d\theta+[\alpha,A_{t}]+[\alpha,\Phi_{t}]d\theta+[A_{t},\varphi]d\theta\right)$

which is equal to $\dfrac{d}{dt}\tilde{F}_{t}.$ This completes the proof of
Lemma 3.2.5. ∎

Now, if we set

$\psi=k\int_{0}^{1}f(\tilde{\alpha},\tilde{F}_{t})dt$

then

$\displaystyle d\psi$ $\displaystyle=D\psi\qquad\text{ (by Lemma
\ref{L:dpsi=Dpsi}) }$
$\displaystyle=k\int_{0}^{1}f(D_{t}\tilde{\alpha},\tilde{F}_{t})dt$
$\displaystyle=k\int_{0}^{1}f(\frac{d}{dt}\tilde{F}_{t},\tilde{F}_{t})dt$
$\displaystyle=\int_{0}^{1}\frac{d}{dt}f(\tilde{F}_{t})dt$
$\displaystyle=f(\tilde{F}_{1})-f(\tilde{F}_{0}).$

So $s_{2k-1}^{P}(f)$ is independent of choice of connection and Higgs field.

∎

It remains only to prove that $s_{2k-1}^{P}(f)$ is the pull-back of $\tau(f)$
by $\operatorname{hol}_{\Phi}.$ For this we follow the argument in [11] that
will give us a formula for $f(\tilde{F})$ that we can use to calculate
$s_{2k-1}^{\Omega\widetilde{P}}(f)$ for a loop bundle
$\Omega\widetilde{P}\xrightarrow{\Omega G}\Omega(M\times S^{1})$ and then we
can use Lemma 3.1.1 to generalise to a general $\Omega G$-bundle.

If we start with the $G$-bundle $\widetilde{P}\to M\times S^{1}$ we can pull-
back by the evaluation map $\operatorname{ev}\colon[0,1]\times\Omega(M\times
S^{1})\to(M\times S^{1})$ to get a trivial bundle
$\operatorname{ev}^{*}\widetilde{P}$ over $[0,1]\times\Omega(M\times S^{1}).$
A section is given by

$h\colon[0,1]\times\Omega(M\times
S^{1})\to\operatorname{ev}^{*}\widetilde{P};\quad(t,\gamma)\mapsto\hat{\gamma}(t),$

where $\hat{\gamma}$ is the horizontal lift of $\gamma.$ If $\tilde{A}$ is the
connection in $\widetilde{P}$ we can pull it back to
$\operatorname{ev}^{*}\widetilde{P}$ and then back to
$[0,1]\times\Omega(M\times S^{1})$ to obtain

$\tilde{A}^{\prime}:=h^{*}\operatorname{ev}^{*}\tilde{A}.$

We can calculate the curvature $\tilde{F}$ of $\tilde{A}$ and pull it back by
$\operatorname{ev}$ to $[0,1]\times\Omega(M\times S^{1})$ and because this is
a product manifold we can decompose it into parts with a $dt$ and parts
without a $dt.$ Under this decomposition, we have

$\operatorname{ev}^{*}\tilde{F}=-\frac{\partial}{\partial
t}\tilde{A}^{\prime}\wedge dt+\tilde{F}^{\prime},$

where we call the component without a $dt$ $\tilde{F}^{\prime}$ since if we
view the form $\tilde{A}^{\prime}$ for fixed $t_{0}$ as a connection form on
$\Omega(M\times S^{1})$ then its curvature is $\tilde{F}^{\prime}$ evaluated
at $t_{0}.$

Now, we want to calculate $\int_{S^{1}}f(\tilde{F})$ and using Lemma 3.1.2 we
have for a general $\Omega G$-bundle $P\to M,$

$\displaystyle\int_{S^{1}}f(\tilde{F})$
$\displaystyle=\eta^{*}\int_{S^{1}}\operatorname{ev}^{*}f(\tilde{F})$
$\displaystyle=\eta^{*}\int_{S^{1}}f(\operatorname{ev}^{*}\tilde{F}).$

So we wish to calculate explicitly
$\int_{S^{1}}f(\operatorname{ev}^{*}\tilde{F}).$ If we view the circle as the
interval $[0,1]$ with endpoints identified, then we can write

$\int_{S^{1}}f(\operatorname{ev}^{*}\tilde{F})=\int_{[0,1]}f(\operatorname{ev}^{*}\tilde{F})$

and so we have

$\displaystyle k\int_{S^{1}}f(\nabla\Phi,F)d\theta$
$\displaystyle=\eta^{*}\int_{S^{1}}f(\operatorname{ev}^{*}\tilde{F})$
$\displaystyle=\eta^{*}\int_{[0,1]}f(-\frac{\partial}{\partial
t}\tilde{A}^{\prime}\wedge dt+\tilde{F}^{\prime})$
$\displaystyle=\eta^{*}\int_{[0,1]}f(\tilde{F}^{\prime})-k\eta^{*}\int_{[0,1]}f(-\frac{\partial}{\partial
t}\tilde{A}^{\prime},\tilde{F}^{\prime})dt$
$\displaystyle=-k\eta^{*}\int_{[0,1]}f(-\frac{\partial}{\partial
t}\tilde{A}^{\prime},\tilde{F}^{\prime})dt.$

Using the formula
$\tilde{F}^{\prime}=d\tilde{A}^{\prime}+\frac{1}{2}[\tilde{A}^{\prime},\tilde{A}^{\prime}],$
we can write this as:

$-k\eta^{*}\left\\{\int_{[0,1]}f(\partial\tilde{A}^{\prime},d\tilde{A}^{\prime})dt\right.\\\
\left.+(k-1)\frac{1}{2}\int_{[0,1]}f(\partial\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt+\ldots\right.\\\
...+\binom{k-1}{k-2}\left(\frac{1}{2}\right)^{k-2}\int_{[0,1]}f(\partial\tilde{A}^{\prime},d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
\left.+\left(\frac{1}{2}\right)^{k-1}\int_{[0,1]}f(\partial\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\right\\}$

where we have written $\partial\tilde{A}^{\prime}$ for
$\partial\tilde{A}^{\prime}/\partial t.$ Thus we need to work with the general
term

$\binom{k-1}{i}\left(\frac{1}{2}\right)^{i}\int_{[0,1]}f(\partial\tilde{A}^{\prime},\underbrace{d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime}}_{k-i-1},\underbrace{[\tilde{A}^{\prime},\tilde{A}^{\prime}],\ldots,[\tilde{A}^{\prime},\tilde{A}^{\prime}]}_{i})dt.$

To deal with these terms we shall use integration by parts and the
$ad$-invariance of $f.$ Thus we need to know in detail how $ad$-invariance
works.

######  Lemma 3.2.6.

Let $\varphi_{1},\ldots,\varphi_{k}$ be ${\mathfrak{g}}$-valued forms of
degree $q_{1},\ldots,q_{k}$ respectively. Then if $A$ is a
${\mathfrak{g}}$-valued $p$-form, we have

$f([\varphi_{1},A],\varphi_{2},\ldots,\varphi_{k})\\\
=f(\varphi_{1},[A,\varphi_{2}],\ldots,\varphi_{k})+(-1)^{pq_{2}}f(\varphi_{1},\varphi_{2},[A,\varphi_{3}],\ldots,\varphi_{k})+\ldots\\\
\ldots+(-1)^{p(q_{2}+\ldots
q_{k-1})}f(\varphi_{1},\ldots,\varphi_{k-1},[A,\varphi_{k}]).$

###### Proof.

We can expand $\varphi_{i}$ as $\varphi_{i}=\varphi_{i,j}\omega_{i}^{j}$ for
$\varphi_{i,j}\in{\mathfrak{g}}$ and $\omega_{i}^{j}$ a $q_{i}$-form. Then we
have

$f(\varphi_{1},\ldots,\varphi_{k})=f(\varphi_{1,j_{1}},\ldots,\varphi_{k,j_{k}})\omega_{1}^{j_{1}}\wedge\ldots\wedge\omega_{k}^{j_{k}}.$

Now if $A$ is a ${\mathfrak{g}}$ valued $p$-form and we write
$A=A_{i}\alpha^{i}$ as above, then

$f([A,\varphi_{1}],\varphi_{2},\ldots,\varphi_{k})\\\
\phantom{f}=f([A_{i},\varphi_{1,j_{1}}],\varphi_{2,j_{2}},\ldots,\varphi_{k,j_{k}})\alpha^{i}\wedge\omega_{1}^{j_{1}}\wedge\ldots\wedge\omega_{k}^{j_{k}}\\\
\phantom{f}=f(\varphi_{1,j_{1}},[\varphi_{2,j_{2}},A_{i}],\ldots,\varphi_{k,j_{k}})(-1)^{p(q_{1}+q_{2})}\omega_{1}^{j_{1}}\wedge\omega_{2}^{j_{2}}\alpha^{i}\wedge\ldots\wedge\omega_{k}^{j_{k}}\\\
+f(\varphi_{1,j_{1}},\varphi_{2,j_{2}},[\varphi_{3,j_{3}},A_{i}],\ldots,\varphi_{k,j_{k}})(-1)^{p(q_{1}+q_{2}+q_{3})}\omega_{1}^{j_{1}}\wedge\omega_{2}^{j_{2}}\wedge\omega_{3}^{j_{3}}\wedge\alpha^{i}\wedge\ldots\wedge\omega_{k}^{j_{k}}\\\
\ldots+f(\varphi_{1,j_{1}},\varphi_{2,j_{2}},\ldots,[\varphi_{k,j_{k}},A_{i}])(-1)^{p(q_{1}+q_{2}+\ldots+q_{k})}\omega_{1}^{j_{1}}\wedge\omega_{2}^{j_{2}}\wedge\ldots\wedge\omega_{k}^{j_{k}}\wedge\alpha^{i}$

That is,

$f([A,\varphi_{1}],\varphi_{2},\ldots,\varphi_{k})\\\
=(-1)^{pq_{1}}f(\varphi_{1},[\varphi_{2},A],\ldots,\varphi_{k})+(-1)^{p(q_{1}+q_{2})}f(\varphi_{1},\varphi_{2},[\varphi_{3},A],\ldots,\varphi_{k})+\ldots\\\
\ldots+(-1)^{p(q_{1}+\ldots+q_{k})}f(\varphi_{1},\ldots,\varphi_{k-1},[\varphi_{k},A]),$

which we can write as:

$f([\varphi_{1},A],\varphi_{2},\ldots,\varphi_{k})\\\
=f(\varphi_{1},[A,\varphi_{2}],\ldots,\varphi_{k})+(-1)^{pq_{2}}f(\varphi_{1},\varphi_{2},[A,\varphi_{3}],\ldots,\varphi_{k})+\ldots\\\
\ldots+(-1)^{p(q_{2}+\ldots
q_{k-1})}f(\varphi_{1},\ldots,\varphi_{k-1},[A,\varphi_{k}]).$

∎

We are now in a position to prove

######  Proposition 3.2.7.

$s_{2k-1}^{P}(f)=\operatorname{hol}_{\Phi}^{*}\tau(f).$

###### Proof.

To calculate the general term given above, we integrate by parts in the
$\Omega(M\times~{}S^{1})$ and $t$ directions giving

$\int_{[0,1]}f_{i}dt=\int_{[0,1]}f(d\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
+i\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},d[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-d\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt$

and

$\int_{[0,1]}f_{i}dt=f(\tilde{A}^{\prime}_{1},d\tilde{A}^{\prime}_{1},\ldots,d\tilde{A}^{\prime}_{1},[\tilde{A}^{\prime}_{1},\tilde{A}^{\prime}_{1}])-f(\tilde{A}^{\prime}_{0},d\tilde{A}^{\prime}_{0},\ldots,d\tilde{A}^{\prime}_{0},[\tilde{A}^{\prime}_{0},\tilde{A}^{\prime}_{0}])\\\
-(k-1-i)\int_{[0,1]}f(\tilde{A}^{\prime},\partial
d\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-i\int_{[0,1]}f(\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},\partial[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt$

where we have written $f_{i}$ for the integrand of the general term given
earlier. Combining these gives

$(k-i)\int_{[0,1]}f_{i}dt=f_{i,1}-f_{i,0}-i\int_{[0,1]}f(\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},\partial[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
+i(k-1-i)\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},d[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-(k-1-i)d\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt$

where we have written $f_{i,1}$ and $f_{i,0}$ for $f_{i}$ evaluated at $t=1$
and $0$ respectively. Using $ad$-invariance, the term on the middle line
simplifies as follows:

$\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},d[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
=2\int_{[0,1]}f([d\tilde{A}^{\prime},\tilde{A}^{\prime}],\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
=2\int_{[0,1]}f(d\tilde{A}^{\prime},[\tilde{A}^{\prime},\partial\tilde{A}^{\prime}],\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-2\int_{[0,1]}f(d\tilde{A}^{\prime},\partial\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}],d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
+2(k-2-i)\int_{[0,1]}f(d\tilde{A}^{\prime},\partial\tilde{A}^{\prime},\tilde{A}^{\prime},[\tilde{A}^{\prime},d\tilde{A}^{\prime}],d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
=\int_{[0,1]}f(d\tilde{A}^{\prime},\partial[\tilde{A}^{\prime},\tilde{A}^{\prime}],\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-2\int_{[0,1]}f(\partial\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-(k-2-i)\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},d[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt$

and so

$(k-1-i)\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},d[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
=\int_{[0,1]}f(\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},\partial[\tilde{A}^{\prime},\tilde{A}^{\prime}],[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt\\\
-2\int_{[0,1]}f(\partial\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt.$

Inserting this into the formula for $\int f_{i}dt$ gives

$(k-i)\int_{[0,1]}f_{i}dt=f_{i,1}-f_{i,0}-2i\int_{[0,1]}f_{i}dt\\\
-(k-1-i)d\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt$

and hence

$(k+i)\int_{[0,1]}f_{i}dt\\\
=f_{i,1}-f_{i,0}-(k-1-i)d\int_{[0,1]}f(\partial\tilde{A}^{\prime},\tilde{A}^{\prime},d\tilde{A}^{\prime},\ldots,d\tilde{A}^{\prime},[\tilde{A}^{\prime},\tilde{A}^{\prime}])dt.$

So we have the following expression for $s_{2k-1}^{P}(f):$

$k\int_{S^{1}}f(\nabla\Phi,F)d\theta\\\
=-k\eta^{*}\left\\{\sum_{i=0}^{k-1}\binom{k-1}{i}\left(\frac{1}{2}\right)^{i}\frac{1}{k+i}\left(f_{i,1}-f_{i,0}-(k-i-1)dc_{i}\vphantom{\tilde{f}}\right)\right\\}$

where $c_{i}$ is the last integral in the equation above (with $i$
$[\tilde{A}^{\prime},\tilde{A}^{\prime}]$’s).

Now since $\tilde{A}^{\prime}_{0}=0$ and
$h(0,\gamma)=h(1,\gamma)\operatorname{hol}(\gamma)$ (where $h$ is the section
from earlier), we have that

$\tilde{A}^{\prime}_{0}=ad(\operatorname{hol}^{-1})\tilde{A}^{\prime}_{1}+\operatorname{hol}^{-1}d\operatorname{hol}$

and so

$\tilde{A}^{\prime}_{1}=-d\operatorname{hol}\operatorname{hol}^{-1}.$

Therefore we have that $f_{i,0}=0$ and we can calculate $f_{i,1}$ in terms of
$f_{0,1}$ as follows:

$\displaystyle f_{0,1}$
$\displaystyle=f(\tilde{A}^{\prime}_{1},d\tilde{A}^{\prime}_{1})$
$\displaystyle=f(-d\operatorname{hol}\operatorname{hol}^{-1},d(-d\operatorname{hol}\operatorname{hol}^{-1}))$
$\displaystyle=(-1)^{k}\left(\frac{1}{2}\right)^{k-1}\operatorname{hol}^{*}f(\Theta,[\Theta,\Theta])$

and in general,

$\displaystyle f_{i,1}$
$\displaystyle=f(\tilde{A}^{\prime}_{1},d\tilde{A}^{\prime}_{1},\ldots,d\tilde{A}^{\prime}_{1},[\tilde{A}^{\prime}_{1},\tilde{A}^{\prime}_{1}])$
$\displaystyle=(-1)^{k-i}\left(\frac{1}{2}\right)^{k-1-i}\operatorname{hol}^{*}f(\Theta,[\Theta,\Theta])$
$\displaystyle=(-1)^{i}2^{i}f_{0,1}$

using the fact that
$d(-d\operatorname{hol}\operatorname{hol}^{-1})=-\frac{1}{2}[d\operatorname{hol}\operatorname{hol}^{-1},d\operatorname{hol}\operatorname{hol}^{-1}].$

Therefore we have

$k\int_{S^{1}}f(\nabla\Phi,F)d\theta\\\
=\left(-\frac{1}{2}\right)^{k-1}k\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{(-1)^{i}}{k+i}\operatorname{hol}_{\Phi}^{*}f(\Theta,[\Theta,\Theta])\\\
+k\sum_{i=0}^{k-i}\binom{k-1}{i}\left(\frac{1}{2}\right)^{i}\frac{1}{k+i}(k-i-1)dc_{i}.$

We have seen already that the coefficient above is equal to the coefficient in
the definition of the transgression map:

$k\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{(-1)^{i}}{k+i}=\frac{k!(k-1)!}{(2k-1)!}.$

So we see that the pull-back of the transgression of $f$ is cohomologous to
the string $(2k-1)$-class.

∎

Combining Propositions 3.2.2, 3.2.4 and 3.2.7, we have the following Theorem

######  Theorem 3.2.8.

The diagram

$\textstyle{H^{2k}(BG)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{C\text{-}W_{\widetilde{P}}}$$\scriptstyle{\tau}$$\textstyle{H^{2k}(M\times
S^{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\int_{S^{1}}}$$\textstyle{H^{2k-1}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{hol}_{\Phi}^{*}}$$\textstyle{H^{2k-1}(M)}$

commutes. Furthermore, the composition map

$H^{2k}(BG)\to H^{2k-1}(M)$

associates to any invariant polynomial its string $(2k-1)$-class, which is a
characteristic class.

###  3.3 The universal string class for
$L^{\scriptscriptstyle{\vee}}G$-bundles

We would now like to return to the study of the free loop group. In this
section, we shall give a partial generalisation of the results in the previous
section. However, we shall be working with a slightly different group than in
the rest of this thesis. For the remainder of this chapter we shall be
considering the group of smooth maps from the interval $[0,2\pi]$ into $G$
whose endpoints are coincident. This group shall be denoted
$L^{\scriptscriptstyle{\vee}}G.$ Note that $LG\subseteq
L^{\scriptscriptstyle{\vee}}G.$ We also have the based version
$\Omega^{\scriptscriptstyle{\vee}}G$ of this group consisting of maps
$[0,2\pi]\to G$ such that the endpoints are mapped to the identity in $G.$

We will give a classifying theory for $L^{\scriptscriptstyle{\vee}}G$ bundles
and present a calculation for the string class of the universal
$L^{\scriptscriptstyle{\vee}}G$-bundle.

