2]$ and $\mathcal{K}_{2}\times[1/2,1]$ are
$\mathcal{K}$-bundle over $X\times[0,1/2]$ and $X\times[1/2,1]$ respectively.
Then we can glue $\mathcal{K}_{1}\times[0,1/2]$ and
$\mathcal{K}_{2}\times[1/2,1]$ at $\\{1/2\\}\times X$ via $\lambda$ and get a
$\mathcal{K}$-bundle $\mathcal{K}_{0}$ over $X\times[0,1]$. Now we give the
main theorem of this section.

######  Theorem 3.5.

Let $X$ be a locally finite $CW$-complex. There exists an isomorphism
$\mathcal{F}:K^{g}(X,\alpha)\rightarrow K^{geo}(X,\mathcal{A})$.

To make the proof easier to read, we give the basic idea first. The idea is
that we transform spinor bundles in $D$-cycles to $\mathcal{K}$-bundles over
$M\times[0,1]$. Then we use this $\mathcal{K}$-bundles to define homotopies in
$\alpha$-twisted spinc-manifolds and also define $F$ below. And we reverse the
procedure to prove that $F$ is surjective. The injectivity of $F$ is
essentially implied by the compatibility of $F$ with $\sim$. Now we start the
proof.

###### Proof.

Let $[x]$ be a class in $K^{geo}(X,\mathcal{A})$ and $(M^{g},E,\phi,S)$ be a
$D$-cycle which represents $[x]$. By definition $S$ is a spinor bundle of
Cliff${}^{+}_{\mathbb{C}(TM)}\otimes\phi^{\ast}\mathcal{A}^{op}$. And we
denote the chosen isomorphism between $\mathcal{K}(S)$ and
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\phi^{\ast}\mathcal{A}^{op}$ by $\lambda$.
We define $F([x])$ in $K^{g}(X,\alpha)$ to be a class represented by
$(M,\phi,\upsilon,\eta,E)$, in which

  * •

$M$ is the manifold of $M^{g}$ forgetting Riemiannian structure, $\phi$ and
$E$ are the same map and bundle in the $D$-cycle;

  * •

$\upsilon$ is the classifying map of the stable normal bundle of $M$;

  * •

$\eta$ is a homotopy between $W_{3}\circ\upsilon$ and $\alpha\circ\phi$.

We only need to explain how $\eta$ is constructed from $(M^{g},E,\phi,S)$. By
Lemma 3.4 we know that there exists a canonical Hilbert bundle $V$ over $M$
and a canonical isomorphism $c$ between
$\mathcal{K}(V)\cong\phi^{\ast}\mathcal{A}\otimes\phi^{\ast}\mathcal{A}^{op}$.
Combine $c$ and $h$ we get an isomorphism between
$\mathcal{K}(S)\otimes\mathcal{A}$ and
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\mathcal{K}(V)$. According to the
discussion before the theorem we can obtain a $\mathcal{K}$-bundle
$\mathcal{W}$ over $M\times I$ such that
$\mathcal{W}_{M\times\\{0\\}}\cong\mathcal{K}(S)\otimes\mathcal{A}$ and
$\mathcal{W}_{M\times\\{1\\}}\cong
Cliff^{+}_{\mathbb{C}}(TM)\otimes\mathcal{K}(V)$. Since $BPU(\mathbb{H})$ is
also a classifying space of $\mathcal{K}$-bundles, therefore we can get that
there exists an $\eta:X\times[0,1]\rightarrow K(\mathbb{Z},3)$ such that
$(\eta\circ(\phi\times Id))^{\ast}(\mathfrak{K})$ is isomorphic to
$\mathcal{W}$. Moreover, we get two maps $\alpha_{0}$,
$\alpha_{1}:X\rightarrow K(\mathbb{Z},3)$ by restricting $\eta$ to
$X\times\\{0\\}$ and $X\times\\{1\\}$ respectively, which gives the following
isomorphisms:

$(\alpha_{0}\circ\phi)^{\ast}(\mathfrak{K})\cong\mathcal{K}(S)\otimes\mathcal{A},\
\ (\alpha_{1}\circ\phi)^{\ast}(\mathfrak{K})\cong
Cliff^{+}_{\mathbb{C}}(TM)\otimes\mathcal{K}(V)$ (3.2)

Since different choices of $\eta$ are homotopic to each other, so they
represent the same class in $K^{g}(X,\alpha)$ via bordism relation. To show
that $F$ is well defined, we still need to check that it is compatible with
the relations which define $\sim$.

  * •

From the definition of $F$ we can see

$\displaystyle
F([(M^{g},E_{1},\phi,S)]\cup[(M^{g},E_{2},\phi,S)])=[(M,E_{1}\oplus
E_{2},\phi,S)];$ $\displaystyle F([M^{g},E_{1},\phi,S])\cup
F([M^{g},E_{2},\phi,S])=[(M,E_{1}\oplus E_{2},\phi,S)]$

i.e $F$ respects the disjoint union relation.

  * •

Let $(M^{g},E,\phi,S)$ be a bordism between $(M^{g}_{0},E_{0},\phi_{0},S_{0})$
and $(M^{g}_{1},E_{1},\phi_{1},\\\ S_{1})$. Denote the associated isomorphisms
of the spinor bundles by $h$, $h_{0}$ and $h_{1}$. Denote the chosen
representing cycles of the image of their homology classes under $F$ by
$(M,\phi,\upsilon,\eta,E)$, $(M_{0},\phi_{0},\upsilon_{0},\eta_{0},E_{0})$ and
$(M_{1},\phi_{1},\upsilon_{1},\\\ \eta_{1},E_{1})$ respectively. Choosing a
tubular neighborhood of $M_{0}$ in $M$ we can get that $TM|_{M_{0}}\cong
TM_{0}\otimes\underline{\mathbb{R}}$, so we get the stable normal bundle of
$M_{0}$ agrees with the restriction of the stable normal bundle of $M$ to
$M_{0}$ i.e $\upsilon|_{M_{0}}$ is homotopic to $\upsilon_{0}$. Similarly we
can get $\upsilon_{M_{1}}$ is homotopic to $\upsilon_{1}$.

Let $\mathcal{W}$, $\mathcal{W}_{0}$ and $\mathcal{W}_{1}$ be the three
$\mathcal{K}$-bundles over $M\times[0,1]$, $M_{0}\times[0,1]$ and
$M_{1}\times[0,1]$ respectively. The isomorphism $c$ in Lemma 3.4 is natural
and $h|_{\mathcal{K}(S_{i})}=h_{i}$ ($i=0$, $1$), so we get that
$\mathcal{W}|_{M_{i}\times[0,1]}$ is isomorphic to $\mathcal{W}_{i}$ ($i=0$,
$1$). This implies that $\eta|_{M\times\\{i\\}}$ is homotopic to $\eta_{i}$
($i=0$, $1$). So $F$ is compatible with bordism.

  * •

Use the notion in Section $2.1$ and denote
$F([(\hat{M}^{g},S^{+}_{V}\otimes\rho^{\ast}E,\phi\circ\rho,\rho^{\ast}S)$ by
$[\hat{M},S^{+}_{V}\otimes\rho^{\ast}E,\phi\circ\rho,\upsilon^{\prime},\eta^{\prime}]$.
We only need to prove that $\eta^{\prime}$ is homotopic the $\eta\circ\pi$.
The is implied by the observation that the corresponding $\mathcal{K}$-bundles
$\mathcal{V}^{\prime}$ and $\mathcal{V}^{\prime\prime}$ over
$\hat{M}\times[0,1]$ are isomorphic.

Obviously $F$ maps a trivial $D$-cycle to a trivial geometric twisted
$K$-cycle, so we get that $F$ is a well defined injective homomorphism. The
left is to show that $F$ is surjective. For any class $[y]\in
K^{g}(X,\alpha)$, choose a geometric cycle $(M,\phi,\upsilon,\eta,E)$ to
represent it. $\eta$ induces a $\mathcal{K}$-bundle over $M\times[0,1]$, which
we denote by $\mathcal{X}$. By the definition of $\eta$ we have
$\mathcal{X}|_{M\times\\{0\\}}\cong$
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\mathcal{K}$,
$\mathcal{X}|_{M\times\\{1\\}}\cong(\alpha\circ\phi)^{\ast}(\mathfrak{K})$.
Let $l$ be an isomorphism between
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\mathcal{K}$ and
$(\alpha\circ\phi)^{\ast}(\mathfrak{K})$ and $\mathcal{L}$ be a
$\mathcal{K}$-bundle constructed via gluing
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\mathcal{K}\times[0,1/2]$ and
$(\alpha\circ\phi)^{\ast}(\mathfrak{K})\times[1/2,1]$ by $l$ at
$M\times\\{1/2\\}$. Then we have that $\mathcal{L}$ is isomorphic
$\mathcal{X}$ since they have the same Dixmier-Dourady class. Combining $l$
with the canonical isomorphism in Lemma 3.4, we get a spinor bundle $S$ for
Cliff${}^{+}_{\mathbb{C}}(TM)\otimes\phi^{\ast}\mathcal{A}^{op}$. The
$D$-cycle $(M^{g},E,\phi,S)$ satisfies that $F([M^{g},E,\phi,S])=[y]$. ∎

######  Remark 3.6.

The above proposition shows that geometric $K$-cycles in [2] and $D$-cycles in
[20] are equivalent to each other. Therefore we can choose one of them to
construct the topological $T$-duality for geometric twisted $K$-homology.
Moreover, $K^{g}(X,\alpha)\cong K^{a}(X,\alpha)$ imples that
$K^{geo}(X,\alpha)$ is also isomorphic to $K^{a}(X,\alpha)$.

