# Many finite-dimensional lifting bundle gerbes are torsion.††thanks: This
document is released under a Creative Commons Zero license
creativecommons.org/publicdomain/zero/1.0/

David Michael Roberts orcid.org/0000-0002-3478-0522

(April 26, 2021)

###### Abstract

Many bundle gerbes constructed in practice are either infinite-dimensional, or
finite-dimensional but built using submersions that are far from being fibre
bundles. Murray and Stevenson proved that gerbes on simply-connected
manifolds, built from finite-dimensional fibre bundles with connected fibres,
always have a torsion $DD$-class. In this note I prove an analogous result for
a wide class of gerbes built from principal bundles, relaxing the requirements
on the fundamental group of the base and the connected components of the
fibre, allowing both to be nontrivial. This has consequences for possible
models for basic gerbes, the classification of crossed modules of finite-
dimensional Lie groups, the coefficient Lie-2-algebras for higher gauge theory
on principal 2-bundles, and finite-dimensional twists of topological
$K$-theory.

##  1 Introduction

A bundle gerbe [Mur96] is a geometric object that sits over a given space or
manifold $X$ classified by elements of $H^{3}(X,\mathbb{Z})$, in the same way
that (complex) line bundles on $X$ are classified by elements of
$H^{2}(X,\mathbb{Z})$. And, just as line bundles on manifolds have connections
giving rise to _curvature_ , a 2-form, giving a class in $H^{3}_{dR}(X)$,
bundle gerbes have a notion of geometric ‘connection’ data, with curvature a
3-form and hence a class in $H^{3}_{dR}(X)$. Since de Rham cohomology sees
only the non-torsion part of integral cohomology, bundle gerbes that are
classified by torsion classes in $H^{3}$ are thus trickier in one sense to
‘see’ geometrically. The problem is compounded by the fact that bundle gerbes
with the same 3-class may look wildly different, as the correct notion of
equivalence is much coarser than isomorphism. Thus different constructions
that lead to the same class are still of interest, due to the flexibility this
introduces.

One wide class of bundle gerbes—so-called _lifting bundle gerbes_ —arise from
the following data. Given a Lie group $G$, a principal $G$-bundle $P\to X$,
and a central extension

$U(1)\to\widehat{G}\to G$ (1.1)

of Lie groups, there is a bundle gerbe on $X$ that is precisely the
obstruction to the extension of a $\widehat{G}$-bundle lifting $P$. One can
see the cohomology class corresponding to the bundle gerbe as analogous to the
class $w_{2}\in H^{2}(M,\mathbb{Z}/2)$ obstructing the lifting of the frame
bundle $F(M)$ to be a spin bundle.

For a non-trivial example, one can consider the central extension $U(1)\to
U(n)\to PU(n)$, and $P$ a principal $PU(n)$-bundle. The lifting bundle gerbe
associated to such a $PU(n)$-bundle has torsion class in
$H^{3}(X,\mathbb{Z})$. Conversely, by a result of Serre, and published by
Grothendieck [Gro68, I, Théorème 1.6], every torsion class in
$H^{3}(X,\mathbb{Z})$ is associated to at least one lifting bundle gerbe of
this form.111It is in fact nontrivial to find which ranks $n$ this is possible
for, given a specific torsion element

In the other direction, one can consider the extension $U(1)\to
U(\mathcal{H})\to PU(\mathcal{H})$ of infinite-dimensional groups, where
$\mathcal{H}\simeq L^{2}[0,1]$, and lifting bundle gerbes of principal
$PU(\mathcal{H})$-bundles. These bundle gerbes are infinite-dimensional, and
_every_ class in $H^{3}(X,\mathbb{Z})$ can be realised by _some_ lifting
bundle gerbe of this form. There are also constructions of bundle gerbes on
compact, simply-connected, simple Lie groups $G$ that use lifting bundle
gerbes for the infinite-dimensional Kac–Moody central extensions
$U(1)\to\widehat{\Omega_{k}G}\to\Omega G$, and these are also non-torsion
gerbes.

Between these examples, then, one might wonder when an a priori _given_
finite-dimensional lifting bundle gerbe on $X$ has a torsion class in
$H^{3}(X,\mathbb{Z})$. That is, given a finite-dimensional central extension
(1.1) and a principal $G$-bundle $P\to X$, is the lifting bundle gerbe
classified by a torsion class? The answer is: not always, as there are
explicit and easy examples where it is non-torsion, for instance over
$X=S^{2}\times S^{1}$ (cf Example 4.4 below). However, under a mild condition
on $X$, it _is_ true.

Murray’s original paper had a claim about a sufficient condition for a finite-
dimensional bundle gerbe to be torsion, though the proof had a subtle error.
Murray and Stevenson later [MS11] gave a correct proof of a slightly stronger
result, demanding (i) simple-connectivity of the base manifold $X$, and (ii)
that the fibres of a certain submersion $Y\to X$ (part of the bundle gerbe
data) are connected. This result was not specifically about lifting bundle
gerbes, but it suffices to prove that lifting gerbes for $G$-bundles on
simply-connected spaces $X$, where $G$ is a _connected_ Lie group, are
torsion.

However, the proof in _loc. cit._ is more general, and doesn’t use anything
specific about lifting bundle gerbes, which are somewhat more rigid than the
general case. In this note I shall prove the following result:

######  Theorem 1.1.

Given any connected manifold $X$ with finite fundamental group, any central
extension (1.1) of finite-dimensional Lie groups, and any principal $G$-bundle
$P\to X$, the corresponding lifting bundle gerbe is torsion.

In fact, the proof suffices to give a stronger result, which applies to
topological spaces, assuming a group theoretical fact about homomorphisms
$\pi_{1}(X)\to\pi_{0}(G)$. We shall give this result below as Theorem 3.7, and
some applications in the final section.

#### Acknowledgements

I thank Konrad Waldorf for asking me whether all finite-dimensional lifting
bundle gerbes are torsion, inspiring this note, and Maxime Ramzi for patient,
extensive discussions. David Baraglia and MathOverflow
users222https://mathoverflow.net/q/389703 also provided helpful information.

This work is supported by the Australian Research Council’s _Discovery
Projects_ funding scheme (grant number DP180100383), funded by the Australian
Government.

##  2 Preliminaries

We first recall the definition of a bundle gerbe.

######  Definition 2.1 ([Mur96]).

A bundle gerbe on a manifold $M$ consists of the following data:

  * •

A surjective submersion $Y\to M$.

  * •

A principal $U(1)$-bundle $E\to Y^{[2]}:=Y\times_{M}Y$.

  * •

An isomorphism
$\mu\colon\operatorname{pr}_{12}^{*}E\otimes_{U(1)}\operatorname{pr}_{23}^{*}E\to\operatorname{pr}_{13}^{*}E$
of $U(1)$-bundles333The tensor product here is defined, for arbitrary
$U(1)$-bundles $E,F$ on $X$, to be $(E\times_{X}F)/U(1)$, where $U(1)$ acts
via the anti-diagonal action. on $Y^{[3]}:=Y\times_{M}Y\times_{M}Y$, called
the bundle gerbe multiplication.