####  3.3.1 Classification of $L^{\scriptscriptstyle{\vee}}G$-bundles

In order to extend the ideas from the previous section (namely, calculating
the string class of the universal $L^{\scriptscriptstyle{\vee}}G$-bundle) we
need a model for $EL^{\scriptscriptstyle{\vee}}G.$ To construct this we view
$L^{\scriptscriptstyle{\vee}}G$ as the semi-direct product
$\Omega^{\scriptscriptstyle{\vee}}G\rtimes G.$ The group multiplication is
given by

$(\gamma_{1},g_{1})(\gamma_{2},g_{2})=(g_{2}^{-1}\gamma_{1}g_{2}\gamma_{2},g_{1}g_{2})$

and the isomorphism between $\Omega^{\scriptscriptstyle{\vee}}G\rtimes G$ and
$L^{\scriptscriptstyle{\vee}}G$ is

$\displaystyle\Omega^{\scriptscriptstyle{\vee}}G\rtimes G$
$\displaystyle\xrightarrow{\sim}L^{\scriptscriptstyle{\vee}}G;\quad(\gamma,g)\mapsto
g\gamma.$

On the level of Lie algebras, the isomorphism is

$\displaystyle\Omega^{\scriptscriptstyle{\vee}}{\mathfrak{g}}\rtimes{\mathfrak{g}}$
$\displaystyle\xrightarrow{\sim}L^{\scriptscriptstyle{\vee}}{\mathfrak{g}};\quad(\xi,X)\mapsto
X+\xi.$

We therefore need a model for the universal
$\Omega^{\scriptscriptstyle{\vee}}G\rtimes G$-bundle. For this, we shall take
the product of the universal $\Omega^{\scriptscriptstyle{\vee}}G$-bundle and
the universal $G$-bundle. A model for the universal
$\Omega^{\scriptscriptstyle{\vee}}G$-bundle is given by the space of maps from
the interval $[0,2\pi]$ into $G,$ denoted $P^{\scriptscriptstyle{\vee}}G.$ The
based loop group $\Omega^{\scriptscriptstyle{\vee}}G$ acts on this space by
right multiplication and evaluation at the endpoint of a path gives a locally
trivial $\Omega^{\scriptscriptstyle{\vee}}G$-bundle
$P^{\scriptscriptstyle{\vee}}G\to G.$ As our study of
$\Omega^{\scriptscriptstyle{\vee}}G$ will be confined to this section, we
shall refer to $P^{\scriptscriptstyle{\vee}}G$ as the _path fibration_ without
any risk of confusion. $P^{\scriptscriptstyle{\vee}}G$ is contractible since
any path $p$ can be homotopied to the identity path by the map

$h\colon I\times P^{\scriptscriptstyle{\vee}}G\to
P^{\scriptscriptstyle{\vee}}G;\quad(t,p)\mapsto(\theta\mapsto p(t\theta)).$

Therefore the path fibration is a model for the universal
$\Omega^{\scriptscriptstyle{\vee}}G$-bundle. So, for our model for
$EL^{\scriptscriptstyle{\vee}}G$ we shall take the space
$P^{\scriptscriptstyle{\vee}}G\times EG$ which is contractible since
$P^{\scriptscriptstyle{\vee}}G$ and $EG$ are both contractible. This is acted
on by $\Omega^{\scriptscriptstyle{\vee}}G\rtimes G:$

$(p,x)(\gamma,g)=(g^{-1}pg\gamma,xg)$

where $xg$ is the right action of $G$ on $EG.$ This action is free (since $G$
acts on $EG$ freely) and transitive on fibres (since the action on $EG$ is
transitive and the equation $g^{-1}p_{1}g\gamma=p_{2}$ can always be solved)
and so $P^{\scriptscriptstyle{\vee}}G\times EG$ is a model for
$EL^{\scriptscriptstyle{\vee}}G$ and $BL^{\scriptscriptstyle{\vee}}G$ is equal
to $(P^{\scriptscriptstyle{\vee}}G\times
EG)/(\Omega^{\scriptscriptstyle{\vee}}G\rtimes G).$ In fact, if we consider
the map

$(P^{\scriptscriptstyle{\vee}}G\times
EG)/(\Omega^{\scriptscriptstyle{\vee}}G\rtimes G)\to(G\times
EG)/G;\quad[p,x]\mapsto[p(2\pi),x],$

where $[h,x]=[g^{-1}hg,xg],$ we can see this is well-defined, since

$[p,x]=[g^{-1}pg\gamma,xg]\mapsto[g^{-1}p(2\pi)g\gamma(2\pi),xg]=[p(2\pi),x].$

Furthermore, this is onto, as the projection $P^{\scriptscriptstyle{\vee}}G\to
G$ is onto, and 1–1, for if we consider two elements
$[p,x],\,[q,y]\in(P^{\scriptscriptstyle{\vee}}G\times
EG)/(\Omega^{\scriptscriptstyle{\vee}}G\rtimes G)$ such that
$[p(2\pi),x]=[q(2\pi),y]$ we have $y=xg$ and $q(2\pi)=g^{-1}p(2\pi)g.$ That
is, the paths $q$ and $g^{-1}pg$ have the same endpoint. Therefore, the path
$g^{-1}p^{-1}gq$ is actually a (based) loop. And since
$q=g^{-1}pg(g^{-1}p^{-1}gq),$ we have

$\displaystyle[q,y]$ $\displaystyle=[g^{-1}pg\gamma,xg]$
$\displaystyle=[p,x],$

where $\gamma=g^{-1}p^{-1}gq\in\Omega^{\scriptscriptstyle{\vee}}G.$ Thus we
have a diffeomorphism between $BL^{\scriptscriptstyle{\vee}}G$ and $(G\times
EG)/G$ (or simply $G\times_{G}EG$). Note that this allows us to calculate the
cohomology of $BL^{\scriptscriptstyle{\vee}}G$ as the equivariant cohomology
of $G$ (with its adjoint action). That is,

$H(BL^{\scriptscriptstyle{\vee}}G)=H_{G}(G).$

Given an $L^{\scriptscriptstyle{\vee}}G$-bundle $P\to M$ we can write down the
classifying map of this bundle as follows. Choose a Higgs field, $\Phi,$ for
$P.$ Then define the map $f\colon P\to P^{\scriptscriptstyle{\vee}}G\times EG$
by

$f(q)=(\operatorname{hol}_{\Phi}(q),f_{G}(q)),$

where $\operatorname{hol}_{\Phi}$ is the Higgs field holonomy and $f_{G}$ is
the classifying map for the $G$-bundle associated to $P$ by the projection
$L^{\scriptscriptstyle{\vee}}G\to G$ given by mapping a loop to its
start/endpoint (or equivalently, the projection
$\Omega^{\scriptscriptstyle{\vee}}G\rtimes G\to G$). That is, $f(q)=(p,x)$
where $p^{-1}\partial p=\Phi(q)$ and $x$ is $f_{G}$ applied to the image of
$q$ in $P\times_{L^{\scriptscriptstyle{\vee}}G}G.$ It is easy to see that this
is equivariant with respect to the $L^{\scriptscriptstyle{\vee}}G$ action and
hence descends to a map $M\to BL^{\scriptscriptstyle{\vee}}G$ since if
$(\gamma,g)\in\Omega^{\scriptscriptstyle{\vee}}G\rtimes G$ then

$\displaystyle f(q(g\gamma))$
$\displaystyle=(\operatorname{hol}_{\Phi}(q(g\gamma)),f_{G}(q)g)$

and so $f$ is equivariant in the $EG$ slot (by virtue of the fact that $f_{G}$
is a classifying map) and also in the $P^{\scriptscriptstyle{\vee}}G$ slot
since if $\operatorname{hol}_{\Phi}(q)=p$ then

$\displaystyle\Phi(q(g\gamma))$
$\displaystyle=ad((g\gamma)^{-1})\Phi(q)+(g\gamma)^{-1}\partial(g\gamma)$
$\displaystyle=ad((g\gamma)^{-1})\Phi(q)+\gamma^{-1}\partial\gamma$

and

$\displaystyle(p(\gamma,g))^{-1}\partial(p(\gamma,g))$
$\displaystyle=(g^{-1}pg\gamma)^{-1}\partial(g^{-1}pg\gamma)$
$\displaystyle=\gamma^{-1}g^{-1}p^{-1}g(g^{-1}\partial
pg\gamma+g^{-1}pg\partial\gamma)$
$\displaystyle=ad((g\gamma)^{-1})p^{-1}\partial p+\gamma^{-1}\partial\gamma$

and so
$\operatorname{hol}_{\Phi}(q(g\gamma))=p(\gamma,g)=\operatorname{hol}_{\Phi}(q)(g\gamma).$

####  3.3.2 The universal string class

Now that we have a model for the universal
$L^{\scriptscriptstyle{\vee}}G$-bundle we would like to calculate its string
class according to Theorem 2.4.1. So far everything we have said works on the
topological level. In order to use Theorem 2.4.1 however, the first thing we
need is a connection on $P^{\scriptscriptstyle{\vee}}G\times EG.$ Now,
$P^{\scriptscriptstyle{\vee}}G$ is already a smooth manifold. In order to
define a smooth structure and find a connection on $EG$ we use the results in
[36, 37]. As long as the dimension of the base of the $G$-bundle $P\to M$ is
less than or equal to $n$ this gives a construction of a smooth bundle
$EG_{n}\to BG_{n}$ with connection which is a model for the universal
$G$-bundle. From now on we assume therefore that the dimension of the base of
our $L^{\scriptscriptstyle{\vee}}G$-bundle is fixed (and less than or equal to
$n$ for some $n$).

To define a connection we need to know what a vertical vector looks like.
Consider the vector in $T_{(p,x)}(P^{\scriptscriptstyle{\vee}}G\times
EG_{n})=T_{p}P^{\scriptscriptstyle{\vee}}G\times T_{x}EG_{n}$ generated by the
Lie algebra element
$(\xi,X)\in\Omega^{\scriptscriptstyle{\vee}}{\mathfrak{g}}\rtimes{\mathfrak{g}}:$

$\displaystyle\iota_{(p,x)}(\xi,X)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}((1-tX)p(1+tX)(1+t\xi),xe^{tX})$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(t(-Xp+pX+p\xi),xe^{tX})$
$\displaystyle=(p(X-ad(p^{-1})X+\xi),\iota_{x}(X)).$

Note that the $P^{\scriptscriptstyle{\vee}}G$ part of a vertical vector is a
vector field along $p$ that ends at $p(2\pi)(X-ad(p(2\pi)^{-1})X)$ (since
$\xi$ is a based loop). We will assume that we have a connection in $EG_{n}$
since this is always possible by the discussion above. Call this connection
$a.$ So to find the horizontal part of a vector $(V,W)\in
T_{p}P^{\scriptscriptstyle{\vee}}G\times T_{x}EG_{n}$ we need a vector field
along $p$ that ends at $V(2\pi)-p(2\pi)(X-ad(p(2\pi)^{-1})X)$ (since then
$V-\\{\text{this vector}\\}$ will end at the right point to be vertical).
Consider the vector field

$\left(\frac{\theta}{2\pi}\right)p\left\\{ad(p^{-1})\left(V(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right)\right\\}.$

If we define the horizontal projection of $(V,W),h(V,W),$ to be the vector
field above together with the horizontal component of $W$ (that is,
$hW=W-\iota_{x}(a(W))$), then we have an invariant splitting of the tangent
space at each point in $P^{\scriptscriptstyle{\vee}}G\times EG_{n}.$ This is
easily verified: Since the $EG_{n}$ part has a connection, we need only check
the $P^{\scriptscriptstyle{\vee}}G$ part. First calculate the right action on
the vector above (which we will call $hV$ even though technically the part of
the connection on $P^{\scriptscriptstyle{\vee}}G$ is not actually a connection
itself):

$\left(hV(\gamma,g)\right)_{(g^{-1}pg\gamma,xg)}\\\
=\left(\frac{\theta}{2\pi}\right)g^{-1}p\left\\{ad(p^{-1})\left(V(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right)\right\\}g\gamma.$

Compare this with the horizontal projection of a vector $V^{\prime}$ at
$(p,x)(\gamma,g)=(g^{-1}pg\gamma,xg):$

$hV^{\prime}_{(g^{-1}pg\gamma,xg)}\\\
\phantom{hV^{\prime}_{(g^{-1}pg\gamma}}=\left(\frac{\theta}{2\pi}\right)g^{-1}pg\gamma\left\\{ad(g^{-1}p^{-1}g\gamma)^{-1}\left(V^{\prime}(2\pi)g^{-1}p(2\pi)^{-1}g\right.\right.\\\
\left.\left.-ad(g^{-1}p(2\pi)g)a(W^{\prime})+a(W^{\prime})\right)\right\\}\\\
\phantom{hV^{\prime}_{(g^{-1}pg\gamma}}=\left(\frac{\theta}{2\pi}\right)g^{-1}p\left\\{ad(p^{-1})g\left(V^{\prime}(2\pi)g^{-1}p(2\pi)^{-1}g\right.\right.\\\
\left.\left.-ad(g^{-1})ad(p(2\pi))ad(g)a(W^{\prime})+a(W^{\prime})\right)g^{-1}\right\\}g\gamma\\\
\phantom{hV^{\prime}_{(g^{-1}pg\gamma}}=\left(\frac{\theta}{2\pi}\right)g^{-1}p\left\\{ad(p^{-1})g\left(V^{\prime}(2\pi)g^{-1}p(2\pi)^{-1}g\right.\right.\\\
\left.\left.-ad(g^{-1})ad(p(2\pi))ad(g)ad(g^{-1})a(W)+ad(g^{-1})a(W)\right)g^{-1}\right\\}g\gamma$

(for $W=W^{\prime}g^{-1}$)

$\phantom{hV^{\prime}_{(g^{-1}pg\gamma}}=\left(\frac{\theta}{2\pi}\right)g^{-1}p\left\\{ad(p^{-1})\left(gV^{\prime}(2\pi)g^{-1}p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right)\right\\}g\gamma\\\
\phantom{hV^{\prime}_{(g^{-1}pg\gamma}}=\left(\frac{\theta}{2\pi}\right)g^{-1}p\left\\{ad(p^{-1})\left(V(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right)\right\\}g\gamma\\\
$

(for $V=V^{\prime}(\gamma,g)^{-1},$ so that
$V^{\prime}(2\pi)=g^{-1}V(2\pi)g)$).

So we see that the push forward of the vector $hV$ is horizontal (at
$(g^{-1}pg\gamma,xg)$) and conversely the vector $hV^{\prime}$ is the push
forward of a horizontal vector at $(p,x).$ Thus we have defined a horizontal
splitting of $T_{(p,x)}(P^{\scriptscriptstyle{\vee}}G\times EG_{n})$ for each
$(p,x).$ To find the connection form for this connection we need to recover
the Lie algebra element $(\xi,X)$ from the vector $(V,W).$ We know that the
vector

$v(V,W)\\\
=\left(V-\left(\frac{\theta}{2\pi}\right)p\left\\{ad(p^{-1})\left(V(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right)\right\\},a(W)\right)$

is the vertical component of $(V,W)$ and that the vertical vector generated by
$(\xi,X)\in\Omega^{\scriptscriptstyle{\vee}}{\mathfrak{g}}\rtimes{\mathfrak{g}}$
looks like

$(p(X-ad(p^{-1})X+\xi),\iota_{x}(X)).$

Thus to recover $\xi$ from $v(V,W)$ we just subtract $p(a(W)-ad(p^{-1})a(W))$
and, writing $A$ for the part of the connection on
$P^{\scriptscriptstyle{\vee}}G,$ we have

$A(V,W)=\\\
p^{-1}V-\left(\frac{\theta}{2\pi}\right)ad(p^{-1})\left\\{V(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right\\}\\\
-(a(W)-ad(p^{-1})a(W)).$

Therefore, the connection form $(A,a)$ is given by

$(A,a)=\left(\Theta-\left(\frac{\theta}{2\pi}\right)ad(p^{-1})\left\\{\operatorname{ev}_{2\pi}^{*}\hat{\Theta}-ad(p(2\pi))a+a\right\\}-\left(a-ad(p^{-1})a\right),a\right)$

where $\Theta$ is the Maurer-Cartan form, $\hat{\Theta}$ is the right Maurer-
Cartan form and $\operatorname{ev}_{2\pi}\colon
P^{\scriptscriptstyle{\vee}}G\to G$ is evaluation at the endpoint of a path.
It can be easily checked that this form satisfies the conditions for a
connection. It will be useful later on to write this as a form valued in
$L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}.$ To do this we use the
isomorphism of Lie algebras given in section 3.3.1. The connection form
becomes

$A_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}=\Theta-\left(\frac{\theta}{2\pi}\right)ad(p^{-1})\left\\{\operatorname{ev}_{2\pi}^{*}\hat{\Theta}-ad(p(2\pi))a+a\right\\}+ad(p^{-1})a.$

To calculate the string class we will need the curvature of this connection
and a Higgs field. As usual, the curvature (as an
$L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}$-valued form) is given by the
formula

$F_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}=DA_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}$

where $D$ is the covariant exterior derivative. So we have

$F_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}((V,W),(V^{\prime},W^{\prime}))=\tfrac{1}{2}A_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}([h(V,W),h(V^{\prime},W^{\prime})]).$

Now,

$[h(V,W),h(V^{\prime},W^{\prime})]=([hV,hV^{\prime}],[hW,hW^{\prime}])\\\
=\left(\left[\left(\frac{\theta}{2\pi}\right)p\left\\{ad(p^{-1})\left(V(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W)+a(W)\right)\right\\},\right.\right.\\\
\left(\left.\frac{\theta}{2\pi}\right)p\left\\{ad(p^{-1})\left(V^{\prime}(2\pi)p(2\pi)^{-1}-ad(p(2\pi))a(W^{\prime})+a(W^{\prime})\right)\right\\}\right],\\\
\left.\vphantom{\frac{\theta}{2\pi}}[hW,hW^{\prime}]\right)$

and calculating just the first slot gives

$p\left(\frac{\theta}{2\pi}\right)^{2}ad(p^{-1})\left\\{[V(2\pi)p(2\pi)^{-1},V^{\prime}(2\pi)p(2\pi)^{-1}]\right.\\\
-[V(2\pi)p(2\pi)^{-1},ad(p(2\pi))a(W^{\prime})]+[V(2\pi)p(2\pi)^{-1},a(W^{\prime})]\\\
-[ad(p(2\pi))a(W),V^{\prime}(2\pi)p(2\pi)^{-1}]+ad(p(2\pi))[a(W),a(W^{\prime})]\\\
-[ad(p(2\pi))a(W),a(W^{\prime})]+[a(W),V^{\prime}(2\pi)p(2\pi)^{-1}]\\\
\left.-[a(W),ad(p(2\pi))a(W^{\prime})]+[a(W),a(W^{\prime})]\right\\}.$

This yields

$F_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}=\\\
\left(\left(\frac{\theta}{2\pi}\right)^{2}-\left(\frac{\theta}{2\pi}\right)\right)ad(p^{-1})\left\\{\tfrac{1}{2}[\operatorname{ev}_{2\pi}^{*}\hat{\Theta},\operatorname{ev}_{2\pi}^{*}\hat{\Theta}]-[\operatorname{ev}_{2\pi}^{*}\hat{\Theta},ad(p(2\pi)^{-1})a]+\tfrac{1}{2}[a,a]\right.\\\
\left.+[\operatorname{ev}_{2\pi}^{*}\hat{\Theta},a]-[ad(p(2\pi))a,a]+[a,a]\right\\}-\left(\frac{\theta}{2\pi}\right)ad(p^{-1})(f-ad(p(2\pi))f)+ad(p^{-1})f$

where $f$ is the curvature of $a.$

The other piece of data we need to calculate the string class is a Higgs field
for $EL^{\scriptscriptstyle{\vee}}G.$ Define the map $\Phi\colon
P^{\scriptscriptstyle{\vee}}G\times
EG_{n}\to\Omega^{\scriptscriptstyle{\vee}}{\mathfrak{g}}\rtimes{\mathfrak{g}}$
by

$\Phi(p,x)=(p^{-1}\partial p,0).$

Or, as a map to $L^{\scriptscriptstyle{\vee}}{\mathfrak{g}},$

$\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}(p,x)=p^{-1}\partial p.$

Then by the calculation at the end of section 3.3.1 we see that
$\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}$ is a Higgs field for
$P^{\scriptscriptstyle{\vee}}G\times EG_{n}.$ Next we need to calculate

$\nabla\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}=d\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}+[A_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}},\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}]-\partial
A_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}.$

We have

$\displaystyle d\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}(V,W)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}(pe^{t\xi})$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(e^{-t\xi}p^{-1}\partial(pe^{t\xi}))$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(e^{-t\xi}p^{-1}\partial
pe^{t\xi}+e^{-t\xi}\partial e^{t\xi})$ $\displaystyle=p^{-1}\partial p\xi-\xi
p^{-1}\partial p+\partial\xi,$

for $V=\frac{d}{dt}\big{|}_{0}\,p\exp(t\xi).$ That is,

$d\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}=[\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}},\Theta]+\partial\Theta.$

So

$\nabla\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}=[\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}},\Theta]+\partial\Theta\\\
+\left[\Theta-\left(\frac{\theta}{2\pi}\right)ad(p^{-1})\left\\{\operatorname{ev}_{2\pi}^{*}\hat{\Theta}-ad(p(2\pi))a+a\right\\}+ad(p^{-1})a,\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}\right]\\\
-\partial\left(\Theta-\left(\frac{\theta}{2\pi}\right)ad(p^{-1})\left\\{\operatorname{ev}_{2\pi}^{*}\hat{\Theta}-ad(p(2\pi))a+a\right\\}+ad(p^{-1})a\right)$
$\phantom{\nabla\Phi_{L^{\scriptscriptstyle{\vee}}{\mathfrak{g}}}}=\frac{1}{2\pi}ad(p^{-1})\left\\{\operatorname{ev}_{2\pi}^{*}\hat{\Theta}-ad(p(2\pi))a+a\right\\}.\\\
$

So the string class for $P^{\scriptscriptstyle{\vee}}G\times EG_{n}$ is

$-\frac{1}{4\pi^{2}}\int_{S^{1}}\left\langle\left(\frac{\theta^{2}}{4\pi^{2}}-\frac{\theta}{2\pi}\right)\left(\tfrac{1}{2}[\operatorname{ev}_{2\pi}^{*}\hat{\Theta},\operatorname{ev}_{2\pi}^{*}\hat{\Theta}]-[\operatorname{ev}_{2\pi}^{*}\hat{\Theta},ad(p(2\pi)^{-1})a]\right.\right.\\\
+\tfrac{1}{2}[a,a]\left.+[\operatorname{ev}_{2\pi}^{*}\hat{\Theta},a]-[ad(p(2\pi))a,a]+[a,a]\right)\\\
-\left(\frac{\theta}{2\pi}\right)(f-ad(p(2\pi))f)+f,\left.\frac{1}{2\pi}\left(\operatorname{ev}_{2\pi}^{*}\hat{\Theta}-ad(p(2\pi))a+a\right)\right\rangle$
$=-\frac{1}{8\pi^{2}}\left\langle-\tfrac{1}{3}\left(\tfrac{1}{2}[\hat{\Theta},\hat{\Theta}]-[\hat{\Theta},ad(p(2\pi)^{-1})a]\right.\right.\\\
+\tfrac{3}{2}[a,a]\left.+[\hat{\Theta},a]-[ad(p(2\pi))a,a]\right)\\\
+ad(p(2\pi))f+f,\left.\left(\hat{\Theta}-ad(p(2\pi))a+a\right)\right\rangle.$

##  Chapter 4 String structures for $LG\rtimes S^{1}$-bundles

Thus far we have discussed central extensions of both the loop group (in
chapter 2) and the based loop group (in chapter 3). The loop group $LG$ has a
natural action of the circle given by rotating loops. In this chapter, we
shall consider the more general case where we allow rotations of the loops in
$LG.$ That is, we shall be working with the semi-direct product $LG\rtimes
S^{1}.$ This group arises when we consider a natural generalisation of the
caloron correspondence from section 2.5. There we showed that a $G$-bundle
over $M\times S^{1}$ corresponds to an $LG$-bundle over $M$. If we allow the
base space of the $G$-bundle to be a non-trivial $S^{1}$-bundle (rather than
$M\times S^{1}$) we obtain not an $LG$-bundle but an $LG\rtimes S^{1}$-bundle.
If, further, we consider a non-trivial $S^{1}$ fibre bundle (instead of a
principal bundle), we obtain an $LG\rtimes\operatorname{Diff}(S^{1})$-bundle.

