$\psi|_{A^{\sigma}}$ is an extremal lower semi-continuous trace on the fixed
point algebra $A^{\sigma}$. Then
$\Gamma(\pi_{\psi}(A)^{\prime\prime})=\operatorname{Sp}_{\alpha}\left(\pi_{\psi}(A)^{\prime\prime}\right)$
where $\alpha_{t}:=\sigma^{\prime\prime}_{-\beta t}$.

###### Proof.

This follows from the definitions and Theorem 4.3.1 if we show that the fixed
point algebra $N:=(\pi_{\psi}(A)^{\prime\prime})^{\alpha}$ of $\alpha$ on
$\pi_{\psi}(A)^{\prime\prime}$ is a factor. Set
$\tau_{\psi}=\psi|_{A^{\sigma}}$. Consider a projection $e$ in the centre of
$N$. Define $\tau_{e}:(A^{\sigma})^{+}\to[0,\infty]$ and
$\tau_{1-e}:(A^{\sigma})^{+}\to[0,\infty]$ by

$\tau_{e}(a):=\psi^{\prime\prime}(e\pi_{\psi}(a))$

and

$\tau_{1-e}(a):=\psi^{\prime\prime}((1-e)\pi_{\psi}(a)),$

respectively. Then
$\tau_{e}(a^{*}a)\leq\psi^{\prime\prime}\circ\pi_{\psi}(a^{*}a)=\psi(a^{*}a)$
for all $a\in A^{\sigma}$ by Lemma 4.2.16. Hence $\tau_{e}$ is densely defined
since $\psi$ is. Since $\psi^{\prime\prime}$ is lower semi-continuous so is
$\tau_{e}$. Since $e$ is central in $N$ and is fixed by
$\sigma^{\prime\prime}$ it follows from Lemma 4.2.15 that

$\displaystyle\tau_{e}(a^{*}a)=\psi^{\prime\prime}(\pi_{\psi}(a)^{*}e\pi_{\psi}(a))=\psi^{\prime\prime}(e\pi_{\psi}(a)\pi_{\psi}(a)^{*}e)=\tau_{e}(aa^{*})$

for all $a\in A^{\sigma}$. Thus $\tau_{e}$ is a trace on $A^{\sigma}$ if it is
not zero. Since it is dominated by $\psi$ on $A^{\sigma}$ it follows that
there is a scalar $c_{e}\geq 0$ such that $\tau_{e}=c_{e}\tau_{\psi}$.
Similarly, there is a scalar $c_{1-e}\geq 0$ such that
$\tau_{1-e}=c_{1-e}\tau_{\psi}$.

######  Observation 10.3.2.

$\pi_{\psi}(A^{\sigma}\cap\mathcal{M}_{\psi})$ is $\sigma$-weakly dense in
$N$.

Before we prove Observation 10.3.2, let us see how it leads to the desired
conclusion: Assume $c_{e}\neq 0$ and $c_{1-e}\neq 0$. Let $d\in
A^{\sigma}\cap\mathcal{M}_{\psi}$. Then $N\ni
x\mapsto\psi^{\prime\prime}(e\pi_{\psi}(d)x\pi_{\psi}(d^{*}))$ is a normal
linear functional on $N$ (of norm $\leq c_{e}\psi(dd^{*})$). By Observation
10.3.2 there is a net $\\{a_{i}\\}$ in $A^{\sigma}\cap\mathcal{M}_{\psi}$ such
that $\lim_{i\to\infty}\pi_{\psi}(a_{i})=1-e$ in the $\sigma$-weak topology.
It follows that,

$\displaystyle
0=\psi^{\prime\prime}(e\pi_{\psi}(d)(1-e)\pi_{\psi}(d^{*}))=\lim_{i\to\infty}\psi^{\prime\prime}(e\pi_{\psi}(d)\pi_{\psi}(a_{i})\pi_{\psi}(d^{*}))$
$\displaystyle=\lim_{i\to\infty}\tau_{e}(da_{i}d^{*})=\frac{c_{e}}{c_{1-e}}\lim_{i\to\infty}\tau_{1-e}(da_{i}d^{*})$
$\displaystyle=\frac{c_{e}}{c_{1-e}}\psi^{\prime\prime}((1-e)\pi_{\psi}(d)(1-e)\pi_{\psi}(d^{*}))=\frac{c_{e}}{c_{1-e}}\psi^{\prime\prime}((1-e)\pi_{\psi}(dd^{*})).$

This equation and the faithfulness of $\psi^{\prime\prime}$ implies that
$(1-e)\pi_{\psi}(dd^{*})=0$. Since $d\in A^{\sigma}\cap\mathcal{M}_{\psi}$ was
arbitrary it follows from Observation 10.3.2 that $1-e=0$. By symmetry we find
also that $e=0$, a contradiction. Hence either $c_{e}=0$ or $c_{1-e}=0$. Note
that $c_{e}=c_{1-e}=0$ is impossible since
$\tau_{e}(a)+\tau_{1-e}(a)=\psi^{\prime\prime}(\pi_{\psi}(a))=\psi(a)$ for
$a\in(A^{\sigma})^{+}$ and $\psi|_{A^{\sigma}}$ is not zero. If $c_{e}\neq 0$
we have that $c_{1-e}=0$ and hence $\tau_{1-e}=0$. The faithfulness of
$\psi^{\prime\prime}$ implies then that $(1-e)\pi_{\psi}(a)=0$ for all $a\in
A^{\sigma}$ and Observation 10.3.2 gives the conclusion that $e=1$. Likewise
the assumption $c_{1-e}\neq 0$ gives that $e=0$. It follows that $0$ and $1$
are the only central projections in $N$; i.e. $N$ is a factor.

Now let’s establish the observation: Let $x\in N$. Since $\pi_{\psi}(A)$ is
$\sigma$-weakly dense in $\pi_{\psi}(A)^{\prime\prime}$ and
$\mathcal{M}_{\psi}$ is norm-dense in $A$ there is a net $\\{b_{i}\\}$ in
$\mathcal{M}_{\psi}$ such that $\lim_{i\to\infty}\pi_{\psi}(b_{i})=x$ in the
$\sigma$-weak topology. Let $Q_{0}:A\to A^{\sigma}$ be the conditional
expectation given by

$Q_{0}(a)=\frac{1}{p}\int_{0}^{p}\sigma_{t}(a)\ \mathrm{d}t,$

where $p$ is the period of $\sigma$, cf. Section 8.2. Let
$a\in\mathcal{M}_{\psi}^{+}$. An application of Theorem 1.2.1 and Lebesques
theorem on monotone convergence shows that

$\psi\left(Q_{0}(a)\right)=\frac{1}{p}\int_{0}^{p}\psi(\sigma_{t}(a))\
\mathrm{d}t=\psi(a),$

implying that $Q_{0}(b_{i})\in\mathcal{M}_{\psi}$ for all $i$. Since
$Q_{0}(b_{i})\in A^{\sigma}\cap\mathcal{M}_{\psi}$ it suffices to show that
$\lim_{i\to\infty}\pi_{\psi}(Q_{0}(b_{i}))=x$ in the $\sigma$-topology. For
this let $\omega$ be a normal functional. Then

$\omega(\pi_{\psi}(Q_{0}(b_{i})))=\frac{1}{p}\int_{0}^{p}\omega\left(\sigma^{\prime\prime}_{t}(\pi_{\psi}(b_{i}))\right)\
\mathrm{d}t$

which converges to
$\frac{1}{p}\int_{0}^{p}\omega\left(\sigma^{\prime\prime}_{t}(x)\right)\
\mathrm{d}t$. Since $\sigma^{\prime\prime}_{t}=\alpha_{-\frac{t}{\beta}}$, we
have that

$\frac{1}{p}\int_{0}^{p}\omega\left(\sigma^{\prime\prime}_{t}(x)\right)\
\mathrm{d}t=\omega(x),$

and we conclude that
$\lim_{i\to\infty}\omega(\pi_{\psi}(Q_{0}(b_{i})))=\omega(x)$. Since $\omega$
was arbitrary this means that $\lim_{i\to\infty}\pi_{\psi}(Q_{0}(b_{i}))=x$ in
the $\sigma$-topology. ∎

######  Example 10.3.3.

(Example 8.2.16 continued.) The flow $\sigma^{\rho}$ from Example 8.2.16 is
periodic and has a fixed point algebra which is a UHF algebra when $\rho\neq
0$. Since the trace state of a UHF algebra is unique, Lemma 10.3.1 applies to
show that for the $\frac{\log 2}{\rho}$-KMS state $\psi$ for $\sigma^{\rho}$,
the $\Gamma$-invariant of $\pi_{\psi}(A)^{\prime\prime}$ is

$\Gamma(\pi_{\psi}(A)^{\prime\prime})=\operatorname{Sp}_{\alpha}(\pi_{\psi}(A)^{\prime\prime}),$

where $\alpha_{t}={\sigma^{\rho}}^{\prime\prime}_{-\beta t}$ and
$\beta=\frac{\log 2}{\rho}$. We seek therefore to calculate
$\operatorname{Sp}_{\alpha}(\pi_{\psi}(A)^{\prime\prime})$.

It follows from Observation 8.2.17 and the description of the eigenspaces
$V(k)$, given in Example 8.2.16 that $\bigcup_{k\in\mathbb{Z}}V(k)$ spans a
norm dense subspace of $A$ and it follows therefore that
$\bigcup_{k\in\mathbb{Z}}\pi_{\psi}(V(k))$ spans a $\sigma$-weakly dense
subspace of $\pi_{\psi}(A)^{\prime\prime}$. It follows that when $f\in
L^{1}(\mathbb{R})$ the condition

$\alpha_{f}(\pi_{\psi}(A)^{\prime\prime})=\\{0\\}$

will hold if and only if

$\int_{\mathbb{R}}f(t){\sigma^{\rho}}^{\prime\prime}_{-\beta t}(x)\
\mathrm{d}t=0$

for all $x\in\pi_{\psi}(V(k))$ and all $k\in\mathbb{Z}$. If
$y:=V_{a}V_{b}^{*}$ where $a\in\\{0,1\\}^{n+k}$ and $b\in\\{0,1\\}^{n}$ where
$n,k\geq 0$, we find that

${\sigma^{\rho}}^{\prime\prime}_{-\beta
t}\circ\pi_{\psi}(y)=\pi_{\psi}\circ\sigma^{\rho}_{-\beta
t}(y)=e^{-ik\beta\rho t}\pi_{\psi}(y).$

Note that $\pi_{\psi}(y)\neq 0$ since
$\pi_{\psi}(V_{a}^{*})\pi_{\psi}(y)\pi_{\psi}(V_{b})=\pi_{\psi}(1)=1$. Hence

$\alpha_{f}(\pi_{\psi}(V(k)))=0$

if and only if $\int_{\mathbb{R}}f(t)e^{-ik\beta\rho t}\ \mathrm{d}t=0$. A
similar argument applies when $k<0$ and we find therefore that
$\alpha_{f}(\pi_{\psi}(A)^{\prime\prime})=\\{0\\}$ if and only if
$\hat{f}(-k\beta\rho)=0$ for all $k\in\mathbb{Z}$, where $\hat{f}$ is the
Fourier transform of $f$. Thus

$\mathbb{Z}\beta\rho\subseteq Z(f)$

for all $f\in L^{1}(\mathbb{R})$ such that
$\alpha_{f}(\pi_{\psi}(A)^{\prime\prime})=\\{0\\}$. If
$t\in\mathbb{R}\backslash\mathbb{Z}\beta\rho$ there is a $g\in
L^{1}(\mathbb{R})$ such that $\hat{g}(\mathbb{Z}\beta\rho)=\\{0\\}$ and
$\hat{g}(t)=1$, cf. Theorem 2.6.2 in [Ru3]. Then
$\alpha_{g}(\pi_{\psi}(A)^{\prime\prime})=\\{0\\}$, but $t\notin Z(g)$. It
follows therefore from (10.3.1) that
$\operatorname{Sp}_{\alpha}(\pi_{\psi}(A)^{\prime\prime})=\mathbb{Z}\beta\rho=\mathbb{Z}\log
2$. It follows that $\psi$ is of type $III_{\frac{1}{2}}$, regardless of which
$\rho\neq 0$ we consider; a fact which is not surprising because the KMS state
is the same for all $\rho\neq 0$.

####  10.3.2 On the factor type of the KMS states of an ITPFI flow

Let $m\in\mathbb{N},\ m\geq 2$, and let $h=h^{*}\in M_{m}(\mathbb{C})$ be a
selfadjoint $m\times m$ matrix. We denote in the following the tensor product
$M_{m}(\mathbb{C})\otimes M_{m}(\mathbb{C})\otimes\cdots\otimes
M_{m}(\mathbb{C})$ with $k$ tensor factors by

$M_{m}(\mathbb{C})^{\otimes k}.$

As in Section 6.5.1 we can consider the UHF algebra
$\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$ and the flow $\sigma$ defined on
$\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$ defined such that

$\sigma_{t}\circ\phi_{k,\infty}=\phi_{k,\infty}\circ\sigma^{k}_{t},$ (10.3.2)

where $\phi_{k,\infty}:M_{m}(\mathbb{C})^{\otimes
k}\to\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$ is the canonical map and
$\sigma^{k}$ is the flow on $M_{m}(\mathbb{C})^{\otimes k}$ defined such that

$\sigma^{k}_{t}(a_{1}\otimes\cdots\otimes a_{k})=e^{ith}a_{1}e^{-ith}\otimes
e^{ith}a_{2}e^{-ith}\otimes\cdots\otimes e^{ith}a_{k}e^{-ith}$

on simple tensors. The flow $\sigma$ is trivial when $h$ is a scalar multiple
of $1$. When this is not the case, i.e. $h\notin\mathbb{R}1$, we say that
$\sigma$ is a _stationary ITPFI flow_ generated by $h$.

As shown in Section 6.5.1 there is a unique $\beta$-KMS state $\omega_{\beta}$
for $\sigma$ for each $\beta\in\mathbb{R}$. We seek here to determine the
factor type of these KMS states by calculating the $\Gamma$-invariant of
Connes for the factors

$R_{\beta}:=\pi_{\omega_{\beta}}(\otimes_{k=1}^{\infty}M_{m}(\mathbb{C}))^{\prime\prime}.$

######  Lemma 10.3.4.

The fixed point algebra $R_{\beta}^{\sigma^{\prime\prime}}$ for the flow
$\sigma^{\prime\prime}$ on $R_{\beta}$ is a factor.

###### Proof.

Choose an orthonormal basis $\\{\xi_{i}\\}_{i=1}^{m}$ for $\mathbb{C}^{m}$
consisting of eigenvectors for $h$. For each $n\in\mathbb{N}$ define a unitary
$V_{n}\in M_{m}(\mathbb{C})^{2n}$ such that

$V_{n}(\xi_{i_{1}}\otimes\xi_{i_{2}}\otimes\cdots\otimes\xi_{i_{2n}})=\xi_{i_{n+1}}\otimes\xi_{i_{n+2}}\otimes\cdots\
\otimes\xi_{i_{2n}}\otimes\xi_{i_{1}}\otimes\cdots\otimes\xi_{i_{n}}$

for $(i_{1},i_{2},\cdots,i_{2n})\in\\{1,2,\cdots,m\\}^{2n}$. Note that

$\operatorname{Ad}V_{n}(a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{2n})=a_{n+1}\otimes a_{n+2}\otimes\cdots\otimes a_{2n}\otimes
a_{1}\otimes\cdots\otimes a_{n}$ (10.3.3)

for $a_{1},a_{2},\cdots,a_{n}\in M_{m}(\mathbb{C})$ and that
$\operatorname{Ad}V_{n}\circ\sigma^{2n}_{t}=\sigma^{2n}_{t}\circ\operatorname{Ad}V_{n}$.
Set $U_{n}:=\phi_{2n,\infty}(V_{n})$. The decisive properties of $U_{n}$ are
that $U_{n}$ is $\sigma$-invariant, and

$U_{n}xU_{n}^{*}y=yU_{n}xU_{n}^{*}$ (10.3.4)

and

$\omega_{\beta}(U_{n}xU_{n}^{*}y)=\omega_{\beta}(x)\omega_{\beta}(y)$ (10.3.5)

for all $x,y\in\phi_{n,\infty}(M_{m}(\mathbb{C})^{\otimes n})$. The first
property, (10.3.5), follows from (10.3.3), and the second, (10.3.5), follows
from the description of $\omega_{\beta}$ given in Section 6.5.1.

Let $x\in\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$. We shall need the following

######  Observation 10.3.5.

$\lim_{n\to\infty}\pi_{\omega_{\beta}}(U_{n}xU_{n}^{*})=\omega_{\beta}(x)1$ in
the weak operator topology.

To establish this note that since
$\bigcup_{k}\Lambda_{\omega_{\beta}}(\phi_{k,\infty}((M_{m}(\mathbb{C})^{\otimes
k})))$ is dense in $H_{\omega_{\beta}}$ it suffices to show that

$\lim_{n\to\infty}\left<\pi_{\omega_{\beta}}(U_{n}xU_{n}^{*})\Lambda_{\omega_{\beta}}(a),\Lambda_{\omega_{\beta}}(b)\right>=\omega_{\beta}(x)\omega_{\beta}(b^{*}a)$

for $a,b\in\phi_{k,\infty}((M_{m}(\mathbb{C})^{\otimes k}))$. Similarly,
$\bigcup_{k}\phi_{k,\infty}((M_{m}(\mathbb{C})^{k})$ is norm-dense in
$\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$ so it suffices to do this when
$x\in\phi_{l,\infty}(M_{m}(\mathbb{C})^{\otimes l}),\ l\geq k$. Using (10.3.4)
and (10.3.5) we find for $n\geq l$,

$\displaystyle\left<\pi_{\omega_{\beta}}(U_{n}xU_{n}^{*})\Lambda_{\omega_{\beta}}(a),\Lambda_{\omega_{\beta}}(b)\right>=\omega_{\beta}\left(b^{*}U_{n}xU_{n}^{*}a\right)$
$\displaystyle=\omega_{\beta}\left(U_{n}xU_{n}^{*}b^{*}a\right)=\omega_{\beta}(x)\omega_{\beta}(b^{*}a),$

proving Observation 10.3.5.

