$\overline{\psi}_{p_{k+1}}({\mathcal{K}}X_{p_{k+1}})$ is nuclear by the base
case we have that $\overline{B}_{F_{k+1}}$ is nuclear. Inducing on $k$ gives
that $\overline{B}_{F}=\overline{B}_{F_{n}}$ is nuclear. ∎

In the amenably controlled case, and by combining with Theorem 6.4, we can
deduce nuclearity/exactness of ${\mathcal{N}}{\mathcal{T}}(X)$ from
nuclearity/exactness of $A$, and conversely. The exactness equivalence passes
to $A\times_{X}P$, however this fails for nuclearity even for
$P=\mathbb{Z}_{+}$ due to a counterexample of Ozawa in [29]. In [24] it is
shown that $A\times_{X}P$ is nuclear if and only if the embedding
$A\hookrightarrow A\times_{X}P$ is nuclear when $(G,P)$ is a quasi-lattice
controlled by $(\mathbb{Z}^{n},\mathbb{Z}_{+}^{n})$ that satisfies a
minimality condition. In fact this holds for any quotient in-between the
Toeplitz-Nica-Pimsner and the covariance algebra. Here we generalize to
controlled maps by amenable weak right LCM-inclusions. Recall that in the
amenably controlled case the reduced C*-algebras are universal.

######  Theorem 6.12.

Let $\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ be a controlled map
between weak right LCM-inclusions with ${\mathcal{G}}$ amenable and let $X$ be
a compactly aligned product system over $P$ with coefficients in $A$. Let
$(\pi,t)$ be an equivariant injective Nica-covariant representation of $X$.
Then $A$ is exact if and only if $\mathrm{C}^{*}(\pi,t)$ is exact.

######  Proof..

We are going to introduce new product systems from $X$. Therefore in order to
make a distinction we will write $\overline{E}^{X}_{P}$ for the faithful
conditional expectation on the Fock C*-algebra ${\mathcal{T}}_{\lambda}(X)$ of
$X$.

If $\mathrm{C}^{*}(\pi,t)$ is exact then so is $A$, since exactness passes to
C*-subalgebras. For the converse by Theorem 6.4 we have that $X$ is amenable
and thus $\mathrm{C}^{*}(\pi,t)$ is a quotient of
${\mathcal{T}}_{\lambda}(X)$. Hence it suffices to show that
${\mathcal{T}}_{\lambda}(X)$ is exact. In view of Theorem 6.11 it suffices to
show that $\overline{E}_{P}^{X}\otimes_{\max}{\operatorname{id}}_{D}$ is
faithful on ${\mathcal{T}}_{\lambda}(X)\otimes_{\max}D$ for all C*-algebras
$D$.

Towards this end let the product system $Y=\\{Y_{p}\\}_{p\in P}$ be defined by

$Y_{p}:=\overline{\overline{t}_{p}(X_{p})\odot
D}^{\otimes_{\max}}\subseteq{\mathcal{T}}_{\lambda}(X)\otimes_{\max}D.$

That $Y$ is a product system follows by that $X$ is so. Since $X$ is compactly
aligned we have that

$Y_{p}Y_{p}^{*}Y_{q}Y_{q}^{*}\subseteq\overline{\overline{\psi}_{p}({\mathcal{K}}X_{p})\overline{\psi}_{q}({\mathcal{K}}X_{q})\odot
D}^{\otimes_{\max}}=\overline{\overline{\psi}_{w}({\mathcal{K}}X_{w})\odot
D}^{\otimes_{\max}}=[Y_{w}Y_{w}^{*}]$

for $wP=pP\cap qP$, with the understanding that
$Y_{p}Y_{p}^{*}Y_{q}Y_{q}^{*}=(0)$ when $p$ and $q$ have no common right
common multiple. Thus by Proposition 4.7 we get that $Y$ is a compactly
aligned product system over $P$ with coefficients in $A$.

Again by Theorem 6.4 we have that $Y$ is amenable. Our goal is to show that
the identity representation on $Y$ is faithful on
${\mathcal{N}}{\mathcal{T}}(Y)\simeq{\mathcal{T}}_{\lambda}(Y)$, and thus we
have that

${\mathcal{N}}{\mathcal{T}}(Y)\simeq{\mathcal{T}}_{\lambda}(Y)\simeq{\mathcal{T}}_{\lambda}(X)\otimes_{\max}D.$

We then derive that the faithful conditional expectation
$\overline{E}^{Y}_{P}$ on ${\mathcal{T}}_{\lambda}(Y)$ coincides with
$\overline{E}^{X}_{P}\otimes_{\max}{\operatorname{id}}_{D}$ and the proof will
be completed. We will invoke Theorem 6.4.

First we see that the identity representation is ${\mathcal{G}}$-equivariant.
Indeed we have that $(\overline{\pi},\overline{t})$ admits a coaction
$\overline{\delta}_{\mathcal{G}}$ of ${\mathcal{G}}$ and thus we have an
equivariant $*$-homomorphism

$\overline{\delta}_{\mathcal{G}}\otimes_{\max}{\operatorname{id}}_{D}\colon{\mathcal{T}}_{\lambda}(X)\otimes_{\max}D\longrightarrow({\mathcal{T}}_{\lambda}(X)\otimes\mathrm{C}^{*}({\mathcal{G}}))\otimes_{\max}D.$

By amenability of ${\mathcal{G}}$ and associativity of the maximal tensor
product we get that

$\displaystyle({\mathcal{T}}_{\lambda}(X)\otimes\mathrm{C}^{*}({\mathcal{G}}))\otimes_{\max}D$
$\displaystyle\simeq{\mathcal{T}}_{\lambda}(X)\otimes_{\max}\mathrm{C}^{*}({\mathcal{G}})\otimes_{\max}D$
$\displaystyle\simeq({\mathcal{T}}_{\lambda}(X)\otimes_{\max}D)\otimes_{\max}\mathrm{C}^{*}({\mathcal{G}})\simeq({\mathcal{T}}_{\lambda}(X)\otimes_{\max}D)\otimes\mathrm{C}^{*}({\mathcal{G}})$

and thus we deduce that
$\overline{\delta}_{\mathcal{G}}\otimes_{\max}{\operatorname{id}}_{D}$ is a
coaction of ${\mathcal{G}}$ on ${\mathcal{T}}_{\lambda}(X)\otimes_{\max}D$. By
construction
$\overline{\delta}_{\mathcal{G}}\otimes_{\max}{\operatorname{id}}_{D}$
satisfies the coaction identity with aligned fibers in the sense that

$[{\mathcal{T}}_{\lambda}(X)\otimes_{\max}D]_{g}=\overline{[{\mathcal{T}}_{\lambda}(X)]_{g}\odot
D}^{\otimes_{\max}}.$

Secondly let $F\subseteq P$ be $\vee$-closed finite set and let
$k^{\prime}_{p}\in{\mathcal{K}}Y_{p}$ such that $\sum_{p\in
F}{\operatorname{id}}(k^{\prime}_{p})=0$. For every state
$\phi\in{\mathcal{S}}(D)$ we have the completely contractive map

${\operatorname{id}}\otimes_{\max}\phi\colon[Y_{p}Y_{p}^{*}]\otimes_{\max}D\longrightarrow\overline{\psi}_{p}({\mathcal{K}}X_{p});\overline{\psi}_{p}(k_{p})\otimes
d\mapsto\phi(d)\overline{\psi}_{p}(k_{p}).$

Therefore we derive

$\sum_{p\in
F}({\operatorname{id}}\otimes_{\max}\phi)(k^{\prime}_{p})=({\operatorname{id}}\otimes_{\max}\phi)(\sum_{p\in
F}k^{\prime}_{p})=0.$

Note here that this is a relation in ${\mathcal{T}}_{\lambda}(X)$ with every
$({\operatorname{id}}\otimes_{\max}\phi)(k^{\prime}_{p})\in\overline{\psi}_{p}({\mathcal{K}}X_{p})$.
Thus if $p_{0}$ is a minimal element in $F$ such that
$({\operatorname{id}}\otimes_{\max}\phi)(k^{\prime}_{p_{0}})\neq 0$ then we
get

