$W=\pi_{2}^{-1}(K)\bigcap\underline{f}^{-1}(\omega)$ where $K$ is an open set
in $Y$ and $\omega$ is an open set in $E$. Now since $E$ is a bundle there
exists $\epsilon>0$ and $V$ neighbourhood of $f^{*}(g)$ such that
$V\subseteq_{\epsilon}\omega$. Now using the fact that by definition
$\underline{f}$ is isometric on each fibre, we have
$g\in\pi_{2}^{-1}(K)\bigcap\underline{f}^{-1}(V)\subseteq_{\epsilon}W$. So the
pullback of a bundle in $\mathsf{Top}$ is again a bundle.

Before continuing let us describe the morphisms in the fibred category in the
case $\mathcal{M}=\mathrm{k}\text{-}\mathrm{CompMet}$: suppose that
$E\xrightarrow[]{\pi_{1}}X$ and $E^{{}^{\prime}}\xrightarrow[]{\pi_{2}}Y$ are
two bundles a morphism from $E$ to $E^{{}^{\prime}}$ is a pair $(f,h)$ where
is a continuous map from $X$ to $Y$ and $h$ is a map in
$\mathrm{Bun}(\mathcal{M},X)$, equivalently such morphism can be defined to be
a pair $(f,h^{{}^{\prime}})$ where $h^{{}^{\prime}}$ is a continuous map from
$E$ to $E^{{}^{\prime}}$ such that the following diagram commutes :

${E}$${E^{{}^{\prime}}}$${X}$${Y}$$\scriptstyle{\pi_{1}}$$\scriptstyle{h^{{}^{\prime}}}$$\scriptstyle{\pi_{2}}$$\scriptstyle{f}$

And such that for every $y$ $h^{{}^{\prime}}|_{\pi_{1}^{-1}(x)}$ is a
contraction.

Now we want to extend the same construction when
$\mathcal{M}=\mathrm{Comp}_{\mathfrak{L}}$ and also the case
$\mathcal{M}=\mathrm{Comp}_{\mathfrak{L},\mathbb{T}}$. So suppose we are given
a continuous first order signature $\mathrm{Comp}_{\mathfrak{L}}$ and a
theory(family of axioms) $\mathbb{T}$. For each sort $S$ we already know that
the pullback of $E_{S}$ is going to be a bundle of complete metric spaces
bounded by $k_{S}$, it remains to show that for every function symbol
$g^{E^{{}^{\prime}}}$ the global function defined from
$E^{{}^{\prime}}_{S_{1}}\times_{Y}\dots\times_{Y}E^{{}^{\prime}}_{S_{n}}$ to
$E^{{}^{\prime}}_{S^{{}^{\prime}}}$ is continuous and for every relation
symbol $\phi$ the global relation defined on
$E^{{}^{\prime}}_{S_{1}}\times_{Y}\dots\times_{Y}E^{{}^{\prime}}_{S_{n}}$ is
upper semi-continuous, the proof of both those facts follows exactly the proof
that the global distance function is upper semi-continuous. First starting
with a function symbol $g$:

${{E^{{}^{\prime}}_{S_{1}}\times_{Y}\dots\times_{Y}E^{{}^{\prime}}_{S_{n}}}}$${{E_{S_{1}}\times_{X}\dots
E_{S_{n}}}}$${{E^{{}^{\prime}}_{S^{{}^{\prime}}}}}$${{E_{S^{{}^{\prime}}}}}$$\scriptstyle{\langle\underline{f}_{S_{1}}\circ\pi_{2,S_{1}},\dots,\underline{f}_{S_{n}}\circ\pi_{2,S_{n}}\rangle}$$\scriptstyle{g^{E^{{}^{\prime}}}}$$\scriptstyle{g^{E}}$$\scriptstyle{\underline{f}_{S^{{}^{\prime}}}}$

The map $g^{E^{{}^{\prime}}}$ is the unique map that exists because of the
universal property of $E^{{}^{\prime}}_{S^{{}^{\prime}}}$ being a pullback in
$\mathsf{Top}$ Hence it’s continuous (here $\pi_{2,S_{i}}$ is the projection
maps of
$E^{{}^{\prime}}_{S_{1}}\times_{Y}\dots\times_{Y}E^{{}^{\prime}}_{S_{n}}$ onto
$E^{{}^{\prime}}_{S_{i}}$. Now for relation symbols, suppose that we have a
relation symbol $\phi$ then the global relation function
$\phi^{E^{{}^{\prime}}}$ for the bundle $E^{{}^{\prime}}$ is the composition
$\phi^{E}\circ\langle
f^{*}_{S_{1}}\circ\pi_{4,S_{1}},\dots,f^{*}_{S_{n}}\circ\pi_{4,S_{n}}\rangle$:

${{E^{{}^{\prime}}_{S_{1}}\times_{Y}\dots\times_{Y}E^{{}^{\prime}}_{S_{n}}}}$${{E_{S_{1}}\times_{X}\dots
E_{S_{n}}}}$${V}$$\scriptstyle{\langle\underline{f}_{S_{1}}\circ\pi_{2,S_{1}},\dots,\underline{f}_{S_{n}}\circ\pi_{2,S_{n}}\rangle}$$\scriptstyle{\phi^{E}}$

Here $V$ is a compact interval of $\mathbb{R}$ (i.e of the form $[a,b]$ where
$a,b$ are reals) equipped with the left order topology. Thus we get that
$\phi^{E^{{}^{\prime}}}$ is upper semi-continuous.

Now the fact that the construction $E\mapsto E^{{}^{\prime}}$ where
$E^{{}^{\prime}}$ is the pullback along $f:Y\rightarrow X$, is a contravariant
pseudo-functor comes from the fact that the pullback along $f\circ
f^{{}^{\prime}}$ is the pullback along $f^{{}^{\prime}}$ of the pullback along
$f$ up to natural isomorphism. ∎

We define $\mathrm{Bun}$ to be the fibreed category for this pseudofunctor
from $\mathrm{CompHaus}$ to $\mathsf{Cat}$. We are going to replace the
category $\mathrm{Bun}$ with the equivalent category
$\mathrm{Bun^{{}^{\prime}}}$, in which for every sort $S$ the bundle $E^{S}$
as a set is equal (not just isomorphic) to $\coprod_{x\in X}E^{S}_{x}$ and in
which the projection is defined by sending $(x,g)\in E$ to $x$. The reason we
did this is because this is going to force the pullback along the identity to
be just the same bundle, also it forces the pullback along $f\circ
f^{{}^{\prime}}$ to be the pullback along $f^{{}^{\prime}}$ of the pullback
along $f$ not just up to isomorphism, hence this forces the assignment
$X\mapsto\mathrm{Bun}^{{}^{\prime}}/X$ to be a functor and not just a
pseudofunctor. And we are going to rename $\mathrm{Bun}^{{}^{\prime}}$ to
$\mathrm{Bun}$ (since they are essentially the same). Now we claim the
following result:

######  Theorem 7.2.

the functors defined by $\mathcal{L}$ on each fibre extends to a Cartesian
equivalence of categories between the fibreed category (which we denoted by
$\mathrm{Bun}$) and the category $\mathsf{CompHaus}_{\mathcal{M}}$.

###### Proof.

For the category of bundles over $X$ $\mathrm{Bun}/X$ let us denote by
$\mathcal{L}_{X}$ the equivalence of categories between $\mathrm{Bun}/X$ and
$\mathrm{Left\ ultrafunctors}(X,\mathcal{M})$ and suppose that we have a
continuous function $f$ from $Y$ to $X$ we want to show that the following
diagram commutes :

${\mathrm{Bun}/Y}$${{\mathrm{Left\
ultrfunctors}(Y,\mathcal{M}))}}$${\mathrm{Bun}/X}$${{\mathrm{Left\
ultrfunctors}(X,\mathcal{M}))}}$$\scriptstyle{f^{*}}$$\scriptstyle{-\ \circ
f}$$\scriptstyle{\mathcal{L}_{Y}}$$\scriptstyle{\mathcal{L}_{X}}$

Here $f^{*}$ sends a bundle over $Y$ to the bundle over $X$ obtained by
pulling back in $\mathsf{Top}$ sortwise. Suppose that we have a left
ultrafunctor $\mathcal{F}$ from $Y$ to $\mathcal{M}$. First suppose that
$\mathcal{M}=\mathrm{k}\text{-}\mathrm{CompMet}$ :

${E^{{}^{\prime}}}$${E}$${Y}$${X}$$\scriptstyle{\pi_{2}}$$\scriptstyle{\underline{f}}$$\scriptstyle{\pi_{1}}$$\scriptstyle{f}$

The set $E^{{}^{\prime}}=\coprod_{y\in Y}E_{f(y)}$ admits two bundle
topologies with the same projection map $\pi_{2}$, the first being the
pullback topology, and the second being the topology resulting from the left
ultrafunctors $f\circ\mathcal{F}$, we now show they coincide:

Before that let us introduce a notation convention: Let $E=\coprod_{x\in
X}M_{x}$ then for any element $g\in M_{x}$, we are going to denote the element
$(x,g)\in E$ by $g^{(x)}$.

Now suppose that $\mu$ is an ultrafilter on $E^{{}^{\prime}}=\coprod_{y\in
Y}\mathcal{F}(f(y)$ with the pullback topology that converges to some point
$g^{(y)}$. First we have that $\pi_{2}\mu$ converges $y$ by definition of the
pullback topology, now suppose that
$\sigma_{f\pi_{2}\mu}(g^{(}f(y)))=(b_{x})_{x\in X}$, then we get that
$\sigma_{\pi_{2}\mu}=(b_{f(y)})_{y\in Y}$ (We are going to show this result
but after this proof).

Now notice the following $\coprod_{y\in
Y}B(b_{f(y)}),\epsilon)\in\mu\iff\coprod_{y\in Y}B(b_{x},\epsilon)$. Hence
$\mu$ converges to $g^{(}y)$ in the topology resulting from the left
ultrafunctors $f\circ\mathcal{F}$. On the other hand suppose that $\mu$
converges to $g^{(y)}$ in the topology resulting from the left ultrafunctors
$f\circ\mathcal{F}$, first we get that $\pi_{2}\mu$ converges to $\pi_{2}(g)$
Also $\pi_{1}\underline{f}\mu$ converges to $\pi_{1}\underline{f}g$ and using
$\coprod_{y\in Y}B(b_{f(y}),\epsilon)\in\mu\iff\coprod_{y\in
Y}B(b_{x},\epsilon)$ we get that $f^{*}\mu$ converges to $g^{(f(y))}$. This
implies that the ultrafilter $\mu$ converges to $g$ in the pullback topology.

Now we turn to the case where $\mathcal{M}=\mathrm{CompMet}_{\mathfrak{L}}$ or
$\mathcal{M}=\mathrm{CompMet}_{\mathfrak{L},\mathbb{T}}$, in this case we know
that for each sort the two topologies on the bundles corresponding to each
sort agree as well as the projection maps. So the family of functors
$\\{\mathcal{L}_{X}\\}_{X\in\mathcal{S}}$ (where $\mathcal{S}$ is the set of
sorts) defines an equivalence of categories between the functors composition
with $f$ of left ultrafunctors and pullback along $f$ of bundles, or in other
words defines a Cartesian equivalence of categories between $\mathrm{Bun}$ and
$\mathrm{CompHaus}_{\mathcal{M}}$. ∎

Now we turn back to showing the result that we used: Suppose that we have a
map of sets $f$ from $Y$ to $X$ where $X$ is a compact Hausdorff space and
suppose that have a left ultrafunctor $\mathcal{F}$ from $X$ to $\mathcal{M}$.
And suppose that $\mu$ is an ultrafilter on $Y$ that converges to $y$.