Let $\omega^{\prime\prime}_{\beta}$ be the normal extension of
$\omega_{\beta}$ to $R_{\beta}$. Since $\omega_{\beta}$ is a $\beta$-KMS state
and $U_{n}$ is $\sigma$-invariant we have that
$\omega_{\beta}(U_{n}xU_{n}^{*})=\omega_{\beta}(x)$ for all
$x\in\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$, implying that
$\omega^{\prime\prime}_{\beta}\circ\operatorname{Ad}\pi_{\omega_{\beta}}(U_{n})=\omega^{\prime\prime}_{\omega_{\beta}}$.
Hence when $p$ is a central projection in $R_{\beta}^{\sigma^{\prime\prime}}$
and $x\in\otimes_{k=1}^{\infty}M_{m}(\mathbb{C})$ we find that

$\displaystyle\omega^{\prime\prime}_{\beta}(p\pi_{\omega_{\beta}}(x))=\omega^{\prime\prime}_{\beta}\left(\pi_{\omega_{\beta}}(U_{n})p\pi_{\omega_{\beta}}(x)\pi_{\omega_{\beta}}(U_{n})^{*}\right)$
$\displaystyle=\omega^{\prime\prime}_{\beta}\left(p\pi_{\omega_{\beta}}(U_{n})\pi_{\omega_{\beta}}(x)\pi_{\omega_{\beta}}(U_{n})^{*}\right)$

for all $n$ since $\pi_{\omega_{\beta}}(U_{n})\in
R_{\beta}^{\sigma^{\prime\prime}}$. By using that
$\omega^{\prime\prime}_{\beta}$ is continuous for the weak operator topology
on norm-bounded sets, it follows from Observation 10.3.5 that
$\lim_{n\to\infty}\omega^{\prime\prime}_{\beta}\left(p\pi_{\omega_{\beta}}(U_{n})\pi_{\omega_{\beta}}(x)\pi_{\omega_{\beta}}(U_{n})^{*}\right)=\omega_{\beta}(x)\omega^{\prime\prime}_{\beta}(p)$.
Thus

$\omega^{\prime\prime}_{\beta}(p\pi_{\omega_{\beta}}(x))=\omega_{\beta}(x)\omega^{\prime\prime}_{\beta}(p).$

Since $\pi_{\omega_{\beta}}(\otimes_{k=1}^{\infty}M_{m}(\mathbb{C}))$ is
$\sigma$-weakly dense in $R_{\beta}$ and
$\omega^{\prime\prime}\circ\pi_{\omega_{\beta}}=\omega_{\beta}$ we conclude
that
$\omega^{\prime\prime}_{\beta}(py)=\omega^{\prime\prime}_{\beta}(y)\omega^{\prime\prime}_{\beta}(p)$
for all $y\in R_{\beta}$. Inserting $1-p$ for $y$ we see that
$\omega^{\prime\prime}_{\beta}(1-p)\omega^{\prime\prime}_{\beta}(p)=0$ which
implies that $p=0$ or $p=1$ since $\omega^{\prime\prime}_{\beta}$ is faithful
by Lemma 4.2.4. It follows that $R_{\beta}^{\sigma^{\prime\prime}}$ is a
factor. ∎

######  Corollary 10.3.6.

For $\beta\neq 0$, $\Gamma(R_{\beta})=\operatorname{Sp}_{\alpha}(R_{\beta})$
where $\alpha_{t}=\sigma^{\prime\prime}_{-\beta t}$ for all $t\in\mathbb{R}$.

###### Proof.

Thanks to Lemma 10.3.4 this follows from the definition of $\Gamma(R_{\beta})$
and Theorem 4.3.1. ∎

######  Lemma 10.3.7.

$\Gamma(R_{\beta})=\beta\overline{G_{h}}$ where ${G_{h}}$ is the additive
group generated by the numbers

$\\{\lambda-\mu:\ \lambda,\mu\ \text{are eigenvalues of $h$}\\}$

and $\overline{G_{h}}$ denotes it closure in $\mathbb{R}$.

###### Proof.

$\Gamma(R_{\beta})=\operatorname{Sp}_{\alpha}(R_{\beta})$ by Corollary 10.3.6.
To calculate the spectrum $\operatorname{Sp}_{\alpha}(R_{\beta})$ let $f\in
L^{1}(\mathbb{R})$. Let $\\{e_{i,j}\\}_{i,j=1}^{m}$ be a set of matrix units
in $M_{m}(\mathbb{C})$ such that $e_{ii}$ is the orthogonal projection onto
$\xi_{i}$. Since $\xi_{i}$ is an eigenvector for $h$ there are real numbers

$\lambda_{i}\in\mathbb{R},\ i\in\\{1,2,\cdots,m\\},$

such that $h\xi_{i}=\lambda_{i}\xi_{i}$. Let $k\in\mathbb{N}$ and
$i_{l},j_{l}\in\\{1,2,\cdots,m\\}$, $l=1,2,\cdots,k$. Since

$\operatorname{Ad}e^{ith}(e_{i,j})=e^{it(\lambda_{i}-\lambda_{j})}e_{i,j}$

we find that

$\sigma^{k}_{t}(e_{i_{1},j_{1}}\otimes e_{i_{2},j_{2}}\otimes\cdots\otimes
e_{i_{k},l_{k}})=e^{it\chi}e_{i_{1},j_{1}}\otimes
e_{i_{2},j_{2}}\otimes\cdots\otimes e_{i_{k},l_{k}},$

where $\chi:=\sum_{r=1}^{k}(\lambda_{i_{r}}-\lambda_{j_{r}})$. For any normal
functional $\mu$ of $R_{\beta}$,

$\displaystyle\mu\left(\alpha_{f}(\pi_{\omega_{\beta}}(\phi_{k,\infty}(e_{i_{1},j_{1}}\otimes
e_{i_{2},j_{2}}\otimes\cdots\otimes e_{i_{k},l_{k}})))\right)$
$\displaystyle=\int_{\mathbb{R}}f(t)\mu\circ\pi_{\omega_{\beta}}\left(\sigma_{-\beta
t}(\phi_{k,\infty}(e_{i_{1},j_{1}}\otimes e_{i_{2},j_{2}}\otimes\cdots\otimes
e_{i_{k},l_{k}})\right)\ \mathrm{d}t$
$\displaystyle=\left(\int_{\mathbb{R}}f(t)e^{-i\beta\chi t}\
\mathrm{d}t\right)\mu\circ\pi_{\omega_{\beta}}(\phi_{k,\infty}(e_{i_{1},j_{1}}\otimes
e_{i_{2},j_{2}}\otimes\cdots\otimes e_{i_{k},l_{k}})).$

Note that $\pi_{\omega_{\beta}}(\phi_{k,\infty}(e_{i_{1},j_{1}}\otimes
e_{i_{2},j_{2}}\otimes\cdots\otimes e_{i_{k},l_{k}}))\neq 0$ since
$\pi_{\omega_{\beta}}\circ\phi_{k,\infty}$ is faithful. Hence
$\mu\left(\alpha_{f}(\pi_{\omega_{\beta}}(\phi_{k,\infty}(e_{i_{1},j_{1}}\otimes
e_{i_{2},j_{2}}\otimes\cdots\otimes e_{i_{k},l_{k}})))\right)=0$ if and only
if $\hat{f}(-\beta\chi)=\int_{\mathbb{R}}f(t)e^{-i\beta\chi t}\
\mathrm{d}t=0$. Since
$\bigcup_{k}\pi_{\omega_{\beta}}\circ\phi_{k,\infty}(M_{m}(\mathbb{C})^{\otimes
k})$ is $\sigma$-weakly dense in $R_{\beta}$ it follows that
$\alpha_{f}(R_{\beta})=\\{0\\}$ if and only if

$\hat{f}(-\beta\chi)=\int_{\mathbb{R}}f(t)e^{-i\beta\chi t}\ \mathrm{d}t=0$

for all $\chi\in G_{h}$. Hence $\beta G_{h}\subseteq Z(f)$ when
$\alpha_{f}(R_{\beta})=\\{0\\}$, implying that the additive subgroup $\beta
G_{h}$ of $\mathbb{R}$ is in $\operatorname{Sp}_{\alpha}(R_{\beta})$. Since
$\operatorname{Sp}_{\alpha}(R_{\beta})=\Gamma(R_{\beta})$ is a closed subset
of $\mathbb{R}$ it follows that the closure $\beta\overline{G_{g}}$ of $\beta
G_{h}$ is contained in $\Gamma(R_{\beta})$. If
$r\in\mathbb{R}\backslash\beta\overline{G_{h}}$ there is a function $f\in
L^{1}(\mathbb{R})$ such that $\hat{f}(r)\neq 0$ and $\hat{f}(\beta
G_{h})=\\{0\\}$; see Theorem 2.6.2 in [Ru3]. By the preceding calculation the
last property implies $\alpha_{f}(R_{\beta})=\\{0\\}$ and then the first
implies that $r\notin\operatorname{Sp}_{\alpha}(R_{\beta})$. ∎

Note the $G_{h}$ is dense in $\mathbb{R}$ if and only if there are eigenvalues
$\lambda,\mu,\lambda^{\prime},\mu^{\prime}$ for $h$ such that
$\lambda^{\prime}\neq\mu^{\prime}$ and

$\frac{\lambda-\mu}{\lambda^{\prime}-\mu^{\prime}}$

is irrational. When this is not the case there is a $\kappa_{h}>0$ such that

$G_{h}=\overline{G_{h}}=\mathbb{Z}\kappa_{h}.$

######  Corollary 10.3.8.

Let $\sigma$ be a stationary $ITPFI$ flow generated by the selfadjoint matrix
$h$, cf. (10.3.2).

  * $\bullet$

The $\beta$-KMS state for $\sigma$ is of type $III$ for all $\beta\neq 0$.

  * $\bullet$

When $G_{h}$ is dense in $\mathbb{R}$ the $\beta$-KMS state for $\sigma$ is of
type $III_{1}$ for all $\beta\neq 0$.

  * $\bullet$

When $G_{h}$ is not dense in $\mathbb{R}$ and $\beta\neq 0$, the $\beta$-KMS
state for $\sigma$ is of type $III_{\lambda_{\beta}}$ where
$\lambda_{\beta}=e^{-|\beta|\kappa_{h}}$.

######  Example 10.3.9.

Consider a stationary $ITPFI$ flow generated by the selfadjoint matrix $h\in
M_{2}(\mathbb{C})$. $h$ has two eigenvalue $\lambda_{1}$ and $\lambda_{2}$ is
this case. Set $\kappa_{h}=|\lambda_{1}-\lambda_{2}|$. By Corollary 10.3.8 the
$\beta$-KMS state for $\sigma$ is of type $III_{\lambda_{\beta}}$ where
$\lambda_{\beta}=e^{-|\beta|\kappa_{h}}$ when $\beta\neq 0$.

###  10.4 KMS-invariants for cocycle-conjugacy of flows

Let $A$ and $B$ be unital $C^{*}$-algebras, $\sigma$ a flow on $A$ and
$\sigma^{\prime}$ a flow on $B$. $\sigma$ is _conjugate_ to $\sigma^{\prime}$
when there is a $*$-isomorphim $\alpha:B\to A$ such that
$\sigma_{t}\circ\alpha=\alpha\circ\sigma^{\prime}_{t}$ for all
$t\in\mathbb{R}$.

As in Section 9.1 we denote by $S^{\sigma}_{\beta}$ the set of $\beta$-KMS
states for $\sigma$.

######  Lemma 10.4.1.

Let $\sigma$ be a flow on the unital $C^{*}$-algebra $A$ and $\sigma^{\prime}$
a flow on the unital $C^{*}$-algebra $B$. Assume $\alpha:B\to A$ is a
$*$-isomorphism such that
$\sigma_{t}\circ\alpha=\alpha\circ\sigma^{\prime}_{t}$ for all
$t\in\mathbb{R}$. For $\beta\in\mathbb{R}$ the map
$\psi\mapsto\psi\circ\alpha$ is an affine homeomorphism from
$S^{\sigma}_{\beta}$ onto $S^{\sigma^{\prime}}_{\beta}$, and
$\pi_{\psi}(A)^{\prime\prime}$ is isomorphic to
$\pi_{\psi\circ\alpha}(B)^{\prime\prime}$ for all $\psi\in
S^{\sigma}_{\beta}$.

###### Proof.

It follows from Lemma 2.4.8 the map $\psi\mapsto\psi\circ\alpha$ is a
bijection from $S^{\sigma}_{\beta}$ onto $S^{\sigma^{\prime}}_{\beta}$ and it
is clearly a homeomorphism with respect to the weak* topologies. Furthermore,
for every $\psi\in S^{\sigma}_{\beta}$ there is a unitary
$W:H_{\psi\circ\alpha}\to H_{\psi}$ defined such that
$W\Lambda_{\psi\circ\alpha}(b)=\Lambda_{\psi}(\alpha(b))$. It is easy to check
that $W\pi_{\psi\circ\alpha}(b)=\pi_{\psi}(\alpha(b))W$ for all $b\in B$ which
implies the claim, $\pi_{\psi}(A)^{\prime\prime}\ \simeq\
\pi_{\psi\circ\alpha}(B)^{\prime\prime}$. ∎

Let $A$ be a unital $C^{*}$-algebra. Let $\sigma$ and $\sigma^{\prime}$ be
flows on $A$. Following [Ki1] we say that $\sigma^{\prime}$ is a _cocycle
perturbation_ of $\sigma$ when there is a continuous unitary representation
$u=(u_{t})_{t\in\mathbb{R}}$ of $\mathbb{R}$ in $A$ such that
$u_{s}\sigma_{s}(u_{t})=u_{s+t}$ and
$\sigma^{\prime}_{t}=\operatorname{Ad}u_{t}\circ\sigma_{t}$ for all
$s,t\in\mathbb{R}$. Let $B$ be another unital $C^{*}$-algebra and let
$\sigma^{\prime\prime}$ be a flow on $B$. Still following [Ki1] we say that
$\sigma^{\prime\prime}$ is _cocycle conjugate_ to $\sigma$ when
$\sigma^{\prime\prime}$ is conjugate to a cocycle perturbation of $\sigma$,
i.e. if when there is an isomorphism $\phi:B\to A$ and a continuous unitary
representation $u=(u_{t})_{t\in\mathbb{R}}$ of $\mathbb{R}$ in $A$ such that
$u_{s}\sigma_{s}(u_{t})=u_{s+t}$ and
$\phi\circ\sigma^{\prime\prime}_{t}\circ\phi^{-1}=\operatorname{Ad}u_{t}\circ\sigma_{t}$
for all $s,t\in\mathbb{R}$.

######  Theorem 10.4.2.

Let $A$ and $B$ be unital $C^{*}$-algebras, $\sigma$ a flow on $A$ and
$\sigma^{\prime\prime}$ a flow on $B$. Assume that $\sigma^{\prime\prime}$ is
cocycle conjugate to $\sigma$ and let $\beta\in\mathbb{R}$. It follows that
there is a homeomorphism $\eta:S^{\sigma}_{\beta}\to
S^{\sigma^{\prime\prime}}_{\beta}$ such that $\eta(\omega)$ is extremal if and
only if $\omega$ is extremal. Furthermore, when $\omega\in S^{\sigma}_{\beta}$
is extremal there is a von Neumann algebra factor $N$ and projections $e,f\in
N$ such that $\pi_{\omega}(A)^{\prime\prime}\simeq eNe$ and
$\pi_{\eta(\omega)}(B)^{\prime\prime}\simeq fNf$.