$P_{p_{0}}({\operatorname{id}}\otimes_{\max}\phi)(k^{\prime}_{p_{0}})P_{p_{0}}=P_{p_{0}}({\operatorname{id}}\otimes_{\max}\phi)(\sum_{p\in
F}k^{\prime}_{p})P_{p_{0}}=0,$

where $P_{p_{0}}\colon{\mathcal{F}}X\to X_{p_{0}}$ is the canonical
projection. However the compression to $X_{p_{0}}$ is a faithful
$*$-representation on $\overline{\psi}_{p_{0}}({\mathcal{K}}X_{p_{0}})$, and
thus we get the contradiction that
$({\operatorname{id}}\otimes_{\max}\phi)(k^{\prime}_{p_{0}})=0$. Continuing
inductively we deduce that
$({\operatorname{id}}\otimes_{\max}\phi)(k^{\prime}_{p})=0$ for all $p\in F$
(one by one for fixed $\phi$). As this holds for all $\phi$ and the family
$\\{{\operatorname{id}}\otimes_{\max}\phi\\}_{\phi\in{\mathcal{S}}(D)}$
separates $[Y_{p}Y_{p}^{*}]\otimes_{\max}D$ we get that $k^{\prime}_{p}=0$ for
all $p\in F$. Hence the assumptions of Theorem 6.4 hold for $Y$ and the proof
is completed. ∎

######  Theorem 6.13.

Let $\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ be a controlled map
between weak right LCM-inclusions with ${\mathcal{G}}$ amenable and let $X$ be
a compactly aligned product system over $P$ with coefficients in $A$. Let
$(\pi,t)$ be an equivariant injective Nica-covariant representation of $X$.
Then $A\hookrightarrow\mathrm{C}^{*}(\pi,t)$ is nuclear if and only if
$\mathrm{C}^{*}(\pi,t)$ is nuclear.

######  Proof..

It is clear that if $\mathrm{C}^{*}(\pi,t)$ is nuclear then
$A\hookrightarrow\mathrm{C}^{*}(\pi,t)$ is nuclear. Let us prove the converse.
By Theorem 6.4 we have that
${\mathcal{T}}_{\lambda}(X)\simeq{\mathcal{N}}{\mathcal{T}}(X)$ and so
$(\pi,t)$ promotes to a $*$-representation of ${\mathcal{T}}_{\lambda}(X)$.
Due to amenability
$\mathrm{C}^{*}({\mathcal{G}})=\mathrm{C}^{*}_{\lambda}({\mathcal{G}})$ is
nuclear (and so the minimal and the maximal tensor product coincide). Let
$\delta\colon\mathrm{C}^{*}(\pi,t)\to\mathrm{C}^{*}(\pi,t)\otimes\mathrm{C}^{*}_{\lambda}({\mathcal{G}})$
be the coaction of ${\mathcal{G}}$ and let $E=({\operatorname{id}}\otimes
E_{\mathcal{G}})\delta$ be the faithful conditional expectation induced on
$\mathrm{C}^{*}(\pi,t)$ by the faithful conditional expectation
$E_{\mathcal{G}}$ of $\mathrm{C}^{*}_{\lambda}({\mathcal{G}})$. Let $D$ be any
C*-algebra. Associativity of $\otimes_{\max}$ and nuclearity of
$\mathrm{C}^{*}_{\lambda}({\mathcal{G}})$ yields

$D\otimes_{\max}\mathrm{C}^{*}(\pi,t)\otimes_{\max}\mathrm{C}^{*}_{\lambda}({\mathcal{G}})\simeq(D\otimes_{\max}\mathrm{C}^{*}(\pi,t))\otimes\mathrm{C}^{*}_{\lambda}({\mathcal{G}})$

and so
${\operatorname{id}}_{D}\otimes_{\max}{\operatorname{id}}\otimes_{\max}E_{\mathcal{G}}=({\operatorname{id}}_{D}\otimes_{\max}{\operatorname{id}})\otimes
E_{\mathcal{G}}$ is faithful on
$D\otimes_{\max}\mathrm{C}^{*}(\pi,t)\otimes_{\max}\mathrm{C}^{*}_{\lambda}({\mathcal{G}})$.
Hence

${\operatorname{id}}_{D}\otimes_{\max}E:=({\operatorname{id}}_{D}\otimes_{\max}{\operatorname{id}}\otimes_{\max}E_{\mathcal{G}})({\operatorname{id}}_{D}\otimes_{\max}\delta)$

is a faithful conditional expectation of
$D\otimes_{\max}\mathrm{C}^{*}(\pi,t)$ on $D\otimes_{\max}B_{\mathcal{P}}$.
Therefore we have the following commutative diagram

$\textstyle{\mathrm{C}^{*}(\pi,t)\otimes_{\max}D\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{C}^{*}(\pi,t)\otimes
D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{{\mathcal{P}}}\otimes_{\max}D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{{\mathcal{P}}}\otimes
D}$

where the vertical arrows are faithful conditional expectations. Hence it
suffices to show that if $\pi\colon A\to B_{{\mathcal{P}}}$ is nuclear then
the fixed point algebra $B_{{\mathcal{P}}}$ is nuclear. As the latter is an
inductive limit, it suffices to show that nuclearity of $\pi$ in
$B_{{\mathcal{P}}}$ induces nuclearity of the embedding
$B_{\vartheta^{-1}(F)}\hookrightarrow B_{\vartheta^{-1}({\mathcal{P}})}$ for
every finite $\vee$-closed ${\mathcal{F}}\subseteq{\mathcal{P}}$. We will
actually show nuclearity of the embedding

$B_{\vartheta^{-1}({\mathcal{F}})}\hookrightarrow
B_{\vartheta^{-1}({\mathcal{F}}\cdot{\mathcal{P}})}\subseteq
B_{\vartheta^{-1}({\mathcal{P}})},$

where we write

$\vartheta^{-1}({\mathcal{F}}\cdot{\mathcal{P}})=\\{pP\mid\vartheta(p)\in{\mathcal{F}}\\}.$

First we remark that $B_{\mathcal{F}}$ contains a c.a.i. for
$B_{{\mathcal{F}}\cdot{\mathcal{P}}}$. Indeed let $(e_{i})$ be a c.a.i. for
$B_{\mathcal{F}}$ so that $\lim_{i}e_{i}\psi_{p}(k_{p})=\psi_{p}(k_{p})$ for
every $p\in\vartheta^{-1}({\mathcal{F}})$. Consequently
$\lim_{i}e_{i}t_{p}(\xi_{p})=t_{p}(\xi_{p})$ for every
$p\in\vartheta^{-1}({\mathcal{F}})$ and thus

$\lim_{i}e_{i}t_{p}(\xi_{p})t_{r}(\xi_{r})t_{s}(\xi_{s})^{*}=t_{p}(\xi_{p})t_{r}(\xi_{r})t_{s}(\xi_{s})^{*}\text{
for all }r,s\in P.$

Thus $\lim_{i}e_{i}\psi_{p,q}(k_{p,q})=\psi_{p,q}(k_{p,q})$ for every
$p,q\in\vartheta^{-1}({\mathcal{F}}\cdot P)$.

Now fix a finite $\vee$-closed ${\mathcal{F}}$. By using maximal elements we
can write ${\mathcal{F}}$ in levels, i.e.,

${\mathcal{F}}=\\{h_{11},\cdots,h_{1n_{1}},h_{21},\cdots,h_{2n_{2}},\cdots,h_{m1},\cdots,h_{mn_{m}}\\},$

such that every

${\mathcal{F}}_{i}:=\\{h_{i1},\dots,h_{in_{i}}\\}\text{ with }i\in\\{1,\dots
m\\},$

consists of the maximal elements of
${\mathcal{F}}\setminus\cup_{j=1}^{i-1}{\mathcal{F}}_{j}$ and
${\mathcal{F}}_{1}$ consists of the maximal elements of ${\mathcal{F}}$.