  * •

This isomorphism needs to satisfy an associativity condition, namely that the
diagram444I have suppressed some canonical isomorphisms for clarity; all
tensor products are over $U(1)$

$\textstyle{E_{12}\otimes E_{23}\otimes
E_{34}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{id}\times\mu}$$\scriptstyle{\mu\times\operatorname{id}}$$\textstyle{E_{12}\otimes
E_{24}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu}$$\textstyle{E_{13}\otimes
E_{34}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu}$$\textstyle{E_{14}}$

of isomorphisms of $U(1)$-bundles on $Y^{[4]}$ needs to commute. Here
$E_{ij}=\operatorname{pr}_{ij}^{*}E$, with $\operatorname{pr}_{ij}\colon
Y^{[4]}\to Y^{[2]}$

A bundle gerbe will be denoted by $(E,Y)$, the other data implicit. A bundle
gerbe on a _topological space_ is defined the same way, except in that case we
require instead that $Y\to M$ admits local sections.

Here is the only type of example we will need. Fix a locally trivial central
extension $U(1)\to\widehat{G}\to G$ of Lie groups.555That is, ignoring the
group structures on $\widehat{G}$ and $G$, it is a principal $U(1)$-bundle
Recall that the multiplication map of $\widehat{G}$ induces an isomorphism
$\operatorname{pr}_{1}^{*}\widehat{G}\otimes_{U(1)}\operatorname{pr}_{2}^{*}\widehat{G}\simeq
m^{*}\widehat{G}$ over $G\times G$ where $m\colon G\times G\to G$ is the
multiplication map.

######  Example 2.2 ([Mur96]).

Let $P\to X$ be a principal $G$-bundle. The lifting bundle gerbe associated to
this bundle (and the fixed central extension) is given by the data:

  * •

The submersion is $P\to X$. Recall that the action map induces an isomorphism
$P\times G\simeq P\times_{X}P$ _over $X$_ we will use this silently from now
on.

  * •

The $U(1)$-bundle is $P\times\widehat{G}\to P\times G$.

  * •

Using the isomorphism $P\times G\times G\to P^{[3]}$,
$(p,g,h)\mapsto(p,pg,pgh)$, the multiplication is given by the composite

$\operatorname{pr}_{12}^{*}(P\times\widehat{G})\otimes_{U(1)}\operatorname{pr}_{23}^{*}(P\times\widehat{G})\simeq
P\times\operatorname{pr}_{1}^{*}\widehat{G}\otimes_{U(1)}\operatorname{pr}_{2}^{*}\widehat{G}\simeq
P\times m^{*}\widehat{G}\simeq\operatorname{pr}_{13}^{*}(P\times\widehat{G})$

  * •

The associativity condition follows from associativity in the group
$\widehat{G}$.

Given a bundle gerbe on a space $X$, and a map $f\colon X^{\prime}\to X$,
there is a bundle gerbe on $X^{\prime}$ given by the data $f^{*}Y\to
X^{\prime}$ and $(p^{[2]})^{*}E$, where $p\colon f^{*}Y\to Y$ is the
projection. Moreover, the maps

$(p^{[2]})^{*}E\to E,\quad(f^{*}Y)^{[2]}\to Y^{[2]},\quad p\colon f^{*}Y\to Y$

are compatible with all the bundle gerbe structure. This is the pullback of
$(E,Y)$ by $f$, denoted $f^{*}(E,Y)$. If $(P\times\widehat{G},P)$ is a lifting
bundle gerbe, then
$f^{*}(P\times\widehat{G},P)\simeq(f^{*}P\times\widehat{G},f^{*}P)$.

More generally, given bundle gerbes $(F,Z)$ and $(E,Y)$ on $X$ one can take a
map $g\colon Z\to Y$ commuting with the projections to $X$, and a map $k\colon
F\to E$ of $U(1)$-bundles covering $g^{[2]}$ as in the diagram

$\textstyle{F\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z^{[2]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g^{[2]}}$$\textstyle{Y^{[2]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X}$

such that the bundle gerbe multiplications are respected. The data of $g$ and
$k$ is then a morphism of bundle gerbes, denoted $(k,g)\colon(F,Z)\to(E,Y)$
Note that in this case, we have $F\simeq(g^{[2]})^{*}E$, as $k$ is a map of
principal bundles. An _isomorphism_ $(E,Y)\simeq(F,Z)$ of bundle gerbes on $X$
will be meant in the strictest possible sense, namely where $g$ and $k$ are
isomorphisms.

######  Definition 2.3.

Given a bundle gerbe $(E,Y)$, a _stable trivialisation_ consists of a
principal $U(1)$-bundle $T\to Y$ and an isomorphism
$\operatorname{pr}_{1}^{*}T^{*}\times\operatorname{pr}_{2}T\simeq E$ of
bundles on $Y^{[2]}$, making an isomorphism of bundle gerbes along with
$\operatorname{id}_{Y}$.

######  Example 2.4.

Given a lifting bundle gerbe associated to $P\to X$ and $U(1)\to\widehat{G}\to
G$, a stable trivialisation is equivalent data to a principal
$\widehat{G}$-bundle $\widehat{P}\to X$ lifting $P$: the quotient map
$\widehat{P}\to P$ is the required principal $U(1)$-bundle, and vice versa.

There is a general notion of tensor product of a pair of bundle gerbes,
analogous to the tensor product of $U(1)$-bundles, but we will not need it
here. However, the notion of a power of a single given bundle gerbe is easier
to describe, namely if $(E,Y)$ is a bundle gerbe, then $(E,Y)^{\otimes
n}:=(E^{\otimes n},Y)$ is also a bundle gerbe, using the tensor powers of the
$U(1)$-bundle $E$.

######  Example 2.5.

Let $P\to X$ and $U(1)\to\widehat{G}\to G$ be the data necessary to build a
lifting bundle gerbe. Then $(P\times\widehat{G},P)^{\otimes
n}\simeq(P\times\widehat{G}^{\otimes n},P)$, where
$U(1)\to\widehat{G}^{\otimes n}\to G$ is the $n$-fold central product of
$\widehat{G}$ with itself, which is the same as the $n$-fold tensor power of
the underlying $U(1)$-bundle over $G$.

Associated to each bundle gerbe $(E,Y)$ on $X$ there is a class $DD(E,Y)\in
H^{3}(X,\mathbb{Z})$—the Dixmier–Douady, or $DD$-class—that satisfies the
following properties (see [Mur96], and [MS00] for the last item).

######  Lemma 2.6.

  1. 1.

Given $f\colon X^{\prime}\to X$, then $f^{*}DD(E,Y)=DD(f^{*}(E,Y))$;

  2. 2.

Given a morphism $(k,g)\colon(F,Z)\to(E,Y)$ of bundle gerbes on $X$, then
$DD(F,Z)=DD(E,Y)$;

  3. 3.

For all integers $n$, $DD((E,Y)^{\otimes n})=n\cdot DD(E,Y)$, where a negative
tensor power involves the dual $U(1)$-bundle;

  4. 4.

We have $DD(E,Y)=0$ if and only if the bundle gerbe $(E,Y)$ has a stable
trivialisation;

  5. 5.

Given a central extension $U(1)\to\widehat{G}\to G$, classified by $\alpha\in
H^{3}(BG,\mathbb{Z})\simeq H^{2}(BG,\underline{U(1)})$, and a principal
$G$-bundle classified by some map $\chi\colon X\to BG$, the lifting bundle
gerbe has class in $H^{3}(X,\mathbb{Z})$ corresponding to $\chi^{*}\alpha$.