Our goal is to show the following diagram commutes :

${\mathcal{F}(\int_{X}xd\int_{Y}\delta_{f(y)}d\mu)}$${\int_{X}\mathcal{F}(x)df\mu}$${\int_{Y}\mathcal{F}(f(y))d\mu}$$\scriptstyle{\sigma_{f\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\Delta_{\mu,f}}$

Which by our description of the ultraproduct diagonal map(see section) means
that if $g\in\mathcal{F}(f(y))$ then $\sigma_{f\mu}(g)=(b_{y})_{y\in Y}$ then
$\sigma_{\mu}(g)=(b_{f(y)})$.

Now to show we use the following diagram:

${{\mathcal{F}(\int_{X}xd\int_{Y}\delta_{f(y)}d\mu)}}$${{\int_{X}\mathcal{F}(x)df\mu}}$${{\mathcal{F}(\int_{Y}\int_{X}xd\delta_{f(y)}d\mu)=\mathcal{F}(\int_{Y}f(y)d\mu)}}$${{\int_{Y}\mathcal{F}(\int_{X}xd\delta_{f(y)})d\mu}}$${{\int_{Y}\int_{X}\mathcal{F}(x)d\delta_{f(y)}d\mu}}$${{\int_{Y}\mathcal{F}(f(y))d\mu}}$$\scriptstyle{\sigma_{f\mu}}$$\scriptstyle{\mathrm{id}}$$\scriptstyle{\Delta_{\delta_{f(\bullet)},\mu}}$$\scriptstyle{\Delta_{f,\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\int_{Y}\sigma_{\delta_{f(y)}}d\mu}$$\scriptstyle{\int_{Y}\mathcal{F}(\epsilon_{X,f(y})d\mu=\mathrm{id}}$$\scriptstyle{\int_{Y}\epsilon_{X,f(y)}d\mu}$

The upper diagram commutes by axiom $(2)$ of [19, definition 1.4.1] and the
lower diagram commutes by axiom $(1)$ of [19, definition 1.4.1].

##  8 Examples

At this point, it is important for us to give examples of our constructions of
bundles and show that they correspond to the already existing notions of
continuous families of metric structures. But first, we need to explain how to
axiomatise some structures in continuous model theory.

###  8.1 Banach bundles

####  8.1.1 Axiomatisation of Banach spaces

The signature of Banach spaces includes a sort for each ball of radius $n$,
inclusion symbols between sorts and additional symbols for the
$\mathbb{K}$-vector space structure

  1. 1.

A constant symbol $0$ (a function symbol) with formal range $D_{1}$, this
symbol should be interpreted as the $0$ of the vector space (we can get rid of
that symbol since we can get $0$ by multiplication by $0$, strictly speaking,
we must also check whether including this symbol or not is going to affect the
continuous model-theoretic ultraproduct since we want to be able to recover
the usual definition of ultraproduct of Banach spaces or any related
structure).

  2. 2.

For each natural number $n\geq 1$ we define a sort $D_{n}$ this sort should be
interpreted as the closed ball of radius $n$.

  3. 3.

For every pair of sorts we define a function symbol $+_{n,m}$ which has formal
domain $D_{n}\times D_{m}$ and formal range $D_{n+m}$ and should be
interpreted as the addition, the modulus of continuity of this symbol is
$2\mathrm{Id}$.

  4. 4.

For every sort pair of sorts $D_{n}$ and $D_{m}$ such that $n<m$ we define a
function symbol $\iota_{n,m}$, which should be interpreted as the inclusion of
the ball $D_{n}$ inside the ball $D_{m}$, this symbol is of uniform continuity
modulus the identity function.

  5. 5.

For every sort $D_{n}$ and every $k\in\mathbb{C}$, we define a function symbol
$m_{n,k}$, this function should be interpreted as the multiplication, the
formal domain of this symbol is $S_{n}$ and the formal range is $S_{m}$ where
$m=\lceil|k|.n\rceil$, the modulus of this symbol is $k\mathrm{id}$.

  6. $6^{{}^{\prime}}$

If we want to have isometries between Banach space, we are going to add the
following relation symbols $k_{n}$ with domain $D_{n}$, which takes values in
the interval $[0,n]$, with uniform continuity modulus the identity function.
$k_{n}(x)$ should be interpreted as $n-||x||$. Note that these symbols have
not been introduced in literature before.

Now we are going to list the necessary axioms informally, here $||x||$ means
$d(x,0)$ (notice that it is possible to make the norm an additional function
symbol (sortwise) and add axioms ensuring that the distance and the norm
define the same metric):

  1. 1.

$\mathbb{K}$ vector space axioms ($\mathbb{K}$ is by default $\mathbb{C}$
unless it’s indicated to be $\mathbb{R}$)

  2. 2.

Norm axioms: which are axioms ensuring that the norm (which is defined for
each sort $||x||=d(x,0)$) is a norm.

  3. 3.

Axioms that ensure that the inclusion function is compatible with distance,
addition, and additive inverse.

  4. 4.

Axioms ensuring that each $D_{n}$ is interpreted as the ball of radius $n$,
these are $(\mathrm{Sup}_{x\in D_{1}}(||x||\mathbin{\ooalign{\hss\raise
4.30554pt\hbox{.}\hss\cr$-$}}1)$, and $\mathrm{Sup}_{x\in D_{n}}Inf_{y\in
D_{1}}(d(x,\iota_{1,n}(y))\mathbin{\ooalign{\hss\raise
4.30554pt\hbox{.}\hss\cr$-$}}(||x||\mathbin{\ooalign{\hss\raise
4.30554pt\hbox{.}\hss\cr$-$}}1))$. (what the last axiom is telling us
informally is that if some element $x$ in $B_{n}$ has norm less than or equal
to $1$, iff there exists an element $y$ in $B_{1}$ such $\iota_{1,n}(y)=x$,
see [12]).

  5. $5^{^{\prime}}$

Axiom ensuring that the new symbol $k_{n}$ is interpreted as $n-||x||$,
formally speaking this axiom should be $\mathrm{Sup}_{x\in
D_{n}}|(k_{n}(x)-(n-||x||))|$.

As stated before, axiomatising Banach spaces without this newly introduced
symbol leads to the category of Banach space with contractions, while
axiomatising Banach spaces with the newly introduced symbol $k_{n}$, will
force maps to be isometries, and hence this is going to lead to two different
notions of bundles of Banach spaces, which turned out to be already existing
in the literature.

####  8.1.2  Definition Banach bundles

This definition is the one present in [18], and we are going to be calling it
a semi-continuous bundle of Banach spaces.

We say that a triple $(E,X,\pi)$ defines a bundle of Banach space, where $E$
and $X$ are topological spaces ($X$ is usually required to be Hausdorff, in
our work we studied the case where the $X$ is compact Hausdorff) and
$\pi:E\rightarrow X$ is a function required to satisfy the following
conditions:

  1. 1.

For every $x$, $\pi^{-1}(x)$ is a Banach space.

  2. 2.

$\pi$ is continuous and open.

  3. 3.

scalar multiplication from $\mathbb{K}\times E$ to $E$, addition from
$E\times_{X}E$ to $E$ are continuous.

  4. 4.

norm $||\ ||$ from $E$ to $[0,\infty)$ is upper semi-continuous (it is not
hard to see that in the presence of the other axioms, this is equivalent to
saying that the distance from $E\times_{X}E$ to $[0,\infty)$ is upper semi-
continuous)

  5. 5.

for any $x\in X$, if we call $\mathcal{N}_{x}$ the set of all open
neighbourhoods of $x$, then $\\{\coprod_{y\in
U}B(0_{y},r)\\}_{r>0,U\in\mathcal{N}_{x}}$ is neighbourhood basis at $0_{x}$.

First notice that axiom $3$ can be replaced with the following apparently
weaker axiom $3^{*}$:

  6. $3^{*}$

for each $k\in\mathbb{K}$ the function from $E$ to $E$ defined by
multiplication by $k$ is continuous, also addition from $E\times_{X}E$ to $E$
is continuous.

An unnecessary condition is imposed in the definition [18] which is requiring
the map $x\mapsto 0_{x}$ to be continuous (we can deduce this easily from the
condition $5$).

Another unnecessary condition required in [18] is the requirement that the
subspace topology agrees with the Banach space topology on each bundle. The
argument for dropping it can be found in [10, proposition 1.3] (notice that
the argument uses the fact that the norm is continuous, but this can be easily
replaced by the requirement that the norm is upper semi-continuous since the
neighbourhood filter of $0\in[0,\infty)$ is the same in the left order
topology and the usual topology), also the argument uses a different
equivalent version of axiom $5$.

In what follows a section (or local section) from $U\subseteq X$ to $E$, here
$U$ is open, means a continuous map such that $\sigma\circ f=\mathrm{Id}_{U}$,
such section is called global if $U=X$.

In [18] the definition above is called a pre-bundle, an additional condition
is imposed in [18] in order to obtain the definition of bundle: for every
$f\in E$, and for every $\epsilon>0$ there exists a local section $\gamma$
such that $||\gamma(\pi(f))-f||<\epsilon$.

A bundle for which every element has a global section that hits it is called a
full bundle in [18], another name for this property is a bundle with enough
cross sections. A good result is that every pre-bundle over a locally
paracompact space is a full bundle (so when it comes to our work in which we
studied bundles over a compact Hausdorff, every pre-bundle is a full bundle).
This is due to a result by Douady and Dal Soglio-Hérault which can be found in
the appendix of [11].

In what follows we are going to call a semi-continuous Banach bundle a triple
$(E,X,\pi)$ satisfying these five conditions.

Next we state the following theorem regarding this definition of bundles:

######  Theorem 8.1.

Let $(E,X,\pi)$ be a bundle and suppose that $\mu$ is an ultrafilter on $E$
such that $\pi\mu$ converges to $y$ and suppose that $\gamma$ is a section
such that for some $\gamma(y)=f$ then the set $\\{\coprod_{x\in
U}B(\gamma(x),r)\\}_{U\in N_{y},r>0}$ is a basis for the neighbourhood system
at $f$, here $N_{y}$ is the set of all open neighbourhoods at $y$.

###### Proof.

Take the homeomorphism from $E$ to itself defined by $g\mapsto
g+\gamma(\pi(g))$, and use axiom $5$ in the first definition. ∎

Before continuing we should note that there is an alternative way to state
axiom $5$ above:

  1. $5^{*}$

Suppose that $(b_{i})$ is a net such that $||b_{i}||\rightarrow 0$, and such
that $\pi(b_{i})\rightarrow x$ then $(b_{i})$ converges to $0_{x}$ (we can
write this axiom in ultrafilter terms as follows if $\mu$ is a ultrafilter on
$E$ such that $||\ ||\mu$ converges to $0\in[0,\infty)$ and $\pi\mu$ converges
to $x$ then $\mu$ converges to $0_{x}$).

Here we should note that when we say that $||\ ||\mu$ (or $||b_{i}||$)
converges to $0\in[0,\infty)$, we are either equipping $[0,\infty)$ with the
left order topology, or with the usual topology because we reminder the reader
that the neighbourhood filter of $0$ is the same in these two topologies. We
show that the axioms $5$ and $5^{*}$ are equivalent (in the presence of the
other four axioms):

Let $E$ be a bundle satisfying axioms
$1\mathrm{-}2\mathrm{-}3\mathrm{-}4\mathrm{-}5^{*}$. We need to check the
axiom which states that the set $\\{\coprod_{y\in
U}B(0_{x},r)\\}_{U\in\mathcal{N}_{x},r>0}$, is a neighbourhood basis for
$0_{x}$, to show that we can use the lemma 3.4, towards this let $V$ be an
open nhood of $0_{x}$ and let $\mu$ be an ultrafilter on $E$. If
$\\{\coprod_{x\in U}B(\gamma(x),r)\\}_{U\in N_{y},r>0}\\}\subseteq\mu$ this
would imply that $||\ ||\mu$ converges to $0$ and that $\pi\mu$ converges to
$x$. and hence $\mu$ converges to $0_{x}$, but this implies that $V\in\mu$,
and hence by 3.4 there exists $r>0$ and $U$ open neighbourhood of $x$ such
that $\coprod_{y\in U}B(0_{y},\epsilon)\subseteq U$.