We now proceed by induction. For the base case let $h\in{\mathcal{P}}$ and
consider the space

$Y_{h}:=\sum_{p\in\vartheta^{-1}(h)}t_{p}(X_{p}).$

By using condition (A2) of Definition 5.1 we can equip $Y_{h}$ with the
$A$-valued bilinear map defined by

$\left\langle
y_{h},y^{\prime}_{h}\right\rangle:=y_{h}^{*}y^{\prime}_{h}\in\pi(A)\text{ for
all }y_{h},y^{\prime}_{h}\in Y_{h}.$

Then each $Y_{h}$ becomes a C*-correspondence over $A$, since $\pi$ is
faithful. The embedding
$Y_{h}\hookrightarrow\big{[}Y_{h}B_{\vartheta^{-1}({\mathcal{P}})}\big{]}$ and
nuclearity of $\pi(A)\hookrightarrow B_{\vartheta^{-1}({\mathcal{P}})}$ imply
nuclearity of the embedding

$B_{\vartheta^{-1}(h)}=[Y_{h}Y_{h}^{*}]\hookrightarrow\big{[}Y_{h}B_{\vartheta^{-1}({\mathcal{P}})}Y_{h}^{*}\big{]}=B_{\vartheta^{-1}(h{\mathcal{P}})},\text{
for all }h\in{\mathcal{P}},$

by [29, Proposition B.8]. Maximality of the $h_{1j}$ in ${\mathcal{F}}$ yields
that the $h_{1j}{\mathcal{P}}$ are minimal in $\\{h{\mathcal{P}}\mid
h\in{\mathcal{F}}\\}$ with respect to inclusions. As ${\mathcal{F}}$ is
$\vee$-closed we have that $h_{1j}{\mathcal{P}}\cap
h_{1j^{\prime}}{\mathcal{P}}=\emptyset$ for $j\neq j^{\prime}$. Hence the
C*-algebras $B_{\vartheta^{-1}(h_{1j})}$ are orthogonal and thus the embedding

$B_{{\mathcal{F}}_{1}}=\operatornamewithlimits{\sum^{\oplus}}_{j=1}^{n_{1}}B_{\vartheta^{-1}(h_{1j})}\hookrightarrow\sum_{j=1}^{n_{1}}B_{\vartheta^{-1}(h_{1j}{\mathcal{P}})}\subseteq
B_{\vartheta^{-1}({\mathcal{F}}_{1}\cdot{\mathcal{P}})}$

is nuclear. For the inductive hypothesis suppose that we have shown that the
embedding $B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}\hookrightarrow
B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}$ is nuclear for

${\mathcal{F}}^{\prime}=\\{h_{11},\cdots,h_{1n_{1}},\cdots
h_{i1},\cdots,h_{ij}\\}$

for some $j\in\\{1,\dots,n_{i}\\}$. If $j<n_{i}$ then set $h:=h_{i(j+1)}$; if
$j=n_{i}$ then set $h=h_{(i+1)1}$. We will show that the embedding

$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}\hookrightarrow
B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}\quad\text{for}\quad{\mathcal{F}}^{\prime\prime}:={\mathcal{F}}^{\prime}\cup\\{h\\}$

is nuclear. By construction $B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$ is an
ideal in $B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}$ and
$B_{\vartheta^{-1}(F^{\prime\prime})}=B_{\vartheta^{-1}(h)}+B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$;
thus

${\raisebox{1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}\simeq{\raisebox{1.00006pt}{$B_{\vartheta^{-1}(h)}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}(h)}\cap
B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}.$

Likewise $B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}$ is an
ideal of $B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}$.
From the base case we have nuclearity of the map

$B_{\vartheta^{-1}(h)}\hookrightarrow
B_{\vartheta^{-1}(h{\mathcal{P}})}\subseteq
B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}.$

By applying Lemma 6.9 on the commutative diagram of short exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\raisebox{1.00006pt}{$B_{\vartheta^{-1}(h)}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}(h)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\raisebox{1.00006pt}{$B_{\vartheta^{-1}(h)}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}(h)}\cap
B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot
P)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\raisebox{1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}$}\right.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

we get that the right vertical arrow is nuclear, i.e., the map

${\raisebox{1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}\simeq{\raisebox{1.00006pt}{$B_{\vartheta^{-1}(h)}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}(h)}\cap
B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}\longrightarrow{\raisebox{1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}$}\right.}$

is nuclear. Let $(e_{i})\subseteq B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$
be a c.a.i. for
$B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}$, and note that

$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}\cdot e_{i}\subseteq
B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}\cdot
B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}=B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}.$

Using the inductive hypothesis and Lemma 6.10 on the commutative diagram of
short exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\raisebox{1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime})}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime})}$}\right.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\raisebox{1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime\prime}\cdot{\mathcal{P}})}$}\left/\,\raisebox{-1.00006pt}{$B_{\vartheta^{-1}({\mathcal{F}}^{\prime}\cdot{\mathcal{P}})}$}\right.}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

we derive that the middle vertical arrow is nuclear, as required. This
concludes the inductive step. Now by using induction we derive that
$B_{\vartheta^{-1}({\mathcal{F}})}\hookrightarrow
B_{\vartheta^{-1}({\mathcal{F}}\cdot{\mathcal{P}})}$ is nuclear, and the proof
is complete. ∎

##  7\. Saturated controlled maps

###  7.1. A product system re-parametrization

Let $\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ be a controlled map
between weak right LCM-inclusions and let $X$ be a compactly aligned product
system over $P$ with coefficients in $A$. We can then define the
C*-correspondence

$Y_{h}:=\operatornamewithlimits{\sum^{\oplus}}_{p\in\vartheta^{-1}(h)}X_{p}\text{
for all }h\in{\mathcal{P}}.$

One is tempted to consider the family $Y=\\{Y_{h}\\}_{h\in{\mathcal{P}}}$ and
associate its C*-algebras with those of $X$. However it is not clear that $Y$
is in general a product system (let alone compactly aligned). Nevertheless
this happens for controlled maps that satisfy one extra condition.

######  Definition 7.1.

Let $\vartheta\colon(G,P)\rightarrow({\mathcal{G}},{\mathcal{P}})$ be a
controlled map of weak right LCM-inclusions. We say that $\vartheta$ is
_saturated_ if for any $h\in{\mathcal{P}}$ and
$t\in\vartheta^{-1}(h{\mathcal{P}})$ there exists an $s\in P$ with
$\vartheta(s){\mathcal{P}}=h{\mathcal{P}}$ and $t\in sP$.

######  Remark 7.2.

In particular, saturated maps satisfy the following property:

  1. (A3)

If $z\in{\mathcal{P}}^{*}$ then there exists an $x\in P^{*}$ such that
$\vartheta(x)=z$.

Indeed, we apply the saturation property for $z\in{\mathcal{P}}^{*}$ and
$t=e_{G}\in\vartheta^{-1}(z{\mathcal{P}})$ to obtain an $x\in P$ with
$e_{G}\in xP$. Hence we get that $P=xP$ giving that $x\in P^{*}$.

The following provides a good supply of saturated controlled maps. Recall that
a pair $(G,P)$ is a _total order_ if $G=P^{-1}\cup P$ and $P^{-1}\cap
P=\\{e_{G}\\}$. It is clear that total orders, being lattices, form weak right
LCM-inclusions.

######  Proposition 7.3.

Let $(G,P)$ be an abelian total order. For $n\in\mathbb{N}\cup\\{\infty\\}$
consider the free product $(\Asterisk_{i=1}^{n}G,\Asterisk_{i=1}^{n}P)$ of $n$
copies of $(G,P)$. Then the map

$\vartheta\colon(\Asterisk_{i=1}^{n}G,\Asterisk_{i=1}^{n}P)\longrightarrow(G,P);(g_{1})_{i_{1}}(g_{2})_{i_{2}}\dots(g_{k})_{i_{k}}\mapsto
g_{1}+g_{2}+\cdots+g_{k}$

is a saturated controlled map.

######  Proof..

For condition (A1) of Definition 5.1, if
$\overline{p},\overline{q}\in\Asterisk_{i=1}^{n}P$ with
$\overline{p}(\Asterisk_{i=1}^{n}P)\bigcap\overline{q}(\Asterisk_{i=1}^{n}P)\neq\emptyset$
then the freeness construction implies that either
$\overline{p}\leq\overline{q}$ or $\overline{q}\leq\overline{p}$.

For condition (A2) of Definition 5.1 suppose that $\overline{p},\overline{q}$
have a right LCM and they satisfy
$\vartheta(\overline{p})=\vartheta(\overline{q})$. Without loss of generality
assume that
$\overline{r}=\overline{p}^{-1}\overline{q}\in\Asterisk_{i=1}^{n}P$. Then
$\vartheta(\overline{r})=0$. If
$\overline{r}=(r_{1})_{i_{1}}\cdots(r_{k})_{i_{k}}$ then
$r_{1}+\cdots+r_{k}=0$ giving that $r_{k}\in-P\cap P=\\{0\\}$. Inductively we
get that $r_{1}=\cdots=r_{k}=0$ and so $\overline{p}=\overline{q}$.