We shall say a bundle gerbe $(E,Y)$ is _torsion_ if $DD(E,Y)$ is a torsion
element, and it is bundle gerbes of this form that are the main focus of this
note. The reader can rephrase all of the results of this paper without
mentioning gerbes, if desired, using the following easy result.

######  Corollary 2.7.

Given a principal $G$-bundle $P\to X$ and a central extension
$U(1)\to\widehat{G}\to G$, the associated lifting bundle gerbe
$(P\times\widehat{G},P)$ is torsion if an only if there some $n>0$ such that
the $G$-bundle $P$ lifts to a principal $\widehat{G}^{\otimes n}$-bundle.

##  3 Main results

This first, relatively short, proof serves to illustrate the idea of the more
complex proof of Theorem 3.7.

###### Proof of Theorem 1.1.

Let us fix a lifting bundle gerbe $(E,P)$ associated to a principal $G$-bundle
$P\to X$ and an extension $U(1)\to\widehat{G}\to G$. Let $G_{0}$ denote the
connected component of the identity. We can induce a principal
$\pi_{0}(G)$-bundle on $X$ by defining $Q:=P/G_{0}$, where we use the fact
$\pi_{0}(G)=G/G_{0}$. Let $\pi\colon\tilde{X}\to X$ denote the universal
covering space of $X$, and recall that for any covering space on $X$, the
pullback to $\tilde{X}$ is trivialisable. Hence if we form the covering space
$\pi^{*}Q\to\tilde{X}$, it is trivialisable. From this it follows that the
structure group of the $G$-bundle $\tilde{P}:=\pi^{*}P\to\tilde{X}$ reduces to
$G_{0}$. Thus we can find a subbundle $P^{\prime}\subset\tilde{P}$ that on
fibres looks like the inclusion $G_{0}\hookrightarrow G$.

If we denote by $\widehat{G}_{0}\subset\widehat{G}$ the preimage of $G_{0}$,
then we can form the lifting bundle gerbe on $\tilde{X}$ associated to
$P^{\prime}\to\tilde{X}$ and the central extension $U(1)\to\widehat{G}_{0}\to
G_{0}$. By the construction of a lifting bundle gerbe, we get that there is a
morphism of bundle gerbes on $\tilde{X}$:

$\textstyle{P^{\prime}\times\widehat{G}_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P\times\widehat{G}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P^{\prime}\times
G_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P\times
G\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P}$

Thus the DD-class $DD(P^{\prime})$ of the lifting gerbe of $P^{\prime}$ equal
to $DD(\tilde{P})$.

However, the central extension $U(1)\to\widehat{G}_{0}\to G_{0}$ is classified
by an element of $\alpha\in H^{3}(BG_{0},\mathbb{Z})$. This cohomology group
is pure torsion [Bor53, Lemme 26.1], where we use the fact that the cohomology
of the classifying space of a connected Lie group is isomorphic to the
cohomology of the classifying space of the maximal compact subgroup. More
precisely, the torsion subgroup of $H^{*}(BG_{0},\mathbb{Z})$ is exactly the
kernel of the restriction map $H^{*}(BG_{0},\mathbb{Z})\to
H^{*}(BT,\mathbb{Z})$ [Fes81], where $T\subseteq G_{0}$ is a maximal torus of
$G_{0}$ (and $H^{*}(BT,\mathbb{Z})$ is torsion-free). Now $DD(P^{\prime})$ is
the image of $\alpha$ under $\chi^{*}\colon H^{3}(BG_{0},\mathbb{Z})\to
H^{3}(\tilde{X},\mathbb{Z})$, where $\chi\colon\tilde{X}\to BG_{0}$ is a
classifying map. Thus the lifting gerbe associated to $P^{\prime}$ is torsion,
hence so is the lifting gerbe (on $\tilde{X}$) associated to $\tilde{P}$.

Now we can apply the following lemma, as $\pi_{1}(X)$ is finite, and conclude
that the lifting gerbe associated to $P$ is also torsion, since
$DD(\tilde{P})=\pi^{*}DD(P)$. ∎

######  Lemma 3.1.

Given a $k$-sheeted covering space $\pi\colon Y\to X$, and a class $c\in
H^{n}(X,\mathbb{Z})$, if $\pi^{*}c\in H^{n}(Y,\mathbb{Z})$ is torsion, then
$c$ is torsion.

###### Proof.

The composite
$H^{n}(X,\mathbb{Z})\otimes\mathbb{Q}\xrightarrow{\pi^{*}}H^{n}(Y,\mathbb{Z})\otimes\mathbb{Q}\xrightarrow{\pi_{*}}H^{n}(X,\mathbb{Z})\otimes\mathbb{Q}$
is multiplication by $k$, which is invertible, so the pushforward map
$\pi_{*}$ is a retraction of $\pi^{*}$, hence the latter is injective.

Then given a class $c\in H^{n}(X,\mathbb{Z})$, if $\pi^{*}c$ is torsion, then
$\pi^{*}c\otimes\mathbb{Q}$ is zero, hence $c\otimes\mathbb{Q}$ is zero, hence
$c$ is a torsion class. ∎

######  Remark 3.2.

The proof of Theorem 3.6 of [MS11] actually tells us a tiny bit more than is
claimed. What it shows is that instead of demanding the base space $M$ is
simply-connected, it is sufficient to ask that the composite $\pi_{3}(M)\to
H_{3}(M)\to H_{3}(M)/\mathrm{tors}$ is surjective; this is because
$\operatorname{Hom}(H_{3}(M)/\mathrm{tors},\mathbb{Z})\simeq\operatorname{Hom}(H_{3}(M),\mathbb{Z})$,
and it is the latter group that is used in the proof.

Further, the previous Lemma could be used to instead only ask that there is a
finite-sheeted covering space $\tilde{M}\to M$ with $\pi_{3}(\tilde{M})\to
H_{3}(\tilde{M})/\mathrm{tors}$ a surjective map.

Returning to the case of lifting gerbes, we can make a sharper claim, leaving
the smooth category altogether. We first need a preparatory technical lemma.

######  Lemma 3.3.

Let $K$ be a discrete group such that $H_{3}(K,\mathbb{Z})$,
$\operatorname{Ext}(H_{2}(K,\mathbb{Z}),\mathbb{Z})$ and $H_{1}(K,A)$ are
torsion, where $A$ is any $K$-module with underlying abelian group
$A\simeq\mathbb{Z}^{n}$. If $G$ is a finite-dimensional Lie group with
$\pi_{0}(G)\simeq K$, then $H^{3}(BG,\mathbb{Z})$ is torsion.