Now suppose that we have a bundle satisfying axioms
$1\mathrm{-}2\mathrm{-}3\mathrm{-}4\mathrm{-}5$, we need to check that axiom
$5^{*}$ holds, so suppose that $\mu$ is an ultrafilter on $E$ such that $||\
||\mu$ converges to $0\in[0,\infty)$ and $\pi\mu$ converges to $x$, both these
conditions implies that for any $r>0$ and $U$ open neighbourhood of $x$
$\coprod_{y\in U}B(0_{y},r)\in\mu$ then $\mu$ converges to $0_{x}$ since
$\\{\coprod_{y\in U}B(0_{y},r)\\}_{U\in\mathcal{N}_{x}}\\}$ is a basis of the
neighbourhood filter of $x$.

There is another definition of Banach bundles given in [11]. In that
definition, the norm function is required to be continuous instead of being
just semi-continuous. And we are going to call such a bundle a continuous
Banach bundle. Note that in that definition the bundle space is required to be
Hausdorff, but this requirement can be dropped provided the base space is
Hausdorff (see [16, 16.4]), our work provides proof of this fact when the base
space is compact Hausdorff.

####  8.1.3 Relating the definition of Banach bundles to our work

##### Semi-continuous bundles

Now we should explain how to relate the concept of semi-continuous Banach
bundles as defined in [18], to the bundles of models for the continuous model
theory of Banach spaces (the classic definition not including the symbol
$k_{n}$). The idea is clear: giving a family of bundles of balls
$(E_{n})_{n\in\mathbb{n}}$ over $X$ (a bundle for which every fibre is the
ball of radius $n$ of the Banach space) which is the notion of bundles
corresponding to the continuous model theory of Banach spaces, we can
construct a bundle of Banach space as introduced by Hoffman by taking
$E=\bigcup_{n}E_{n}$ equipped with the final topology along the inclusion
maps, on the other hand, given a bundle in the definition of Hoffman we can
easily recover the bundle in our definition by defining $E_{n}=\\{f\in
E\mid||f||\leq n\\}$.

######  Theorem 8.2.

Let $X$ be a compact Hausdorff space, then there exists an equivalence of
categories of Banach bundles over $X$, and that of bundles of models of the
continuous model of Banach spaces over $X$.

###### Proof.

Let us explain further to the reader what we are trying to do, we already have
an equivalence of ultracategories between dissections of Banach spaces and the
category of Banach spaces, we want to extend this equivalence to the level of
bundles and we claim that our notion of bundles as developed in section 3
through 6 (the bundles of the continuous theory of Banach spaces, each of
which is a family of sorted bundles $(E_{n})_{n\in\mathbb{N}}$, where each
fibre is exactly the ball of radius $n$ of the Banach space, satisfying
certain axioms), and bundles as a single topological space, as defined above.

Suppose that we have a family of sorted bundles $(E_{n})_{n\in\mathbb{N}}$,
take the topological space $E=\bigcup_{n=1}^{\infty}E_{n}$ (equipped with the
final topology along the inclusion maps i.e the colimit of
$E_{1}\xhookrightarrow{}E_{2}\xhookrightarrow{}\dots E_{n}\dots$). This space
clearly satisfies the fact that the projection $\pi$ and scalar multiplication
by any $K\in\mathbb{K}$ are continuous by the universal property of the
$\bigcup_{n=1}^{\infty}E_{n}$, also the continuity of the global distance
function, and addition follows from the universal property of the final
topology, as well as the fact that directed colimits commute with pullbacks in
$\mathsf{Top}$. The fact that $\pi$ is open follows from the fact that an open
set $V$ in $E$ can be written as $V=\bigcup_{n=1}^{\infty}V\bigcap E_{n}$
hence $\pi(V_{=}\bigcap_{n=1}^{\infty}\pi(V\bigcap E_{n})$, and hence since
the restriction of $\pi$ to every sorted bundle is open then $\pi$ is open. So
we have shown that the bundle $E=\bigcup_{n\in\mathbb{N}}E_{n}$ satisfies
axioms $1\text{-}2\text{-}3^{*}\text{-}4$.

Finally, let us show that the bundle $E=\bigcup_{n\in\mathbb{N}}$ in our
definition satisfies axiom $5^{*}$. Let $\mu$ be an ultrafilter on $E$ such
that $\pi\mu$ converges to $x$, and also suppose that for any $||\ ||\mu$
converges to $0$.

We know that every bundle of the continuous theory corresponds to a left
ultrafunctor $\mathcal{F}$ from $X$ to the ultracategory $\mathsf{Ban}_{1}$
i.e Banach spaces with contractions, this can be done by regarding the
following composition:

${X}$${{\mathsf{Ban}_{\mathrm{diss}}}}$${{\mathsf{Ban}_{\mathrm{1}}}}$$\scriptstyle{(\mathcal{F}_{n})_{n\in\mathbb{N}}}$$\scriptstyle{\simeq}$

Here $\simeq$ is the equivalence of ultracategories between
$\mathsf{Ban}_{\mathrm{diss}}$, the category of dissections of Banach spaces
and $\mathsf{Ban}_{1}$ the category of Banach spaces, and
$(\mathcal{F}_{n})_{n\in\mathbb{N}}$ is the family of left ultrafunctors to
$n\text{-}\mathsf{CompMet}$ which defines a left ultrafunctor from $X$ to
$\mathsf{Ban}_{\mathrm{diss}}$, thus by construction $\mathcal{F}$ (defined on
objects by $\mathcal{F}(x)=E_{x}$) satisfies the commutativity of the
following diagram:

${{{E_{x}}_{n}}}$${{\int_{X}{E_{y}}_{n}d\mu\simeq(\int_{X}E_{y}d\mu)_{n}}}$${{E_{x}}}$${{\int_{X}E_{y}d\mu}}$$\scriptstyle{\sigma^{(n)}_{\pi\mu}}$$\scriptstyle{\sigma_{\pi\mu}}$

Now we know the topology of $E$ has $\epsilon$ thin nhood basis for arbitrary
small $\epsilon$, we can deduce that $\mu$ restricts to an ultrafilter on
$E_{1}$.Thus since $\sigma_{\pi\mu}(0_{x})=(0_{y})_{y\in X}$ (because it’s a
Banach spaces map), we can deduce by the diagram above that
$\sigma_{\pi\mu}^{(1)}=(0_{y})_{y\in X}$. Now using semi-continuity of the
norm and the fact that $||\ ||\mu$ converges to $0$, we get that
$\coprod_{x\in U}B(0_{y},\epsilon)\in\mu$ for any $U$ open neighbourhood of
$x$ and $\epsilon>0$, thus $\mu$ converges to $0_{x}$ (this follows from the
definition of topology associated to a left ultrafunctor 3.1, and we know that
every bundle of metric spaces bounded my $n$ over $X$, comes from a left
ultrafunctor from $X$ to $n\text{-}\mathsf{CpmpMet}$) thus we showed that our
definition satisfies axiom $5^{*}$ which we showed to be equivalent to axiom
$5$.

Now suppose that we have a bundle of Banach spaces in the definition above
$(E,\pi,X)$, we claim that $(E_{n})_{n\in\mathbb{N}}$, where each
$E_{n}=\\{f\in B\mid||f||\leq n\\}$ equipped with the subspace topology, is a
bundle of the continuous of Banach spaces, first of all for each $B_{n}$ the
global distance function is upper semi-continuous and the restriction of $\pi$
to each $B_{n}$ is continuous. Now let us show that for each $B_{n}$ axiom
$(3)$ of 5.1 is satisfied:

Since the base space is compact Hausdorff then the bundle $E$ has enough cross
sections. Let $W$ be an open set and let $f\in W$, we know that by 8.1 there
exists a set of the form $\coprod_{y\in U}B(\sigma(y),\epsilon)$ such that
$\coprod_{y\in U}B(\sigma(y),\epsilon)\subseteq W$. Now we get $\coprod_{y\in
U}B(\sigma(y),\epsilon/2)\subseteq_{\epsilon/2}W$ here $U$ is some open
neighbourhood of $\pi(f)$. The final thing is to justify why the sets of the
form $\coprod_{y\in U}B(\sigma(y),\epsilon)$ are open. To answer this notice
that they are the image of the sets of form $\coprod_{y\in U}B(0,\epsilon)$ by
the homeomorphism defined in the proof of 8.1, and these are open by semi-
continuity of the norm.

Now to show that $\pi|_{E_{n}}$ is open let $O$ be an open set in $E_{n}$.
Define $E_{n}^{\mathrm{o}}$ to be $\\{f\in E\mid||f||<n)$, this set is open by
upper semi-continuity of the norm. Let $x\in\pi(O)$ Take $f\in
O\bigcap\pi^{-1}(x)$. Since the subspace topology of $\pi^{-1}(x)$ agrees with
the metric topology of $\pi^{-1}(x)$ (this result follows from axiom $(5)$)
then $O\bigcap\pi^{-1}(x)$ is an open set in the metric topology of
$\pi^{-1}(x)\bigcap E_{n}$, which is the closed ball $B(0_{x},n)$ in the
Banach space $\pi^{-1}(x)$. this means that there exists a sequence of
elements $(y_{i})$ of $O$ that converges into $f$ such that $||y_{i}||<n$ for
every $i$, this means that for any $\pi(O)=\pi(O\bigcap
E_{n}^{\mathrm{o}})=\pi(O^{{}^{\prime}}\bigcap E_{n}^{\mathrm{o}})$ which is
open since $\pi$ is open.

Finally, we have that the function from $X$ to $E_{1}$ defined by $x\mapsto
0_{x}$ is continuous (as we stated before this can be deduced from axiom $5$
of the definition of Banach bundles), and for any $n,m$ the inclusion of
$B_{n}$ inside $B_{m}$ is continuous. So the collection
$(E_{n})_{n\in\mathbb{N}}$ is a bundle of structures of the language of Banach
spaces as we defined it in 5.1, where each fibre is a model of the theory of
Banach space, so this is a bundle of the continuous theory of Banach spaces as
we defined it in 6.1. So far we have shown that the nested union of every
family of sorted bundles as defined above, and vice versa. We need to check
that these two processes (which are obviously functorial) are inverses, given
a bundle of the continuous model theory $(E_{n})_{n\in\mathbb{N}}$, it is
clear that the topology of each $E_{n}$ is the subspace topology inside
$\bigcup_{n\in\mathbb{N}}E_{n}$, on the other hand, suppose that we are given
a Banach bundle $E$, we want to show that its topology is the final topology
of the colimit of
$E_{1}\xhookrightarrow{}E_{2}\xhookrightarrow{}\dots\xhookrightarrow{}E_{n}\dots$,
by the universal property of the colimit the topology of the colimit is finer
that of $E$, on the other hand let $\mu$ be a converging ultrafilter on $E$,
since $E$ has a basis of some $\epsilon$-thin nhood by 3.5, there exists $n$,
such that $B_{n}\in\mu$, which shows the colimit topology is coarser than that
of $B$ (notice that this is just a generalisation of the argument that shows
the topology of any normed space $M$ is the colimit of
$M_{1}\xhookrightarrow{}M_{2}\xhookrightarrow{}\dots\xhookrightarrow{}M_{n}\dots$).

∎

Before continuing let us state a useful lemma that also follows from the last
argument, which extends the result 3.3.

