is balanced covered by any $\Lambda$-faithful representation.

  4. 4.

In Lemma 4.21 we show that in the presence of appropriate coactions, we can
lift a balanced covering to a covering of the entire algebras.

  5. 5.

Finally, we show in Theorem 4.22 that there is a unique tight,
$\Lambda$-faithful representation with a normal gauge coaction, and that this
representation is co-universal for $\Lambda$-faithful coacting representations
in the sense of [3, Theorem 5.3].

####  4.1.1 The Ultrafilter Representation

In this subsection, we construct the ultrafilter representation $f$ of a
$P$-graph $\Lambda$, and demonstrate that it is $\Lambda$-faithful and tight.
This construction follows the one in [3, Section 3] and is included here for
completeness.

######  Definition 4.1.

Let $(G,P)$ be a WQLO group, and $\Lambda$ a finitely aligned $P$-graph.
Recall that $\Lambda$ has a partial order given by $\alpha\leq\beta$ if
$\beta\in\alpha\Lambda$.

A filter of $\Lambda$ is a nonempty subset $U\subseteq\Lambda$ such that:

  1. (F1)

If $\mu\in U$ and $\lambda\leq\mu$, then $\lambda\in U$

  2. (F2)

If $\mu,\nu\in U$, there exists a $\lambda\in U$ such that $\mu\leq\lambda$
and $\nu\leq\lambda$.

Given a filter $U$, it is nonempty so it contains some $\mu$. By $F1$,
$r(\mu)\in U$, and by $F2$ since no distinct vertices have a common extension,
then $r(\mu)$ is the only vertex in $U$. Therefore, for any $\mu,\nu\in U$,
$r(\mu)=r(\nu)$, so we write $r(U)$ for this unique vertex.

A ultrafilter of $\Lambda$ is a maximal filter. We denote the set of filters
by $\widehat{\Lambda}$ and the set of ultrafilters by
$\widehat{\Lambda}_{\infty}$.

######  Lemma 4.2.

Every filter is contained in an ultrafilter.

###### Proof.

The proof is a standard Zorn’s Lemma argument: fix a filter $U$, and let
$\widehat{\Lambda}_{U}=\\{V\in\widehat{\Lambda}:U\subseteq V\\}$. We will show
that $\widehat{\Lambda}_{U}$ contains a maximal element by Zorn’s lemma, and
that this maximal element of $\widehat{\Lambda}_{U}$ is also maximal in
$\widehat{\Lambda}$.

To apply Zorn’s Lemma, we must check that
$\widehat{\Lambda}_{U}=\\{V\in\widehat{\Lambda}:U\subseteq V\\}$ is nonempty
and that chains in $\widehat{\Lambda}_{U}$ have an upper bound in
$\widehat{\Lambda}_{U}$. The former is immediate since
$U\in\widehat{\Lambda}_{U}$. If $\mathcal{C}$ is a chain in
$\widehat{\Lambda}_{U}$, then let $W=\bigcup_{V\in\mathcal{C}}V$, which we
will show is a filter. To check F1, if $\mu\in W$ and $\nu\leq\mu$, then
$\mu\in V$ for some $V\in C$, so $\nu\in V$ and thus $\nu\in W$. To check F2,
if $\mu,\nu\in W$, then $\mu\in V_{1}$, $\nu\in V_{2}$ for some
$V_{1},V_{2}\in\mathcal{C}$. Since $\mathcal{C}$ is a chain, either
$V_{1}\subseteq V_{2}$ or $V_{2}\subseteq V_{1}$. Without loss of generality
assuming the former, we have that $\mu,\nu\in V_{2}$, so there is some
$\lambda\in V_{2}\subseteq W$ with $\mu,\nu\leq\lambda$ as desired. Since $W$
also contains $U$, then $W\in\widehat{\Lambda}_{U}$, so by Zorn’s Lemma
$\widehat{\Lambda}_{U}$ contains a maximal element $U_{\infty}$.

Finally, we will show that $U_{\infty}$ is an ultrafilter, meaning that we
will check that $U_{\infty}$ is maximal in $\widehat{\Lambda}$. If it were
not, there would be some $V\supsetneq U_{\infty}\supseteq U$, but then we’d
have $V\in\widehat{\Lambda}_{U}$, so $U_{\infty}$ was not maximal in
$\widehat{\Lambda}_{U}$, a contradiction. Thus $U_{\infty}$ is indeed maximal
in $\widehat{\Lambda}$, so $U_{\infty}\in\widehat{\Lambda}_{\infty}$ as
desired.

∎

######  Lemma 4.3.

Let $U$ be an ultrafilter, and suppose $\mu\in U$, $E\subset\mu\Lambda$, and
that $E$ is exhaustive for $\mu\Lambda$. Then $E\cap U$ is nonempty.

###### Proof.

We split into two cases: U is finite and U is infinite. If $U$ is finite, it
has a greatest element $\omega$ by F2. Since $\mu\leq\omega$, then
$\omega\in\mu\Lambda$, and since $E$ is exhaustive for $\mu\Lambda$ there is
some $\alpha\in E$ with $MCE(\alpha,\omega)\neq\emptyset$. That is, we can
find some $\beta\in MCE(\alpha,\omega)$, so define
$U^{\prime}=\\{\nu:\nu\leq\beta\\}$. Then $U^{\prime}$ is a filter containing
$U$, and since $U$ was an ultrafilter then $U^{\prime}=U$. Thus $\beta\in U$
and by F1, $\alpha\leq\beta\in U$, as desired.

Suppose instead that $U$ is infinite. Since $\Lambda$ is countable, so is $U$,
so let give $U$ an enumeration $U=\\{\nu_{1},\nu_{2},...\\}$ and without loss
of generality suppose $\mu=\nu_{1}$. By the F2 property, for each
$n\in\mathbb{N}$, we may choose an $\eta_{n}$ such that
$\eta_{n-1}\leq\eta_{n}$ if $n>1$ and $\nu_{i}\leq\eta_{n}$ for all $1\leq
i\leq n$. Then in particular $\\{\eta_{n}\\}_{n\in\mathbb{N}}$ is a rising
sequence of elements of $U$ such that $\eta_{n}\in\mu\Lambda$ for all $n$, and
by the F1 property $\lambda\in U$ if and only if $\lambda\leq\eta_{n}$ for
some $n$. Finally, we may assume that the sequence
$\\{\eta_{n}\\}_{n\in\mathbb{N}}$ is strictly increasing by passing to a
subsequence.

Now, for each $\eta_{n}$, since $\eta_{n}\in\mu\Lambda$ and $E$ is exhaustive
is $\mu\Lambda$, there is an $\alpha_{n}\in E$ such that $\eta_{n}$ has a
common extension with $\alpha_{n}$. But $E$ is finite, so one $\alpha\in E$
must appear infinitely many times in the sequence
$\\{\alpha_{n}\\}_{n\in\mathbb{N}}$. Thus there are infinitely many $n$ with
$MCE(\eta_{n},\alpha)$ nonempty. Our goal now is to show that $\alpha\in U$ by
constructing a potentially bigger ultrafilter that contains $\alpha$, but
since $U$ is already an ultrafilter it must be that $\alpha$ is already in
$U$.

To this end, for $i\leq j$, if $\lambda\in MCE(\eta_{j},\alpha)$, we can
factorize $\lambda=\zeta\zeta^{\prime}$ where $d(\zeta)=d(\eta_{i})\vee
d(\alpha)$, which implies $\zeta\in MCE(\eta_{i},\alpha)$, and therefore
$\lambda\in\zeta\Lambda$ for some $\zeta\in MCE(\eta_{i},\alpha)$. Thus

$MCE(\eta_{j},\alpha)=\bigcup_{\zeta\in
MCE(\eta_{i},\alpha)}\left[MCE(\eta_{j},\alpha)\cap\zeta\Lambda\right].$

Considering $i$ as fixed but taking the union over all $j\geq i$, we have that

$\displaystyle\bigcup_{j\geq i}MCE(\eta_{j},\alpha)$ $\displaystyle=$
$\displaystyle\bigcup_{j\geq i}\left(\bigcup_{\zeta\in
MCE(\eta_{i},\alpha)}\left[MCE(\eta_{j},\alpha)\cap\zeta\Lambda\right]\right)$
$\displaystyle=$ $\displaystyle\bigcup_{\zeta\in
MCE(\eta_{i},\alpha)}\left(\bigcup_{j\geq
i}MCE(\eta_{j},\alpha)\cap\zeta\Lambda\right)$

Since infinitely many $MCE(\eta_{j},\alpha)$ are nonempty and the sequence
$\\{\eta_{n}\\}_{n\in\mathbb{N}}$ is strictly increasing, the lefthand side is
infinite, and thus the righthand side must have one of the terms
$\bigcup_{j\geq i}MCE(\eta_{j},\alpha)\cap\zeta\Lambda$ being infinite. In
particular, when $i=1$, there is some $\zeta_{1}$ such that
$MCE(\eta_{j},\alpha)\cap\zeta_{1}\Lambda$ is infinite. Then we may repeat a
version of this argument: we have

$\bigcup_{j\geq 2}MCE(\eta_{j},\alpha)\cap\zeta_{1}\Lambda=\bigcup_{\zeta\in
MCE(\eta_{2},\alpha)}\left(\bigcup_{j\geq
2}MCE(\eta_{j},\alpha)\cap\zeta\Lambda\cap\zeta_{1}\Lambda\right)$

so one of the terms $\bigcup_{j\geq
2}MCE(\eta_{j},\alpha)\cap\zeta\Lambda\cap\zeta_{1}$ must be infinite, so we
take $\zeta_{2}$ to be the element of $MCE(\eta_{2},\alpha)$ giving that
infinite term. Next, since $\zeta_{1}\leq\zeta_{2}$ then
$\zeta_{2}\Lambda\cap\zeta_{1}=\zeta_{2}\Lambda$, and we may repeat this
process inductively.

In this way we create a sequence $\\{\zeta_{n}\\}_{n\in\mathbb{N}}$ with
$\zeta_{n}\in MCE(\eta_{n},\alpha)$ and $\zeta_{n-1}\leq\zeta_{n}$ for all
$n$. Finally, let $U_{\infty}=\\{\lambda:\lambda\leq\zeta_{n}\text{ for some
}n\in\mathbb{N}\\}$. Then it is immediate to verify that $U_{\infty}$ is a
filter. Furthermore, if $\lambda\in U$, then $\lambda\leq\eta_{n}$ for some
$n$, so $\lambda\leq\eta_{n}\leq\zeta_{n}$, so $\lambda\in U_{\infty}$. That
is, $U\subseteq U_{\infty}$, but $U$ was an ultrafilter, so it must be that
$U=U_{\infty}$. Since $\alpha\leq\zeta_{1}$, then $\alpha\in U_{\infty}=U$, as
desired.

In either case, we have shown that there is some $\alpha\in E\cap U$. ∎

The following result is [3, Lemma 3.4]:

######  Lemma 4.4.

Let $(G,P)$ be a WQLO group, $\Lambda$ a finitely-aligned $P$-graph,
$\lambda\in\Lambda$ and let $U$ and $V$ be filters. If $r(U)=s(\lambda)$ and
if $\lambda\in V$, we define

$\lambda\cdot U:=\bigcup_{\mu\in U}\\{\alpha\leq\lambda\mu\\}\text{ and }$

$\lambda^{*}\cdot V:=\\{\mu\in\Lambda:\lambda\mu\in V\\}$

Then $\lambda\cdot U$ and $\lambda^{*}\cdot V$ are filters and if $U$ and $V$
are ultrafilters, then so are $\lambda\cdot U$ and $\lambda^{*}\cdot V$.
Finally, $\lambda^{*}\cdot(\lambda\cdot U)=U$ and
$\lambda\cdot(\lambda^{*}\cdot V)=V$.

We are now ready to define our $\Lambda$-faithful tight representation by
letting $\Lambda$ act by translation on its set of ultrafilters:

######  Lemma 4.5.

Let $(G,P)$ be a WQLO group, and $\Lambda$ a finitely-aligned $P$-graph. For
$\lambda\in\Lambda$, define an operator
$f_{\lambda}\in\mathcal{B}(\ell^{2}(\widehat{\Lambda}_{\infty}))$ by

$f_{\lambda}e_{U}=\begin{cases}e_{\lambda\cdot U}&\text{ if
}r(U)=s(\lambda)\\\ 0&\text{ otherwise}\end{cases}.$

Then $f$ is a representation of $\Lambda$ which is $\Lambda$-faithful and
tight. We call $f$ the ultrafilter representation.

###### Proof.

Defining the operators $f_{\lambda}$ as given above, a typical inner product
argument shows that $f_{\lambda}^{*}e_{V}=\begin{cases}e_{\lambda^{*}\cdot
V}&\text{ if }\lambda\in V\\\ 0&\text{ otherwise}\end{cases}$.

To check the T1 operator, it is immediate that if $v\in\Lambda^{0}$, then
$f_{v}$ is a projection, namely a projection onto the subspace spanned by the
set of ultrafilters containing $v$. To show that the
$\\{f_{v}\\}_{v\in\Lambda^{0}}$ are orthogonal, first note that this family
commutes, and fix distinct $v,w\in\Lambda^{0}$. For any ultrafilter $U$,
$r(U)$ is the unique vertex contained in $U$, so either $v\not\in U$ or
$w\not\in U$. In the former case, $f_{w}f_{v}e_{U}=0$ and in the latter case
$f_{v}f_{w}e_{U}=0$, so in either case $f_{v}f_{w}=f_{w}f_{v}=0$.

The T2 operator is immediate.

The T3 operator is immediate from the previous lemma.

For the T4 operator, observe that $f_{\lambda}f_{\lambda}^{*}$ is projection
onto the ultrafilters containing $\lambda$. Therefore,
$f_{\mu}f_{\mu}^{*}f_{\nu}f_{\nu}^{*}$ is projection onto the ultrafilters
containing $\mu$ and $\nu$. By the F2 and F1 properties, such an ultrafilter
would contain some $\lambda\in MCE(\mu,\nu)$. In fact, it would contain
exactly one such term, since if it contained two distinct
$\lambda_{1},\lambda_{2}\in MCE(\mu,\nu)$, by the F2 property it would contain
a common extension of $\lambda_{1}$ and $\lambda_{2}$, but no such extension
exists by the uniqueness of factorizations. Thus the set of ultrafilters
containing $\mu$ and $\nu$ is precisely the disjoint union of the ultrafilters
containing a $\lambda\in MCE(\mu,\nu)$. As operators, this is to say that
$f_{\mu}f_{\mu}^{*}f_{\nu}f_{\nu}^{*}=\displaystyle\sum_{\lambda\in
MCE(\mu,\nu)}f_{\lambda}f_{\lambda}^{*}$.

Thus $f$ is indeed a representation.

To show that $f$ is $\Lambda$-faithful, fix some $\lambda\in\Lambda$. Since
$\\{s(\lambda)\\}$ is a filter, then by Lemma 4.2, there is an ultrafilter $U$
containing $s(\lambda)$, in which case $r(U)=s(\lambda)$. Then,
$f_{\lambda}\cdot e_{U}=e_{\lambda\cdot U}\neq 0$, so $f_{\lambda}\neq 0$.

To show that $f$ is tight, fix some $\mu\in\Lambda$ and $E\subset\mu\Lambda$
which is finite and exhaustive for $\mu\Lambda$. Let
$B=\displaystyle\prod_{\alpha\in
E}(f_{\mu}f_{\mu}^{*}-f_{\alpha}f_{\alpha}^{*})$ be the corresponding bolt
(see Definition 4.10). We’ll now show that for any ultrafilter $U$,
$Be_{U}=0$, and this implies that $B=0$ as desired.

First, if $U$ does not contain $\mu$, then it also does not contain any of the
$\alpha\in E$, so $f_{\mu}f_{\mu}^{*}e_{U}=0=f_{\alpha}f_{\alpha}^{*}e_{U}$,
and thus $Be_{U}=0$.

If instead $U$ does contain $\mu$, by the previous lemma, then $U$ contains
one of the $\alpha\in E$.

Then, since $\mu,\alpha\in U$,
$f_{\mu}f_{\mu}^{*}e_{U}=e_{U}=f_{\alpha}f_{\alpha}^{*}U$, so
$Be_{U}=\displaystyle\prod_{\alpha\in
E}(f_{\mu}f_{\mu}^{*}-f_{\alpha}f_{\alpha}^{*})e_{U}=0$.

We have now shown that $Be_{U}=0$ for all ultrafilters $U$, so $B=0$ as an
operator in $\mathcal{B}(\ell^{2}(\widehat{\Lambda}_{\infty}))$. Thus $f$ is
tight.

∎

Since the ultrafilter representation is $\Lambda$-faithful and tight, it is
natural to ask if it also has a gauge coaction. The answer is often no, as we
will show in the next example.

######  Example 4.6.

Let $(G,P)$ be a WQLO group, and suppose that $P$ is directed, meaning that
any two elements of $P$ have a common upper bound. Examples of directed
positive cones include $\mathbb{N}^{k}$ and any total order. Then letting
$\Lambda=P$, $\Lambda$ itself is a filter, so $\Lambda$ is the unique
ultrafilter. Thus $\widehat{\Lambda}_{\infty}=\\{\Lambda\\}$, so the
ultrafilter representation is given by $f_{\mu}=1$ for all $\mu\in\Lambda$. In
particular, the ultrafilter representation fails to have a gauge coaction if
there are distinct $\mu,\nu\in\Lambda=P$, since such a coaction $\delta$
cannot send $f_{\mu}=f_{\nu}$ to the distinct elements $f_{\mu}\otimes
U_{d(\mu)}$ and $f_{\mu}\otimes U_{d(\nu)}$.

####  4.1.2 Balanced Algebras and Balanced Coverings

We will now discuss a particularly nice AF subalgebra of a $P$-graph algebra.
Our approach in this section mimics [3, Section 4], although we use slightly
different notation.

######  Lemma 4.7.

Let $(G,P)$ be a WQLO group, $\Lambda$ a finitely-aligned $P$-graph, and $t$ a
representation of $\Lambda$. Then
$\mathcal{B}(t)=\overline{\operatorname{span}}\\{t_{\mu}t_{\nu}^{*}:d(\mu)=d(\nu)\\}$
is a closed $*$-subalgebra of $C^{*}(t)$. We call $\mathcal{B}(t)$ the
balanced (sub)algebra.