###### Proof.

For the rest of the proof, unspecified coefficients for cohomology should be
taken as $\mathbb{Z}$. Given a Lie group as in the statement, there is a short
exact sequence of Lie groups $G_{0}\to G\xrightarrow{\pi}K$, where $G_{0}$ is
the connected component of the identity. This gives a fibration $BG_{0}\to
BG\xrightarrow{B\pi}BK$, where the fibre $BG_{0}$ is connected. There is a
(first quadrant) Serre spectral sequence for this fibration given by

$E_{2}^{s,t}=H^{s}(K,\mathcal{H}^{t}(BG_{0}))\Longrightarrow H^{s+t}(BG),$

where $\mathcal{H}^{t}(BG_{0})$ is the $K$-module arising from the local
system on $BK$ with fibre over $x\in BK$ given by $H^{t}(B\pi^{-1}(x))\simeq
H^{t}(BG_{0})$. Note in particular that
$\mathcal{H}^{0}(BG_{0})\simeq\mathbb{Z}$ carries the trivial $K$-action,
since it can be identified with the constant $\mathbb{Z}$-valued functions on
$BG_{0}$. Recall that $H^{k}(BG_{0})$ is finitely generated and is torsion for
odd $k$ [Bor53].

|  |
---|---|---
$\mathcal{H}^{3}(BG_{0})^{K}$$\textstyle{\ast}$$E_{2}$$\textstyle{\mathcal{H}^{2}(BG_{0})^{K}}$$H^{1}(K,\mathcal{H}^{2}(BG_{0}))$$\textstyle{\ast}$$\textstyle{0}$$\textstyle{0}$$0$$\textstyle{0}$$\textstyle{\Big{\uparrow}\
t}$$\textstyle{\ast}$$\textstyle{\ast}$$\textstyle{\ast}$$H^{3}(K)$$\textstyle{\longrightarrow\
s}$ Figure 1: The $E_{2}$ page of the Serre spectral sequence associated to
the fibre sequence $BG_{0}\to BG\to BK$. Boxed entries lie on the line
$s+t=3$.

Figure 1 shows part of the $E_{2}$ page, making the substitutions that
$H^{0}(K,A)=V^{K}$, the invariants of the $K$-action on the module $A$, and
that $H^{1}(BG_{0})=H_{1}(BG_{0})=0$, as $G_{0}$ is connected, giving the
zeroes in the 1-row. The asterisk entries are not needed.

Note that by the universal coefficient theorem there is a short exact sequence

$0\to\operatorname{Ext}(H_{2}(K),\mathbb{Z})\to
H^{3}(K)\to\operatorname{Hom}(H_{3}(K),\mathbb{Z})\to 0,$

hence if $H_{3}(K)$ is torsion, the right term vanishes, and so $H^{3}(K)$ is
torsion, by the assumption on the Ext group.

The short exact sequence $H^{2}(BG_{0})_{\mathrm{tors}}\hookrightarrow
H^{2}(BG_{0})\to H^{2}(BG_{0})/\mathrm{tors}$ of groups is also a short exact
sequence of $K$-modules, so there is a long exact sequence which reads in part

$\cdots\to H^{1}(K,\mathcal{H}^{2}(BG_{0})_{\mathrm{tors}})\to
H^{1}(K,\mathcal{H}^{2}(BG_{0}))\to
H^{1}(K,\mathcal{H}^{2}(BG_{0})/\mathrm{tors})\to\cdots$

As $\mathcal{H}^{2}(BG_{0})_{\mathrm{tors}}$ is finite, the left term is
torsion, and so it suffices to know that
$H^{1}(K,\mathcal{H}^{2}(BG_{0})/\mathrm{tors})$ is torsion to conclude that
$H^{1}(K,\mathcal{H}^{2}(BG_{0}))$ a torsion group.

There is a version of the universal coefficient theorem (UCT) for cohomology
with local coefficients, that generally only holds under slightly special
extra assumptions. Notice that while everything here is in terms of group
cohomology of $K$, this is the same as the singular cohomology of $BK$ Finding
this UCT was set as an exercise by Spanier [Spa66, Ch. 5, Exercise J.4], and
can be completed by using [CE56, Ch VI, Theorem 3.3a] under the special
assumption (3a’) in _loc. cit._.666The formulation I use here was given by
Oscar Randal-Williams at https://math.stackexchange.com/q/4135432 (version:
2021-05-11). The special assumptions are satisfied by the setup here, as we
are working over the integers, and where the local coefficient system
$\mathcal{H}^{2}(BG_{0})/\mathrm{tors}$ has fibres that are finitely generated
free abelian groups. The latter fact means we can also rely on the local
system being reflexive, in that it is naturally isomorphic to its double dual.

The upshot is, there is a short exact sequence

$0\to\operatorname{Ext}(H_{0}(K,A^{*}),\mathbb{Z})\to
H^{1}(K,A)\to\operatorname{Hom}(H_{1}(K,A^{*}),\mathbb{Z})\to 0$

where $A^{*}=\operatorname{Hom}_{\mathbb{Z}}(A,\mathbb{Z})$ is the dual
module, also with underlying abelian group isomorphic to $\mathbb{Z}^{n}$.
Then $H_{0}(K,A^{*})$ is a quotient of $A^{*}$, hence is finitely generated,
and so $\operatorname{Ext}(H_{0}(K,A^{*}),\mathbb{Z})$ is torsion (as it only
sees the torsion subgroup of $H_{0}(K,A^{*})$, which is finite). Thus if
$H_{1}(K,A^{*})$ is torsion, the right term vanishes, and so $H^{1}(K,A)$ is
torsion, as needed, where we take $A=\mathcal{H}^{2}(BG_{0})/\mathrm{tors}$.

We thus know that all the boxed groups are torsion groups, and so the groups
$E_{\infty}^{s,t}$ for $s+t=3$, being subquotients of them, are also torsion.
Further, this implies that $H^{3}(BG,\mathbb{Z})$, being an iterated extension
of torsion group, is torsion, as needed. ∎

Note that nontrivial coefficients for $H_{1}$ are essential in this result,
since the action of $K$ on $\mathcal{H}^{2}(BG_{0})$ is nontrivial in general.

######  Example 3.4.

For the group $G=U(1)\rtimes\operatorname{Aut}((U(1))$, the
$\operatorname{Aut}(U(1))\simeq\\{\pm 1\\}$-action on
$\mathcal{H}^{2}(BU(1))\simeq\mathbb{Z}$ is nontrivial, as the inversion map
on $U(1)$ induces a map on $BU(1)$ that pulls back the Chern class of the
universal line bundle to its negative. More generally, given a group $K$ with
a surjection $K\to\\{\pm 1\\}$, we get a nontrivial semidirect product
$U(1)\rtimes K$ where $\mathcal{H}^{2}(BU(1))$ is a nontrivial $K$-module.

It is not immediately obvious what the action can be for our class of examples
appearing in Example 4.3, so this is not a redundant assumption.

######  Remark 3.5.

Given more specific information about the torsion in
$H^{*}(BG_{0},\mathbb{Z})$, for instance if it vanishes, or is all $P$-torsion
for some class $P$ of primes, and $K$ is a $P$-torsion group, then more
information about what sort of torsion appears in $H^{3}(BG,\mathbb{Z})$ is
available.

Recall that a _locally finite group_ is a discrete group that is the filtered
colimit of finite groups, or, equivalently, the direct limit of its finite
subgroups. For instance, an infinite direct sum of finite groups is locally
finite.

######  Corollary 3.6.

Let $K$ be a locally finite group such that
$\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ is torsion. If $G$ is a finite-
dimensional Lie group with $\pi_{0}(G)\simeq K$, then $H^{3}(BG,\mathbb{Z})$
is torsion.

###### Proof.

The key fact we need is that homology of groups commutes with taking filtered
colimits, and this is true even with nontrivial coefficients, assuming the
coefficients are also expressed as a direct limit. For a locally finite group
$K$ acting on an abelian group $A$, we can restrict the action on $A$ to
finite subgroups $K_{\alpha}\subset K$, and so then $A$ as a $K$-module is the
colimit of $K_{\alpha}$-modules.