Next we verify that $\vartheta$ is saturated. To this end let

$\overline{p}=(p_{1})_{i_{1}}(p_{2})_{i_{2}}\dots(p_{k})_{i_{k}},$

and let $h\in{\mathcal{P}}$ with

$h\leq\vartheta(\overline{p})=p_{1}+p_{2}+\dots+p_{k}.$

Let $\ell\in\\{1,\dots,k\\}$ be the smallest index so that $h\leq
p_{1}+p_{2}+\dots+p_{\ell}$. Set

$h^{\prime}=\begin{cases}h&\text{if }\ell=1,\\\
h-(p_{1}+p_{2}+\dots+p_{\ell-1})&\text{otherwise},\end{cases}$

and notice that $h^{\prime}\in P$ with $h^{\prime}\leq p_{\ell}$. Let

$\overline{q}=\begin{cases}(h^{\prime})_{i_{1}}&\text{if }\ell=1,\\\
(p_{1})_{i_{1}}(p_{2})_{i_{2}}\cdots(p_{\ell-1})_{i_{\ell-1}}(h^{\prime})_{i_{\ell}}&\text{otherwise.}\end{cases}$

Then $\overline{q}\leq\overline{p}$ and $\vartheta(\overline{q})=h$, as
desired. ∎

######  Example 7.4.

A second example comes from types of semi-direct products. Let $(G,P)$ and
$(H,S)$ be quasi-lattice ordered groups and let an action $\alpha\colon
H\to\operatorname{Aut}(G)$ such that $\alpha|_{S}\colon
S\to\operatorname{Aut}(P)$ restricts to automorphisms of $P$. Then we can form
the semi-direct products $G\rtimes_{\alpha}H$ and $P\rtimes_{\alpha}S$ with
respect to the relations $\alpha_{h}(g)h=hg$. The condition on $\alpha$ makes
$P\cdot S$ a subsemigroup of the semi-direct product, and in [24] it is shown
that the pair $(G\rtimes_{\alpha}H,P\rtimes_{\alpha}S)$ is quasi-lattice
ordered. Now suppose that $(G,P)$ admits an abelian controlled map
$\vartheta_{1}$ in $({\mathcal{G}}_{1},{\mathcal{P}}_{1})$ and $(H,S)$ admits
an abelian controlled map $\vartheta_{2}$ in $({\mathcal{G}}_{2},P_{2})$. In
order for the semi-direct product to inherit the obvious controlled map on
$({\mathcal{G}}_{1}\oplus{\mathcal{G}}_{2},{\mathcal{P}}_{1}\oplus{\mathcal{P}}_{2})$
it is necessary that $\alpha$ is $\vartheta_{1}$-invariant in the sense that
$\vartheta_{1}\alpha_{h}=\vartheta_{1}$ for all $h\in H$. We can then define
the homomorphism

$\vartheta\colon(G\rtimes_{\alpha}H,P\rtimes_{\alpha}S)\longrightarrow({\mathcal{G}}_{1}\oplus{\mathcal{G}}_{2},{\mathcal{P}}_{1}\oplus{\mathcal{P}}_{2})\textup{
such that }\vartheta(gh)=(\vartheta_{1}(g),\vartheta_{2}(h)).$

We claim that if $\vartheta_{1}$ and $\vartheta_{2}$ are saturated, then so is
$\vartheta$. Suppose that
$\vartheta(gh)=(\vartheta_{1}(g),\vartheta_{2}(h))\geq(m,\ell)$. Then there
are $s_{1},r_{1}\in G$ and $s_{2},r_{2}\in H$ such that

$g=s_{1}r_{1},\vartheta_{1}(s_{1})=m\quad\text{and}\quad
h=s_{2}r_{2},\vartheta(s_{2})=\ell.$

It follows that $\vartheta(s_{1}s_{2})=(m,\ell)$ and
$gh=s_{1}s_{2}\alpha_{s_{2}}^{-1}(r_{1})r_{2}$.

The following examples show that surjectivity is not enough to render a
controlled map saturated.

######  Example 7.5.

Take the free quasi-lattice on two symbols $a,b$ and take $\vartheta$ be its
abelianization map. Then for $ab$ and $(0,1)\in\mathbb{Z}^{2}$ we have that
$\vartheta(ab)=(1,1)\geq(0,1)$. However $\\{b\\}=\vartheta^{-1}((0,1))$ and
$ab\not\geq b$. (Although, Proposition 7.3 induces a saturated map on free
quasi-lattices.)

######  Example 7.6.

Consider the Baumslag-Solitar group $B(3,3)=\left\langle a,b\mid
a^{3}b=ba^{3}\right\rangle$. Recall that every element $x\in B(3,3)$ admits a
unique normal form

$x=a^{p_{1}}b^{\varepsilon_{1}}a^{p_{2}}\cdots
a^{p_{k}}b^{\varepsilon_{k}}a^{p_{k+1}}\text{ with
}p_{1},\dots,p_{k}\in\\{0,1,2\\},p_{k+1}\in\mathbb{Z},k\in\mathbb{Z}_{+}.$

Let $B_{+}(3,3)$ be its sub-semigroup generated by $a,b$. It follows that if
$x$ is in its normal form as above then

$x=a^{p_{1}}b^{\varepsilon_{1}}a^{p_{2}}\cdots
a^{p_{k}}b^{\varepsilon_{k}}a^{p_{k+1}}\in B_{+}(3,3)\quad\text{if and only
if}\quad\varepsilon_{1},\dots,\varepsilon_{k}=1,p_{k+1}\geq 0.$

By [44, Theorem 2.11] we have that the pair $(B(3,3),B_{+}(3,3))$ is a quasi-
lattice ordered group. In [24] it is shown that the abelianization gives a
surjective controlled map

$\vartheta\colon(B(3,3),B_{+}(3,3))\longrightarrow(\mathbb{Z}^{2},\mathbb{Z}_{+}^{2});a^{p_{1}}ba^{p_{2}}b\cdots
a^{p_{k}}ba^{p_{k+1}}\mapsto(p_{1}+\cdots+p_{k+1},k).$

However this map is not saturated. Take $t=a^{2}b$ and $h=(1,1)$ so that

$\vartheta(t)=(2,1)\in(1,1)+\mathbb{Z}_{+}^{2}.$

We have that $\vartheta^{-1}(1,1)=\\{ab,ba\\}$ and thus these are the only
choices for a possible $s$ with $\vartheta(s)=(1,1)$ and $s\leq t$. However we
see that

$(ab)^{-1}t=b^{-1}ab\notin
B_{+}(3,3)\quad\text{and}\quad(ba)^{-1}t=a^{-1}b^{-1}a^{2}b\notin B_{+}(3,3).$

######  Theorem 7.7.

Let $\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ be a saturated
controlled map between weak right LCM-inclusions. Let $X$ be a (resp.
injective) compactly aligned product system over $P$ with coefficients in $A$
and let

$Y_{h}:=\operatornamewithlimits{\sum^{\oplus}}_{p\in\vartheta^{-1}(h)}X_{p}\quad\text{for}\quad
h\in{\mathcal{P}}.$

Then the collection $Y=\\{Y_{h}\\}_{h\in{\mathcal{P}}}$ is a (resp. injective)
compactly aligned product system over ${\mathcal{P}}$ with coefficients in $A$
such that ${\mathcal{T}}_{\lambda}(X)^{+}\simeq{\mathcal{T}}_{\lambda}(Y)^{+}$
with

${\mathcal{T}}_{\lambda}(X)\simeq{\mathcal{T}}_{\lambda}(Y)\quad\text{and}\quad
A\times_{X,\lambda}P\simeq A\times_{Y,\lambda}{\mathcal{P}},$

by $*$-homomorphisms that preserve the inclusions $X_{p}\mapsto
Y_{\vartheta(p)}$ for all $p\in P$. These $*$-isomorphisms further lift to
$*$-isomorphisms

${\mathcal{N}}{\mathcal{T}}(X)\simeq{\mathcal{N}}{\mathcal{T}}(Y)\quad\text{and}\quad
A\times_{X}P\simeq A\times_{Y}{\mathcal{P}},$

that preserve the inclusions $X_{p}\hookrightarrow Y_{\vartheta(p)}$ for all
$p\in P$.