###### Proof.

It is immediate that $\mathcal{B}(t)$ is a closed $*$-invariant subspace, so
it suffices to check that it is closed under multiplication. To this end,
suppose $t_{\mu}t_{\nu}^{*},t_{\alpha}t_{\beta}^{*}$ satisfy $d(\mu)=d(\nu)$
and $d(\alpha)=d(\beta)$. Then,

$\displaystyle(t_{\mu}t_{\nu}^{*})(t_{\alpha}t_{\beta}^{*})$ $\displaystyle=$
$\displaystyle(t_{\mu}t_{\nu}^{*})(t_{\nu}t_{\nu}^{*}t_{\alpha}t_{\alpha}^{*})(t_{\alpha}t_{\beta}^{*})$
$\displaystyle=$
$\displaystyle(t_{\mu}t_{\nu}^{*})(\displaystyle\sum_{\lambda\in
MCE(\nu,\alpha)}t_{\lambda}t_{\lambda}^{*})(t_{\alpha}t_{\beta}^{*})$
$\displaystyle=$ $\displaystyle\displaystyle\sum_{\lambda\in
MCE(\nu,\alpha)}t_{\mu(\nu^{-1}\lambda)}t_{\beta(\alpha^{-1}\lambda)}^{*}$

where $\nu^{-1}\lambda$ denotes the unique path such that
$\nu(\nu^{-1}\lambda)=\lambda$ and similarly $\alpha^{-1}\lambda$ denotes the
unique path such that $\alpha(\alpha^{-1}\lambda)=\lambda$. Then note that

$d(\mu(\nu^{-1}\lambda))=d(\mu)d(\nu)^{-1}d(\lambda)=d(\lambda)=d(\alpha)d(\beta)^{-1}d(\lambda)=d(\alpha(\beta^{-1}\lambda))$

so the product is in $\mathcal{B}(t)$ as desired. ∎

In the following lemma, we rephrase Lemma 2.26 for the context of a gauge
coaction on a $P$-graph $C^{*}$-algebra:

######  Lemma 4.8.

Let $(G,P)$ be a WQLO group, $\Lambda$ a finitely-aligned $P$-graph, $t$ a
representation of $\Lambda$ and suppose $t$ has a gauge coaction $\delta$.
Then there is a conditional expectation
$\Phi_{t}:C^{*}(t)\rightarrow\mathcal{B}(t)$ such that

$\Phi_{t}(t_{\mu}t_{\nu}^{*})=\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ otherwise}\end{cases}.$

###### Proof.

Let $A=C^{*}(t)$. For any $\mu,\nu\in\Lambda$, observe that
$\delta(t_{\mu}t_{\nu}^{*})=t_{\mu}t_{\nu}^{*}\otimes U_{d(\mu)d(\nu)^{-1}}$,
so $t_{\mu}t_{\nu}^{*}\in A_{d(\mu)d(\nu)^{-1}}$. Then $t_{\mu}t_{\nu}^{*}\in
A_{e}$ if and only if $d(\mu)=d(\nu)$, so $\mathcal{B}(t)=A_{e}$.

Let $\Phi_{t}:C^{*}(t)\rightarrow\mathcal{B}(t)$ be the conditional
expectation arising from Lemma 2.26. If $d(\mu)=d(\nu)$, then
$t_{\mu}t_{\nu}^{*}\in A_{e}$, and $\Phi_{t}$ fixes its range space, so
$\Phi_{t}(t_{\mu}t_{\nu}^{*})=t_{\mu}t_{\nu}^{*}$. If $d(\mu)\neq d(\nu)$,
then $t_{\mu}t_{\nu}^{*}\in A_{g}$ for $g=d(\mu)d(\nu)^{-1}\neq e$, and by
Lemma 2.26, $\Phi_{t}$ vanishes on all $A_{g}$ for $g\neq e$, so
$\Phi_{t}(t_{\mu}t_{\nu}^{*})=0$, as desired.

∎

Recall from Definition 2.27 that $\Phi_{t}$ is a faithful conditional
expectation if and only if $\delta$ is a normal coaction.

######  Definition 4.9.

Let $(G,P)$ be a WQLO group, $\Lambda$ a $P$-graph, and $s,t$ two
representations of $\Lambda$. Let us say that a balanced covering is a
(necessarily surjective) $*$-homomorphism
$\psi^{t}_{s}:\mathcal{B}(t)\rightarrow\mathcal{B}(s)$ given by

$\psi^{t}_{s}(t_{\mu}t_{\nu}^{*})=s_{\mu}s_{\nu}^{*}$

for all $\mu,\nu\in\Lambda$ with $d(\mu)=d(\nu)$.

If such a $*$-homomorphism exists, we will write $t\geq_{bal}s$. It is
immediate that $\geq_{bal}$ is reflexive and transitive, but it may not be a
partial ordering since it may not be antisymmetric (that is, there may be two
representations whose balanced algebras are isomorphic, but which are not
canonically isomorphic as representations). If $t\geq_{bal}s$ and
$s\geq_{bal}t$, we will write $t\cong_{bal}s$, and say that their balanced
algebras are canonically isomorphic.

When there may be ambiguity between a balanced covering a canonical covering,
we will write $\geq_{rep}$ to clarify that we mean a canonical covering of the
full $P$-graph $C^{*}$-algebra. Note that a canonical covering gives rise to a
balanced covering by restricting the canonical covering to the balanced
algebra. That is, $t\geq_{rep}s$ implies $t\geq_{bal}s$. The converse can
fail, but we prove a partial converse in Lemma 4.21.

####  4.1.3 The Kernel of $\psi^{\mathcal{T}}_{t}$

This section is devoted to proving the following fact which is our Theorem
4.15: letting $\mathcal{T}$ denote the Toeplitz representation from Definition
2.47, then for any representation $t$, $\ker\psi^{\mathcal{T}}_{t}$ is
generated (as an ideal) by the “bolts” (see Definition 4.10) and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it contains. This is our analogue of
[3, Theorem 4.9], although slightly generalized to allow the
$\ker\psi^{\mathcal{T}}_{t}$ to contain
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$.

This result is highly involved in the sense that the proof is many times
longer than the statement. The key to the argument is to show that the
property “$\ker\psi^{\mathcal{T}}_{t}\cap A$ is generated by the bolts and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it contains” is preserved under
direct limits, and that this property is true for a dense collection of
subalgebras $A\subseteq\mathcal{B}(\mathcal{T})$. This will take several
lemmas to establish.

######  Definition 4.10.

Let $(G,P)$ be a WQLO group, let $\Lambda$ be a finitely aligned $P$-graph and
let $t$ be a representation of $\Lambda$. We will say that a bolt in
$C^{*}(t)$ is an element of the form

$\displaystyle\prod_{\alpha\in
E}(t_{\mu}t_{\mu}^{*}-t_{\alpha}t_{\alpha}^{*})$

where $\mu\in\Lambda$ and $E\subset\mu\Lambda$ is finite and exhaustive for
$\mu\Lambda$. We say it is a proper bolt if $E\subset s(\mu)\Lambda\setminus
s(\mu)$.

The term “bolt” comes from a tortured metaphor: a representation is tight if
all of its bolts are fastened down (i.e. equal to 0).

######  Lemma 4.11.

Let $(G,P)$ be a WQLO group, let $\Lambda$ be a finitely aligned $P$-graph,
and let $t$ a representation of $\Lambda$.

Let $\\{A_{n}\\}_{n\in\mathbb{N}}$ be an increasing sequence of subalgebras of
$\mathcal{B}(\mathcal{T})$, and let
$A=\overline{\bigcup_{n\in\mathbb{N}}A_{n}}$. Suppose that for each
$n\in\mathbb{N}$, $\ker\psi^{\mathcal{T}}_{t}\cap A_{n}$ is generated by the
bolts and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it contains. Then
$\ker\psi^{\mathcal{T}}_{t}\cap A$ is generated by the bolts and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it contains.

###### Proof.

Let $J=\ker\psi^{\mathcal{T}}_{t}\cap A$ which is an ideal in $A$, and let $I$
denote the ideal generated by the bolts and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ contained in $J$. Certainly
$I\subseteq J$, and it suffices to show $I=J$.

By [5, Lemma III.4.1],

$J=\overline{\bigcup_{n\in\mathbb{N}}(J\cap A_{n})}$

But note that $J\cap A_{n}=\ker\psi^{\mathcal{T}}_{t}\cap A\cap
A_{n}=\ker\psi^{\mathcal{T}}_{t}\cap A_{n}$, so $J\cap A_{n}$ is generated by
the bolts and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it contains. Such a
bolt or $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ is certainly in $J\supseteq
J\cap A_{n}$, and thus $J\cap A_{n}\subseteq I$, and taking the union and
closure, we have

$I\supseteq\overline{\bigcup_{n\in\mathbb{N}}(J\cap A_{n})}=J$

so $I=J$ as desired.

∎

We will now build towards our dense family of subalgebras.

######  Definition 4.12.

Let $(G,P)$ be a WQLO group, let $\Lambda$ be a finitely aligned $P$-graph,
and let $S\subseteq\Lambda$. We will say that $S$ is MCE closed if for all
$\mu,\nu\in S$, $MCE(\mu,\nu)\subseteq S$.

We will say that $S$ is substitution closed if for all $\mu,\nu\in S$ with
$d(\mu)=d(\nu)$ and $s(\mu)=s(\nu)$, and for all $\alpha\in\Lambda$, we have
that $\mu\alpha\in S$ implies $\nu\alpha\in S$.

For $S\subseteq\Lambda$, we will write $D(S)=\\{d(\mu):\mu\in S\\}$.

As we will see in the next lemma, the condition that $S$ is MCE closed and
substitution closed is the correct condition in order to make
$A_{S}=\overline{\operatorname{span}}\\{\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}:\mu,\nu\in
S,d(\mu)=d(\nu)\\}$ a subalgebra of the balanced algebra.

######  Lemma 4.13.

Let $(G,P)$ be a WQLO group and let $\Lambda$ be a finitely-aligned $P$-graph.
For each $S\subseteq\Lambda$, let
$A_{S}=\overline{\operatorname{span}}\\{\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}:\mu,\nu\in
S,d(\mu)=d(\nu)\\}$. Then:

  1. 1.

If $S$ is MCE closed and substitution closed, then $A_{S}$ is a closed
∗-subalgebra of $\mathcal{B}(\mathcal{T})$.

  2. 2.

If $S$ is MCE closed and substitution closed, and $D(S)=\\{d(\mu):\mu\in S\\}$
contains a minimal element $m$, then $S^{\prime}=\\{\mu\in S:d(\mu)\neq m\\}$
is MCE closed and substitution closed.

  3. 3.

Let $S,D(S),m,$ and $S^{\prime}$ be as above. For any representation $t$ of
$\Lambda$, if $\mu,\nu\in\Lambda^{m}\cap S=\\{\lambda\in\Lambda\cap
S:d(\lambda)=m\\}$ with $s(\mu)=s(\nu)$, then
$t_{\mu}t_{\nu}^{*}\in\psi^{\mathcal{T}}_{t}(A_{S^{\prime}})$ implies that
either $t_{\mu}=0$ or there is a finite set $E\subset\mu\Lambda\cap
S^{\prime}$ which is exhaustive for $\mu\Lambda$ such that the bolt
$B=\displaystyle\prod_{\alpha\in
E}(t_{\mu}t_{\mu}^{*}-t_{\alpha}t_{\alpha}^{*})$ is equal to $0$.

###### Proof.

(1): It is immediate that $A_{S}$ is a closed $*$-invariant subspace of
$\mathcal{B}(\mathcal{T})$. It then suffices to check that it is closed under
multiplication. If
$\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*},\mathcal{T}_{\alpha}\mathcal{T}_{\beta}^{*}\in
A_{S}$, then

$\displaystyle(\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*})(\mathcal{T}_{\alpha}\mathcal{T}_{\beta}^{*})$
$\displaystyle=$
$\displaystyle(\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*})\displaystyle\sum_{\lambda\in
MCE(\nu,\alpha)}\mathcal{T}_{\lambda}\mathcal{T}_{\lambda}^{*}(\mathcal{T}_{\alpha}\mathcal{T}_{\beta}^{*})$
$\displaystyle=$ $\displaystyle\displaystyle\sum_{\lambda\in
MCE(\nu,\alpha)}\mathcal{T}_{\mu(\nu^{-1}\lambda)}\mathcal{T}_{\alpha(\beta^{-1}\lambda)}^{*}.$

Note that since $S$ is MCE closed, then each $\lambda\in MCE(\nu,\alpha)$ is
in $S$, and since $S$ is substitution closed, then
$\mu(\nu^{-1}\lambda),\alpha(\beta^{-1}\lambda)\in S$. Thus
$(\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*})(\mathcal{T}_{\alpha}\mathcal{T}_{\beta}^{*})\in
A_{S}$, as desired.

(2): Given $\mu,\nu\in S^{\prime}$, $MCE(\mu,\nu)\subseteq S$ since $S$ is MCE
closed, and for $\lambda\in MCE(\mu,\nu)$, we have $d(\lambda)\geq d(\mu)\neq
m$, so $d(\lambda)\neq m$, so $\lambda\in S^{\prime}$. Similarly, given
$\mu,\mu\alpha,\nu\in S^{\prime}$ with $d(\mu)=d(\nu)$, we have that
$\nu\alpha\in S$ since $S$ is substitution closed, but
$d(\nu\alpha)=d(\mu\alpha)\neq m$, so $\nu\alpha\in S^{\prime}$. Thus
$S^{\prime}$ is MCE closed, and substitution closed.

(3): Suppose that
$t_{\mu}t_{\nu}^{*}\in\psi^{\mathcal{T}}_{t}(A_{S^{\prime}})=\overline{\operatorname{span}}\\{t_{\alpha}t_{\beta}^{*}:\alpha,\beta\in
S^{\prime},d(\alpha)=d(\beta)\\}$. Then there is an element of
$\operatorname{span}\\{t_{\alpha}t_{\beta}^{*}:\alpha,\beta\in
S^{\prime},d(\alpha)=d(\beta)\\}$ within a distance of $1$ of
$t_{\mu}t_{\nu}^{*}$, which is to say there are $c_{i}\in\mathbb{C}$ and
$\alpha_{i},\beta_{i}\in S^{\prime}$ with $d(\alpha_{i})=d(\beta_{i})$ such
that writing
$x=t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{i=1}^{n}c_{i}t_{\alpha_{i}}t_{\beta_{i}}^{*}$,
then $\left\|{x}\right\|<1$.

Now, let $y=t_{\mu}t_{\mu}^{*}x$, we have
$\left\|{y}\right\|\leq\left\|{t_{\mu}}\right\|^{2}\cdot\left\|{x}\right\|<1$,
and

$\displaystyle y$ $\displaystyle=$ $\displaystyle
t_{\mu}t_{\mu}^{*}\left(t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{i=1}^{n}c_{i}t_{\alpha_{i}}t_{\beta_{i}}^{*}\right)$
$\displaystyle=$ $\displaystyle
t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{i=1}^{n}c_{i}t_{\mu}t_{\mu}^{*}t_{\alpha_{i}}t_{\beta_{i}}^{*}$
$\displaystyle=$ $\displaystyle
t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{i=1}^{n}\displaystyle\sum_{\lambda\in
MCE(\alpha_{i},\mu)}c_{i}t_{\lambda}t_{\beta(\alpha^{-1}\lambda)}$
$\displaystyle=$ $\displaystyle
t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{j=1}^{N}c_{j}t_{\lambda_{j}}t_{\eta_{j}}$

where the last line is simply relabelling the previous line. Note that the sum
is still indeed finite since $\Lambda$ is finitely aligned. Also, since $S$ is
substitution closed and MCE closed, then $\lambda_{j},\eta_{j}\in S$ for all
$j$. Since $d(\lambda_{j})\geq d(\alpha_{i})\neq m$ and $m$ was minimal, then
$d(\lambda_{j})\neq m$ and thus $\lambda_{j}\in S^{\prime}$. Also,
$\lambda_{j}$ is a common extension of $\mu$ and some $\alpha_{i}$, so
$\lambda_{j}\in\mu\Lambda$.

Now let $E=\\{\lambda_{1},...,\lambda_{N}\\}$, which is finite and satisfies
$E\subseteq\mu\Lambda\cap S^{\prime}$ by our previous remarks. Suppose for the
sake of contradiction that $E$ were not exhaustive for $\mu\Lambda$. Then
there would be some $\gamma\in\mu\Lambda$ such that
$MCE(\gamma,\lambda_{i})=\emptyset$ for all $\alpha_{i}\in E$. Then,
$\left\|{(t_{\gamma}t_{\gamma}^{*})y}\right\|\leq\left\|{t_{\gamma}}\right\|^{2}\left\|{y}\right\|<1$,
and

$\displaystyle(t_{\gamma}t_{\gamma}^{*})y$ $\displaystyle=$
$\displaystyle(t_{\gamma}t_{\gamma}^{*})(t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{j=1}^{N}c_{j}t_{\lambda_{j}}t_{\eta_{j}})$
$\displaystyle=$ $\displaystyle t_{\gamma}t_{\gamma}^{*}t_{\mu}t_{\nu}^{*}$

since each $MCE(\lambda_{i},\gamma)=\emptyset$ for $1\leq i\leq N$. Next,
since $\gamma\in\mu\Lambda$, then $\gamma=\mu\gamma^{\prime}$ for some
$\gamma^{\prime}\in\Lambda$, so we may simplify this expression to

$(t_{\gamma}t_{\gamma}^{*})(t_{\mu}t_{\nu}^{*})=t_{\mu\gamma^{\prime}}t_{\gamma^{\prime}}^{*}t_{\mu}^{*}t_{\mu}t_{\nu}^{*}=t_{\mu\gamma^{\prime}}t_{\nu\gamma^{\prime}}^{*}$

which is a partial isometry. Since partial isometries have norm either 0 or 1,
and we’ve seen it has norm less than 1, we have that
$0=t_{\mu\gamma^{\prime}}t_{\nu\gamma^{\prime}}^{*}$, and since
$s(\mu\gamma^{\prime})=s(\mu)=s(\nu)=s(\nu\gamma^{\prime})$, we have
$0=t_{s(\mu\mu_{1})}=t_{s(\mu)}=t_{\mu}^{*}t_{\mu}=t_{\mu}$, showing that
either $E$ is exhaustive or $t_{\mu}=0$, as desired.