Now since the group homology of a finite group is torsion, the homology groups
$H_{n}(K,\mathbb{Z})$, $n=1,3$ of a locally finite group are also torsion. By
the observation about homology with nontrivial module coefficients, we can
even conclude that $H_{1}(K,A)$ is torsion for $A$ a $K$-module.

Thus if we also know that $\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ is
torsion, then we have all the hypotheses for Lemma 3.3, and so we can conclude
that $H^{3}(BG,\mathbb{Z})$ as required. ∎

With Lemma 3.3 in hand, we can prove the sharper version of the main result.

######  Theorem 3.7.

Fix a finite-dimensional Lie group $G$ and a connected topological space $X$
such that there is a finite-sheeted covering space $Y\to X$, with fundamental
group satisfying the following condition: every homomorphism
$\pi_{1}(Y)\to\pi_{0}(G)$ factors through a group $K$ satisfying the algebraic
properties in Lemma 3.3. Then for any central extension $U(1)\to\widehat{G}\to
G$, and principal $G$-bundle $P\to X$, the associated lifting gerbe is
torsion.

###### Proof.

The proof is similar to that for Theorem 1.1, except instead of reducing the
structure group to $G_{0}$, we proceed as follows.

Fix a central extension $U(1)\to\widehat{G}\to G$. First note that the lifting
bundle gerbe for the universal $G$-bundle $EG\to BG$ has $DD$-class given by
the the class in $H^{3}(BG,\mathbb{Z})$ classifying the central extension, by
Lemma 2.6.5 applied to the identity map of $BG$.

Then, given a group $K$ satisfying the hypothesis of Lemma 3.3, and a
homomorphism $K\to\pi_{0}(G)$, the pullback group
$G_{K}:=K\times_{\pi_{0}(G)}G$ has a central extension $U(1)\to
K\times_{\pi_{0}}\widehat{G}\to G_{K}$. Since $H^{3}(G_{K},\mathbb{Z})$ is
torsion (as $\pi_{0}(G_{K})\simeq K$) all lifting gerbes associated to
$G_{K}$-bundles are torsion. Further, the lifting gerbe of the pulled-back
$G$-bundle $BG_{K}\times_{BG}EG\to BG_{K}$ is also torsion. Hence if the
classifying map of a $G$-bundle factors through $BG_{K}$, even up to homotopy,
then the $G$-bundle has torsion lifting gerbe.

Now fix a $G$-bundle $P\to X$, take the covering space $Y\to X$ as in the
statement of the theorem, and the classifying map $Y\to X\to BG$. Now there is
a diagram commuting (up to homotopy)

---
$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{BG\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\pi_{1}(Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\pi_{0}(G)}$$\textstyle{BK\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

where $K$ is as in the theorem statement. Since $BG_{K}$ is the (homotopy)
pullback $BK\times_{B\pi_{0}(G)}BG$, the classifying map $Y\to BG$ factors
through $BG_{K}$, up to homotopy. Thus the lifting gerbe for $Y\times_{X}P\to
Y$ is the pullback of the torsion lifting gerbe on $BG_{K}$, and hence is
torsion.

The rest of the proof of Theorem 1.1 applies, to arrive at the conclusion that
$DD(P)$ is torsion. ∎

######  Remark 3.8.

The hypothesis on homomorphisms $\pi_{1}(Y)\to\pi_{0}(G)$ is stated so that
one recovers both the cases $K=\pi_{1}(Y)$ and $K\subseteq\pi_{0}(G)$, which
are what will presumably occur most often in practice.

##  4 Examples and applications

### Examples

Theorem 3.7 applies fairly trivially when $G$ is a finite-dimensional Lie
group with $\pi_{0}(G)$ finite, so the interesting cases are when $\pi_{0}(G)$
is infinite.

######  Corollary 4.1.

If $X$ is a topological space with a locally finite fundamental group
$\pi=\pi_{1}(X)$, with $\operatorname{Ext}\left(H_{2}(\pi),\mathbb{Z}\right)$
torsion, and $G$ is a finite-dimensional Lie group, then every lifting gerbe
arising from a $G$-bundle on $X$ is torsion.

Similarly, if $G$ is a finite-dimensional Lie group with $\pi=\pi_{0}(G)$
locally finite with $\operatorname{Ext}\left(H_{2}(\pi),\mathbb{Z}\right)$
torsion, then for any topological space $X$, every lifting gerbe associated to
a $G$-bundle on $X$ and central extension of $G$ is torsion.

The main difficulty thus lies in arranging that the $\operatorname{Ext}$ group
is torsion. But there are certainly infinite groups where this is so.

######  Lemma 4.2.

Let $K$ be a locally finite group given by a countable direct limit of finite
abelian groups $K_{\alpha}$, the exponents of which are bounded, but whose
orders are unbounded. Then $K$ is an infinite group satisfying the hypotheses
of Lemma 3.3.

###### Proof.

This implies $K$ has finite exponent, namely the least common multiple of the
exponents of the groups $K_{\alpha}$, and also implies that
$H_{2}(K,\mathbb{Z})\simeq\operatorname*{colim}_{\alpha}H_{2}(K_{\alpha},\mathbb{Z})$
is torsion, since each $H_{2}(K_{\alpha},\mathbb{Z})$ is finitely generated
and torsion, hence finite. Further, the torsion in
$H_{2}(K_{\alpha},\mathbb{Z})$ is bounded in terms of the exponent of
$K_{\alpha}$, so there is a uniform bound on the torsion as $\alpha$ varies.

Using the fact that a countable directed diagram has a cofinal subsequence
$(A_{n})_{n\in\mathbb{N}}$, the colimit defining $K$ can be assumed without
loss of generality to be sequential. We can then consider the exact sequence
(all homology with integer coefficients)

$0\to\mathop{\lim\nolimits^{1}}\operatorname{Hom}(H_{2}(K_{n}),\mathbb{Z})\to\operatorname{Ext}(H_{2}(K),\mathbb{Z})\to\lim_{\mathbb{N}}\operatorname{Ext}(H_{2}(K_{n}),\mathbb{Z})\to
0.$

Now $\operatorname{Hom}(H_{2}(K_{n}),\mathbb{Z})=0$ for all $n$, as
$H_{2}(K_{n},\mathbb{Z})$ is finite, hence the $\mathop{\lim\nolimits^{1}}$
term is trivial. Now we use the fact that the sequence
$H_{2}(K_{n},\mathbb{Z})$ has bounded exponent, hence
$\operatorname{Ext}(H_{2}(K_{n}),\mathbb{Z})$ has bounded torsion, so that
$\lim_{\mathbb{N}}\operatorname{Ext}(H_{2}(K_{n}),\mathbb{Z})$ is still
torsion. This, finally, gives us that
$\operatorname{Ext}(H_{2}(K,\mathbb{Z}),\mathbb{Z})$ is torsion, and hence we
can apply Corollary 3.6 to conclude. ∎

######  Example 4.3.

For instance, consider the infinite direct sum $\bigoplus_{n\in\mathbb{N}}A$
for any finite abelian group $A$. This is locally finite, as it is the colimit
of the sugroups given by finite direct sums, and its exponent is same as that
of $A$.

More generally, if $K_{0}\subset K_{1}\subset K_{2}\subset\cdots$ is a
countable increasing sequence of finite abelian groups with eventually
constant exponent $e$, then $\bigcup_{n}K_{n}$ is locally finite with exponent
$e$.