######  Lemma 8.1.

Let $X$ be a compact Hausdorff space and let $\mathcal{F}$ be a left
ultrafunctor from $X$ to $\mathsf{Ban}_{1}$, and let $E$ the corresponding
semi-continuous Banach bundle, then a set $V\subseteq E$ is open, if for every
$\mu$ ultrafilter on $X$ such that $\mu$ converges to $x\in\pi(V)$, and every
$f\in V\bigcap\pi^{-1}(x)$ if $\sigma_{\mu}(f)=(b_{x})_{x\in X}$, then there
exists $U\in\mu$ and $\epsilon>0$, such that $\coprod_{x\in
X}B(b_{x},\epsilon)\subseteq U$.

##### Continuous bundles

######  Theorem 8.3.

Continuous Bundles over $X$ are the bundles of the theory of Banach space as
defined above with the new function symbol and the new corresponding axiom.

###### Proof.

Since we added new relation symbols $k_{n}$, upper semi-continuity in the
global function corresponding to these symbols, implies lower semi-continuity
in norm on each $E_{n}$, which in term implies lower semi-continuity of the
norm on the semi-continuous bundle $E=\bigcup_{n=1}^{\infty}E_{n}$. And thus
the norm function is continuous), hence we obtain continuous Banach bundles.

∎

Now we show that the bundle space of continuous Banach bundles is always
Hausdorff, for that we first show the following theorem:

######  Theorem 8.4.

Let $E$ be a bundle of models of continuous model theory, which is a single
topological space (this could be just the simple case where we have metric
spaces bounded by $n$, or Banach spaces where we can form the bundle by taking
the union bundles of balls and equipping it with the final topology along the
inclusions), then the bundle is Hausdorff iff the maps $\sigma_{\mu}$ of the
left ultrafunctor associated with the bundle are injective.

###### Proof.

In this proof, we use the left ultrafunctor bundle equivalence, so in
particular we show the theorem above for bundles constructed using the functor
$\mathcal{L}$ from left ultrafunctors to bundles.

We start the proof as follows: let $\mu$ be an ultrafilter converging to $f$
and $g$, suppose that $\sigma_{\pi\mu}(f)=(a_{x})_{x\in X}$ and also suppose
that $\sigma_{\pi\mu}(g)=(b_{x})_{x\in X}$. We already now that for any
$\epsilon>0$ we have $\coprod_{x\in X}(b_{x},\epsilon)\in\mu$, also we have
that $\coprod_{x\in X}B(a_{x},\epsilon)\in\mu$, but this simply implies that
there exists a set $U\in\pi\mu$ such that for every $x\in U$
$d(b_{x},a_{x})<\epsilon$, thus $d((b_{x}),(a_{x}))\leq\epsilon$, and since
$\epsilon$ is arbitrary this implies that $(a_{x})=(b_{x})$ and since
$\sigma_{\mu}$ is injective we deduce that $f=g$ and hence the bundle topology
is Hausdorff. On the other hand suppose that the bundle is Hausdorff, let
$\mu$ be an ultrafilter on $X$ converging to $x$, and suppose that
$\sigma_{\mu}(f)=\sigma_{\mu}(g)$, we need to show that $f=g$, suppose
$\sigma_{\mu}(f)=(a_{x})_{x\in X}$. Take the set
$(\coprod_{x\in\mu}B(a_{x},\epsilon))_{\epsilon>0,A\in\mu}$, this is clearly a
filter basis, and thus extends to an ultrafilter, this ultrafilter converges
to $f$ and to $g$ and thus since $E$ is Hausdorff, we get that $f=g$.

∎

Now we know that in the cases of Banach spaces with isometries, the maps
$\sigma_{\mu}$ are isometries, hence injective, hence the bundle is Hausdorff.

###  8.2 Bundles of C* algebras

The signature of $C^{*}$ algebras is built on that of Banach spaces. So we
require on top Banach spaces signature , these additional symbols:

  * •

For every sort $D_{n}$ we define a function symbol $*_{n}$ from $D_{n}$ to
$D_{n}$, the modulus of this symbol is the identity function.

  * •

For every pair of sorts $D_{n},D_{m}$ we define a function symbol
$\mathrm{dot}_{n,m}$ with formal domain $D_{n}\times D_{m}$ and formal range
$D_{n.m}$, the modulus of this symbol is $(n+m)\mathrm{Id}$ (we are of course
going to be writing $xy$ instead of $\mathrm{dot}_{n,m}(x,y)$.

Of course we require $*$ operation and multiplication axioms including in the
case of $\mathrm{C}^{*}$ algebras $||x||=||x^{*}||$ (which formally stated is
an infinite family of axiom for every sort of the form: $sup_{x\in D_{n}}|\
||x||-||x^{*}|||$), the $\mathrm{C}^{*}$ identity, which can be stated as an
infinite family of axioms of the form $\mathrm{Sup}_{x\in D_{n}}(|\
||x^{*}x||-||x||^{2}|)$, and of course the fact that $(x^{*})^{*}=x$ (again
axiomatised with an infinite family of axioms), and axioms ensuring that the
algebra is a Banach algebra (for example we need:$||xy||\leq||x||.||y||$, this
can be written formally by $\mathrm{Sup}_{x\in D_{n}}\mathrm{Sup}_{y\in
D_{n}}||xy||\mathbin{\ooalign{\hss\raise
4.30554pt\hbox{.}\hss\cr$-$}}||x||.||y||$). For a detailed account of the
axioms see [12].

A semi-continuous bundle of C* algebra is a semi-continuous bundle of Banach
spaces such that every fibre is a $\mathrm{C}^{*}$ algebra, and such that the
global multiplication and $*$ maps are continuous [13, 24, 22].

This concept is equivalent to $C_{0}(X)$ algebras as defined in [5, 2, 22]
(sometimes called $C(X)$ algebra) (a $C_{0}(X)$ algebra $A$ is defined to be
an inclusion $\iota$ of $C_{0}(X)$ inside $\mathcal{Z}(\mathcal{M}(A))$ such
that $C_{0}(X)A$ is dense in $A$), a detailed account of this equivalence can
be found in [24, Appendix C] or [22], an important detail to note is that $A$
is the $\mathrm{C}^{*}$ algebra of continuous sections to the “toplogical”
bundle, so in particular the space $A$ can be used to define the left
ultrastructure on the left ultrafunctor corresponding to the bundle. Also it
should be noted that semi-continuous bundles over $X$ are equivalent to
continuous functions from $\mathrm{Prim}(A)$ to $X$ where $A$ is a
$\mathrm{C}^{*}$ algebra [24].

Similarly, we may require the global norm function to be continuous, so we can
get continuous bundles of $C^{*}$ algebras as defined in [7, 22], this turns
out to be equivalent to continuous fields of $\mathrm{C}^{*}$ algebras as
defined in [6, CH.10] (for this equivalence see [7]), and to $C_{0}(X)$
algebras satisfying that for each $a\in A$ the map $N(a)$ on $X$ defined by
$x\mapsto||a_{x}||$ is continuous, here $a_{x}$ is the image of $a$ in the
quotient $A/\mathcal{I}_{x}$, where $\mathcal{I}_{x}$ is the ideal of $A$
generated by elements $\\{\iota(f)\mid f(x)=0\\}$ [2], equivalently these are
$C_{0}(X)$ algebras satisfying that $\mathrm{Res}_{\iota}:\
\mathrm{Spec}(A)\rightarrow\mathrm{Spec}(C_{0}(X))\simeq
X:\mathrm{ker}(\sigma)\mapsto\mathrm{ker}(\bar{\sigma})\circ\iota$ is open
[22] (here $\bar{\sigma}$ is the extension of the representation $\sigma$ to
$\mathcal{M}(A)$).

These definitions fit our framework since we want every map corresponding to a
function symbol to be continuous. Of course, when dealing with continuous
$\mathrm{C}^{*}$ bundles we should add the additional relation symbols
$(k_{n})_{n\in\mathbb{N}}$ as we did with Banach spaces, in other words, semi-
continuous bundles corresponds to the usual axiomatisation of $\mathrm{C}^{*}$
algebras which has $*$ homomorphisms as morphisms, while continuous bundles
arise from the axiomatisation of $C^{*}$ algebras with the additional symbols
$(k_{n})_{n\in\mathbb{N}}$ which gives injective $*$ homomorphisms as
morphisms.

###  8.3 Bundles of Hilbert spaces

The axiomatisation of Hilbert spaces is also built upon that of Banach spaces,
there are two different ways, one should give us Hilbert spaces with
isometries and the other with contractions. If we want isometries, we add a
family of symbols for the real and imaginary part of the inner product with
specific axioms ensuring it’s an inner product, but if we want the maps of
models to be just contraction, this can be done by adding the parallelogram
law as an axiom to the axioms of Banach spaces with contractions, the
parallelogram law can be stated as the following axiom: $\mathrm{Sup}_{y\in
D}\mathrm{Sup}_{x\in D}(|\ ||x-y||^{2}+||x+y||^{2}-2||x||^{2}-||y||^{2}|)$.

A bundle of Hilbert spaces is a continuous bundle of Banach spaces, where each
fibre is a Hilbert space, more precisely We say that a triple $(E,X,\pi)$
defines a bundle of Banach space, where $E$ and $X$ are topological spaces
($X$ is usually required to be Hausdorff, in our work we studied the case
where the $X$ is compact Hausdorff) and $\pi:E\rightarrow X$ is a function
required to satisfy the following conditions:

  1. 1.

For every $x$, $\pi^{-1}(x)$ is a Hilbert space

  2. 2.

$\pi$ is continuous and open

  3. 3.

scalar multiplication from $\mathbb{K}\times E$ to $E$, addition from
$E\times_{X}E$ to $E$ are continuous

  4. 4.

norm $||\ ||$ from $E$ to $[0,\infty)$ is continuous

  5. 5.

for any $x\in X$, if we call $\mathcal{N}_{x}$ the set of all open
neighbourhoods of $x$, then $\\{\coprod_{y\in
U}B(0_{y},r)\\}_{r>0,U\in\mathcal{N}_{x}}$ is neighbourhood basis at $0_{x}$.

this corresponds exactly to the bundle of the model theory of Hilbert space
with isometries, which is the theory of Hilbert spaces with real and imaginary
parts of the inner product symbols (for each sort).

To see why we necessarily get continuity of the norm, notice that in a Hilbert
space $||x||=\sqrt{\langle x,x\rangle}$, Now by our requirement for a bundle
the function $E\times_{X}E$ to $[0,k]$ defined by $\Re(\langle x,y\rangle)$ is
upper semi-continuous, but this implies that it is also lower semi-continuous
since the map defined by $\Re(\langle-x,y\rangle)=-\Re(\langle x,y\rangle)$ is
upper semi-continuous (using the fact that multiplying by $-1$ is continuous),
now this implies necessarily that $\Re(\langle x,y\rangle)$ is continuous
hence the norm is continuous.

If we want Bundles of Hilbert spaces with semi-continuous norm, we need to use
the second axiomatisation of Hilbert spaces (with the parallelogram identity
as an axiom), which corresponds to the category of Hilbert spaces with
contractions (these bundles are to our best knowledge not explored in
literature).

###  8.4 Tracially continuous W* bundles

Now we move to a more subtle case which is tracially continuous
$\mathrm{W}^{*}$ bundles, or bundles of tracial von Neumann algebras, we use
the axiomatisation of tracial von Neumann algebras as present in [14], in
which the ultraproduct corresponds to the Ocneanu ultraproduct[15]. The
interesting dichotomy in this axiomatisation is the fact that sorts are
interpreted as bounded operator unit balls with the $||\ ||_{2}$ topology.

We want to show the following theorem

######  Theorem 8.5.

There is an equivalence of categories, between tracially continuous
$\mathrm{W}^{*}$ bundles over $X$, and the bundle of models of the continuous
model theory of tracial von Neumann algebras over $X$.