Finally, we wish to show that $B=0$, where $B$ is the bolt
$B=\displaystyle\prod_{\lambda\in
E}(t_{\mu}t_{\mu}^{*}-t_{\lambda}t_{\lambda}^{*})$. To this end, we will make
a similar argument that $By$ is a partial isometry of norm less than 1, so it
must be 0. The latter is immediate:
$\left\|{By}\right\|\leq\left\|{B}\right\|\cdot\left\|{y}\right\|<1$. Now let
us simplify the expression $By$:

$\displaystyle By$ $\displaystyle=$ $\displaystyle
B(t_{\mu}t_{\nu}^{*}-\displaystyle\sum_{j=1}^{N}c_{j}t_{\lambda_{j}}t_{\eta_{j}})$
$\displaystyle=$ $\displaystyle
Bt_{\mu}t_{\nu}^{*}-\displaystyle\sum_{j=1}^{N}c_{j}\left(\displaystyle\prod_{k=1}^{N}(t_{\mu}t_{\mu}^{*}-t_{\lambda_{k}}t_{\lambda_{k}}^{*})\right)t_{\lambda_{j}}t_{\eta_{j}})$
$\displaystyle=$ $\displaystyle
Bt_{\mu}t_{\nu}^{*}-\displaystyle\sum_{j=1}^{N}c_{j}\left(\displaystyle\prod_{k\neq
j}^{N}(t_{\mu}t_{\mu}^{*}-t_{\lambda_{k}}t_{\lambda_{k}}^{*})\right)\left((t_{\mu}t_{\mu}^{*}-t_{\lambda_{j}}t_{\lambda_{j}}^{*})t_{\lambda_{j}}\right)t_{\eta_{j}})$
$\displaystyle=$ $\displaystyle
Bt_{\mu}t_{\nu}^{*}-\displaystyle\sum_{j=1}^{N}0$ $\displaystyle=$
$\displaystyle Bt_{\mu}t_{\nu}^{*}$

where the main simplification occurs since
$(t_{\mu}t_{\mu}^{*}-t_{\lambda_{j}}t_{\lambda_{j}}^{*})t_{\lambda_{j}}=t_{\lambda_{j}}-t_{\lambda_{j}}=0$.

Thus $By=Bt_{\mu}t_{\nu}^{*}$, so it is again a partial isometry of norm
strictly less than 1, and thus $By=0$. Finally, since $Bt_{\mu}t_{\nu}^{*}=0$,
then

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(Bt_{\mu}t_{\nu}^{*})(Bt_{\mu}t_{\nu}^{*})^{*}$ $\displaystyle=$
$\displaystyle Bt_{\mu}t_{\nu}^{*}t_{\nu}t_{\mu}^{*}B$ $\displaystyle=$
$\displaystyle Bt_{\mu}t_{\mu}^{*}B$

and recalling that $B$ is a subprojection of $t_{\mu}t_{\mu}^{*}$, we have
that $0=B$, completing the proof of (3).

∎

The next lemma shows that a sufficient collection of $A_{S}$ have the property
that $A_{S}\cap\ker\psi^{\mathcal{T}}_{t}$ is generated by the bolts and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$s it contains.

######  Lemma 4.14.

Let $(G,P)$ be a WQLO group and let $\Lambda$ be a finitely-aligned $P$-graph.
For each $S\subseteq\Lambda$ which is MCE closed and substitution closed, let
$A_{S}=\overline{\operatorname{span}}\\{\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}:\mu,\nu\in
S,d(\mu)=d(\nu)\\}$. Let $t$ be a representation of $\Lambda$, and let
$\psi^{\mathcal{T}}_{t}:\mathcal{B}(\mathcal{T})\rightarrow\mathcal{B}(t)$ be
the balanced covering. If $D(S)=\\{d(\mu):\mu\in S\\}$ is finite, then
$\ker\psi^{\mathcal{T}}_{t}\cap A_{S}$ is generated (as an ideal) by the bolts
and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$s it contains.

###### Proof.

Given a $MCE$-closed and substitution closed subset $S\subseteq\Lambda$, let
$J_{S}$ denote $\ker\psi^{\mathcal{T}}_{t}\cap A_{S}$, which is an ideal in
$A_{S}$. Let $I_{S}$ denote the ideal in $A_{S}$ generated by the bolts and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$s contained in $J_{S}$. Then certainly
$I_{S}\subseteq J_{S}$, and our claim is equivalent to proving that
$J_{S}/I_{S}=0$.

We proceed by induction on $|D(S)|$. In the base case of $|D(S)|=0$, the claim
is trivial since $A_{S}=J_{S}=I_{S}=0$.

In the inductive case, suppose that $|D(S)|>0$, and that the claim is true for
all $D^{\prime}$ with $|D^{\prime}|<|D(S)|$. Since $D(S)$ is finite, it has
some minimal element $m$. First consider the case where $S^{m}=\\{\mu\in
S:d(\mu)=m\\}$ is finite. Note that $S^{m}$ is itself substitution closed and
$MCE$-closed and that $A_{S}=A_{S^{\prime}}+A_{S^{m}}$.

Fix some $x+I_{S}\in J_{S}/I_{S}$, in which case $x\in J_{S}$. Since $x\in
J_{S}\subseteq A_{S}=A_{S^{\prime}}+A_{S^{m}}$, we may write
$x=x_{m}+x^{\prime}$, where $x_{m}\in A_{S^{m}}$ and $x^{\prime}\in
A_{S^{\prime}}$. Additionally, since $A_{S^{m}}$ is finite dimensional, then
we may write
$x_{m}=\displaystyle\sum_{i=1}^{n}c_{i}\mathcal{T}_{\mu_{i}}\mathcal{T}_{\nu_{i}}^{*}$
where $d(\mu_{i})=d(\nu_{i})=m$ for all $i$. There are potentially many such
representations of $x+I_{S}$ as $x_{m}+x^{\prime}+I_{S}$. We will assume
without loss of generality we have chosen the decomposition with the fewest
terms in the sum
$x_{m}=\displaystyle\sum_{i=1}^{n}c_{i}\mathcal{T}_{\mu_{i}}\mathcal{T}_{\nu_{i}}^{*}$,
and our goal is to show that in fact there are zero terms in the sum and thus
$x_{m}=0$.

To this end, suppose for the sake of contradiction that
$x_{m}=\displaystyle\sum_{i=1}^{n}c_{i}\mathcal{T}_{\mu_{i}}\mathcal{T}_{\nu_{i}}^{*}$
has a nonzero summand, and without loss of generality that it is the first
summand. That is, suppose
$c_{1}\mathcal{T}_{\mu_{1}}\mathcal{T}_{\nu_{1}}^{*}\neq 0$.

Now,

$\displaystyle(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*})x(\mathcal{T}_{\nu_{1}}\mathcal{T}_{\nu_{1}}^{*})$
$\displaystyle=$
$\displaystyle(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*})\left(\displaystyle\sum_{i=1}^{n}c_{i}\mathcal{T}_{\mu_{i}}\mathcal{T}_{\nu_{i}}^{*}+x^{\prime}\right)(\mathcal{T}_{\nu_{1}}\mathcal{T}_{\nu_{1}}^{*})$
$\displaystyle=$ $\displaystyle
c_{1}\mathcal{T}_{\mu_{1}}\mathcal{T}_{\nu_{1}}^{*}+(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*})x^{\prime}(\mathcal{T}_{\nu_{1}}\mathcal{T}_{\nu_{1}}^{*}).$

Since $x\in J_{S}\subseteq\ker\psi^{\mathcal{T}}_{t}$, by applying
$\ker\psi^{\mathcal{T}}_{t}$ to both sides, we have that

$0=c_{i}t_{\mu_{1}}t_{\nu_{1}}^{*}+\psi^{\mathcal{T}}_{t}((\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*})x^{\prime}(\mathcal{T}_{\nu_{1}}\mathcal{T}_{\nu_{1}}^{*}))$

so solving for $t_{\mu_{1}}t_{\nu_{1}}^{*}$, we have

$t_{\mu_{1}}t_{\nu_{1}}^{*}=\psi^{\mathcal{T}}_{t}(\frac{-1}{c_{1}}(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*})x^{\prime}(\mathcal{T}_{\nu_{1}}\mathcal{T}_{\nu_{1}}^{*}))$

but $x^{\prime}\in A_{S^{\prime}}$, and $A_{S^{m}}A_{S^{\prime}}\subseteq
A_{S^{\prime}}$, so the righthand side is an element of
$\psi^{\mathcal{T}_{t}}(A_{S^{\prime}})$. Now by part (3) of Lemma 4.13, we
have either $t_{\mu_{1}}=0$ or there is a finite exhaustive set
$E\subset{\mu_{1}}\Lambda$ with $d(\alpha)\in S\setminus\\{m\\}$ for all
$\alpha\in E$, and such that the bolt $B=\displaystyle\prod_{\alpha\in
E}(t_{\mu_{1}}t_{\mu_{1}}^{*}-t_{\alpha}t_{\alpha}^{*})$ is equal to $0$.

In the former case, $\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*}\in I_{S}$,
so taking
$y=c_{1}(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*})\mathcal{T}_{\mu_{1}}\mathcal{T}_{\nu_{1}}^{*}=c_{1}\mathcal{T}_{\mu_{1}}\mathcal{T}_{\nu_{1}}^{*}$,
we have $y\in I_{S}$, so $x+I_{S}=(x_{m}-y)+x^{\prime}+I_{S}$ is a
decomposition with one fewer terms in the sum $x_{m}-y$, a contradiction. In
the latter case, $\displaystyle\prod_{\alpha\in
E}(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*}-\mathcal{T}_{\alpha}\mathcal{T}_{\alpha}^{*})$
is a bolt in $\ker\pi^{\mathcal{T}}_{t}$, so
$B_{1}=\displaystyle\prod_{\alpha\in
E}(\mathcal{T}_{\mu_{1}}\mathcal{T}_{\mu_{1}}^{*}-\mathcal{T}_{\alpha}\mathcal{T}_{\alpha}^{*})\in
I_{S}$. Let $y=c_{1}B_{1}\mathcal{T}_{\mu_{1}}\mathcal{T}_{\nu_{1}}^{*}$, and
then we may write $x+I_{S}=x_{m}-y+x^{\prime}+I_{S}$ which will have zeroed
out the $c_{1}\mathcal{T}_{\mu_{1}}\mathcal{T}_{\nu_{1}}^{*}$ summand in the
$x_{m}$ term (while possibly adding more terms to the $x^{\prime}$ part, which
is acceptable), again a contradiction.

In either case, we found a contradiction with $x_{m}$ being the representative
with the fewest summands. Thus $x_{m}$ had no terms in its summation, and thus
$x_{m}=0$. That is to say, $x=x_{m}+x^{\prime}=x^{\prime}$, where
$x^{\prime}\in A_{S^{\prime}}$, and therefore $x\in
A_{S^{\prime}}\cap\ker^{\mathcal{T}}_{t}=J_{S^{\prime}}$. By the inductive
hypothesis, since $D^{\prime}=\\{d(\mu):\mu\in
S^{\prime}\\}=D(S)\setminus\\{m\\}$ has fewer terms than $D(S)$, we have that
$J_{S^{\prime}}=I_{S^{\prime}}$ so $x+I_{S}\in I_{S^{\prime}}+I_{S}=I_{S}$, so
$J_{S}\subseteq I_{S}$, as desired.

Now in the case that $S^{m}$ is not finite, we let $F_{n}$ be an increasing
sequence of finite subsets of $S^{m}$ with $S^{m}=\bigcup F_{n}$, in which
case $S^{\prime}\cup F_{n}$ is MCE closed and substitution-closed, so the
previous case applies to $A_{S^{\prime}\cup F_{n}}$. But
$A_{S}=\overline{\bigcup_{n=1}^{\infty}A_{S^{\prime}\cup F_{n}}}$, and so the
general case follows from Lemma 4.11. ∎

We can now add a second dash of Lemma 4.11 to get our desired result:

######  Theorem 4.15.

Let $(G,P)$ be a WQLO group and let $\Lambda$ be a finitely-aligned $P$-graph.
Let $t$ be a representation of $\Lambda$, and
$\psi^{\mathcal{T}}_{t}:\mathcal{B}(\mathcal{T})\rightarrow\mathcal{B}(t)$ be
the balanced covering. Then $\ker\psi^{\mathcal{T}}_{t}$ is generated (as an
ideal) by the bolts and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$s it contains.

###### Proof.

As above, for any $S\subseteq\Lambda$, let $D(S)=\\{d(\mu):\mu\in\Lambda\\}$.
Note that taking a substitution preserves the degree of a path, and taking
minimal common extensions of paths can take joins of degrees (that is, replace
paths of length $p$ and $q$ with paths of length $p\vee q$). Therefore, if
$F\subseteq\Lambda$ and $S$ is the set of all paths in $\Lambda$ obtained by
taking substitutions and MCEs of $F$, then $D(S)\subseteq\\{p_{1}\vee
p_{2}\vee...\vee p_{n}:p_{i}\in D(F)\\}$. In particular, if $F$ is finite,
then $|D(S)|\leq 2^{|F|}<\infty$, so any finite $F\subseteq\Lambda$ is
contained in a set $S\subseteq\Lambda$ which is MCE closed and substitution
closed and for which $D(S)$ is finite.

Since $\Lambda$ is countable, we can enumerate the elements as
$\lambda_{1},\lambda_{2},...$. For each $n\in\mathbb{N}$, let
$F_{n}=\\{\lambda_{1},...,\lambda_{n}\\}$, and let $S_{n}$ denote the set of
all paths in $\Lambda$ obtained by taking substitutions and MCEs of $F_{n}$.

Then $S_{n}$ is substitution closed and MCE closed and by the above argument,
$D(S_{n})<\infty$. Therefore, by Lemma 4.14, $\ker\psi^{\mathcal{T}}_{t}\cap
A_{S_{n}}$ is generated by the bolts and
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it contains.

But recall that
$\mathcal{B}(\mathcal{T})=\overline{\operatorname{span}}\\{\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}:d(\mu)=d(\nu)\\}$,
and for any $\mu,\nu\in\Lambda$ with $d(\mu)=d(\nu)$, since we have an
enumeration, there exists some $m,n\in\mathbb{N}$ with
$\mu=\lambda_{m},\nu=\lambda_{n}$, so $\mu,\nu\in F_{\max(m,n)}\subset
S_{\max(m,n)}$, so $\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}\in
A_{S_{\max(m,n)}}$. Thus
$\mathcal{B}(\mathcal{T})=\overline{\bigcup_{n\in\mathbb{N}}A_{S_{n}}}$.

Then by Lemma 4.11,
$\ker\psi^{\mathcal{T}}_{t}\cap\mathcal{B}(\mathcal{T})=\ker\psi^{\mathcal{T}}_{t}$
is generated by the bolts and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ it
contains. ∎

The above result classifies ideals in the balanced algebra
$\mathcal{B}(\mathcal{T})$, and when combined with the Factors Through Theorem
(Lemma 2.8) shows that two balanced algebras are isomorphic if and only if
they have the same set of bolts and range projections which are set equal to
0. We make use of this fact in the following lemma:

######  Lemma 4.16.

Let $(G,P)$ be a WQLO group, $\Lambda$ a $P$-graph, and let $s$ and $t$ be two
representations of $(G,P)$. Then

  1. 1.

If $t$ is tight and $s$ is $\Lambda$-faithful, then $t\leq_{bal}s$.

  2. 2.

If $t$ and $s$ are both tight and $\Lambda$-faithful, then $t\cong_{bal}s$.

###### Proof.

For the first claim, let $\mathcal{T}$ denote the universal representation of
$\Lambda$. Then, there are balanced coverings
$\psi^{\mathcal{T}}_{s}:\mathcal{B}(\mathcal{T})\rightarrow\mathcal{B}(s)$ and
$\psi^{\mathcal{T}}_{t}:\mathcal{B}(\mathcal{T})\rightarrow\mathcal{B}(t)$. We
wish to use the Factors Through Theorem (Lemma 2.8) to conclude there is a
balanced covering $\psi^{s}_{t}:\mathcal{B}(s)\rightarrow\mathcal{B}(t)$ given
by $s_{\mu}s_{\nu}^{*}\rightarrow t_{\mu}t_{\nu}^{*}$, which we can do if and
only if $\ker\psi^{\mathcal{T}}_{s}\subseteq\ker\psi^{\mathcal{T}}_{t}$.

By Theorem 4.15, $\ker\psi^{\mathcal{T}}_{s}$ and $\ker\psi^{\mathcal{T}}_{t}$
are generated by the bolts and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ they
contain. Since $s$ is $\Lambda$-faithful, its kernel contains no
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$ (and possibly some bolts), and since
$t$ is tight, its kernel contains every bolt (and possibly some
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$). Thus
$\ker\psi^{\mathcal{T}}_{s}\subseteq\ker\psi^{\mathcal{T}}_{t}$, so
$t\leq_{bal}s$ as desired.

For the second claim, by applying (1) twice, we have $t\leq_{bal}s$ and
$s\leq_{bal}t$, so $t\cong_{bal}s$. ∎

####  4.1.4 Normalizations of Coactions

We will now touch briefly on the normalization of a coaction. For a more
thorough introduction, the reader is directed to [6, Appendix A.7]

######  Remark 4.17.

Given a discrete coaction $(A,G,\delta)$, define $j_{A}:A\rightarrow A\otimes
C^{*}_{r}(G)$ by $j_{A}=(\operatorname{id}_{A}\otimes\pi^{U}_{L})\circ\delta$,
which is a $*$-homomorphism.