##  4 Properties of geometric twisted K-homology

In this section we prove that geometric twisted $K$-homology satisfies
Elienberg-Steenrod axioms of homology theory. We first give a simple lemma.

######  Lemma 4.1.

If $f:Y\rightarrow X$ is a continuous map, then $f$ induces a homomorphism
$f_{\ast}:K^{g}(Y,\alpha\circ f)\rightarrow K^{g}(X,\alpha)$.

###### Proof.

Given a twisted geometric $K$-cycle $(M,\phi,\upsilon,\eta,E)$ over
$(Y,\alpha\circ f)$, we define $f_{\ast}$ by

$f_{\ast}(M,\phi,\upsilon,\eta,E)=(M,f\circ\phi,\upsilon,\eta,E)$ (4.1)

We need to check that $f_{\ast}$ is compatible with disjoint union, bordism
and spinc-vector bundle modification.

  * •

Given two geometric $K$-cycles $(M,\phi,\upsilon,\eta,E_{i})$ ($i=1,2$), we
have

$f_{\ast}((M,\phi,\upsilon,\eta,E_{1})\cup(M,\phi,\upsilon,\eta,E_{2}))=(M,\phi\circ
f,\upsilon,\eta,E_{1}\oplus E_{2})$

  * •

If $(M,\phi,\upsilon,\eta,E)$ is a bordism between
$(M_{1},\phi_{1},\upsilon_{1},\eta_{1},E_{1})$ and
$(M_{2},\phi_{2},\upsilon_{2},\\\ \eta_{2},E_{2})$, then clearly
$(M,f\circ\phi,\upsilon,\eta,E)$ gives a bordism between
$(M_{1},f\circ\phi_{1},\upsilon_{1},\eta_{1},E_{1})$ and
$(M_{2},f\circ\phi_{2},\upsilon_{2},\eta_{2},E_{2})$.

  * •

Since $f((\hat{M},\phi\circ\rho,\upsilon\circ\rho,\eta\circ(\rho\times
Id),\rho^{\ast}E\otimes S^{+}_{V}))$ is
$(\hat{M},f\circ\phi\circ\rho,\upsilon\circ\rho,\eta\circ(\rho\times
Id),\rho^{\ast}E\otimes S^{+}_{V})$, which is exactly a spinc-vector bundle
modification of $(M,f\circ\phi,\upsilon,\eta,E)$, so we get that $f_{\ast}$
respects the spinc-vector bundle modification relation.

∎

######  Theorem 4.2 (Homotopy).

If $f:Y\rightarrow X$ is a homotopy equivalence, then the induced map
$f_{\ast}:K^{g}_{\ast}(Y,\alpha\circ f)\rightarrow K^{g}_{\ast}(X,\alpha)$ is
an isomorphism.

###### Proof.

We first show that if $g:Y\rightarrow X$ is homotopic to $f$, then
$K^{g}_{\ast}(Y,\alpha\circ f)\cong K^{g}_{\ast}(Y,\alpha\circ g)$. Let
$H:Y\times[0,1]\rightarrow X$ be a homotopy from $f$ to $g$ i.e $H(y,0)=f(y)$
and $H(y,1)=g(y)$. Given a twisted geometric $K$-cycle
$(M,\phi,\upsilon,\eta,E)$ over $(Y,\alpha\circ f)$, we get a twisted
geometric $K$-cycle over $(Y,\alpha\circ g)$ as follows:
$(M,\phi,\upsilon,\eta^{\prime},E)$. Here
$\eta^{\prime}:M\times[0,1]\rightarrow K(\mathbb{Z},3)$ is defined by

$\displaystyle\eta^{\prime}(m,t)=\begin{cases}\eta(m,2t)&0\leq t\leq 1/2\\\
\alpha\circ H(m,2t-1)\circ(\phi\times Id)&1/2\leq t\leq 1\end{cases}$

It is not hard to check that the above map is compatible with the disjoint
union, bordism and spinc-vector bundle modification. We skip the details here
since they are similar to the proof of the above lemma. Therefore we get a
homomorphism $H_{\ast}$ from $K^{g}_{\ast}(Y,\alpha\circ f)$ to
$K^{g}_{\ast}(Y,\alpha\circ g)$. Similarly we can get the inverse of
$H_{\ast}$ by using $H(1-t,y)$ as a homotopy from $g$ to $f$. So we get that
$H_{\ast}$ is an isomorphism. Clearly, we have that $f_{\ast}=g_{\ast}\circ
H_{\ast}$. Let $q:X\rightarrow Y$ be a homotopy inverse of $f:Y\rightarrow X$
i.e $f\circ q$ is homotopic to $id_{X}$ and $q\circ f$ is homotopic to
$id_{Y}$. Denote the associated homotopies by $H_{1}$ and $H_{2}$
respectively. Then we have that $(f\circ q)_{\ast}=(H_{1})_{\ast}$ and
$(q\circ f)_{\ast}=(H_{2})_{\ast}$. Since $(H_{1})_{\ast}$ and
$(H_{2})_{\ast}$ are both isomorphisms, we get that $f_{\ast}$ is an
isomorphism as well. ∎

######  Theorem 4.3 (Excision).

Let $(X,Y)$ be a pair of locally finite $CW$-spaces, $U$ is an open set of $X$
such that $\overline{U}\in Y$. Then we the inclusion

$i:(X-U,Y-U)\hookrightarrow(X,Y)$

induces an isomorphism

$K^{g}_{\ast}(X-U,Y-U;\alpha\circ i)\cong K^{g}_{\ast}(X,Y;\alpha)$ (4.2)

###### Proof.

By Lemma 4.1 the inclusion $i$ induces a homomorphism $i_{\ast}$. And moreover
$i_{\ast}$ is injective by its definition. The remainder is to prove that
$i_{\ast}$ is surjective. For any $y\in K^{g}_{\ast}(X,Y;\alpha)$, we choose a
geometric cycle $(M,\phi,\upsilon,\eta,E)$ to represent it. We choose a point
$p_{0}\in U$. By the Urysohn’s Lemma, there exists a continuous function
$f:X\rightarrow[0,1]$ such that $f(p_{0})=0$ and $f(x)=1$ for any $x\in X-U$.
Then $f\circ\phi$ is a continuous function from $M$ to $[0,1]$. Let $W$ be a
sub-manifold of $M\times[0,1]$ defined by
$W=\\{(t,(f\circ\phi)^{-1}([t,1])|t\in[0,1]\\}$. Let $\pi:W\rightarrow M$ be
the canonical projection. For each slice $W_{t}=(f\circ\phi)^{-1}([t,1])$ in
$W$, we denote the inclusion of $W_{t}$ to $M$ by $i_{t}$. Define
$\eta^{\prime}:W\times[0,1]$ as follows

$\eta^{\prime}(m,s,t)=\eta\circ(i_{s}\times id)(m,t)$

Here $(m,s)\in W$ and $t\in[0,1]$ Then the $\alpha$-twisted spinc-manifold
$(W,\phi\circ\pi,\upsilon\circ\pi,\eta^{\prime})$, which gives a bordism
between $(M,\phi,\upsilon,\eta,E)$ and $(f^{-1}(X-U),\phi\circ j,\upsilon\circ
j,\eta\circ(j\times Id),j^{\ast}E)$. Here $j:f^{-1}(X-U)\hookrightarrow M$ is
the inclusion. While $(f^{-1}(X-U),\phi\circ j,\upsilon\circ
j,\eta\circ(j\times Id),j^{\ast}E)$ is also a twisted geometric $K$-cycle over
$(X-U,Y-U;\alpha\circ i)$, whose image under $i_{\ast}$ is equivalent to
$(M,\phi,\upsilon,\eta,E)$. ∎

######  Remark 4.4.

In the above proof we assume each slice $W_{t}$ to be a sub-manifold of $M$,
this is not true in general. But here we assume that both $X$ and $U$ are
homotopic to finite $CW$-complexes, so we can always find a smooth manifold
which is homotopy equivalent to $W_{t}$. This makes the proof still works for
general cases.