######  Proof..

Let $A$ act on both the left and right of each $Y_{h}$ with
$h\in{\mathcal{P}}$ via the usual multiplication of operators. By using
condition (A2) of Definition 5.1 we can equip $Y_{h}$ with the $A$-valued
bilinear map defined by

$\left\langle y_{h},y^{\prime}_{h}\right\rangle:=y_{h}^{*}y^{\prime}_{h}\in
A\subseteq{\mathcal{T}}_{\lambda}(X)\text{ for all }y_{h},y^{\prime}_{h}\in
Y_{h}.$

Then each $Y_{h}$ becomes a C*-correspondence over $A$. Since
$\ker\varphi_{Y_{h}}=\bigcap_{p\in\vartheta^{-1}(h)}\ker\varphi_{X_{p}}$ we
have that every $Y_{h}$ is injective when every $X_{p}$ is so.

We now show that $Y:=\\{Y_{h}\\}_{h\in{\mathcal{P}}}$ is a product system.
Since $[Y_{h}Y_{g}]\simeq Y_{h}\otimes_{A}Y_{g}$ we have to show that

$[Y_{h}Y_{g}]=Y_{hg}\text{ for all }h,g\in{\mathcal{P}}.$

As $(\overline{\pi},\overline{t})$ is a Toeplitz representation we have that
$Y_{h}Y_{g}\subseteq Y_{hg}$ for all $h,g\in{\mathcal{P}}$. For the reverse
inclusion, let $p\in P$ with $\vartheta(p)=hg$ and we will show that
$\overline{t}_{p}(X_{p})\in[Y_{h}Y_{g}]$. Since $\vartheta$ is saturated there
are $q,q^{\prime}\in P$ such that

$p=qq^{\prime}\quad\text{and}\quad\vartheta(q){\mathcal{P}}=h{\mathcal{P}}.$

We can write $\vartheta(q)=hz$ for some $z\in{\mathcal{P}}^{*}$ and let $w\in
P^{*}$ with $\vartheta(w)=z$ by condition (A3) of the saturation property.
Since $\vartheta(q)\vartheta(q^{\prime})=\vartheta(p)=hg$ it follows that
$\vartheta(q^{\prime})=z^{-1}g$. We thus conclude that

$\vartheta(qw^{-1})=h\quad\text{and}\quad\vartheta(wq^{\prime})=g.$

Recall that $X_{w}$ satisfies
$[\overline{t}_{w^{-1}}(X_{w^{-1}})\overline{t}_{w}(X_{w})]=\overline{\pi}(A)$.
By taking elementary vectors we get the required

$\displaystyle\overline{t}_{p}(X_{p})$
$\displaystyle=[\overline{t}_{q}(X_{q})\overline{t}_{q^{\prime}}(X_{q^{\prime}})]=[\overline{t}_{q}(X_{q})\overline{\pi}(A)\overline{t}_{q^{\prime}}(X_{q^{\prime}})]$
$\displaystyle=[\overline{t}_{q}(X_{q})\overline{t}_{w^{-1}}(X_{w^{-1}})\overline{t}_{w}(X_{w})\overline{t}_{q^{\prime}}(X_{q^{\prime}})]$
$\displaystyle\subseteq[\overline{t}_{qw^{-1}}(X_{qw^{-1}})\overline{t}_{wq^{\prime}}(X_{wq^{\prime}})]\subseteq[Y_{\vartheta(q)\vartheta(w^{-1})}Y_{\vartheta(w)\vartheta(q^{\prime})}]=[Y_{h}Y_{g}].$

Next we show that $Y$ is compactly aligned. Let $h,h^{\prime}\in{\mathcal{P}}$
and take $p\in\vartheta^{-1}(h)$ and $q\in\vartheta^{-1}(h^{\prime})$. If
$h\vee h^{\prime}=\infty$ then $p\vee q=\infty$ as well for all
$p\in\vartheta^{-1}(h)$ and $q\in\vartheta^{-1}(h^{\prime})$, and so

$Y_{h}^{*}Y_{h^{\prime}}=\sum_{p\in\vartheta^{-1}(h),q\in\vartheta^{-1}(h^{\prime})}\overline{t}_{p}(X_{p})^{*}\overline{t}_{q}(X_{q})=(0).$

On the other hand, if $h\vee h^{\prime}<\infty$ and $p\vee q<\infty$ for
$p\in\vartheta^{-1}(h)$ and $q\in\vartheta^{-1}(h^{\prime})$, then

$\overline{t}_{p}(X_{p})^{*}\overline{t}_{q}(X_{q})\subseteq[\overline{t}_{p^{-1}w}(X_{p^{-1}w})\overline{t}_{q^{-1}w}(X_{q^{-1}w})^{*}].$

Since $w=px=qy$ we have that
$\vartheta(p^{-1}w)=h^{-1}\vartheta(w)=(h^{\prime})^{-1}\vartheta(w)=\vartheta(q^{-1}w)$
and also $\vartheta(w)=h\vee h^{\prime}$. Hence

$Y_{h}^{*}Y_{h^{\prime}}=\sum_{p\in\vartheta^{-1}(h),q\in\vartheta^{-1}(h^{\prime}),p\vee
q<\infty}\overline{t}_{p}(X_{p})^{*}\overline{t}_{q}(X_{q})\subseteq[Y_{h^{-1}(h\vee
h^{\prime})}Y_{h^{\prime}(h\vee h^{\prime})}].$

Thus Proposition 4.7 gives that $Y$ is compactly aligned.

By definition we have that ${\mathcal{F}}X\simeq{\mathcal{F}}Y$ (by grouping
together summands with the same $\vartheta$-image), and therefore we have that
${\mathcal{T}}_{\lambda}(X)\simeq{\mathcal{T}}_{\lambda}(Y)$ and that
${\mathcal{T}}_{\lambda}(X)^{+}\simeq{\mathcal{T}}_{\lambda}(Y)^{+}$. Notice
that these identifications are ${\mathcal{G}}$-compatible. By applying
Proposition 5.9 and Theorem 4.14 we then get

$\displaystyle A\times_{X,\lambda}P$
$\displaystyle\simeq\mathrm{C}^{*}_{\textup{env}}({\mathcal{T}}_{\lambda}(X)^{+},G,\overline{\delta}_{G})\simeq\mathrm{C}^{*}_{\textup{env}}({\mathcal{T}}_{\lambda}(X)^{+},{\mathcal{G}},\overline{\delta}_{\mathcal{G}})\simeq\mathrm{C}^{*}_{\textup{env}}({\mathcal{T}}_{\lambda}(Y)^{+},{\mathcal{G}},\overline{\delta}_{\mathcal{G}})\simeq
A\times_{Y,\lambda}{\mathcal{P}}.$

The second part of the proof is treated likewise. First note that any
representation of $X$ lifts to a representation of $Y$ in a unique way, as
every fiber of $Y$ is spanned independently by the corresponding fibers of
$X$. Applying similar arguments as above for a representation $(\pi,t)$ in the
place of the Fock representation we see that this correspondence preserves
Nica-covariant representations. Hence we get that
${\mathcal{N}}{\mathcal{T}}(X)\simeq{\mathcal{N}}{\mathcal{T}}(Y)$.