In this chapter then, we will calculate the obstruction to lifting a principal
$LG\rtimes S^{1}$-bundle $P$ to a principal $\widehat{LG\rtimes S^{1}}$-bundle
$\widehat{P}.$ In section 4.2 we will construct a correspondence for
$LG\rtimes S^{1}$-bundles in analogy with the caloron correspondence from
chapter 2. This will be used to prove a theorem which extends Theorem 2.5.3
relating the string class and the first Pontrjagyn class. In section 4.3 we
shall consider the lifting problem for the more general case where we allow
general (orientation preserving) diffeomorphisms of the loops in $LG,$ that
is, principal bundles with structure group
$LG\rtimes\operatorname{Diff}(S^{1}).$

###  4.1 The string class of an $LG\rtimes S^{1}$-bundle

In this section we present a formula for the obstruction to lifting a
principal $LG\rtimes S^{1}$-bundle $P$ to a principal $\widehat{LG\rtimes
S^{1}}$-bundle $\widehat{P},$ which we call the _string class_ of $P.$ We
shall follow the methods of [35], outlined in section 2.4. In section 4.1.2 we
will give another method for calculating the 3-curvature of a lifting bundle
gerbe, first presented in [18], and apply this to the problem of the string
class of an $LG\rtimes S^{1}$-bundle.

####  4.1.1 The string class via lifting bundle gerbes

Let $LG\rtimes S^{1}$ be the semi-direct product, whose multiplication is
given by

$(\gamma_{1},\phi_{1})(\gamma_{2},\phi_{2})=(\gamma_{1}\rho_{\phi_{1}}(\gamma_{2}),\phi_{1}+\phi_{2}),$

where $\rho_{\phi}(\gamma)(\theta)=\gamma(\theta-\phi).$ For convenience, let
us record some facts about the Lie algebra of $LG\rtimes S^{1}$ here. The
bracket on the Lie algebra $L{\mathfrak{g}}\rtimes i{\mathbb{R}}$ is given by

$[(\xi,x),(\zeta,y)]=([\xi,\zeta]-x\partial\zeta+y\partial\xi,0)$

and the adjoint action of $LG\rtimes S^{1}$ on $L{\mathfrak{g}}\rtimes
i{\mathbb{R}}$ is

$ad(\gamma,\phi)(\xi,x)=\left(ad(\gamma)\rho_{\phi}(\xi)+x\,\partial\gamma\gamma^{-1},x\right).$

##### The central extension of $LG\rtimes S^{1}$

Recall from section 2.4 that in order to perform calculations involving the
lifting bundle gerbe, we needed an explicit construction of the central
extension of $LG.$ This was given following the construction in section 2.3 in
terms of a pair of differential forms satisfying a certain compatibility
condition. Namely, a pair $(R,\alpha),$ where $R$ is a closed, integral 2-form
on $LG$ and $\alpha$ is a 1-form on $LG\times LG,$ satisfying the conditions
$\delta R=d\alpha$ and $\delta\alpha=0.$ In a similar manner, for what follows
we will require an explicit construction of the central extension of
$LG\rtimes S^{1}.$ Note, however, that the construction in section 2.3 only
works for ${\mathcal{G}}$ a simply connected Lie group. This is because in
order to construct the extension given the pair $(R,\alpha)$ we used the fact
that a flat bundle over a simply connected base has a section satisfying
certain conditions. This allowed us to find a $U(1)$-bundle $P$ over
${\mathcal{G}}$ such that $\delta P\to{\mathcal{G}}\times{\mathcal{G}}$ was
trivial and had a section which defined the multiplication on the central
extension.111See the discussion in section 2.3.1. However, even though the
semi-direct product $LG\rtimes S^{1}$ is not simply connected we can modify
the construction from section 2.3 slightly to cover this case [35]. This
involves replacing the 2-form $R$ with a differential character [12] for the
bundle $\widehat{{\mathcal{G}}}\to{\mathcal{G}}$. That is, we add to our pair
$(R,\alpha)$ a homomorphism $h\colon Z_{1}({\mathcal{G}})\to U(1)$ satisfying

$h(\partial\sigma)=\exp\left(\int_{\sigma}R\right)$

for every two-cycle $\sigma$ in ${\mathcal{G}}.$ We also require the
compatibility condition

$(\delta h)(\gamma)=\exp\left(\int_{\gamma}\alpha\right)$

for every closed one-cycle $\gamma$ in ${\mathcal{G}}\times{\mathcal{G}}.$

Therefore, we need to find a triple of objects $(R,\alpha,h)$ as above. Note
first that

$H^{2}(LG\rtimes S^{1})\simeq H^{2}(LG).$

To see this, we observe that as $LG\rtimes S^{1}=LG\times S^{1}$ as a space,
the Künneth formula (see [2]) gives

$H^{2}(LG\rtimes S^{1})\simeq H^{2}(LG)\otimes H^{0}(S^{1})\oplus
H^{1}(LG)\otimes H^{1}(S^{1}),$

since $H^{2}(S^{1})=0.$ Now, $H^{0}(S^{1})\simeq
H^{1}(S^{1})\simeq{\mathbb{R}},$ so we have

$H^{2}(LG\rtimes S^{1})\simeq H^{2}(LG)\oplus H^{1}(LG).$

Recall, however, that $LG\simeq\Omega G\times G$ as a space, and so
$\pi_{1}(LG)=\pi_{2}(G)\times\pi_{1}(G).$ Therefore, as $G$ is simply
connected, so is $LG,$ and thus $H_{1}(LG,{\mathbb{R}})=0$ by the Hurewicz
Theorem (see for example [20]). Therefore, by the Universal Coefficient
Theorem (see for example [27]) $H^{1}(LG)=0,$ and so $H^{2}(LG\rtimes
S^{1})\simeq H^{2}(LG).$ Thus, we take as the 2-form $R,$ the pull-back of the
form from section 2.4 to $LG\rtimes S^{1}.$ That is,

$R=\frac{i}{4\pi}\int_{S^{1}}\langle\Theta,\partial\Theta\rangle\,d\theta.$

Note that since we are integrating over the circle, this expression is
unchanged when each term is rotated by a fixed angle. That is,

$\frac{i}{4\pi}\int_{S^{1}}\langle\rho_{\phi}(\Theta),\partial\rho_{\phi}(\Theta)\rangle\,d\theta=\frac{i}{4\pi}\int_{S^{1}}\langle\Theta,\partial\Theta\rangle\,d\theta$

Now, to find $\alpha$ we need to calculate $\delta
R=\pi_{1}^{*}R-m^{*}R+\pi_{2}^{*}R,$ where as before, $\pi_{i}$ is the
projection $LG\rtimes S^{1}\times LG\rtimes S^{1}\to LG\rtimes S^{1}$ which
omits the $i^{\text{th}}$ factor and $m$ is the multiplication defined above.
As in chapter 2, $\pi_{i}^{*}R$ is given by

$\frac{i}{4\pi}\int_{S^{1}}\langle\pi_{i}^{*}\Theta,\partial\pi_{i}^{*}\Theta\rangle\,d\theta$

and so it remains to calculate $m^{*}R.$ For this, note that a tangent vector
to $LG\rtimes S^{1}$ at the point $(\gamma,\phi)$ can be written as
$(\gamma,\phi)(\xi,x)=(\gamma\rho_{\phi}(\xi),x_{\phi})$ for some $(\xi,x)\in
L{\mathfrak{g}}\rtimes i{\mathbb{R}}$ by using the left multiplication to
transport elements of the Lie algebra to the point $(\gamma,\phi).$ Therefore,
we can calculate $m^{*}R$ by noting that

$m^{*}R((\gamma_{1}\rho_{\phi_{1}}(\xi_{1}),x_{1\phi_{1}}),(\gamma_{2}\rho_{\phi_{2}}(\xi_{2}),x_{2\phi_{2}}))=R(m_{*}((\gamma_{1}\rho_{\phi_{1}}(\xi_{1}),x_{1\phi_{1}}),(\gamma_{2}\rho_{\phi_{2}}(\xi_{2}),x_{2\phi_{2}})))$

and calculating the push-forward of $m.$ We have

$m_{*}((\gamma_{1}\rho_{\phi_{1}}(\xi_{1}),x_{1\phi_{1}}),(\gamma_{2}\rho_{\phi_{2}}(\xi_{2}),x_{2\phi_{2}}))\\\
=\frac{d}{dt}\bigg{|}_{0}\left(\gamma_{1}(1+t\xi_{1}^{\rho_{1}})\rho_{(\phi_{1}+tx_{1})}(\gamma_{2})\rho_{(\phi_{1}+tx_{1})}(1+t\xi_{2}^{\rho_{2}})),\phi_{1}+\phi_{2}+t(x_{1}+x_{2})\right),$

where we have written (for example) $\xi_{1}^{\rho_{1}}$ for
$\rho_{\phi_{1}}(\xi_{1})$. As the multiplication on the $S^{1}$ factor is not
twisted, the second slot above will give $x_{1}+x_{2}.$ Thus it suffices to
calculate the first slot only. Using the fact that

$\frac{d}{dt}\bigg{|}_{0}\rho_{(\phi_{1}+tx_{1})}(\gamma_{2})=-x_{1}\rho_{\phi_{1}}(\partial\gamma_{2}),$

we have

$\displaystyle\frac{d}{dt}\bigg{|}_{0}$
$\displaystyle\left(\gamma_{1}(1+t\xi_{1}^{\rho_{1}})\rho_{(\phi_{1}+tx_{1})}(\gamma_{2})\rho_{(\phi_{1}+tx_{1})}(1+t\xi_{2}^{\rho_{2}})\right)$
$\displaystyle=\gamma_{1}\xi_{1}^{\rho_{1}}\gamma_{2}^{\rho_{1}}+\gamma_{1}\gamma_{2}^{\rho_{1}}\xi_{2}^{\rho_{2}}-x_{1}\gamma_{1}\partial\gamma_{2}^{\rho_{1}}$
$\displaystyle=\gamma_{1}\gamma_{2}^{\rho_{1}}\rho_{(\phi_{1}+\phi_{2})}\left((\gamma_{2}^{-1}\xi_{1}\gamma_{2}^{\vphantom{-1}})^{\rho_{2}^{-1}}+\xi_{2}-x_{1}(\gamma_{2}^{-1}\partial\gamma_{2}^{\vphantom{-1}})^{\rho_{2}^{-1}}\right).$

Therefore, $m^{*}R$ evaluated on the pairs of tangent vectors
$\left((\gamma_{1},\phi_{1})(\xi_{1},x_{1}),(\gamma_{2},\phi_{2})(\xi_{2},x_{2})\right)$
and
$\left((\gamma_{1},\phi_{1})(\zeta_{1},y_{1}),(\gamma_{2},\phi_{2})(\zeta_{2},y_{2})\right)$
is given by

$\frac{i}{4\pi}\int_{S^{1}}\left\langle(ad(\gamma_{2}^{-1})\xi_{1})^{\rho_{2}^{-1}}+\xi_{2}-x_{1}(\gamma_{2}^{-1}\partial\gamma_{2}^{\vphantom{-1}})^{\rho_{2}^{-1}},\partial\left((ad(\gamma_{2}^{-1})\zeta_{1})^{\rho_{2}^{-1}}+\zeta_{2}-y_{1}(\gamma_{2}^{-1}\partial\gamma_{2}^{\vphantom{-1}})^{\rho_{2}^{-1}}\right)\right\rangle\,d\theta,$

where we have used the fact that the integral is unchanged by rotating
everything by $\rho_{(\phi_{1}+\phi_{2})}^{-1}.$ Expanding this, we have

$\frac{i}{4\pi}\int_{S^{1}}\left\langle(ad(\gamma_{2}^{-1})\xi_{1}),\partial(ad(\gamma_{2}^{-1})\zeta_{1})\right\rangle+\left\langle\xi_{2},\partial\zeta_{2}\right\rangle\\\
+x_{1}y_{1}\left\langle(ad(\gamma_{2}^{-1})Z_{2}),\partial(ad(\gamma_{2}^{-1})Z_{2})\right\rangle\\\
+\left\langle(ad(\gamma_{2}^{-1})\xi_{1}),\partial\zeta_{2}^{\rho_{2}}\right\rangle+\left\langle\xi_{2}^{\rho_{2}},\partial(ad(\gamma_{2}^{-1})\zeta_{1})\right\rangle\\\
-y_{1}\left\langle\xi_{2}^{\rho_{2}},\partial(ad(\gamma_{2}^{-1})Z_{2})\right\rangle-
x_{1}\left\langle(ad(\gamma_{2}^{-1})Z_{2}),\partial\zeta_{2}^{\rho_{2}}\right\rangle\\\
-y_{1}\left\langle(ad(\gamma_{2}^{-1})\xi_{1}),\partial(ad(\gamma_{2}^{-1})Z_{2})\right\rangle\\\
-x_{1}\left\langle(ad(\gamma_{2}^{-1})Z_{2}),\partial(ad(\gamma_{2}^{-1})\zeta_{2})\right\rangle
d\theta,$

where as before $Z$ is the function $\gamma\mapsto\partial\gamma\gamma^{-1}$
and, again, we have used the rotation invariance of the integral. Using the
$ad$-invariance of the Killing form and integration by parts, along with the
identity from section 2.4,

$\partial\left(ad(\gamma^{-1})X\right)=ad(\gamma^{-1})[X,Z]+ad(\gamma^{-1})\partial
X$

for a vector $X\in L{\mathfrak{g}},$ this simplifies to

$\frac{i}{4\pi}\int_{S^{1}}\left\langle[\xi_{1},\zeta_{1}],Z_{2}\right\rangle+\left\langle\xi_{1},\partial\zeta_{1}\right\rangle+\left\langle\xi_{2},\partial\zeta_{2}\right\rangle\\\
+\left\langle
ad(\gamma_{2}^{-1})\xi_{1},\partial\zeta_{2}^{\rho_{2}}\right\rangle-\left\langle\partial\xi_{2}^{\rho_{2}},ad(\gamma_{2}^{-1})\zeta_{1}\right\rangle-
x_{1}\left<Z_{2},\partial\zeta_{1}\right\rangle+y_{1}\left\langle\partial\xi_{1},Z_{2}\right\rangle\\\
-x_{1}\left\langle
ad(\gamma_{2}^{-1})Z_{2},\partial\zeta_{2}^{\rho_{2}}\right\rangle+y_{1}\left\langle\partial\xi_{2}^{\rho_{2}},ad(\gamma_{2}^{-1})Z_{2}\right\rangle
d\theta,$

or simply

$m^{*}R=\frac{i}{4\pi}\int_{S^{1}}\left\langle[\Theta_{1},\Theta_{1}],Z_{2}\right\rangle+\left\langle\Theta_{1},\partial\Theta_{1}\right\rangle+\left\langle\Theta_{2},\partial\Theta_{2}\right\rangle\\\
+2\left\langle
ad(\gamma_{2}^{-1})\Theta_{1},\partial\Theta_{2}^{\rho_{2}}\right\rangle-2\left\langle\mu_{1}ad(\gamma_{2}^{-1})Z_{2},\partial\Theta_{2}^{\rho_{2}}\right\rangle-2\left\langle\mu_{1}Z_{2},\partial\Theta_{1}\right\rangle
d\theta,$

where $\mu$ represents the Maurer-Cartan form on $S^{1}$. Therefore, we have

$\delta
R=\frac{i}{2\pi}\int_{S^{1}}-\tfrac{1}{2}\left\langle[\Theta_{1},\Theta_{1}],Z_{2}\right\rangle-\left\langle
ad(\gamma_{2}^{-1})\Theta_{1},\partial\Theta_{2}^{\rho_{2}}\right\rangle\\\
+\left\langle\mu_{1}ad(\gamma_{2}^{-1})Z_{2},\partial\Theta_{2}^{\rho_{2}}\right\rangle+\left\langle\mu_{1}Z_{2},\partial\Theta_{1}\right\rangle
d\theta.$

Recall from section 2.4 that for the loop group case, the form $\alpha$ such
that $d\alpha=\delta R$ is given by

$\alpha=\frac{i}{2\pi}\int_{S^{1}}\left\langle\Theta_{1},Z_{2}\right\rangle
d\theta.$

When evaluated on the vector $(\gamma_{1}\xi_{1},\gamma_{2}\xi_{2})$ tangent
to the point $(\gamma_{1},\gamma_{2})\in LG\times LG,$ $\alpha$ is given by

$\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi_{1},\partial\gamma_{2}^{\vphantom{-1}}\gamma_{2}^{-1}\right\rangle
d\theta.$

Consider the generalisation of this form to $LG\rtimes S^{1}\times LG\rtimes
S^{1}.$ That is, define $\alpha_{1}$ as

$\alpha_{1}(\gamma_{1}\xi^{\rho_{1}},x_{1\phi_{1}},\gamma_{2}\xi^{\rho_{2}},x_{2\phi_{2}})=\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi_{1},\partial\gamma_{2}^{\vphantom{-1}}\gamma_{2}^{-1}\right\rangle,$

or

$\alpha_{1}=\frac{i}{2\pi}\int_{S^{1}}\big{\langle}\Theta_{1}^{\rho_{1}^{-1}},Z_{2}\big{\rangle}\,d\theta.$

We can calculate the derivative of this form via

$d\alpha_{1}(X,Y)=\tfrac{1}{2}\left\\{X(\alpha_{1}(Y))-Y(\alpha_{1}(X))-\alpha_{1}([X,Y])\right\\},$

for tangent vectors $X$ and $Y.$ Thus we need to calculate

$\displaystyle(\gamma_{1}\xi_{1}^{\rho_{1}},$ $\displaystyle
x_{1\phi_{1}},\gamma_{2}\xi_{2}^{\rho_{2}},x_{2\phi_{2}})\left(\alpha_{1}(\gamma_{1}\zeta_{1}^{\rho_{1}},y_{1\phi_{1}},\gamma_{2}\zeta_{2}^{\rho_{2}},y_{2\phi_{2}})\right)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}\frac{i}{2\pi}\int_{S^{1}}\left\langle\zeta_{1},\partial(\gamma_{2}(1+t\xi_{2}^{\rho_{2}}))(1-t\xi_{2}^{\rho_{2}})\gamma_{2}^{-1}\right\rangle
d\theta$
$\displaystyle=\frac{i}{2\pi}\int_{S^{1}}\left\langle\zeta_{1},ad(\gamma_{2})\partial\xi_{2}^{\rho_{2}}\right\rangle
d\theta,$

and

$\displaystyle\alpha_{1}([(\gamma_{1}\xi_{1}^{\rho_{1}},x_{1\phi_{1}}),$
$\displaystyle(\gamma_{1}\zeta_{1}^{\rho_{1}},y_{1\phi_{1}})],[(\gamma_{2}\xi_{2}^{\rho_{2}},x_{2\phi_{2}}),(\gamma_{2}\zeta_{2}^{\rho_{2}},y_{2\phi_{2}})])$
$\displaystyle=\frac{i}{2\pi}\int_{S^{1}}\left\langle[(\xi_{1},x_{1}),(\zeta_{1},y_{1})],\partial\gamma_{2}^{\vphantom{-1}}\gamma_{2}^{-1}\right\rangle
d\theta$
$\displaystyle=\frac{i}{2\pi}\int_{S^{1}}\left\langle[\xi_{1},\zeta_{1}],\partial\gamma_{2}^{\vphantom{-1}}\gamma_{2}^{-1}\right\rangle-\left\langle
x_{1}\partial\zeta_{1}-y_{1}\partial\xi_{1},\partial\gamma_{2}^{\vphantom{-1}}\gamma_{2}^{-1}\right\rangle
d\theta.$

Therefore, we have

$d\alpha_{1}=\frac{1}{2\pi}\int_{S^{1}}-\tfrac{1}{2}\left\langle[\Theta,\Theta],Z_{2}^{\vphantom{-1}}\right\rangle-\left\langle
ad(\gamma_{2}^{-1})\Theta_{1},\partial\Theta_{2}^{\rho_{2}}\right\rangle+\left\langle\mu_{1}Z_{2},\partial\Theta_{1}^{\vphantom{-1}}\right\rangle
d\theta.$

Note that $\delta R$ does not equal $d\alpha_{1}.$ However,

$\delta
R-d\alpha_{1}=\frac{i}{2\pi}\int_{S^{1}}\left\langle\mu_{1}ad(\gamma_{2}^{-1})Z_{2},\partial\Theta_{2}^{\rho_{2}}\right\rangle
d\theta.$

Using the identity

$ad(\gamma)\partial\Theta^{\rho}=dZ,$

we see that

$\delta
R-d\alpha_{1}=\frac{i}{2\pi}\int_{S^{1}}\left\langle\mu_{1}Z_{2},dZ_{2}\right\rangle
d\theta.$

Now, if we define

$\alpha_{2}=-\frac{i}{4\pi}\int_{S^{1}}\left\langle\mu_{1}Z_{2},Z_{2}\right\rangle
d\theta,$

then

$\displaystyle d\alpha_{2}$
$\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle\mu_{1}dZ_{2},Z_{2}\right\rangle+\left\langle\mu_{1}Z_{2},dZ_{2}\right\rangle
d\theta$
$\displaystyle=\frac{i}{2\pi}\int_{S^{1}}\left\langle\mu_{1}Z_{2},dZ_{2}\right\rangle
d\theta$ $\displaystyle=\delta R-d\alpha_{1}.$