Another important axiom in Eilenberg-Steenrod axioms is the long exact
sequence. Before moving on to the long exact sequence of geometric twisted
$K$-homology, we first introduce a notion.

######  Definition 4.5.

A twist $\alpha:X\rightarrow K(\mathbb{Z},3)$ is called representable if there
exists an oriented real vector bundle $V$ over $X$ such that
$W_{3}(V)=[\alpha]$. Here $[\alpha]$ is the pullback of the generator of
$H^{3}(K(\mathbb{Z}),3)$ along $\alpha$.

######  Theorem 4.6 (Six-term exact sequence).

Let $Y$ be a sub-space of $X$ and $i$ be the inclusion map from $Y$ to $X$.
Let $\alpha$ be a representable twist over $X$. Then we have a six-term exact
sequence:

$K^{g}_{0}(Y,\alpha\circ
i)$$K^{g}_{0}(X,\alpha)$$K^{g}_{0}(X,Y;\alpha)$$K^{g}_{1}(X,Y;\alpha)$$K^{g}_{1}(X,\alpha)$$K^{g}_{1}(Y,\alpha\circ
i)$$i_{\ast}$$j_{\ast}$$\delta$$i_{\ast}$$j_{\ast}$$\delta$

Here the boundary operator is given by

$\delta([M,\phi,\upsilon,\eta,E])=[(\partial M,\phi|_{\partial
M},\upsilon|_{\partial M},\eta|_{\partial M\times[0,1]},E|_{\partial M})]$
(4.3)

To prove this theorem, we first prove two lemmas as a preparation.

######  Lemma 4.7.

Let $\theta=(M,\iota,\upsilon,\eta,E)$ be a geometric $K$-cycle over
$(X,\alpha)$ and $E_{i}$ ($i=1$, $2$) be spinc-vector bundles over $M$ with
even dimensional fibers. Denote the vector bundle modification of $\theta$
with a spinc-vector bundle $F$ by $\theta_{F}$. Then we have that
$\theta_{E_{1}\oplus E_{2}}$ is bordant to
$(\theta_{E_{1}})_{p_{1}^{\ast}E_{2}}$, in which $p_{1}$ is the projection
from the sphere bundle $S(E_{1}\otimes\underline{\mathbb{R}})$ to $M$.

###### Proof.

Assume the dimension of the fiber of $E_{i}$ is $n_{i}$ and write
$\theta_{E_{1}\oplus E_{2}}$ and $(\theta_{E_{1}})_{p_{1}^{\ast}E_{2}}$
explicitly as $(V,\iota_{V},\upsilon_{V},\eta_{V},E_{V})$ and
$(W,\iota_{W},\upsilon_{W},\eta_{W},E_{W})$ respectively. Then the fibers of
$V$ and $W$ are $S^{n_{1}+n_{2}}$ and $S^{n_{1}}\times S^{n_{2}}$
respectively. We can embed both of the two bundles over $M$ into the vector
bundle $E_{1}\oplus E_{2}\oplus\mathbb{R}$ as follows. First we choose a
Riemannian metric over $E_{1}\oplus E_{2}\oplus\mathbb{R}$ and embed $V$ into
$E_{1}\oplus E_{2}\oplus\mathbb{R}$ as the standard unit sphere bundle. We
embed $W$ into $E_{1}\oplus E_{2}\oplus\underline{\mathbb{R}}$ such that in
each fiber over $p\in M$ it is embedded as follows:

$((x,s),(y,t))\mapsto((5-t)(x,s),y)$

in which $x\in(E_{1})_{p}$, $y\in(E_{2})_{p}$ and
$s,t\in(\underline{\mathbb{R}})_{p}$. A careful tells us that this indeed
induces an embedding of $W$ into $E_{1}\oplus
E_{2}\oplus\underline{\mathbb{R}}$ and we still denote the image of the
embedding by $W$. We embed $V$ into $E_{1}\oplus
E_{2}\oplus\underline{\mathbb{R}}$ with a scaling such that the radius of each
fiber is 10. Denote the standard disk bundle with radius 10 of $E_{1}\oplus
E_{1}\oplus\underline{\mathbb{R}}$ by $D^{n_{1}+n_{2}+1}(10)$ and the solid
torus bundle bounds by $V$ by $\bar{V}$. Then we have that
$D^{n_{1}+n_{2}+1}(10)-\bar{V}$ (which we denote by $Z$) gives rise to a
bordism between $V$ and $W$. And $(Z,\iota\circ
p_{Z},\upsilon_{Z},\eta\circ(p_{Z}\times id),E_{Z})$ gives a bordism between
$(V,\iota_{V},\upsilon_{V},\eta_{V},E_{V})$ and
$(W,\iota_{W},\upsilon_{W},\eta_{W},E_{W})$. ∎

######  Lemma 4.8.

Let $\alpha$ be a representable twist over $X$ and
$(M,\iota,\upsilon,\eta,E_{M})$ be a geometric cycle over $(X,\alpha)$. Let
$(\partial M,\iota_{\partial M},\\\ \upsilon_{\partial M},\eta_{\partial
M},E_{\partial M})$ be the restriction to the boundary of $M$. If a spinc-
vector bundle modification with vector bundle $E$ of $(\partial
M,\iota_{\partial M},\upsilon_{\partial M},\eta_{\partial M},E_{\partial M})$
is bordant to trivial cycle, then there exists a spinc-vector bundle $V$ over
$\partial M$ such that the spinc-vector bundle modification with $V$ is
bordant to the trivial cycle and $V$ can be extended to a spinc-vector bundle
over $M$.

###### Proof.

Denote the spinc-vector bundle modification of $(\partial M,\iota_{\partial
M},\upsilon_{\partial M},\eta_{\partial M},E_{\partial M})$ with vector bundle
$E$ by $(Q,\iota_{\partial M}\circ\pi,\upsilon_{Q},\eta_{Q},E_{Q})$, which is
bordant to the trivial cycle via a bordism of
$(W,\iota_{W},\upsilon_{W},\eta_{W},E_{W})$. There exists a normal bundle $F$
over $W$, whose restriction to $Q$ is also a normal bundle of $TQ$. On the
other hand, by the construction of spinc-vector bundle modification we can
observe that there exists a normal bundle of $TQ$ such that it is isomorphic
to the pullback (along $\pi$) of the direct sum of a normal bundle of
$T\partial M$ (which we denote by $N(T\partial M)$) and a normal bundle of $E$
(which we denote by $N(E)$). Consider the spinc-modification of
$(W,\iota_{W},\upsilon_{W},\eta_{W},E_{W})$ with
$F\oplus\iota_{W}^{\ast}(\mathbf{V})$ (here $\mathbf{V}$ is the vector bundle
over $X$ with $W_{3}(V)=[\alpha]$). It gives a bordism from the spinc-
modification of $(Q,\iota_{\partial M}\circ\pi,\upsilon_{Q},\eta_{Q},E_{Q})$
with $(F\oplus\iota_{W}^{\ast}(\mathbf{V}))_{|Q}$ and the trivial cycle.
According to Lemma 4.7 and the observation before we can see that the spinc-
modification of $(Q,\iota_{\partial M}\circ\pi,\upsilon_{Q},\eta_{Q},E_{Q})$
with $(F\oplus\iota_{W}^{\ast}(\mathbf{V}))_{|Q}$ is bordant to the spinc-
modification of $(\partial M,\iota_{\partial M},\upsilon_{\partial
M},\eta_{\partial M},E_{\partial M})$ with $E\oplus N(T\partial M)\oplus
N(E)\oplus\iota_{W}^{\ast}(\mathbf{V})_{|Q}$. While $E\oplus N(E)$ is trivial,
we get that the spinc-modification of $(\partial M,\iota_{\partial
M},\upsilon_{\partial M},\eta_{\partial M},E_{\partial M})$ with $N(T\partial
M)\oplus\iota_{\partial M}^{\ast}(\mathbf{V})$ is bordant to a trivial cycle.
The normal bundle on the boundary can be extended to the whole manifold
obviously and $\iota_{\partial M}^{\ast}(\mathbf{V})$ can be extended to a
vector bundle $\iota_{M}^{\ast}(\mathbf{V})$ So we get our statement. ∎

Now we start the proof of Theorem 4.6.

###### Proof.

We will show the exactness at $K^{g}_{0}(X,\alpha)$, $K^{g}_{0}(X,Y;\alpha)$
and $K^{g}_{1}(Y,\alpha)$. The proof of the rest part is similarly.