Let $A^{n}=j_{A}(A)\cong A/\ker j_{A}$. Then by Lemma A.55 and Definition A.56
of [6] there is a coaction $(A^{n},G,\delta^{n})$ which is normal, and such
that $j_{A}$ is $\delta$-$\delta^{n}$ equivariant, meaning that
$\delta^{n}\circ j_{A}=(j_{A}\otimes\operatorname{id}_{G})\circ\delta$. We
call $(A^{n},G,\delta^{n})$ the normalization of $(A,G,\delta)$.

Note that [6] uses a definition of normality which is equivalent to ours, but
not identical. Readers may wish to read [17, Lemma 1.4] for a proof of the
equivalence of the two definitions.

The following lemma is analogous to Lemma 2.50 but refers to a “normalization”
process for representations instead of a “coactionization” process. Note that
while coactionization made the representation larger with respect to
$\leq_{rep}$, normalization will make the representation smaller.

######  Lemma 4.18.

Let $(G,P)$ be a WQLO group, and $\Lambda$ a $P$-graph. Let $t$ be a
representation of $\Lambda$ with a gauge coaction $\delta$. Then,

  1. 1.

Let $j_{A}:C^{*}(t)\rightarrow C^{*}(t)^{n}$ denote the quotient map taking
the cosystem $(C^{*}(t),G,\delta)$ to its normalization
$(C^{*}(t)^{n},G,\delta^{n})$, and for $\lambda\in\Lambda$ let
$\tilde{t}_{\lambda}:=j_{A}(t_{\lambda})$. Then $\tilde{t}$ is a
representation of $\Lambda$.

  2. 2.

There is a canonical covering $t\mapsto\tilde{t}$.

  3. 3.

The coaction $\delta^{n}$ on $C^{*}(\tilde{t})$ is normal and is a gauge
coaction.

  4. 4.

$t$ is canonically isomorphic to $\tilde{t}$ if and only if $\delta$ is a
normal coaction.

  5. 5.

There is a balanced covering
$\psi^{t}_{\tilde{t}}:\mathcal{B}(t)\rightarrow\mathcal{B}(\tilde{t})$, which
is a balanced isomorphism.

  6. 6.

$\tilde{t}$ is $\Lambda$-faithful (respectively, tight) if and only if $t$ is.

######  Definition 4.19.

We call the representation $\tilde{t}$ the given in the previous lemma the
normalization of $t$.

###### Proof of 4.18.

(1) The fact that $\tilde{t}_{\lambda}$ is a representation is immediate since
it is the image of a representation under a homomorphism.

(2) $j_{A}$ is a canonical covering.

(3) $\delta^{n}$ is normal by construction. To show it is a gauge coaction,
recall from Definition A.56 and Lemma A.55 of [6] that the canonical covering
$j_{A}$ is $\delta$-$\delta^{n}$ equivariant, meaning that $\delta^{n}\circ
j_{A}=(j_{A}\otimes\operatorname{id}_{G})\circ\delta$. Then, for any
$\mu\in\Lambda$,

$\displaystyle\delta^{n}(\tilde{t}_{\mu})$ $\displaystyle=$
$\displaystyle\delta^{n}(j_{A}(t_{\mu}))$ $\displaystyle=$
$\displaystyle(j_{A}\otimes\operatorname{id}_{G})\circ\delta(t_{\mu})$
$\displaystyle=$
$\displaystyle(j_{A}\otimes\operatorname{id}_{G})(t_{\mu}\otimes U_{d(\mu)})$
$\displaystyle=$ $\displaystyle\tilde{t}_{\mu}\otimes U_{d(\mu)}$

as desired.

(4) By [17, Lemma 1.4], a coaction is normal (in the sense of its conditional
expectation $\Phi$ being faithful) if and only if $j_{A}$ is injective, which
is equivalent to $j_{A}$ being a canonical isomorphism.

(5) The balanced covering $\psi^{t}_{\tilde{t}}$ exists because it is a
restriction of the canonical covering $j_{A}$ from part (2). Since it is a
balanced covering, it is surjective, so it suffices to show
$\psi^{t}_{\tilde{t}}$ is injective. To this end, recall that
$\psi^{t}_{\tilde{t}}=j_{A}\mid_{\mathcal{B}(t)}$ and that
$j_{A}:=(\operatorname{id}_{A}\otimes\pi^{U}_{L})\circ\delta$. Now for
$x\in\mathcal{B}(t)$ we have $\delta(x)=x\otimes U_{e}$, so
$j_{A}(x)=((\operatorname{id}_{A}\otimes\pi^{U}_{L})\circ\delta)(x)=x\otimes
L_{e}=x\otimes 1$. Then since $\left\|{\cdot}\right\|$ is a $C^{*}$-cross
norm,
$\left\|{j_{A}(x)}\right\|=\left\|{x}\right\|\cdot\left\|{1}\right\|=\left\|{x}\right\|$,
so $x=0$ if and only if $\psi^{t}_{\tilde{t}}(x)=j_{A}(x)=0$, as desired.

(6) Having proven (5), this is immediate: since tightness is a relation among
elements of the balanced algebra, $t$ is tight if and only if $\tilde{t}$ is
tight. Similarly, $t$ is $\Lambda$-faithful if and only if every
$t_{\lambda}\neq 0$ if and only if every $t_{\lambda}^{*}t_{\lambda}\neq 0$,
which is a relation among elements of the balanced algebra, so $t$ is
$\Lambda$-faithful if and only if $\tilde{t}$ is.

∎

######  Corollary 4.20.

Let $(G,P)$ be a WQLO group, and let $\Lambda$ be a finitely aligned
$P$-graph. For any $\Lambda$-faithful, tight representation $t$ of $\Lambda$,
the normalization of the coactionization of $t$ is a $\Lambda$-faithful, tight
representation with a normal gauge coaction.

###### Proof.

Recall that $t^{\prime}$ denotes the coactionization of $t$ as in Lemma 2.50
and let $\widetilde{t^{\prime}}$ denoting the normalization of $t^{\prime}$ as
in Lemma 4.18. By those two lemmas, $\widetilde{t^{\prime}}$ has a normal
gauge coaction, and since $t$ is $\Lambda$-faithful and tight, so is
$\widetilde{t^{\prime}}$. ∎

####  4.1.5 The Co-Universal Algebra

The following result shows that a balanced covering extends to a covering of
the entire algebras in the context of a normal coaction:

######  Lemma 4.21.

Let $(G,P)$ be a WQLO group, let $\Lambda$ be a finitely-aligned $P$-graph,
and let $s$ and $t$ be two representations of $\Lambda$. If $s\leq_{bal}t$,
$s$ and $t$ have gauge coactions, and the gauge coaction on $s$ is normal,
then $s\leq_{rep}t$.

###### Proof.

Our argument will make use of several functions, organized according to the
following diagram (which will commute):

${C^{*}(\mathcal{T})}$${C^{*}(t)}$${\mathcal{B}(t)}$${C^{*}(s)}$${\mathcal{B}(s)}$$\scriptstyle{\pi^{\mathcal{T}}_{t}}$$\scriptstyle{\pi^{\mathcal{T}}_{s}}$$\scriptstyle{\Phi_{t}}$$\scriptstyle{\exists\pi^{t}_{s}}$$\scriptstyle{\psi^{t}_{s}}$$\scriptstyle{\Phi_{s}}$

Here $\mathcal{T}$ denotes the universal representation of $\Lambda$, and
$\pi^{\mathcal{T}}_{t}$ and $\pi^{\mathcal{T}}_{s}$ are the canonical
coverings of $C^{*}(t)$ and $C^{*}(s)$ respectively. We wish to show the
existence of the canonical covering $\pi^{t}_{s}$. The algebras
$\mathcal{B}(t)$ and $\mathcal{B}(s)$ are the balanced subalgebras as in Lemma
4.7 and the maps $\Phi_{t}:C^{*}(t)\rightarrow\mathcal{B}(t)$ and
$\Phi_{s}:C^{*}(s)\rightarrow\mathcal{B}(s)$ are the conditional expectations
arising from Lemma 4.8. From that lemma, these conditional expectations
satisfy

$\Phi_{t}(t_{\mu}t_{\nu}^{*})=\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ otherwise}\end{cases}\text{ and similarly
}\Phi_{s}(s_{\mu}s_{\nu}^{*})=\begin{cases}s_{\mu}s_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ otherwise}\end{cases}.$

Also note that by Definition 2.27, $\Phi_{s}$ is faithful because the gauge
coaction on $s$ is normal. Finally, $\psi^{t}_{s}$ is the balanced covering
that exists because $t\geq_{bal}s$.

We will now verify that this diagram commutes. That is, we will show that

$\psi^{t}_{s}\circ\Phi_{t}\circ\pi^{\mathcal{T}}_{t}=\Phi_{s}\circ\pi^{\mathcal{T}}_{s}.$

Since
$C^{*}(\mathcal{T})=\overline{\operatorname{span}}\\{\mathcal{T}_{\mu}\mathcal{T}_{\nu}:\mu,\nu\in\Lambda\\}$
and all the maps are linear and continuous, it suffices to check that the two
functions agree on each $\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}$. To this end,
we fix some $\mu,\nu\in\Lambda$, and we have

$\displaystyle\psi^{t}_{s}\circ\Phi_{t}\circ\pi^{\mathcal{T}}_{t}(\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*})$
$\displaystyle=$ $\displaystyle\psi^{t}_{s}\circ\Phi_{t}(t_{\mu}t_{\nu}^{*})$
$\displaystyle=$
$\displaystyle\psi^{t}_{s}\left(\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ otherwise}\end{cases}\right)$ $\displaystyle=$
$\displaystyle\begin{cases}s_{\mu}s_{\nu}^{*}&\text{ if }d(\mu)=d(\nu)\\\
0&\text{ otherwise}\end{cases}$ $\displaystyle=$
$\displaystyle\Phi_{s}(s_{\mu}s_{\nu}^{*})$ $\displaystyle=$
$\displaystyle\Phi_{s}\circ\pi^{\mathcal{T}}_{s}(\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*})$

as desired.

We will now show that
$\ker\pi^{\mathcal{T}}_{t}\subseteq\ker\pi^{\mathcal{T}}_{s}$. To this end,
suppose that $x\in\ker\pi^{\mathcal{T}}_{t}$. Then
$x^{*}x\in\ker\pi^{\mathcal{T}}_{t}$, so

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\psi^{t}_{s}\circ\Phi_{t}\circ\pi^{\mathcal{T}}_{t}(x^{*}x)$
$\displaystyle=$ $\displaystyle\Phi_{s}\circ\pi^{\mathcal{T}}_{s}(x^{*}x)$

and since $\pi^{\mathcal{T}}_{s}(x^{*}x)\geq 0$ and $\Phi_{s}$ is faithful,
then we must have that $\pi^{\mathcal{T}}_{s}(x^{*}x)=0$, so
$\pi^{\mathcal{T}}_{s}(x)=0$, and thus $x\in\ker\pi^{\mathcal{T}}_{s}$ as
desired.

Thus by Lemma 2.8 (the Factors Through Theorem), there exists a map
$\pi^{t}_{s}$ satisfying
$\pi^{\mathcal{T}}_{s}=\pi_{s}^{t}\circ\pi^{\mathcal{T}}_{t}$, from which it
is immediate that $\pi^{t}_{s}$ is a canonical covering. Thus $s\leq_{rep}t$.

∎

Now we may prove the main theorem of this section, which is a slight
generalization of [3, Theorem 5.3] from the context of quasi-lattice ordered
groups to weakly quasi-lattice ordered groups. We have also rephrased the
result:

######  Theorem 4.22.

Let $(G,P)$ be a WQLO group, and let $\Lambda$ be a finitely aligned
$P$-graph. If $s$ is a $\Lambda$-faithful, tight representation with a normal
gauge coaction, and $t$ is a $\Lambda$-faithful gauge-coacting representation,
then $s\leq t$.

In particular, there is a unique representation $S$ of $\Lambda$ which is
$\Lambda$-faithful, tight, and has a normal gauge coaction, and $S\leq t$ for
any $\Lambda$-faithful gauge-coacting representation $t$.

We write $C^{*}_{min}(\Lambda)$ for $C^{*}(S)$, and call
$C^{*}_{min}(\Lambda)$ the co-universal algebra of the graph.

###### Proof.

Suppose that $s$ was a $\Lambda$-faithful, tight representation with a normal
gauge coaction, and that $t$ was any $\Lambda$-faithful gauge-coacting
representation. Since $s$ is tight and $t$ is $\Lambda$-faithful, by Lemma
4.16, $s\leq_{bal}t$. Then since $s$ has a normal gauge coaction and $t$ has a
gauge coaction, by Lemma 4.21, $s\leq_{rep}t$.

To show the existence of such an $S$, recall that $f$ denotes the ultrafilter
representation from Lemma 4.5, which is $\Lambda$-faithful and tight by that
lemma. By Corollary 4.20, $S:=\widetilde{f^{\prime}}$ is a $\Lambda$-faithful,
tight representation with a normal gauge coaction.

If $t$ were another $\Lambda$-faithful, tight representation that had a normal
gauge coaction, then by the first part of the claim $t\leq_{rep}S$ and
$S\leq_{rep}t$, so $t\cong_{rep}S$. That is, $S$ is unique up to canonical
isomorphism.

∎

###  4.2 The Tight Algebra

The following result is a relatively straightforward combination of the
properties of $C^{*}_{min}(\Lambda)$ and the construction of a universal
algebra (Lemma 2.9).

######  Proposition 4.23.

Let $(G,P)$ be a WQLO group, and $\Lambda$ a finitely-aligned $P$-graph.

  1. 1.

There is a tight representation $T$ of $\Lambda$ which is universal for tight
representations. We will denote the algebra generated by this representation
as $C^{*}_{tight}(\Lambda)$.

  2. 2.

There is a canonical covering $\pi_{S}^{T}:C^{*}_{tight}(\Lambda)\rightarrow
C^{*}_{min}(\Lambda)$.

  3. 3.

$T$ is $\Lambda$-faithful.

  4. 4.

$T$ has a gauge coaction $\delta$.

  5. 5.

The gauge coaction $\delta$ on $C^{*}_{tight}(\Lambda)$ is normal if and only
if $\pi_{S}^{T}$ is an isomorphism, in which case $C^{*}_{tight}(\Lambda)\cong
C^{*}_{min}(\Lambda)$.

  6. 6.

If $\delta$ is normal, there is a gauge-invariant uniqueness theorem of this
form: if $t$ is another $\Lambda$-faithful, gauge coacting, tight
representation, then $C^{*}_{min}(\Lambda)$ is canonically isomorphic to
$C^{*}(t)$.

###### Proof.

For (1), the relators (T1)-(T4) and the tightness condition are all polynomial
relations in the generators, and $(T1)$ and $(T3)$ together imply that all the
generators are partial isometries, so by Lemma 2.9, there is a universal
$C^{*}$-algebra for such tight representations. Let us denote this
representation by $T$ and the algebra it generates by
$C^{*}_{tight}(\Lambda)$.

For (2), by Theorem 4.22, the representation $S$ which generates
$C^{*}_{min}(\Lambda)$ is a tight representation, so by the universality of
$T$, there is a canonical covering
$\pi^{T}_{S}:C^{*}_{tight}(\Lambda)=C^{*}(T)\rightarrow
C^{*}(S)=C^{*}_{min}(\Lambda)$.

For (3), by Theorem 4.22(1), $S$ is $\Lambda$-faithful, so each $S_{\lambda}$
is nonzero, and since $\pi^{T}_{S}(T_{\lambda})=S_{\lambda}$, then each
$T_{\lambda}$ is nonzero as well.

For (4), let $T^{\prime}$ denote the coactionization of $T$ given by
Proposition 2.50. By part (7) of that proposition, since $T$ is tight, then
$T^{\prime}$ is also tight, and therefore by the universality of $T$, there is
a canonical covering $T_{\lambda}\mapsto
T_{\lambda}^{\prime}=T_{\lambda}\otimes U_{d(\lambda)}$, which is the desired
gauge coaction.

For (5), if $S\cong T$, then the gauge coaction on $T$ is normal since the
gauge coaction on $S$ is normal. Conversely, if the gauge coaction on $T$ is
normal, then by the uniqueness part of Theorem 4.22, the canonical covering
$\pi_{S}^{T}$ is an isomorphism.

For (6), suppose that $\delta$ is normal and $t$ is a $\Lambda$-faithful gauge
coacting, tight representation. Then by the universality of
$C^{*}_{tight}(\Lambda)=C^{*}(T)$, $T\geq_{rep}t$ (here we are using the
partial order notation of Lemma 2.48). Similarly, by the co-universality of
$C^{*}_{min}(\Lambda)=C^{*}(S)$, $t\geq_{rep}S$. Finally, by (5) since
$\delta$ is normal, then $T\cong S$. Putting these together, we have that
$S\cong T\geq_{rep}t\geq_{rep}S$, so $t\cong S$.

∎

This is a pleasant result if $\delta$ is normal, but there are examples
arising from non-amenable groups where $\delta$ is not normal:

######  Lemma 4.24.

Let $(G,P)$ be a group with a total ordering (meaning $G=P\cup P^{-1}$ in
addition to $P\cap P^{-1}=\\{1\\}$), and let $\Lambda=P$, thought of as a
small category with one object and with degree functor
$\operatorname{id}_{P}$. Then any tight representation $t$ of $\Lambda=P$
extends uniquely to a unitary representation $\bar{t}$ of $G$ by
$\bar{t}_{g}=\begin{cases}t_{g}&\text{ if }g\in P\\\ t_{g^{-1}}^{*}&\text{ if
}g\in P^{-1}\end{cases}$.

Therefore, $C^{*}_{tight}(\Lambda)\cong C^{*}(G)$, and in particular the gauge
coaction on $C^{*}_{tight}(\Lambda)$ is normal if and only if $G$ is amenable.

###### Proof.

Let $(G,P)$ be a group with a total ordering, and let $\Lambda=P$, thought of
as a small category with the natural degree functor. Let $t$ be a tight
representation of $P$.