Thus $\alpha$ is given by

$\alpha=\frac{i}{2\pi}\int_{S^{1}}\big{\langle}\pi_{2}^{*}\Theta^{\rho^{-1}}\\!-\tfrac{1}{2}\pi_{2}^{*}\mu\,\pi_{1}^{*}Z,\pi_{1}^{*}Z\big{\rangle}\,d\theta,$

and $\delta R=d\alpha.$ One can also easily check that $\delta\alpha=0.$
Notice that the $2$-form $R$ is left invariant and the $1$-form $\alpha$ is
left invariant in the first slot. To find the homomorphism $h\colon
Z_{1}(LG\rtimes S^{1})\to U(1)$ we note that since $\pi_{1}(LG\rtimes
S^{1})={\mathbb{Z}}$ any cycle $a\in Z_{1}(LG\rtimes S^{1})$ can be written as
$n\gamma+\partial\sigma,$ for some two-cycle $\sigma,$ where $\gamma$ is the
generator of $H_{1}(LG\rtimes S^{1}),$ a loop around the $S^{1}$ factor. It is
easy to see that the integral of $\alpha$ over the generators of
$H_{1}(LG\rtimes S^{1}\times LG\rtimes S^{1})$ vanishes, that is,

$\int_{\gamma_{1}}\alpha=0=\int_{\gamma_{2}}\alpha$

for $\gamma_{1},\gamma_{2}$ loops around the first and second $S^{1}$ factors
respectively. This suggests that we define

$h(a)=h(\partial\sigma)=\exp\left(\int_{\sigma}R\right).$

This is well defined since if
$a=n\gamma+\partial\sigma=n\gamma+\partial\sigma^{\prime}$ then
$\partial(\sigma-\sigma^{\prime})=0$ and so
$\displaystyle\int_{\sigma-\sigma^{\prime}}R\in 2\pi i{\mathbb{Z}}$ (since $R$
is integral). Because the integral of $\alpha$ over the generators of
$H_{1}(LG\rtimes S^{1}\times LG\rtimes S^{1})$ vanishes, it is easy to check
that for any one-cycle $\gamma$ we have

$(\delta h)(\gamma)=\exp\left(\int_{\gamma}\alpha\right).$

Thus we have proven

######  Proposition 4.1.1.

The triple $(R,\alpha,h)$ as above determines a central extension of the semi-
direct product $LG\rtimes S^{1}.$

##### A connection for the lifting bundle gerbe

Now that we have a construction of the central extension of $LG\rtimes S^{1},$
the next step is to write down a bundle gerbe connection for the lifting
bundle gerbe. Recall from section 2.4 that if $P$ is an $LG\rtimes
S^{1}$-bundle and $\nu$ is a connection on the central extension
$\widehat{LG\rtimes S^{1}}$ thought of as a bundle over $LG\rtimes S^{1}$ then
a bundle gerbe connection is given by $\tau^{*}\nu-\epsilon,$ where $\epsilon$
is some $1$-form on $P^{[2]}$ satisfying $\delta\epsilon=\tau^{*}\alpha.$ In
the $LG$ case, this form was given by

$\epsilon=\frac{i}{2\pi}\int_{S^{1}}\langle\pi_{2}^{*}A,\tau^{*}Z\rangle\,d\theta,$

where $A$ is a connection on $P.$ As mentioned in section 2.4, it is possible
to write $\epsilon$ in general in terms of $\alpha$ [43]. We shall now
demonstrate how to do this. Let $P$ be a ${\mathcal{G}}$-bundle with
connection $A.$ Recall that $A$ satisfies

$\pi_{1}^{*}A=ad(\tau_{12}^{-1})\pi_{2}^{*}A+\tau_{12}^{*}\Theta.$

For tangent vectors $(X_{1},X_{2},X_{3})$ at $(p_{1},p_{2},p_{3})\in P^{[3]},$
we can calculate

$(\delta\alpha)_{(1,\tau_{12},\tau_{23})}(A(X_{1}),\tau_{12}(X_{1},X_{2}),\tau_{23}(X_{2},X_{3}))=\\\
\phantom{(\delta\alpha)}\alpha_{(\tau_{12},\tau_{23})}(\tau_{12}(X_{1},X_{2}),\tau_{23}(X_{2},X_{3}))\\\
-\alpha_{(\tau_{12},\tau_{23})}(m_{*}(A(X_{1}),\tau_{12}(X_{1},X_{2})),\tau_{23}(X_{2},X_{3}))\\\
+\alpha_{(1,\tau_{12}\tau_{23})}(A(X_{1}),m_{*}(\tau_{12}(X_{1},X_{2}),\tau_{23}(X_{2},X_{3})))\\\
-\alpha_{(1,\tau_{12})}(A(X_{1}),\tau_{12}(X_{1},X_{2})).$

Notice that the first term above is actually $\tau^{*}\alpha.$ Since
$\delta\alpha=0,$ we have

$(\tau^{*}\alpha)_{(p_{1},p_{2},p_{3})}(X_{1},X_{2},X_{3})=\\\
\alpha_{(\tau_{12},\tau_{23})}(m_{*}(A(X_{1}),\tau_{12}(X_{1},X_{2})),\tau_{23}(X_{2},X_{3}))\\\
-\alpha_{(1,\tau_{12}\tau_{23})}(A(X_{1}),m_{*}(\tau_{12}(X_{1},X_{2}),\tau_{23}(X_{2},X_{3})))\\\
+\alpha_{(1,\tau_{12})}(A(X_{1}),\tau_{12}(X_{1},X_{2})).$

Now, if we define $\epsilon$ in terms of $\alpha$ and $A$ as

$\epsilon_{(p_{1},p_{2})}(X_{1},X_{2})=\alpha_{(1,\tau_{12})}(A(X_{1}),\tau_{12}(X_{1},X_{2}))$

then we have

$(\delta\epsilon)_{(p_{1},p_{2},p_{3})}(X_{1},X_{2},X_{3})=\\\
\alpha_{(1,\tau_{23})}(A(X_{2}),\tau_{23}(X_{2},X_{3}))-\alpha_{(1,\tau_{13})}(A(X_{1}),\tau_{13}(X_{1},X_{3}))\\\
+\alpha_{(1,\tau_{12})}(A(X_{1}),\tau_{12}(X_{1},X_{2})).$

Using the fact that $\tau_{13}=\tau_{12}\tau_{23},$ we see

$\alpha_{(1,\tau_{13})}(A(X_{1}),\tau_{13}(X_{1},X_{2}))=\alpha_{(1,\tau_{12}\tau_{23})}(A(X_{1}),m_{*}(\tau_{12}(X_{1},X_{2}),\tau_{23}(X_{2},X_{3})))$

and since $\alpha$ is left invariant in the first slot, and using the equation
above relating $A(X_{1})$ and $A(X_{2}),$ we have

$\displaystyle\alpha_{(1,\tau_{23})}($ $\displaystyle
A(X_{2}),\tau_{23}(X_{2},X_{3}))$
$\displaystyle=\alpha_{(\tau_{12},\tau_{23})}(\tau_{12}A(X_{2}),\tau_{23}(X_{2},X_{3}))$
$\displaystyle=\alpha_{(\tau_{12},\tau_{23})}(\tau_{12}(ad(\tau_{12}^{-1})A(X_{1})+\tau_{12}^{-1}(\tau_{12}(X_{1},X_{2}))),\tau_{23}(X_{2},X_{3}))$
$\displaystyle=\alpha_{(\tau_{12},\tau_{23})}(\tau_{12}ad(\tau_{12}^{-1})A(X_{1})+\tau_{12}(X_{1},X_{2}),\tau_{23}(X_{2},X_{3})),$

which equals

$\alpha_{(\tau_{12},\tau_{23})}(m_{*}(A(X_{1}),\tau_{12}(X_{1},X_{2})),\tau_{23}(X_{2},X_{3})).$

Thus we have $\delta\epsilon=\tau^{*}\alpha.$

Consider now the $LG\rtimes S^{1}$-bundle $P.$ Choose a connection $(A,a)$ for
$P,$ where $A$ and $a$ are $1$-forms on $P$ with values in $L{\mathfrak{g}}$
and $i{\mathbb{R}}$ respectively. Note that $a$ is a connection for the
associated $S^{1}$-bundle $P/LG$ whereas $A$ is not a connection form. In
fact, if $X$ is a tangent vector to $P,$ we have

$\displaystyle(A,a)(X(\gamma,\phi))$
$\displaystyle=ad(\gamma,\phi)^{-1}(A(X),a(X))$
$\displaystyle=\left(\rho_{\phi^{-1}}\left(ad(\gamma^{-1})A(X)-a(X)\gamma^{-1}\partial\gamma\right),a(X)\right),$

and so $A$ does not have the correct transformation properties to be a
connection.222Notice the similarity with the treatment of connections for the
universal $\Omega G\rtimes G$-bundle in section 3.3. Given $(A,a)$ then, we
can write down the $1$-form $\epsilon\in\Omega^{1}(P^{[2]})$ as above:

$\epsilon=\frac{i}{2\pi}\int_{S^{1}}\left\langle\pi_{2}^{*}A-\tfrac{1}{2}\pi_{2}^{*}a\,\tau^{*}Z,\tau^{*}Z\right\rangle
d\theta.$

It is easy to check that $\delta\epsilon=\tau^{*}\alpha$ and so we have that
$\tau^{*}\nu-\epsilon$ is a connection for the lifting bundle gerbe. Of
course, as in section 2.4 we are concerned with finding a curving for this
bundle gerbe and so we are really interested in calculating the curvature of
this connection, given by $\tau^{*}R-d\epsilon.$ Recall that for a connection
$A$ on a ${\mathcal{G}}$-bundle, we have the formula

$\pi_{1}^{*}A=ad(\tau^{-1})\pi_{2}^{*}A+\tau^{*}\Theta.$

In the case where ${\mathcal{G}}=LG\rtimes S^{1},$ the formula relating
$\pi_{1}^{*}(A,a)=(A_{2},a_{2})$ and $\pi_{2}^{*}(A,a)=(A_{1},a_{1})$ is

$(A_{2},a_{2})=\left(\rho_{\tau_{S^{1}}}^{-1}\left(ad(\tau_{LG}^{-1})A_{1}-a_{1}\tau_{LG}^{-1}\partial\tau_{LG}^{\vphantom{-1}}\right)+\tau_{LG}^{*}(\rho_{\tau_{S^{1}}}^{-1}(\Theta)),a_{1}+\tau_{S^{1}}^{*}\mu\right)$

where we have written the difference map $\tau$ as $(\tau_{LG},\tau_{S^{1}}).$
That is, $\tau_{LG}$ is the $LG$ part of $\tau$ and $\tau_{S^{1}}$ is the
circle part. From now on, we will simply write $\tau$ and assume that it is
clear from the context which part we mean. In particular, then, we have

$\tau^{*}\rho_{\tau}^{-1}(\Theta)=A_{2}-\rho_{\tau}^{-1}\left(ad(\tau^{-1})A_{1}+a_{1}\tau^{-1}\partial\tau\right).$

Note that here we have used the fact that the Maurer-Cartan form on $LG\rtimes
S^{1}$ is not the pair $(\Theta,\mu)$ but in fact includes a rotation of
$\Theta.$ So at the point $(\gamma,\phi),$ it is given by
$(\rho_{\phi^{-1}}(\Theta),\mu).$

We can use this to calculate $\tau^{*}R-d\epsilon.$ Writing $A^{\rho}$ for
$\rho(A)$ and so on, as before, we have

$\tau^{*}R=\frac{i}{4\pi}\int_{S^{1}}\langle\tau^{*}\Theta,\partial\tau^{*}\Theta\rangle
d\theta\\\ \phantom{\tau^{*}R}=\frac{i}{4\pi}\int_{S^{1}}\langle
A_{2}^{\rho}-ad(\tau^{-1})A_{1}+a_{1}\tau^{-1}\partial\tau,\partial(A_{2}^{\rho}-ad(\tau^{-1})A_{1}+a_{1}\tau^{-1}\partial\tau)\rangle
d\theta\\\ \phantom{\tau^{*}R}=\frac{i}{4\pi}\int_{S^{1}}\langle
A_{2},\partial A_{2}\rangle-2\langle
A_{2}^{\rho},\partial(ad(\tau^{-1})A_{1})\rangle+2\langle
A_{2}^{\rho},a_{1}\partial(\tau^{-1}\partial\tau)\rangle\\\ +\langle
ad(\tau^{-1})A_{1},\partial(ad(\tau^{-1})A_{1})\rangle-2\langle
ad(\tau^{-1})A_{1},a_{1}\partial(\tau^{-1}\partial\tau)\rangle\\\ +\langle
a_{1}\tau^{-1}\partial\tau,a_{1}\partial(\tau^{-1}\partial\tau)\rangle
d\theta\\\ \phantom{\tau^{*}R}=\frac{i}{4\pi}\int_{S^{1}}\langle
A_{2},\partial A_{2}\rangle-2\langle
A_{2}^{\rho},\partial(ad(\tau^{-1})A_{1})\rangle+2\langle
A_{2}^{\rho},a_{1}\partial(\tau^{-1}\partial\tau)\rangle\\\ +\langle
ad(\tau^{-1})A_{1},\partial(ad(\tau^{-1})A_{1})\rangle-2\langle
ad(\tau^{-1})A_{1},a_{1}\partial(\tau^{-1}\partial\tau)\rangle d\theta.$

The last term vanishes since $a_{1}\wedge a_{1}=0.$ For $d\epsilon$ we have:

$d\epsilon=\frac{i}{2\pi}d\int_{S^{1}}\langle
A_{1}-\tfrac{1}{2}a_{1}\tau^{*}Z,\tau^{*}Z\rangle d\theta\\\
\phantom{d\epsilon}=\frac{i}{2\pi}\int_{S^{1}}\langle
dA_{1},\tau^{*}Z\rangle-\langle A_{1},d(\tau^{*}Z)\rangle-\tfrac{1}{2}\langle
da_{1}\tau^{*}Z,\tau^{*}Z\rangle+\langle a_{1}\tau^{*}Z,d(\tau^{*}Z)\rangle
d\theta\\\ $

and using the fact that
$d(\tau^{*}Z)=ad(\tau)\partial(\tau^{*}\Theta^{\rho}),$

$\phantom{d\epsilon}=\frac{i}{2\pi}\int_{S^{1}}\langle
dA_{1},\tau^{*}Z\rangle-\langle
A_{1},ad(\tau)\partial(\tau^{*}\Theta^{\rho})\rangle\\\ -\tfrac{1}{2}\langle
da_{1}\tau^{*}Z,\tau^{*}Z\rangle+\langle
a_{1}\tau^{*}Z,ad(\tau)\partial(\tau^{*}\Theta^{\rho})\rangle d\theta\\\
\phantom{d\epsilon}=\frac{i}{2\pi}\int_{S^{1}}\langle
dA_{1},\tau^{*}Z\rangle\\\ -\langle
A_{1},ad(\tau)\partial(A_{2}^{\rho}-ad(\tau^{-1})A_{1}+a_{1}\tau^{-1}\partial\tau)\rangle-\tfrac{1}{2}\langle
da_{1}\tau^{*}Z,\tau^{*}Z\rangle\\\ +\langle
a_{1}\tau^{*}Z,ad(\tau)\partial(A_{2}^{\rho}-ad(\tau^{-1})A_{1}+a_{1}\tau^{-1}\partial\tau)\rangle
d\theta\\\ \phantom{d\epsilon}=\frac{i}{2\pi}\int_{S^{1}}\langle
dA_{1},\tau^{*}Z\rangle-\langle A_{1},ad(\tau)\partial
A_{2}^{\rho}\rangle+\langle
A_{1},ad(\tau)\partial(ad(\tau^{-1})A_{1})\rangle\\\ -\langle
A_{1},a_{1}ad(\tau)\partial(\tau^{-1}\partial\tau)\rangle-\tfrac{1}{2}\langle
da_{1}\tau^{*}Z,\tau^{*}Z\rangle\\\ +\langle a_{1}\tau^{*}Z,ad(\tau)\partial
A_{2}^{\rho}\rangle-\langle
a_{1}\tau^{*}Z,ad(\tau)\partial(ad(\tau^{-1})A_{1})\rangle d\theta.$

Therefore,

$\tau^{*}R-d\epsilon=\frac{i}{4\pi}\int_{S^{1}}\langle A_{2},\partial
A_{2}\rangle-2\langle dA_{1},\tau^{*}Z\rangle-\langle
A_{1},ad(\tau)\partial(ad(\tau^{-1})A_{1})\rangle\\\ +2\langle
a_{1}\tau^{-1}\partial\tau,\partial(ad(\tau^{-1})A_{1})\rangle+\langle
da_{1}\tau^{*}Z,\tau^{*}Z\rangle d\theta,$

using the $ad$ invariance of the Killing form and integration by parts. Then,
using the identity from before,

$\partial(ad(\tau^{-1})A)=ad(\tau^{-1})[A,\tau^{*}Z]+ad(\tau^{-1})\partial A,$

yields

$\tau^{*}R-d\epsilon=\frac{i}{4\pi}\int_{S^{1}}\langle A_{2},\partial
A_{2}\rangle-2\langle dA_{1},\tau^{*}Z\rangle-\langle
A_{1},[A_{1},\tau^{*}Z]\rangle-\langle A_{1},\partial A_{1}\rangle\\\
+2\langle\tau^{*}Za_{1},[A_{1},\tau^{*}Z]\rangle+2\langle
a_{1}\tau^{*}Z,\partial A_{1}\rangle+\langle da_{1}\tau^{*}Z,\tau^{*}Z\rangle
d\theta\\\ \phantom{\tau^{*}R-d\epsilon}=\frac{i}{4\pi}\int_{S^{1}}\langle
A_{2},\partial A_{2}\rangle-\langle A_{1},\partial A_{1}\rangle-2\langle
dA_{1},\tau^{*}Z\rangle-\langle[A_{1},A_{1}],\tau^{*}Z\rangle\\\
+2\langle\tau^{*}Za_{1},\partial A_{1}\rangle+\langle
da_{1}\tau^{*}Z,\tau^{*}Z\rangle d\theta.$

Note now that if $(F,f)$ is the curvature of the connection $(A,a)$ then we
have

$\displaystyle(F,f)(X,Y)$
$\displaystyle=(dA(X,Y)+\tfrac{1}{2}[(A,a)(X),(A,a)(Y)],da(X,Y))$
$\displaystyle=(dA(X,Y)+\tfrac{1}{2}([A(X),A(Y)]-a(X)\partial
A(Y)+a(Y)\partial A(X)),da(X,Y)).$

That is,

$(F,f)=(dA+\tfrac{1}{2}[A,A]-a\wedge\partial A,da).$

Therefore, the formula above for $\tau^{*}R-d\epsilon$ reads

$\tau^{*}R-d\epsilon=\frac{i}{4\pi}\int_{S^{1}}\left\langle\pi_{1}^{*}A,\partial\pi_{1}^{*}A\right\rangle-\left\langle\pi_{2}^{*}A,\partial\pi_{2}^{*}A\right\rangle-2\left\langle\pi_{2}^{*}F-\tfrac{1}{2}\pi_{2}^{*}f\,\tau^{*}Z,\tau^{*}Z\right\rangle
d\theta.$

##### A curving for the lifting bundle gerbe

Recall that in order to find the 3-curvature of the lifting bundle gerbe, and
hence a representative for the image in real cohomology of the Dixmier-Douady
class, we need a curving for $\tau^{*}\widehat{LG\rtimes S^{1}}.$ That is,
some 2-form $B$ on $P$ such that $\delta B=\tau^{*}R-d\epsilon.$ Note that
$\delta=\pi_{1}^{*}-\pi_{2}^{*}$ and

$\tau^{*}R-d\epsilon=\delta\left(\frac{i}{4\pi}\int_{S^{1}}\left\langle
A,\partial A\right\rangle
d\theta\right)-\frac{i}{2\pi}\int_{S^{1}}\left\langle\pi_{2}^{*}F-\tfrac{1}{2}\pi_{2}^{*}f\,\tau^{*}Z,\tau^{*}Z\right\rangle
d\theta.$

To deal with the second term above, we use a similar method to the one in
section 2.4. Namely, we will need a Higgs field for the $LG\rtimes
S^{1}$-bundle $P.$

######  Definition 4.1.2.

A _Higgs field_ for $P$ is a map $\Phi\colon P\to L{\mathfrak{g}}$ satisfying

$\Phi(p(\gamma,\phi))=\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\Phi(p)+\gamma^{-1}\partial\gamma\right).$

We shall explain the geometric significance of this map in section 4.2. As in
the $LG$ case, Higgs fields exist for $LG\rtimes S^{1}$-bundles. Note that the
condition above implies that a Higgs field $\Phi$ satisfies

$\pi_{1}^{*}\Phi=\rho_{\tau}^{-1}\left(ad(\tau^{-1})\pi_{2}^{*}\Phi+\tau^{-1}\partial\tau\right)$

or simply,

$ad(\tau)\Phi_{2}^{\rho}=\Phi_{1}+\tau^{*}Z.$

Using this, the second term in $\tau^{*}R-d\epsilon$ becomes

$\frac{i}{2\pi}\int_{S^{1}}\left\langle
F_{1}-\tfrac{1}{2}f_{1}\,\tau^{*}Z,ad(\tau)\Phi_{2}^{\rho}-\Phi_{1}\right\rangle
d\theta.$