  * •

For any $[y]\in K^{g}_{0}(Y,\alpha)$, we choose a twisted geometric $K$-cycle
$(M_{0},\phi_{0},\upsilon_{0},\\\ \eta_{0},E_{0})$ to represent it. Its image
under $j_{\ast}\circ i_{\ast}$ is
$(M_{0},i\circ\phi_{0},\upsilon_{0},\eta_{0},E_{0})$, which is cobordant to
trivial $K$-cycle relative $Y$ in $X$. Therefore we have $j_{\ast}\circ
i_{\ast}([y])$ is trivial. Assume that $[x]\in K^{g}_{0}(X,\alpha)$ and
$j^{\ast}([x])=0$. We still choose a twisted geometric $K$-cycle
$(M_{1},\phi_{1},\upsilon_{1},\eta_{1},E_{1})$ to represent $[x]$. Since
$j^{\ast}([x])$ is trivial, we obtain that if we do several times of spinc-
vector bundle modification for
$(M_{1},j\circ\phi_{1},\upsilon_{1},\eta_{1},E_{1})$ relative to $Y$ in $X$
i.e if we denote the results of $Spin^{c}$-vector bundle modifications by
$(\hat{M_{1}},j\circ\phi_{1}\circ\rho,\upsilon_{1}^{\prime},\eta_{1}^{\prime},E_{1}^{\prime})$,
then
$(\hat{M_{1}},j\circ\phi_{1}\circ\rho,\upsilon_{1}^{\prime},\eta_{1}^{\prime},E_{1}^{\prime})$
satisfies that $(j\circ\phi_{1}\circ\rho(\hat{M_{1}}))\subset Y$, which also
implies that $j\circ\phi_{1}(M)\subset Y$. Therefore we get that $[x]\in$ im
$i_{\ast}$.

  * •

First of all we need to point out that $\delta$ is well defined i.e it is
compatible with disjoint union, bordism and spinc-vector bundle modification.
It is a tedious check from the definition of $\delta$, which we skip here. By
the definition of $\delta$ we can see that $\delta\circ j_{\ast}=0$. To show
that $ker\delta\subset imj_{\ast}$, we choose a twisted geometric $K$-cycle
$(M_{2},\phi_{2},\upsilon_{2},\eta_{2},E_{2})$ such that
$\delta[(M_{2},\phi_{2},\upsilon_{2},\eta_{2},E_{2})]$ is trivial i.e several
times of spinc-vector bundle modification for $(\partial M_{2},\phi|_{\partial
M_{2}},\upsilon|_{\partial M_{2}},\eta|_{\partial M_{2}},E|_{\partial M_{2}})$
is cobordant to trivial $K$-cycle over $Y$. By Lemma 4.8 we can assume that
each spinc-vector bundle over the boundary of a manifold can be extended to a
spinc-vector bundle over the whole manifold, therefore we get that if we do
the spinc-vector bundle modifications for
$(M_{2},\phi_{2},\upsilon_{2},\eta_{2},E_{2})$, then the resulting twisted
spinc-manifold is bordant to a twisted spinc-manifold without boundary over
$X$. Finally we obtain that $(M_{2},\phi_{2},\upsilon_{2},\eta_{2},E_{2})$ is
equivalent to a twisted geometric $K$-cycle whose underling twisted spinc-
manifold is closed, which implies that
$[(M_{2},\phi_{2},\upsilon_{2},\eta_{2},\\\ E_{2})]$ lies in the image of
$j_{\ast}$.

  * •

Let $(M_{3},\phi_{3},\upsilon_{3},\eta_{3},E_{3})$ be a geometric
$\alpha$-twisted $K$-cycle over $(X,Y;\alpha)$. Then $[(\partial
M_{3},\phi_{3}|_{\partial M_{3}},\upsilon_{3}|_{\partial
M_{3}},\eta_{3}|_{\partial M_{3}},E_{3}|_{\partial M_{3}})$ is clearly bordant
to a trivial twisted geometric $K$-cycle over $X$ i.e $i_{\ast}\circ\delta=0$.
Let $[(M_{4},\phi_{4},\upsilon_{4},\eta_{4},E_{4})]\in K^{g}_{1}(Y,\alpha\circ
i)$ be a class which lies in the kernel of $i_{\ast}$. A similar strategy
leads us to get that the underling twisted spinc-manifold of several times of
spinc-vector bundle modification of
$[(M_{4},\phi_{4},\upsilon_{4},\eta_{4},E_{4})]$ is a boundary of a twisted
spinc-manifold over $X$ , from which we can easily get that
$[(M_{4},\phi_{4},\upsilon_{4},\eta_{4},E_{4})]$ belongs to the image of
$\delta$.

∎

######  Remark 4.9.

The condition of representability of the twist is essential for the proof
here. In general, a twist is not always representable. We leave the six-term
exact sequence of geometric twisted $K$-homology for general twists as a
further question to be investigated.

######  Theorem 4.10 (Additivity).

Let $(X_{i})_{i\in I}$ be a family of locally finite $CW$-complexes and
$\alpha_{i}:X_{i}\rightarrow K(\mathbb{Z},3)$ be a twist over $X_{i}$ for each
$i$. Moreover, we require that there exists an $\alpha_{i}$-twisted spinc-
manifold over $X_{i}$ for each $i$. Denote $X$ to be the disjoint union of
$X_{i}$ and $\alpha$ is a twist over $X$ such that the restriction of $X$ to
each $X_{i}$ is $\alpha_{i}$. Then we have the following isomorphism:

$K^{g}_{\ast}(X,\alpha)\cong\oplus_{i}K^{g}_{\ast}(X_{i},\alpha_{i})$ (4.4)

The proof is not difficult from the definition, so we skip it here for
simplicity. The above theorem implies the Mayer-Vietoris sequence and the
Milnor’s $\underleftarrow{lim}^{1}$-exact sequence of geometric twisted
$K$-homology.

######  Theorem 4.11 (Mayer-Vietoris sequence).

Assume two open set $U$ and $V$ of $X$ satisfies $X=U\bigcup V$ and the twist
$\alpha$ is representable, we have the Mayer-Vietoris sequence of twisted
$K$-homology:

$K_{1}^{g}(X,\alpha)$$K_{0}^{g}(U\bigcap V,\alpha_{U\bigcap
V})$$K_{0}^{g}(U,\alpha_{U})\oplus
K_{0}^{g}(V,\alpha_{V})$$K_{1}^{g}(U,\alpha_{U})\oplus
K_{1}^{g}(V,\alpha_{V})$$K_{1}^{g}(U\bigcap V,\alpha_{U\bigcap
V})$$K_{0}^{g}(X,\alpha)$$\delta$${j_{U}}_{\ast}\oplus{j_{V}}_{\ast}$${i_{U}}_{\ast}-{i_{V}}_{\ast}$$\delta$${j_{U}}_{\ast}\oplus{j_{V}}_{\ast}$${i_{U}}_{\ast}-{i_{V}}_{\ast}$

###### Proof.

Let $Z$ be the disjoint union of $U$ and $V$, $Y$ be $U\cap V$. Then consider
the six-term exact sequence for the pair $(Z,Y)$ and use the excision
isomorphism $K^{g}_{\ast}(Z,Y;\alpha)\cong K^{g}_{\ast}(X,\alpha)$ we can get
the Mayer-Vietoris sequence for twisted geometric $K$-homology groups. ∎

######  Theorem 4.12 ($\underleftarrow{lim}^{1}$-exact sequence).

Given a countable $CW$-complex $X$ and a representable twist $\alpha$ over $X$
such that there exists an $\alpha$-twisted spinc-manifold over $X$. We denote
the $n$-skeleton of $X$ to be $X_{n}$ and the inclusion $X_{n}\hookrightarrow
X$ to be $i_{n}$. Let $\alpha_{n}=\alpha\circ i_{n}$. Then we have the
following exact sequence

$1\rightarrow\underleftarrow{lim}^{1}K^{g}_{\ast-1}(X_{n},\alpha_{n})\rightarrow
K^{g}_{\ast}(X,\alpha)\rightarrow\underleftarrow{lim}K^{g}_{\ast}(X_{n},\alpha_{n})\rightarrow
1$ (4.5)

The proof is standard and the same as the proof of Milnor’s
$\underleftarrow{lim}^{1}$-sequence of homology theory, which can be found in
chapter 19 of [15].

##  5 The charge map is an isomorphism

In Section $8$ of [2] they propose a question:

Is the charge map $h$ (see (2.14)) an isomorphism?

In this section we show this is true for countable $CW$-complexes by
considering the following diagram.