Finally the $*$-isomorphisms $A\times_{X,\lambda}P\simeq
A\times_{Y,\lambda}{\mathcal{P}}$ gives an injective map

$[A\times_{X}P]_{pq^{-1}}\simeq[A\times_{X,\lambda}P]_{pq^{-1}}\hookrightarrow[A\times_{Y,\lambda}{\mathcal{P}}]_{\vartheta(pq^{-1})}\simeq[A\times_{Y}{\mathcal{P}}]_{\vartheta(pq^{-1})}.$

Therefore we get a commutative diagram

$\textstyle{{\mathcal{N}}{\mathcal{T}}(X)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Phi}$$\scriptstyle{q_{X}}$$\textstyle{{\mathcal{N}}{\mathcal{T}}(Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q_{Y}}$$\textstyle{A\times_{X}P\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Psi}$$\textstyle{A\times_{Y}{\mathcal{P}}}$

where the upper horizontal arrow is a $*$-isomorphism. Since the ideals of
strong covariance relations are induced, it suffices to show that

$\ker\Phi
q_{Y}\bigcap[{\mathcal{N}}{\mathcal{T}}(X)]_{e_{\mathcal{G}}}\subseteq\ker
q_{X}.$

Equivalently that $\Psi$ is faithful on the ${\mathcal{G}}$-fixed point
algebra defined on $A\times_{X}P$. However this follows by Corollary 5.8 as
$\Psi$ is by definition faithful on $A$. ∎

Theorem 7.7 gives a very clear picture for the covariance algebras of a
product system over a free product order of the form
$(\Asterisk_{i=1}^{n}G,\Asterisk_{i=1}^{n}P)$ for an abelian total order
$(G,P)$. It is well-known that the Cuntz C*-algebra ${\mathcal{O}}_{n}$,
$n\in\mathbb{N}$, can be viewed as either the Nica-Cuntz-Pimsner C*-algebra of
the trivial product system over the free semigroup on $n$ generators or as the
Cuntz-Pimsner C*-algebra of the C*-correspondence
$(\mathbb{C}^{n},\mathbb{C})$. Our next result generalizes this fact to
arbitrary product systems over the free semigroup.

######  Corollary 7.8.

Let $X$ be a compactly aligned product system over the free semigroup
$\mathbb{F}^{+}_{n}=\left\langle i_{1},\dots,i_{n}\right\rangle$. Then
$A\times_{X}\mathbb{F}^{+}_{n}\simeq{\mathcal{O}}_{Y}$ for the
C*-correspondence
$Y=\operatornamewithlimits{\sum^{\oplus}}\limits_{j=1,\dots,n}X_{i_{j}}$.

###  7.2. Reversible product systems and total orders

An application of Burns-Hale Theorem [6] asserts that $G$ admits a total order
if and only if for every non-trivial finitely-generated subgroup $H$ of $G$
there exists a totally ordered $L$ and a non-trivial homomorphism $H\to L$. If
$L=\mathbb{Z}$ then the group is called _left indicable_. There are plenty of
abelian total orders. Examples include $\mathbb{R}^{2}$ with the
lexicographical order and $\mathbb{Z}^{2}$ with the semigroup given by the
half-plane defined by any line through the origin with irrational slope.
Conrad’s Theorem asserts that if $(G,P)$ is a total order and $G$ is
Archimedean then $G$ embeds in $\mathbb{R}$ so that $P$ embeds in
$\mathbb{R}^{+}$ [9]. Here we say that $G$ is _Archimedean_ if whenever
$e_{G}<x<y$, there exists an $n\in\mathbb{N}$ such that $y<x^{n}$. We refer
the reader to [8] for an exposition of these results.

There are not many ways for a total order to be controlled by an _abelian_
total order.

######  Proposition 7.9.

Let $(G,P)$ be a total order and let
$\vartheta_{\textup{ab}}\colon(G,P)\to(G_{\textup{ab}},P_{\textup{ab}})$ be
the abelianization map. Then the following are equivalent:

  1. (i)

There is a controlled map
$\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ where
$({\mathcal{G}},{\mathcal{P}})$ is an abelian total order.

  2. (ii)

$\vartheta_{\textup{ab}}^{-1}(0)\cap P=\\{e_{G}\\}$.

  3. (iii)

The abelianization map is a controlled map and
$(G_{\textup{ab}},P_{\textup{ab}})$ is a total order.

If any (and thus all) of the above holds then the abelianization is a
saturated controlled map.

######  Proof..

If item (i) holds then $\vartheta$ factors through the abelianization. Since
$\vartheta^{-1}(e_{\mathcal{G}})\cap P=\\{e_{G}\\}$ being a controlled map,
then $\vartheta_{\textup{ab}}^{-1}(0)\cap P=\\{e_{G}\\}$ as well.

Assume that item (ii) holds and we will show that
$(G_{\textup{ab}},P_{\textup{ab}})$ is a total order. First we clearly have
that

$-P_{\textup{ab}}\cup P_{\textup{ab}}=\vartheta_{\textup{ab}}(P^{-1}\cup
P)=G_{\textup{ab}}.$

Next suppose that $-P_{\textup{ab}}\cap P_{\textup{ab}}\neq\\{0\\}$ so that
there are $h,g\in P_{\textup{ab}}$ with $h+g=0$. As the abelianization map is
surjective there are $p,q\in P$ with $pq=e_{G}$ with $\vartheta(p)=h$ and
$\vartheta(q)=g$. As $(G,P)$ is a total order we derive that $p=q=e_{G}$ and
thus $h=g=0$. Next we show that $\vartheta_{\textup{ab}}$ satisfies conditions
(A1) and (A2) of Definition 5.1. Let $p,q\in P$. Then either $p\leq q$ or
$q\leq p$ and condition (A1) follows. For condition (A2) suppose without loss
of generality that $p\leq q$ with
$\vartheta_{\textup{ab}}(p)=\vartheta_{\textup{ab}}(q)$. Then $q=ps$ for $s\in
P\cap\vartheta_{\textup{ab}}^{-1}(0)$. Then $s=e_{G}$ and so $p=q$.

If item (iii) holds then clearly item (i) holds, concluding the equivalences
between all items.

For the saturation property let a $t\in P$ and an $h\in P_{\textup{ab}}$ such
that $\vartheta_{\textup{ab}}(t)=h+h^{\prime}$. Take an
$s\in\vartheta_{\textup{ab}}^{-1}(h)$ since the abelianization map is
surjective. Then either $s\leq t$ or $s>t$. But if $s>t$ then
$h=\vartheta_{\textup{ab}}(s)>\vartheta_{\textup{ab}}(t)$ which is a
contradiction. Thus we must have that $s\leq t$. ∎

######  Remark 7.10.

There are exact total orders for which the abelianization map is not
controlled. An example is given by the Klein bottle group

$\mathbb{K}:=\langle x,y\mid x^{-1}yx=y^{-1}\rangle=\langle x,y\mid
x=yxy\rangle$

with the total order induced by the semigroup $\mathbb{K}^{+}$ generated by
$x,y$ in $\mathbb{K}$. It is not hard to see that $\mathbb{K}^{+}$ induces a
total order on $\mathbb{K}$, being left indicable (or since $\mathbb{K}$ is
the extension $\mathbb{Z}\rtimes\mathbb{Z}$). Alternatively one can see that
every element in $\mathbb{K}$ is written (uniquely) in the form $x^{m}y^{n}$
for $m,n\in\mathbb{Z}$ and we take cases: if $m,n\geq 0$ then
$x^{m}y^{n}\in\mathbb{K}$; if $m\geq 1$ and $n\leq 0$ then have that
$x^{m}y^{n}=x^{m-1}y^{-n}x\in\mathbb{K}^{+}$; if $m=0$ and $n\leq 0$ then
$x^{m}y^{n}=y^{n}\in(\mathbb{K}^{+})^{-1}$. By symmetry these cover all cases.
We see that $\vartheta_{\textup{ab}}(yxy)=\vartheta_{\textup{ab}}(x)$ and so
$e_{\mathbb{K}}\neq
y^{2}\in\mathbb{K}^{+}\cap\vartheta^{-1}_{\textup{ab}}(0)$. In fact we have
that $\mathbb{K}_{\textup{ab}}=\mathbb{Z}\times\mathbb{Z}_{2}$ and
$\mathbb{K}^{+}_{\textup{ab}}=\mathbb{Z}^{+}\times\mathbb{Z}_{2}$ and thus it
does not define a total order as
$-\mathbb{K}_{\textup{ab}}^{+}\cap\mathbb{K}^{+}_{\textup{ab}}=\mathbb{Z}_{2}$.

######  Definition 7.11.

Let $(G,P)$ be a total order and let $X$ be a product system over $P$ with
coefficients in $A$. We say that $X$ is a reversible product system if every
$X_{p}$ is a Hilbert bimodule in $A\times_{X,\lambda}P$, i.e., if
$A\times_{X,\lambda}P=\mathrm{C}^{*}(\pi,t)$ then
$t_{p}(X_{p})t_{p}(X_{p})^{*}\subseteq A$ for all $p\in P$.