Tracially continuous $\mathrm{W}^{*}$ bundles are defined ([23], [1], [8]) as
a unital inclusion of $C(X)\xhookrightarrow{}\mathcal{Z}(A)$ where $A$ is a
$\mathrm{C}^{*}$ algebra, together with a $\mathrm{C}^{*}$ conditional
expectation $E$ from $A$ to $C(X)$ satisfying the following properties:

  1. 1.

$E(a_{1}a_{2})=E(a_{2}a_{1})$

  2. 2.

$E(a^{*}a)=0$ iff $a=0$

  3. 3.

the unit ball of $A$ is complete with respect to the $2$-norm defined by
$||a||=||E(a^{*}a)^{1/2}||_{C(X)}$

However in [8], the authors showed a theorem allowing us to express a bundle
as a topological space $B$ over $X$ (which they called the topological
bundles) satisfying the following axioms:

  * (i)

(Global) Addition $B\times_{X}B\rightarrow B$ is continuous.

  * (ii)

(Global) Scalar multiplication, viewed as a map $\mathbb{C}\times B\rightarrow
B$ is continuous.

  * (iii)

The global $*$ operation viewed as a map $B\rightarrow B$, is continuous.

  * (iv)

The map $X\rightarrow B$ which sends $x$ to the to the additive identity
$0_{x}$ of $B_{x}$ is continuous and so is the analogous map $X\rightarrow B$
which sends $x$ to the to the multiplicative identity $1_{x}$ of $B_{x}$.

  * (v)

The map $\tau:B\rightarrow\mathbb{C}$ which restricts to the corresponding
trace on each fibre is continuous and so is the map
$\|\cdot\|_{2}:B\rightarrow\mathbb{C}$ arising from combining the 2-norms from
each fibre.

  * (vi)

A net $(b_{\lambda})\subseteq B$ converges to $0_{x}$ whenever
$\pi(b_{\lambda})\rightarrow x$ and $\|b_{\lambda}\|_{2}\rightarrow 0$.

  * (vii)

Multiplication, viewed as a map $B\times_{p}B\rightarrow B$, is continuous on
$\|\cdot\|$-bounded subsets.

  * (viii)

The restriction $\pi|_{B_{1}}:B_{1}\rightarrow X$ is open.

Before continuing we can immediately notice that in the presence of the other
axioms, and as in every other definition of metric bundles, axiom
$\mathrm{(ii)}$ can replaced with

  * (ii’)

for every scalar $k$, the map $B\xrightarrow[]{k}B$, defined by scalar
multiplication by $k$ is continuous.

Also we can notice that the axiom $\mathrm{(v)}$ can be replaced with either
one of the following equivalent (in the presence of other axioms) axioms:

  * (v’)

The maps $\tau:\ B\rightarrow\mathbb{C}$ which restricts to the corresponding
trace on each fibre is continuous

  * (v”)

The maps $\|\cdot\|_{2}\ :\ B\rightarrow\mathbb{C}$ arising from combining the
2-norms from each fibre is continuous

To see why notice that $\tau(a)=(1/4)\sum_{k=0}^{3}i^{k}||a+i^{k}.1||_{2}$,
here $i$ is the root of $x^{2}+1$.

We also prefer to write axiom $(\mathrm{vi})$ in a filter language for our
convenience:

  * (vi’)

an ultrafilter $\mu$ on $B$ converges to $0_{x}$ iff $\pi\mu$ converges to $x$
and $||\ ||_{2}\mu$ converges to $0$.

Showing that $(\mathrm{vi})$ implies $(\mathrm{vi}^{{}^{\prime}})$ is
straightforward, for the other direction we use lemma 3.4, we omit the
details.

##### Proof description

  1. 1.

Showing that the $\bf{GNS}$ construction from the category of tracial von
Neumann algebras to the category of Hilbert spaces is a left ultrafunctor.

  2. 2.

Starting from a topological bundle as defined by [8], we construct a bundle of
models of the continuous model theory of tracial von Neumann algebras. this
bundle looks like $(E_{n})_{n\geq 1}$ where each bundle is a bundle of balls
of operator norm radius $n$ each equipped with the $2$-norm on the tracial von
Neumann algebra, and hence using the left ultrafunctor-bundle of models
equivalence we get a left ultrafunctor from $X$ to the category of tracial von
Neumann algebras. Let us call this process $\mathcal{F}$.

  3. 3.

starting from a left ultrafunctor from $X$ to the category of tracial von
Neumann algebras, we construct a topological bundle as follows, we take the
composition with the $\mathrm{GNS}$ construction to construct a Hilbert bundle
and then we construct the topological $\mathrm{W}^{*}$ bundle by equipping the
disjoint union of fibres with the initial topology along the inclusion map
defined by the $\mathbf{GNS}$ construction on each fibre. Let us call this
process $\mathcal{G}$.

At this point, we want to show that the two constructions $\mathcal{F}$ and
$\mathcal{G}$ are inverse of each other.

  4. 4.

We show that given a left ultrafunctor, then every bundle of operator norm
balls $E_{n}$ equipped with the $2$-norm, coming from the corresponding bundle
of models $(E_{n})_{n>1}$ is homeomorphic inside its image inside the Hilbert
bundle.

  5. 5.

We show that given a topological bundle $E$, then after constructing the
associated left ultrafunctor and composing with the $\mathbf{GNS}$
construction, $E$ is going to be homeomorphic onto its image inside the
Hilbert bundle.

This result shows that the two processes are really inverses, to see why,
starting from a topological bundle $E$, if $\mathcal{F}(E)=F$, where the
corresponding bundle of models to $E(F)$ is $(E_{n})_{n>1}$, now the
corresponding topological bundle $\mathcal{G}\mathcal{F}(E)$ is homeomorphic
to $E$ using the result $5$ above, on the other hand, suppose that we have a
left ultrafunctor $F$, with corresponding bundle of models $(E_{n})$, then
take $\mathcal{G}(E)$ we have shown that each $E_{n}$ as well as
$\mathcal{G}(E)$ are homeomorphic onto their images inside the Hilbert bundle
and hence the bundle of models corresponding to $\mathcal{G}(E)$ is going to
be just the collection of operator norm balls of $\mathcal{G}(E)$ inside the
Hilbert bundle, hence by result $4$ the bundle of models of tracial von
Neumann algebras $(\mathcal{G}(E))_{n\geq 1}$ and $(E_{n})_{n\geq 1}$ are
isomorphic and hence the left ultrafunctors $\mathcal{F}\mathcal{G}(F)$ and
$F$ are isomorphic.

#### The proof

Before starting our proof let us briefly explain the constructions done by
Evington and Pennig[8], given a $\mathrm{W}^{*}$ bundle $\mathcal{M}$ over
$X$, one defines the fibres of the topological bundle by taking
$\mathcal{M}_{x}=\mathcal{M}/I_{x}$, where $\mathcal{I}_{x}=\\{a\mid
E(a^{*}a)(x)=0\\}$. Now the bundle topology on the space $\coprod_{x\in
X}M_{x}$ is defined to be the topology generated by the basic open sets of the
form $\coprod_{x\in X}\mathsf{B}(a(x),\epsilon)$, for $a$ continuous bounded
section of the projection map. On the other hand, given a topological bundle,
one may define an inclusion $C(X)\xhookrightarrow{}A$ where $A$ is
$\mathrm{C}^{*}$ algebra of all bounded, continuous sections to the projection
map, and we define the conditional expectation by $E(a)(x)=\tau_{x}(a(x))$.

Now we get to see why the case of relating the bundles of the continuous model
theory of tracial von Neumann algebras to that of the corresponding concept
above is more subtle, let us look at the trivial case where $X=\\{*\\}$, in
that case the bundle reduces to a von Neumann algebra with the $||\ ||_{2}$
norm topology, and such topology is $\mathbf{not}$ the inductive limit
topology of the operator norm balls with the $||\ ||_{2}$ norm. So the
question becomes, for an arbitrary compact Hausdorff space $X$, to find a
suitable topology on the nested union of bundles of bounded operator norm
balls (with the $||\ ||_{2}$ topology fibre-wise ). In order to do so, we will
attempt to construct a bundle of Hilbert spaces for which every fibre is the
$\mathrm{GNS}$ Hilbert space corresponding to the tracial von Neumann algebra.
Before that, we start by stating a necessary theorem:

######  Theorem 8.6.

The $\bf{GNS}$ construction is a left ultrafunctor from the category of
tracial von Neumann algebras to that of Hilbert spaces (with isometries).

###### Proof.

Now let $(B_{i},\phi_{i})_{i\in I}$ be a family tracial von Neumann algebras
and let $\mu$ be an ultrafilter on $I$, and define the map
$\sigma^{{}^{\prime}}_{\mu}\
:\mathcal{H}_{\phi_{\mu}}\rightarrow\int_{I}\mathcal{H}_{\phi_{i}}d\mu$ by
sending $\widehat{(a_{i})}$ to $(\widehat{a_{i}})$. Of course we need to make
sure that such construction is well defined, to do that notice that
$||\widehat{(a_{i})}||_{2,\mu}=\tau_{\mu}(\widehat{(a_{i})}^{*}(\widehat{(a_{i})}))=\lim_{\mu}\tau_{i}(a_{i}^{*}a_{i})=\lim_{\mu}||a_{i}||_{2,i}$
which is by definition the norm in $\int_{I}\mathcal{H}_{\phi(i)}d\mu$, this
shows that such construction is an isometry and hence well defined. Now
showing that the $\mathrm{GNS}$ construction is a left ultrafunctor is mostly
trivial, and requires, for axiom $(3)$ of left ultrafunctor axioms, using the
description of the categorical Fubini transform that we explained in 2.1.

######  Note.

We defined $\mathbf{GNS}$ only on objects, but its definition on morphisms is
clear. Notice that maps between tracial von Neumann algebras (seen as models
of their continuous model as defined in [14]) are $2$-norm isometries (this
follows from a similar argument to that that showed that maps of the
continuous model theory of Hilbert spaces with an inner product symbol are
isometries), and hence the induced maps between the $\mathbf{GNS}$ Hilbert
spaces are isometries as we want them to be (since we want to work with
continuous Hilbert bundles).

∎

##### Every topological bundle defines a bundle of models

Let $B$ be a topological $\mathrm{W}^{*}$ bundle, first we claim that
$(B_{n})_{n\geq 1}$ is a sorted bundle (that means that each $B_{i}$ is a
bundle corresponding to a sort in the continuous first order axiomatisation of
tracial von Neumann algebras appearing in [14]), which in turn implies that
it’s a left ultrafunctor from $X$ to the ultracategory of tracial von Neumann
algebras. The first thing we need is that the restriction of the projection to
each sorted ball is open but this is just axiom $(\mathrm{vii})$, the next
thing we are going to show is that the global functions corresponding to
relation and function symbols appearing in the axiomatisation [14] are
continuous, the continuity for the $1$ and $0$ symbols for each sorted bundle
follows from axioms $(\mathrm{iv})$, continuity of scalar multiplication with
appropriate source and target sorted bundle follows from axiom
$(\mathrm{ii})$, continuity of addition from appropriate fibre product of
sorted bundles follows from axiom $(\mathrm{(i)})$, while continuity of
subtractions follows from continuity of addition and multiplication by $-1$,
continuity of $*$ operation on each sorted bundle follows from axiom
$(\mathrm{iii})$, continuity of multiplication on sorted norm balls is just
axiom $(\mathrm{vii})$ (reminder that the sorted bundles are by definition
operator norm bounded on each fibre). Now the continuity of $2$-norm and trace
operations on each sorted bundle follows from $(v)$. So by our equivalence of
sorted bundles and left ultrafunctors we can deduce that every bundle in the
definition of [8] defines a left ultrafunctor.