###### Proof.

By assumption there is an isomorphism $\phi:B\to A$ such that
$\sigma^{\prime}_{t}:=\phi\circ\sigma^{\prime\prime}_{t}\circ\phi^{-1}$ is a
flow $\sigma^{\prime}$ on $A$ which is a cocycle perturbation of $\sigma$. By
Lemma 10.4.1 there is an affine homeomorphism
$\eta^{\prime}:S^{\sigma^{\prime}}_{\beta}\to
S^{\sigma^{\prime\prime}}_{\beta}$ such that
$\pi_{\eta^{\prime}(\omega)}(B)^{\prime\prime}\simeq\pi_{\omega}(A)^{\prime\prime}$
when $\omega\in S^{\sigma^{\prime}}_{\beta}$ is extremal. Since
$\sigma^{\prime}$ is a cocycle perturbation of $\sigma$ there is a continuous
unitary representation $u=(u_{t})_{t\in\mathbb{R}}$ of $\mathbb{R}$ in $A$
such that $u_{s}\sigma_{s}(u_{t})=u_{s+t}$ and
$\sigma^{\prime}_{t}=\operatorname{Ad}u_{t}\circ\sigma_{t}$ for all
$s,t\in\mathbb{R}$. As in the proof of Theorem 6.3.1 we consider now the flow
$\gamma$ on $M_{2}(A)$ given by

$\gamma_{t}\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right):=\left(\begin{matrix}\sigma_{t}(a)&\sigma_{t}(b)u_{-t}\\\
u_{t}\sigma_{t}(c)&\sigma^{\prime}_{t}(d)\end{matrix}\right)$

and the embeddings $\iota_{i}:A\to M_{2}(A),\ i=1,2$, of $A$ into the
canonical orthogonal corners of $M_{2}(A)$. Set

$p_{1}=\left(\begin{smallmatrix}1&0\\\ 0&0\end{smallmatrix}\right)\
\text{and}\ p_{2}=\left(\begin{smallmatrix}0&0\\\
0&1\end{smallmatrix}\right).$

It follows from Theorem 3.1.8 that
$\psi\mapsto\psi(p_{1})^{-1}\psi\circ\iota_{1}$ is a bijection $\mu_{1}$ from
$S^{\gamma}_{\beta}$ onto $S^{\sigma}_{\beta}$ and
$\psi\mapsto\psi(p_{2})^{-1}\psi\circ\iota_{2}$ is a bijection $\mu_{2}$ from
$S^{\sigma}_{\beta}$ onto $S^{\sigma^{\prime}}_{\beta}$. Note also that
$\mu_{1}$ and $\mu_{2}$ both are continuous for the weak* topologies and
preserve extremality. Hence

$\eta^{\prime\prime}:=\mu_{2}\circ\mu_{1}^{-1}:S^{\sigma}_{\beta}\to
S^{\sigma^{\prime}}_{\beta}$

is a continuous bijection which takes extremal $\beta$-KMS states for $\sigma$
to extremal $\beta$-KMS states for $\sigma^{\prime}$. Define
$W_{i}:H_{\mu_{i}(\psi)}\to H_{\psi}$ such that

$W_{i}\eta_{\mu_{i}(\psi)}(a)=\psi(p_{i})^{-\frac{1}{2}}\Lambda_{\psi}(\iota_{i}(a))$

for all $a\in A$, and note that $W_{i}$ is a unitary from $H_{\mu_{i}(\psi)}$
onto $H^{i}_{\psi}$. Then

$W_{i}\pi_{\mu_{i}(\psi)}(x)\Lambda_{\mu_{i}(\psi)}(a)=\pi_{\psi}(\iota_{i}(x))W_{i}\Lambda_{\mu_{i}(\psi)}(a)$

for all $x,a\in A$ and hence

$W_{i}\pi_{\mu_{i}(\psi)}(x)W_{i}^{*}|_{H_{\psi}^{i}}=\pi_{\psi}(\iota_{i}(x))|_{H_{\psi}^{i}}=\pi_{\psi}(p_{i})\pi_{\psi}(\iota_{i}(x))\pi_{\psi}(p_{i})|_{H_{\psi}^{i}}$

for all $x\in A$, where
$H_{\psi}^{i}:=\overline{\Lambda_{\psi}(\iota_{i}(A))}$. Since

$p_{i}\iota_{i}(A)p_{i}=p_{i}M_{2}(A)p_{i},$

it follows that
$\pi_{\mu_{i}(\psi)}(A)^{\prime\prime}\simeq\pi_{\psi}(p_{i})\pi_{\psi}(A)^{\prime\prime}\pi_{\psi}(p_{i})$.
Hence $\eta:=\eta^{\prime}\circ\eta^{\prime\prime}:S^{\sigma}_{\beta}\to
S^{\sigma^{\prime\prime}}_{\beta}$ is a continuous bijection with the stated
properties.

∎

######  Corollary 10.4.3.

Let $A$ and $B$ be unital $C^{*}$-algebras, $\sigma$ a flow on $A$ and
$\sigma^{\prime}$ a flow on $B$. Assume that $\sigma$ and $\sigma^{\prime}$
are cocycle conjugate. There is a homeomorphism $\eta:S^{\sigma}_{\beta}\to
S^{\sigma^{\prime}}_{\beta}$ such that $\eta(\omega)$ is extremal if and only
if $\omega$ is extremal.

Let $\omega\in S^{\sigma}_{\beta}$ be extremal. Then $\omega$ is of factor
type $I$ if and only if $\eta(\omega)$ is of factor type $I$, $\omega$ is of
factor type $II$ if and only if $\eta(\omega)$ is of factor type $II$ and for
$\lambda\in[0,1]$, $\omega$ is of factor type $III_{\lambda}$ if and only if
$\eta(\omega)$ is of factor type $III_{\lambda}$.

###### Proof.

This follows from Theorem 10.4.2 plus the following facts. Let $e$ be a non-
zero projection in the von Neumann algebra factor. Then

  * (i)

$N$ is of type $I$ if and only if $eNe$ is of type $I$,

  * (ii)

$N$ is of type $II$ if and only if $eNe$ is of type $II$,

  * (iii)

$N$ is of type $III_{\lambda}$ if and only if $eNe$ is of type $III_{\lambda}$
for any $\lambda\in[0,1]$.

The two first facts, (i) and (ii), are well-known, although their proofs are
hard to find. In textbooks they are mostly relegated to the exercises. See
e.g. E.4.18 in [SZ]. The last fact (iii) follows from Corollaire 3.2.8 (b) of
[Co1]. ∎

######  Example 10.4.4.

Consider the stationary ITPFI flows generated by two selfadjoint matrices
$h,h^{\prime}\in M_{2}(\mathbb{C})$ with eigenvalues $\lambda_{1},\lambda_{2}$
and $\lambda^{\prime}_{1},\lambda^{\prime}_{2}$, respectively. It follows by
combining Example 10.3.9 with Corollary 10.4.3 that these two flows are not
cocycle conjugate unless
$|\lambda_{1}-\lambda_{2}|=|\lambda^{\prime}_{1}-\lambda^{\prime}_{2}|$. Note
this holds if and only if there is a real number $r\in\mathbb{R}$ and a
unitary $W\in M_{2}(\mathbb{C})$ such that
$W(h-r1)W^{*}\in\\{h^{\prime},-h^{\prime}\\}$.

######  Notes and remarks 10.4.5.

The material in this chapter is new in the sense that the author has not seen
it in any book or paper. But most if not all arguments are straightforward
applications of well-known facts for normal flows on von Neumann algebras, and
the content can safely be considered to be folklore.

##  Appendix A Integration in Banach spaces

Let $X$ be a Banach space, $M$ a locally compact Hausdorff space and $\mu$ a
regular measure on $M$. Let $f:M\to X$ be a continuous function such that

$\int_{M}\left\|f(x)\right\|\ \mathrm{d}\mu(x)<\infty.$ (A.0.1)

In this appendix we describe a way to define the integral

$\int_{M}f(x)\ \mathrm{d}\mu(x),$

and establish some of its properties that are used in the main text.

###  A.1 The definition

We consider pairs $(\mathcal{U},\epsilon)$ where $\mathcal{U}$ is a finite
collection of mutually disjoint Borel sets $U$ of $M$ and $\epsilon>0$ is a
positive number such that

  * •

$\bigcup_{U\in\mathcal{U}}$ is pre-compact; that is, its closure is compact,

  * •

$\int_{M\backslash(\bigcup_{U\in\mathcal{U}}U)}\ \left\|f(x)\right\|\
\mathrm{d}\mu(x)\leq\epsilon$, and

  * •

$\sup_{x,y\in
U}\left\|f(x)-f(y)\right\|\left(\mu\left(\bigcup_{U\in\mathcal{U}}U\right)+1\right)\leq\epsilon$
for all $U\in\mathcal{U}$.

The collection of such pairs will be denoted by $\mathcal{I}$. We consider
$\mathcal{I}$ as a pre-ordered set where
$(\mathcal{U},\epsilon)\leq(\mathcal{V},\delta)$ means that

  * •

$\delta\leq\epsilon$,

  * •

$\bigcup_{U\in\mathcal{U}}\subseteq\bigcup_{V\in\mathcal{V}}V$, and

  * •

for every $V\in\mathcal{V}$ there is an element $U^{\prime}\in\mathcal{U}$
such that $V\cap(\bigcup_{U\in\mathcal{U}}U)\subseteq U^{\prime}$.

######  Lemma A.1.1.

$\mathcal{I}$ is a directed set.

###### Proof.

Let $(\mathcal{U},\epsilon),(\mathcal{V},\delta)\in\mathcal{I}$. Set
$\epsilon^{\prime}=\min\\{\epsilon,\delta\\}$. We choose a compact set
$K\subseteq M$ such that

$\left(\overline{\bigcup_{U\in\mathcal{U}}U}\right)\cup\left(\overline{\bigcup_{V\in\mathcal{V}}V}\right)\subseteq
K$

and

$\int_{M\backslash K}\|f(x)\|\ \mathrm{d}\mu(x)\leq\epsilon^{\prime}.$

The sets

$K\backslash\left({\bigcup_{U\in\mathcal{U}}U}\cup{\bigcup_{V\in\mathcal{V}}V}\right),$
$U\cap V,\ \ \ \ U\in\mathcal{U},\ V\in\mathcal{V},$
$U\backslash\bigcup_{V\in\mathcal{V}}V,\ \ \ \ U\in\mathcal{U}$

and

$V\backslash\bigcup_{U\in\mathcal{U}}U,\ \ \ \ V\in\mathcal{V},$

constitute a partition $\mathcal{P}$ of $K$ into Borel sets. Since $f$ is
continuous and $K$ is compact there is a finite Borel partition

$K=\bigcup_{W\in\mathcal{W}}W$

of $K$ subordinate to $\mathcal{P}$ such that

$\sup_{x,y\in W}\left\|f(x)-f(y)\right\|(\mu(K)+1)\leq\epsilon^{\prime}$

for all $W\in\mathcal{W}$. Then
$(\mathcal{U},\epsilon)\leq(\mathcal{W},\epsilon^{\prime})$ and
$(\mathcal{V},\delta)\leq(\mathcal{W},\epsilon^{\prime})$. ∎

For $(\mathcal{U},\epsilon)\in\mathcal{I}$, we denote by
$\mathcal{S}(\mathcal{U},\epsilon)$ the set of elements $S$ in $X$ of the form

$S=\sum_{U\in\mathcal{U}}f(i(U))\mu(U)$

for some function $i:\mathcal{U}\to M$ with the property that $i(U)\in U$ for
all $U\in\mathcal{U}$.

######  Lemma A.1.2.

There is an element $I\in X$ with the property that for every $\delta>0$ there
is a $(\mathcal{U},\epsilon)\in\mathcal{I}$ such that

$\left\|I-S\right\|\leq\delta$

for all $S\in S(\mathcal{V},\epsilon^{\prime})$ when
$(\mathcal{U},\epsilon)\leq(\mathcal{V},\epsilon^{\prime})$.

###### Proof.

It follows from (A.0.1) that for each $n\in\mathbb{N}$ there is a compact
subset $K_{n}\subseteq M$ such that

$\int_{M\backslash K_{n}}\|f(x)\|\ \mathrm{d}\mu(x)\leq\frac{1}{n}.$

Since $f$ is continuous and $K_{n}$ compact there is a finite partition
$\mathcal{U}_{n}$ of $K_{n}$ into mutually disjoint Borel subsets $U$ of
$K_{n}$ such that $\|f(x)-f(y)\|(\mu(K_{n})+1)\leq\frac{1}{n}$ for all $x,y\in
U$ and all $U\in\mathcal{U}_{n}$. Then
$(\mathcal{U}_{n},\frac{1}{n})\in\mathcal{I}$. We proceed inductively, as we
can, to arrange that
$(\mathcal{U}_{n},\frac{1}{n})\leq(\mathcal{U}_{n+1},\frac{1}{n+1})$ in
$\mathcal{I}$ for all $n$. For each $n$ choose
$S_{n}\in\mathcal{S}(\mathcal{U}_{n},\frac{1}{n})$. Let
$(\mathcal{V},\delta)\geq(\mathcal{U}_{n},\frac{1}{n})$ and consider an
element

$S=\sum_{V\in\mathcal{V}}f(j(V))\mu(V)\in\ \mathcal{S}(\mathcal{V},\delta).$

Then

$\displaystyle
S=\sum_{V\in\mathcal{V}}f(j(V))\mu\left(V\backslash\bigcup_{U\in\mathcal{U}_{n}}U\right)+\sum_{V\in\mathcal{V},U\in\mathcal{U}_{n}}f(j(V))\mu\left(V\cap
U\right).$

Note that

$\displaystyle\left\|\sum_{V\in\mathcal{V}}f(j(V))\mu\left(V\backslash\bigcup_{U\in\mathcal{U}_{n}}U\right)\right\|\leq\sum_{V\in\mathcal{V}}\left\|f(j(V))\right\|\mu\left(V\backslash\bigcup_{U\in\mathcal{U}_{n}}U\right)$
$\displaystyle=\sum_{V\in\mathcal{V}}\int_{V\backslash\bigcup_{U\in\mathcal{U}_{n}}U}\left\|f(j(V))\right\|\
\mathrm{d}\mu(x)$
$\displaystyle\leq\sum_{V\in\mathcal{V}}\int_{V\backslash\bigcup_{U\in\mathcal{U}_{n}}U}\left(\left\|f(x)\right\|+\frac{\delta}{\mu\left(\bigcup_{W\in\mathcal{V}}W\right)+1}\right)\
\mathrm{d}\mu(x)$
$\displaystyle\leq\delta+\int_{M\backslash\bigcup_{U\in\mathcal{U}_{n}}U}\left\|f(x)\right\|\
\mathrm{d}\mu(x)\leq\delta+\frac{1}{n}\leq\frac{2}{n}.$

Let $i_{n}:\mathcal{U}_{n}\to X$ be the function defining $S_{n}$; that is,

$S_{n}=\sum_{U\in\mathcal{U}_{n}}f(i_{n}(U))\mu(U).$

Then

$\displaystyle\left\|S_{n}-\sum_{V\in\mathcal{V},U\in\mathcal{U}_{n}}f(j(V))\mu\left(V\cap
U\right)\right\|$
$\displaystyle=\left\|\sum_{V\in\mathcal{V},U\in\mathcal{U}_{n}}f(i_{n}(U))\mu\left(V\cap
U\right)-\sum_{V\in\mathcal{V},U\in\mathcal{U}_{n}}f(j(V))\mu\left(V\cap
U\right)\right\|$
$\displaystyle\leq\sum_{V\in\mathcal{V},U\in\mathcal{U}_{n}}\left\|f(i_{n}(U))-f(j(V))\right\|\mu\left(V\cap
U\right)$
$\displaystyle\leq\sum_{V\in\mathcal{V},U\in\mathcal{U}_{n}}\left(\frac{1}{n\left(\mu(K_{n})+1\right)}+\frac{\delta}{\mu\left(\bigcup_{V\in\mathcal{V}}V\right)+1}\right)\mu\left(V\cap
U\right)$ $\displaystyle\leq\frac{1}{n}+\delta\leq\frac{2}{n}.$

It follows that $\left\|S-S_{n}\right\|\leq\frac{4}{n}$. In particular, the
sequence $\\{S_{n}\\}$ is Cauchy in $X$ and we set $I=\lim_{n\to\infty}S_{n}$.
Then $I$ has stated property: Let $\delta>0$ be given. There is an
$N\in\mathbb{N}$ such that $\frac{4}{n}\leq\frac{\delta}{2}$ for all $n\geq N$
and $\left\|S_{n}-I\right\|\leq\frac{\delta}{2}$ for all $n\geq N$. If
$(\mathcal{V},\epsilon^{\prime})\geq(\mathcal{U}_{N},\frac{1}{N})$ the
calculations above imply that
$\left\|S-I\right\|\leq\left\|S-S_{N}\right\|+\frac{\delta}{2}\leq\frac{4}{N}+\frac{\delta}{2}\leq\delta$
for all $S\in\mathcal{S}(\mathcal{V},\epsilon^{\prime})$. ∎

The element $I$ of Lemma A.1.2 will be denoted by

$\int_{M}f(x)\ \mathrm{d}\mu(x).$

By construction there is a sequence $\\{S_{n}\\}$ in $X$ such that

$\lim_{n\to\infty}S_{n}=\int_{M}f(x)\ \mathrm{d}\mu(x)$

and for each $n$ there is a finite collection of mutually disjoint Borel sets
$\\{U_{i}:\ i=1,2,\cdots,N_{n}\\}$ in $M$ and elements $x_{i}\in U_{i},\
i=1,2,\cdots,N_{n}$, such that the sum

$I_{n}:=\sum_{i=1}^{N_{n}}f(x_{i})\mu(U_{i})$

has the properties that

$\left\|I_{n}-\int_{M}f(x)\ \mathrm{d}\mu(x)\right\|\leq\frac{1}{n}$

and

$\left\|I_{n}\right\|\leq\int_{M}\left\|f(x)\right\|\
\mathrm{d}\mu(x)+\frac{1}{n}.$

In particular, it follows that

$\left\|\int_{M}f(x)\ \mathrm{d}\mu(x)\right\|\leq\int_{M}\left\|f(x)\right\|\
\mathrm{d}\mu(x).$ (A.1.1)

###  A.2 Miscellaneous

######  Lemma A.2.1.

Let $X$ and $Y$ be Banach spaces, $D\subseteq X$ a subspace and $L:D\to Y$ a
linear map. Let $M$ be a locally compact Hausdorff space and $\mu$ a regular
Borel measure on $M$. Let $f:M\to X$ be a continuous function. Assume that

  * •

$L:D\to Y$ is closed,

  * •

$f(x)\in D$ for all $x\in M$,

  * •

$M\ni x\mapsto L(f(x))$ is continuous,

  * •

$\int_{M}\left\|f(x)\right\|\ \mathrm{d}x<\infty$, and

  * •

$\int_{M}\left\|L(f(x)\right\|\ \mathrm{d}x<\infty$.