In the other direction, here is a counterexample to Theorem 3.7 that was used
in [MS11]. The specific hypothesis of [MS11, Theorem 3.6] that is
violated—connectivity of $G$—is not one required for the version of the
theorem here, but it still breaks our hypotheses in other ways.

######  Example 4.4 ([Bry93, §4.1] and [Joh03, §3.5]).

Take a principal $U(1)$-bundle $Q\to X$ and a function $X\to U(1)$, with
classes $\alpha\in H^{2}(X,\mathbb{Z})$ and $\beta\in H^{1}(X,\mathbb{Z})$.
The bundle gerbe classified by the cup product $\alpha\cup\beta$ is given by
the lifting bundle gerbe corresponding to the central extension

$U(1)\to(U(1)\times\mathbb{Z})\tilde{\times}U(1)\to U(1)\times\mathbb{Z}$

with product
$(n_{1},z_{1};w_{1})\cdot(n_{2},z_{2};w_{2})=(n_{1}+n_{2},z_{1}z_{2};w_{1}w_{2}z_{1}^{n_{2}})$
and the principal $(U(1)\times\mathbb{Z})$-bundle
$Q\times_{X}f^{*}\mathbb{R}\to X$.777Treating $\mathbb{R}\to U(1)$ as a
principal $\mathbb{Z}$-bundle. Note that there is a typo the definition of the
product in [MS11], we use the definition in [Bry93, Equation (4-10)]

Then if $X=M\times S^{1}$, then every every finite-index subgroup
$H<\pi_{1}(M\times S^{1})\simeq\pi_{1}(M)\times\mathbb{Z}$ admits a surjection
onto a finite-index subgroup of $\mathbb{Z}$, for example the one induced by
the composite $Y\to M\times S^{1}\xrightarrow{\operatorname{pr}_{2}}S^{1}$
Thus for every finite-sheeted covering space $Y\to M\times S^{1}$, every
nontrivial homomorphism
$\phi\colon\pi_{1}(Y)\to\pi_{0}(U(1)\times\mathbb{Z})=\mathbb{Z}$ fails to
factor through a subgroup satisfying Lemma 3.3, as
$\operatorname{im}(\phi)\simeq\mathbb{Z}$ and
$H_{1}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$.

If $H^{2}(M,\mathbb{Z})$ contains non-torsion classes, then by Künneth we have
$H^{3}(M\times S^{1},\mathbb{Z})\simeq H^{2}(M,\mathbb{Z})\otimes
H^{1}(S^{1},\mathbb{Z})\oplus H^{3}(M,\mathbb{Z})\otimes
H^{0}(S^{1},\mathbb{Z})$, which means there are cup product classes that are
non-torsion and corresponding to finite-dimensional lifting gerbes by the
above construction.

However, what happens if we are in the situation where we know $K$ is locally
finite, and so Corollary 3.6 has a chance of applying?

######  Example 4.5.

Even if $K$ is locally finite, if there is a non-torsion class in
$H^{3}(K,\mathbb{Z})$, this gives rise to a nontrivial central extension
$U(1)\to\widehat{K}\to K$. Given any semidirect product $G=G_{0}\rtimes K$, we
can form the pullback $G\times_{K}\widehat{K}$, which is then a nontrivial
central extension of $G$, and in fact is classified by a non-torsion class in
$H^{3}(BG,\mathbb{Z})$, using the fact $H^{3}(K,\mathbb{Z})\to
H^{3}(BG,\mathbb{Z})$ has a retraction induced by the section of $G\to K$.

How might one get non-trivial classes in $H^{3}(K,\mathbb{Z})$ for a locally
finite group $K$? If there is some nontrivial
$c\in\operatorname{Ext}(H_{2}(K),\mathbb{Z})$, then using the exact sequence

$0=\operatorname{Hom}(H_{2}(K),\mathbb{R})\to\operatorname{Hom}(H_{2}(K),\mathbb{R}/\mathbb{Z})\xrightarrow{\simeq}\operatorname{Ext}(H_{2}(K),\mathbb{Z})\to\operatorname{Ext}(H_{2}(K),\mathbb{R})=0$

we get a nontrivial homomorphism $\phi_{c}\colon
H_{2}(K)\to\mathbb{R}/\mathbb{Z}\simeq U(1)$ (here using the fact $\mathbb{R}$
is divisible, hence an injective group, and $H_{2}(K)$ is torsion). Further,
since $H^{2}(K,H_{2}(K))\to\operatorname{Hom}(H_{2}(K),H_{2}(K))$ has a
section, by the universal coefficient theorem for group homology, the identity
map on $H_{2}(K)$ coresponds to a nontrivial central extension
$H_{2}(K)\to\widehat{K}\to K$ classified by class in $H^{2}(K,H_{2}(K))$. We
can then induce a central extension of $K$ by $U(1)$ using the homomorphism
$\phi_{c}$, and central extensions by $U(1)$ are classified by
$H^{3}(K,\mathbb{Z})$. All of these constructions are induced by
homomorphisms, so all up we have an injective homomorphism

$\operatorname{Ext}(H_{2}(K),\mathbb{Z})\simeq\operatorname{Hom}(H_{2}(K),U(1))\hookrightarrow
H^{2}(K,U(1))\xrightarrow{\simeq}H^{3}(K,\mathbb{Z})\to H^{3}(BG,\mathbb{Z}),$

and so non-torsion elements in $\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ are a
source of non-torsion elements in $H^{3}(BG,\mathbb{Z})$.

Now the $E_{3}$ page of the spectral sequence in Figure 1 has a potentially
nontrivial differential $d_{3}^{0,2}\colon\mathcal{H}^{2}(BG_{0})^{K}\to
H^{3}(K,\mathbb{Z})$. Moreover $\operatorname{im}(d_{3}^{0,2})$ is the kernel
of the edge homomorphism $H^{3}(K,\mathbb{Z})\to H^{3}(BG,\mathbb{Z})$, since
$H^{3}(K,\mathbb{Z})/\operatorname{im}(d_{3}^{0,2})\simeq
E_{\infty}^{3,0}\subset H^{3}(BG,\mathbb{Z})$. Without more specific analysis
of the differential $d_{3}^{0,2}$, though, it is not immediately obvious that
a non-torsion element remains non-torsion in $H^{3}(BG,\mathbb{Z})$, in
general.

However, some simple examples can be calculated directly, for instance:

######  Example 4.6.

Take $K$ some locally finite group with a surjection $\alpha\colon K\to\\{\pm
1\\}$, and define $G=U(1)\rtimes K$ as in Example 3.4, so that
$\mathcal{H}^{2}(BU(1))^{K}=\\{0\\}$, and so
$\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ injects into $H^{3}(BG)$.

We can even give a concrete example of a locally finite group $K$ where
$\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ has lots of non-torsion elements,
giving non-torsion elements in $H^{3}(U(1)\rtimes K)$. This is despite
$\pi_{0}(U(1)\rtimes K)$ being purely torsion, in contrast to
$U(1)\times\mathbb{Z}$ in Example 4.4.