${K^{geo}(X,\mathcal{A})}$${K^{top}(X,\mathcal{A})}$${K^{g}(X,\alpha)}$${K^{a}(X,\alpha)}$${F}$${h}$${\mu}$${\eta}$
(5.1)

Here $\mu$ is the analytic index map in [20] and
$\eta:K^{top}_{\ast}(X,\mathcal{A})\rightarrow K^{a}_{\ast}(X,\mathcal{A})$ is
the natural map in [2] , which is defined as follows:

$\eta(M,\phi,\sigma)=\phi_{\ast}(PD(\sigma))$ (5.2)

Moreover, we know that $\mu$ is an isomorphisms for compact manifolds and
$\eta$ is an isomorphism for locally finite $CW$-complexes. And we proved that
$F$ is an isomorphism for any locally finite $CW$-complexes. If we can show
that diagram (5.1) is commutative, then we will get that $h$ must also be an
isomorphism.

######  Proposition 5.1.

For any locally finite $CW$-complex $X$, the diagram (5.1) is commutative.

###### Proof.

If we write the formula of the map in diagram (5.1) for a $D$-cycle
$(M,E,\iota,S)$ over $(X,\alpha)$ using the notation before, we get

$\displaystyle\eta\circ h(M,E,\iota,S)=\iota_{\ast}\circ\rho_{\ast}\circ
PD\circ\chi\circ s_{!}([E])$ (5.3) $\displaystyle\mu\circ
F(M,E,\iota,S)=\iota_{\ast}\circ\eta_{\ast}\circ I_{\ast}\circ PD([E])$ (5.4)

Therefore the commutativity of diagram (5.1) is equivalent to

$\iota_{\ast}\circ\rho_{\ast}\circ PD\circ\chi\circ
s_{!}=\iota_{\ast}\circ\eta_{\ast}\circ I_{\ast}\circ PD$ (5.5)

i.e equivalent to the commutativity of the following diagram:

${K^{0}(E)}$${K_{\ast}(M,W_{3}\circ\tau)}$${K^{\ast}(\hat{M},W_{3}\circ\upsilon\circ\varrho))}$${K_{\ast}(M,W_{3}\circ\upsilon)}$${K^{\ast}(\hat{M},-\alpha\circ\iota\circ\rho)}$${K_{\ast}(\hat{M},\alpha\circ\iota\circ\rho)}$${K_{\ast}(M,\alpha\circ\iota)}$${s_{!}}$${PD}$${\chi}$${PD}$${I_{\ast}}$${\eta_{\ast}}$${\rho_{\ast}}$
(5.6)

Here $PD:K^{\ast}(\hat{M},(\iota\circ\rho)^{\ast}(\mathcal{A}^{op})\rightarrow
K_{\ast}(\hat{M},(\iota\circ\rho)(\mathcal{A}))$ is the $Poincar\acute{e}$
duality map between twisted $K$-theory and twisted $K$-homology in [9] and
[19]. By the naturality of Poincar$\check{e}$ duality (one can see Corollary
$3.8$ in [8]), we have that the commutative diagram below

${K^{0}(M)}$${K_{\ast}(M,W_{3}\circ\tau)}$${K^{0}(\hat{M},W_{3}\circ\upsilon\circ\rho)}$${K_{\ast}(\hat{M},W_{3}\circ\tau\circ\rho)}$${s_{\ast}}$${\rho_{\ast}}$${s_{!}}$${PD}$${PD}$
(5.7)

The twist of the lower right $K$-homology group is $W_{3}\circ\tau\circ\rho$
since $\hat{M}$ admits a spinc-structure. Since $\rho\circ s=Id$, we have
$\rho_{\ast}\circ s_{\ast}=id$, therefore we can get

$PD=\rho_{\ast}\circ s_{\ast}\circ PD=\rho_{\ast}\circ PD\circ s_{!}$ (5.8)

To prove the whole proposition, we first give the following lemma.

######  Lemma 5.2.

Denote the map on twisted $K$-homology groups induced by changing twist by
$\tilde{\chi}:K_{\ast}(\hat{M},W_{3}\circ\tau\circ\rho)\rightarrow
K_{\ast}(\hat{M},\alpha\circ\iota\circ\rho)$. Then the following diagram is
commutative

${K^{\ast}(\hat{M},W_{3}\circ\upsilon\circ\rho)}$${K_{\ast}(\hat{M},W_{3}\circ\tau\circ\rho)}$${K_{\ast}(M,W_{3}\circ\tau)}$${K^{\ast}(\hat{M},-\alpha\circ\iota\circ\rho)}$${K_{\ast}(\hat{M},\alpha\circ\iota\circ\rho)}$${K_{\ast}(M,\alpha\circ\iota)}$${\chi}$${PD}$${PD}$${\tilde{\chi}}$${\rho_{\ast}}$${\rho_{\ast}}$${\eta_{\ast}\circ
I_{\ast}}$ (5.9)

If we combine the above lemma and diagram (5.7), we can get that diagram (5.1)
is commutative. ∎

######  Corollary 5.3.

If $X$ is an smooth manifold and $\mathcal{A}$ is a twisting on $X$, then the
charge map $h:K^{geo}_{\ast}(X,\alpha)\rightarrow K^{top}_{\ast}(X,\alpha)$ is
an isomorphism.

###### Proof of (5.2).

  * •

Step 1 We prove the first square in diagram ((5.9)) is commutative. First of
all, we review the definition of Poincar$\acute{e}$ duality
$PD:K^{\ast}(\hat{M},\mathcal{A})\rightarrow
K_{\ast}(\hat{M},\mathcal{A}^{op})$ for twisted $K$-theory in Lemma $2.1$ of
[9]:

$PD(x)=\sigma_{C(\hat{M},\mathcal{A}^{op})}(x)$ (5.10)

Here
$\sigma_{C(\hat{M},\mathcal{A}^{op})}:KK(C(\hat{M},\mathcal{A})\hat{\otimes}A,B)\rightarrow
KK(A,C(\hat{M},\mathcal{A}^{op})\hat{\otimes}B)$ is a canonical isomorphism
for any $C^{\ast}$-algebra $A$ and $B$, which is given by tensoring with
$C(\hat{M},\mathcal{A}^{op})$. If we choose $A$ and $B$ both to be
$\mathbb{C}$ and $C(\hat{M},\mathcal{A})$ to be
$C(\hat{M},(W_{3}\circ\upsilon\circ\rho)^{\ast}(\mathfrak{K}))$, then we get
the Poincar$\acute{e}$ duality $PD$ on the top of the first square . If we
choose $A$ and $B$ to be $\mathbb{C}$ and $C(\hat{M},\mathcal{A})$ to be
$C(\hat{M},(-\alpha\circ\iota\circ\rho)^{\ast}(\mathfrak{K}))$, then we get
the Poincar$\acute{e}$ duality $PD$ on the bottom of the first square. Then
the commutativity of the first square follows from that $PD$ is natural over
$C(\hat{M},\mathcal{A}^{op})$.

  * •

Step 2 From the definition of $\tilde{\chi}$ , we know that it is induced by
changing the twist from $W_{3}\circ\tau\circ\rho$ to
$\alpha\circ\iota\circ\rho$ using the trivialization given by the spinor
bundle $\rho^{\ast}S$ and the canonical trivialization of $TM\oplus\upsilon$.
On the other hand, we know that $I_{\ast}$ is the changing twist map induced
by the canonical trivialization of $TM\oplus\upsilon$ and $\eta_{\ast}$ is the
map induced by the homotopy $\eta$, which is induced by the spinor bundle $S$.
So the commutativity of the second square follows from that the changing twist
map is natural.

∎

In section 4, we proved that the geometric twisted $K$-homology group defined
in [2] is homotopy invariant and each finite $CW$-complex is homotopy
equivalent to a smooth manifold. Therefore we have that the charge map is an
isomorphism for any finite $CW$-complex.

##  6 Geometric twisted K-cycles via bundle gerbes

In this section we give another definition of geometric twisted $K$-homology
via bundle gerbes. First of all, we give the definition of bundle gerbes
first.

######  Definition 6.1.

Given a $CW$-complex $B$, a bundle gerbe over $B$ is a pair $(P,Y)$, where
$\pi:Y\rightarrow B$ is a locally split map and $P$ is a principal
$U(1)$-bundle over $Y\times_{M}Y$ with an associative product, i.e for every
point $(y_{1},y_{2}),(y_{2},y_{3})\in Y\times_{M}Y$, there is an isomorphism

$P_{(y_{1},y_{2})}\otimes_{\mathbb{C}}P_{(y_{2},y_{3})}\rightarrow
P_{(y_{1},y_{3})}$ (6.1)

and the following diagram commutes

${P_{(y_{1},y_{2})}\otimes P_{(y_{2},y_{3})}\otimes
P_{(y_{3},y_{4})}}$${P_{(y_{1},y_{3})}\otimes
P_{(y_{3},y_{4})}}$${P_{(y_{1},y_{2})}\otimes
P_{(y_{2},y_{4})}}$${P_{(y_{1},y_{4})}}$ (6.2)

For every principal $U(1)$-bundle $J$ over $B$ we can define a bundle gerbe
$\delta(J)$ by $\delta(J)_{(y_{1},y_{2})}=J_{y_{1}}\otimes J^{\ast}_{y_{2}}$.
The product is induced by the pairing between $J^{\ast}$ and $J$. A bundle
gerbe $(P,Y)$ is called trivial if there is a hermitian line bundle $J$ such
that $P\cong\delta(J)$. In this case, $J$ is called a trivialization of
$(P,Y)$. For every bundle gerbe $(P,Y)$ over $M$ we can associate a third
integer cohomology class $d(P)\in H^{3}(M,\mathbb{Z})$ to describe the non-
triviality of the bundle gerbe which we call Dixmier-Douady class (see [16]).