It follows that $\int_{M}f(x)\ \mathrm{d}\mu(x)\in D$ and
$L\left(\int_{M}f(x)\ \mathrm{d}\mu(x)\right)=\int_{M}L(f(x))\
\mathrm{d}\mu(x)$.

###### Proof.

Consider the function $M\to X\oplus Y$ defined by

$M\ni x\mapsto(f(x),L(f(x)).$

By applying Lemma A.1.2 to this function we obtain sequences $I_{n}$ in $D$
and $J_{n}$ in $Y$ such that $L(I_{n})=J_{n}$ and

$\lim_{n\to\infty}(I_{n},J_{n})=\left(\int_{M}f(x)\
\mathrm{d}\mu(x),\int_{M}L(f(x))\ \mathrm{d}\mu(x)\right)$

in $X\oplus Y$. This gives the conclusion because $L$ is closed. ∎

######  Lemma A.2.2.

Let $X$ be a Banach space, $M$ a locally compact Hausdorff space and $\mu$ a
regular measure on $M$. Let $\\{f_{n}\\}$ be a sequence of continuous
functions $f_{n}:M\to X$ and $f:M\to X$ a continuous function such that
$\lim_{n\to\infty}f_{n}(x)=f(x)$ for all $x\in M$. Assume that there is
function $h\in L^{1}(M,\mu)$ such that

$\left\|f_{n}(x)\right\|\leq h(x)\ \ \forall x\in M.$

Then $\lim_{n\to\infty}\int_{M}f_{n}(x)\ \mathrm{d}\mu(x)=\int_{M}f(x)\
\mathrm{d}\mu(x)$ in $X$.

###### Proof.

Note that all the integrals $\int_{M}f_{n}(x)\ \mathrm{d}\mu(x)$ and
$\int_{M}f(x)\ \mathrm{d}\mu(x)$ are defined since $\int_{M}h(x)\
\mathrm{d}\mu(x)<\infty$. It follows from (A.1.1) that

$\left\|\int_{M}f_{n}(x)\ \mathrm{d}\mu(x)-\int_{M}f(x)\
\mathrm{d}\mu(x)\right\|\leq\int_{M}\left\|f_{n}(x)-f(x)\right\|\
\mathrm{d}\mu(x).$

An application of Lebesgue’s theorem on dominated convergence shows that
$\lim_{n\to\infty}\int_{M}\left\|f_{n}(x)-f(x)\right\|\ \mathrm{d}\mu(x)=0$. ∎

The setting of the next lemma is that from Section 2.1.1. In particular, it
deals with the smoothing operators defined in (2.1.9).

######  Lemma A.2.3.

Let $k\in\mathbb{N}$, $a\in A$. For every $\epsilon>0$ there are numbers
$\lambda_{i}\in[0,1]$ and $t_{i}\in\mathbb{R}$, $i=1,2,\cdots,n$, such that
$\sum_{i=1}^{n}\lambda_{i}=1$ and

$\left\|R_{k}(a)-\sum_{i=1}^{n}\lambda_{i}\sigma_{t_{i}}(a)\right\|\leq\epsilon.$

###### Proof.

Define the continuous function $f:\mathbb{R}\to A\oplus\mathbb{C}$ such that

$f(t)=\left(\sqrt{\frac{k}{\pi}}e^{-kt^{2}}\sigma_{t}(a),\
\sqrt{\frac{k}{\pi}}e^{-kt^{2}}\right).$

Let $\delta\in]0,1[$ be so small that

$\frac{\delta}{1-\delta}(\|a\|+1)\leq\epsilon.$

By applying Lemma A.1.2 to $f$ we get numbers $s_{i}\geq 0$ and
$t_{i}\in\mathbb{R}$, $i=1,2,\cdots,n$, such that

$\left\|\sum_{i=1}^{n}\sqrt{\frac{k}{\pi}}e^{-kt_{i}^{2}}\sigma_{t_{i}}(a)s_{i}-R_{k}(a)\right\|\leq\delta$

and

$\left|\sum_{i=1}^{n}\sqrt{\frac{k}{\pi}}e^{-kt_{i}^{2}}s_{i}-1\right|\leq\delta.$

Set $x;:=\sum_{i=1}^{n}\sqrt{\frac{k}{\pi}}e^{-kt_{i}^{2}}s_{i}$ and

$\lambda_{i}:=x^{-1}\sqrt{\frac{k}{\pi}}e^{-kt_{i}^{2}}s_{i}.$

Then $\sum_{i=1}^{n}\lambda_{i}=1$ and

$\displaystyle\left\|R_{k}(a)-\sum_{i=1}^{n}\lambda_{i}\sigma_{t_{i}}(a)\right\|$
$\displaystyle\leq\left\|x^{-1}R_{k}(a)-R_{k}(a)\right\|+\left\|x^{-1}R_{k}(a)-x^{-1}\sum_{i=1}^{n}\sqrt{\frac{k}{\pi}}e^{-kt_{i}^{2}}\sigma_{t_{i}}(a)s_{i}\right\|$
$\displaystyle\leq\left|x^{-1}-1\right|\left\|R_{k}(a)\right\|+x^{-1}\delta\leq\frac{\delta}{1-\delta}\|a\|+\frac{\delta}{1-\delta}\leq\epsilon.$

∎

##  Appendix B Flows on compact operators

Let $\mathbb{K}$ denote the $C^{*}$-algebra of compact operators on the
infinite dimensional separable Hilbert space $\mathbb{H}$. In this appendix we
prove the following folklore theorem.

######  Theorem B.0.1.

Let $\sigma$ be a flow on $\mathbb{K}$. There is a strongly continuous unitary
representation $U$ of $\mathbb{R}$ on $\mathbb{H}$ such that

$\sigma_{t}(a)=U_{t}aU_{t}^{*}$

for all $a\in\mathbb{K}$ and all $t\in\mathbb{R}$.

By Stone’s theorem, cf. e.g. Theorem 5.6.36 of [KR], this theorem has the
following corollary.

######  Corollary B.0.2.

Let $\sigma$ be a flow on $\mathbb{K}$. There is a (possible unbounded) self-
adjoint operator $H$ on $\mathbb{H}$ such that

$\sigma_{t}(a)=e^{itH}ae^{-itH}$

for all $t\in\mathbb{R}$ and all $a\in\mathbb{K}$.

An important step of the proof of Theorem B.0.1 depends on a cohomological
fact which is isolated in the following subsection.

####  B.0.1 Cohomology

Let $\mathbb{T}$ be the circle group; $\mathbb{T}=\left\\{z\in\mathbb{C}:\
|z|=1\right\\}$. A continuous function
$\lambda:\mathbb{R}\times\mathbb{R}\to\mathbb{T}$ satisfying

  * (i)

$\lambda(s,t)\lambda(s+t,u)=\lambda(t,u)\lambda(s,t+u)$ for all
$s,t,u\in\mathbb{R}$,

is called a $2$-cocycle, and it is said to be normalized when

  * (ii)

$\lambda(0,t)=\lambda(t,0)=1$ for all $t\in\mathbb{R}$.

The set of normalized $2$-cocycles is an abelian group under pointwise
multiplication. Any continuous function $\mu:\mathbb{R}\to\mathbb{T}$ with
$\mu(0)=1$ gives rise to a normalized $2$-cocycle $\lambda_{\mu}$ such that

$\lambda_{\mu}(s,t)=\mu(s)\mu(t)\overline{\mu(s+t)},$

and such a $2$-cocycle is called a coboundary. The coboundaries constitute a
subgroup of the normalized $2$-cocycles and the quotient group is denoted by
$H^{2}(\mathbb{R},\mathbb{T})$. For the proof of Theorem B.0.1 it is crucial
that this group is trivial.

######  Theorem B.0.3.

$H^{2}(\mathbb{R},\mathbb{T})=0$, i.e. every normalized $2$-cocycle is a
coboundary.

###### Proof.

Let $\lambda$ be a normalized $2$-cocycle. By continuity of $\lambda$ there is
$\epsilon>0$ such that $|\lambda(s,t)-1|\leq 2^{-\frac{1}{2}}$ when
$s,t\in[0,\epsilon]$. Set

$\lambda^{\prime}(s,t):=\lambda(\epsilon s,\epsilon t).$

Then $\lambda^{\prime}$ is a normalized $2$-cocycle and $\lambda$ is a
coboundary if and only if $\lambda^{\prime}$ is. We may therefore assume that

$|\lambda(s,t)-1|\leq 2^{-\frac{1}{2}}$ (B.0.1)

when $s,t\in[0,1]$. There is a unique function $\mu:\mathbb{R}\to\mathbb{T}$
such that $\mu(0)=1$ and

$\mu(t+n)=\mu(n)\lambda(n,t),\ \ \forall n\in\mathbb{Z},\ t\in[0,1].$ (B.0.2)

Note that $\mu$ is continuous and that

$\mu(t)=\lambda(0,t)=1,\ \ \forall t\in[0,1].$ (B.0.3)

Define $\lambda^{1}:\mathbb{R}\times\mathbb{R}\to\mathbb{T}$ such that

$\lambda^{1}(s,t):=\overline{\mu(s+t)}\mu(s)\mu(t)\lambda(s,t)\ .$

Then $\lambda^{1}$ is a normalized $2$-cocycle and it follows from (B.0.2) and
(B.0.3) that

$\lambda^{1}(n,t)=1,\ \ \forall t\in[0,1],\ \forall n\in\mathbb{Z}.$ (B.0.4)

Since $\lambda^{1}$ is a $2$-cocycle,

$\lambda^{1}(n,1)\lambda^{1}(n+1,t)=\lambda^{1}(1,t)\lambda^{1}(n,1+t)$

and it follows therefore by induction starting with (B.0.4), that
$\lambda^{1}(n,t)=1$ for all $n\in\mathbb{Z}$ and all $t\geq 0$. Similarly, by
using that

$\lambda^{1}(n,-1)\lambda^{1}(n-1,t)=\lambda^{1}(-1,t)\lambda^{1}(n,t-1),$

it follows that $\lambda^{1}(n,t)=1$ for all $n\in\mathbb{Z}$ and all $t\leq
0$. It follows therefore that

$\lambda^{1}(n+s,t)=\lambda^{1}(n,s)\lambda^{1}(n+s,t)=\lambda^{1}(s,t)\lambda^{1}(n,s+t)=\lambda^{1}(s,t)$

for all $s,t\in\mathbb{R}$, showing that

$s\mapsto\lambda^{1}(s,t)$

is $1$-periodic for all $t\in\mathbb{R}$. When $t\in[1,2]$, it follows from
the equations (B.0.1), (B.0.2) and (B.0.3) that

$|\mu(t)-1|=|\mu(1+(t-1))-1|=|\lambda(1,t-1)-1|\leq 2^{-\frac{1}{2}},$

which combined with (B.0.3) shows that

$|\mu(t)-1|\leq 2^{-\frac{1}{2}},\ \ \forall t\in[0,2].$

Combined with (B.0.3) this shows that

$\left|\lambda^{1}(s,t)-\lambda(s,t)\right|=\left|{\mu(s+t)}-1\right|\leq
2^{-\frac{1}{2}}$

when $s,t\in[0,1]$. By using (B.0.1) this implies that

$\left|\lambda^{1}(s,t)-1\right|\leq 2\cdot 2^{-\frac{1}{2}}=\sqrt{2}$ (B.0.5)

when $s,t\in[0,1]$. Since $\lambda^{1}$ is $1$-periodic in the first variable
it follows that (B.0.5) holds for all $s\in\mathbb{R}$ and all $t\in[0,1]$.
Let $\chi:\mathbb{T}\backslash\\{-1\\}\to]-\pi,\pi[$ be the inverse of
$]-\pi,\pi[\ni x\mapsto e^{ix}$ and set

$y(s,t):=\chi(\lambda^{1}(s,t))$

when $t\in[0,1]$. It follows from the cocycle identity (i) that

$y(s,t)+y(s+t,u)=y(t,u)+y(s,t+u)$ (B.0.6)

for all $s\in\mathbb{R}$ and all $t,u\in[0,1]$ such that $t+u\leq 1$. Define
$\alpha:[0,1]\to\mathbb{R}$ such that

$\alpha(t):=\int_{0}^{1}y(x,t)\ \mathrm{d}x.$

Then $\alpha$ is continuous. Since $y$ is $1$-periodic in the first variable
it follows that

$\int_{0}^{1}y(s+t,u)\ \mathrm{d}s=\int_{0}^{1}y(s,u)\ \mathrm{d}s$

for all $t\in\mathbb{R}$ and all $u\in[0,1]$. By integrating the equation
(B.0.6) with respect to $s$ it follows therefore that

$y(t,u)=\alpha(t)+\alpha(u)-\alpha(t+u)$

for all $t,u\in[0,1]$ with $t+u\leq 1$. Let $\mu^{1}:\mathbb{R}\to\mathbb{T}$
be a continuous function such that $\mu^{1}(x)=e^{i\alpha(x)}$ when
$x\in[0,1]$. Then

$\mu^{1}(t)\mu^{1}(u)\overline{\mu^{1}(t+u)}=\lambda^{1}(t,u)$

when $t,u\in[0,1]$ and $t+u\leq 1$. Set

$\lambda^{2}(s,t):=\overline{\mu^{1}(s)}\overline{\mu^{1}(t)}\mu^{1}(s+t)\lambda^{1}(s,t)$

for all $s,t\in\mathbb{R}$. Then $\lambda^{2}$ differ from $\lambda^{1}$ by a
coboundary and it suffices therefore now to show that $\lambda^{2}$ is a
coboundary. For this observe that $\lambda^{2}(s,t)=1$ for all
$s,t\in[0,\frac{1}{2}]$. Set

$\lambda^{3}(s,t):=\lambda^{2}\left(\frac{s}{2},\frac{t}{2}\right).$

Then $\lambda^{3}$ is a normalized $2$-cocycle and it suffices to show that
$\lambda^{3}$ is a coboundary. Note that $\lambda^{3}(s,t)=1$ for
$s,t\in[0,1]$. By repeating the part of the argument above which constructed
$\lambda^{1}$ from $\lambda$ we get a continuous function
$\mu^{2}:\mathbb{R}\to\mathbb{T}$ such that $\mu^{2}(0)=1$ and such that the
$2$-cocycle $\lambda^{4}$ given by

$\lambda^{4}(s,t):=\overline{\mu^{2}(s+t)}\mu^{2}(s)\mu^{2}(t)\lambda^{3}(s,t)\
$

is $1$-periodic in the first variable and

$\lambda^{4}(n,t)=1,\ \ \forall t\in[0,1],\ \forall n\in\mathbb{Z}.$

But this time, because $\lambda^{3}(1,t)=1$ for all $t\in[0,1]$ we get in
addition that $\mu^{2}(t)=1$ for all $t\in[0,2]$, and hence

$\lambda^{4}(s,t)=\lambda^{3}(s,t)=1$

for all $s,t\in[0,1]$. By periodicity in the first variable this implies that
$\lambda^{4}(s,t)=1$ for all $s\in\mathbb{R}$ when $t\in[0,1]$. The cocycle
equation (i) implies then, by induction, that $\lambda^{4}(s,t)=1$ for all
$s\in\mathbb{R}$ and all $t\geq 0$. Let $t\geq 0$. Inserting $-t$ for $u$ in
(i) yields now that

$\lambda^{4}(s+t,-t)=\lambda^{4}(t,-t)$

for all $s\in\mathbb{R}$. Hence $\lambda^{4}(x,-t)$ does not depend on $x$ and
we conclude that $\lambda^{4}(x,-t)=\lambda^{4}(0,-t)=1$ for all
$x\in\mathbb{R}$. Thus $\lambda^{4}$ is constant $1$ and $\lambda^{3}$ is
therefore a coboundary.

∎

_Proof of Theorem B.0.1:_

Let $e_{ij},\ i,j=1,2,3,\cdots$, be a generating set of matrix units in
$\mathbb{K}$. That is,

  * •

$e_{jj},j=1,2,3,\cdots$, are one-dimensional projections,

  * •

$\lim_{N\to\infty}\sum_{j=1}^{N}e_{jj}=1$ in the strong operator topology,

  * •

$e_{ij}e_{kl}=\begin{cases}e_{il},\ &\ k=j\\\ 0,\ &\ k\neq j,\end{cases}$ for
all $i,j,k,l$, and

  * •

$e_{ij}^{*}=e_{ji}$ for all $i,j$.

Then $\mathbb{K}$ is the norm-closure of the linear span of
$\\{e_{ij}\\}_{i,j\in\mathbb{N}}$.

######  Observation B.0.4.

There is a norm continuous path $V(t),t\in\mathbb{R}$, of unitaries in
$\mathbb{C}1+\mathbb{K}$ such that $V(t)e_{11}V(t)^{*}=\sigma_{t}(e_{11})$ for
all $t\in\mathbb{R}$.