We will first show that $t_{p}$ is a unitary for each $p\in P$. For all
$p,q\in P$, $\max(p,q)\geq p,q$, so $p$ and $q$ have a common extension. That
is, every $p\in P$ is exhaustive for $P$, so by tightness, we have that
$1=t_{e}t_{e}^{*}=t_{p}t_{p}^{*}$ for all $p\in P$. Since also
$t_{p}t_{p}^{*}=t_{s(p)}=t_{e}=1$ by the (T3) relator, this says that the
$t_{p}$s are unitaries.

Now, extend $t$ to $\bar{t}$ on all of $G=P\cup P^{-1}$ by
$\bar{t}_{g}=\begin{cases}t_{g}&\text{ if }g\in P\\\ t_{g^{-1}}^{*}&\text{ if
}g\in P^{-1}\end{cases}$, or equivalently $\bar{t}_{p}=t_{p}$ and
$\bar{t}_{p^{-1}}=t_{p}^{*}$ for $p\in P$. We wish to show that $\bar{t}$ is a
representation of all of $G$. To this end, it suffices to check that it is a
group homomorphism, and since it extends a representation of $P$, it suffices
to check that multiplication is correct for one element in $P$ and one element
in $P^{-1}$.

To this end, suppose $p,q\in P$. We have several cases depending on which
products are positive. If $1\leq p^{-1}q$, then

$\bar{t}_{p^{-1}}\bar{t}_{q}=t_{p}^{*}t_{q}=t_{p}^{*}t_{p}t_{p^{-1}q}=t_{p^{-1}q}=\bar{t}_{p^{-1}q}$

and if instead $1\geq p^{-1}q$, then $1\leq q^{-1}p$, so

$\bar{t}_{p^{-1}}\bar{t}_{q}=t_{p}^{*}t_{q}=(t_{q}t_{q^{-1}p})^{*}t_{q}=t_{q^{-1}p}^{*}t_{q}^{*}t_{q}=t_{q^{-1}p}^{*}=\bar{t}_{p^{-1}q}$

so in either case $\bar{t}_{p^{-1}}\bar{t}_{q}=\bar{t}_{p^{-1}q}$.

For products the other way, if $pq^{-1}\geq 1$, then

$\bar{t}_{p}\bar{t}_{q^{-1}}=t_{p}t_{q}^{*}=t_{pq^{-1}}t_{q}t_{q}^{*}=t_{pq^{-1}}=\bar{t}_{pq^{-1}}$

while if $pq^{-1}\leq 1$, then $qp^{-1}\geq 1$, so

$\bar{t}_{p}\bar{t}_{q^{-1}}=t_{p}t_{q}^{*}=t_{p}(t_{qp^{-1}}t_{p})^{*}=t_{p}t_{p}^{*}t_{qp^{-1}}^{*}=t_{qp^{-1}}^{*}=\bar{t}_{pq^{-1}}$

so in either case $\bar{t}_{p}\bar{t}_{q^{-1}}=\bar{t}_{pq^{-1}}$. Thus
$\bar{t}$ is indeed a representation of $G$.

To show that $\bar{t}$ is the unique extension of $t$ to a unitary
representation of $G$, observe that if $\tilde{t}$ were another such
representation, then for all $p\in P$,
$1=\tilde{t}_{p}\tilde{t}_{p^{-1}}=t_{p}\tilde{t}_{p^{-1}}$, so since $t_{p}$
is a unitary we get that $t_{p}^{*}=\tilde{t}_{p^{-1}}$, so
$\tilde{t}=\bar{t}$.

Thus the algebra generated by the universal tight representation $T$ is the
algebra generated by the universal group representation. That is,
$C^{*}_{tight}(\Lambda)=C^{*}(G)$, and the gauge coaction on
$C^{*}_{tight}(\Lambda)$ is the standard coaction $\delta_{G}$ on $C^{*}(G)$
given in Example 2.24.

Now for normality, by Lemma 2.28, if $G$ is amenable, then all of its
coactions are amenable, and in particular $\delta_{G}$ is amenable.
Conversely, if the gauge coaction on $C^{*}_{tight}(\Lambda)\cong C^{*}(G)$ is
normal, then considering $\\{\mathbb{C}U_{g}\\}_{g\in\mathbb{C}}$ as a
topological grading over $G$ with conditional faithful expectation, by [9,
Proposition 3.7], $C^{*}(G)$ is naturally isomorphic to
$C^{*}_{r}(\\{\mathbb{C}U_{g}\\}_{g\in\mathbb{C}})=C^{*}_{r}(G)$, so the full
and reduced $C^{*}$-algebras are isomorphic and thus $G$ is amenable.

∎

######  Corollary 4.25.

If $(G,P)=(F_{2},R)$ where $R$ a total ordering on $G$ such as the Magnus
expansion given in Section 3.2 of [4], then $C^{*}_{tight}(\Lambda)\not\cong
C^{*}_{min}(\Lambda)$. Since both $C^{*}_{tight}(\Lambda)$ and
$C^{*}_{min}(\Lambda)$ are $\Lambda$-faithful, tight, gauge coacting
representations, there is no gauge-invariant uniqueness theorem for this
$R$-graph.

###### Proof.

By the previous lemma, $C^{*}_{tight}(\Lambda)\cong C^{*}(G)$, and by
Proposition 4.23 $C^{*}_{tight}(\Lambda)$ is a tight, $\Lambda$-faithful
representation of $\Lambda$ with a gauge coaction. Since $G=F_{2}$ is not
amenable, the gauge coaction is not normal. But by part (5) of 4.23, since the
gauge coaction is not normal, then $\pi^{T}_{S}$ is not an isomorphism, so
$C^{*}_{min}(\Lambda)$ is another tight, $\Lambda$-faithful representation of
$\Lambda$ with a gauge coaction, but which is not canonically isomorphic to
$C^{*}_{tight}(\Lambda)$.

∎

The reader may recognize the ordered group $(F_{2},R)$ from Example 3.6
earlier in this manuscript, where we used it as an example of an ordering that
had no reduction to an amenable group. This recognition may ignite a spark of
hope: perhaps the existence of a reduction to an amenable group prevents the
obstruction seen in the previous lemma and guarantees that $\delta$ is normal.
We shall see in the next section that this is indeed the case.

###  4.3 Reductions and Representations

In this section, we develop the notion that if $(G,P)$ reduces to $(H,Q)$,
then $P$-graphs are $Q$-graphs and they have the same representations.

######  Remark 4.26.

In this section, we will be considering a category $\Lambda$ as a graph with
respect to more than one semigroup (ex: as a $P$-graph and as a $Q$-graph). In
cases where there may be confusion, we will write $C^{*}_{min}(\Lambda,P)$ and
$C^{*}_{tight}(\Lambda,P)$ to indicate the semigroup $P$, and so on for other
semigroups.

First we will show that if $\Lambda$ is a $P$-graph, and $(G,P)$ has reduction
$(H,Q)$, then $\Lambda$ is also a $Q$-graph in a natural way.

######  Lemma 4.27.

Let $(G,P)$ be an ordered group, and let $(H,Q)$ be a reduction of $(G,P)$
with degree functor $d^{P}_{Q}:P\rightarrow Q$. Then a $P$-graph $\Lambda$
with degree functor $d_{P}^{\Lambda}$ is also a $Q$-graph with degree functor
$d=d^{P}_{Q}\circ d_{P}^{\Lambda}$.

###### Proof.

Certainly, $d$ maps $\Lambda$ into $Q$, so, it suffices to check that
$\Lambda$ has unique factorization with respect to $d$.

Let $\lambda\in\Lambda$, and $q\in Q$ satisfy $q\leq d(\lambda)$. We must show
that there is a unique $\alpha,\beta\in\Lambda$ such that
$\lambda=\alpha\beta$ and $d(\alpha)=q$.

For existence, since $d_{P}^{\Lambda}(\lambda)\in P$ is a path of length
$d^{P}_{Q}(d_{P}^{\Lambda}(\lambda))=d(\lambda)\geq q$, then by unique
factorization of $d^{P}_{Q}$, there exist unique $p,p_{1}\in P$ such that
$pp_{1}=d_{P}^{\Lambda}(\lambda)$ and $d^{P}_{Q}(p)=q$. Then $p\leq
d_{P}^{\Lambda}(\lambda)$, so by unique factorization of $d_{P}^{\Lambda}$,
there exist unique $\alpha,\beta\in\Lambda$ such that $\alpha\beta=pp_{1}$ and
$d_{P}^{\Lambda}(\alpha)=p$. Then
$d(\alpha)=d^{P}_{Q}(d_{P}^{\Lambda}(\alpha))=d^{P}_{Q}(p)=q$, so the desired
$\alpha,\beta$ exist.

For uniqueness, if $\alpha^{\prime},\beta^{\prime}\in\Lambda$ were two other
paths with $\lambda=\alpha^{\prime}\beta^{\prime}$ and $d(\alpha^{\prime})=q$,
then
$d_{P}^{\Lambda}(\lambda)=d_{P}^{\Lambda}(\alpha^{\prime})d_{P}^{\Lambda}(\beta^{\prime})$
would be a factorization of $d_{P}^{\Lambda}(\lambda)$ where the first term
has degree $d^{P}_{Q}(d_{P}^{\Lambda}(\alpha^{\prime}))=d(\alpha^{\prime})=q$,
so by uniqueness of the factorization of $d^{P}_{Q}$, we have that
$d_{P}^{\Lambda}(\alpha^{\prime})=p$. Then by the uniqueness of the
factorization $d_{P}^{\Lambda}$, we have $\alpha^{\prime}=\alpha$, as desired.

∎

Next, we will show that whether $\Lambda$ is regarded as a $P$-graph or a
$Q$-graph, it has the same representations.

######  Lemma 4.28.

Let $(G,P)$ and $(H,Q)$ be two WQLO groups, and $\Lambda$ a small category
with two functors $d^{\Lambda}_{P}:\Lambda\rightarrow P$ and
$d^{\Lambda}_{Q}:\Lambda\rightarrow Q$ such that $\Lambda$ is a $P$-graph with
respect to $d^{\Lambda}_{P}$ and a $Q$-graph with respect to
$d^{\Lambda}_{Q}$. Then,

  1. 1.

$\Lambda$ is finitely-aligned as a $P$-graph if and only if it is finitely-
aligned as a $Q$-graph.

  2. 2.

The (T1)-(T4) relators are the same for $\Lambda$ independently of being a
$P$-graph or a $Q$-graph.

  3. 3.

A representation is tight independently of $\Lambda$ being a $P$-graph or a
$Q$-graph.

###### Proof.

For (1), recall that $MCE(\mu,\nu)$ depends only on the category structure of
$\Lambda$ by definition, so it is finite regardless of the degree functor
given to $\Lambda$.

Similarly for (2) and (3), the T1-T4 and tightness relations as written only
depend on the category structure of $\Lambda$.

∎

######  Corollary 4.29.

Let $(G,P)$ and $(H,Q)$ be two WQLO groups, and $\Lambda$ a small category
with two functors $d^{\Lambda}_{P}:\Lambda\rightarrow P$ and
$d^{\Lambda}_{Q}:\Lambda\rightarrow Q$ such that $\Lambda$ is a finitely-
aligned $P$-graph with respect to $d^{\Lambda}_{P}$ and a finitely-aligned
$Q$-graph with respect to $d^{\Lambda}_{Q}$. Then $C^{*}_{tight}(\Lambda,P)$
is canonically isomorphic to $C^{*}_{tight}(\Lambda,Q)$.

###### Proof.

Let $a$ and $b$ be the universal tight representations of $\Lambda$ as a
$P$-graph and $Q$-graph, respectively, so that
$C^{*}_{tight}(\Lambda,P)=C^{*}(a)$ and $C^{*}_{tight}(\Lambda,Q)=C^{*}(b)$.

By the previous Lemma, both the $a_{\lambda}$s and the $b_{\lambda}$s satisfy
the equivalent T1-T4 and tightness relators, so by their respective universal
properties, there are canonical covers taking $a_{\lambda}\mapsto b_{\lambda}$
and $b_{\lambda}\mapsto a_{\lambda}$, which are inverses, and hence canonical
isomorphisms. ∎

######  Lemma 4.30.

Let $\varphi:(G,P)\rightarrow(H,Q)$ be a reduction of WQLO groups. Let
$\Lambda$ be a finitely-aligned $P$-graph, and $t$ a representation of
$\Lambda$. If $t$ has a gauge coaction by $G$, then it has a gauge coaction by
$H$ (when $\Lambda$ inherits the structure of a $Q$-graph as described in
4.27).

In particular, if $\delta$ is the gauge coaction by $G$, then
$\epsilon:=(\operatorname{id}_{C^{*}(t)}\otimes\bar{\varphi})\circ\delta$ is
the gauge coaction by $H$, where $\bar{\varphi}:C^{*}(G)\rightarrow C^{*}(H)$
is given by $U_{g}\mapsto V_{\varphi(g)}$ .

###### Proof.

First let us fix some notation. Let $\\{U_{g}\\}_{g\in G}$ and
$\\{V_{h}\\}_{h\in H}$ denote the universal unitary representations of $G$ and
$H$ respectively, so that $C^{*}(G)$ and $C^{*}(H)$ are the closed span of the
$U_{g}$ and $V_{h}$, respectively. Let $\delta$ denote the gauge coaction on
$C^{*}(t)$, $d^{\Lambda}_{P}$ the degree function on $\Lambda$ as a $P$-graph,
and recall that the degree functor on $\Lambda$ as a $Q$-graph was
$d^{\Lambda}_{Q}=\varphi\circ d^{\Lambda}_{P}$.

Then we have that $g\mapsto V_{\varphi(g)}$ is a unitary representation of
$G$, so by the universal property of $C^{*}(G)$, there is a $*$-homomorphism
$\bar{\varphi}:C^{*}(G)\rightarrow C^{*}(H)$ given by $U_{g}\mapsto
V_{\varphi(g)}$. We will prove that there is a gauge coaction of $H$ on
$C^{*}(t)$ given by
$\epsilon:=(\operatorname{id}_{C^{*}(t)}\otimes\bar{\varphi})\circ\delta$.
First note that by Lemma 2.14, there is such a function $\epsilon$, it is a
$*$-homomorphism, and it maps $C^{*}(t)\rightarrow C^{*}(t)\otimes C^{*}(H)$.

But now observe that

$\displaystyle\epsilon(t_{\lambda})$ $\displaystyle=$
$\displaystyle(\operatorname{id}_{A}\otimes\bar{\varphi})\circ\delta(t_{\lambda})$
$\displaystyle=$
$\displaystyle(\operatorname{id}_{A}\otimes\bar{\varphi})(t_{\lambda}\otimes
U_{d_{P}^{\Lambda}(\lambda)})$ $\displaystyle=$ $\displaystyle
t_{\lambda}\otimes V_{\varphi(d_{P}^{\Lambda}(\lambda))}$ $\displaystyle=$
$\displaystyle t_{\lambda}\otimes V_{d^{\Lambda}_{Q}(\lambda)}$

so by Proposition 2.50(3), $\epsilon$ is a gauge coaction by $H$, as desired.
∎

###  4.4 Gauge-Invariant Uniqueness for Ordered Groups which Reduce to
Amenable Groups

The following is the mathematical core of our gauge-invariant uniqueness
theorem:

######  Proposition 4.31.

Let $(G,P)$ be a WQLO group and suppose there is a reduction to an amenable
group $\varphi:(G,P)\rightarrow(H,Q)$. Then for any finitely-aligned $P$-graph
$\Lambda$, the following are canonically isomorphic:

  1. a.

$C^{*}_{tight}(\Lambda,P)$.

  2. b.

$C^{*}_{tight}(\Lambda,Q)$.

  3. c.

$C^{*}_{min}(\Lambda,P)$.

  4. d.

$C^{*}_{min}(\Lambda,Q)$.

  5. e.

$C^{*}(t)$, where $t$ is any representation of $\Lambda$ as a $P$-graph which
is tight, $\Lambda$-faithful, and which has a gauge coaction by $G$.

###### Proof.

For the sake of abbreviation, let us denote the representations generating
$C^{*}_{tight}(\Lambda,P)$, $C^{*}_{tight}(\Lambda,Q)$,
$C^{*}_{min}(\Lambda,P),$ and $C^{*}_{min}(\Lambda,Q)$ as $A,$ $B,$ $C,$ and
$D$, respectively. Our argument is organized by the following diagram:

${C^{*}(A)=C^{*}_{tight}(\Lambda,P)}$${C^{*}(B)=C^{*}_{tight}(\Lambda,Q)}$${C^{*}(C)=C^{*}_{min}(\Lambda,P)}$${C^{*}(D)=C^{*}_{min}(\Lambda,Q)}$$\scriptstyle{\pi_{B}^{A}}$$\scriptstyle{\pi_{C}^{A}}$$\scriptstyle{\pi_{D}^{B}}$$\scriptstyle{\pi^{C}_{D}}$

All arrows here denote canonical coverings.

The map $\pi_{B}^{A}$ exists and is an isomorphism by Corollary 4.29.

The maps $\pi_{C}^{A}$ and $\pi_{D}^{B}$ are the canonical coverings of
Proposition 4.23(2). Since $H$ is amenable, then by Lemma 2.28 the gauge
coaction by $H$ on $C^{*}_{tight}(\Lambda,Q)$ is normal, and so $\pi_{D}^{B}$
is a canonical isomorphism by Proposition 4.23 (5).

Combining these two, we have that $A\cong B\cong D$.

The map $\pi_{D}^{C}$ arises from the work established in the previous lemmas:
by Lemma 4.30 $C^{*}(C)$ carries a gauge coaction by $H$. Since also $C$ is
$\Lambda$-faithful by 4.22, then by the minimality of
$C^{*}_{min}(\Lambda,Q)$, there is a canonical covering
$\pi_{D}^{C}:C^{*}_{min}(\Lambda,P)\rightarrow C^{*}_{min}(\Lambda,Q)$.

Now, with the order notation of Lemma 2.48, we have that
$A\geq_{rep}C\geq_{rep}D\cong B\cong A$, so $A\cong B\cong C\cong D$. That is,
(a.)-(d.) are canonically isomorphic.