It follows that reversible product systems consist of Hilbert bimodules. The
converse holds also for injective product systems, as in this case every
strongly covariant representation is Katsura-covariant fiberwise.

######  Proposition 7.12.

Let $(G,P)$ be a total order and let $X$ be a product system over $P$ with
coefficients in $A$. Suppose that every $X_{p}$ is injective. If $(\pi,t)$ is
a strongly covariant representation of $X$ then $(\pi,t_{p})$ is a covariant
representation of $X_{p}$, in the sense of Katsura, for every $p\in P$.

Therefore an injective product system $X$ is reversible if and only if every
$X_{p}$ is a Hilbert bimodule.

######  Proof..

Fix $p\in P$ and $a\in A$ such that
$\varphi_{p}(a)=k_{p}\in{\mathcal{K}}X_{p}$. In view of strong covariance of
Proposition 4.9 and Katsura covariance we have to show that

$\left[\pi_{F}(a)+\psi_{p,F}(k_{p})\right]_{X_{F}}=0\quad\text{for}\quad
F=\\{e,p\\},$

where

$X_{F}=\oplus_{r\in P}X_{r}I_{r^{-1}(r\vee F)}.$

Let $r\in P$ with $r=ps$ for some $s\in P$. Then for every
$\xi_{r}=\xi_{p}\xi_{s}\in X_{r}$ and $b\in I_{r^{-1}(r\vee F)}$ we have that

$\pi_{F}(a)\xi_{r}b=(\varphi_{p}(a)\xi_{p})\xi_{s}b=(k_{p}\xi_{p})\xi_{s}b=\psi_{F,p}(k_{p})\xi_{r}b.$

Now suppose that $r<p$. Then by construction $\psi_{F,p}(k_{p})\xi_{r}b=0$ and
we have to show that $\pi_{F}(a)\xi_{r}b=0$ as well. To this end it suffices
to show that

$I_{r^{-1}(r\vee F)}:=I_{r^{-1}K_{\\{r,e\\}}}\bigcap
I_{r^{-1}K_{\\{r,p\\}}}=(0).$

Since $r<p$ we have that $r\notin K_{\\{r,p\\}}\subseteq pP$ while $p\in
K_{\\{r,p\\}}$. Therefore $r^{-1}p\neq e_{G}$ and so

$I_{r^{-1}K_{\\{r,p\\}}}=\bigcap_{t\in
K_{\\{r,p\\}}}\ker\varphi_{r^{-1}t}\subseteq\ker\varphi_{r^{-1}p}=(0),$

and the proof is complete. ∎

In the case of $(G,P)=(\mathbb{Z},\mathbb{Z}_{+})$, the following result was
established in [23].

######  Proposition 7.13.

Let $(G,P)$ be a total order and let $X$ be a product system over $P$ with
coefficients in $A$. Then $X$ is a reversible product system if and only if
the tensor algebra ${\mathcal{T}}_{\lambda}(X)^{+}$ is Dirichlet in
$A\times_{X,\lambda}P$.

######  Proof..

Fix $(\pi,t)$ be a faithful representation of $A\times_{X,\lambda}P$. Suppose
first that $X$ is a reversible product system so that
$t_{p}(X_{p})t_{p}(X_{p})^{*}\subseteq\pi(A)$ for all $p\in P$. We will show
that

$A\times_{X,\lambda}P=\overline{\operatorname{span}}\\{t_{s}(X_{s})+t_{r}(X_{r})^{*}\mid
s,r\in P\\}.$

Let $s,r\in P$. If $rs^{-1}\in P$ then we have that

$t_{s}(X_{s})t_{r}(X_{r})^{*}\subseteq\big{[}t_{s}(X_{s})t_{s}(X_{s})^{*}t_{rs^{-1}}(X_{rs^{-1}})^{*}\big{]}\subseteq\big{[}\pi(A)t_{rs^{-1}}(X_{rs^{-1}})^{*}\big{]}=t_{rs^{-1}}(X_{rs^{-1}})^{*}.$

If $sr^{-1}\in P$ then we have that

$t_{s}(X_{s})t_{r}(X_{r})^{*}\subseteq\big{[}t_{sr^{-1}}(X_{sr^{-1}})t_{r}(X_{r})t_{r}(X_{r})^{*}\big{]}\subseteq\big{[}t_{sr^{-1}}(X_{sr^{-1}})\pi(A)\big{]}=t_{sr^{-1}}(X_{sr^{-1}}).$

Hence

$A\times_{X,\lambda}P=\overline{\operatorname{span}}\\{t_{s}(X_{s})t_{r}(X_{r})^{*}\mid
s,r\in
P\\}\subseteq\overline{\operatorname{span}}\\{t_{s}(X_{s})+t_{r}(X_{r})^{*}\mid
s,r\in P\\}\subseteq A\times_{X,\lambda}P,$

and so ${\mathcal{T}}_{\lambda}(X)^{+}$ is Dirichlet in
$A\times_{X,\lambda}P$.

Conversely, assume that ${\mathcal{T}}_{\lambda}(X)^{+}$ is Dirichlet in
$A\times_{X,\lambda}P$ and let $E$ be the conditional expectation induced by
the coaction of $G$ on $A\times_{X,\lambda}P$. Then
$E({\mathcal{T}}_{\lambda}(X)^{+})=\pi(A)$ and

$E(A\times_{X,\lambda}P)=E(\overline{{\mathcal{T}}_{\lambda}(X)^{+}+({\mathcal{T}}_{\lambda}(X)^{+})^{*}})=\pi(A).$

Thus for each $p\in P$ we have that $t_{p}(X_{p})t_{p}(X_{p})^{*}\subseteq
E(A\times_{X,\lambda}P)=\pi(A)$ as desired. ∎

The next corollary squares with the fact that Popescu’s non-commutative disc
algebra is not Dirichlet. Recall that for abelian coactions the C*-envelope of
a cosystem coincides with the usual C*-envelope of the ambient operator
algebra.

######  Corollary 7.14.

Let $\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ be a saturated
controlled map between weak right LCM-inclusions and suppose that
$({\mathcal{G}},{\mathcal{P}})$ is an abelian total order. Let $X$ be an
injective product system over $P$ with coefficients in $A$. Then
${\mathcal{T}}_{\lambda}(X)^{+}$ is Dirichlet if and only if every strongly
covariant representation $(\pi,t)$ of $X$ satisfies
$t_{p}(X_{p})t_{q}(X_{q})^{*}\subseteq A$ whenever
$\vartheta(p)=\vartheta(q)$.

######  Proof..

By Theorem 6.1, and since the controlling pair is abelian, the C*-envelope of
${\mathcal{T}}_{\lambda}(X)^{+}$ is $A\times_{X}P$. For the injective $X$, let
$Y$ be the injective product system over ${\mathcal{P}}$ with coefficients in
$A$ constructed in Theorem 7.7. By construction we see that $Y_{h}$ with
$h\in{\mathcal{P}}$ is a Hilbert bimodule if and only if
$t_{p}(X_{p})t_{q}(X_{q})^{*}\subseteq A$ for all $p,q\in\vartheta^{-1}(h)$.
By applying Remark 2.8, Theorem 6.1, Theorem 7.7, Proposition 7.12, and
Proposition 7.13 we have that the Fock tensor algebra
${\mathcal{T}}_{\lambda}(X)^{+}$ is Dirichlet in $A\times_{X}P$, if and only
if ${\mathcal{T}}_{\lambda}(Y)^{+}$ is Dirichlet in
$A\times_{Y}{\mathcal{P}}$, if and only if every $Y_{h}$ with
$h\in{\mathcal{P}}$ is a Hilbert bimodule, if and only if
$t_{p}(X_{p})t_{q}(X_{q})^{*}\subseteq\pi(A)$ whenever
$\vartheta(p)=\vartheta(q)=h$ for all $h\in{\mathcal{P}}$. ∎

The next theorem shows that, for weak right LCM-inclusions that are controlled
by total orders in a saturated way, reversible product systems produce all
possible covariance algebras.