_Proof of Observation B.0.4._ Choose $\delta>0$ such that
$\left\|\sigma_{t}(e_{11})-e_{11}\right\|<1$ when $|t|\leq\delta$. Set

$x_{t}=e_{11}\sigma_{t}(e_{11})+(1-e_{11})(1-\sigma_{t}(e_{11})).$

Then

$\left\|1-x_{t}\right\|=\left\|(2e_{11}-1)(e_{11}-\sigma_{t}(e_{11})\right\|\leq\left\|e_{11}-\sigma_{t}(e_{11}\right\|<1$

and hence $x_{t}$ is invertible when $|t|\leq\delta$. Then

$V(t):=(x_{t}^{*}x_{t})^{-1/2}x_{t}^{*}$

is a unitary when $|t|\leq\delta$. Note that $[-\delta,\delta]\ni t\mapsto
V(t)$ is norm-continuous and that $V(0)=1$. Since
$e_{11}x_{t}=e_{11}\sigma_{t}(e_{11})=x_{t}\sigma_{t}(e_{11})$ it follows that
$x_{t}^{*}e_{11}=\sigma_{t}(e_{11})x_{t}^{*}$, and hence $x_{t}^{*}x_{t}$
commutes with $\sigma_{t}(e_{11})$. So does $(x_{t}^{*}x_{t})^{-1/2}$ and thus
$V(t)e_{11}=\sigma_{t}(e_{11})V(t)$ for $t\in[-\delta,\delta]$. For
$t\in[\delta,2\delta]$, set

$V(t)=\sigma_{t-\delta}(V(\delta))V(t-\delta).$

Then $[0,2\delta]\ni t\mapsto V(t)$ is norm-continuous, $V(t)$ is a unitary in
$\mathbb{C}1+\mathbb{K}$ for all $t\in[0,2\delta]$ and
$V(t)e_{11}V(t)^{*}=\sigma_{t}(e_{11})$ for all $t\in[0,2\delta]$. This
construction can be continued. If $[0,n\delta]\ni t\mapsto V(t)$ is norm-
continuous, $V(t)$ is a unitary in $\mathbb{C}1+\mathbb{K}$ for all
$t\in[0,n\delta]$ and $V(t)e_{11}V(t)^{*}=\sigma_{t}(e_{11})$ for all
$t\in[0,n\delta]$, set

$V(t)=\sigma_{t-n\delta}(V(n\delta))V(t-n\delta)$

for $t\in[n\delta,(n+1)\delta]$. In this way we get a norm-continuous path
$V(t),t\in[0,\infty)$, of unitaries in $\mathbb{C}1+\mathbb{K}$ such that
$V(0)=1$ and $V(t)e_{11}V(t)^{*}=\sigma_{t}(e_{11})$ for all $t\in[0,\infty)$.
To define $V(t)$ for $t\leq 0$, proceed in the same way starting with $V(t),\
t\in[-\delta,0]$, or set $V(t)=\tilde{\sigma}_{t}(V(-t)^{*})$, where
$\tilde{\sigma}_{t}\in\operatorname{Aut}(\mathbb{C}1+\mathbb{K})$ is the
automorphism extending $\tilde{\sigma}_{t}$. ∎

Let $\psi\in H$. For $n\in\mathbb{N}$, set

$W_{n}(t):=\sum_{j=1}^{n}\sigma_{t}(e_{j1})V(t)e_{1j}.$

Since

$\left\|\sum_{j=n}^{m}\sigma_{t}(e_{j1})V(t)e_{1j}\psi\right\|^{2}=\sum_{j=n}^{m}\left\|\sigma_{t}(e_{j1})V(t)e_{1j}\psi\right\|^{2}\leq\sum_{j=n}^{m}\left\|e_{jj}\psi\right\|^{2}$

when $n\leq m$, we see that $\left\|W_{n}(t)\right\|\leq 1$, and that the sum

$W(t):=\sum_{j=1}^{\infty}\sigma_{t}(e_{j1})V(t)e_{1j}=\lim_{n\to\infty}W_{n}(t)$

converges in the strong operator topology, uniformly in $t$, to an operator
$W(t)$ of norm $\leq 1$. In fact,

$\displaystyle
W(t)W(t)^{*}=\lim_{n\to\infty}\lim_{m\to\infty}W_{n}(t)W_{m}(t)^{*}$
$\displaystyle=\lim_{n\to\infty}\lim_{m\to\infty}\sum_{j=1}^{n}\sum_{k=1}^{m}\sigma_{t}(e_{j1})V(t)e_{1j}e_{k1}V(t)^{*}\sigma_{t}(e_{1k})$
$\displaystyle=\lim_{n\to\infty}\lim_{m\to\infty}\sum_{j=1}^{\min\\{n,m\\}}\
\sigma_{t}(e_{j1})V(t)e_{11}V(t)^{*}\sigma_{t}(e_{1j})$
$\displaystyle=\lim_{n\to\infty}\lim_{m\to\infty}\sum_{j=1}^{\min\\{n,m\\}}\
\sigma_{t}(e_{j1})\sigma_{t}(e_{11})\sigma_{t}(e_{1j})$
$\displaystyle=\lim_{n\to\infty}\lim_{m\to\infty}\sum_{j=1}^{\min\\{n,m\\}}\sigma_{t}\left(e_{jj}\right)=1,$

where the last equality follows from the fact that
$\sum_{j=1}^{\infty}\sigma_{t}(e_{jj})=1$ with convergence in the strong
operator topology. Similarly, $W(t)^{*}W(t)=1$, showing that $W(t)$ is a
unitary. Since the sum defining $W(t)$ converges uniformly in $t$ with respect
to the strong operator topology it follows from the continuity of $V$ that
$\mathbb{R}\ni t\mapsto W(t)$ is continuous for the strong operator topology.
For $k,l\in\mathbb{N}$,

$\displaystyle
W(t)e_{kl}W(t)^{*}=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\sigma_{t}(e_{j1})V(t)e_{1j}e_{kl}e_{i1}V(t)^{*}\sigma_{t}(e_{1i})$
$\displaystyle=\sigma_{t}(e_{k1})V(t)e_{11}V(t)^{*}e_{1l}=\sigma_{t}(e_{kl}),$

showing that

$W(t)aW(t)^{*}=\sigma_{t}(a)\ \forall a\in\mathbb{K},\ \forall
t\in\mathbb{R}.$ (B.0.7)

In particular, since $\sigma_{t+s}=\sigma_{t}\circ\sigma_{s}$, it follows that

$W(t+s)aW(t+s)^{*}=W(t)W(s)aW(s)^{*}W(t)^{*}$

for all $s,t,a$. Thus $W(s)W(t)W(t+s)^{*}$ is a unitary which commutes with
all elements of $\mathbb{K}$, and it must therefore be a scalar multiple of
$1$. It follows that there is a continuous function
$\lambda:\mathbb{R}\times\mathbb{R}\to\mathbb{T}$ such that

$W(s)W(t)W(t+s)^{*}=\lambda(s,t)1\ \ \forall t,s\in\mathbb{R}.$ (B.0.8)

Note that

$\displaystyle\lambda(s,t)\lambda(s+t,u)1=W(s)W(t)W(u)W(s+t+u)^{*}$
$\displaystyle=W(s)W(t)W(u)W(t+u)^{*}W(t+u)W(s+t+u)^{*}$
$\displaystyle=\lambda(t,u)W(s)W(t+u)W(s+t+u)^{*}=\lambda(t,u)\lambda(s,t+u)1.$

Furthermore, $\lambda(0,t)=\lambda(t,0)=1$ for all $t\in\mathbb{R}$. Thus
$\lambda$ is a normalized $2$-cocycle and by Theorem B.0.3 there is a
continuous function $\mu:\mathbb{R}\to\mathbb{T}$ such that $\mu(0)=1$ and
$\lambda(s,t)=\overline{\mu(s)}\overline{\mu(t)}\mu(s+t)$ for all
$s,t\in\mathbb{R}$. Set $U_{t}=\mu(t)W(t)$. ∎

######  Notes and remarks B.0.5.

The proof of Theorem B.0.3 is taken from [BuR]. According to that source the
result itself, that $H^{2}(\mathbb{R},\mathbb{T})=0$, is due Iwasawa.

##  Appendix C Multipliers

Let $A$ be a $C^{*}$-algebra. We denote by $M(A)$ the set of maps $m:A\to A$
(_a priori not even linear_) for which there is another map $m^{\ast}:A\to A$
such that

$m(a)^{\ast}b=a^{\ast}m^{\ast}(b),\ a,b\in A.$ (C.0.1)

######  Lemma C.0.1.

Every element $m$ of $M(A)$ is a bounded linear operator on $A$ and
$m^{\ast}\in M(A)$ with $(m^{\ast})^{\ast}=m$.

###### Proof.

We first observe that the following implication holds for any element $a\in A$
:

$ax=0\ \forall x\in A\Rightarrow a=0.$ (C.0.2)

Indeed, if $ax=0$ for all $x\in A$, then in particular
$\|a\|^{2}=\|aa^{\ast}\|=0$ which implies that $a=0$.

To show that $m$ is linear, let $a,b\in A$ and $\lambda\in\mathbb{C}$. Then

$\displaystyle(m(\lambda a+b)-\lambda m(a)-m(b))^{\ast}x=(m(\lambda
a+b)^{\ast}-\overline{\lambda}m(a)^{\ast}-m(b)^{\ast})x=$
$\displaystyle(\lambda
a+b)^{\ast}m^{\ast}(x)-\overline{\lambda}a^{\ast}m^{\ast}(x)-b^{\ast}m^{\ast}(x)=0$

for all $x\in A$, so by (C.0.2) we have that $m(\lambda a+b)=\lambda
m(a)+m(b)$; i.e. $m$ is linear.

That $m$ is bounded follows from the principle of uniform boundedness in the
following way. For each $a\in A$ with $\|a\|\leq 1$, define $S_{a}:A\to A$ by
$S_{a}(b)=m(a)^{\ast}b$. Each $S_{a}$ is a bounded linear operator
($\|S_{a}\|\leq\|m(a)\|$) and for every fixed $b\in A$ we have that
$\|S_{a}(b)\|=\|m(a)^{\ast}b\|=\|a^{\ast}m^{\ast}(b)\|\leq\|m^{\ast}(b)\|$.
Therefore the principle of uniform boundedness implies that there is an
$M<\infty$ such that $\|S_{a}\|\leq M$ for all $a$ with $\|a\|\leq 1$. Hence
$\|m(a)^{\ast}b\|=\|S_{a}(b)\|\leq M$ for all $a,b\in A$ with $\|a\|\leq
1,\|b\|\leq 1$. In particular, whenever $\|a\|\leq 1$ and $m(a)\neq 0$ we have
that

$\|m(a)\|=\|m(a)^{\ast}\frac{m(a)}{\|m(a)\|}\|\leq M.$

It follows that $\|m\|\leq M$. Finally, by applying the involution $*$ to the
identity $m(a)^{\ast}b=a^{\ast}m^{\ast}(b)$ we get immediately that
$m^{\ast}\in M(A)$ and that in fact $(m^{\ast})^{\ast}=m$. ∎

######  Proposition C.0.2.

$M(A)$ is a $C^{\ast}$-algebra with the involution $m\mapsto m^{\ast}$ and the
norm $\|m\|=\sup\ \\{\|m(a)\|:\|a\|\leq 1\\}$. The product in the algebra is
the composition of operators on $A$ and the identity operator is in $M(A)$
(and is therefore a unit of $M(A)$).

###### Proof.

We leave it to the reader to check that $M(A)$ is a subalgebra of the algebra
of bounded linear operators on $A$. That is, you should check that $\lambda
m+n,mn\in M(A)$ when $m,n\in M(A)$ and $\lambda\in\mathbb{C}$. It is also
trivial that the identity operator is in $M(A)$.

Let us next check the $C^{\ast}$-identity. Since $\|mn\|\leq\|m\|\|n\|$
because we are dealing with the operator norm, we see immeditaly that
$\|m^{\ast}m\|\leq\|m^{\ast}\|\|m\|$. Let $a,b\in A,\|a\|\leq 1,\|b\|\leq 1$.
Then

$\|m(a)^{\ast}b\|=\|a^{\ast}m^{\ast}(b)\|\leq\|m^{\ast}(b)\|\leq\|m^{\ast}\|.$

When we set $b=\frac{m(a)}{\|m(a)\|}$ in this inequality we get that
$\|m(a)\|\leq\|m^{\ast}\|$ (when $m(a)\neq 0$). The conclusion is that
$\|m\|\leq\|m^{\ast}\|,\ m\in M(A)$. When we substitute $m^{\ast}$ into this
inequality we get the converse version, $\|m^{\ast}\|\leq\|m\|$, implying that
$\|m^{\ast}\|=\|m\|$. Now, for any $a\in A$ with $\|a\|\leq 1$ we have the
estimate

$\|m(a)\|^{2}=\|m(a)^{\ast}m(a)\|=\|a^{\ast}m^{\ast}m(a)\|\leq\|m^{\ast}m\|,$

implying that

$\|m\|^{2}\leq\|m^{\ast}m\|.$

All in all we have the following inequalities

$\|m\|^{2}\leq\|m^{\ast}m\|\leq\|m^{\ast}\|\|m\|=\|m\|^{2}$

from which the $C^{\ast}$-identity
$\left\|m\right\|^{2}=\left\|m^{*}m\right\|$ follows.

To show that $M(A)$ is complete with respect to the operator norm, let
$\\{m_{k}\\}$ be a Cauchy sequence in $M(A)$. For each $a\in A,\
k,l\in\mathbb{N}$, we have that

$\|m_{k}a-m_{l}a\|\leq\|m_{k}-m_{l}\|\|a\|\ .$

It follows that $\\{m_{k}a\\}$ is a Cauchy sequence in $A$ for all $a\in A$.
Since $\|m_{k}^{*}-m_{l}^{*}\|=\|(m_{k}-m_{l})^{*}\|=\|m_{k}-m_{l}\|$ for all
$k,l$, we see that also $\\{m_{k}^{*}\\}$ is a Cauchy sequence in $M(A)$.
Consequently, $\\{m_{k}^{*}a\\}$ is a Cauchy sequence in $A$ for all $a\in A$.
We can therefore define maps $m,n:A\to A$ by $ma=\lim_{k\to\infty}m_{k}a,\
na=\lim_{k\to\infty}m_{k}^{*}a,\ a\in A$. Note that

$(ma)^{*}b\ =\ \lim_{k\to\infty}(m_{k}a)^{*}b\ =\
\lim_{k\to\infty}a^{*}m_{k}^{*}b\ =\ a^{*}nb$

for all $a,b\in A$, proving that $m\in M(A)$ (and that $m^{*}=n$). To show
that $\lim_{k\to\infty}m_{k}=m$ in $M(A)$, let $\epsilon>0$. There is a
$N\in\mathbb{N}$ such that

$\|m_{k}a-m_{l}a\|\leq\|m_{k}-m_{l}\|\leq\epsilon$ (C.0.3)

for all $k,l\geq N$ and all $a\in A$ with $\|a\|\leq 1$. If we let $k$ tend to
infinity we get from (C.0.3) that

$\|m_{k}a-ma\|\leq\epsilon$

for all $k\geq N$ and all $a\in A$ with $\|a\|\leq 1$. Hence
$\|m_{k}-m\|\leq\epsilon$ for all $k\geq N$. ∎

The $C^{*}$-algebra $M(A)$ is called the multiplier algebra  of $A$. Every
element $b\in A$ defines a multiplier $m_{b}\in M(A)$ such that

$m_{b}(a):=ba.$

######  Lemma C.0.3.

The map $b\mapsto m_{b}$ is an isometric $*$-homomorphism mapping $A$ onto a
closed two-sided ideal in $M(A)$.

###### Proof.

Left to the reader. ∎

As is customary we identify $A$ with its image in $M(A)$ under this embedding.

Besides the norm topology the multiplier algebra carries the _strict topology_
which is the topology defined by the semi-norms $\|\ \cdot\ \|_{a},a\in A$,
where

$\left\|m\right\|_{a}:=\left\|ma\right\|+\left\|m^{*}a\right\|.$

######  Lemma C.0.4.

$A$ is dense in $M(A)$ with respect to the strict topology. In fact, every
element $m\in M(A)$ is the limit in the strict topology of a net $\\{a_{i}\\}$
in $A$ with $\|a_{i}\|\leq\|m\|$ for all $i$.

###### Proof.

Let $m\in M(A)$ and let $\\{u_{i}\\}$ be an approximate unit in $A$. Then
$mu_{i}\in A$ and for any $a\in A$ we find that

$\lim_{i\to\infty}\left\|ma-
mu_{i}a\right\|\leq\lim_{i\to\infty}\left\|m\right\|\left\|a-u_{i}a\right\|=0$

and

$\lim_{i\to\infty}\left\|am-amu_{i}\right\|=0$

since $am\in A$. Hence $\lim_{i\to\infty}mu_{i}=m$ in the strict topology. Set
$a_{i}:=mu_{i}$. ∎

######  Lemma C.0.5.

Let $A$ be a $C^{*}$-algebra, $\mathbb{H}$ a Hilbert space and $\pi:A\to
B(\mathbb{H})$ a $*$-representation of $A$ on $\mathbb{H}$. Assume that $\pi$
is non-degenerate; i.e. $\pi(A)\mathbb{H}$ spans a dense subspace of
$\mathbb{H}$. There is a $*$-homomorphism $\overline{\pi}:M(A)\to
B(\mathbb{H})$ such that

  * •

$\overline{\pi}$ extends $\pi$, and

  * •

when $\\{m_{i}\\}_{i\in I}$ is a norm-bounded net in $M(A)$ converging to
$m\in M(A)$ in the strict topology, then
$\lim_{i\to\infty}\overline{\pi}(m_{i})=\overline{\pi}(m)$ in the strong
operator topology.