For (e.), we have that $A\geq_{rep}t\geq_{rep}C\cong A$ by the respective
universality and co-universality of $C^{*}(A)$ and $C^{*}(C)$, so $A\cong
t\cong C$. ∎

One nice outcome is that this result is independent of the amenable group
$(H,Q)$ and the reduction $\varphi$, and simply requires that one exist. In
particular, we have a gauge-invariant uniqueness theorem for WQLO groups which
have a reduction to an amenable group:

###### Proof.

Let $T$ be the universal tight representation as in Proposition 4.23 (1), so
$C^{*}(T)\cong C^{*}_{tight}(\Lambda)$. By that theorem, $T$ is
$\Lambda$-faithful, tight, and has a gauge coaction. It is also universal for
tight representations.

By Proposition 4.31, $C^{*}(T)\cong C^{*}_{min}(\Lambda)$, so $T$ is co-
universal for representations which are $\Lambda$-faithful and have gauge
coaction.

To see that $T$ is the unique representation (up to canonical isomorphism)
which is $\Lambda$-faithful, tight, and has a gauge coaction, if there were
such another representation $t$, it would be covered by $T$ (since $t$ is
tight, and $T$ is universal for tight representations) and it would cover $T$
(since $t$ is $\Lambda$-faithful and has a gauge coaction and $T$ is co-
universal for such representations). Thus $t\cong T$.

∎

######  Remark 4.32.

In the case that $(G,P)$ reduces to an amenable group, we believe that the
$C^{*}$-algebra generated by this unique $\Lambda$-faithful, tight, gauge
coacting representation deserves the title of _the_ $C^{*}$-algebra of the
graph. We will write $C^{*}(\Lambda)$ for this algebra.

###  4.5 Additional Results in the case of a Strong Reduction

In this section we will consider what further results can be provided about a
$P$-graph algebra under the assumption that $(G,P)$ strongly reduces to an
amenable group.

######  Lemma 4.33.

Let $(G,P)$ be a WQLO group, and suppose $(G,P)$ has a strong reduction to an
amenable group. Then every gauge coaction on a finitely-aligned $P$-graph is
normal.

###### Proof.

Let $\varphi:(G,P)\rightarrow(H,Q)$ be the strong reduction where $H$ is
amenable. Fix some finitely-aligned $P$-graph $\Lambda$ and a representation
$s$ of $\Lambda$ which has a gauge coaction by $G$.

Then by Lemma 4.30, $s$ has a gauge coaction by $H$. For sake of clarity, let
$\Phi_{s}$ and $\Omega_{s}$ denote the conditional expectations on $C^{*}(s)$
as in Lemma 2.26 which respectively arise from its $G$-coaction and
$H$-coaction.

By Lemma 2.28, since $H$ is amenable, then $\Omega_{s}$ is faithful. Our
desired result is that $\Phi_{s}$ is faithful.

To this end, it suffices to show that $\Omega_{s}=\Phi_{s}$ as maps on
$C^{*}(s)$, and since
$C^{*}(s)=\overline{\operatorname{span}}\\{s_{\alpha}s_{\beta}^{*}:\alpha,\beta\in\Lambda\\}$
and both functions are continuous and linear, it suffices to check that
$\Omega_{s}=\Phi_{s}$ on a single term $s_{\alpha}s_{\beta}^{*}$.

Recall that by Lemma 2.26, we have

$\Phi_{s}(s_{\alpha}s_{\beta}^{*})=\begin{cases}s_{\alpha}s_{\beta}^{*}&\text{
if }d(\alpha)=d(\beta)\\\ 0&\text{ if }d(\alpha)\neq d(\beta)\end{cases}$

and

$\Omega_{s}(s_{\alpha}s_{\beta}^{*})=\begin{cases}s_{\alpha}s_{\beta}^{*}&\text{
if }\varphi(d(\alpha))=\varphi(d(\beta))\\\ 0&\text{ if
}\varphi(d(\alpha))\neq\varphi(d(\beta))\end{cases}.$

But $\varphi$ is a strong reduction, so for the positive elements
$d(\alpha),d(\beta)\in P$, we have $d(\alpha)=d(\beta)$ if and only if
$\varphi(d(\alpha))=\varphi(d(\beta))$.

Thus $\Omega_{s}=\Phi_{s}$ on each spanning element, so they are equal as
functions, and thus $\Phi_{s}$ is faithful, so $s$ is normal.

∎

######  Lemma 4.34.

Let $(G,P)$ be a WQLO group with a strong reduction onto an amenable group,
let $\Lambda$ be a $P$-graph, and let $s$ be a representations of $(G,P)$. If
$s$ is $\Lambda$-faithful, has a gauge coaction, and every proper bolt in $s$
is nonzero, then $s\cong\mathcal{T}$.

###### Proof.

Since $\mathcal{T}$ covers $s$, it suffices to show that
$\ker\pi^{\mathcal{T}}_{s}=0$.

To this end, consider the diagram

${C^{*}(\mathcal{T})}$${\mathcal{B}(\mathcal{T})}$${C^{*}(s)}$${\mathcal{B}(s)}$$\scriptstyle{\Phi_{\mathcal{T}}}$$\scriptstyle{\pi^{\mathcal{T}}_{s}}$$\scriptstyle{\psi^{\mathcal{T}}_{s}}$$\scriptstyle{\Phi_{s}}$

which commutes by a routine computation on the spanning elements
$\mathcal{T}_{\mu}\mathcal{T}_{\nu}^{*}$.

Fix some $x\in\ker\pi^{\mathcal{T}}_{s}=0$. Then
$\pi^{\mathcal{T}}_{s}(x^{*}x)=0$, so

$0=\Phi_{s}(\pi^{\mathcal{T}}_{s}(x^{*}x))=\psi^{\mathcal{T}}_{s}(\Phi_{\mathcal{T}}(x^{*}x)).$

Now by Theorem 4.15, since $\ker\psi^{\mathcal{T}}_{s}$ is generated by the
bolts and $\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$s it contains, and by the
hypotheses on $s$ this kernel contains no proper bolts or
$\mathcal{T}_{\mu}\mathcal{T}_{\mu}^{*}$, then we have
$\ker\psi^{\mathcal{T}}_{s}=0$, and thus $\Phi_{\mathcal{T}}(x^{*}x)=0$.

By Lemma 4.33, $\Phi_{\mathcal{T}}$ is faithful, so $x^{*}x=0$ and thus $x=0$.
Thus $\ker\pi_{s}^{\mathcal{T}}=\\{0\\}$, as desired. ∎

######  Corollary 4.35.

Let $\Lambda$ be a finitely-aligned $\mathbb{N}^{2}*\mathbb{N}$-graph. Then
any gauge coaction by $G=\mathbb{Z}^{2}*\mathbb{Z}$ is automatically normal.

###### Proof.

Recall that $(\mathbb{Z}^{2}*\mathbb{Z},\mathbb{N}^{2}*\mathbb{N})$ strongly
reduces to $(\mathbb{Z}^{2}\wr\mathbb{Z},\mathbb{N}^{2}\wr\mathbb{N})$, which
is amenable. Then by Lemma 4.33, the gauge coaction is normal.

∎

######  Lemma 4.36.

Let $(G,P)$ be a WQLO group and suppose that $(G,P)$ has a strong reduction to
an amenable group. Let $\Lambda$ be a finitely-aligned $P$-graph, and let $t$
be a representation of $\Lambda$. Then the following are equivalent:

  1. 1.

$t$ has a gauge coaction by $G$

  2. 2.

There is a bounded linear map $\Phi_{t}:C^{*}(t)\rightarrow\mathcal{B}(t)$
satisfying

$\Phi_{t}(t_{\mu}t_{\nu}^{*})=\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ if }d(\mu)\neq d(\nu)\end{cases}$

###### Proof.

$(1\Rightarrow 2)$ If $t$ has a gauge coaction, then such a map $\Phi_{t}$
exists by Lemma 4.8.

$(2\Rightarrow 1)$ If such a $\Phi_{t}$ exists, then by Theorem 19.1 and
Definition 19.2 of [10], writing
$B_{g}=\overline{\operatorname{span}}\\{t_{\mu}t_{\nu}^{*}:d(\mu)d(\nu)^{-1}=g\\}$,
we have that $\mathcal{B}=\\{B_{g}\\}_{g\in G}$ are linearly independent and
form a topological grading of $C^{*}(t)$.

In particular, $\mathcal{B}$ is a Fell bundle, and we can form the cross-
sectional algebra $C^{*}(\mathcal{B})$ as in Remark 2.21. Each graded
component $B_{g}$ is naturally embedded in $C^{*}(\mathcal{B})$, so for each
$\lambda\in\Lambda$, let $s_{\lambda}$ denote the operator arising from
embedding $t_{\lambda}\in B_{d(\lambda)}$ into $C^{*}(\mathcal{B})$.

But one may immediately check the T1-T4 relators to check that
$\\{s_{\lambda}\\}_{\lambda\in\Lambda}$ is a representation of $\Lambda$ in
$C^{*}(\mathcal{B})$. Note that since the
$\\{s_{\lambda}\\}_{\lambda\in\Lambda}$ generate each fiber $B_{g}$, then they
generate all of $C^{*}(\mathcal{B})$, so $C^{*}(\mathcal{B})=C^{*}(s)$.

Our goal is now to show that $s$ has a gauge coaction and that $s\cong t$. For
the former, by [17, Proposition 3.3], there is a coaction $\delta$ on
$C^{*}(\mathcal{B})$ satisfying

$\delta(b)=b\otimes U_{g}\text{ for all }b\in B_{g},g\in G.$

In particular, since $s_{\lambda}\in B_{d(\lambda)}$, then
$\delta(s_{\lambda})=s_{\lambda}\otimes U_{d(\lambda)}$, so $\delta$ is a
gauge coaction on $s$.

Now we must show that $s\cong t$. By [10, Theorem 19.5], since $C^{*}(t)$ is a
topologically graded $C^{*}$-algebra, with grading $\mathcal{B}$ there is a
commutative diagram of surjective $*$-homomorphisms

${C^{*}(\mathcal{B})=C^{*}(s)}$${C^{*}_{r}(\mathcal{B})}$${C^{*}(t)}$$\scriptstyle{L}$$\scriptstyle{\pi^{s}_{t}}$$\scriptstyle{\psi}$

where $L$ denotes the regular representation from [10, Definition 17.6] (in
that definition, this map is called $\Lambda$, but we have changed its name to
avoid confusion with the $P$-graph $\Lambda$).

Recall that $C^{*}(\mathcal{B})$ is topologically graded with conditional
expectation $\Phi_{s}$. Since $G$ strongly reduces to an amenable group, then
by Lemma 4.33, the gauge action $s$ is normal and thus the conditional
expectation $\Phi_{s}$ is faithful. But by [10, Proposition 19.6], the kernel
of the regular representation $L$ is given by

$\ker(L)=\\{x\in C^{*}(\mathcal{B}):\Phi_{s}(x^{*}x)=0\\}$

and since $\Phi_{s}$ is faithful, then $\ker L=\\{0\\}$. Since $L$ is
injective and $L=\psi\circ\pi_{s}^{t}$, then $\pi_{s}^{t}$ is injective, so
$s\cong t$, so $t$ has a gauge coaction as desired.

∎

The reader should note that if such a bounded linear map exists, then there is
a gauge coaction, so that bounded linear map is the conditional expectation of
Lemma 2.26.

The next result shows that if you represent a $P$-graph on a Hilbert space in
such a way that the Hilbert space also has a “$P$-grading” which interacts
with the grading on $\Lambda$ in a natural way, then this representation has a
gauge coaction.

######  Lemma 4.37.

Let $(G,P)$ be a WQLO group and suppose that $(G,P)$ has a strong reduction to
an amenable group. Let $\Lambda$ be a finitely-aligned $P$ graph. Let $t$ be a
representation of $\Lambda$ in $\mathcal{B}(\mathcal{H})$ for some Hilbert
space $\mathcal{H}$, and suppose that $\mathcal{H}=\bigoplus_{p\in
P}\mathcal{H}_{p}$ such that
$t_{\mu}\mathcal{H}_{p}\subseteq\mathcal{H}_{d(\mu)p}$ for all
$\mu\in\Lambda$, $p\in P$. Then $t$ has a gauge coaction.

###### Proof.

By Lemma 4.36, it suffices to show that there is a bounded linear map
$\Phi_{t}:C^{*}(t)\rightarrow\mathcal{B}(t)$ satisfying

$\Phi_{t}(t_{\mu}t_{\nu}^{*})=\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ if }d(\mu)\neq d(\nu)\end{cases}.$

For each $p\in P$, let $Q_{p}\in\mathcal{B}(\mathcal{H})$ denote the
projection onto $\mathcal{H}_{p}$. Then define

$\Phi_{t}(x)=\displaystyle\sum_{\begin{subarray}{c}p\in P\\\ WOT\text{
limit}\end{subarray}}Q_{p}xQ_{p}$

which is bounded and converges in the weak operator topology because the
$\\{Q_{p}\\}_{p\in P}$ are pairwise orthogonal projections. We then wish to
show that

$\Phi_{t}(t_{\mu}t_{\nu}^{*})=\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ if }d(\mu)\neq d(\nu)\end{cases}$

for all $\mu,\nu\in\Lambda$.

To this end, fix some $q\in P$, $h_{q}\in\mathcal{H}_{q}$, and
$x\in\mathcal{H}$. Then we have that

$\displaystyle\langle\Phi_{t}(t_{\mu}t_{\nu}^{*})h_{q},x\rangle$
$\displaystyle=$ $\displaystyle\displaystyle\sum_{\begin{subarray}{c}p\in P\\\
WOT\text{ limit}\end{subarray}}\langle
Q_{p}t_{\mu}t_{\nu}^{*}Q_{p}h_{q},x\rangle$ $\displaystyle=$
$\displaystyle\langle Q_{q}t_{\mu}t_{\nu}^{*}Q_{q}h_{q},x\rangle$
$\displaystyle=$ $\displaystyle\langle Q_{q}t_{\mu}t_{\nu}^{*}h_{q},x\rangle$

since $Q_{p}h_{q}=0$ if $p\neq q$. Since our choice of $x\in\mathcal{H}$ was
arbitrary, we may unbind the $x$ term, so we have shown that
$\Phi_{t}(t_{\mu}t_{\nu}^{*})h_{q}=Q_{q}t_{\mu}t_{\nu}^{*}h_{q}$.

But note that $t_{\mu}t_{\nu}^{*}h_{q}\in\mathcal{H}_{d(\mu)d(\nu)^{-1}q}$ and
$q=d(\mu)d(\nu)^{-1}q$ if and only if $d(\mu)=d(\nu)$, so

$\Phi_{t}(t_{\mu}t_{\nu}^{*})h_{q}=Q_{q}t_{\mu}t_{\nu}^{*}h_{q}=\begin{cases}t_{\mu}t_{\nu}^{*}h_{q}&\text{
if }d(\mu)=d(\nu)\\\ 0&\text{ if }d(\mu)\neq d(\nu)\end{cases}.$

Since this is true for each $q\in P$ and $h_{q}\in\mathcal{H}_{q}$, and the
$\mathcal{H}_{q}$ span all of $\mathcal{H}$, we may unbind the $h_{q}$ to get
that

$\Phi_{t}(t_{\mu}t_{\nu}^{*})=\begin{cases}t_{\mu}t_{\nu}^{*}&\text{ if
}d(\mu)=d(\nu)\\\ 0&\text{ if }d(\mu)\neq d(\nu)\end{cases}$

as desired.

∎

######  Corollary 4.38.

Let $(G,P)$ be a WQLO group and suppose that $(G,P)$ has a strong reduction to
an amenable group. Let $\Lambda$ be a finitely-aligned $P$-graph. Then
$L\cong\mathcal{T}$, where $L$ denotes the regular representation of $\Lambda$
and $\mathcal{T}$ denotes the Toeplitz representation of $\Lambda$.

###### Proof.

To show that $L$ is universal, by Lemma 4.34 it suffices to show that $L$ it
is $\Lambda$-faithful, has a gauge coaction, and that every proper bolt is
nonzero. By Remark 2.45, $L$ is $\Lambda$-faithful and its proper bolts are
nonzero.

To show that $L$ has a gauge coaction, we will appeal to Lemma 4.37. Recall
that $L$ is a representation in $\mathcal{B}(\ell^{2}(\Lambda))$ by

$L_{\mu}e_{\alpha}=\begin{cases}e_{\mu\alpha}&\text{ if }s(\mu)=r(\alpha)\\\
0&\text{ if }s(\mu)\neq r(\alpha)\end{cases}.$

Therefore, writing $\Lambda^{p}=\\{\mu\in\Lambda:d(\mu)=p\\}$, we can see that
$\ell^{2}(\Lambda)=\bigoplus_{p\in P}\ell^{2}(\Lambda^{p})$, and a simple
calculation shows that
$L_{\mu}\ell^{2}(\Lambda^{p})\subseteq\ell^{2}(\Lambda^{d(\mu)p})$. Therefore,
by Lemma 4.37, $L$ has a gauge coaction.

Thus by by Lemma 4.34, $L\cong\mathcal{T}$.

∎

The following result shows that if $\Lambda$ has no “infinite paths”, then the
ultrafilter representation is the universal tight representation.

######  Lemma 4.39.

Let $(G,P)$ be a WQLO group and suppose that $(G,P)$ has a strong reduction to
an amenable group. Let $\Lambda$ be a finitely-aligned $P$-graph. Recall that
$\Lambda$ has a partial order given by $\alpha\leq\beta$ if
$\beta\in\alpha\Lambda$. Suppose that with respect to this partial order,
every chain in $\Lambda$ has an upper bound in $\Lambda$. Then the ultrafilter
representation $f$ of $\Lambda$ is the universal tight representation of
$\Lambda$.

###### Proof.

By our gauge-invariant uniqueness theorem for $P$-graphs, it suffices to show
that $f$ is $\Lambda$-faithful, tight, and has a gauge coaction. The former
two properties are shown in Lemma 4.5. It then suffices to use Lemma 4.37 to
show that $f$ has a gauge coaction.

To this end, we wish to show that the ultrafilters in $f$ correspond to
maximal paths in $\Lambda$. More formally, for each path $\alpha\in\Lambda$,
let $U_{\alpha}=\\{\mu\in\Lambda:\mu\leq\alpha\\}$. It is straightforward to
check that $U_{\alpha}$ is a filter, and that it is an ultrafilter if and only
if $\alpha$ is maximal. We wish to show that all ultrafilters in $\Lambda$ are
of the form $U_{\alpha}$.