##### Every left ultrafunctor defines a topological bundle

Let $X$ be a compact Hausdorff and suppose that we have a left ultrafunctor
$\mathcal{F}$ from $X$ to the ultractegory of tracial von Neumann algebras
(reminder that, by composing with $\mathbf{GNS}$ we get a left ultrafunctor
from $X$ to $\mathsf{Hilb}$, which as we saw earlier defines a Hilbert bundles
as defined in [11]), now we claim that the space $\coprod_{x\in
X}\mathcal{F}(x)$ equipped with the initial topology of its inclusion in
$\coprod_{x\in X}(\bf{GNS}\circ\mathcal{F})(x)$ is a topological
$\mathrm{W}^{*}$ bundle where $\coprod_{x\in X}(\bf{GNS}\circ\mathcal{F})(x)$
is regarded as bundle of Hilbert spaces by the adequate topology resulting
from the left ultrafunctor bundle equivalence, to show that we need to show
the topological bundle axioms :

In this proof we denote $2$-norm ball by $\mathsf{B}(a,r)$ (these can be
subsets of tracial von Neumann algebras or their Hilbert $2$-norm completion),
while if $B$ is a von Neumann algebra bundle we denote by $B_{n}$ the subset
of $B$ of all elements with operator norm less than or equal to $n$, we are
also going to denote by $\sigma_{\mu}$ the left ultrastructure of
$\mathcal{F}$, $\sigma^{{}^{\prime}}_{\mu}$ the left ultrastructure of the
$\mathbf{GNS}$ and by $\sigma^{{}^{\prime\prime}}_{\mu}$ the left
ultrastructure of $\mathbf{GNS}\circ\mathcal{F}$ .

Axioms $(\mathrm{i})$ and $(\mathrm{ii})$ are Hilbert bundle properties so
they extend to subspaces (reminder that what we claim that $\coprod_{x\in
X}\mathcal{F}(x)$ is a W* bundle when equipped with the initial topology of
its inclusion by the Hilbert bundle whose fibres are the GNS construction of
each $\mathcal{F}(x)$). Now we turn to axiom $(\mathrm{iii})$, let us call the
$B=\coprod_{x\in X}\mathcal{F}(x)$ with projection map $\pi$, and
$B^{{}^{\prime}}=\coprod_{x\in X}(\bf{GNS}\circ\mathcal{F})(x)$ with
projection map $\pi^{{}^{\prime}}$, and let us call $\ \widehat{}\ $ the
inclusion map (so the inclusion of $a$ is $\widehat{a}$ for example).

Let $\mu$ be an ultrafilter on $B$ and let $\widehat{\mu}$ be the ultrafilter
$\ \widehat{}\ \mu$ ( which is the pushforward of $\mu$ by the map $\
\widehat{}\ $ ), and suppose that $\mu$ converges to $a$, we want to show that
$*\mu$ converges to $a^{*}$ or equivalently $\widehat{*\mu}$ converges to
$\widehat{a^{*}}$ (because we defined the topology on $B$ to be the initial
topology by the map that sends $a\in B_{i}$ to
$\widehat{a}\in\mathcal{H}_{\phi_{i}}$). We have by definition that
$\pi\widehat{*\mu}=\pi^{{}^{\prime}}\widehat{\mu}$ converges to $\pi(a)$, now
suppose that $\sigma_{\pi\mu}(a)=(b_{x})_{x\in X}$ then let us remind how the
map $\sigma^{{}^{\prime\prime}}_{\pi\mu}$ which corresponds to left
ultrastructure of the composition of $\mathcal{F}$ with the GNS construction,
is constructed. We define
$\sigma^{{}^{\prime\prime}}_{\pi\mu}=\sigma^{{}^{\prime}}_{\pi\mu}\circ\sigma_{\pi\mu}$.
Then $\sigma^{{}^{\prime\prime}}_{\mu}(\widehat{a})=(\widehat{b_{x}})_{x\in
X}$, now since $\sigma_{\pi\mu}(a)=(b_{x})_{x\in X}$ and since
$\sigma_{\pi\mu}$ is a $*$ homomorphism, then
$\sigma_{\pi\mu}(a^{*})=(b^{*}_{x})_{x\in X}$ then we get that
$\sigma^{{}^{\prime\prime}}_{\pi\mu}(\widehat{b^{*}})=(\widehat{b^{*}_{x}})_{x\in
X}$. We want to show that for any $\epsilon>0\ $ $\coprod_{x\in
X}\mathsf{B}(b^{*}_{x},\epsilon)\in\widehat{*\mu}$ but we already have that
$\coprod_{x\in X}\mathsf{B}(\widehat{b_{x}},\epsilon)\in\widehat{\mu}$ (by
definition of the topology of Hilbert bundle coming from left ultrafunctor)
which implies that $\coprod_{x\in X}\mathsf{B}(b_{x},\epsilon)\in\mu$, which
implies that $\coprod_{x\in X}\mathsf{B}(b^{*}_{x},\epsilon)\in*\mu$, which in
turn implies that $\coprod_{x\in
X}\mathsf{B}(\widehat{b^{*}_{x}},\epsilon)\in\widehat{*\mu}$.

Now we get into axiom $\mathrm{(iv)}$, we already have the continuity of the
$0$ selection by a property of Hilbert bundles, for the $1$ selection we do a
same proof as in the case of adjoint by noticing that
$\sigma^{{}^{\prime\prime}}_{\mu}(\widehat{1_{x}})=(\widehat{1_{y}})_{y\in
Y}$. Axiom $\mathrm{(v^{{}^{\prime\prime}})}$ is true for any Hilbert bundle.
Now we showed that axiom $\mathrm{(vi^{\prime})}$ holds for Banach bundles and
so it holds in our case.

For axiom $\mathrm{(vii)}$, we should first remind that the $||\ ||_{2}$ norm
satisfies the following inequality $||ab||_{2}\leq||a||\ ||b||_{2}$ (because
left multiplication on the Hilbert space is a representation of the von
Neumann algebra) also we have $||ab||_{2}\leq||b||\ ||a||_{2}$ (because
$||ab||_{2}=||b^{*}a^{*}||_{2}$ since the state is tracial). Now let $\mu$ be
an ultrafilter on $B\times_{X}B$ such that there exists some $n$ such that
$B_{n}\times_{X}B_{n}\in\mu$, that converges to $(a,b)$, which is equivalent
to saying that $\widehat{\mu}$ converges to $(\widehat{a},\widehat{b})$ (here
$\widehat{\mu}$ is the pushforword of $\mu$ by the map $\langle\ \widehat{},\
\widehat{}\ \rangle$ ). Let $\epsilon>0$ and suppose that
$\sigma_{\pi\pi_{1}\mu}(a)=(a_{x})_{x\in X}$ and that
$\sigma_{\pi\pi_{1}\mu}(b)=(b_{x})_{x\in X}$. Now using the fact that
$\sigma_{\pi\pi_{1}\mu}$ is a $*$ homomorphism and hence a contraction we get
that there exists a set $X^{{}^{\prime}}\in\pi\pi_{1}\mu$ such that for any
$x\in X^{{}^{\prime}}$ $||a_{x}||\leq a+\epsilon^{{}^{\prime\prime}}\leq
n+\epsilon^{{}^{\prime\prime}}$ where $\epsilon^{{}^{\prime\prime}}$ is to be
chosen later. Now we get that that, using the definition of topology
associated to a left ultrafunctor that $\coprod_{x\in
X^{{}^{\prime}}}\mathsf{B}(\widehat{b_{x}},\epsilon^{{}^{\prime}})\in\widehat{\pi_{2}\mu}$
and $\coprod_{x\in
X^{{}^{\prime}}}\mathsf{B}(\widehat{a_{x}},\epsilon^{{}^{\prime}})\in\widehat{\pi_{1}\mu}$
for some $\epsilon^{{}^{\prime}}$ that we are going to choose later to make
the proof work. Now take any $f$ such that $\widehat{f}\in\coprod_{x\in
X^{{}^{\prime}}}\mathsf{B}(\widehat{a_{x}},\epsilon^{{}^{\prime}})\bigcap\widehat{B_{n}}$
and any $g$ such that $\widehat{g}\in\coprod_{x\in
X^{{}^{\prime}}}\mathsf{B}(\widehat{a_{x}},\epsilon^{{}^{\prime}})\bigcap\widehat{B_{n}}$
and such that $f,g$ in the same fibre, we have that
$||a_{x}b_{x}-fg||_{2}\leq||a_{x}||\ ||b_{x}-g||_{2}+||b||\
||a_{x}-f||_{2}\leq 2(n+\epsilon^{{}^{\prime\prime}})\epsilon^{{}^{\prime}}$,
so we choose $\epsilon^{{}^{\prime}}$ and $\epsilon^{{}^{\prime\prime}}$ such
that $2(n+\epsilon^{{}^{\prime\prime}})\epsilon^{{}^{\prime}}\leq\epsilon$.
Now we know that $\coprod_{x\in
X^{{}^{\prime}}}\mathsf{B}(\widehat{a_{x}},\epsilon^{{}^{\prime}})\bigcap\widehat{B_{n}}\times_{X}\coprod_{x\in
X^{{}^{\prime}}}\mathsf{B}(\widehat{b_{x}},\epsilon^{{}^{\prime}})\bigcap\widehat{B_{n}}\in\widehat{\mu}$,
and hence $\coprod\mathsf{B}(\widehat{ab},\epsilon)\in\widehat{.\mu}$, and
hence $\widehat{.\mu}$ converges to $\widehat{ab}$, and thus $.\mu$ converges
to $ab$.

Now we turn to the last axiom $\mathrm{(vii)}$, let $V=\widehat{B_{1}}\bigcap
W$ be an open set in the subspace topology of $\widehat{B_{1}}$ (remember that
we are equipping $B_{1}$ with the subspace topology of its image
$\widehat{B_{1}}$), here $W$ is an open set in the topology of the Hilbert
bundle, let $\mu$ be an ultrafilter on $X$ that converges to some
$x\in\pi(B_{1}\bigcap W)$, then there exists $a\in(B_{1}\bigcap W)$ in the
fibre over $x$, we may assume without loss of generality that $||a||<1$, why
is that ? since $W$ is open then its intersection with any fibre is open in
that fibre (in a Hilbert bundle the subspace topology agrees with the metric
topology on each fibre) then there exists some $\delta$ such that if
$||a-a^{{}^{\prime}}||_{2}<\delta$, we have then $a^{{}^{\prime}}\in W_{x}$,
let us take the element $a^{{}^{\prime}}=(1-\delta)a$ then
$||a^{{}^{\prime}}||<||a||\leq 1$ and also $a^{{}^{\prime}}\in W_{x}$, so we
can always pick $||a||<1$. Suppose that $\sigma_{\mu}(a)=(c_{x})_{x\in X}$
which implies that
$\sigma^{{}^{\prime\prime}}_{\mu}(\widehat{a})=(\widehat{c_{x}})_{x\in X}$. By
the fact that $\sigma^{{}^{\prime\prime}}_{\mu}$ is contractive ($*$
homomorphism), we may deduce that $||(c_{x})||\leq||a||<1$.

Now since $W$ is open there exists $U_{1}\in\mu$ and $\epsilon>0$ such that
$\coprod_{x\in U}\mathsf{B}(\widehat{c_{x}},\epsilon)\subseteq W$ (using 8.1),
Now the fact that $||(c_{x})||\leq||a||$ means that for any
$\epsilon^{{}^{\prime}}$ there exits a set $U_{2}\in\mu$ such that for any
$y\in U_{2}$, we have $||c_{y}||\leq||a||+\epsilon^{{}^{\prime}}$, we choose
$\epsilon^{{}^{\prime}}$ such that $||a||+\epsilon^{{}^{\prime}}<1$. Now we
get that $U_{1}\bigcap U_{2}\subseteq\pi(W\bigcap B_{1})$, this implies that
$\pi(W\bigcap B_{1})\in\mu$ thus $\pi(W\bigcap B_{1})$ is open (by the
ultrafilter characterisation of open sets in topological spaces).