######  Definition 6.2.

Two bundle gerbes $(P,Y)$ and $(Q,Z)$ are stable isomorphic to each other if
there is a trivialization of $p_{1}^{\ast}(P)\otimes p_{2}^{\ast}(Q)^{\ast}$.
Here $p_{1}:Y\times_{B}Z\rightarrow Y$ and $p_{2}:Y\times_{B}Z\rightarrow Z$
are the natural projections. And the trivialization is called a stable
isomorphism between $(P,Y)$ and $(Q,Z)$.

The following theorem gives the relation between stable isomorphism classes
and Dixmier-Douady classes.

######  Theorem 6.3.

Two bundle gerbes are stable isomorphic iff their Dixmier-Douady classes are
equal to each other. Moreover, the Dixmier-Douady class defines a bijection
between stable isomorphic classes of bundle gerbes over $M$ and
$H^{3}(M,\mathbb{Z})$.

Now we give another definition which is important in our construction of
geometric twisted $K$-homology.

######  Definition 6.4.

Let $(P,Y)$ be a bundle gerbe over $B$. A finite dimensional hermitian bundle
$E$ over $Y$ is called a $(P,Y)$-module if there is a complex vector bundle
isomorphism

$\phi:P\otimes\pi_{1}^{\ast}(E)\cong\pi_{2}^{\ast}(E)$

which is compatible with the bundle gerbe product, i.e the following diagram
is commutative

${P_{(y_{1},y_{2})}\otimes P_{(y_{2},y_{3})}\otimes
E_{y_{3}}}$${P_{(y_{1},y_{3})}\otimes E_{y_{3}}}$${P_{(y_{1},y_{2})}\otimes
E_{y_{2}}}$${E_{y_{1}}}$

where $\pi_{i}$ ($i=1,2$) are two projections from $Y\times_{B}Y$ to $Y$.
Moreover, the Grothendieck group of isomorphism classes of bundle gerbe module
over $(P,Y)$ is called the $K$-group of $(P,Y)$.

Now we give the construction of geometric twisted $K$-cycles

######  Definition 6.5.

Let $B$ be a space , $H\in H^{3}_{tor}(X,\mathbb{Z})$ and $(P,Y)$ be a bundle
gerbe over $B$ with $-H$ as Dixmier-Douady class. A geometric twisted
$K$-cycle is a triple $(M,f,E)$ where

  * •

$M$ is a compact spinc-manifold;

  * •

$f:M\rightarrow B$ is continuous;

  * •

$[E]$ is an isomorphism class of $f^{\ast}(P,Y)$-module.

We denote the whole twisted $K$-cycles over $(B,H)$ by $\Gamma_{(P,Y)}(B)$

To give twisted $K$-homology group, we also need to define an equivalence
$\sim$ on these geometric cycles as follows

  * •

Direct sum-disjoint union For any two cycles $(M,f,E_{1})$ and $(M,f,E_{2})$
over $(B,(P,Y))$, then we have

$(M,f,[E_{1}])\cup(M,f,[E_{2}])\sim(M,f,[E_{1}]+[E_{2}])$ (6.3)

  * •

Bordism Given two cycles $(M_{0},f_{0},E_{0})$ and $(M_{1},f_{1},E_{1})$ over
$(B,(P,Y))$, if there exists a cycle $(M,f,E)$ over $(B,(P,Y))$ such that

$\partial M=-M_{0}\cup M_{1}$ (6.4)

and $E_{\partial M}=-[E_{0}]\cup[E_{1}]$, then

$(M_{0},f_{0},E_{0})\sim(M_{1},f_{1},E_{1})$ (6.5)

  * •

Spinc-vector bundle modification Let $(M,f,E)$ be a geometric twisted
$K$-cycle over $(B,(P,Y))$ and an even dimensional spinc-vector bundle $V$
over $M$. Let $\hat{M}$ to be the sphere bundle of
$V\oplus\underline{\mathbb{R}}$. Denote the bundle map by
$\rho:\hat{M}\rightarrow M$ and the positive spinor bundle of $T^{v}(\hat{M})$
by $S_{V}^{+}$. The vector bundle $S_{V}^{+}\otimes\rho^{\ast}(E)$ over
$\hat{M}$ is a $(\rho\circ f)^{\ast}(P,Y)$ module. Then

$(M,f,[E])\sim(\hat{M},\rho\circ f,[S_{V}^{+}\otimes\rho^{\ast}(E)])$ (6.6)

######  Definition 6.6.

For any space $B$ and bundle gerbe $(P,H)$ over $B$. We define
$K^{gg}_{i,(P,Y)}(B,H)=\Gamma_{i}(B,(P,Y))/\sim$ ($i$=$0$, $1$). The parity
depends on the dimension of the spinc-manifold in a twisted $K$-cycle.

######  Proposition 6.7.

If $(P,Y)$ and $(Q,Z)$ are two bundle gerbes over $B$ with the same Dixmier-
Douady class $-H$, then we have $K^{gg}_{i,(P,Y)}(B,H)$ is isomorphic to
$K^{gg}_{i,(Q,Z)}(B,H)$.

###### Proof.

Let $R$ be a stable isomorphism between $(P,Y)$ and $(Q,Z)$ i.e a
trivialization of $p_{1}^{\ast}(P)\otimes p_{2}^{\ast}(Q)$. Without loss of
generality we can just assume that $Z=Y$. Otherwise we can consider the bundle
gerbe $(p_{1}^{\ast}P,Y\times_{B}Z)$ and $(p_{2}^{\ast}Q,Y\times_{B}Z)$
instead. Let $(M,f,[E])\in\Gamma^{i}(B,(P,Y))$. Since $Q\cong P\otimes R$,
therefore $(M,f,[E]\otimes L_{R})$ (here $L_{R}$ is the natural associated
line bundle of $R$) is a twisted geometric $K$-cycle over $(B,(Q,Y))$. So we
get a homomorphism from $\Gamma(B,(P,Y))$ to $\Gamma(B,(Q,Z))$, which we
denote by $r$. A tedious check tells us that $r$ respects disjoint union,
bordism and spinc-bundle modification. Therefore $r$ induces a homomorphism
from $K_{i}^{gg}(B,(P,Y))$ to $K_{i}^{gg}(B,(Q,Y))$. If we change the roles of
$(P,Y)$ and $(Q,Z)$ in the above construction, then we get an inverse of $r$.
Therefore $r$ is an isomorphism. ∎

Let $(P,Y)$ be a bundle gerbe over $B$ with Dixmier-Douady class $H$.
According to Proposition 6.4 in [5], the $i$-th $K$-group of bundle gerbe
$(P,Y)$ is isomorphic to the $i$-th twisted $K$-group $K^{i}(X,-H)$. Then the
definitions of $K^{gg}_{\ast}(X,H)$ and $K^{top}_{\ast}(X,\alpha)$ implies the
following proposition:

######  Proposition 6.8.

Let $X$ be a finite $CW$-complex and $H\in H^{3}_{torsion}(X,\mathbb{Z})$.
Then we have

$K^{gg}_{i}(X,(P,Y))\cong K^{top}_{i}(X,\alpha)\cong K^{geo}_{i}(X,\alpha)$
(6.7)

##  7 Some constructions about geometric twisted K-cycles

First of all we review notions about topological $T$-duality in [7]. Let
$\pi:P\rightarrow B$ and $\hat{\pi}:\hat{P}\rightarrow B$ be principal
$S^{1}$-bundles over $B$ and $H\in H^{3}(P,\mathbb{Z}),\hat{H}\in
H^{3}(\hat{P},\mathbb{Z})$. Denote the associated line bundles of $P$ and
$\hat{P}$ by $E$ and $\hat{E}$ respectively. Let $V=E\bigoplus\hat{E}$ and
$r:S(V)\rightarrow B$ be the unit sphere bundle of $V$.

######  Definition 7.1.

A class $\mathbf{Th}\in H^{3}(S(V),\mathbb{Z})$ is called Thom class if
$r_{!}(\mathbf{Th})=1\in H^{0}(B,\mathbb{Z})$.