  * •

When $\pi$ is also injective, $\overline{\pi}$ is a $*$-isomorphism of $M(A)$
onto

$\left\\{m\in B(\mathbb{H}):\ m\pi(A)\subseteq\pi(A),\
\pi(A)m\subseteq\pi(A)\right\\}.$

###### Proof.

Let $m\in M(A)$. It follows from Lemma C.0.4 that there is a net $\\{a_{i}\\}$
in $A$ such that $\lim_{i\to\infty}a_{i}=m$ in the strict topology and
$\|a_{i}\|\leq\|m\|$ for all $i$. Since $\pi$ is non-degenerate it follows
that $\\{\pi(a_{i})\\}$ converges in the strong operator topology and we set
$\overline{\pi}(m):=\lim_{i\to\infty}\pi(a_{i})$. It is straightforward to
check that this gives a well-defined $*$-homomorphism $\overline{\pi}:M(A)\to
B(\mathbb{H})$ with the properties specified in first two items. Assume then
that $\pi$ is injective. If $m\in M(A)$ and $\overline{\pi}(m)=0$ we have for
each $a\in A$ that $\pi(ma)=\overline{\pi}(m)\pi(a)=0$ and hence that $ma=0$.
It follows that $m=0$, proving that $\overline{\pi}$ is injective. Let
$m^{\prime}\in B(\mathbb{H})$ and assume that
$m^{\prime}\pi(A)\subseteq\pi(A),\ \pi(A)m^{\prime}\subseteq\pi(A)$. We can
then define $m:A\to A$ such that

$m(a):=\pi^{-1}\left(m^{\prime}\pi(a)\right)$

and $m^{\\#}:A\to A$ such that

$m^{\\#}(a)=\pi^{-1}\left({m^{\prime}}^{*}\pi(a)\right).$

Since $m(a)^{*}b=a^{*}m^{\\#}(b)$ we see that $m\in M(A)$ and $m^{*}=m^{\\#}$.
Since $\overline{\pi}(m)\pi(a)=\pi(ma)=m^{\prime}\pi(a)$ for all $a\in A$, it
follows that $\overline{\pi}(m)=m^{\prime}$. ∎

######  Remark C.0.6.

Note that any automorphism $\alpha\in\operatorname{Aut}A$ of $A$ extends to an
automorphism of $M(A)$ defined such that

$\alpha(m)a:=\alpha(m\alpha^{-1}(a))\ \ \forall a\in A.$

We denote the extension by $\alpha$ again. By Lemma C.0.4 and Lemma C.0.5 the
extension is the only extension of $\alpha$ to $M(A)$ which is continuous for
the strict topology on norm-bounded sets. In particular, when
$\sigma=(\sigma_{t})_{t\in\mathbb{R}}$ is a flow on $A$ we have an associated
one-parameter group $(\sigma_{t})_{t\in\mathbb{R}}$ of automorphisms of
$M(A)$, but this extension is not in general a flow since the map
$t\mapsto\sigma_{t}(m)$ need not be continuous in norm when $m\in M(A)$. It is
however continuous with respect to the strict topology.

##  Appendix D Traces and K-theory

###  D.1 Pairing traces with $K_{0}$

In the following we fix a trace $\tau:A^{+}\to[0,\infty]$ on $A$, as defined
in Definition 2.3.3. Set

$\mathcal{M}^{+}_{\tau}:=\\{a\in A^{+}:\ \tau(a)<\infty\\}$

and

$\mathcal{M}_{\tau}:=\operatorname{Span}\\{a\in A^{+}:\ \tau(a)<\infty\\}.$

######  Lemma D.1.1.

$\mathcal{M}_{\tau},\ \mathcal{M}^{+}_{\tau}$ and $\tau$ have the following
properties.

  * •

$\mathcal{M}_{\tau}$ is a dense two-sided $*$-invariant ideal in $A$.

  * •

$\mathcal{M}_{\tau}\cap A^{+}=\mathcal{M}_{\tau}^{+}$.

  * •

$\tau:\mathcal{M}^{+}_{\tau}\to[0,\infty)$ has a unique extension to a linear
map $\tau|_{\mathcal{M}_{\tau}}:\mathcal{M}_{\tau}\to\mathbb{C}$.

###### Proof.

Set $\mathcal{N}_{\tau}=\left\\{a\in A:\ \tau(a^{*}a)<\infty\right\\}$. The
arguments from the proof of Lemma 1.0.1 show that $\mathcal{N}_{\tau}$ is a
left ideal and
$\mathcal{M}_{\tau}=\operatorname{Span}\mathcal{N}_{\tau}^{*}\mathcal{N}_{\tau}$.
The trace property of $\tau$ implies that
$\mathcal{N}_{\tau}^{*}=\mathcal{N}_{\tau}$ and hence $\mathcal{N}_{\tau}$ is
also a right ideal. It follows therefore that $\mathcal{M}_{\tau}$ is a two-
sided ideal in $A$. The second item follows as in the proof of Lemma 1.0.1 and
implies that $\mathcal{M}_{\tau}$ is dense in $A$. The third is a
straightforward consequence of the additivity of $\tau$, as in Lemma 1.2.4. ∎

Let $A^{\dagger}$ be the $C^{*}$-algebra obtained from $A$ be adjoining a unit
to $A$. Thus as a vector space $A^{\dagger}$ is just $A\oplus\mathbb{C}$, and
the product and involution are given by

$(a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu)$

and

$(a,\lambda)^{*}=(a^{*},\overline{\lambda}).$

For every subset $X\subseteq A$, set

$X^{\dagger}:=\left\\{(a,\lambda)\in A^{\dagger}:\ a\in X,\
\lambda\in\mathbb{C}\right\\}.$

In particular,

$\mathcal{M}_{\tau}^{\dagger}:=\left\\{(a,\lambda)\in A^{\dagger}:\
a\in\mathcal{M}_{\tau},\ \lambda\in\mathbb{C}\right\\}.$

Define $\tau^{\dagger}:\mathcal{M}_{\tau}^{\dagger}\to\mathbb{C}$ such that

$\tau^{\dagger}(a,\lambda)=\tau|_{\mathcal{M}_{\tau}}(a).$

Since $\tau^{\dagger}$ is linear on $\mathcal{M}_{\tau}^{\dagger}$ we can
consider the tensor product map

$\tau^{\dagger}\otimes\operatorname{Tr}_{n}:\
M_{n}(\mathcal{M}_{\tau}^{\dagger})\to\mathbb{C},$

where $\operatorname{Tr}_{n}$ denotes the standard trace on
$M_{n}(\mathbb{C})$; the sum of the diagonal entries. Let
$P_{n}(\mathcal{M}_{\tau}^{\dagger})$ be the set of projections in the
$*$-algebra $M_{n}(\mathcal{M}_{\tau}^{\dagger})$ and set

$P_{\infty}(\mathcal{M}_{\tau}^{\dagger}):=\bigcup_{n}P_{n}(\mathcal{M}_{\tau}^{\dagger}).$

We aim to establish the following

######  Theorem D.1.2.

$K_{0}(A^{\dagger})=\left\\{[e]-[f]:\ e,f\in
P_{\infty}(\mathcal{M}_{\tau}^{\dagger})\right\\}$

and there is a homomorphism $\tau^{\dagger}_{*}:\
K_{0}(A^{\dagger})\to\mathbb{R}$ such that

$\tau^{\dagger}_{*}([e]-[f])=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e)-\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f)$

when $e,f\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$.

####  D.1.1 Proof of Theorem D.1.2

The proof uses the following series of lemmas. Recall that a _strictly
positive_ element of a $C^{*}$-algebra $A$ is a positive element $a\in A^{+}$
such $\omega(a)>0$ for all non-zero positive functionals $\omega$ on $A$. A
$C^{*}$-algebra is _$\sigma$ -unital_ when it contains a strictly positive
element.

######  Lemma D.1.3.

A separable $C^{*}$-algebra is $\sigma$-unital.

###### Proof.

When $A$ is a separable $C^{*}$-algebra there is a dense sequence
$\\{a_{n}\\}_{n=1}^{\infty}$ in $\\{a\in D^{+}:\ 0\leq a\leq 1\\}$. Then

$a:=\sum_{n=1}^{\infty}2^{-n}a_{n}$

is strictly positive. ∎

######  Lemma D.1.4.

Let $D$ be a $\sigma$-unital $C^{*}$-algebra. There is a sequence
$\\{d_{n}\\}_{n=1}^{\infty}$ in $D$ such that

  * •

$0\leq d_{n}\leq 1,\ \ \forall n$,

  * •

$d_{n}d_{n+1}=d_{n},\ \ \forall n$, and

  * •

$\lim_{n\to\infty}d_{n}a=a,\ \ \forall a\in D$.

###### Proof.

Let $a_{0}$ be a strictly positive element of $A$. Let $f_{n}$ be the
continuous function $f_{n}:[0,\infty)\to[0,1]$ such that

$f_{n}(t)=\begin{cases}0,&\ t\in[0,\frac{1}{n+1}]\\\ \text{linear},&\
t\in\left[\frac{1}{n+1},\frac{1}{n}\right],\\\ 1,&\
t\geq\frac{1}{n}.\end{cases}$

Set $d_{n}:=f_{n}(a_{0})$. The first item holds since $0\leq f_{n}\leq 1$ and
the second because $f_{n}f_{n+1}=f_{n}$. To establish the third, assume for a
contradiction that there is an element $x\in D$ for which $(1-f_{n}(a_{0}))x$
does not converge to $0$. Then

$\sup_{\omega}\omega((1-f_{n}(a_{0}))xx^{*}(1-f_{n}(a_{0}))$

does not converge to $0$ when we take the supremum over all states $\omega$ of
$A$. It follows that there is an $\epsilon>0$ and a sequence
$n_{1}<n_{2}<n_{3}<\cdots$ in $\mathbb{N}$ such that for each $k$ there is a
state $\omega_{k}$ with

$\omega_{k}((1-f_{n_{k}}(a_{0}))xx^{*}(1-f_{n_{k}}(a_{0})))\geq\epsilon.$

Since the unit ball of $A^{*}$ is weak* compact there is a weak* condensation
point $\mu$ of the set of functionals defined by

$A\ni a\mapsto\omega_{k}((1-f_{n_{k}}(a_{0}))a(1-f_{n_{k}}(a_{0}))),\ \
k\in\mathbb{N}.$

Then $\mu(xx^{*})\geq\epsilon$, and hence $\mu$ is a non-zero positive
functional on $A$. However, $\lim_{n\to\infty}(1-f_{n}(t))t=0$ uniformly for
$t$ in the spectrum of $a_{0}$, and hence
$\lim_{n\to\infty}(1-f_{n}(a_{0}))a_{0}=0$. It follows that $\mu(a_{0})=0$,
contradicting the strict positivity of $a_{0}$. ∎

Consider now a separable $C^{*}$-subalgebra $D$ of $A$, and let

${\bf d}:=\\{d_{n}\\}_{n=1}^{\infty}$

be a sequence in $D$ with the properties specified in Lemma D.1.4. Set

$D_{d_{n}}:=\left\\{a\in D:\ ad_{n}=d_{n}a=a\right\\}.$

Then $D_{d_{n}}$ is a $C^{*}$-subalgebra of $D$ and

$D_{d_{n}}\subseteq d_{n}Dd_{n}\subseteq D_{d_{n+1}}\subseteq
d_{n+1}Dd_{n+1}.$

In particular,

$\bigcup_{n=1}^{\infty}d_{n}Dd_{n}=\bigcup_{n=1}^{\infty}D_{d_{n}}.$

Set

$D({\bf d}):=\bigcup_{n=1}^{\infty}D_{d_{n}}.$

Then $D({\bf d})$ is a $*$-subalgebra of $D$ and it is dense in $D$ since
$\lim_{n\to\infty}d_{n}ad_{n}=a$ for all $a\in D$.

######  Lemma D.1.5.

$D({\bf d})\subseteq\mathcal{M}_{\tau}$.

###### Proof.

Let $n\in\mathbb{N}$ and consider an element $a\in D^{+}$. Set
$d:=d_{n}ad_{n}$. It suffices to show that $d\in\mathcal{M}_{\tau}$. Since
$\tau$ is densely defined there is an $x\in A^{+}\cap\mathcal{M}_{\tau}$ such
that $\left\|x-d_{n+1}\right\|\leq\frac{1}{2}$. Using
$d_{n+1}\sqrt{d}=\sqrt{d}$ we find that

$\frac{1}{2}d\leq\sqrt{d}(1+(x-d_{n+1}))\sqrt{d}=\sqrt{d}(d_{n+1}+(x-d_{n+1}))\sqrt{d}=\sqrt{d}x\sqrt{d}.$

Note that $\sqrt{d}x\sqrt{d}\in\mathcal{M}_{\tau}$ since $\mathcal{M}_{\tau}$
is a two-sided ideal by Lemma D.1.1. It follows therefore from the estimate
above that $\tau(d)\leq 2\tau(\sqrt{d}x\sqrt{d})<\infty$. Hence
$d\in\mathcal{M}_{\tau}$. ∎

######  Lemma D.1.6.

$\tau^{\dagger}(xy)=\tau^{\dagger}(yx)$ for all
$x,y\in\mathcal{M}_{\tau}^{\dagger}$.

###### Proof.

Write $x=(a,\lambda),\ y=(b,\mu)$ where $a,b\in\mathcal{M}_{\tau},\
\lambda,\mu\in\mathbb{C}$. Then
$\tau^{\dagger}(xy)=\tau|_{\mathcal{M}_{\tau}}(ab+\lambda b+\mu a)$ while
$\tau^{\dagger}(yx)=\tau|_{\mathcal{M}_{\tau}}(ba+\lambda b+\mu a)$. It
suffices therefore to show that
$\tau|_{\mathcal{M}_{\tau}}(ab)=\tau|_{\mathcal{M}_{\tau}}(ba)$, which follows
from the polarization identities

$ab=\frac{1}{4}\sum_{k=1}^{4}i^{k}(b+i^{k}a^{*})^{*}(b+i^{k}a^{*})$

and

$ba=\frac{1}{4}\sum_{k=1}^{4}i^{k}(b+i^{k}a^{*})(b+i^{k}a^{*})^{*}.$

∎

######  Corollary D.1.7.

$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(xy)=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(yx)$
for all $x,y\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$.

######  Lemma D.1.8.

Let $I$ be a right ideal (not necessarily closed) in the $C^{*}$-algebra $A$.
If $x\in M_{n}(I^{\dagger})$ is invertible in $M_{n}(A^{\dagger})$ then
$x^{-1}\in M_{n}(I^{\dagger})$.

###### Proof.

Write $x=(a,k)$ and $x^{-1}=(b,m)$ where $a\in M_{n}(I)$, $b\in M_{n}(A)$ and
$k,m\in M_{n}(\mathbb{C})$. Since $xx^{-1}=1$ we get the equations
$ab+kb+am=0$ and $km=1$, implying that $b=k^{-1}(-am-ab)\in M_{n}(I)$. ∎

######  Lemma D.1.9.

Let $e,f$ be projections in $M_{n}(\mathcal{M}_{\tau}^{\dagger})$, and assume
that there is a partial isometry $v\in M_{n}(A^{\dagger})$ such that
$e=vv^{*}$ and $f=v^{*}v$. Then
$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e)=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f)$.

###### Proof.

Choose a separable $C^{*}$-algebra $D\subseteq A$ such that $e,f,v\in
M_{n}(D^{\dagger})$ and choose in $D$ a sequence $\\{d_{n}\\}_{n=1}^{\infty}$
with the properties specified in Lemma D.1.4. For each $\epsilon>0$ we can
then find $N\in\mathbb{N}$ and projections $e^{\prime},f^{\prime}\in
M_{n}(D_{d_{N}}^{\dagger})$ such that
$\left\|e-e^{\prime}\right\|\leq\epsilon$ and $\|f-f^{\prime}\|\leq\epsilon$,
cf. Lemma 2.2.7 and Lemma 2.2.8 in [Th5]. Set $u:=(2e^{\prime}-1)(2e-1)+1$.
Note that

$\displaystyle\left\|1-\frac{1}{2}u\right\|=\left\|(2e^{\prime}-1)(e^{\prime}-e)\right\|\leq\|e^{\prime}-e\|,$

implying that $u$ is invertible if $\epsilon<1$. Since
$ue=2e^{\prime}e=e^{\prime}u$, we see that $ueu^{-1}=e^{\prime}$. Note that
$u\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$. It follows therefore from Lemma
D.1.8 that $u^{-1}\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$. Since
$e^{\prime}\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$ by Lemma D.1.5 it follows
now from Corollary D.1.7 that
$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e^{\prime})=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e)$.
We observe that $u\in M_{n}(D^{\dagger})$, which implies that $u\in
M_{n}((\mathcal{M}_{\tau}\cap D)^{\dagger})$ and hence that $u^{-1}\in
M_{n}((\mathcal{M}_{\tau}\cap D)^{\dagger})$ by Lemma D.1.8. Similarly,
$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f^{\prime})=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f)$
and $sfs^{-1}=f^{\prime}$ for some invertible element $s\in
M_{n}((\mathcal{M}_{\tau}\cap D)^{\dagger})$ with $s^{-1}\in
M_{n}((\mathcal{M}_{\tau}\cap D)^{\dagger})$. We aim to show that
$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e^{\prime})=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f^{\prime})$.
Set $x:=f^{\prime}sv^{*}u^{-1}e^{\prime},\ y:=e^{\prime}uvs^{-1}f^{\prime}$.
Then $x,y\in M_{n}(D^{\dagger})$ and

$\displaystyle
xy=f^{\prime}sv^{*}u^{-1}e^{\prime}uvs^{-1}f^{\prime}=f^{\prime}sv^{*}evs^{-1}f^{\prime}$
$\displaystyle=f^{\prime}sv^{*}vv^{*}vs^{-1}f^{\prime}=f^{\prime}sfs^{-1}f^{\prime}=f^{\prime}.$