Since $(F,f)$ is a curvature, it satisfies

$\pi_{1}^{*}(F,f)=ad(\tau^{-1})\pi_{2}^{*}(F,f).$

That is, $f_{2}=f_{1}$ and

$F_{2}=\rho_{\tau}^{-1}\left(ad(\tau^{-1})F_{1}-f_{1}\tau^{-1}\partial\tau\right),$

or

$ad(\tau)F_{2}^{\rho}=F_{1}-f_{1}\tau^{*}Z.$

Using this, we have

$\displaystyle\frac{i}{2\pi}\int_{S^{1}}$ $\displaystyle\left\langle
F_{1}-\tfrac{1}{2}f_{1}\,\tau^{*}Z,ad(\tau)\Phi_{2}^{\rho}-\Phi_{1}\right\rangle
d\theta$ $\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle
F_{1}+ad(\tau)F_{2}^{\rho},ad(\tau)\Phi_{2}^{\rho}-\Phi_{1}\right\rangle
d\theta$ $\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle
F_{1},ad(\tau)\Phi_{2}^{\rho}\right\rangle-\left\langle
F_{1},\Phi_{1}\right\rangle+\left\langle
F_{2},\Phi_{2}\right\rangle-\left\langle
ad(\tau)F_{2}^{\rho},\Phi_{1}\right\rangle d\theta$
$\displaystyle=\delta\left(\frac{i}{4\pi}\int_{S^{1}}\left<F,\Phi\right\rangle
d\theta\right)+\frac{i}{4\pi}\int_{S^{1}}\left\langle
F_{1},ad(\tau)\Phi_{2}^{\rho}\right\rangle-\left\langle
ad(\tau)F_{2}^{\rho},\Phi_{1}\right\rangle d\theta.$

Note, however, that the second integral above simplifies further

$\displaystyle\frac{i}{4\pi}\int_{S^{1}}$ $\displaystyle\left\langle
F_{1},ad(\tau)\Phi_{2}^{\rho}\right\rangle-\left\langle
ad(\tau)F_{2}^{\rho},\Phi_{1}\right\rangle d\theta$
$\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle
ad(\tau)F_{2}+f_{1}\tau^{*}Z,ad(\tau)\Phi_{2}^{\rho}\right\rangle-\left\langle
F_{1}-f_{1}\tau^{*}Z,\Phi_{1}\right\rangle d\theta$
$\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\langle F_{2},\Phi_{2}\rangle-\langle
F_{1},\Phi_{1}\rangle+\langle
f_{1}\tau^{*}Z,ad(\tau)\Phi_{2}^{\rho}+\Phi_{1}\rangle d\theta$
$\displaystyle=\delta\left(\frac{i}{4\pi}\int_{S^{1}}\left\langle
F,\Phi\right\rangle d\theta\right)+\frac{i}{4\pi}\int_{S^{1}}\left\langle
f_{1}\tau^{*}Z,2\Phi_{1}+\tau^{*}Z\right\rangle d\theta$
$\displaystyle=\delta\left(\frac{i}{4\pi}\int_{S^{1}}\left\langle
F,\Phi\right\rangle d\theta\right)+\frac{i}{4\pi}\int_{S^{1}}2\left\langle
f_{1}\tau^{*}Z,\Phi_{1}\right\rangle+\left\langle
f_{1}\tau^{*}Z,\tau^{*}Z\right\rangle d\theta.$

Therefore, $\tau^{*}R-d\epsilon$ is equal to

$\delta\left(\frac{i}{4\pi}\int_{S^{1}}\left\langle A,\partial
A\right\rangle-2\left\langle F,\Phi\right\rangle
d\theta\right)-\frac{i}{4\pi}\int_{S^{1}}2\left\langle
f_{1}\tau^{*}Z,\Phi_{1}\right\rangle+\left\langle
f_{1}\tau^{*}Z,\tau^{*}Z\right\rangle d\theta.$

So it is enough to find a $B_{2}\in\Omega^{2}(P)$ such that

$\delta B_{2}=\frac{i}{4\pi}\int_{S^{1}}2\left\langle
f_{1}\tau^{*}Z,\Phi_{1}\right\rangle+\left\langle
f_{1}\tau^{*}Z,\tau^{*}Z\right\rangle d\theta.$

Consider, then, the form

$\frac{i}{4\pi}\int_{S^{1}}\left\langle\Phi,f\Phi\right\rangle d\theta.$

We have

$\delta\left(\frac{i}{4\pi}\int_{S^{1}}\left\langle\Phi,f\Phi\right\rangle
d\theta\right)\\\
\phantom{\delta(\int_{S^{1}}\langle\Phi}=\frac{i}{4\pi}\int_{S^{1}}\left\langle\Phi_{2},f_{2}\Phi_{2}\right\rangle-\left\langle\Phi_{1},f_{1}\Phi_{1}\right\rangle
d\theta\\\
\phantom{\delta(\int_{S^{1}}\langle\Phi}=\frac{i}{4\pi}\int_{S^{1}}\left\langle
ad(\tau^{-1})(\Phi_{1}+\tau^{*}Z),f_{1}ad(\tau^{-1})(\Phi_{1}+\tau^{*}Z)\right\rangle-\left\langle\Phi_{1},f_{1}\Phi_{1}\right\rangle
d\theta\\\
\phantom{\delta(\int_{S^{1}}\langle\Phi}=\frac{i}{4\pi}\int_{S^{1}}\left\langle\Phi_{1},f_{1}\Phi_{1}\right\rangle+\left\langle\tau^{*}Z,f_{1}\Phi_{1}\right\rangle+\left\langle\Phi_{1},f_{1}\tau^{*}Z\right\rangle\\\
+\langle\tau^{*}Z,f_{1}\tau^{*}Z\rangle-\left\langle\Phi_{1},f_{1}\Phi_{1}\right\rangle
d\theta\\\
\phantom{\delta(\int_{S^{1}}\langle\Phi}=\frac{i}{4\pi}\int_{S^{1}}2\left\langle
f_{1}\tau^{*}Z,\Phi_{1}\right\rangle+\left\langle
f_{1}\tau^{*}Z,\tau^{*}Z\right\rangle d\theta.\\\ $

Therefore, a curving for the lifting bundle gerbe is given by

$B=\frac{i}{4\pi}\int_{S^{1}}\left\langle A,\partial A\right\rangle-2\langle
F+\tfrac{1}{2}f\Phi,\Phi\rangle\,d\theta.$

##### The string class of an $LG\rtimes S^{1}$-bundle

The last step now that we have found a curving for the lifting bundle gerbe is
to calculate the $3$-curvature $H=dB.$ Then $H/2\pi i$ is integral and
represents the real image of the Dixmier-Douady class of
$\tau^{*}\widehat{LG\rtimes S^{1}}$ (and hence the obstruction to lifting
$P$). We have

$\displaystyle dB$ $\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle
dA,\partial A\right\rangle-\left\langle A,d(\partial
A)\right\rangle-2\left\langle dF,\Phi\right\rangle-2\left\langle
F,d\Phi\right\rangle-\left\langle
d\Phi,f\Phi\right\rangle-\left\langle\Phi,fd\Phi\right\rangle d\theta$
$\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle dA,\partial
A\right\rangle+\left\langle\partial A,dA\right\rangle-2\left\langle
dF,\Phi\right\rangle-2\left\langle F,d\Phi\right\rangle-2\left\langle
d\Phi,f\Phi\right\rangle d\theta$
$\displaystyle=\frac{i}{2\pi}\int_{S^{1}}\left\langle dA,\partial
A\right\rangle-\left\langle dF,\Phi\right\rangle-\left\langle
F,d\Phi\right\rangle-\left\langle d\Phi,f\Phi\right\rangle d\theta.$

To proceed further, we require the Bianchi identity for $(F,f).$ Note that

$d(F,f)=([dA,A]-f\wedge\partial A+a\wedge\partial(dA),d^{2}a).$

In particular, this means that

$dF=[F,A]-f\wedge\partial A+a\wedge\partial F,$

since

$\displaystyle[F,A]$
$\displaystyle=[dA,A]+\tfrac{1}{2}[[A,A],A]-[a\wedge\partial A,A]$
$\displaystyle=[dA,A]-[a\wedge\partial A,A],$

and

$\displaystyle a\wedge\partial F$
$\displaystyle=a\wedge\partial(dA)+\tfrac{1}{2}a\wedge\partial[A,A]-a\wedge\partial(a\wedge\partial
A)$ $\displaystyle=a\wedge\partial(dA)+[a\wedge\partial A,A].$

Using this, and the fact that $\int_{S^{1}}\langle[A,A],\partial A\rangle
d\theta$ and $\langle a\wedge\partial A,\partial A\rangle$ both vanish (so
that $\int_{S^{1}}\langle dA,\partial A\rangle d\theta=\int_{S^{1}}\langle
F,\partial A\rangle d\theta$), the expression for $dB$ becomes

$dB=\frac{i}{2\pi}\int_{S^{1}}\left\langle F,\partial
A\right\rangle-\left\langle[F,A]-f\wedge\partial A+a\wedge\partial
F,\Phi\right\rangle-\left\langle F,d\Phi\right\rangle-\left\langle
d\Phi,f\Phi\right\rangle d\theta\\\
\phantom{dB}=\frac{i}{2\pi}\int_{S^{1}}\left\langle F,\partial
A\right\rangle-\left\langle[F,A],\Phi\right\rangle+\left\langle
f\wedge\partial A,\Phi\right\rangle-\left\langle a\wedge\partial
F,\Phi\right\rangle-\left\langle F,d\Phi\right\rangle-\left\langle
d\Phi,f\Phi\right\rangle d\theta\\\
\phantom{dB}=\frac{i}{2\pi}\int_{S^{1}}\left\langle F+f\Phi,\partial
A\right\rangle-\left\langle F,[A,\Phi]\right\rangle-\left\langle
a\wedge\partial F,\Phi\right\rangle-\left\langle F+f\Phi,d\Phi\right\rangle
d\theta\\\ \phantom{dB}=\frac{i}{2\pi}\int_{S^{1}}\left\langle
F+f\Phi,\partial A\right\rangle-\left\langle
F,[A,\Phi]\right\rangle+\left\langle F,a\partial\Phi\right\rangle-\left\langle
F+f\Phi,d\Phi\right\rangle d\theta\\\
\phantom{dB}=\frac{i}{2\pi}\int_{S^{1}}\left\langle F+f\Phi,\partial
A-[A,\Phi]+a\partial\Phi-d\Phi\right\rangle d\theta.\\\ $

Where the last line follows from the fact that $\int_{S^{1}}\langle
f\Phi,a\partial\Phi\rangle d\theta$ and $\langle f\Phi,[A,\Phi]\rangle$ both
vanish. If we define the covariant derivative of $\Phi$ by

$\nabla\Phi=d\Phi+[A,\Phi]-\partial A-a\partial\Phi,$

then one can easily check that it is (twisted) equivariant for the adjoint
action. That is,

$\nabla\Phi(X(\gamma,\phi))=\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\nabla\Phi(X)\right),$

for any tangent vector $X.$ The same is true for the quantity $F+f\Phi,$ and
so using the $ad$-invariance of the Killing form and the rotation invariance
of the integral, Lemma 3.2.3 implies that $H=dB$ descends to a form on $M.$
Thus we have proven

######  Theorem 4.1.3.

Let $P\to M$ be a principal $LG\rtimes S^{1}$-bundle and let $\Phi$ be a Higgs
field for $P$ and $(A,a)$ be a connection for $P$ with curvature $(F,f).$ Then
the string class of $P,$ that is, the obstruction to lifting $P$ to an
$\widehat{LG\rtimes S^{1}}$-bundle, is represented in de Rham cohomology by

$-\frac{1}{4\pi^{2}}\int_{S^{1}}\langle F+f\Phi,\nabla\Phi\rangle d\theta,$

where

$\nabla\Phi=d\Phi+[A,\Phi]-\partial A-a\partial\Phi.$

####  4.1.2 Reduced splittings for lifting bundle gerbes

In this section we shall present an alternative method for finding the curving
of a lifting bundle gerbe and show how to apply this to the problem above.
This method uses _reduced splittings_ and was first introduced by Gomi [18].

In [4] Brylinski considers the problem of lifting a principal
${\mathcal{G}}$-bundle $P$ to a $\widehat{{\mathcal{G}}}$-bundle
$\widehat{P},$ for which he uses a _bundle splitting_. He relates the
obstruction class to the _scalar curvature_ of a certain connection on
$\widehat{P}.$ In [18] Gomi phrases this in such a way that he can use the
theory of lifting bundle gerbes in order to calculate the obstruction class.
We shall begin by briefly outlining Brylinski’s results before describing the
reduced splittings of Gomi.

Let ${\mathcal{G}}$ be a Lie group with central extension
$\widehat{{\mathcal{G}}}.$ If $\mathfrak{G}$ and $\widehat{\mathfrak{G}}$ are
the Lie algebras of ${\mathcal{G}}$ and $\widehat{{\mathcal{G}}}$ respectively
then we have an extension of Lie algebras

$0\to i{\mathbb{R}}\to\widehat{\mathfrak{G}}\to\mathfrak{G}\to 0.$

We can define an action of ${\mathcal{G}}$ on $\widehat{\mathfrak{G}}$ by
lifting the adjoint action of ${\mathcal{G}}$ on its Lie algebra. That is, we
define
$ad\colon{\mathcal{G}}\times\widehat{\mathfrak{G}}\to\widehat{\mathfrak{G}}$
by

$ad(g)\hat{\xi}=ad(\hat{g})\hat{\xi},$

where $\hat{\xi}\in\widehat{\mathfrak{G}}$ and
$\hat{g}\in\widehat{{\mathcal{G}}}$ projects to $g\in{\mathcal{G}}.$ This is
well-defined since $U(1)$ acts trivially on $\widehat{\mathfrak{G}}$ and any
two lifts of $g$ differ by an element of $U(1).$ Consider now a principal
${\mathcal{G}}$-bundle $P.$ We can write down an exact sequence of vector
bundles associated to $P$ as follows. Let
$\operatorname{Ad}_{{\mathfrak{g}}}(P)$ denote the adjoint bundle of $P$ where
${\mathcal{G}}$ acts on the Lie algebra ${\mathfrak{g}}.$ For example,
$\operatorname{Ad}_{\mathfrak{G}}(P)$ is the usual adjoint bundle of $P$ and
$\operatorname{Ad}_{i{\mathbb{R}}}(P)=P\times_{ad}i{\mathbb{R}}\simeq M\times
i{\mathbb{R}}.$ Since ${\mathcal{G}}$ acts via the adjoint action on the exact
sequence above, we have an exact sequence of vector bundles

$0\to\operatorname{Ad}_{i{\mathbb{R}}}(P)\to\operatorname{Ad}_{\mathfrak{G}}(P)\to\operatorname{Ad}_{\widehat{\mathfrak{G}}}(P)\to
0.$

This means that $\operatorname{Ad}_{\mathfrak{G}}(P)$ is isomorphic to the
direct sum of $M\times i{\mathbb{R}}$ and
$\operatorname{Ad}_{\widehat{\mathfrak{G}}}(P).$ A choice of isomorphism is
called a _bundle splitting_. That is,

######  Definition 4.1.4 ([4]).

A _bundle splitting_ of $P$ is a vector bundle map

$L\colon\operatorname{Ad}_{\widehat{\mathfrak{G}}}(P)\to\operatorname{Ad}_{i{\mathbb{R}}}(P)$

which is the identity on the (trivial) subbundle
$\operatorname{Ad}_{i{\mathbb{R}}}(P).$

As mentioned above, Brylinski uses the notion of scalar curvature to calculate
the obstruction to the existence of a lift of $P.$ This is essentially the
$i{\mathbb{R}}$ part of the curvature of a connection on $\widehat{P}.$ More
precisely,

######  Definition 4.1.5 ([4]).

Let $\hat{A}$ be a connection on $\widehat{P}$ with curvature $\hat{F},$
viewed as a $2$-form on $M$ with values in
$\operatorname{Ad}_{\widehat{\mathfrak{G}}}(\widehat{P})\simeq\operatorname{Ad}_{\widehat{\mathfrak{G}}}(P).$
Let $L$ be a bundle splitting of $P.$ The _scalar curvature_ of $\hat{A}$ is
the $i{\mathbb{R}}$-valued $2$-form

$K=L\circ\hat{F}.$

To see how this is related to the obstruction class, let $\\{U_{\alpha}\\}$ be
a good cover of $M$ over which $P$ is trivial. Then there exists a lift
$\widehat{P}_{\alpha}$ of $P|_{U_{\alpha}}\to U_{\alpha}.$ Choose a connection
$A_{\alpha}$ on $P|_{U_{\alpha}}$ and let $K_{\alpha}$ be the scalar curvature
of a connection $\hat{A}_{\alpha}$ on $\widehat{P}_{\alpha}$ which is
compatible with $A_{\alpha}$ in the sense that the pull-back of $A_{\alpha}$
to $\widehat{P}_{\alpha}$ coincides with the image of $\hat{A}_{\alpha}$ in
$\mathfrak{G}.$ That is,

$f^{*}A_{\alpha}=p(\hat{A}_{\alpha}),$

where $f$ is the bundle map $\widehat{P}_{\alpha}\to P|_{U_{\alpha}}$ and $p$
is the projection $\widehat{{\mathcal{G}}}\to{\mathcal{G}}.$ Brylinski’s
result, then, is that the (real image of the) obstruction class restricted to
$U_{\alpha}$ coincides with the derivative of the scalar curvature,
$dK_{\alpha}.$

As mentioned, Gomi’s results interpolate between the method described above
and the theory of lifting bundle gerbes which we have used extensively. He
utilises so-called reduced splittings to write down a formula for the curving
of the lifting bundle gerbe associated to a lifting problem and relates the
curving to the scalar curvature. In the case where a splitting of the Lie
algebra of $\widehat{\mathfrak{G}}$ has been specified, reduced splittings are
equivalent to bundle splittings. To describe Gomi’s results, let us assume we
have chosen a splitting of the Lie algebra $\widehat{\mathfrak{G}}$ as
$\mathfrak{G}\oplus i{\mathbb{R}}.$

######  Definition 4.1.6 ([18]).

The _group cocycle_ for the central extension $\widehat{{\mathcal{G}}}$ is the
map $\sigma\colon{\mathcal{G}}\times\mathfrak{G}\to i{\mathbb{R}}$ defined by

$\sigma(g,\xi)=ad(g)(\xi,0)-(ad(g)\xi,0),$

where $ad(g)$ acts on $\widehat{\mathfrak{G}}$ as described above.

The group cocycle gives information about the multiplication in
$\widehat{{\mathcal{G}}}$ in the same way as the $1$-form $\alpha$ which we
used. In fact, as we shall see, to apply Gomi’s results to the case where
${\mathcal{G}}$ is either the loop group $LG$ or the semi-direct product
$LG\rtimes S^{1},$ we shall give $\sigma$ in terms of $\alpha.$

######  Definition 4.1.7 ([18]).

A _reduced splitting_ for a principal ${\mathcal{G}}$-bundle $P$ is a map
$\ell\colon P\times\mathfrak{G}\to i{\mathbb{R}}$ which is linear in the
second factor and satisfies

$\ell(p,\xi)=\ell(pg,ad(g^{-1})\xi)+\sigma(g^{-1},\xi).$

The relation between reduced splittings and bundle gerbe curvings is given by
the following theorem.

######  Theorem 4.1.8 ([18]).

Let $F$ be the curvature of a connection $A$ on $P$ and $\ell$ be a reduced
splitting for $P.$ Define a $2$-form $\kappa$ on $P$ by
$\kappa_{p}=\ell(p,F).$ Then a curving for the lifting bundle gerbe associated
to the lifting problem for $P$ is given by

$B=\frac{1}{2}\omega(A,A)+\kappa,$

where
$\omega(\xi,\zeta)=[(\xi,0),(\zeta,0)]_{\widehat{\mathfrak{G}}}-([\xi,\zeta]_{\mathfrak{G}},0)$
is the cocycle classifying the central extension.

To connect this with Brylinski’s work, Gomi proves the following theorem
relating the curving and the scalar curvature.