We conclude by stating the following theorems:

######  Theorem 8.7.

Let $\mathcal{F}$ be a leftultrafunctor from $X$ to the category of tracial
von Neumann algebras, and let $(E_{n})_{n\in\mathbb{N}}$ be the corresponding
family of sorted bundles then any sorted bundle $E_{n}$ is homeomorphic onto
its image by the $\mathrm{GNS}$ construction on each fibre inside the Hilbert
bundle.

###### Proof.

The proof easily follows from the fact that for any ultrafilter $\mu$ on the
base space $X$, we have the following: if $\sigma_{\mu}(b)=(c_{x})_{x\in X}$
then $\coprod_{x\in X}\mathsf{B}(c_{x},\epsilon)\in\mu\iff\coprod_{x\in
X}\mathsf{B}(\widehat{c_{x}},\epsilon)\in\widehat{\mu}$ ∎

######  Theorem 8.8.

let $E$ be a topological $\mathrm{W}^{*}$ bundle, then $E$ is homeomorphic
onto its image by the $\mathrm{GNS}$ on each fibre inside the Hilbert bundle.

###### Proof.

Let $E$ be a topological $W^{*}$ bundle over $X$, the Hilbert bundle is
constructed by showing that the set of sorted bundles $(E_{n})$ corresponds to
a left ultrafunctor from $X$ to the category of tracial von Neumann algebras
and then by composing with the $\mathrm{GNS}$ left ultrafunctor we obtained a
left ultrafunctor from $X$ to $\mathsf{Hilb}$ which corresponds to a bundle of
Hilbert spaces.

Let $(b_{\alpha})$ be a net on $E$ converging to $b$, let us call $\pi(b)=x$.
By the equivalence established by [8], this bundle can be regarded as a
certain inclusion of $C(X)$ inside the centre of $A$, the $\mathrm{C}^{*}$
algebra of bounded, continuous section over the topological bundle, and each
fibre in this case is isomorphic to the quotient of $A$ by $\mathcal{I}_{x}$,
all this implies that there exists a bounded, continuous section $a$, such
that $a(x)=b$, Now the net $(a(\pi(b_{\alpha}))-b_{\alpha})$ converges to
$0_{x}$, this implies that $\pi(a(\pi(b_{\alpha}))-b_{\alpha})$ converges to
$x$ and the norm of $||(a(\pi(b_{\alpha}))-b_{\alpha}||_{2}$ converges to $0$,
now since the $\mathbf{GNS}$ construction on each fibre is $2$-norm isometric,
we get by the axiom $5$ of the definition of Banach bundles, that
$\widehat{(a(\pi(b_{\alpha}))}-\widehat{b_{\alpha}}$ )converges to
$\widehat{0_{x}}$, now we show that $\widehat{(a(\pi(b_{\alpha}))}$ converges
to $\widehat{b}$, we know that for sorted bundles the inclusion is
homeomorphism, so it’s enough to show that the net $(a(\pi(b_{\alpha}))$ is
operator norm bounded, but this net is operator norm bounded by $||a||$ so we
have that.

On the other hand suppose that $\widehat{(b_{\alpha})}$ converges to
$\widehat{b}$, we wish to show that $(b_{\alpha})$ converges to to $b$, in
order to do that we summon the continuous section $a$ to do the reverse of the
last argument, we know that
$\widehat{(a(\pi(b_{\alpha}))}=\widehat{(a(\pi^{{}^{\prime}}(\widehat{b_{\alpha}}))}$
(here $\pi^{{}^{\prime}}$ is the projection map on the Hilbert bundle)
converges to $\widehat{a(x)}=\widehat{b}$, hence we may deduce that
$\widehat{(a(\pi(b_{\alpha}))}-\widehat{b_{\alpha}}$ converges to
$\widehat{0_{x}}$.

Now $\ \ \widehat{}\ \ $ is a $2$-norm isometry on each fibre and
$\pi(a(\pi(b_{\alpha}))-b_{\alpha})=\pi^{{}^{\prime}}(\widehat{a(\pi(b_{\alpha}))}-\widehat{b_{\alpha}})$
converges to $x$, this implies that $(a(\pi(b_{\alpha}))-b_{\alpha}$ converges
to $0_{x}$. Now we use the fact that $(a(\pi(b_{\alpha}))$ is operator norm
bounded to deduce that $(a(\pi(b_{\alpha}))$ converges to $a(x)=b$ and hence
$(b_{\alpha})$ converges to $b$. ∎

So the last two theorems indicate that the two processes we showed first are
inverses of each other as we have explained in the proof description. We
finish by stating a nice theorem/conclusion to this section:

######  Theorem 8.9.

Let $E$ be a topological $\mathrm{W^{*}}$ bundle then there exists a Hilbert
bundle whose fibres are the corresponding $\mathrm{GNS}$ constructions for
each trace, and such that the subspace topology of the Hilbert bundle of the
subspace which equals on each fibre the image of the tracial von Neumann
algebra by the $\mathrm{GNS}$ construction, is homeomorphic to the
$\mathrm{W}^{*}$ bundle.

###  8.5 Note regarding the examples

The reader may notice that in the definition of bundles existing in the
literature, we can always define a category of bundles by defining a morphism
of bundles between $(E^{{}^{\prime}},Y,\pi^{{}^{\prime}})$ and $(E,X,\pi)$ to
be a continuous map $f$ and $f^{{}^{\prime}}$ such that the following diagram
commutes:

${{E{{}^{\prime}}}}$${E}$${Y}$${Y}$$\scriptstyle{\pi^{{}^{\prime}}}$$\scriptstyle{f^{{}^{\prime}}}$$\scriptstyle{\pi}$$\scriptstyle{f}$

Now it’s not difficult to see that with the equivalence we showed is
functorial. We already know that bundles of models form a topological stack
over the category of compact Hausdorff spaces, with Cartesian lifts given by
pullback in $\mathsf{Top}$ sortwise, this allows us to extend the result to
all bundles above, indeed the argument is easy for Banach, Hilbert and
$\mathrm{C}^{*}$ bundles, since we know that in that case $E=\varinjlim
E_{n}$, where $(E_{n})_{n\geq 1}$ are the sorted bundles, and taking pullbacks
commutes with colimits, for $\mathrm{W}^{*}$ bundles, the argument is more
subtle as usual since it’s not true in general $E=\varinjlim E_{n}$, so let us
make clear what we want to show:

######  Theorem 8.10.

Suppose that $(E,X,\pi)$ is a $\mathrm{W}^{*}$ bundle and let $f:\
Y\rightarrow X$ to be a map of compact Hausdorff topological spaces then the
following are true:

  * •

The pullback along $f$ is a $\mathrm{W}^{*}$ bundle.

  * •

this pullback is the Cartesian lift over $f$ (in other words it corresponds to
the composition of the left ultrafunctor corresponding to the bundle with $f$
(regarded as a left ultrafunctor)).

###### Proof.

let $F$ be the left ultrafunctor corresponding to the bundle $E$ we know that
the Hilbert bundle resulting from composition $\mathbf{GNS}\circ F\circ f$ is
the pullback along $f$ of the Hilbert bundle resulting from the composition
$\mathbf{GNS}\circ F$, let us call this resulting bundle $H^{{}^{\prime}}$,
let $E^{{}^{\prime}}$ be the $\mathrm{W}^{*}$ bundle corresponding to
composition $F\circ f$, $E^{{}^{\prime}}$ is homeomorphic onto its image in
$H^{{}^{\prime}}$, and $E^{{}^{\prime}}$ as a set is the pullback of $E$ (the
justification of this statement is that at the level of each sort
$E^{{}^{\prime}}_{n}$ is the pullback of $E_{n}$ as we have shown), but since
pullbacks commute with taking subspaces, then $E^{{}^{\prime}}$ is the
pullback of $E$ in $\mathsf{Top}$. ∎

##  9 Application: Another proof of Lurie’s result

Now we use this already developed theory to find another proof of Lurie’s
result of equivalence between Sheaves of sets (where the site is
$\mathcal{O}(X)$ where $X$ is compact Hausdorff) and left ultrafunctors from
$X$ to $\sf{Set}$:

######  Theorem 9.1.

Let $X$ be a compact Hausdorff space then there is an equivalence of
categories between $\mathrm{Sh}(X)$ and the category
$\mathrm{Leftultrafunctor}(X,\mathsf{Set})$

Before proving this result notice that the category $\sf{Set}$ is equivalent
to the category of discrete metric spaces. Now let us axiomatise discrete
metric spaces using the previous theory: The language of discrete metric space
is mono-sorted with an upper bound to distance the constant $1$, with no
function symbols, and no relation symbols (unless you want to count the
distance as a relation symbol). If we call $S$ the single sort we get that the
set $\mathbb{T}$ of axioms for discreet metric spaces contains only one
sentence $\mathrm{Sup}_{x\in S}\mathrm{Sup}_{y\in
S}\mathrm{min}(d(x,y),|1-d(x,y)|)$ which translates to the fact that the
metric is discrete.

Now we want to show that the bundle $E_{s}$ is an étale space over $X$,
remember that this is equivalent to saying that the diagonal map:

${{E_{S}}}$${{E_{S}\times_{X}E_{S}}}$$\scriptstyle{\Delta}$

is open (which is equivalent to saying that the diagonal of
$E_{S}\times_{X}E_{S}$ is open since the diagonal map is an embedding). It
follows from upper semi-continuity of the distance function on $E_{s}$ that
for any $\ epsilon>0$ the set $\\{(v,v^{{}^{\prime}})\in E\times_{X}E\
d_{\pi(v)}(v,v^{{}^{\prime}})<\epsilon\ \\}$ is open, so if we take
$\epsilon<1$ we get the diagonal of $E\times_{X}E$ (since the metric on each
fibre is discrete).

On the other hand suppose that we have an étale space $(E,X,\pi)$ then let us
prove that in this case we get a bundle of discrete metric spaces: The three
axioms are easily verifiable: for axiom $(1)$ let $(e,f)\in E\times_{X}E$, the
case where $e\neq f$ is trivial, thus let us suppose that $e=f$, in that case
we know that there exists a neighbourhood $U$ of $e$, such that $\pi(U)\simeq
U$, now take the neighbourhood $U\times_{X}U$ of the $(e,e)$, for any
$(g,h)\in U\times_{X}U$, we have that $g=h$ and hence $d(g,h)=0$, and hence
distance is upper semi-continuous. For axiom(2) $\pi$ is continuous by
definition and also it’s known that $\pi$ is open([21]). For axiom(3) let
$e\in E$ and let $W$ be a neighbourhood of $E$ since $e\in E$ there exits some
neighbourhood $U$ of $e$ such that if $\pi(U)\simeq U$ (via $\pi|_{U}$), now
take $V=U\bigcap W$ and any $0<\epsilon<1$, we claim that $V_{\epsilon}=V$ and
it’s easy to see why.

These maps extend to morphisms since morphisms of étale spaces over $X$ and
maps of bundles of discrete metric spaces are defined the same way.

So we get an equivalence between étale spaces and bundles of sets (seen as
discrete metric spaces), now we already know that étale spaces are sheaves of
sets on $X$, on the other hand we also know that bundles of discrete metric
spaces are equivalent to left ultrafunctors from $X$ to the category of
discrete metric spaces which is equivalent to that of sets (we should also
mention that the equivalence between discrete metric space and $\sf{Set}$ is
an equivalence of ultracategories (preserves the ultraproduct)).

######  Note.

let $E$ be an étale space over $X$, then we have a good description of the
left ultrastructure of the associated left ultrafunctor, since by definition
the étale space has enough local sections. So if $\mu$ converges to $x\in X$,
then $\sigma_{\mu}(a)=(f(y))_{y\in U}$, where $f$ is the local homeomorphism
that hits $a\in E_{x}$.