Similarly, $yx=e^{\prime}$. Since
$f^{\prime}={f^{\prime}}^{*}f^{\prime}=y^{*}x^{*}xy\leq\|x\|^{2}y^{*}y$ it
follows that $y^{*}y$ is positive and invertible in
$f^{\prime}M_{n}(D^{\dagger})f^{\prime}$. Taking the inverse in that algebra,
set

$w:=y(y^{*}y)^{-1/2}.$

Then $w^{*}w=(y^{*}y)^{-1/2}y^{*}y(y^{*}y)^{-1/2}=f^{\prime}$ and hence, in
particular, $ww^{*}$ is a projection. Note that

$\displaystyle
e^{\prime}=yxx^{*}y^{*}\leq\|x\|^{2}yy^{*}=\|x\|^{2}y(y^{*}y)^{-1/2}yy^{*}(y^{*}y)^{-1/2}y^{*}$
$\displaystyle\leq\|x\|^{2}\|y\|^{2}y(y^{*}y)^{-1/2}(y^{*}y)^{-1/2}y^{*}=\|x\|^{2}\|y\|^{2}ww^{*}.$

Then

$0\leq(1-ww^{*})e^{\prime}(1-ww^{*})\leq\|x\|^{2}\|y\|^{2}(1-ww^{*})ww^{*}(1-ww^{*})=0,$

implying that $e^{\prime}=ww^{*}e^{\prime}$. On the other hand,

$\displaystyle
e^{\prime}y=e^{\prime}uvs^{-1}f^{\prime}=ueu^{-1}uvs^{-1}f^{\prime}=uevs^{-1}f^{\prime}$
$\displaystyle=uvv^{*}vs^{-1}f^{\prime}=uvs^{-1}f^{\prime}=e^{\prime}uvs^{-1}f^{\prime}=y,$

implying that $e^{\prime}ww^{*}=ww^{*}$, and hence that $e^{\prime}=ww^{*}$.
Since $w\in M_{n}(D^{\dagger})$ and $D({\bf d})$ is dense in $D$ there is a
$k>N$ and an element $z\in M_{n}(D_{d_{k}}^{\dagger})$ such that
$\left\|z{z}^{*}-e^{\prime}\right\|<\frac{1}{2}$ and
$\left\|{z}^{*}z-f^{\prime}\right\|<\frac{1}{2}$. Since
$e^{\prime},f^{\prime},z\in M_{n}(D_{d_{k}}^{\dagger})$ it follows then from
Lemma 2.2.7 and Lemma 2.2.8 in [Th5] that there is a partial isometry
$w^{\prime}\in M_{n}(D_{d_{k}}^{\dagger})$ such that
$w^{\prime}{w^{\prime}}^{*}=e^{\prime}$ and
${w^{\prime}}^{*}w^{\prime}=f^{\prime}$. Since $w^{\prime}\in
M_{n}(\mathcal{M}_{\tau}^{\dagger})$ by Lemma D.1.5 it follows from Corollary
D.1.7 that
$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e^{\prime})=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f^{\prime})$.
∎

######  Lemma D.1.10.

Let $e\in M_{n}(A^{\dagger})$ be a projection. There is a projection
$e^{\prime}\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$ and a partial isometry
$v\in M_{n}(A^{\dagger})$ such that $vv^{*}=e$ and $v^{*}v=e^{\prime}$.

###### Proof.

Let $D\subseteq A$ be a separable $C^{*}$-subalgebra such that $e\in
M_{n}(D^{\dagger})$ and let $\\{d_{n}\\}_{n=1}^{\infty}$ be a sequence in $D$
with the properties specified in Lemma D.1.4. Since $D({\bf d})$ is a
$*$-algebra which is dense in $D$ and since $M_{n}(D_{d_{k}}^{\dagger})$ is a
$C^{*}$-algebra it follows from Lemma 2.2.7 and Lemma 2.2.8 in [Th5] that
there is a $k$, a projection $e^{\prime}\in M_{n}(D_{d_{k}}^{\dagger})$ and a
partial isometry $v\in M_{n}(A^{\dagger})$ such that $vv^{*}=e$ and
$v^{*}v=e^{\prime}$. This completes the proof because
$M_{n}(D_{d_{k}}^{\dagger})\subseteq M_{n}(\mathcal{M}_{\tau}^{\dagger})$ by
Lemma LABEL:18-11-22g.

∎

_Proof of Theorem D.1.2:_ The identity

$K_{0}(A^{\dagger})=\left\\{[e]-[f]:\ e,f\in
P_{\infty}(\mathcal{M}_{\tau}^{\dagger})\right\\}$

follows from the definition of $K_{0}(A^{\dagger})$ and Lemma D.1.10. To prove
that $\tau^{\dagger}_{*}$ is well-defined assume that
$e,f,e^{\prime},f^{\prime}$ are projections in
$M_{n}(\mathcal{M}_{\tau}^{\dagger})$ such that
$[e]-[f]=[e^{\prime}]-[f^{\prime}]$ in $K_{0}(A^{\dagger})$. There is then a
projection $r\in P_{\infty}(A^{\dagger})$ such that $e\oplus f^{\prime}\oplus
r$ and $e^{\prime}\oplus f\oplus r$ are Murray-von Neumann equivalent in
$M_{N}(A^{\dagger})$ for some $N\in\mathbb{N}$. By Lemma D.1.10 we may assume
that $r\in P_{\infty}(\mathcal{M}_{\tau}^{\dagger})$. It follows then from
Lemma D.1.9 that

$\tau^{\dagger}\otimes\operatorname{Tr}_{N}(e\oplus f^{\prime}\oplus
r)=\tau^{\dagger}\otimes\operatorname{Tr}_{N}(e^{\prime}\oplus f\oplus r)$

and hence by linearity that

$\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e)-\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f)=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e^{\prime})-\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f^{\prime}).$

∎

By definition $K_{0}(A)$ is a subgroup of $K_{0}(A^{\dagger})$ so the map
$\tau^{\dagger}_{*}$ of Theorem D.1.2 gives rise, by restriction, to a
homomorphism

$\tau^{\dagger}_{*}:K_{0}(A)\to\mathbb{R}$

such that

$\tau^{\dagger}_{*}([e]-[f])=\tau^{\dagger}\otimes\operatorname{Tr}_{n}(e)-\tau^{\dagger}\otimes\operatorname{Tr}_{n}(f)$

when $e,f\in M_{n}(\mathcal{M}_{\tau}^{\dagger})$ and $[e]-[f]\in K_{0}(A)$.
Let $\iota:K_{00}(A)\to K_{0}(A)$ be the canonical homomorphism. Set

$\tau_{*}:=\tau^{\dagger}_{*}\circ\iota:\ K_{00}(A)\to\mathbb{R}.$

To obtain more information about $\tau_{*}$ we need the following lemmas.

For each $k\in\mathbb{N}$ define $\tau_{k}:M_{k}(A)^{+}\to[0,\infty]$ such
that

$\tau_{k}(a)=\sum_{i=1}^{k}\tau(a_{kk})$

when $a=(a_{ij})_{i,j=1}^{k}\in M_{k}(A)^{+}$.

######  Lemma D.1.11.

$\tau_{k}$ is a trace on $M_{k}(A)$.

###### Proof.

The two first items of Definition 2.3.3 clearly hold. To check the third note
that

$\displaystyle\tau_{k}(a^{*}a)=\sum_{i=1}^{k}\tau(\sum_{j=1}^{k}a_{ji}^{*}a_{ji})=\sum_{i=1}^{k}\sum_{j=1}^{k}\tau(a_{ji}^{*}a_{ji})$
$\displaystyle=\sum_{i=1}^{k}\sum_{j=1}^{k}\tau(a_{ji}a_{ji}^{*})=\sum_{j=1}^{k}\tau(\sum_{i=1}^{k}a_{ji}a_{ji}^{*})=\tau_{k}(aa^{*}).$

To check the last item of Definition 2.3.3 let $a\in M_{k}(A)$. Since
$\mathcal{M}_{\tau}$ is dense in $A$ we can approximate $a$ with $b\in
M_{k}(\mathcal{M}_{\tau})$. Then $b^{*}b$ approximates $a^{*}a$ and

$\tau_{k}(b^{*}b)=\sum_{i=1}^{k}\tau(\sum_{j=1}^{k}b_{ji}^{*}b_{ji})<\infty.$

∎

######  Lemma D.1.12.

Let $e$ be a projection in $M_{k}(A)$. Then $e\in M_{k}(\mathcal{M}_{\tau})$.

###### Proof.

We use the trace $\tau_{k}$ from Lemma D.1.11. Since $\tau_{k}$ is densely
defined there is an $x\in\mathcal{M}_{\tau_{k}}$ such that
$\left\|e-x\right\|<\frac{1}{2}$. Then

$\frac{1}{2}e\leq e\left(1+(x-e)\right)e=exe\in\mathcal{M}_{\tau_{k}}.$

Hence $\tau_{k}(e)\leq 2\tau_{k}(exe)<\infty$. Let $e=(e_{ij})_{i,j=1}^{k}$.
Since $\tau_{k}(e^{*}e)=\tau_{k}(e)<\infty$ we find that

$\sum_{i=1}^{k}\sum_{j=1}^{k}\tau(e_{ji}^{*}e_{ji})=\tau(e^{*}e)<\infty,$

showing that $e_{ji}\in\mathcal{N}_{\tau}$ for all $i,j$, in the notation from
the proof of Lemma D.1.1. Then
$e_{ij}=\sum_{l=1}^{k}e_{il}e_{lj}\in\operatorname{Span}\mathcal{N}_{\tau}\mathcal{N}_{\tau}=\operatorname{Span}\mathcal{N}_{\tau}^{*}\mathcal{N}_{\tau}=\mathcal{M}_{\tau}$
for all $i,j$.

∎

It follows then from Lemma D.1.12 and Theorem D.1.2 that

$\tau_{*}([e]-[f])=\tau|_{\mathcal{M}_{\tau}}\otimes\operatorname{Tr}_{n}(e)-\tau|_{\mathcal{M}_{\tau}}\otimes\operatorname{Tr}_{n}(f)$
(D.1.1)

when $e$ and $f$ are projections in $M_{n}(A)$. In particular, it follows that
$\tau_{*}$ is a positive homomorphism in the sense that

$\tau_{*}([e])\geq 0$

when $e\in P_{\infty}(A)$.

######  Notes and remarks D.1.13.

This section is based on ideas of Connes and Elliott in the guise given to
them in [Th6].

####  D.1.2 Pairing traces with $K_{0}$ for AF-algebras

######  Lemma D.1.14.

Let $A$ be an AF-algebra and $\varphi:K_{0}(A)\to\mathbb{R}$ a positive
homomorphism. There is a densely defined lower semi-continuous trace $\tau$ on
$A$ such that $\varphi=\tau_{*}$.

###### Proof.

Let $p_{1}\leq p_{2}\leq p_{3}\leq\cdots$ be a sequence of projections in $A$
such that $\lim_{n\to\infty}p_{n}a=a$ for all $a\in A$. Then $p_{n}Ap_{n}$ is
a unital AF-algebra and by Corollary 3.3.24 of [Th5] there is a unique bounded
positive trace $\tau_{n}$ on $p_{n}Ap_{n}$ such that
${\tau_{n}}_{*}=\varphi\circ{j_{n}}_{*}$ on $K_{0}(p_{n}Ap_{n})$, where
$j_{n}:p_{n}Ap_{n}\to A$ is the inclusion. Let $e=e^{*}=e^{2}\in p_{n}Ap_{n}$.
Then

$\tau_{n+1}(e)={\tau_{n+1}}_{*}([e])=\varphi\circ{j_{n+1}}_{*}([e])=\varphi\circ{j_{n}}_{*}([e])={\tau_{n}}_{*}([e])=\tau_{n}(e),$

implying that $\tau_{n+1}|_{p_{n}Ap_{n}}=\tau_{n}$. Let $a\in A^{+}$. Then

$\displaystyle\tau_{n}(p_{n}ap_{n})=\tau_{n+1}(p_{n}ap_{n})=\tau_{n+1}(p_{n}p_{n+1}ap_{n+1}p_{n})=\tau_{n+1}(p_{n+1}ap_{n+1}p_{n})$
$\displaystyle=\tau_{n+1}(\sqrt{p_{n+1}ap_{n+1}}p_{n}\sqrt{p_{n+1}ap_{n+1}})\leq\tau_{n+1}(p_{n+1}ap_{n+1}).$

Define $\tau:A^{+}\to[0,\infty]$ such that

$\tau(a)=\lim_{n\to\infty}\tau_{n}(p_{n}ap_{n})=\sup_{n}\tau_{n}(p_{n}ap_{n}).$

Then $\tau$ is lower semi-continuous and

$\displaystyle\tau(a^{*}a)=\lim_{n\to\infty}\tau_{n}(p_{n}a^{*}ap_{n})=\lim_{n\to\infty}\lim_{l\to\infty}\tau_{n}(p_{n}a^{*}p_{l}ap_{n})$
$\displaystyle=\lim_{n\to\infty}\lim_{l\to\infty}\tau_{l}(p_{l}p_{n}a^{*}p_{l}ap_{n}p_{l})=\lim_{n\to\infty}\lim_{l\to\infty}\tau_{l}(p_{l}ap_{n}a^{*}p_{l})$
$\displaystyle=\lim_{n\to\infty}\tau(ap_{n}a^{*})=\tau(aa^{*}),$

showing that $\tau$ is a trace. When $e=e^{*}=e^{2}\in p_{n}Ap_{n}$,
$\tau(e)=\tau_{n}(e)=\varphi([e])$. It follows that $\tau_{*}=\varphi$.

∎

######  Lemma D.1.15.

Let $\tau_{i},\ i=1,2$, be lower semi-continuous traces on the AF-algebra $A$.
Assume that ${\tau_{1}}_{*}={\tau_{2}}_{*}$ on $K_{0}(A)$. It follows that
$\tau_{1}=\tau_{2}$.

###### Proof.

Let $\\{p_{n}\\}$ be an approximate unit in $A$ consisting of projections.
When $e$ is a projection in $p_{n}Ap_{n}$,
$\tau_{1}(e)={\tau_{1}}_{*}([e])={\tau_{2}}_{*}([e])=\tau_{2}(e)$ and it
follows therefore from Corollary 3.3.24 in [Th5] that
$\tau_{1}|_{p_{n}Ap_{n}}=\tau_{2}|_{p_{n}Ap_{n}}$. Let $a\in A$. Since
$\tau_{1}$ and $\tau_{2}$ are both lower semi-continuous,
$\lim_{n\to\infty}\tau_{i}(a^{*}p_{n}a)=\tau_{i}(a^{*}a),\ i=1,2$. Hence

$\displaystyle\tau_{1}(a^{*}a)=\lim_{n\to\infty}\tau_{1}(a^{*}p_{n}a)=\lim_{n\to\infty}\tau_{1}(p_{n}aa^{*}p_{n})$
$\displaystyle=\lim_{n\to\infty}\tau_{2}(p_{n}aa^{*}p_{n})=\lim_{n\to\infty}\tau_{2}(a^{*}p_{n}a)=\tau_{2}(a^{*}a).$

∎

The two last lemmas combine to give the following

######  Theorem D.1.16.

Let $A$ be an AF-algebra. The map $\tau\mapsto\tau_{*}$ is a bijection from
the set of lower semi-continuous traces $\tau$ on $A$ onto the set
$\operatorname{Hom}^{+}(K_{0}(A),\mathbb{R})$ of positive homomorphisms
$K_{0}(A)\to\mathbb{R}$.

##  Appendix E Simple crossed products of AF-algebras

The following is a result of Elliott, [E2].

######  Theorem E.0.1.

Let $G$ be a discrete group and $\alpha:G\to\operatorname{Aut}A$ an action of
$G$ by automorphisms of the AF-algebra $A$. Assume that for all $g\in
G\backslash\\{e\\}$ there are no non-zero order ideals $I$ of $K_{0}(A)$ such
that ${\alpha_{g}}_{*}(x)=x$ for all $x\in I$, and that there are no non-
trivial order ideals of $K_{0}(A)$ that are invariant under all
${\alpha_{g}}_{*},g\in G$. Then $A\rtimes_{\alpha,r}G$ is simple.

For the proof we fix the AF-algebra $A$. We’ll say that an automorphism
$\gamma\in\operatorname{Aut}A$ is _$K$ -outer_ when $\gamma_{*}$ is not the
identity map on any non-zero order ideal of $K_{0}(A)$. Thus one of the
assumptions in Theorem E.0.1 is that $\alpha_{g}$ is $K$-outer when $g\neq e$.

######  Lemma E.0.2.

Let $p$ be a projection in $A$ and $x\in K_{0}(A)^{+}$ such that $0\leq
x\leq[p]$ in $K_{0}(A)$. It follows that there is a projection $f\in A$ such
that $0\leq f\leq p$ and $[f]=x$.

###### Proof.

By definitions there are projections $q,r$ in matrices over $A$, say $q\in
M_{n}(A),\ r\in M_{m}(A)$, such that $x=[q]$ and $[q\oplus r]=[p]$ in
$K_{0}(A)$. By (1) of Lemma 2.3.34 of [Th5] there is a partial partial
isometry $v\in M_{n+m}(A)$ such that $v^{*}v=q\oplus r$ and $vv^{*}=p\oplus
0_{n+m-1}$. Then $f:=v(q\oplus 0_{m})v^{*}$ is a projection in $A$ such that
$[f]=x$ and $f\leq p$. ∎

######  Lemma E.0.3.

Let $F$ be a finite set of $K$-outer automorphisms of $A$. Let $p$ be a non-
zero projection in $A$ and let $\epsilon>0$. There is a non-zero projection
$f\in A$ such that $f\leq p$ and $\left\|f\gamma(f)\right\|\leq\epsilon$ for
all $\gamma\in F$.