######  Theorem 4.1.9 ([18]).

Let $P$ be a principal ${\mathcal{G}}$-bundle and $\widehat{P}$ be a lift of
$P.$ Let $A$ be a connection on $P$ and $\hat{A}$ be a compatible connection
on $\widehat{P}.$ Then the curving can be written as

$B=\pi^{*}K-\tilde{F},$

where $\tilde{F}$ is the curvature of the connection $\hat{A}-f^{*}A$ on
$\widehat{P}$ (for $f\colon\widehat{P}\to P$ the bundle map defining the lift)
and $K$ is the scalar curvature of $\hat{A}.$

We would now like to consider the case where ${\mathcal{G}}=LG\rtimes S^{1}.$
We shall define a reduced splitting for $P$ so we can use Theorem 4.1.8 to
calculate a curving and show that it is in agreement with the results from
section 4.1.1. The group cocycle in this case is given by

$\displaystyle\sigma((\gamma,\phi)^{-1},(\xi,x))$
$\displaystyle=\alpha_{((1,1),(\gamma,\phi))}((\xi,x),(0,0))$
$\displaystyle=\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi-\tfrac{1}{2}x\partial\gamma\gamma^{-1},\partial\gamma\gamma^{-1}\right\rangle
d\theta,$

and we have

######  Proposition 4.1.10.

A reduced splitting for the $LG\rtimes S^{1}$-bundle $P$ is given by

$\ell(p,(\xi,x))=-\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi+\tfrac{1}{2}x\,\Phi(p),\Phi(p)\right\rangle
d\theta,$

where $\Phi$ is a Higgs field for $P.$

###### Proof.

We need only show that it satisfies the transformation property above. We can
calculate

$\ell(p(\gamma,\phi),ad(\gamma,\phi)^{-1}(\xi,x))\\\
\phantom{\ell(p}=-\frac{i}{2\pi}\int_{S^{1}}\left\langle ad(\gamma^{-1})(\xi-
xZ)+\tfrac{1}{2}x\,ad(\gamma^{-1})(\Phi(p)+Z),ad(\gamma^{-1})(\Phi(p)+Z)\right\rangle
d\theta\\\
\phantom{\ell(p}=-\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi,\Phi(p)\vphantom{\tfrac{1}{2}}\right\rangle-\left\langle
xZ,\Phi(p)\vphantom{\tfrac{1}{2}}\right\rangle+\left\langle\tfrac{1}{2}x\,\Phi(p),\Phi(p)\right\rangle+\left\langle\tfrac{1}{2}xZ,\Phi(p)\right\rangle\\\
+\left\langle\xi,Z\vphantom{\tfrac{1}{2}}\right\rangle-\left\langle
xZ,Z\vphantom{\tfrac{1}{2}}\right\rangle+\left\langle\tfrac{1}{2}x\,\Phi(p),Z\right\rangle+\left\langle\tfrac{1}{2}x\,Z,Z\right\rangle
d\theta\\\
\phantom{\ell(p}=-\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi+\tfrac{1}{2}x\,\Phi(p),\Phi(p)\right\rangle+\left\langle
X-\tfrac{1}{2}xZ,Z\right\rangle d\theta\\\
\phantom{\ell(p}=\ell(p,(\xi,x))-\sigma((\gamma,\phi)^{-1},(\xi,x))\\\ $

as required. ∎

Note that in order to use Theorem 4.1.8, we need the cocycle $\omega.$ This is
simply given by the form $R$ which defines the central extension. In
particular,

$\omega((\xi,x),(\zeta,y))=\frac{i}{2\pi}\int_{S^{1}}\left\langle\xi,\partial\zeta\right\rangle
d\theta.$

Therefore, for the curving of the lifting bundle gerbe, Theorem 4.1.8 gives

$\displaystyle B$
$\displaystyle=\frac{1}{2}R((A,a),(A,a))-\frac{i}{2\pi}\int_{S^{1}}\left\langle
F+\tfrac{1}{2}f\,\Phi,\Phi\right\rangle d\theta$
$\displaystyle=\frac{i}{4\pi}\int_{S^{1}}\left\langle A,\partial
A\right\rangle-2\langle F+\tfrac{1}{2}f\,\Phi,\Phi\rangle\,d\theta,$

where as before, $(A,a)$ is a connection on $P$ and $(F,f)$ is its curvature.

###  4.2 Higgs fields, $LG\rtimes S^{1}$-bundles and the string class

Now that we have an explicit formula for the string class of an $LG\rtimes
S^{1}$-bundle $P,$ it is natural to ask whether there is some relation with
the Pontrjagyn class of a $G$-bundle related to $P$ in some way, as was the
case with the string class of an $LG$-bundle presented in chapter 2. In
particular, in section 2.5, following [35], we saw that there was a
correspondence between $LG$-bundles over $M$ (with connection and Higgs field)
and $G$-bundles over $M\times S^{1}$ (with connection) (Propositions 2.5.1 and
2.5.2) and we used this to prove that the string class of $P$ is given by
integrating over the circle the first Pontrjagyn class of the corresponding
$G$-bundle (Theorem 2.5.3). In this section, we shall show there is a
correspondence between $LG\rtimes S^{1}$-bundles over $M$ and $G$-bundles over
$S^{1}$-bundles over $M,$ which holds on the level of connections as well. As
in section 2.5 we shall use this correspondence to prove that the string class
of $P$ is given in terms of the Pontrjagyn class of some $G$-bundle.

####  4.2.1 Higgs fields and $LG\rtimes S^{1}$-bundles

The following correspondence first appeared in [1]. We will present it here in
detail and also extend it to the level of connections.

Suppose that we have a principal $G$-bundle over a principal $S^{1}$-bundle:

$\textstyle{\widetilde{P}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{G}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{S^{1}}$$\textstyle{M}$

We would like to mimic the construction of the $LG$-bundle in section 2.5
where we essentially took loops in $\widetilde{P}$ such that their image in
$M\times S^{1}$ commuted with the obvious $S^{1}$ action on this space. That
is, for a loop $f\in L\widetilde{P}_{m}$ in the fibre above $\\{m\\}\times
S^{1}$ we required that $\tilde{\pi}(f(\theta))=(m,\theta).$ The difference
here is that we cannot choose a global section $M\to Y$ and thus there is no
way of choosing a ‘starting point’ for the loop $\tilde{\pi}(f)\colon S^{1}\to
Y.$ We can, however, still require that the map $\tilde{\pi}(f)$ commutes with
the $S^{1}$ action on $Y$ (which we will write as addition). That is, we can
define

$P=\\{f\colon
S^{1}\to\widetilde{P}\mid\tilde{\pi}(f(\theta+\phi))=\tilde{\pi}(f(\theta))+\phi\\}$

and there is a canonical map $P\to M.$ $P$ is acted on by $LG\rtimes S^{1}:$

$(f(\gamma,\phi))(\theta)=f(\theta+\phi)\gamma(\theta+\phi),$

i.e.

$f(\gamma,\phi)=\rho_{\phi}^{-1}(f\gamma).$

It is a right action since

$\displaystyle f(\gamma_{1},\phi_{1})(\gamma_{2},\phi_{2})$
$\displaystyle=\rho_{(\phi_{1}+\phi_{2})}^{-1}f\rho_{(\phi_{1}+\phi_{2})}^{-1}\gamma_{1}\rho_{\phi_{2}}^{-1}(\gamma_{2})$
$\displaystyle=\rho_{(\phi_{1}+\phi_{2})}^{-1}(f\gamma_{1}\rho_{\phi_{1}}(\gamma_{2}))$
$\displaystyle=f(\gamma_{1}\rho_{\phi_{1}}(\gamma_{2}),\phi_{1}+\phi_{2}).$

It preserves the fibres of $P$ since the $G$ action on $\widetilde{P}$
preserves fibres and the $S^{1}$ action on $Y$ preserves fibres. It is also
free and transitive on fibres and therefore $P\to M$ is a principal $LG\rtimes
S^{1}$-bundle. Note that local triviality of this bundle follows from the
local triviality of $Y$ as follows: Choose a good cover of $M$ and let $U$ be
an open set such that we can find a local section $s\colon U\to Y_{|_{U}}.$
There is a map $P\to Y$ given by $f\mapsto\tilde{\pi}(f(0)).$ If we pull-back
$P$ by $s$ then $s^{*}P\to U$ is trivial (since $U$ is contractible).

Conversely, suppose we are given a principal $LG\rtimes S^{1}$-bundle $P\to
M.$ Following the construction in section 2.5, define

$\widetilde{P}=(P\times G\times S^{1})/LG\rtimes S^{1},$

where $[p,g,\theta]=[p(\gamma,\phi),\gamma(\theta)^{-1}g,\theta-\phi].$ A $G$
action on $\widetilde{P}$ is given by $[p,g,\theta]h=[p,gh,\theta].$ There is
a natural projection from $\widetilde{P}$ to the $S^{1}$-bundle associated to
$P$ via the homomorphism $LG\rtimes S^{1}\to S^{1},$ that is,
$\widetilde{P}\to(P\times S^{1})/LG\rtimes S^{1}\simeq P/LG,$ given by
$\tilde{\pi}([p,g,\theta])=[p,\theta].$ This makes $\widetilde{P}$ into a
principal $G$-bundle. Thus, given the $LG\rtimes S^{1}$-bundle $P\to M$ we can
construct a $G$-bundle over an $S^{1}$-bundle:

$\textstyle{\dfrac{P\times G\times S^{1}}{LG\rtimes
S^{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{G}$$\textstyle{\dfrac{P\times
S^{1}}{LG\rtimes
S^{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{S^{1}}$$\textstyle{M}$

We would like to show that both constructions above are invertible (as we did
for the constructions in the $LG$ case). Assume, then, that we are given an
$LG\rtimes S^{1}$-bundle $P\to M$ and have constructed the $G$-bundle
$\widetilde{P}$ over the $S^{1}$-bundle $P/LG\to M$ as above. Then use the
first correspondence above to form the $LG\rtimes S^{1}$-bundle $P^{\prime}\to
M$ (by taking certain loops in $\widetilde{P}$). So we have

$P^{\prime}=\\{f\colon S^{1}\to(P\times G\times S^{1})/LG\rtimes
S^{1}\mid\tilde{\pi}(f(\theta+\phi))=\tilde{\pi}(f(\theta))+\phi\\}$

and a bundle isomorphism is given by

$P\to P^{\prime};\quad p\mapsto f_{p}=(\theta\mapsto[p,1,\theta]).$

It is easily checked that this map commutes with the $LG\rtimes S^{1}$ action,
for

$\displaystyle p(\gamma,\phi)\mapsto$ $\displaystyle~{}f_{p(\gamma,\phi)}$
$\displaystyle=$ $\displaystyle~{}[p(\gamma,\phi),1,\theta]$ and on the other
hand, $\displaystyle f_{p}(\gamma,\phi)=$
$\displaystyle~{}f_{p}(\theta+\phi)\gamma(\theta+\phi)$ $\displaystyle=$
$\displaystyle~{}[p,1,\theta+\phi]\gamma(\theta+\phi)$ $\displaystyle=$
$\displaystyle~{}[p,\gamma(\theta+\phi),\theta+\phi]$ $\displaystyle=$
$\displaystyle~{}[p(\gamma,\phi),1,\theta].$

So we have that $P\simeq P^{\prime}.$ If, on the other hand, we are given the
$G$-bundle over the $S^{1}$-bundle $\widetilde{P}\to Y\to M$ and have
constructed $P\to M,$ then we can construct $\widetilde{P}^{\prime}\to P/LG\to
M$ and we would like for these bundles to be isomorphic. That is,
$\widetilde{P}^{\prime}\simeq\widetilde{P}$ and $P/LG\simeq Y.$ Firstly,
consider the map $P/LG\simeq P\times_{LG\rtimes S^{1}}S^{1}\to Y$ defined by
$[f,\theta]\mapsto\tilde{\pi}(f(\theta)).$ This is well-defined on equivalence
classes:

$[f,\theta]=[\rho_{\phi}^{-1}(f\gamma),\theta-\phi]\mapsto\tilde{\pi}(f(\theta-\phi+\phi)\gamma(\theta-\phi+\phi))=\tilde{\pi}(f(\theta)).$

It commutes with the $S^{1}$ action on $P\times_{LG\rtimes S^{1}}S^{1}:$

$[f,\theta+\alpha]\mapsto\tilde{\pi}(f(\theta+\alpha))=\tilde{\pi}(f(\theta))+\alpha$

by the definition of $P$ in terms of $\widetilde{P}.$ Thus $P\times_{LG\rtimes
S^{1}}S^{1}\simeq Y.$ For $\widetilde{P}^{\prime}$ and $\widetilde{P}$
consider the bundle map

$\widetilde{P}^{\prime}\to\widetilde{P};\quad[f,g,\theta]\mapsto f(\theta)g.$

This is well-defined:

$[f,g,\theta]=[f(\gamma,\phi),\gamma(\theta)^{-1}g,\theta-\phi]\mapsto
f(\theta-\phi-\phi)\gamma(\theta-\phi+\phi)\gamma(\theta)^{-1}g=f(\theta)g,$

and commutes with the $G$ action:

$[f,g,\theta]h=[f,gh,\theta]\mapsto f(\theta)gh=(f(\theta)g)h.$

Therefore, it is a bundle isomorphism and
$\widetilde{P}^{\prime}\simeq\widetilde{P}.$ Thus we have proven

######  Proposition 4.2.1 ([1]).

There is a bijective correspondence between isomorphism classes of principal
$LG\rtimes S^{1}$-bundles over $M$ and isomorphism classes of principal
$G$-bundles over principal $S^{1}$-bundles over $M.$

As in section 2.5 the correspondences here hold on the level of connections as
well. We shall now describe how to derive the connections corresponding to one
another.

Suppose we are given a connection $\tilde{A}$ on $\widetilde{P}\to Y$ and a
connection $\tilde{a}$ on $Y\to M.$ This amounts to a splitting of the tangent
spaces $T_{\tilde{p}}\widetilde{P}\simeq V_{\tilde{p}}\widetilde{P}\oplus
H_{\tilde{p}}\widetilde{P}$ at each point $\tilde{p}\in\widetilde{P}$ and also
$T_{y}Y\simeq V_{y}Y\oplus H_{y}Y$ at each point $y\in Y.$ Since $P$ is given
by certain loops in $\widetilde{P},$ a vector $X\in T_{f}P$ is really a vector
field along $f$ in $\widetilde{P}.$ So, $X_{\theta}\in
T_{f(\theta)}\widetilde{P}.$ Thus we can use the splittings of the tangent
spaces of $\widetilde{P}$ and $Y$ to define a splitting for the tangent space
to $P$ at $f$ for each $\theta.$ So we have

$\displaystyle T_{f(\theta)}\widetilde{P}$ $\displaystyle\simeq
V_{f(\theta)}\widetilde{P}\oplus H_{f(\theta)}\widetilde{P}$
$\displaystyle\simeq V_{f(\theta)}\widetilde{P}\oplus
V_{\tilde{\pi}(f(\theta))}Y\oplus H_{\tilde{\pi}(f(\theta))}Y$
$\displaystyle\simeq V_{f(\theta)}\widetilde{P}\oplus
V_{\tilde{\pi}(f(\theta))}Y\oplus T_{(\pi_{Y}\circ\tilde{\pi})(f(\theta))}M,$

using the isomorphisms $H_{f(\theta)}\widetilde{P}\simeq
T_{\tilde{\pi}(f(\theta))}Y$ and $H_{\tilde{\pi}(f(\theta))}Y\simeq
T_{(\pi_{Y}\circ\tilde{\pi})(f(\theta))}M.$ We can find the 1-form for this
connection by calculating

$X_{\theta}-\widehat{\widehat{\pi_{*}X_{\theta}}}$

which equals $\iota_{f(\theta)}(A_{f}(X)_{\theta}),$ where
$\pi=\pi_{Y}\circ\tilde{\pi}$ and $\widehat{\widehat{V}}$ is the horizontal
lift of a vector on $M$ first to $Y,$ then to $\widetilde{P}.$ Note that using
the connection on $Y$ we have

$\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))=\tilde{\pi}_{*}X_{\theta}-\widehat{\pi_{*}X_{\theta}},$

and so

$\widehat{\pi_{*}X_{\theta}}=\tilde{\pi}_{*}X_{\theta}-\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta})).$

Lifting everything, we have

$\widehat{\widehat{{\pi_{*}X_{\theta}}}}=\widehat{\tilde{\pi}_{*}X_{\theta}}-\widehat{\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))},$

and thus

$\iota_{f(\theta)}(A_{f}(X)_{\theta})=X_{\theta}-\widehat{\tilde{\pi}_{*}X_{\theta}}+\widehat{\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))}.$

But
$X_{\theta}-\widehat{\tilde{\pi}_{*}X_{\theta}}=\iota_{f(\theta)}(\tilde{A}(X_{\theta}))$
and so we have

$\iota_{f(\theta)}(A_{f}(X)_{\theta})=\iota_{f(\theta)}(\tilde{A}(X_{\theta}))+\widehat{\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))}.$

To make use of this we need to be able to write $A$ as an
$L{\mathfrak{g}}$-valued 1-form and an $i{\mathbb{R}}$-valued 1-form. That is,
$A(X)_{\theta}=(\xi(\theta),x)$ for $\xi\in L{\mathfrak{g}}$ and $x\in
i{\mathbb{R}}.$ To that end, consider the vertical vector $V$ in $T_{f}P$
generated by the Lie algebra element $(\xi,x):$

$\displaystyle V_{\theta}$
$\displaystyle=\frac{d}{dt}_{|_{0}}f(\exp(t\xi),tx)(\theta)$
$\displaystyle=\frac{d}{dt}_{|_{0}}f(\theta+tx)\exp(t\xi(\theta+tx))$
$\displaystyle=\frac{d}{dt}_{|_{0}}\left(f(\theta)+f^{\prime}(\theta)tx\vphantom{{}^{2}}\right)\left(1+t\xi(\theta)+O(t^{2})\right)$
$\displaystyle=\iota_{f(\theta)}(\xi(\theta))+xf^{\prime}(\theta).$

Since $A$ is a connection, it returns the Lie algebra element corresponding to
the vertical part of a vector $X.$ Therefore, we must solve the following
equation for $\xi$ and $x:$

$\iota_{f(\theta)}(\tilde{A}(X_{\theta}))+\widehat{\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))}=\iota_{f(\theta)}(\xi(\theta))+xf^{\prime}(\theta).$

Applying $\tilde{A}$ to both sides gives

$\tilde{A}(X_{\theta})=\xi(\theta)+x\tilde{A}(f^{\prime}(\theta)),$

since
$\widehat{\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))}$
is horizontal with respect to $\tilde{A}.$ Thus, we have

$\xi(\theta)=\tilde{A}(X_{\theta}-xf^{\prime}(\theta)).$

Taking instead, $\tilde{\pi}_{*}$ of both sides gives

$\iota_{\tilde{\pi}(f(\theta))}(\tilde{a}(\tilde{\pi}_{*}X_{\theta}))=x\,\tilde{\pi}_{*}f^{\prime}(\theta),$

since the vectors $\iota_{f(\theta)}(\tilde{A}(X_{\theta}))$ and
$\iota_{f(\theta)}(\xi(\theta))$ are vertical in $\widetilde{P}$. Then
applying $\tilde{a}$ to both sides yields

$\tilde{a}(\tilde{\pi}_{*}X_{\theta})=x\,\tilde{a}(\tilde{\pi}_{*}f^{\prime}(\theta)).$

So (with a slight abuse of notation) we can write the connection form on $P$
as

$(A,a)_{f}(X)_{\theta}=(\tilde{A}(X_{\theta}-a(X)f^{\prime}(\theta)),a(X)),$

where $\tilde{A}$ and $\tilde{a}$ are connection forms on $\widetilde{P}$ and
$Y$ respectively and $a(X)$ is given by the formula for $x$ above. Now that we
have the connection on $P$ in this form we can check explicitly that it
satisfies the conditions for a connection. By construction, it satisfies
$(A,a)(\iota_{f}(\xi,x))=(\xi,x)$ and so we just need to check that
$(A,a)(X(\gamma,\phi))=ad(\gamma,\phi)^{-1}(A,a)(X).$ Recall that the adjoint
action of $LG\rtimes S^{1}$ on its Lie algebra is given by

$ad(\gamma,\phi)^{-1}(\xi,x)=\left(\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\xi-\gamma^{-1}\partial\gamma\,x\right),x\right)$

and so

$ad(\gamma,\phi)^{-1}(A,a)(X)_{\theta}=\left(\rho_{\phi}^{-1}(ad(\gamma^{-1})\tilde{A}(X_{\theta}-a(X)f^{\prime}(\theta))-\gamma^{-1}\partial\gamma\,a(X)),a(X)\right).$

On the other hand, the action of $LG\rtimes S^{1}$ on the tangent vector $X$
is

$X(\gamma,\phi)=\rho_{\phi}^{-1}(X\gamma).$

Therefore,

$\displaystyle(A,a)(X(\gamma,\phi))_{\theta}$
$\displaystyle=\left(\tilde{A}(X(\gamma,\phi)_{\theta}-a(X(\gamma,\phi))\partial(f(\theta+\phi)\gamma(\theta+\phi))),a(X(\gamma,\phi))\right)$
$\displaystyle=\left(\tilde{A}(\rho_{\phi}^{-1}(X\gamma)_{\theta}-a(\rho_{\phi}^{-1}(X\gamma)_{\theta})\partial(f(\theta+\phi)\gamma(\theta+\phi))),a(\rho_{\phi}^{-1}(X\gamma)_{\theta})\right)$
$\displaystyle\begin{split}&=\left(\tilde{A}(\rho_{\phi}(X\gamma)_{\theta}-a(\rho_{\phi}^{-1}(X\gamma)_{\theta})\\{\partial
f(\theta+\phi)\gamma(\theta+\phi)\right.\\\
&\left.\phantom{(\tilde{A}(\rho_{\phi}^{-1}(X\gamma)_{\theta}-a(\rho_{\phi}^{-1}(X\gamma}+f(\theta+\phi)\partial\gamma(\theta+\phi)\\}),a(\rho_{\phi}^{-1}(X\gamma)_{\theta})\right).\end{split}$

Since $\tilde{A}$ is a connection, we have
$\tilde{A}(\rho_{\phi}^{-1}(X\gamma))_{\theta}=\rho_{\phi}^{-1}(ad(\gamma^{-1})\tilde{A}(X))_{\theta}$
and $\tilde{A}(\partial
f(\theta+\phi)\gamma(\theta+\phi))=\rho_{\phi}^{-1}(ad(\gamma^{-1})\tilde{A}(\partial
f(\theta))).$ Also, since $a$ is $i{\mathbb{R}}$-valued, we have
$a(\rho_{\phi}^{-1}(X\gamma)_{\theta})=a(X).$ Therefore,

$(A,a)(X(\gamma,\phi))_{\theta}=\left(\rho_{\phi}^{-1}(ad(\gamma^{-1})\tilde{A}(X_{\theta}-a(X)\partial
f(\theta)))\right.\\\
\left.-a(X)\tilde{A}(f(\theta+\phi)\partial\gamma(\theta+\phi))),a(X)\vphantom{\tilde{A}}\right).$

But, $f(\theta+\phi)\partial\gamma(\theta+\phi)$ is really shorthand for
$\iota_{f(\theta+\phi)}(\rho_{\phi}^{-1}(\gamma^{-1}\partial\gamma))$ and so

$\tilde{A}(f(\theta+\phi)\partial\gamma(\theta+\phi))=\tilde{A}(\iota_{f(\theta+\phi)}(\rho_{\phi}^{-1}(\gamma^{-1}\partial\gamma)))=\rho_{\phi}^{-1}(\gamma^{-1}\partial\gamma).$

Thus, we have

$(A,a)(X(\gamma,\phi))_{\theta}=\left(\rho_{\phi}^{-1}(ad(\gamma^{-1})\tilde{A}(X_{\theta}-a(X)\partial
f(\theta))-a(X)\gamma^{-1}\partial\gamma),a(X)\vphantom{\tilde{A}}\right),$

as required.