Let $i:P\rightarrow S(V)$ and $\hat{i}:\hat{P}\rightarrow S(V)$ be inclusion
of principal $S^{1}$-bundle into $S^{3}$-bundle.

######  Definition 7.2.

We say that $((P,H),(\hat{P},\hat{H}))$ is a $T$-dual pair over $B$ or $(P,H)$
and $(\hat{P},\hat{H})$ are $T$-dual to each other if there exists a Thom
class $\mathbf{Th}\in H^{3}(S(V),\mathbb{Z})$ such that

$H=i^{\ast}(\mathbf{Th}),\ \ \hat{H}=\hat{i}^{\ast}(\mathbf{Th})$ (7.1)

Given a $T$-dual pair $((P,H),(\hat{P},\hat{H}))$ over $B$, there exists a
$T$-duality isomorphism between the associated twisted $K$-groups, which is
given by

$T=\hat{p}_{!}\circ u\circ p^{\ast}:K^{i}(P,H)\rightarrow
K^{i-1}(\hat{P},\hat{H})$ (7.2)

Here $p^{\ast}$ and $\hat{p}_{!}$ are the pullback and pushforward maps
respectively. $u$ is a changing twist map from
$K^{i}(P\times_{B}\hat{P},p^{\ast}(H))$ to
$K^{i}(P\times_{B}\hat{P},\hat{p}^{\ast}(\hat{H}))$. In order to establish the
$T$-duality transformation of geometric twisted $K$-homology we first give the
construction of analogue maps of induced map , wrong way map and changing
twist map for twisted $K$-homology.

  1. 1.

Induced map Assume $f:(X_{1},Y_{1})\rightarrow(X_{2},Y_{2})$ is a continuous
map between two pairs of topological spaces. Then the induced map
$f_{\ast}:K^{g}_{i}(X_{1},Y_{1};\alpha\circ f)\rightarrow
K^{g}_{i}(X_{2},Y_{2};\alpha)$ is defined as follows:

$f_{\ast}([M,\phi,\upsilon,\eta,E])=[(M,\phi\circ f,\upsilon,\eta,E)]$ (7.3)

  2. 2.

Wrong way map Let $f:P\rightarrow N$ be a $K$-oriented map between smooth
manifolds and $\alpha:N\rightarrow K(\mathbb{Z},3)$ be a twist over $N$. we
define the wrong way map $f^{!}:K^{g}_{i-1}(N,\alpha)\rightarrow
K^{g}_{i}(P,\alpha\circ f)$ to be

$\pi^{!}([M,\phi,\upsilon,\eta,E])=[(\tilde{M},\tilde{\phi},\tilde{\upsilon},\tilde{\eta},\tilde{\pi}^{\ast}(E))]$
(7.4)

Here $\tilde{M}$ is the fiber product $M\times_{N}M$, $\tilde{\upsilon}$ is
the stable normal bundle of $\tilde{M}$ and $\tilde{\eta}$ is induced by
$\eta$.

######  Remark 7.3.

As $f$ is $K$-oriented, therefore $W_{3}(\tilde{\upsilon}\oplus
f^{\ast}(\upsilon))$ is trivial, which implies that
$W_{3}\circ\tilde{\upsilon}$ is homotopic to $\alpha\circ f\circ\tilde{\phi}$
via a homotopy $\lambda$. $\tilde{\eta}:\tilde{M}\times[0,1]\rightarrow
K(\mathbb{Z},3)$ is given by the combination of $\lambda$ and $\eta$ as
follows:

$\displaystyle\tilde{\eta}(x,t)=\begin{cases}\lambda(x,2t),&0\leq t\leq
1/2;\\\ \eta\circ(f^{\prime}\times id)(x,2t-1),&1/2\leq t\leq 1.\end{cases}$

in which $f^{\prime}:\tilde{M}\rightarrow M$ is the canonical projection to
$M$. Therefore we get that
$(\tilde{M},\tilde{\phi},\tilde{\upsilon},\tilde{\eta},\tilde{\pi}^{\ast}(E))$
is a twisted geometric cycle over $(P,\alpha\circ f)$. In particular, when $f$
is the bundle map for a principal $S^{1}$-bundle, $\tilde{M}$ is the pullback
$S^{1}$-bundle along $f$.

  3. 3.

Changing twist map Let $P\rightarrow B$ be a principal $\mathbb{T}^{2}$-bundle
over $B$. If we have two twist $\alpha_{1},\alpha_{2}:P\rightarrow
K(\mathbb{Z},3)$ and their associated cohomology classes are the same, then
$\alpha_{1}$ and $\alpha_{2}$ are homotopic. Choose a homotopy $h$ such that
the restriction of $h$ to each fiber of $P$ corresponds to the cohomology
class $\theta\cup\hat{\theta}\in H^{2}(P_{b},\mathbb{Z})$. Here $\theta$ and
$\hat{\theta}$ are generators of the first cohomology group of the two copies
of $S^{1}$ of a fiber. Then for a geometric cycle $\delta$ of $(X,\alpha_{1})$
one can define the changing twist map $u:K^{g}_{i}(P,\alpha_{1})\rightarrow
K^{g}_{i}(P,\alpha_{2})$ as follows:

$u([M,\phi,\upsilon,\eta,E])=[(M,\phi,\upsilon,\hat{\eta},E)]$ (7.5)

Here $\hat{\eta}$ is induced by the following diagram:

$W_{3}\circ\upsilon$$\alpha_{1}\circ\phi$$\alpha_{2}\circ\phi$$\eta$$h\circ(\phi\times
id)$

More explicitly, the $\hat{\eta}$ is given by the composition of $\eta$ and
$h\circ(\phi\times id)$, which we denote by $(h\circ(\phi\times id))\ast\eta$

$\displaystyle(h\circ(\phi\times
id))(\eta)(x,t)=\begin{cases}\eta(x,2t),&0\leq t\leq 1/2;\\\ h\circ(\phi\times
id)(x,2t-1),&1/2\leq t\leq 1.\end{cases}$

######  Lemma 7.4.

The induced map, wrong way map and changing twist map above are all compatible
with disjoint union, bordism and spinc-vector bundle modification.

###### Proof.

We have proved the induced map part in Lemma 4.1 and it is not hard to see
that they all respect the disjoint union. We do the rest here.

  * •

Let $(M,\phi,\upsilon,\eta,E)$ be a bordism between
$(M_{1},\phi_{1},\upsilon_{1},\eta_{1},E_{1})$ and $(M_{2},\phi_{2},$
$\upsilon_{2},\eta_{2},E_{2})$ over $X$. Denote
$p^{!}(M_{i},\phi_{i},\upsilon_{i},\eta_{i},E_{i})$ by
$(\tilde{M_{i}},\tilde{\phi_{i}},\tilde{\upsilon_{i}},\tilde{\eta_{i}},\tilde{\pi}^{\ast}E_{i})$.
Since the boundary of a pullback space is the pullback of the original
boundary and the stable normal bundle of a principal $S^{1}$-bundle is
isomorphic to the stable normal bundle of the base space, we get that
$(\tilde{M},\tilde{\phi},\\\
\tilde{\upsilon},\tilde{\eta},\tilde{\pi}^{\ast}E)$ gives a bordism between
$(\tilde{M_{1}},\tilde{\phi_{1}},\tilde{\upsilon_{1}},\tilde{\eta_{1}},\tilde{\pi}^{\ast}E_{1})$
and $(\tilde{M_{2}},\tilde{\phi_{2}},\\\
\tilde{\upsilon_{2}},\tilde{\eta_{2}},\tilde{\pi}^{\ast}E_{2})$.

Use the notation above.
$p^{!}(\hat{M},\phi\circ\rho,\upsilon^{\prime},\eta^{\prime},S^{+}_{V}\otimes\rho^{\ast}E)$
is
$(\tilde{\hat{M}},\phi\circ\rho\circ\tilde{\hat{\pi}},\tilde{\upsilon^{\prime}},\tilde{\eta^{\prime}},$
$\tilde{\pi}^{\ast}(S^{+}_{V}\otimes\rho^{\ast}E))$. On the other hand,
$\tilde{\pi}^{\ast}V$ is a spinc-vector bundle over $\tilde{M}$. The
associated spinc-vector bundle modification of
$(\tilde{M},\tilde{\phi},\tilde{\upsilon},\tilde{\eta},\tilde{\pi}^{\ast}E)$
is that
$(\tilde{\hat{M}},\tilde{\phi}\circ\tilde{\rho},\upsilon^{\prime\prime},\tilde{\eta}^{\prime},\tilde{\pi}^{\ast}(S^{+}_{V}\otimes\rho^{\ast}E))$.
The maps in the above twisted geometric $K$-cycles are shown in the following
diagram:

${\tilde{\hat{M}}}$${\tilde{M}}$${\hat{M}}$${M}$${\tilde{\rho}}$${\tilde{\hat{\pi}}}$${\tilde{\pi}}$${\rho}$
(7.6)

By the commutativity, we have that
$\phi\circ\rho\circ\tilde{\hat{\pi}}=\tilde{\phi}\circ\tilde{\rho}$. Moreover,
$\tilde{\upsilon^{\prime}}$ and $\upsilon^{\prime\prime}$ are homotopic
because they are both classifying maps of the stable normal bundle of
$\tilde{\hat{M}}$. The coincidence of $\tilde{\upsilon^{\prime}}$ and
$\upsilon^{\prime\prime}$ implies that $\tilde{\eta^{\prime}}$ and
$\tilde{\eta}^{\prime}$ are homotopic to each other. So the wrong way map
respects the spinc-vector bundle modification relation.