Fix some ultrafilter $U$. If $U$ is finite, it must contain a maximal element
$\alpha$, in which case $U=U_{\alpha}$. If instead $U$ is infinite, since
$U\subseteq\Lambda$ is countable, we may enumerate
$U=\\{\alpha_{n}\\}_{n\in\mathbb{N}}$. Inductively define a sequence
$\\{\beta_{n}\\}_{n\in\mathbb{N}}\subseteq U$ by $\beta_{1}=\alpha_{1}$, and
for $n>1$, let $\beta_{n}$ be a common upper bound of $\beta_{n-1}$ and
$\alpha_{n}$ which is guaranteed to exist by the $F2$ property of filters.
Note that $U=\bigcup_{n=1}^{\infty}U_{\beta_{n}}$.

Then $\\{\beta_{n}\\}_{n\in\mathbb{N}}$ is a chain in $\Lambda$, so by our
hypothesis, it has an upper bound $\beta\in\Lambda$. Then $U_{\beta}\supseteq
U_{\beta_{n}}$ for all $n$, and since $U=\bigcup_{n=1}^{\infty}U_{\beta_{n}}$,
then $U_{\beta}\supseteq U$, and since $U$ is an ultrafilter, $U=U_{\beta}$,
as desired.

Therefore, we can partition the set of ultrafilters by the degree of their
unique maximal element. That is, writing
$\widehat{\Lambda}_{\infty}^{p}=\\{U_{\alpha}\in\widehat{\Lambda}_{\infty}:d(\alpha)=p\\}$,
we have $\widehat{\Lambda}_{\infty}=\bigsqcup_{p\in
P}\widehat{\Lambda}_{\infty}^{p}$, and therefore

$\ell^{2}(\widehat{\Lambda}_{\infty})=\bigoplus_{p\in
P}\ell^{2}(\widehat{\Lambda}_{\infty}^{p}).$

Finally, a simple calculation shows that
$f_{\mu}\ell^{2}(\widehat{\Lambda}_{\infty}^{p})\subseteq\ell^{2}(\widehat{\Lambda}_{\infty}^{d(\mu)p})$,
so by Lemma 4.37, $f$ has a gauge coaction, as desired.

∎

######  Example 4.40.

Let $G=BS(1,2)=\langle c,t|tc=c^{2}t\rangle$, and let $P=\\{c,t\\}^{*}$. Note
that $BS(1,2)$ is amenable, and therefore has a strong reduction to an
amenable group (itself). Define a $P$-graph $\Lambda$ as in this diagram:

${v_{1}}$${v_{2}}$${v_{3}}$${v_{4}}$${v_{5}}$$\scriptstyle{t_{1}}$$\scriptstyle{c_{1}}$$\scriptstyle{t_{2}}$$\scriptstyle{c_{2}}$$\scriptstyle{c_{3}}$

so
$\Lambda=\\{v_{1},v_{2},v_{3},v_{4},v_{5},c_{1},c_{2},c_{3},t_{1},t_{2},c_{2}t_{1},c_{3}c_{2},c_{3}c_{2}t_{1}=c_{1}t_{2}\\}$,
where the range and source maps are as indicated in the diagram, the operation
is concatenation of paths, and the degrees are given by $d(v_{i})=1$,
$d(c_{i})=c$, $d(t_{i})=t$, and so on.

Since $\Lambda$ is finite, all chains have an upper bound, so by Lemma 4.39,
the ultrafilter representation is the universal tight representation and the
set of ultrafilters is in bijection with the maximal paths in $\Lambda$. Thus
$\widehat{\Lambda}_{\infty}=\\{v_{1},c_{1},t_{1},c_{2}t_{1},c_{3}c_{2}t_{1}=t_{2}c_{1}\\}$,
so $C^{*}(f)$ is represented on a 5-dimension vector space and
$C^{*}(f)\subseteq M_{5}$, the space of 5-by-5 matrices. A direct calculation
with respect to the basis
$\\{e_{v_{1}},e_{c_{1}},e_{t_{1}},e_{c_{2}t_{1}},e_{c_{3}c_{2}t_{1}}=e_{t_{2}c_{1}}\\}$
shows that

$\displaystyle f_{v_{1}}$ $\displaystyle=$ $\displaystyle E_{11}$
$\displaystyle f_{c_{1}}$ $\displaystyle=$ $\displaystyle E_{21}$
$\displaystyle f_{t_{1}}$ $\displaystyle=$ $\displaystyle E_{31}$
$\displaystyle f_{c_{2}t_{1}}$ $\displaystyle=$ $\displaystyle E_{41}$
$\displaystyle f_{t_{2}c_{1}}$ $\displaystyle=$ $\displaystyle E_{51}$

where $E_{ij}$ denotes the matrix consisting of all 0s except for a single 1
in the $i$th row and $j$th column (a matrix unit). Since these matrix units
generate $M_{5}$, then $C^{*}(f)=C^{*}(\Lambda)=M_{5}$.

It is notable that the results in this section apply only to groups with
strong reductions, instead of any reduction. The only “bottleneck” for this is
Lemma 4.33, which says that gauge coactions are normal when $(G,P)$ strongly
reduces to an amenable group. We conjecture that a (non-strong) reduction to
an amenable group would suffice:

######  Conjecture 4.41.

Let $(G,P)$ be a WQLO group, let $\Lambda$ be a finitely aligned $P$-graph,
and suppose $(G,P)$ reduces to an amenable group. Then every gauge coaction on
a representation of $\Lambda$ is normal.

Therefore, the other results in this section only require the hypothesis that
$(G,P)$ reduces to an amenable group.

###  4.6 Applications to Kirchberg Algebras in the UCT Class

######  Definition 4.42.

A $C^{*}$-algebra $A$ is called a Kirchberg algebra if it is purely infinite,
simple, nuclear, and separable.

A $C^{*}$-algebra is said to be in the UCT class if it satisfies the
hypotheses of the universal coefficient theorem. (For a more thorough
treatment, the reader is directed to the introduction of [1]).

######  Proposition 4.43.

Let $A$ be a Kirchberg algebra in the UCT class. Then there is a finitely
aligned $\mathbb{N}^{2}*\mathbb{N}$-graph such that $A$ is stably isomorphic
to $C^{*}(\Lambda)$.

###### Proof.

By [3, Corollary 6.3], there is a finitely aligned
$\mathbb{N}^{2}*\mathbb{N}$-graph $\Lambda$ such that $A$ is stably isomorphic
to $C^{*}_{min}(\Lambda)$. But since $\mathbb{N}^{2}*\mathbb{N}$ strongly
reduces to an amenable group by Corollary 3.30, then
$C^{*}_{min}(\Lambda)=C^{*}(\Lambda)$ by the gauge-invariant uniqueness
theorem for $P$-graphs (Theorem 1). Thus $A$ is stably isomorphic to
$C^{*}(\Lambda)$. ∎

##  5 Appendix 1: Tight Representations

In this section, our goal is to justify our use of the term “tight” in the
context of representations of $P$-graphs by showing that a representation is
tight in the sense of Definition 2.44(c) if and only if it corresponds to a
tight representation of a semilattice in the sense of Exel’s definition in
[8].

In this section, we fix $(G,P)$ a WQLO group, $\Lambda$ a finitely-aligned
$P$-graph, and $t:\Lambda\rightarrow C^{*}(t)$ a representation. We will often
write $q_{\lambda}=t_{\lambda}t_{\lambda}^{*}$.

It can be easy to be confused about what symbols denote paths, sets of paths,
and sets of sets of paths. To avoid confusion, we will try to stick to the
convention that greek letters like $\alpha$ denote paths in $\Lambda$, lower
case letters (and $\mu\Lambda$ where $\mu\in\Lambda$) denote subsets of
$\Lambda$, and capital letters denote collections of subsets.

###  5.1 Background

Recall the following definition of the tightness from 2.44(c), slightly
modified to keep with our notation for this section:

######  Definition 5.1.

A representation $t$ of a graph $\Lambda$ is tight if for every
$\mu\in\Lambda$ and finite $z\subseteq\mu\Lambda$ which is exhaustive for
$\mu\Lambda$, we have that

$\displaystyle\prod_{\alpha\in z}(q_{\mu}-q_{\alpha})=0$

where $q_{\lambda}=t_{\lambda}t_{\lambda}^{*}$.

This condition was referred to a tightness before, but in this section will be
referred to as H-tightness to avoid confusion.

The following terminology is from [8, Section 2]:

######  Definition 5.2.

In [8]’s context, a partially ordered set contains a smallest element, denoted
$0$. A semilattice is a partially ordered set where for all $x,y\in X$, the
set $\\{z\in X:z\leq x,y\\}$ has a maximum element, denoted $x\wedge y$. That
is, $x\wedge y$ is the infimum of $x$ and $y$, and every two elements have an
infimum (although it may be $0$).

Given, $x,y\in X$, we say $x$ and $y$ are disjoint and write $x\perp y$ if
there is no nonzero $z\in X$ with $z\leq x,y$. Otherwise we say that $x$ and
$y$ intersect, and write $x\Cap y$.

######  Definition 5.3.

Let $E$ be a semilattice. Given finite $X,Y\subseteq E$, let $E^{X,Y}$ denote
the subset of $E$ given by

$E^{X,Y}=\\{z\in E:z\leq x\ \forall x\in X,z\perp y\ \forall y\in Y\\}.$

Given any subset $F\subseteq E$, we shall say that a subset $Z\subseteq F$ is
a cover for $F$ if, for every nonzero $x\in F$ there exists $z\in Z$ such $x$
and $z$ have a common extension.

######  Definition 5.4.

Let $\sigma:E\rightarrow\mathcal{B}$ be a representation of a semilattice $E$
in a Boolean algebra $\mathcal{B}$. We shall say that $\sigma$ is tight if for
every finite subsets $X,Y\subseteq E$ and for every finite cover $Z$ for
$E^{X,Y}$, one has that

$\bigvee_{z\in Z}\sigma(z)\geq\bigwedge_{x\in X}\sigma(x)\wedge\bigwedge_{y\in
Y}\neg\sigma(y).$

Since the reverse inequality is always true, tightness is equivalent to
equality in this expression.

To avoid confusion with H-tightness, we will refer to this notion of tightness
as E-tightness.

###  5.2 Building our Semilattice

We’ll now define our semilattice $E$ and develop its basic properties.

######  Remark 5.5.

Given $\Lambda$, let
$E(\Lambda)=\\{\bigcup_{i=1}^{n}\alpha_{i}\Lambda:\alpha_{i}\in\Lambda\\}$
denote the collection of finite unions of $\alpha\Lambda$s, ordered under
containment. When $\Lambda$ is clear from context, we will write $E$ for
$E(\Lambda)$. The empty set serves as the zero element for $E$, and two
elements $a,b$ of $E$ have a least lower bound $a\cap b$, which is also in $E$
by finite alignment of $\Lambda$. Thus $E$ is a semilattice.

Recall that $\Lambda$ is given a partial ordering $\leq$ by $\alpha\leq\beta$
if and only if $\beta=\alpha\alpha_{1}$. Note that $\alpha\leq\beta$ if and
only if $\beta\Lambda\subseteq\alpha\Lambda$.

For $a\in E$, we let $m(a)=\\{\alpha\in a:\alpha\text{ minimal in }a\\}$
denote the set of maximal elements of $a$ (with respect to the ordering on
$\Lambda$). For any $a\in E$, we may write
$a=\bigcup_{i=1}^{n}\alpha_{i}\Lambda$ for some $\alpha_{i}\in\Lambda$, so it
follows that $m(a)\subseteq\\{\alpha_{i}\\}_{i=1}^{n}$, so $m(a)$ is finite.
Note also that $m(a)$ is empty if and only if $a$ is empty. Finally, note that
$a=\bigcup_{\alpha\in m(a)}\alpha\Lambda$. So in this way, $m$ provides a
unique representation of each element of $E$, and provides a bijection between
$E$ and the set of finite, pairwise incomparable subsets of $\Lambda$.

If $t$ be a representation of $\Lambda$, let
$q_{\alpha}=t_{\alpha}t_{\alpha}^{*}$ for all $\alpha\in\Lambda$, and let
$\mathcal{B}_{t}\subseteq C^{*}(t)$ be the Boolean algebra generated by $1$
and $\\{q_{\alpha}:\alpha\in\Lambda\\}$, where $q_{\alpha}\wedge
q_{\beta}=q_{\alpha}q_{\beta}$.

Finally, given such a representation $t$, define
$\sigma_{t}:E\rightarrow\mathcal{B}_{t}$ by

$\sigma_{t}:a=\bigcup_{\alpha\in m(a)}\alpha\Lambda\mapsto\bigvee_{\alpha\in
m(a)}t_{\alpha}t_{\alpha}^{*}=\bigvee_{\alpha\in m(a)}q_{\alpha}.$

When the representation $t$ is clear from context, we will write $\sigma$ and
$\mathcal{B}$ for $\sigma_{t}$ and $\mathcal{B}_{t}$, respectively.

###  5.3 Technical Lemmas

######  Lemma 5.6 (Intersection Criterion).

If $a,b\in E$, then $a\Cap b$ if and only if there are $\alpha\in
m(a),\beta\in m(b)$ such that $\alpha\Cap\beta$ (in the sense of having a
common extension in $\Lambda$).

###### Proof.

Observe that $a\cap b=\bigcup_{\alpha\in m(a),\beta\in
m(b)}\alpha\Lambda\cap\beta\Lambda$, so the lefthand side is nonempty if and
only if one of the terms on the righthand side are nonempty. But
$\alpha\Lambda\cap\beta\Lambda$ is nonempty if and only if $\alpha$ has a
common extension with $\beta$. ∎

######  Lemma 5.7.

Let $Z$ be a finite cover of $E^{X,Y}$. Then there is another finite cover
$Z^{\prime}$ of $E^{X,Y}$ such that every element of $Z^{\prime}$ is of the
form $\alpha\Lambda$, and $\bigvee_{z\in Z}z=\bigvee_{z^{\prime}\in
Z^{\prime}}z^{\prime}$.

###### Proof.

Let $B=m(\bigcup_{z\in Z}z)$ and let $Z^{\prime}=\\{\beta\Lambda:\beta\in
B\\}$. Then by construction $B$ and hence $Z^{\prime}$ are finite sets, and by
our initial remark on $m$, we know that $\bigcup_{z\in Z}z=\bigvee_{z\in Z}z$
can be reconstituted from its minimal elements in the sense that
$\bigvee_{z\in Z}z=\bigvee_{z^{\prime}\in Z^{\prime}}z^{\prime}$.

It then suffices to show that $Z^{\prime}$ is a cover of $E^{X,Y}$. To first
show that it is contained in $E^{X,Y}$, we know that if $\beta\Lambda\in
Z^{\prime}$, then $\beta\in B$, so $\beta$ must be minimal in some $z_{0}\in
Z$. Then for all $x\in X$ and $y\in Y$, $\beta\Lambda\leq z_{0}\leq x$, and
$\beta\Lambda\wedge y\leq z_{0}\wedge y=0$, so $\beta\Lambda\leq x$ and
$\beta\Lambda\perp y$ for all $x\in X,y\in Y$. Thus $Z^{\prime}\subseteq
E^{X,Y}$.

To show it covers, fix some nonzero $w\in E^{X,Y}$. Then since $Z$ covers
$E^{X,Y}$, there is a $z\in Z$ such that $z\Cap w$. Then by the intersection
criterion, there is a $\mu\in m(z),\nu\in m(w)$ such that $\mu\Cap\nu$. Since
$\mu\in m(z)$, then $\mu\Lambda\leq\bigcup_{\beta\in B}\beta\Lambda$, so there
is some $\beta\in B$ with $\beta\leq\mu$. Then $\beta\Cap\nu$, so
$\beta\Lambda\Cap w$. Thus $Z^{\prime}$ is a finite cover.

∎

######  Lemma 5.8.

Let $X,Y\subseteq E$ be finite sets. Then there is a finite set
$Y^{\prime}\subseteq E$ such that $E^{X,Y}=E^{X,Y^{\prime}}$ and
$Y^{\prime}\subseteq E^{X,\emptyset}$.

###### Proof.

By the remark in [8] following Definition 2.5, we may assume without loss of
generality, that $X=\\{x\\}$, a singleton. Let $Y^{\prime}=\\{y\wedge x:x\in
X\\}$ which is finite. Then by construction, $y^{\prime}=y\wedge x\leq x$ for
all $y^{\prime}\in Y^{\prime}$, so $Y^{\prime}\subseteq E^{X,\emptyset}$. It
then suffices to show that $E^{X,Y}=E^{X,Y^{\prime}}$.

If $w\in E^{X,Y}$, then $w\leq x$ and for all $y\in Y$, $w\perp y$, so
$w\wedge y=0$, and thus $w\wedge y^{\prime}=w\wedge y\wedge x=0$ for all
$y^{\prime}\in Y^{\prime}$. Thus $w\in E^{X,Y^{\prime}}$.

Conversely, if $w\in E^{X,Y^{\prime}}$, then $w\leq x$ and for all
$y^{\prime}\in Y$, $w\perp y^{\prime}$, so $w\wedge y\wedge x=0$. But $w\leq
x$, so $w\wedge y\wedge x=(w\wedge x)\wedge y=w\wedge y=0$, so $w\in E^{X,Y}$.

Thus $E^{X,Y}=E^{X,Y^{\prime}}$ as desired. ∎

######  Lemma 5.9.

Let $\mu\in\Lambda$ and $B\subseteq\mu\Lambda$ be finite. Then
$q_{\mu}=\bigvee_{\beta\in B}q_{\beta}$ if and only if
$0=\displaystyle\prod_{\beta\in B}(q_{\mu}-q_{\beta})$.