As for the $LG$-bundle case in section 2.5, to define a connection333Of
course, here we need to define a connection on $Y$ as well as on
$\widetilde{P}.$ on $\widetilde{P}$ given the bundle $P$ we need a connection
and Higgs field for $P$. Unlike the case in the previous section, however, in
order to define a connection we require a Higgs field to satisfy a slightly
different condition. Recall that a Higgs field for an $LG\rtimes S^{1}$-bundle
$P$ satisfies

$\Phi(p(\gamma,\phi))=\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\Phi(p)+\gamma^{-1}\partial\gamma\right).$

It will be instructive to define a Higgs field for $P$ given the bundles
$\widetilde{P}\to Y\to M$ now since we will need this later to show that the
construction is invertible. Define then, the map $\Phi\colon P\to
L{\mathfrak{g}}$ by

$\Phi(f)=\tilde{A}(\partial f).$

This is a Higgs field since

$\displaystyle\Phi(f(\gamma,\phi))$
$\displaystyle=\tilde{A}(\rho_{\phi}^{-1}(\partial
f\gamma)+\rho_{\phi}^{-1}(\gamma^{-1}\partial\gamma)$
$\displaystyle=\tilde{A}(\rho_{\phi}^{-1}(\partial
f\gamma))+\iota_{\rho_{\phi}(f)}(\rho_{\phi}^{-1}(\gamma^{-1}\partial\gamma)$
$\displaystyle=ad(\rho_{\phi}^{-1}(\gamma^{-1}))\tilde{A}(\rho_{\phi}^{-1}(\partial
f))+\rho_{\phi}^{-1}(\gamma^{-1}\partial\gamma)$
$\displaystyle=\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\tilde{A}(\partial
f)+\gamma^{-1}\partial\gamma\right)$
$\displaystyle=\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\Phi(f)+\gamma^{-1}\partial\gamma\right).$

To define a connection on $\widetilde{P}=(P\times G\times S^{1})/LG\rtimes
S^{1}$ we need to be able to write a form on $P\times G\times S^{1}$ which is
zero on vertical vectors (with respect to the $LG\rtimes S^{1}$ action) and
invariant under the $LG\rtimes S^{1}$ action (so as to ensure that it is well-
defined). Thus we need to calculate the action of $LG\rtimes S^{1}$ on: the
connection, $(A,a),$ on $P,$ the Higgs field, $\Phi,$ on $P$ and the Maurer-
Cartan forms $\Theta$ and $d\theta$ on $G$ and $S^{1}$ respectively. Then we
can combine these in an invariant way. We can calculate the action of
$(\gamma,\phi)$ on the connection on $P$:

$\displaystyle(\gamma,\phi)^{*}(A,a)(X)$ $\displaystyle=(A,a)(X(\gamma,\phi))$
$\displaystyle=ad(\gamma,\phi)^{-1}(A,a)(X)$
$\displaystyle=\left(\rho_{\phi}^{-1}\left(ad(\gamma^{-1})A(X)-\gamma^{-1}\partial\gamma\,a(X)\right),a(X)\right),$

and we know that the Higgs field satisfies

$\Phi(p(\gamma,\phi))=\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\Phi(p)+\gamma^{-1}\partial\gamma\right),$

and the Maurer-Cartan form on $S^{1}$ is unchanged. To calculate the action on
the Maurer-Cartan form on $G,$ consider a vector $(X,g\zeta,x_{\theta})\in
T_{(p,g,\theta)}(P\times G\times S^{1}).$ We have:

$\displaystyle(X,g\zeta,x_{\theta})(\gamma,\phi)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(\gamma_{X}(t)(\gamma,\phi),\gamma(\theta+tx)^{-1}g\exp(t\zeta),\theta+tx-\phi)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(\gamma_{X}(t)(\gamma,\phi),(\gamma(\theta)^{-1}-\gamma(\theta)^{-1}\partial\gamma(\theta)\gamma(\theta)^{-1}tx)g(1+t\zeta),\theta+tx-\phi)$
$\displaystyle=(X(\gamma,\phi),\gamma(\theta)^{-1}g\zeta-\gamma(\theta)^{-1}\partial\gamma(\theta)\gamma(\theta)^{-1}gx,x)$
$\displaystyle=(X(\gamma,\phi),\gamma(\theta)^{-1}g\\{\zeta-x\,ad(g^{-1})\partial\gamma(\theta)\gamma(\theta)^{-1}\\},x),$

and so

$\displaystyle(\gamma,\phi)^{*}\Theta(g\zeta)$
$\displaystyle=\Theta_{\gamma(\theta)^{-1}g}(\gamma(\theta)^{-1}g\\{\zeta-x\,ad(g^{-1})\partial\gamma(\theta)\gamma(\theta)^{-1}\\})$
$\displaystyle=\zeta-x\,ad(g^{-1})\partial\gamma(\theta)\gamma(\theta)^{-1}.$

Now consider the form on $P\times G\times S^{1}$ given by

$\tilde{A}=ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta).$

This is invariant under the $LG\rtimes S^{1}$ action, for

$\displaystyle(\gamma,\phi)^{*}\tilde{A}_{(p,g,\theta)}$
$\displaystyle(X,g\zeta,x_{\theta})$
$\displaystyle=\tilde{A}_{(p(\gamma,\phi),\gamma(\theta)^{-1}g,\theta+\phi)}(X(\gamma,\phi),\gamma(\theta)^{-1}g\\{\zeta-x\,ad(g^{-1})\partial\gamma(\theta)\gamma(\theta)^{-1}\\},x)$
$\displaystyle=ad(g^{-1}\gamma(\theta))\rho_{\phi}^{-1}\left(ad(\gamma^{-1})A(X)_{\theta-\phi}-\gamma^{-1}\partial\gamma_{\theta-\phi}\,a(X)\right)$
$\displaystyle\qquad\qquad+\zeta-x\,ad(g^{-1})\partial\gamma(\theta)\gamma(\theta)^{-1}$
$\displaystyle\qquad\qquad+ad(g^{-1}\gamma(\theta))\rho_{\phi}^{-1}\left(ad(\gamma^{-1})\Phi(p)_{\theta-\phi}+\gamma^{-1}\partial\gamma_{\theta-\phi}\right)\left(a(X)+x\right)$
$\displaystyle=ad(g^{-1}\gamma(\theta))\left(ad(\gamma^{-1})A(X)_{\theta}-\gamma^{-1}\partial\gamma_{\theta}\,a(X)\right)$
$\displaystyle\qquad\qquad+\zeta-x\,ad(g^{-1})\partial\gamma(\theta)\gamma(\theta)^{-1}$
$\displaystyle\qquad\qquad+ad(g^{-1}\gamma(\theta))\left(ad(\gamma^{-1})\Phi(p)_{\theta}+\gamma^{-1}\partial\gamma_{\theta}\right)\left(a(X)+x\right)$
$\displaystyle=ad(g^{-1})A(X)_{\theta}+\zeta+ad(g^{-1})\Phi(p)_{\theta}\left(a(X)+x\right)$
$\displaystyle=\tilde{A}_{(p,g,\theta)}(X,g\zeta,x_{\theta}),$

by the calculations above. So for it to be well-defined on the quotient space
we just need to check that it vanishes on vertical vectors. The vertical
vector at the point $(p,g,\theta)$ generated by the vector $(\xi,x)$ is

$\displaystyle V$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(p,g,\theta)(\exp(t\xi),tx)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(p(\exp(t\xi),tx),(1-t\xi(\theta))g,\phi-
tx)$ $\displaystyle=(\iota_{p}(\xi,x),-g\,ad(g^{-1})\xi(\theta),-x),$

and so

$\displaystyle\tilde{A}(V)$
$\displaystyle=ad(g^{-1})A(\iota_{p}(\xi,x))_{\theta}-ad(g^{-1})\xi(\theta)+ad(g^{-1})\Phi(p)(a(\iota_{p}(\xi.x))-x)$
$\displaystyle=ad(g^{-1})\xi(\theta)-ad(g^{-1})\xi(\theta)+ad(g^{-1})\Phi(p)(x-x)$
$\displaystyle=0.$

Thus we have defined a $G$-valued 1-form on $\widetilde{P}.$ $\tilde{A}$ is in
fact a connection form, since if we evaluate it on the vertical vector
generated by $\zeta\in{\mathfrak{g}},$ that is,
$\iota_{[p,g,\theta]}(\zeta)=(0,g\zeta,0),$ we get $\tilde{A}(g\zeta)=\zeta$
and further,

$\displaystyle\tilde{A}((X,g\zeta,x_{\theta})h)$
$\displaystyle=\tilde{A}(X,ghh^{-1}\zeta h,x_{\theta})$
$\displaystyle=\left(ad(gh)^{-1}A+ad(h^{-1})\Theta+ad(gh)^{-1}\Phi(a+d\theta)\right)(X,g\zeta,x_{\theta})$
$\displaystyle=ad(h^{-1})\tilde{A}(X,g\zeta,x_{\theta}).$

To define a connection on the $S^{1}$-bundle $P/LG$ we just take the
projection of the $i{\mathbb{R}}$-valued 1-form $a$ which is a connection
form.

What remains to be shown now is that the constructions presented here for
connections on $P,$ $\widetilde{P}$ and $Y$ are invertible. In particular,
suppose we have the $LG\rtimes S^{1}$-bundle $P\to M$ with connection $(A,a)$
and Higgs field $\Phi$ and have constructed $\widetilde{P}\to Y\to M$ with
connections $\tilde{A}$ and $\tilde{a}.$ Then if we construct the
corresponding $LG\rtimes S^{1}$-bundle $P^{\prime}$ (which is isomorphic to
$P$ via the map $f\colon P\to P^{\prime};\,p\mapsto
f_{p}=(\theta\mapsto[p,1,\theta])$) and the connection
$(A^{\prime},a^{\prime})$ for $P^{\prime},$ we would like to show that
$f^{*}(A^{\prime},a^{\prime})=(A,a).$ Note that for the vector $X\in T_{p}P$
we have

$f_{*}X=(X,0,0)\in T_{f_{p}}P^{\prime}.$

Therefore,

$\displaystyle f^{*}(A^{\prime},a^{\prime})(X)$
$\displaystyle=(A^{\prime},a^{\prime})(X,0,0)$
$\displaystyle=(\tilde{A}(X),a^{\prime}(X))$
$\displaystyle=(A(X),a^{\prime}(X))$

by the definition of $A^{\prime}$ in terms of $\tilde{A}$ and $\tilde{A}$ in
terms of $A$ and also $a^{\prime}(X)=\tilde{a}(\tilde{\pi}_{*}X)=a(X).$ On the
other hand, suppose we had the bundles $\widetilde{P}\to Y\to M$ with
connections $\tilde{A}$ and $\tilde{a}$ and constructed $P\to M$ with
connection $(A,a)$ and Higgs field $\Phi(f)=\tilde{A}(\partial f).$ Then we
would like to show that if we construct the bundles $\widetilde{P}\to Y\to M$
with connections $\tilde{A}^{\prime}$ and $\tilde{a}^{\prime},$ we have
$\tilde{A}^{\prime}=f^{*}\tilde{A}$ where
$f\colon\widetilde{P}^{\prime}\xrightarrow{\sim}\widetilde{P}$ is the
isomorphism given by $[f,g,\theta]\mapsto f(\theta)g.$ Note that at the point
$[p,g,\theta]$ we have

$\displaystyle f_{*}(X,g\zeta,x_{\theta})$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}\gamma_{X(\theta+tx)}(t)g\exp(t\zeta)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}\gamma_{X(\theta)}(t)g+\partial\gamma_{X(\theta)}(0)xg+\gamma_{X(\theta)}(0)g\zeta$
$\displaystyle=X(\theta)g+\partial p(\theta)xg+p(\theta)g\zeta$

and therefore,

$\displaystyle f^{*}\tilde{A}(X,g\zeta,x_{\theta})$
$\displaystyle=\tilde{A}(X(\theta)g+\partial p(\theta)xg+p(\theta)g\zeta)$
$\displaystyle=\tilde{A}(X(\theta)g)+x\tilde{A}(\partial p(\theta)g)+\zeta$
$\displaystyle=ad(g^{-1})\tilde{A}(X(\theta))+x\,ad(g^{-1})\tilde{A}(\partial
p(\theta))+\zeta$

while for $\tilde{A}^{\prime}$ we have

$\displaystyle\tilde{A}^{\prime}(X,g\zeta,x_{\theta})$
$\displaystyle=ad(g^{-1})A(X)+\zeta+ad(g^{-1})\Phi(p)(a(X)+x)$
$\displaystyle=ad(g^{-1})\left(\tilde{A}(X)-a(X)\tilde{A}(\partial
p)\right)+\zeta+ad(g^{-1})\tilde{A}(\partial p)(a(X)+x)$
$\displaystyle=f^{*}\tilde{A}(X,g\zeta,x_{\theta}).$

Thus we have proven the analogue of Proposition 2.5.2

######  Proposition 4.2.2.

The correspondence from Proposition 4.2.1 extends to a bijection between
$G$-bundles with connection over $S^{1}$-bundles with connection and
$LG\rtimes S^{1}$-bundles with connection and Higgs field.

####  4.2.2 The string class and the first Pontrjagyn class

Now that we have extended the correspondence from section 2.5, we are in a
position to extend the result concerning the string class and the Pontrjagyn
class (Theorem 2.5.3). Recall that Theorem 2.5.3 extended Killingback’s result
to a general $LG$-bundle $P\to M$ by relating the string class of $P$ to the
first Pontrjagyn class of the corresponding $G$-bundle $\widetilde{P}\to
M\times S^{1}.$ In particular, the string class of $P$ is given by integrating
$p_{1}(\widetilde{P})$ over the circle. We would like now to extend this
further to the case where $P\to M$ is an $LG\rtimes S^{1}$-bundle and
$\widetilde{P}$ is the corresponding $G$-bundle over a circle bundle $Y$ over
$M.$ In this case we find that the string class is given by integrating the
first Pontrjagyn class of $\widetilde{P}$ over the fibre of the circle bundle
$Y$. In particular, we have the following theorem

######  Theorem 4.2.3.

Let $P\to M$ be a principal $LG\rtimes S^{1}$-bundle and $\widetilde{P}\to
Y\to M$ be the corresponding $G$-bundle over an $S^{1}$-bundle. Then the
string class of $P$ is given by the integration over the fibre of the first
Pontrjagyn class of $\widetilde{P}.$ That is,

$s(P)=\int_{S^{1}}p_{1}(\widetilde{P}).$

###### Proof.

We prove this in analogy with the proof of Theorem 2.5.3, that is, by
calculating the integral of the first Pontrjagyn class of $\widetilde{P}.$

Recall that the first Pontrjagyn class is given by

$p_{1}=-\frac{1}{8\pi^{2}}\langle\tilde{F},\tilde{F}\rangle,$

where $\tilde{F}=d\tilde{A}+\tfrac{1}{2}[\tilde{A},\tilde{A}]$ is the
curvature of the connection $\tilde{A}$ corresponding to the pair $(A,\Phi)$
on $P.$ We have

$\tilde{F}=d(ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta))\\\
+\tfrac{1}{2}[ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta),ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta)]\\\
\phantom{\tilde{F}}=d(ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta))\\\
+\tfrac{1}{2}[ad(g^{-1})A,ad(g^{-1})A]+[ad(g^{-1})A,\Theta]+[ad(g^{-1})A,ad(g^{-1})\Phi(a+d\theta)]\\\
+\tfrac{1}{2}[\Theta,\Theta]+[\Theta,ad(g^{-1})\Phi(a+d\theta)]+\tfrac{1}{2}[ad(g^{-1})\Phi(a+d\theta),ad(g^{-1})\Phi(a+d\theta)].$

To calculate $d(ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta))$ we use

$d(ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta))((X,g\xi,x_{\theta}),(Y,g\zeta,y_{\theta}))\\\
\phantom{d(ad}=\tfrac{1}{2}\left\\{(X,g\xi,x_{\theta})(ad(g^{-1})A(Y)_{\theta})-(Y,g\zeta,y_{\theta})(ad(g^{-1})A(Y)_{\theta})\right.\\\
\left.-ad(g^{-1})A([X,Y])_{\theta}\right\\}\\\ +d\Theta\\\
\phantom{d(ad}+\tfrac{1}{2}\left\\{(X,g\xi,x_{\theta})(ad(g^{-1})(a(Y)+y)\Phi(p)_{\theta})\right.\\\
\left.-(Y,g\zeta,y_{\theta})(ad(g^{-1})(a(X)+x)\Phi(p)_{\theta})-ad(g^{-1})[x,y]\Phi(p)_{\theta}\right\\},\\\
$

for tangent vectors $(X,g\xi,x_{\theta})$ and $(Y,g\zeta,y_{\theta})$ at the
point $[p,g,\theta]\in\widetilde{P}.$ For the first term, calculate

$\displaystyle(X,g$ $\displaystyle\xi,x_{\theta})(ad(g^{-1})A(Y)_{\theta})$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}(1-t\xi)g^{-1}A_{\gamma_{p}(t)}(Y)_{(\theta+tx)}g(1+t\xi)$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}g^{-1}A_{\gamma_{p}(t)}(Y)_{(\theta+tx)}g-t\xi
g^{-1}A_{\gamma_{p}(t)}(Y)_{(\theta+tx)}g+g^{-1}A_{\gamma_{p}(t)}(Y)_{(\theta+tx)}gt\xi$
$\displaystyle=\frac{d}{dt}\bigg{|}_{0}g^{-1}A_{\gamma_{p}(t)}(Y)_{\theta}g+g^{-1}\partial
A(Y)_{\theta}xg-\xi g^{-1}A(Y)_{\theta}g+g^{-1}A(Y)_{\theta}g\xi.$

Combining this with the other terms for the first derivative above, we have

$d(ad(g^{-1})A)=ad(g^{-1})dA-ad(g^{-1})\partial A\wedge
d\theta-[\Theta,ad(g^{-1})A].$

For the last term, calculate

$(X,g\xi,x_{1\theta})(ad(g^{-1})(a(Y)+y)\Phi(p)_{\theta})\\\
=\frac{d}{dt}\bigg{|}_{0}(1-t\xi)g^{-1}(a_{\gamma_{p}(t)}(Y)+y)\Phi(\gamma_{p}(t))_{(\theta+tx)}g(1+t\xi)\\\
=\frac{d}{dt}\bigg{|}_{0}g^{-1}(a_{\gamma_{p}(t)}(Y)+y)\Phi(\gamma_{p}(t))_{(\theta+tx)}g\\\
-t\xi g^{-1}(a_{\gamma_{p}(t)}(Y)+y)\Phi(\gamma_{p}(t))_{(\theta+tx)}g\\\
+g^{-1}(a_{\gamma_{p}(t)}(Y)+y)\Phi(\gamma_{p}(t))_{(\theta+tx)}gt\xi\\\
=\frac{d}{dt}\bigg{|}_{0}g^{-1}(a_{\gamma_{p}(t)}(Y)+y)\Phi(p)_{\theta}g+\frac{d}{dt}_{|_{0}}g^{-1}(a(Y)+y)\Phi(\gamma_{p}(t))_{\theta}g\\\
+g^{-1}(a(Y)+y)\partial\Phi(p)_{\theta}xg-\xi
g^{-1}(a(Y)+y)\Phi(p)_{\theta}g\\\ +g^{-1}(a(Y)+y)\Phi(p)_{\theta}g\xi.\\\ $

Subtracting $(Y,g\zeta,y_{\theta})(ad(g^{-1})(a(X)+x)\Phi(p)_{\theta})$ from
this gives

$d(ad(g^{-1})\Phi(a+d\theta))=ad(g^{-1})f\Phi+ad(g^{-1})d\Phi\wedge(a+d\theta)\\\
-[\Theta,ad(g^{-1})(a+d\theta)\Phi]-ad(g^{-1})a\partial\Phi\wedge d\theta.$

We also have $d\Theta=-\tfrac{1}{2}[\Theta,\Theta].$ Therefore, the curvature
of $\widetilde{P}$ is given by

$\tilde{F}=d(ad(g^{-1})A+\Theta+ad(g^{-1})\Phi(a+d\theta))\\\
+\tfrac{1}{2}[ad(g^{-1})A,ad(g^{-1})A]+[ad(g^{-1})A,\Theta]+[ad(g^{-1})A,ad(g^{-1})\Phi(a+d\theta)]\\\
+\tfrac{1}{2}[\Theta,\Theta]+[\Theta,ad(g^{-1})\Phi(a+d\theta)]+\tfrac{1}{2}[ad(g^{-1})\Phi(a+d\theta),ad(g^{-1})\Phi(a+d\theta)]\\\
\phantom{\tilde{F}}=ad(g^{-1})dA-ad(g^{-1})\partial A\wedge
d\theta+ad(g^{-1})f\Phi\\\
+ad(g^{-1})d\Phi\wedge(a+d\theta)-ad(g^{-1})a\partial\Phi\wedge d\theta\\\
+\tfrac{1}{2}[ad(g^{-1})A,ad(g^{-1})A]+[ad(g^{-1})A,ad(g^{-1})\Phi(a+d\theta)].\\\
$

That is,

$\tilde{F}=ad(g^{-1})\left(F+f\Phi+\nabla\Phi\wedge(a+d\theta)\right).$

So the first Pontrjagyn class is

$p_{1}=-\frac{1}{8\pi^{2}}\langle\tilde{F},\tilde{F}\rangle\\\
\phantom{p_{1}}=-\frac{1}{8\pi^{2}}\left\langle
F+f\Phi+\nabla\Phi\wedge(a+d\theta),F+f\Phi+\nabla\Phi\wedge(a+d\theta)\right\rangle\\\
\phantom{p_{1}}=-\frac{1}{8\pi^{2}}\Big{(}\left\langle
F+f\Phi,F+f\Phi\right\rangle-2\left\langle
F+f\Phi,\nabla\Phi\wedge(a+d\theta)\right\rangle\\\
-\left\langle\nabla\Phi\wedge(a+d\theta),\nabla\Phi\wedge(a+d\theta)\right\rangle\Big{)}\\\
\phantom{p_{1}}=-\frac{1}{8\pi^{2}}\Big{(}\left\langle
F+f\Phi,F+f\Phi\right\rangle-2\left\langle F+f\Phi,\nabla\Phi\wedge
a\right\rangle-2\left\langle F+f\Phi,\nabla\Phi\right\rangle
d\theta\Big{)}.\\\ $

Thus, integrating $p_{1}$ over the fibre, we get

$-\frac{1}{4\pi^{2}}\int_{S^{1}}\langle F+f\Phi,\nabla\Phi\rangle d\theta,$

which is the expression from Theorem 4.1.3.