######  Example 4.7.

Define our locally finite group to be

$K:=\bigoplus_{p\text{ prime}}(\mathbb{Z}/p)^{2}$

which is the colimit of the groups
$K_{p}:=(\mathbb{Z}/2)^{2}\oplus(\mathbb{Z}/3)^{2}\oplus\cdots\oplus(\mathbb{Z}/p)^{2}$.
It is a classical result that
$H_{2}((\mathbb{Z}/p)^{2},\mathbb{Z})\simeq\mathbb{Z}/p$, and we get888This
and the ‘classical result’ follow from the formula $H_{2}(G\times
H,\mathbb{Z})\simeq H_{2}(G,\mathbb{Z})\oplus
H_{2}(H,\mathbb{Z})\oplus(G^{ab}\otimes H^{ab})$ and the fact cyclic groups
have vanishing $H_{2}$.

$H_{2}(K,\mathbb{Z})\simeq\bigoplus_{p\text{ prime}}\mathbb{Z}/p.$

Since $\operatorname{Ext}$ sends direct sums to direct products, we have that

$\operatorname{Ext}(H_{2}(K,\mathbb{Z}),\mathbb{Z})\simeq\prod_{p\text{
prime}}\mathbb{Z}/p,$

and so $H^{3}(K,\mathbb{Z})$ and hence $H^{3}(U(1)\rtimes K,\mathbb{Z})$ has
plenty of non-torsion elements.

For the sake of concreteness, one can take the element
$(1,0;1,0;1,0;\ldots)\in\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ and from this
induce a central extension

$U(1)\to\mathcal{T}\to U(1)\rtimes K$

with non-torsion characteristic class in $H^{3}(B(U(1)\rtimes K),\mathbb{Z})$,
and then this class is the $DD$-class of a lifting bundle gerbe following
Lemma 2.6.5.

As a result of this example, we can see that the assumption of torsion
$\operatorname{Ext}(H_{2}(K),\mathbb{Z})$ in Corollary 3.6 is necessary if one
wants to make a statement about all possible finite-dimensional Lie groups
with group of connected components $K$.

### Applications

These results give strong obstructions to finding finite-dimensional lifting
bundle gerbes. Namely, take a manifold or space $X$ satisfying the hypotheses
of Theorem 3.7 with $H^{3}(X,\mathbb{Z})$ torsion-free. Then all finite-
dimensional lifting gerbes on $X$ have a stable trivialisation.

An example of particular focus in the literature considers a generic
connected, compact simple Lie group, not necessarily simply-connected, which
always has nontrivial third integral cohomology.

######  Proposition 4.8.

The only connected, compact simple Lie groups $G$ that admit a nontrivial
lifting bundle gerbe are the groups $PSO(4n)$ for $n>1$, and in this case
there is precisely one, up to stable isomorphism999It is enough to know that
stable isomorphism is equivalent to having the same $DD$-class.

###### Proof.

First recall that for a Lie group as in the proposition, the fundamental group
is finite, so we can apply Theorem 1.1. Then for all $G\neq PSO(4n)$,
$H^{3}(G,\mathbb{Z})\simeq\mathbb{Z}$, hence any lifting gerbe must be
trivial, and for $PSO(4)\simeq SO(3)\times SO(3)$,
$H^{3}\simeq\mathbb{Z}\oplus\mathbb{Z}$, with the same conclusion.

For $H=PSO(4n)$ with $n>1$, as
$H^{3}(PSO(4n),\mathbb{Z})\simeq\mathbb{Z}\oplus\mathbb{Z}/2$, there is only
one nontrivial torsion class, and this _can_ be can realised by a lifting
gerbe, following [Kre18, Remark 5.1]. Namely, there is a nontrivial central
extension

$U(1)\to\widehat{Z}\to\mathbb{Z}/2\times\mathbb{Z}/2=:Z$

where the underlying manifold of $\widehat{Z}$ is
$U(1)\times\mathbb{Z}/2\times\mathbb{Z}/2$, with multiplication arising from
the nontrivial 2-cocycle $\beta\colon Z\times Z\to U(1)$ with
$\beta((n_{1},n_{2}),(m_{1},m_{2}))=(-1)^{n_{1}+m_{2}}$ [GW09]. Then there is
a nontrivial lifting bundle gerbe arising to the
$(\mathbb{Z}/2\times\mathbb{Z}/2)$-bundle $\operatorname{Spin}(4n)\to
PSO(4n)$, and this has torsion $DD$-class in $H^{3}(PSO(4n),\mathbb{Z})$. ∎

The following corollary is immediate, as a basic gerbe on a Lie group is, by
definition, a non-torsion generator of $H^{3}(G,\mathbb{Z})$ [GR04].

######  Corollary 4.9.

No basic gerbe on a connected, compact simple Lie group $G$, not necessarily
simply-connected, can be constructed as a lifting bundle gerbe.

######  Remark 4.10.

Note however, that the pullback of a basic gerbe can be a finite-dimensional
lifting bundle gerbe. For example, the basic gerbe on $SU(2)$ can be
constructed via a carefully constructed submersion, and is not a lifting
gerbe. Its pullback along the Weyl map $U(1)\times SU(2)/U(1)\to SU(2)$ (where
$U(1)\subset SU(2)$ is the diagonal matrices), however, is stably isomorphic
to the Weyl bundle gerbe [BMS20], which _is_ a lifting bundle gerbe.

This theorem as algebraic consequences, as well. Recall that a _crossed
module_ of Lie groups consists of a homomorphism $t\colon\hat{K}\to L$ of Lie
groups such that $\ker(t)\to\hat{K}\to t(\hat{K})=:K$ is a central extension,
$K$ is a normal closed subgroup of $L$, and a lift of the adjoint action of
$L$ on $K$ to $\hat{K}$, such that $\hat{K}\to
L\to\operatorname{Aut}(\hat{K})$ agrees with the adjoint action of $\hat{K}$
on itself. Moreover, $L\to\operatorname{coker}(t)$ is a principal $K$-bundle,
and we can consider the corresponding lifting bundle gerbe.

######  Corollary 4.11.

Let $G$ be a finite-dimensional connected, compact simple Lie group. Then
there is no finite-dimensional crossed module $t\colon\hat{K}\to L$ of Lie
groups with $\ker(t)\simeq U(1)$ and $\operatorname{coker}(t)\simeq G$ whose
associated lifting bundle gerbe is nontrivial. More generally, no
multiplicative bundle gerbe on $G$ can be stably isomorphic to a finite-
dimensional lifting bundle gerbe with non-torsion $DD$-class.

###### Proof.

By Proposition 4.8 if $\hat{K}\to L$ is a crossed module with
$\operatorname{coker}(t)=G$, then the corresponding lifting bundle gerbe must
be torsion, hence we can consider just the case of $G=PSO(4n)$, $n>1$.

The lifting bundle gerbe arising from a crossed module $\hat{K}\to L$ is
_multiplicative_ [Car+05], as it fits into a strict 2-group extension
corresponding to the extension of crossed modules

$\textstyle{U(1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\hat{K}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{coker}(t)}$

of $\operatorname{coker}(t)$ by $\mathbf{B}U(1)$ (as strict Lie 2-groups). By
[Car+05, Proposition 5.2], the $DD$-class of a multiplicative bundle gerbe on
$G$ is in the image of the transgression map $H^{4}(BG,\mathbb{Z})\to
H^{3}(G,\mathbb{Z})$. But by [Hen17, Theorem 6], for any connected compact Lie
group $G$, the restriction map $H^{4}(BG,\mathbb{Z})\to H^{4}(BT,\mathbb{Z})$
induced by a maximal torus $T\subset G$ is injective, hence
$H^{4}(BG,\mathbb{Z})$ is torsion-free, including for $G=PSO(4n)$. Thus the
only class in $H^{3}(PSO(4n),\mathbb{Z})$ represented by some lifting gerbe is
not the $DD$-class of the lifting gerbe arising from a crossed module.