##  10 Bundles of pointed metric spaces

There is a natural notion of ultraproduct of complete pointed metric spaces,
constructed in a similar fashion to the ultraproduct of bounded metric space
as follows: suppose $(M_{i},p_{i})_{i\in I}$ is a family of pointed metric
spaces and suppose that $\mu$ is an ultrafilter on $I$, we define
$\int_{I}M_{i}d\mu$ to be the space of all bounded sequences (with respect to
the point of each space) quotiented by the equivalence relation
$(x_{i})\sim(y_{i})$ iff $\lim_{\mu}d_{i}(x_{i},y_{i})=0$, and by taking as
point for the space the equivalence class of $(p_{i})_{i\in I}$, the same
argument as in the bounded case shows that such space is complete, this allows
the definition of an ultracategory of metric spaces with contractions.

######  Definition 10.1.

We say that $(E,X,\pi)$ where $X$ and $E$ are topological spaces and $\pi:\
E\rightarrow X$ define a bundle of pointed metric spaces if they satisfy the
following set of axioms:

  * •

Axiom(1): The global distance function is upper semi-continuous.

  * •

Axiom(2): $\pi$ is continuous and open.

  * •

Axiom(3): For every open set $W$ and every $f\in W$ there exist an open set
$V$ open neighbourhood of $f$ and $\epsilon>0$ such that
$V\subseteq_{\epsilon}W$.

  * •

Axiom(4): The point selection function $x\mapsto p(x)$ is continuous

Let $\mathsf{Point}_{1}$ denote the category of pointed metric spaces with
contractions.

######  Theorem 10.1.

Let $X$ be a compact Hausdorff space, then there exists an equivalence of
categories of bundles of pointed complete metric spaces over $X$, and left
ultrafunctors from $X$ to $\mathsf{Point}_{1}$.

We are not going to do the proof of this theorem. The proof is just repeating
the steps of the equivalence between left ultrafunctors from $X$ to
$k\text{-}\mathsf{CompMet}$ and bundles of metric spaces bounded by $k$ as was
done in 3.

This equivalence is not part of the examples section, because we don’t know of
any axiomatisation of pointed complete metric spaces in continuous model
theory. Indeed if we try to imitate the Banach space case and define a
language with a sort intended to be interpreted as the closed ball of a radius
$n$ for every $n\in\mathbb{N}$, and a constant symbol for the point of the
space ($p\in B_{1}$), then dissections of pointed metric spaces are not an
elementary class in this language indeed take the following example: for very
$i\in\mathbb{N}$ let $X_{i}=\\{p_{i},x_{i}\\}$ and such that
$d(p_{i},x_{i})=1+1/i$, the dissection of such space is
${X_{i}}_{1}=\\{p_{i}\\}$ and ${X_{i}}_{n}=\\{p_{i},x_{i}\\}$ for $n>1$. Let
$\mu$ be a non-principal ultrafilter on $\mathbb{N}$, then
$(\int{X_{i}}_{1}d\mu)=\\{(p_{i})_{i\in\mathbb{N}}\\}$, but if
$(\int{X_{i}}_{n}d\mu)_{n\in\mathbb{N}}$ was a “model of the continuous theory
of pointed complete spaces”, then $(\int{X_{i}}_{1}d\mu)$ should be
$\\{(p_{i})_{i\in\mathbb{N}},(x_{i})_{i\in\mathbb{N}}\\}$. Then dissections of
pointed spaces in this language are not axiomatisable.

One final thing to note is that when we showed that Banach (Hilbert,
$\mathrm{C}^{*}$, etc) bundles are equivalent to left ultrafunctors we used an
approach based on bundles of completed bounded metric spaces as our main
building block, another viable approach could have been using bundles of
pointed metric spaces as the main ingredient. Indeed some of the results may
have been easier to show, but our approach has the advantage of having a
notion of bundles that works for any continuous theory.

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# Big Bang Nucleosynthesis Initial Conditions: Revisiting Wagoner et al.
(1967)

Charlie Sharpe  Sydney Institute for Astronomy, School of Physics, A28,
The University of Sydney, NSW 2006, Australia  Geraint F. Lewis Sydney
Institute for Astronomy, School of Physics, A28,
The University of Sydney, NSW 2006, Australia  Luke A. Barnes  Western Sydney
University, Locked Bag 1797,
Penrith South, NSW 2751, Australia

###### Abstract

We revisit Wagoner et al. (1967), a classic contribution in the development of
Big Bang Nucleosynthesis. We demonstrate that it presents an incorrect
expression for the temperature of the early universe as a function of time in
the high temperature limit, $T\gtrsim 10^{10}$K. As this incorrect expression
has been reproduced elsewhere, we present a corrected form for the initial
conditions required for calculating the formation of the primordial elements
in the Big Bang.

Cosmology (343)

##  1 Introduction

Big Bang Nucleosynthesis (BBN) is considered an outstanding success and pillar
of modern cosmology, explaining the abundance of primordial elements within
the universe (Tytler et al., 2000; Barnes & Lewis, 2020). Calculating the
output of BBN requires integrating a series of coupled differential equations
representing the sources and sinks of various elements, encompassing cross-
sections that are temperature dependent. Such integrals, of course, depend
upon the initial conditions.

The structure of this research note is as follows; in Section 2, we discuss
the process Wagoner et al. (1967) take in deriving the initial conditions for
the integration of BBN in the early universe. We then discuss the issue with
their results in Section 3 before concluding in Section 4.

##  2 BBN Initial Conditions

In their classic paper, Wagoner et al. (1967) derive the dependence of the
temperature of the early universe on time, in the high temperature limit
$T\gtrsim 10^{10}$K. To do this, they first derive the work-energy equation
for the Friedmann–Lemaître–Robertson–Walker metric. The baryon mass density
$\rho_{b}$ scales as $R^{-3}$ and the neutrino energy density $\rho_{\nu}$
scales as $R^{-4}$, where $R$ is the scale factor of the universe. After some
algebra, they arrive at the following first-order differential equation for
the expansion $R(T)$,

$\frac{dR}{dT}=\frac{-R}{3\left[\rho_{1}(T)+p_{1}(T)/c^{2}\right]}\frac{d\rho_{1}(T)}{dT}$
(1)

where $\rho_{1}=\rho_{e}+\rho_{\gamma}$ is the sum of the energy densities of
electrons and radiation, and $p_{1}=p_{e}+p_{\gamma}$ is the sum of the
pressures of electrons and radiation. They then combine this equation with the
Friedmann equation to arrive at a differential equation for $T(t)$,

$\frac{dT}{dt}=\mp(24\pi
G\rho)^{1/2}\left[\rho_{1}(T)+\frac{p_{1}(T)}{c^{2}}\right]\left[\frac{d\rho_{1}(T)}{dT}\right]^{-1}$
(2)

Now, at such high temperatures ($T\gtrsim 10^{10}$K), electrons are highly
relativistic with energy density $\frac{7}{4}\rho_{\gamma}$, the same as
neutrinos. The coefficient of $\frac{7}{4}$ comes from the number of species
and allowed spin states of each respective particle. This gives
$\rho=\frac{9}{2}\rho_{\gamma}$ and $\rho_{1}=\frac{11}{4}\rho_{\gamma}$.
Hence, assuming an equation of state of $\frac{1}{3}$ for highly relativistic
species, they finally arrive at a relationship between the temperature of the
universe and time (their equation A15),

$T_{9}=(12\pi Ga_{r}c^{-2})^{1/4}\ t^{-1/2}=10.4\ t^{-1/2}$ (3)

Here, $T_{9}$ is the temperature, $G$ is Newton’s gravitational constant,
$a_{r}=4\sigma/c$ is the radiation density constant,
$\sigma=2\pi^{2}k_{b}^{2}/15h^{3}c^{2}$ is the Stefan-Boltzmann constant, and
$c$ is the speed of light. The right-hand expression gives the temperature of
the universe in units of $10^{9}$K when the time, $t$, is measured in seconds
from $t=0$.

There are some interesting observations to be made about Equation 3, but
perhaps the most important is that while the second equality,
$T_{9}=10.4t^{-1/2}$, is correct, the first equality is incorrect. Most
obviously, the dimensions are incorrect. Working in cgs units, the central
expression in Equation 3 has units of s-1K-1, which is clearly inconsistent
with $T_{9}$ having units of kelvin.

##  3 Correcting the Initial Conditions

Wagoner et al. (1967) state that details that are relevant to their derivation
of Equation 3 are provided by Alpher et al. (1953). However, this earlier
paper finds that temperature follows $T_{9}=15.2t^{-1/2}$, which has a
different numerical factor from Equation 3. This discrepancy arises from the
fact that Alpher et al. (1953) do not consider the presence of electrons or
neutrinos at early times, and hence use $\rho=\rho_{1}=\rho_{\gamma}$.

Directly solving Equation 2 we find that, for time in seconds,

$T_{9}=(48\pi a_{r}c^{-2}G)^{-1/4}t^{-1/2}=10.4t^{-1/2}$ (4)

This expression is dimensionally correct, with the central equation having
units K, and yields the same numerical factor as presented in Wagoner et al.
(1967).

##  4 Conclusions

Normally, such a minor typographical error in a paper has no significant
consequences and so can pass unnoted. However, the incorrect expression
presented in Wagoner et al. (1967) has been propagated into other key
publications, and numerical codes for calculating BBN. We highlight two of
these below;

  * •

Kawano (1992) presents a fortran implementation of the integration of the
coupled BBN equations named NUC123111https://github.com/ckald/KAWANO-sterile.
This directly reproduces the expression from Wagoner et al. (1967) as the
initial condition for the integration (their equation D.6). However, an
examination of the fortran source code reveals that the initial time is set to
be t = 1/(const1*t9)**2 !Initial time (Ref 1) where PARAMETER (const1=0.09615)
!Relation between time and temperature, so only the numerical aspect of
Equation 3 is employed.

  * •

Arbey et al. (2020) presents a $C$ implementation of BBN known as
AlterBBN222https://alterbbn.hepforge.org. They too cite Equation 3 in
determining the initial conditions of their integration. Exploring the source
code, the initial time is set to double t=sqrt(12.*pi*G*sigma_SB)/pow(Ti,2.);
and so directly implements the incorrect form of the initial time. We note
that in the natural (energy) units employed in AlterBBN results in a very
early starting time, well into the high temperature limit, and so does not
impact the resulting integration. We have discussed this with the AlterBBN
author(s) and the initial conditions will be corrected in a forthcoming update
(Arbey 2021, priv. comm.).

This work was undertaken as part of Charlie Sharpe’s honours project at the
University of Sydney’s School of Physics.

## References

  * Alpher et al. (1953) Alpher, R. A., Follin, J. W., & Herman, R. C. 1953, Physical Review, 92, 1347, doi: 10.1103/PhysRev.92.1347
  * Arbey et al. (2020) Arbey, A., Auffinger, J., Hickerson, K. P., & Jenssen, E. S. 2020, Computer Physics Communications, 248, 106982, doi: 10.1016/j.cpc.2019.106982
  * Barnes & Lewis (2020) Barnes, L. A., & Lewis, G. F. 2020, The Cosmic Revolutionary’s Handbook: (Or: How to Beat the Big Bang) (Cambridge University Press), doi: 10.1017/9781108762090
  * Kawano (1992) Kawano, L. 1992, Let’s go: Early universe 2. Primordial nucleosynthesis the computer way, NASA STI/Recon Technical Report N
  * Tytler et al. (2000) Tytler, D., O'Meara, J. M., Suzuki, N., & Lubin, D. 2000, Physica Scripta, T85, 12, doi: 10.1238/physica.topical.085a00012
  * Wagoner et al. (1967) Wagoner, R. V., Fowler, W. A., & Hoyle, F. 1967, ApJ, 148, 3, doi: 10.1086/149126