###### Proof.

The result follows from a chain of equivalences:

$\displaystyle q_{\mu}$ $\displaystyle=$ $\displaystyle\bigvee_{\beta\in
B}q_{\beta}$ $\displaystyle 0$ $\displaystyle=$ $\displaystyle
q_{\mu}-\bigvee_{\beta\in B}q_{\beta}\text{ by subtraction}$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle\bigwedge_{\beta\in B}(q_{\mu}-q_{\beta})\text{
by De Morgan's laws}$ $\displaystyle 0$ $\displaystyle=$
$\displaystyle\displaystyle\prod_{\beta\in B}(q_{\mu}-q_{\beta})\text{ by
definition of }\bigwedge.$

∎

###  5.4 The Proof of Equivalence

Here we prove the following proposition. The main interest is the equivalence
of the first and last criteria, and the middle statements are there to ease
the proof.

######  Proposition 5.10.

If $t$ is a representation of $\Lambda$, and $\sigma=\sigma_{t}$ is the
associated representation of the semilattice $E=E(\Lambda)$, then the
following are equivalent:

  1. 1.

$\sigma$ is a tight representation in the $E$-sense.

  2. 2.

$\sigma$ is a tight representation in the $E$-sense when $Y=\emptyset$.

  3. 3.

$\sigma$ is a tight representation in the $E$-sense when $Y=\emptyset$ and
$X=\\{\mu\Lambda\\}$.

  4. 4.

$t$ is a tight representation in the $H$-sense.

###### Proof.

Certainly $(1)\Rightarrow(2)\Rightarrow(3)$. We will prove four non-obvious
directions: $(3)\Rightarrow(4),(4)\Rightarrow(3),(3)\Rightarrow(2)$ and
$(2)\Rightarrow(1)$.

$(3)\Rightarrow(4)$ Fix some $\mu\in\Lambda$. We wish to show that if
$B\subseteq\mu\Lambda$ is exhaustive, then $Z_{B}=\\{\beta\Lambda:\beta\in
B\\}$ is a finite cover of $E^{\\{\mu\Lambda\\},\emptyset}$. In particular,
given such an exhaustive $B$, if $w\in E^{\\{\mu\Lambda\\},\emptyset}$ is
nonzero, then taking $\alpha\in m(w)$, $\alpha\in w\leq\mu\Lambda$, so
$\alpha\in\mu\Lambda$, so since $B$ is exhaustive there is a $\beta\in B$ such
that $\alpha\Cap\beta$, and by the intersection criterion, $\beta\Lambda\Cap
w$. Thus $Z_{B}$ covers.

Then by (3), we have that $\bigvee_{z\in Z_{B}}\sigma(z)=\sigma(\mu\Lambda)$
and substituting $\sigma(z)=\sigma(\beta\Lambda)=q_{\beta}$ and
$\sigma(\mu\Lambda)=q_{\mu}$, we have

$\bigvee_{\beta\in B}q_{\beta}=q_{\mu}$

which is equivalent to H-tightness by Lemma 5.9.

Given such an exhaustive $B$, if $w\in E^{\\{\mu\Lambda\\},\emptyset}$ is
nonzero, then taking $\alpha\in m(w)$, $\alpha\in w\leq\mu\Lambda$, so
$\alpha\in\mu\Lambda$, so since $B$ is exhaustive there is a $\beta\in B$ such
that $\alpha\Cap\beta$, and by the intersection criterion, $\beta\Lambda\Cap
w$. Thus $Z_{B}$ covers.

$(4)\Rightarrow(3)$ Fix some $\mu\in\Lambda$. We wish to show that if
$Z\subseteq E^{\\{\mu\Lambda\\},\emptyset}$ is a finite cover, then
$B_{Z}=\bigcup_{z\in Z}m(z)$ is a finite exhaustive subset of $\mu\Lambda$. In
particular, given $\alpha\in\mu\Lambda$, we have that $\alpha\Lambda\in
E^{\\{\mu\Lambda\\},\emptyset}$, so there is $z\in Z$ such that
$z\Cap\alpha\Lambda$. Then by the intersection property, there is a $\beta\in
m(z)$ such that $\beta\Cap\alpha$. Thus $B_{Z}$ is exhaustive for
$\mu\Lambda$, as desired.

Then, by (4), and Lemma 5.9, we have that

$q_{\mu}=\bigvee_{\beta\in B_{Z}}q_{\beta}.$

Substituting $\sigma(\mu\Lambda)=q_{\mu}$ and
$\sigma(\beta\Lambda)=q_{\beta}$, then regrouping the terms on the right, we
have that

$\displaystyle\sigma(\mu\Lambda)$ $\displaystyle=$
$\displaystyle\bigvee_{\beta\in B_{Z}}\sigma(\beta\Lambda)$ $\displaystyle=$
$\displaystyle\bigvee_{z\in Z}\bigvee_{\beta\in m(z)}\sigma(\beta\Lambda)$
$\displaystyle=$ $\displaystyle\bigvee_{z\in Z}\sigma(\bigvee_{\beta\in
m(z)}\beta\Lambda)$ $\displaystyle=$ $\displaystyle\bigvee_{z\in Z}\sigma(z)$

as desired.

$(3\Rightarrow 2)$ Let $X\subseteq E$ and let $Z$ be a finite cover of
$E^{X,\emptyset}$. By the remark in [8] following Definition 2.5, we may
assume without loss of generality that $X$ is a singleton, since otherwise we
may replace $X$ with $\\{x_{min}\\}$ where $x_{min}=\bigwedge_{x\in X}x$.

Let $Z$ be a finite cover of $E^{X,\emptyset}$. By Lemma 5.7, we may assume
without loss of generality that
$Z=\\{\beta_{1}\Lambda,...,\beta_{n}\Lambda\\}$ for some
$\beta_{1},...,\beta_{n}\in\Lambda$. Let $B=\\{\beta_{1},...\beta_{n}\\}$ and
$A=m(x)$.

Now, for each $\alpha\in A$ and $\beta\in B$, let
$Z_{\alpha,\beta}=\\{\gamma\Lambda:\gamma\in MCE(\alpha,\beta)\\}$. Note that
$Z_{\alpha,\beta}$ is finite and $\gamma\Lambda\in Z_{\alpha,\beta}$ is in
$\alpha\Lambda\cap\beta\Lambda$.

We wish to show that for fixed $\beta\in B$, $\bigcup_{\alpha\in
A}Z_{\alpha,\beta}$ is a cover for $E^{\\{\beta\Lambda\\},\emptyset}$ and for
fixed $\alpha\in A$, that $\bigcup_{\beta\in B}Z_{\alpha,\beta}$ is a cover
for $E^{\\{\alpha\Lambda\\},\emptyset}$. Each is a finite union of finite
sets, hence finite, and is contained in the appropriate space. It then
suffices to show that they are covering.

For the former set, fix some nonzero $w\in E^{\\{\beta\Lambda\\},\emptyset}$.
Since $w$ is nonzero, there is some $\mu\in m(w)$. Then $\mu\Lambda\leq
w\leq\beta\Lambda\leq x$, so $\mu\Lambda\wedge x=\mu\Lambda<\infty$, so
$\mu\Lambda\Cap x$, and thus there is $\alpha\in m(x)$ such that
$\mu\Cap\alpha$. That is, $\alpha$ and $\mu$ have a common extension $\nu$.
Since $\beta$ is a prefix of $\mu$, then $\nu$ is a common extension of
$\alpha$ and $\beta$ as well. Thus some prefix of $\nu$ is a $\gamma\in
MCE(\alpha,\beta)$, so $\nu$ is a common extension of both $\gamma$ and $\mu$.
That is, $\gamma\Lambda\Cap w$, and $\gamma\Lambda\in Z_{\alpha,\beta}$. Thus
$\bigcup_{\alpha\in A}Z_{\alpha,\beta}$ is a cover for
$E^{\\{\beta\Lambda\\},\emptyset}$, as desired.

For the latter set, fix some nonzero $w\in E^{\\{\alpha\Lambda\\},\emptyset}$.
Then we know that $w\leq\alpha\Lambda\leq x$, so $w\in E^{X,\emptyset}$, and
because $Z$ is a cover, there is some $\beta\Lambda\in Z$ with
$w\Cap\beta\Lambda$. Then by the intersection criterion, there is a $\mu\in
m(w)$ such that $\mu\Cap\beta$. That is, $\mu$ and $\beta$ have a common
extension $\nu$. However, $\alpha\leq\mu$ as $w\in
E^{\\{\alpha\Lambda\\},\emptyset}$, so $\nu$ is a common extension of $\beta$
and $\alpha$, and thus has a prefix which is some $\gamma\in
MCE(\beta,\alpha)$. Thus $\nu$ is a common extension of $\mu$ and $\gamma$, so
$w\Cap\gamma\Lambda$, and $\gamma\Lambda\in Z_{\alpha,\beta}$. Thus
$\bigcup_{\beta\in B}Z_{\alpha,\beta}$ is a cover for
$E^{\\{\alpha\Lambda\\},\emptyset}$, as desired.

Now, using (3), the fact that these two sets cover gives us that

$\displaystyle\sigma(\beta\Lambda)$ $\displaystyle=$
$\displaystyle\bigvee_{\gamma\in\bigcup_{\alpha\in
A}Z_{\alpha,\beta}}\sigma(\gamma\Lambda)$ $\displaystyle=$
$\displaystyle\bigvee_{\alpha\in A}\bigvee_{\gamma\in
MCE(\alpha,\beta)}\sigma(\gamma\Lambda)$

and

$\displaystyle\sigma(\alpha\Lambda)$ $\displaystyle=$
$\displaystyle\bigvee_{\gamma\in\bigcup_{\beta\in
B}Z_{\alpha,\beta}}\sigma(\gamma\Lambda)$ $\displaystyle=$
$\displaystyle\bigvee_{\beta\in B}\bigvee_{\gamma\in
MCE(\alpha,\beta)}\sigma(\gamma\Lambda)$

Finally, using the fact that $x=\bigcup_{\alpha\in A}\alpha\Lambda$, we have
that

$\displaystyle\sigma(x)$ $\displaystyle=$ $\displaystyle\bigvee_{\alpha\in
A}\sigma(\alpha\Lambda)$ $\displaystyle=$ $\displaystyle\bigvee_{\alpha\in
A}\left(\bigvee_{\beta\in B}\bigvee_{\gamma\in
MCE(\alpha,\beta)}\sigma(\gamma\Lambda)\right)$ $\displaystyle=$
$\displaystyle\bigvee_{\beta\in B}\left(\bigvee_{\alpha\in
A}\bigvee_{\gamma\in MCE(\alpha,\beta)}\sigma(\gamma\Lambda)\right)$
$\displaystyle=$ $\displaystyle\bigvee_{\beta\in B}\sigma(\beta\Lambda)$
$\displaystyle=$ $\displaystyle\bigvee_{z\in Z}\sigma(z)$

which is the E-tightness condition for $E^{X,\emptyset}$.

$(2\Rightarrow 1)$ Let $X,Y\subseteq E$ be finite, and let $Z$ be a finite
cover of $E^{X,Y}$. By Lemma 5.8, we may assume that $Y$ is a subset of
$E^{X,\emptyset}$. Let $Z^{\prime}=Z\cup Y$. We claim that $Z^{\prime}$ is a
finite cover of $E^{X,\emptyset}$. In particular, given $w\in
E^{X,\emptyset}$, if $w\perp y$ for all $y\in Y$, then $w\in E^{X,Y}$, so
$w\Cap z$ for some $z\in Z\subseteq Z^{\prime}$. If instead $w\not\perp y$ for
some $y$, then $w\Cap y$ for some $y\in Y\subseteq Z^{\prime}$.

Then by (2), we know that

$\bigvee_{z^{\prime}\in Z^{\prime}}\sigma(z^{\prime})=\bigwedge_{x\in
X}\sigma(x)$

and meeting with $\bigwedge_{y\in Y}\neg\sigma(y)$, we have that

$\left(\bigvee_{z^{\prime}\in
Z^{\prime}}\sigma(z^{\prime})\right)\wedge\bigwedge_{y\in
Y}\neg\sigma(y)=\bigwedge_{x\in X}\sigma(x)\wedge\bigwedge_{y\in
Y}\neg\sigma(y)$

Simplifying the lefthand side, we have that

$\displaystyle\left(\bigvee_{z^{\prime}\in
Z^{\prime}}\sigma(z^{\prime})\right)\wedge\bigwedge_{y\in Y}\neg\sigma(y)$
$\displaystyle=$ $\displaystyle\left(\bigvee_{z\in
Z}\sigma(z)\vee\bigvee_{y^{\prime}\in
Y}\sigma(y^{\prime})\right)\wedge\bigwedge_{y\in Y}\neg\sigma(y)$
$\displaystyle=$ $\displaystyle\left(\bigvee_{z\in
Z}\sigma(z)\wedge\bigwedge_{y\in
Y}\neg\sigma(y)\right)\vee\left(\bigvee_{y^{\prime}\in
Y}\sigma(y^{\prime})\wedge\bigwedge_{y\in Y}\neg\sigma(y)\right)$
$\displaystyle=$ $\displaystyle\left(\bigvee_{z\in
Z}\sigma(z)\wedge\bigwedge_{y\in Y}\neg\sigma(y)\right)$ $\displaystyle=$
$\displaystyle\bigvee_{z\in Z}\left(\sigma(z)\wedge\bigwedge_{y\in
Y}\neg\sigma(y)\right)$

Now, for a fixed $z\in Z,y\in Y$, and for any $\alpha\in m(z)$, $\beta\in
m(y)$, we have that $\alpha\perp\beta$, so $q_{\alpha}q_{\beta}=0$. Then
$\sigma(\alpha\Lambda)\wedge\neg\sigma(\beta\Lambda)=q_{\alpha}(1-q_{\beta})=q_{\alpha}$,
so $\sigma(z)\vee\neg\sigma(y)=\sigma(z)$. Thus

$\bigvee_{z\in Z}\left(\sigma(z)\wedge\bigwedge_{y\in
Y}\neg\sigma(y)\right)=\bigvee_{z\in Z}\sigma(z)$

so by transitivity,

$\bigvee_{z\in Z}\sigma(z)=\bigwedge_{x\in X}\sigma(x)\wedge\bigwedge_{y\in
Y}\neg\sigma(y)$

as desired. This completes the proof that (2) implies (1).

∎

## References

  * [1] Selçuk Barlak and Xin Li.  Cartan subalgebras and the UCT problem, II.  Math. Ann., 378(1-2):255–287, 2020.
  * [2] Bruce Blackadar.  Shape theory for $C^{\ast}$-algebras.  Math. Scand., 56(2):249–275, 1985.
  * [3] Nathan Brownlowe, Aidan Sims, and Sean T. Vittadello.  Co-universal $C^{\ast}$-algebras associated to generalised graphs.  Israel J. Math., 193(1):399–440, 2013.
  * [4] Adam Clay and Dale Rolfsen.  Ordered groups and topology, volume 176 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 2016.
  * [5] Kenneth R. Davidson.  $C^{*}$-algebras by example, volume 6 of Fields Institute Monographs.  American Mathematical Society, Providence, RI, 1996.
  * [6] Siegfried Echterhoff, S. Kaliszewski, John Quigg, and Iain Raeburn.  A categorical approach to imprimitivity theorems for $C^{*}$-dynamical systems.  Mem. Amer. Math. Soc., 180(850):viii+169, 2006.
  * [7] Siegfried Echterhoff and John Quigg.  Induced coactions of discrete groups on $C^{*}$-algebras.  Canad. J. Math., 51(4):745–770, 1999.
  * [8] R. Exel.  Tight representations of semilattices and inverse semigroups.  Semigroup Forum, 79(1):159–182, 2009.
  * [9] Ruy Exel.  Amenability for Fell bundles.  J. Reine Angew. Math., 492:41–73, 1997.
  * [10] Ruy Exel.  Partial dynamical systems, Fell bundles and applications, volume 224 of Mathematical Surveys and Monographs.  American Mathematical Society, Providence, RI, 2017.
  * [11] M. Hochster.  Subsemigroups of amenable groups.  Proc. Amer. Math. Soc., 21:363–364, 1969.
  * [12] Eberhard Kaniuth and Keith F. Taylor.  Induced representations of locally compact groups, volume 197 of Cambridge Tracts in Mathematics.  Cambridge University Press, Cambridge, 2013.
  * [13] David W. Kribs and Stephen C. Power.  Free semigroupoid algebras.  J. Ramanujan Math. Soc., 19(2):117–159, 2004.
  * [14] David W. Kribs and Stephen C. Power.  The analytic algebras of higher rank graphs.  Math. Proc. R. Ir. Acad., 106A(2):199–218, 2006.
  * [15] Alex Kumjian and David Pask.  Higher rank graph $C^{\ast}$-algebras.  New York J. Math., 6:1–20, 2000.
  * [16] Roger C. Lyndon and Paul E. Schupp.  Combinatorial group theory.  Classics in Mathematics. Springer-Verlag, Berlin, 2001.  Reprint of the 1977 edition.
  * [17] John C. Quigg.  Discrete $C^{*}$-coactions and $C^{*}$-algebraic bundles.  J. Austral. Math. Soc. Ser. A, 60(2):204–221, 1996.
  * [18] Iain Raeburn.  Graph algebras, volume 103 of CBMS Regional Conference Series in Mathematics.  Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 2005.
  * [19] Iain Raeburn.  On graded $C^{*}$-algebras.  Bull. Aust. Math. Soc., 97(1):127–132, 2018.
  * [20] Camila F. Sehnem.  On $C^{*}$-algebras associated to product systems.  J. Funct. Anal., 277(2):558–593, 2019.
  * [21] Jack Spielberg.  Graph-based models for Kirchberg algebras.  J. Operator Theory, 57(2):347–374, 2007.
  * [22] Jack Spielberg.  Groupoids and $C^{*}$-algebras for categories of paths.  Trans. Amer. Math. Soc., 366(11):5771–5819, 2014.
  * [23] M. Takesaki.  Theory of operator algebras. I, volume 124 of Encyclopaedia of Mathematical Sciences.  Springer-Verlag, Berlin, 2002.  Reprint of the first (1979) edition, Operator Algebras and Non-commutative Geometry, 5.
  * [24] Jun Tomiyama.  On the projection of norm one in $W^{\ast}$-algebras.  Proc. Japan Acad., 33:608–612, 1957.