The more general statement follows as multiplicative bundle gerbes on $G$ are
classified by $H^{4}(BG,\mathbb{Z})$ [Car+05, Proposition 5.2] (see also
[GW09], where the image of the injective map $H^{4}(BG,\mathbb{Z})\to
H^{4}(B\tilde{G},\mathbb{Z})$ is characterised for all possible compact
connected simple Lie groups). ∎

This tells us that any finite-dimensional crossed module $t\colon\hat{K}\to L$
as in Corollary 4.11, there is a principal $U(1)$-bundle $T\to L$ whose
restriction to $K\subset L$ is $\hat{K}\to K$, and there is a $\hat{K}$-action
of $T$ covering the action on $L$ by multiplication via $t$. Note that this
corollary places a constraint on the structure of a crossed module analogous
to how one can have group extensions that are topologically trivial, but still
algebraically nontrivial—there may still be algebraically nontrivial crossed
modules with cokernel $G$.

As Corollary 4.11 constrains the structure of finite-dimensional crossed
modules, it means that higher geometry of principal 2-bundles with strict
structure 2-group (i.e. a crossed module), as in [Wal18] for example, really
must use infinite-dimensional constructions. A connection for such a bundle
takes its values in the truncated $L_{\infty}$-algebra that is the crossed
module of Lie algebras associated to the given crossed module of Lie groups
[Wal18, Definition 5.1.1]. As a result of the preceeding corollary, finite-
dimensional crossed modules $\tau\colon\hat{\mathfrak{k}}\to\mathfrak{l}$ of
Lie algebras where $\operatorname{coker}(\tau)$ is simple are insufficient to
capture all examples of interest.

Finally, recall that twisted $K$-theory is a particular cohomology theory
generalising topological $K$-theory, where one has an extra piece of data, a
_twist_ ; homotopy-theoretically this is a map to the classifying space for
bundles of spectra with fibre the $K$-theory spectrum. The most well-studied
twists101010There are other twists, which arise from bundles of certain self-
absorbing $C^{*}$-algebras arise from a factor of $K(\mathbb{Z},3)$ of this
classifying space, hence when constructing $K$-theory using geometric objects,
geometric objects classified by maps to $K(\mathbb{Z},3)$ are used. As such,
bundle gerbes are one model for twists, as are principal
$PU(\mathcal{H})$-bundles. The latter give rise to lifting bundle gerbes, and
as noted in the introduction, every bundle gerbe is (non-canonically) the
lifting gerbe of some $PU(\mathcal{H})$-bundle. But finite-dimensional bundle
gerbes coming from $PU(n)$-bundles can be used as well, and more generally,
the lifting bundle gerbes of principal $G$-bundles for other groups $G$.

######  Corollary 4.12.

Given a space $X$ as in Theorem 3.7, a finite-dimensional $G$-bundle $P$ and
central extension $\widehat{G}$, every twist $\tau=\tau(P,\widehat{G})$ of the
$K$-theory of $X$ coming from this data is a torsion twist. In particular if
$H^{3}(X,\mathbb{Z})$ is torsion-free, the resulting twisted $K$-theory
$K^{\ast,\tau}(X)$ is just ordinary $K$-theory.

We can view this result as putting strong constraints on the maps

$H^{1}(X,G)\times H^{3}(BG,\mathbb{Z})\to\\{\text{twists of $K$-theory on
$X$}\\}.$

For instance in the case that $\pi_{1}(X)$ is locally finite and
$\operatorname{Ext}(H^{2}(\pi_{1}(X)),\mathbb{Z})$ is torsion, bundles for
arbitrary finite-dimensional Lie groups give _only_ torsion twists.

## References

  * [BMS20] Kimberly E. Becker, Michael K. Murray and Daniel Stevenson  “The Weyl map and bundle gerbes” arXiv:1910.07173 In _J. Geom. Phys._ 149, 2020, pp. 10357219
  * [Bor53] Armand Borel  “Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts”  In _Ann. of Math. (2)_ 57, 1953, pp. 115–207
  * [Bry93] Jean-Luc Brylinski  “Loop Spaces, Characteristic Classes and Geometric Quantisation” 107, Progess in Mathematics  Boston: Birkhäuser-Verlag, 1993
  * [Car+05] Alan Carey et al.  “Bundle Gerbes for Chern-Simons and Wess-Zumino-Witten theories” arXiv:math.DG/0410013 In _Comm. Math. Phys._ 259, 2005, pp. 577–613
  * [CE56] Henri Cartan and Samuel Eilenberg  “Homological algebra”  Princeton University Press, Princeton, N. J., 1956, pp. xv+390
  * [Fes81] Mark Feshbach  “The image of $H^{\ast}(BG,\,{\bf Z})$ in $H^{\ast}(BT,\,{\bf Z})$ for $G$ a compact Lie group with maximal torus $T$”  In _Topology_ 20.1, 1981, pp. 93–95
  * [GR04] Krzysztof Gawędzki and Nuno Reis  “Basic gerbe over non-simply connected compact groups” arXiv:math.DG/307010 In _J. Geom. Phys._ 50.1-4, 2004, pp. 28–55
  * [Gro68] Alexander Grothendieck  “Le groupe de Brauer, I, III, III.”  In _Dix Exposés sur la Cohomologie des Schémas_ Amsterdam: North-Holland, 1968, pp. 46–66
  * [GW09] Krzysztof Gawędzki and Konrad Waldorf  “Polyakov-Wiegmann formula and multiplicative gerbes” arXiv:0908.1130 In _J. High Energy Phys._ , 2009, pp. 07331
  * [Hen17] André Henriques  “The classification of chiral WZW models by $H^{4}_{+}(BG,\mathbb{Z})$” arXiv:1602.02968 In _Lie Algebras, Vertex Operator Algebras, and Related Topics_ 695, Contemporary Mathematics, 2017, pp. 99–122
  * [Joh03] Stuart Johnson  “Constructions with Bundle Gerbes” arXiv:math.DG/0312175, 2003
  * [Kre18] Derek Krepski  “Basic equivariant gerbes on non-simply connected compact simple Lie groups” arXiv:1712.08294 In _J. Geom. Phys._ 133, 2018, pp. 30–41
  * [MS00] Michael K. Murray and Daniel Stevenson  “Bundle gerbes: stable isomorphism and local theory” arXiv:math.DG/9908135 In _J. London Math. Soc. (2)_ 62.3, 2000, pp. 925–937
  * [MS11] Michael K. Murray and Danny Stevenson  “A note on bundle gerbes and infinite-dimensionality” arXiv:1007.4922 In _J. Aust. Math. Soc._ 90.1, 2011, pp. 81–92
  * [Mur96] Michael K. Murray  “Bundle gerbes” arXiv:dg-ga/9407015 In _J. London Math. Soc._ 54.2, 1996, pp. 403–416
  * [Spa66] E.H. Spanier  “Algebraic topology”  McGraw-Hill, New York, 1966
  * [Wal18] Konrad Waldorf  “A global perspective to connections on principal 2-bundles” arXiv:1608.00401 In _Forum Math._ 30.4, 2018, pp. 809–843