######  Theorem 7.15.

Let $\vartheta\colon(G,P)\to({\mathcal{G}},{\mathcal{P}})$ be a saturated
controlled map between weak right LCM-inclusions and suppose that
$({\mathcal{G}},{\mathcal{P}})$ is a total order. Let $X$ be a (resp.
injective) product system over $P$ with coefficients in $A$. Then there exists
a (resp. injective) reversible product system $Z$ over ${\mathcal{P}}$ with
coefficients in a C*-algebra $B$ such that

(7.1) $A\subseteq B\quad\text{and}\quad X_{p}\subseteq Z_{\vartheta(p)}\text{
for all }p\in P,$

that satisfies

(7.2) $A\times_{X}P\simeq B\times_{Z}{\mathcal{P}}\quad\text{and}\quad
A\times_{X,\lambda}P\simeq B\times_{Z,\lambda}{\mathcal{P}},$

by $*$-homomorphisms that preserve the inclusions $X_{p}\hookrightarrow
Z_{\vartheta(p)}$ for all $p\in P$.

######  Proof..

By Theorem 7.7 we can assume that $(G,P)=({\mathcal{G}},{\mathcal{P}})$. Fix
$(\pi,t)$ be a faithful representation of $A\times_{X,\lambda}P$ and let

$B:=B_{P}=\mathrm{C}^{*}(\\{t_{s}(X_{s})t_{s}(X_{s})^{*}\mid s\in P\\}\mbox{
and }Z_{p}:=[t_{p}(X_{p})B]\text{ for all }p\in P\backslash\\{e\\}.$

The trivial C*-correspondence structure on $A\times_{X,\lambda}P$ descends to
a C*-correspondence structure on each $Z_{p}$ over $B$. Notice here that since
$(G,P)$ is totally ordered we automatically have that the product system
$Z=\\{Z_{p}\\}_{p\in P}$ is compactly aligned. Also
$\mathrm{C}^{*}(B,Z)=\mathrm{C}^{*}(\pi,t)$ admits a coaction of $G$ from
$A\times_{X,\lambda}P$. Hence by Theorem 4.13 we have that

$\overline{\operatorname{alg}}\\{B,Z_{p}\mid p\in
P\\}\simeq{\mathcal{T}}_{\lambda}(Z)^{+}.$

By construction

$A\times_{X,\lambda}P=\overline{{\mathcal{T}}_{\lambda}(Z)^{+}+({\mathcal{T}}_{\lambda}(Z)^{+})^{*}},$

thus the cosystem of ${\mathcal{T}}_{\lambda}(Z)^{+}$ over $G$ is Dirichlet in
a C*-cover. This gives at the same time that this C*-cover
$A\times_{X,\lambda}P$ is the C*-envelope of the cosystem
${\mathcal{T}}_{\lambda}(Z)^{+}$ over $G$, and that $Z$ is reversible by
Proposition 7.13. Theorem 4.14 then concludes that

$B\times_{Z,\lambda}P\simeq\mathrm{C}^{*}_{\textup{env}}({\mathcal{T}}_{\lambda}(Z)^{+},G,\overline{\delta}_{G})\simeq
A\times_{X,\lambda}P.$

For the case of the universal C*-algebras we proceed as in Theorem 7.7. That
is first we notice that the $*$-isomorphism between the reduced C*-algebras
implies an embedding of the Fell bundles

$[A\times_{X}P]_{pq^{-1}}\simeq[A\times_{X,\lambda}P]_{pq^{-1}}\hookrightarrow[B\times_{Z,\lambda}{\mathcal{P}}]_{pq^{-1}}\simeq[B\times_{Z}P]_{pq^{-1}}$

which lifts to a $*$-epimorphism $\Psi\colon A\times_{X}P\to B\times_{Z}P$.
Since $X\subseteq Z$ we also have a $*$-epimorphism at the level of the Nica-
Toeplitz-Pimsner algebras and thus the following diagram

$\textstyle{{\mathcal{N}}{\mathcal{T}}(X)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Phi}$$\scriptstyle{q_{X}}$$\textstyle{{\mathcal{N}}{\mathcal{T}}(Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q_{Z}}$$\textstyle{A\times_{X}P\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Psi}$$\textstyle{B\times_{Z}P}$

is commutative, and fixes $X$. Since the ideals of strong covariance relations
are induced, it suffices to show that

$\ker\Phi q_{Z}\bigcap[{\mathcal{N}}{\mathcal{T}}(X)]_{e}\subseteq\ker q_{X}.$

Equivalently that $\Psi$ is faithful on the $G$-fixed point algebra defined on
$A\times_{X}P$, which by definition is $B$. However this follows by the
property of $A\times_{X}P$ as $\Psi|_{A}$ is by construction faithful.

It is left to show that injective of $X$ implies injective of $Z$. By Theorem
7.7 we can still assume that $(G,P)=({\mathcal{G}},{\mathcal{P}})$. To this
end let $p\in P$ and $f\in\ker\varphi_{p}^{Z}$. We need to show that $f=0$.

As $B_{pP}$ is an ideal in $B$ we have that $B=B_{\\{s<p\\}}+B_{pP}$, and let
$f_{1}\in B_{\\{s<p\\}}$ and $f_{2}\in B_{pP}$ such that $f=f_{1}+f_{2}$. Let
$(e_{i})$ be a c.a.i. of $\psi_{p}({\mathcal{K}}X_{p})$ so that

$0=fe_{i}=f_{1}e_{i}+f_{2}e_{i}.$

However $(e_{i})$ is also a c.a.i. for $B_{pP}$ and so

$\lim_{i}f_{1}e_{i}=-\lim_{\lambda}f_{2}e_{i}=f_{2}.$

By Nica-covariance $f_{1}e_{i}\in B_{p}$ for all $i$, and so we have that
$f_{2}\in B_{p}$. Thus we can assume without loss of generality that $f\in
B_{\\{s\leq p\\}}$. As $B_{\\{s\leq p\\}}$ is the inductive limit of $B_{F}$
for $F=\\{p_{1}<p_{2}<\cdots<p_{n}=p\\}$ we may assume that

$f=\sum_{i=1}^{n}\psi_{p_{i}}(k_{p_{i}})\text{ with
}k_{p_{i}}\in{\mathcal{K}}X_{p_{i}}\text{ and }p_{1}<p_{2}<\cdots<p_{n}=p.$

Recall the representation $(\pi_{F},t_{F})$ on $X_{F}=\oplus_{r\in
P}X_{r}I_{r^{-1}(r\vee F)}^{X}$ and we will show that

$\sum_{i=1}^{n}\psi_{F,p_{i}}(k_{p_{i}})|_{X_{F}}=0.$

As $(\pi,t)$ is strongly covariant this will give that $f=0$ by Proposition
4.9. For $r\geq p$ we have that
$f\in\ker\varphi_{p}^{Z}\subseteq\ker\varphi_{r}^{Z}$, and for every
$\eta_{r}\in X_{r}I_{r^{-1}(r\vee F)}^{X}$ we have that $t_{r}(\eta_{r})\in
t_{r}(X_{r})\subseteq Z_{r}$. Hence

$t_{r}(\sum_{i=1}^{n}i_{p_{i}}^{r}(k_{p_{i}})(\eta_{r}))=\sum_{i=1}^{n}\psi_{p_{i}}(k_{p_{i}})t_{r}(\eta_{r})=ft_{r}(\eta_{r})=0.$

As $t$ is isometric we obtain

(7.3) $\sum_{i=1}^{n}\psi_{F,p_{i}}(k_{p_{i}})|_{X_{r}I_{r^{-1}(r\vee
F)}^{X}}=\sum_{i=1}^{n}i_{p_{i}}^{r}(k_{p_{i}})=0,\quad\text{for}\quad r\geq
p.$

On the other hand for $r<p$ we have that $r^{-1}p\neq e_{G}$ and so

$I_{r^{-1}(r\vee F)}^{X}\subseteq
I_{r^{-1}K_{\\{r,p\\}}}^{X}\subseteq\ker\varphi_{{r^{-1}p}}^{X}=(0).$

Hence trivially

(7.4) $\sum_{i=1}^{n}\psi_{F,p_{i}}(k_{p_{i}})|_{X_{r}I_{r^{-1}(r\vee
F)}^{X}}=0,\quad\text{for}\quad r<p.$

By equations (7.3) and (7.4) we have that
$\sum_{i=1}^{n}\psi_{p_{i},F}(k_{p_{i}})|_{X_{F}}=0$, and the proof is
complete. ∎

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