∎

###  4.3 String structures for $LG\rtimes\operatorname{Diff}(S^{1})$-bundles

So far in this chapter we have generalised the results from [35] to include
the possibility of rotating loops. That is, we have worked with the semi-
direct product $LG\rtimes S^{1}.$ We would like to conclude now with a brief
outline of one way in which the results we have seen regarding $LG\rtimes
S^{1}$ lead us to information about a more general situation. Namely, we shall
consider the problem of lifting a bundle whose structure group is the semi-
direct product $LG\rtimes\operatorname{Diff}(S^{1}).$ That is, we shall allow
an action of the orientation preserving diffeomorphisms of the circle on the
loops in $LG.$

The group $\operatorname{Diff}(S^{1})$ has a well known central extension. In
particular, the Lie algebra of this extension is the Virasoro algebra (see for
example [29]). In this section, we would like to consider the central
extension of the semi-direct product above

$U(1)\to\widehat{LG\rtimes\operatorname{Diff}(S^{1})}\to
LG\rtimes\operatorname{Diff}(S^{1}).$

Thus far, we have seen that principal $LG$-bundles over $M$ correspond to
principal $G$-bundles over $M\times S^{1}$ (via the caloron correspondence)
and in the previous section we showed that isomorphism classes of principal
$LG\rtimes S^{1}$-bundles are in bijective correspondence with isomorphism
classes of principal $G$-bundles over principal $S^{1}$-bundles. If instead we
considered a principal $G$-bundle over a general $S^{1}$ fibre bundle444Such
bundles have structure group $\operatorname{Diff}(S^{1})$ and give rise to
principal $\operatorname{Diff}(S^{1})$-bundles in a natural way. we would find
that these bundles correspond to principal
$LG\rtimes\operatorname{Diff}(S^{1})$-bundles.

Now let $R\to M$ be a principal $LG\rtimes\operatorname{Diff}(S^{1})$-bundle.
We are interested in finding the obstruction to lifting this bundle to an
$\widehat{LG\rtimes\operatorname{Diff}(S^{1})}$-bundle $\widehat{R}.$ The
following result, due to Smale, gives us a way of using our previous results
to solve this problem. Namely, we have

######  Theorem 4.3.1 ([42]).

$\operatorname{Diff}(S^{1})$ is homotopy equivalent to $S^{1}.$

This means that if $Y\to M$ is a $\operatorname{Diff}(S^{1})$-bundle then its
transition functions can be chosen to be valued in $S^{1}$ and so $Y$ actually
admits an action of the circle (by identifying $Y$ locally with $S^{1}\times
U$ (for some open subset $U\subseteq M$) and rotating the $S^{1}$ factor).
This makes $Y$ into a principal $S^{1}$-bundle. In particular, then, if we
have a $G$-bundle $\widetilde{P}$ over an $S^{1}$ fibre bundle $Y\to M$ we can
replace the $LG\rtimes\operatorname{Diff}(S^{1})$-bundle in question with an
$LG\rtimes S^{1}$-bundle. That is, $R$ has a reduction to a principal
$LG\rtimes S^{1}$-bundle $P,$ so $R=P\times_{LG\rtimes
S^{1}}LG\rtimes\operatorname{Diff}(S^{1}).$ We can thus give the lift of $R$
in terms of the central extension of $LG\rtimes\operatorname{Diff}(S^{1})$ and
the lift $\widehat{P}$ of $P.$ In particular, we have a bundle map

$\widehat{P}\times_{\widehat{LG\rtimes
S^{1}}}\widehat{LG\rtimes\operatorname{Diff}(S^{1})}\to P\times_{LG\rtimes
S^{1}}LG\rtimes\operatorname{Diff}(S^{1})$

given by

$[\hat{p},\widehat{(\gamma,\varphi)}]\mapsto[p,(\gamma,\varphi)],$

where $\hat{p}$ is a lift of $p$ to $\widehat{P}$ and
$\widehat{(\gamma,\varphi)}$ is a lift of $(\gamma,\varphi)$ to the central
extension of $LG\rtimes\operatorname{Diff}(S^{1}).$ This map commutes with the
homomorphism $\widehat{LG\rtimes\operatorname{Diff}(S^{1})}\to
LG\rtimes\operatorname{Diff}(S^{1})$ and so
$\widehat{P}\times_{\widehat{LG\rtimes
S^{1}}}\widehat{LG\rtimes\operatorname{Diff}(S^{1})}$ is a lift of $R.$

##  Appendix A Infinite-dimensional manifolds and Lie groups

In this thesis we have largely been concerned with the loop group of a compact
group. This is an example of an infinite-dimensional Lie group – specifically,
it is a _Fréchet_ Lie group. In this Appendix we collect some of the basic
results on Fréchet manifolds and Lie groups. We follow closely the expositions
presented in [19], [31] and [39]

###  A.1 Fréchet spaces

We will begin with some basic definitions and examples of the sorts of spaces
we shall be dealing with. An infinite-dimensional manifold, like any manifold,
is a topological space modelled on some sort of Euclidean space. In the case
we are considering, this is a locally convex topological vector space called a
_Fréchet_ space.

######  Definition A.1.1.

A _Fréchet_ space is a complete metrisable Hausdorff locally convex
topological vector space, where by _locally convex_ we mean a space whose
topology is generated from some family of seminorms.111An equivalent
definition of local convexity for a topological vector space is that every
neighbourhood of $0$ contains a neighbourhood which is convex. This is the
definition used in [31].

Perhaps the most immediate example of a Fréchet space is given by any Banach
space. In general, however, there are examples of Fréchet spaces which are not
Banach spaces. The particular example we will consider is the space of all
smooth maps222More generally, the space of smooth sections of a vector bundle
over a compact manifold is also a Fréchet space. We shall restrict our
interest however, to the case of a trivial bundle (that is, the space of all
maps as above) since this covers the case we are really interested in – the
Lie algebra of the loop group, $\operatorname{Map}(S^{1},{\mathfrak{g}}).$
from a compact manifold $X$ into a vector space $V,$ that is, the space
$\operatorname{Map}(X,V).$ We define the topology on this space in terms of a
collection of neighbourhoods of the zero map. (Since this is a topological
vector space this will give the topology completely.) To do this, choose a
small neighbourhood $E$ of $0\in V.$ Then consider an open coordinate chart
$U\subseteq X$ with local coordinates $x_{1},\ldots,x_{m}$ and a compact set
$K\subseteq U.$ We define a family of sub-basic neighbourhoods (for each
choice of coordinate chart, compact set, neighbourhood of $0\in V$ and non-
negative integer $n$)

$N=\\{f\colon X\to V\mid\partial^{k}f/\partial x_{i_{1}}\ldots x_{i_{k}}\in E\
\forall\,x\in K,0\leq k\leq n,i_{j}\in\\{1,\ldots,m\\}\\}.$

Finite intersections of sets of this form give the basic neighbourhoods for
the topology on $\operatorname{Map}(X,V).$

The above example is important for our purposes since the special case of maps
from the circle into the Lie algebra of a compact group $G$ will be the
Fréchet space on which the loop group $LG$ is modelled.

###  A.2 Groups of maps

Now that we have seen an example of a Fréchet space, we can give an example of
an infinite-dimensional manifold modelled on this space. This is the space
$\operatorname{Map}(X,G)$ of smooth maps from a compact manifold $X$ into a
compact Lie group $G$ and it is in fact an example of an infinite-dimensional
Lie group.

To define the coordinate charts for this manifold consider an open
neighbourhood $U$ of the identity in $G$. Using the exponential map, this is
homeomorphic to an open neighbourhood of the identity in ${\mathfrak{g}},$ say
$\tilde{U}.$ The set $\tilde{\mathcal{U}}:=\operatorname{Map}(X,\tilde{U})$ is
then an open neighbourhood of the identity in
$\operatorname{Map}(X,{\mathfrak{g}})$ and an atlas for
$\operatorname{Map}(X,G)$ is given by the open sets $\mathcal{U}f$ (where
$\mathcal{U}:=\operatorname{Map}(X,U)$), which are also homeomorphic to
$\tilde{\mathcal{U}}.$ The case where $X$ is the circle is the loop group
$LG.$

Note that there is a slightly more general example given by taking sections of
a fibre bundle over $X.$ Recall from the previous section that sections of a
vector bundle form a Fréchet space. Given a fibre bundle $Y\xrightarrow{\pi}X$
we can associate to any section $f\colon X\to Y$ a vector bundle over $X,$
called the vertical tangent bundle to $f$ and denoted $T_{\text{vert}}Y_{f},$
whose fibre at $x\in X$ is given by all vertical tangent vectors to $Y$ at
$f(x).$ That is, $T_{\text{vert}}Y_{f}=\\{V\in T_{f(x)}Y\mid\pi_{*}V=0\\}.$
Then the sections of $T_{\text{vert}}Y_{f}\to X$ form a Fréchet space and
there is a diffeomorphism from a neighbourhood of the zero section to a
neighbourhood of the image of $f$ in $Y$ which serves as a coordinate chart.

###  A.3 The path fibration

In chapter 3 we made extensive use of a particular $\Omega G$-bundle called
the _path fibration_. This is a model for the universal $\Omega G$-bundle. In
this section we shall explain why this is in fact a locally trivial $\Omega
G$-bundle. Recall that the total space of the path fibration is defined as

$PG=\\{p:{\mathbb{R}}\to G\mid p(0)=1\text{ and }p^{-1}dp\text{ is
periodic}\\}.$

We can equivalently view this as the space of connections on the trivial
$G$-bundle over the circle, since if $p$ is a path in $G$ as above then
$p^{-1}dp$ is a ${\mathfrak{g}}$-valued 1-form on $S^{1}$ and conversely, each
connection form $A$ on the trivial $G$-bundle over $S^{1}$ uniquely determines
a periodic path by solving the ordinary differential equation $A=p^{-1}dp$
subject to the initial condition $p(0)=1.$ This means that $PG$ is
contractible. Note that when viewed as the space of connections $\Omega G$
acts freely on the right of this space by gauge transformations. Notice also
that if $p$ and $q$ are two paths in the same fibre of the projection
$PG\xrightarrow{\pi}G$ (so $p(2\pi)=q(2\pi)$) then $p^{-1}q$ is a smooth based
loop, since if $f(t)=(p^{-1}q)(t+2\pi)$ then $f$ satisfies the same
differential equation as $p^{-1}q$ and $f(0)=1$ so $f=p^{-1}q$ and thus
$p^{-1}q$ is periodic. This means that $q=p\gamma$ for some $\gamma\in\Omega
G$ and so $PG/\Omega G=G.$

For the local triviality of this bundle, consider an open neighbourhood $U$ of
the identity in $G.$ We can define a map

$U\times\Omega G\xrightarrow{\sim}\pi^{-1}(U);\quad(g,\gamma)\mapsto p,$

where $p(t)=\exp(t\xi)\gamma(t)$ and $\exp(2\pi\xi)=g.$ The inverse of this
map is given by

$p\mapsto(\pi(p),\exp(t\pi(p))^{-1}p).$

This gives us a trivialisation near the identity. To extend this to a local
trivialisation for the entire bundle we consider the open cover $\\{Uh\\}$ for
$h\in G.$ Let $\tilde{h}$ be a path ending at $h$ (that is,
$\pi(\tilde{h})=h$). Then the maps

$Uh\times\Omega G\xrightarrow{\sim}\pi^{-1}(Uh);\quad(g,\gamma)\mapsto p_{h},$

for $p_{h}(t)=\tilde{h}(t)\exp(t\xi)\gamma(t),$ give a local trivialisation
for the path fibration. So we have that the path fibration is a model for the
universal $\Omega G$-bundle.

##  Appendix B Classification of semi-direct product bundles

###  B.1 Classification of semi-direct product bundles

In section 3.3 we gave a model for the universal $L^{\vee}G$-bundle (where
$L^{\vee}G$ is the group of smooth maps $[0,2\pi]\to G$ with coincident
endpoints) by utilising its description as the semi-direct product
$\Omega^{\vee}G\rtimes G.$ Following those ideas we can actually give a
classification theory for general $K\rtimes H$-bundles.

Suppose $K$ and $H$ are Lie groups and we have an action $\varphi\colon
H\to\operatorname{Aut}(K).$ Then we can form the semi-direct product $K\rtimes
H,$ where the multiplication is defined by

$(k_{1},h_{1})(k_{2},h_{2})=(k_{1}\varphi_{h_{1}}(k_{2}),h_{1}h_{2}).$

We can give a model for the classifying space $E(K\rtimes H)$ as follows.
Consider the space $EK\times EH.$ This is contractible, since both $EK$ and
$EH$ are. Suppose we can find a left action of $H$ on $EK.$ That is, some
$\tilde{\varphi}\colon H\to\operatorname{Diff}(EK)$ such that
$\tilde{\varphi}_{h_{1}}\tilde{\varphi}_{h_{2}}=\tilde{\varphi}_{h_{1}h_{2}}.$
Suppose also that this action satisfies

$\tilde{\varphi}_{h}(xk)=\tilde{\varphi}_{h}(x)\varphi_{h}(k)$

for all $x\in EK.$ Then we can define a right action of $K\rtimes H$ on
$EK\times EH$ by

$(x,y)(k,h)=(\tilde{\varphi}_{h^{-1}}(xk),yh),$

where $(x,y)\in EK\times EH.$ This is clearly a right action since

$\displaystyle(\tilde{\varphi}_{h_{1}^{-1}}(xk_{1}),yh_{1})(k_{2},h_{2})$
$\displaystyle=(\tilde{\varphi}_{h_{2}^{-1}}(\tilde{\varphi}_{h_{1}^{-1}}(xk_{1})k_{2}),yh_{1}h_{2})$
$\displaystyle=(\tilde{\varphi}_{(h_{1}h_{2})^{-1}}(xk_{1}\varphi_{h_{1}}(k_{2})),yh_{1}h_{2})$
$\displaystyle=(x,y)(k_{1}\varphi_{h_{1}}(k_{2}),h_{1}h_{2}).$

It is also free and transitive on fibres and so

$\textstyle{EK\times
EH\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(EK\times
EH)/(K\rtimes H)}$

is a model for the universal bundle. To see that $\tilde{\varphi}$ exists,
consider the following construction of $EK$ [41] (see also [14]). Let
$\Delta^{n}$ be the standard $n$-simplex in ${\mathbb{R}}^{n+1}.$ That is,

$\Delta^{n}=\\{(t_{0},\ldots,t_{n})\mid t_{i}\geq 0,\textstyle\sum
t_{i}=1\\}.$

Then

$EK=\bigsqcup_{n\geq 0}\Delta^{n}\times K^{n+1}/\sim,$

where we make the identifications

$\left((t_{0},\ldots,t_{i-1},0,t_{i+1},\ldots,t_{n}),(k_{0},\ldots,k_{n})\right)\sim\left((t_{0},\ldots,t_{n}),(k_{0},\ldots,k_{i-1},1,k_{i+1},\ldots,k_{n})\right).$

Equivalently, we can think of $EK$ as the set of formal linear combinations of
elements of $K:$

$EK=\left\\{\textstyle\sum t_{i}k_{i}\mid t_{i}\geq 0,\textstyle\sum
t_{i}=1,k_{i}\in K\right\\}$

where in any given sum, only finitely many of the $t_{i}$’s are non-zero. Then
$\tilde{\varphi}$ is given by

$\tilde{\varphi}_{h}\left(\textstyle\sum t_{i}k_{i}\right)=\textstyle\sum
t_{i}\varphi_{h}(k_{i}).$

Using this construction, we can also write down a classifying map for any
$K\rtimes H$-bundle $P\xrightarrow{\pi}M.$ For this we will need a
correspondence between these bundles and certain pairs of $K$-bundles and
$H$-bundles. Let us briefly outline this correspondence now. First note that
there is a homomorphism $K\rtimes H\to H$ and so we can form the associated
$H$-bundle $P\times_{K\rtimes H}H\xrightarrow{\pi_{H}}M,$ where
$[p,h]=[p(k^{\prime},h^{\prime}),h^{\prime-1}h],\,$
$[p,h]h^{\prime}=[phh^{\prime}]$ and $\pi_{H}([p,h])=\pi(p).$ Further, there’s
a free action of $K$ on $P$ that identifies $P\times_{K\rtimes H}H$ with
$P/K.$ Namely, $pk=p(k,1).$ Then we have that
$P\xrightarrow{\pi_{K}}P\times_{K\rtimes H}H$ is a principal $K$-bundle.111For
the proof of the local triviality of this bundle, see [23], Proposition 5.5, p
57. Thus, we have constructed a $K$-bundle over an $H$-bundle out of the
$K\rtimes H$-bundle $P$ that we started with. In addition, we have an action
of $H$ on $P$ that covers the $H$ action on $P/K.$ That is, define $ph=p(1,h)$
and then $\pi_{K}(ph)=[p(1,h),1]=[p,h]=[p,1]h=\pi_{K}(p)h.$ This $H$ action
also has the property that
$(ph)k=p(1,h)(k,1)=p(\varphi_{h}(k),h)=(p\varphi_{h}(k))h.$ Therefore, we have
constructed a $K$-bundle with a twisted $H$-equivariant action as above over
an $H$-bundle:

$\textstyle{P\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{K,H}$$\textstyle{P/K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H}$$\textstyle{M}$

In fact, this construction is invertible. That is, given a $K$-bundle over an
$H$-bundle that satisfies the properties above, we can construct a $K\rtimes
H$-bundle. Suppose, then, that we have two Lie groups $K$ and $H$ with an
action $\varphi\colon H\to\operatorname{Aut}(K)$ as above. Suppose also that
we have a principal $K$-bundle $P\xrightarrow{\pi_{K}}P/K$ and a principal
$H$-bundle $P/K\xrightarrow{\pi_{H}}M$ and that there’s an $H$ action on $P$
covering that on $P/K$ and such that $(ph)k=(p\varphi_{h}(k))h.$ We can define
an action of $K\rtimes H$ on $P$ by $p(k,h)=(pk)h.$ This is a right action
since

$\displaystyle p(k_{1},h_{1})(k_{2},h_{2})$
$\displaystyle=(((pk_{1})h_{1})k_{2})h_{2}$
$\displaystyle=(((pk_{1})\varphi_{h_{1}}(k_{2}))h_{1})h_{2}$
$\displaystyle=p(k_{1}\varphi_{h_{1}}(k_{2}),h_{1}h_{2}).$

It is a free action, for suppose that $p(k,h)=p.$ Then $(pk)h=p$ and so
$\pi_{K}((pk)h)=\pi_{K}(p).$ But $\pi_{K}((pk)h)=\pi_{K}(pk)h$ and
$\pi_{K}(pk)=\pi_{K}(p),$ so we have $\pi_{K}(p)h=\pi_{K}(p)$ and therefore
$h=1$ since the $H$ action is free. But if $h=1$ we have that $pk=p$ and so
$k=1.$ We also have that $P/(K\rtimes H)=(P/K)/H=M.$ To see that $P\to M$ is
locally trivial, consider an open set $U\subset M$ over which $P/K$ is
trivial. Then there exists a section $s\colon U\to P/K.$ Since $U$ is
contractible, the pull-back $s^{*}P$ over $U$ is trivial and so there exists a
section $s^{\prime}\colon U\to s^{*}P.$ But a choice of section
$s^{\prime}\colon U\to s^{*}P$ is equivalent to a map $\sigma\colon U\to P$
such that $\pi_{K}(\sigma(x))=s(x).$ That is, such that $\pi(\sigma(x))=x.$ So
$\sigma$ is a local section of $P\to M.$ Therefore, we have that $P\to M$ is a