  * •

Use the above notions. $u((M,\phi,\upsilon,\eta,E))$ still gives a bordism
between $(M_{1},\phi_{1},\upsilon_{1},\hat{\eta_{1}},E_{1})$ and
$(M_{2},\phi_{2},\upsilon_{2},\hat{\eta_{2}},E_{2})$ i.e $u$ respects bordism
equivalence. Since the composition of homotopies are associative up to
homotopy, we get that $u$ respects the spinc-vector bundle modification
relation.

∎

##  8 T-duality for twisted geometric K-homology

######  Theorem 8.1.

Let $B$ be a finite $CW$-complex and $((P,H),(\hat{P},\hat{H}))$ be a $T$-dual
pair over $B$.

${P\times_{B}\hat{P}}$${P}$${\hat{P}}$${B}$${p}$${\hat{p}}$${\pi}$${\hat{\pi}}$

Moreover, we assume that $\alpha:P\rightarrow K(\mathbb{Z},3)$ and
$\hat{\alpha}:\hat{P}\rightarrow K(\mathbb{Z},3)$ satisfy that
$\alpha^{\ast}([\Theta])=H$ and $\hat{\alpha^{\ast}}([\Theta])=\hat{H}$ (Here
$[\Theta]$ is the positive generator of $H^{3}(K(\mathbb{Z},3),\mathbb{Z})$).
Moreover we assume that both $\alpha$ and $[\hat{\alpha}]$ is representable.
Then the map $T=\hat{p}_{\ast}\circ u\circ p^{!}:K_{\ast}^{g}(P,\alpha)\mapsto
K_{\ast+1}^{g}(\hat{P},\hat{\alpha})$ is an isomorphism.

To prove the theorem (8.1) we need the following lemmas.

######  Lemma 8.2.

$T$ is compatible with the boundary operator and the induced map in the Mayer-
Vietoris sequence.

###### Proof.

We first prove the compatibility with the induced map. Assume we have a map
$f:X\mapsto Y$ and we have the associated $T$-duality diagrams over $Y$ and
pullback it to $X$. $f$ induces maps by $F:P_{X}\mapsto P_{Y}$,
$\hat{F}:\hat{P_{X}}\rightarrow\hat{P_{Y}}$ and
$G:P_{X}\times_{X}\hat{P_{X}}\rightarrow P_{Y}\times_{Y}\hat{P_{Y}}$. Then we
have the following identities:

$\displaystyle\hat{F}_{\ast}\circ T_{X}$ $\displaystyle=$
$\displaystyle\hat{F}_{\ast}\circ(\hat{p}_{X})_{\ast}\circ u_{X}\circ
p_{X}^{!}$ (8.1) $\displaystyle=$ $\displaystyle(\hat{p}_{Y})_{\ast}\circ
G_{\ast}\circ u_{X}\circ p_{X}^{!}$ $\displaystyle=$
$\displaystyle(\hat{p}_{Y})_{\ast}\circ u_{Y}\circ(G\circ p_{Y})^{!}\circ
F_{\ast}$ $\displaystyle=$ $\displaystyle T_{Y}\circ F_{\ast}$

Now we turn to the compatibility with the boundary map, in the Mayer-Vietoris
sequence of the boundary operator$\delta:K^{g}_{\ast}(X,\alpha)\rightarrow
K^{g}_{\ast+1}(U\cap V,\alpha\circ i_{U\cap V})$ is given as follows: Choose a
continuous map $f:X\rightarrow[0,1]$ such that $f_{U-U\cap V}$ is $0$ and
$f_{V-U\cap V}$ is $1$. Without loss of generality we assume that
$f\circ\phi:M\rightarrow[0,1]$ is a smooth function and $1/2$ is a regular
point of $f\circ\phi$. For any twisted geometric $K$-cycle
$x=(M,\phi,\upsilon,\eta,E)$, define $\delta x=(f^{-1}(1/2),\phi\circ
i,\upsilon\circ i,\eta\circ(i\times id),i^{\ast}E)$. By this formula, we get
that $\delta$ is compatible with induced map. Also the homotopies
$(h\circ(\phi\circ i\times id))\ast(\eta\circ(i\times id))$ and
$(\eta\ast(h\circ(\phi\times id))\circ(i\times id)$ are homotopic to each
other, which implies that $u\circ\delta=\delta\circ u$. The remainder is to
show that $\hat{p}^{!}\circ\delta=\delta\circ\hat{p}^{!}$. We write both sides
explicitly first:

Given a principal $S^{1}$-bundle $\pi:P\rightarrow B$ and a twisted geometric
cycle $(M,\phi,\upsilon,\eta,E)$ (which we denote by $x$) over $P$

$\displaystyle\hat{p}^{!}\circ\delta
x=(\tilde{(f\circ\phi)^{-1}(1/2)}),\phi\circ\tilde{\pi}\circ
i,\tilde{\upsilon}\circ i,\eta\circ((\tilde{\pi}\circ i)\times
id),(i\circ\tilde{\pi})^{\ast}E);$
$\displaystyle\delta\circ\hat{p}^{!}x=((f\circ\phi\circ\tilde{\pi})^{-1}(1/2),\phi\circ\tilde{\pi}\circ
i,\tilde{\upsilon}\circ i,\eta\circ((\tilde{\pi}\circ i)\times
id),(i\circ\tilde{\pi})^{\ast}E)$

Since $\tilde{f\circ\phi}^{-1}(1/2)$ is exactly
$(f\circ\phi\circ\tilde{\pi})^{-1}(1/2)$, we get that
$\hat{p}^{!}\circ\delta=\delta\circ\hat{p}^{!}$. Finally, we have that

$T\circ\delta=(\hat{p}^{!}\circ u\circ
p_{\ast})\circ\delta=\delta\circ(\hat{p}^{!}\circ u\circ p_{\ast})=\delta\circ
T$

∎

######  Lemma 8.3.

###### Proof of Theorem 8.1.

We do the proof by induction on the number of cells. Assume $X$ is a point,
then $P$ and $\hat{P}$ are both $S^{1}$ and the correspondence space is
$S^{1}\times S^{1}$. Denote $p_{\ast}\circ t^{-1}\circ\hat{p}$ by
$T^{\prime}$. We can see that for any geometric cycle
$(M,\iota,\upsilon,\eta,[E])$, the image of this cycle under the map
$T^{\prime}\circ T$ is $(M\times S^{1}\times
S^{1},\iota\circ\breve{p},\upsilon\circ\breve{p},\breve{\eta},\breve{p}^{\ast}([E]))$.
Here $\breve{p}$ is the projection from $M\times S^{1}\times S^{1}$ to $M$.
One can see that this is cobordant to a trivial $S^{2}$-bundle as spinc-
manifolds. The cobordism can be given by $M\times B$, where $B$ is a solid
torus with a solid disk cut from it. As all of the bundles involved are
trivial, the geometric cycle $(M\times
B,\iota\circ\acute{p},\upsilon\circ\acute{p},\acute{\eta},\acute{p}^{\ast}([E])$
gives the cobordism between the image of $T$ and the spinc-modification i.e
$T^{\prime}\circ T$ is equal to identity in this case. As a consequence we
also get that $T$ is an isomorphism. Assume $T$ is an isomorphism when the
number of cells is $n$,then we adjoin another cell $\sigma_{n+1}$ to $X$ and
we choose open set $U=X\cup\sigma_{n+1}-pt,V=\sigma_{n+1}-\bar{pt}$ and we can
get the Mayer-Vietoris sequence of geometric twisted $K$-homology groups. Then
the conclusion of this theorem follows from induction and the Five-Lemma. ∎

######  Remark 8.4.

The construction of $T$-duality transformation of geometric twisted
$K$-homology can be easily generalized to $T$-dual pairs of higher dimensional
torus bundles.

We end up with this paper with an interesting question. Can we release the
condition in Remark 2.7 and construct geometric twisted $K$-theory for general
twists? An idea is to replace twisted spinc-manifolds in the construction of
geometric $K$-cycles by noncommutative analogue objects. We leave this for
further investigation.

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