###### Proof.

Let $\gamma\in F$. We first prove that ${\gamma}_{*}([p_{0}])\neq[p_{0}]$ for
some projection $p_{0}\in pAp$. To this end set $I^{+}=\left\\{x\in
K_{0}(A)^{+}:\ x\leq n[p]\ \text{for some}\ n\in\mathbb{N}\right\\}$. Then
$I:=I^{+}-I^{+}$ is a non-zero order ideal in $K_{0}(A)$ and thanks to the
assumptions it follows that there is an element $x\in I^{+}$ such that
${\gamma}_{*}(x)\neq x$. Since $(K_{0}(A),K_{0}(A)^{+})$ has the Riesz
decomposition property by Proposition 2.3.41 of [Th5] and since $x\leq n[p]$
for some $n\in\mathbb{N}$ it follows that $x=x_{1}+x_{2}+\cdots+x_{n}$ where
$0\leq x_{i}\leq[p]$ for each $i$. By Lemma E.0.2 there are then projections
$p_{1},p_{2},\cdots,p_{n}$ in $A$ such that $p_{i}\leq p$ and $x_{i}=[p_{i}]$
for all $i$. For at least for one $i$ we have that
${\gamma}_{*}([p_{i}])\neq[p_{i}]$. By taking $p_{0}:=p_{i}$ we have
established the claim .

Since $-p_{0}\leq\gamma(p_{0})-p_{0}\leq\gamma(p_{0})$ we have that
$\left\|\gamma(p_{0})-p_{0}\right\|\leq 1$. Since
${\gamma}_{*}([p_{0}])\neq[p_{0}]$ this implies that
$\left\|\gamma(p_{0})-p_{0}\right\|=1$. Indeed, if not it follows from
Exercise 2.2.19 in [Th5] (or Proposition 4.6.6 in [B]) that
$\left[\gamma(p_{0})\right]=[p_{0}]$ in $K_{0}(A)$; a contradiction. Thus
$\left\|\gamma(p_{0})-p_{0}\right\|=1$ as asserted. Let $\delta>0$. Since $A$
is AF there is a finite dimensional $C^{*}$-subalgebra $B\subseteq A$ and
projections $f,h\in B$ such that $\|f-p_{0}\|\leq\delta$ and
$\|h-\gamma(p_{0})\|\leq\delta$. This follows for example by use of Lemma
2.2.7 of [Th5]. Then $\|f-h\|\geq 1-2\delta$. Since $B$ is finite dimensional
it is a direct sum of matrix algebras and it follows that there is a central
projection $z\in B$ such that $zB\simeq M_{k}(\mathbb{C})$ for some
$k\in\mathbb{N}$ and $\left\|zf-zh\right\|=\|f-h\|\geq 1-2\delta$. We identify
$zB=M_{k}(\mathbb{C})$ and let $zB$ act on the Hilbert space $\mathbb{C}^{k}$
in the standard way. Then

$-1\leq-\left<zh\psi,\psi\right>\leq\left<(zf-
zh)\psi,\psi\right>\leq\left<zf\psi,\psi\right>\leq 1$

when $\psi\in\mathbb{C}^{k}$ and $\|\psi\|=1$. Since $\left\|zf-zh\right\|\geq
1-2\delta$ there is a $\psi\in\mathbb{C}^{k}$ such that $\|\psi\|=1$ and
either $1-3\delta\leq\left<(zf-zh)\psi,\psi\right>\leq 1$ or $-1\leq\left<(zf-
zh)\psi,\psi\right>\leq-1+3\delta$. In the first case

  * $\bullet$

$1-3\delta\leq\left<zf\psi,\psi\right>\leq 1$, and

  * $\bullet$

$\left<zh\psi,\psi\right>\leq 3\delta$

and in the second case

  * $\bullet$

$-1\leq-\left<zh\psi,\psi\right>\leq-1+3\delta$, and

  * $\bullet$

$\left<zf\psi,\psi\right>\leq 3\delta$.

In the first case we consider the one dimensional projections $f_{1}$ and
$f_{1}^{\prime}$ onto $\mathbb{C}zf\psi$ and $\mathbb{C}\psi$, respectively.
Note that $f_{1}\leq zf$. Furthermore, since

$\left\|zf\psi-\psi\right\|^{2}\leq 2-2\left<zf\psi,\psi\right>\leq 6\delta$

we find that

$\displaystyle\left\|f_{1}-f_{1}^{\prime}\right\|\leq
2\left\|\frac{zf\psi}{\|zf\psi\|}-\psi\right\|$ $\displaystyle\leq
2\left(\left\|\frac{zf\psi}{\|zf\psi\|}-zf\psi\right\|+\left\|zf\psi-\psi\right\|\right)$
$\displaystyle\leq
2\left(\left|\frac{1}{\|zf\psi\|}-1\right|+\sqrt{6\delta}\right)\leq
2\left(\frac{3\delta}{1-3\delta}+\sqrt{6\delta}\right):=\delta^{\prime}.$

For any $\phi\in\mathbb{C}^{k}$, $\|\phi\|\leq 1$, we have then that

$\left<f_{1}zhf_{1}\phi,\phi\right>\leq\left<f^{\prime}_{1}zhf^{\prime}_{1}\phi,\phi\right>+2\delta^{\prime}\leq\left<zh\psi,\psi\right>+2\delta^{\prime}\leq
3\delta+2\delta^{\prime},$

proving that $\left\|f_{1}h\right\|=\left\|f_{1}zh\right\|\leq
3\delta+2\delta^{\prime}$. And hence

$\left\|f_{1}\gamma(p_{0})\right\|\leq 4\delta+2\delta^{\prime}.$

Since $f_{1}\leq f$ and $\left\|f-p_{0}\right\|\leq\delta$ it follows that

$\left\|(p_{0}f_{1}p_{0})^{2}-p_{0}f_{1}p_{0}\right\|\leq\|f_{1}p_{0}f_{1}-f_{1}\|\leq\delta.$

Provided $\delta<\frac{1}{4}$ it follows from Lemma 2.2.7 in [Th5] that there
is a projection $q\leq p_{0}$ such that $\|q-p_{0}f_{1}p_{0}\|\leq 2\delta$
and hence $\|q-f_{1}\|\leq 4\delta$ . In particular, if we choose $\delta$
smaller than $1/4$, $q$ will be non-zero since $f_{1}\neq 0$. Furthermore,
$\left\|q\gamma(p_{0})\right\|\leq 8\delta+2\delta^{\prime}$ and hence

$\left\|q\gamma(q)\right\|=\sqrt{\left\|q\gamma(q)q\right\|}\leq\sqrt{\left\|q\gamma(p_{0})q\right\|}=\left\|q\gamma(p_{0})\right\|\leq
8\delta+2\delta^{\prime}.$

In the second case we consider the one dimensional projections $h_{1}$ and
$h_{1}^{\prime}$ onto $\mathbb{C}zh\psi$ and $\mathbb{C}\psi$, respectively.
Arguments identical to the preceding show that $h_{1}\leq zh$ and
$\left\|h_{1}-h_{1}^{\prime}\right\|\leq\delta^{\prime}$ while
$\left\|h_{1}f\right\|\leq 3\delta+2\delta^{\prime}$. Using that
$\left\|h-\gamma(p_{0})\right\|\leq\delta$ and $h_{1}\leq h$ we get also a
projection $r\leq\gamma(p_{0})$ such that $\left\|r-h_{1}\right\|\leq
4\delta$. In this case we set $q:=\gamma^{-1}(r)$. Then $q\leq p_{0}$ and

$\displaystyle\|rp_{0}\|\leq\|rf\|+\delta\leq\|h_{1}f\|+5\delta\leq
8\delta+2\delta^{\prime}.$

Hence
$\left\|q\gamma(q)\right\|=\sqrt{\|\gamma(q)q\gamma(q)\|}\leq\sqrt{\|\gamma(q)p_{0}\gamma(q)\|}=\left\|rp_{0}\right\|\leq
8\delta+2\delta^{\prime}$, as in the first case.

Now take another element $\gamma^{\prime}\in F$. From what we have just shown,
with $p$ replaced by $q$, we get a non-zero projection $q^{\prime}\leq q$ such
that $\left\|q^{\prime}\gamma^{\prime}(q^{\prime})\right\|\leq
8\delta+2\delta^{\prime}$. Since

$\displaystyle\left\|q^{\prime}\gamma(q^{\prime})\right\|=\sqrt{\left\|q^{\prime}\gamma(q^{\prime})q^{\prime}\right\|}\leq\sqrt{\left\|q^{\prime}\gamma(q)q^{\prime}\right\|}=\left\|q^{\prime}\gamma(q)\right\|$
$\displaystyle=\sqrt{\left\|\gamma(q)q^{\prime}\gamma(q)\right\|}\leq\sqrt{\left\|\gamma(q)q\gamma(q)\right\|}=\left\|q\gamma(q)\right\|\leq
8\delta+2\delta^{\prime},$

we can continue this reasoning until the finite set $F$ is exhausted. This
completes the proof provided only that $\delta>0$ is chosen such that
$8\delta+2\delta^{\prime}\leq\epsilon$.

∎

######  Lemma E.0.4.

Let $F$ be a finite set of $K$-outer automorphisms of $A$. Let $B$ be a finite
dimensional $C^{*}$-subalgebra of $A$ and $e\in B$ a non-zero central
projection in $B$. Set $B^{\prime}:=\left\\{a\in A:\ ab=ba\ \forall b\in
B\right\\}$ and let $\epsilon>0$. There is a non-zero projection $f\in
eB^{\prime}$ such that $\left\|f\gamma(f)\right\|\leq\epsilon$ for all
$\gamma$.

###### Proof.

Let $e_{0}$ be a minimal non-zero central projection in $B$ such that
$e_{0}\leq e$. Then $e_{0}B^{\prime}\subseteq eB^{\prime}$ and it suffices
therefore to find $f$ when $e=e_{0}$, i.e. we may assume that $e$ is a minimal
central projection in $B$. This means that $eB$ is a full matrix algebra, say
$eB\simeq M_{k}(\mathbb{C})$. Then $eB$ is generated by $k$ mutually
orthogonal projections $e_{1},e_{2},\cdots,e_{k}$, and a unitary $u$ with the
properties that $ue_{i}u^{*}=e_{i+1}$ for $i\leq k-1$, $ue_{k}u^{*}=e_{1}$ and
$u^{k}=e$. Set $w:=u+(1-e)$, which is a unitary in the algebra $A^{\dagger}$
obtained from $A$ by adjoining a unit. Then $w^{k}=1$, and we set

$F_{1}:=\left\\{\operatorname{Ad}w^{i}\circ\gamma:\ \gamma\in F,\
i=1,2,\cdots,k\right\\}.$

This is a finite set of $K$-outer automorphisms of $A$. Set
$\delta:=k^{-2}\epsilon$. It follows from Lemma E.0.3 that there is non-zero
projection $f_{1}\in A$ such that $f_{1}\leq e_{1}$ and
$\left\|f_{1}\gamma(f_{1})\right\|\leq\delta$ for all $\gamma\in F_{1}$. There
is also a non-zero projection $f_{2}\leq e_{2}$ such that
$\left\|f_{2}\gamma(f_{2})\right\|\leq\delta$ for all $\gamma\in F_{1}$ and a
non-zero projection $f_{1}^{\prime}\leq u^{*}f_{2}u$ such that
$\left\|f_{1}^{\prime}\gamma(f_{1}^{\prime})\right\|\leq\delta$ for all
$\gamma\in F_{1}$. Exchange $f_{1}$ with $f_{1}^{\prime}$ and $f_{2}$ with
$uf_{1}^{\prime}u^{*}$. Then $0\neq f_{i}\leq e_{i}$,
$\left\|f_{i}\gamma(f_{i})\right\|\leq\delta,\ i=1,2$, for all $\gamma\in
F_{1}$, and $uf_{1}u^{*}=f_{2}$. By Lemma E.0.3 there is a non-zero projection
$f_{3}^{\prime}\leq uf_{2}u^{*}$ such that
$\left\|f_{3}^{\prime}\gamma(f_{3}^{\prime})\right\|\leq\delta$ for all
$\gamma\in F_{1}$, and there is a non-zero projection
$f_{1}^{\prime\prime}\leq{u^{*}}^{2}f_{3}^{\prime}u^{2}$ such that
$\left\|f_{1}^{\prime\prime}\gamma(f_{1}^{\prime\prime})\right\|\leq\delta$
for all $\gamma\in F_{1}$ . Exchange $f_{1}$ with $f_{1}^{\prime\prime}$,
$f_{2}$ with $uf_{1}^{\prime\prime}u^{*}$ and set
$f_{3}=u^{2}f_{1}^{\prime\prime}{u^{2}}^{*}$. Then $0\neq f_{i}\leq e_{i}$,
$\left\|f_{i}\gamma(f_{i})\right\|\leq\delta,\ i=1,2,3$, for all $\gamma\in
F_{1}$, and $uf_{i}u^{*}=f_{i+1},\ i=1,2$. This process can be continued and
successfully completed because $u^{k}=e$ to get projections
$f_{i},i=1,2,\cdots,k$, such that $0\neq f_{i}\leq e_{i}$ and
$\left\|f_{i}\gamma(f_{i})\right\|\leq\delta,\ i=1,2,3,\cdots k$, for all
$\gamma\in F_{1}$, and $uf_{i}u^{*}=f_{i+1},\ i=1,2,\cdots,k-1$, while
$uf_{k}u^{*}=f_{1}$. Let $i,j\in\left\\{1,2,3,\cdots,k\right\\}$ and
$\gamma\in F$. There are then $x,y\in\\{1,2,\cdots,k\\}$ such that
$w^{x}f_{j}{w^{x}}^{*}=f_{i}$, $w^{y}={w^{x}}^{*}$, and then

$\left\|f_{i}\gamma(f_{j})\right\|=\left\|w^{x}f_{j}{w^{x}}^{*}\gamma(f_{j})\right\|=\left\|f_{j}{w^{x}}^{*}\gamma(f_{j})w^{x}\right\|=\left\|f_{j}w^{y}\gamma(f_{j}){w^{y}}^{*}\right\|\leq\delta$

since $\operatorname{Ad}w^{y}\circ\gamma\in F_{1}$. Set
$f=\sum_{i=1}^{k}f_{i}$. Then $0\neq f\leq e$ and

$\left\|f\gamma(f)\right\|\leq\sum_{i,j=1}^{k}\left\|f_{i}\gamma(f_{j})\right\|\leq
k^{2}\delta\leq\epsilon.$

By construction $f\leq e$ and $f$ commutes with each $e_{i}$ and with $u$;
thus $f\in eB^{\prime}$. ∎

######  Lemma E.0.5.

Fix $a\in A$ and let $S$ be a finite subset of $A$ and $F$ be a finite set of
$K$-outer automorphisms of $A$. For each $\epsilon>0$ there is a projection
$f\in A$ such that

  * $\bullet$

$\left\|faf\right\|\geq\|a\|-\epsilon$,

  * $\bullet$

$\left\|fb-bf\right\|\leq\epsilon$ for all $b\in S$, and

  * $\bullet$

$\left\|f\gamma(f)\right\|\leq\epsilon$ for all $\gamma\in F$.

###### Proof.

Set $\delta:=\frac{\epsilon}{2}$. Since $A$ is AF there is a finite
dimensional $C^{*}$-subalgebra $B\subseteq A$, an element $b\in B$ and a
subset $T\subseteq B$ such that $\left\|a-b\right\|\leq\delta$ and
$\operatorname{Dist}(s,T)\leq\delta$ for all $s\in S$. There is a minimal non-
zero central projection $e$ in $B$ such that $\|eb\|=\|b\|$. By Lemma E.0.4
there is a non-zero projection $f\in eB^{\prime}$ such that
$\left\|f\gamma(f)\right\|\leq\delta$ for all $\gamma\in F$. Since $eB$ is a
full matrix algebra commuting with $f$, the map $eB\ni d\mapsto df$ is an
injective $*$-homomorphism and hence an isometry. Therefore
$\left\|fbf\right\|=\left\|febf\right\|=\left\|ebf\right\|=\left\|eb\right\|=\left\|b\right\|$.
It follows that
$\|faf\|\geq\left\|fbf\right\|-\delta=\left\|b\right\|-\delta\geq\|a\|-2\delta$.
Similarly, since $f\in B^{\prime}$ commutes with the elements of $T$ we have
that $\left\|fs-sf\right\|\leq 2\delta$ for all $s\in S$. ∎

We are now ready for the _proof of Theorem E.0.1:_ Let $J$ be an ideal in
$A\rtimes_{\alpha,r}G$. Then $J\cap A$ is an $\alpha$-invariant ideal in $A$
and by Theorem 4.1.8 of [Th5] it corresponds to an order ideal $I$ in
$K_{0}(A)$ which is left globally invariant by ${\alpha_{g}}_{*}$ for all
$g\in G$. By assumption this ideal $I$ is either $K_{0}(A)$ or $\\{0\\}$. If
$I=K_{0}(A)$ it follows from Theorem 4.1.8 of [Th5] that $A\cap J=A$. Since
$A$ contains an approximate unit for $A\rtimes_{\alpha,r}G$ by Lemma 7.3.3,
this implies that $J=A\rtimes_{\alpha,r}G$. The proof will be completed by
showing that the second possibility, $I=\\{0\\}$ implies that $J=\\{0\\}$. So
assume that $I=\\{0\\}$. By Theorem 4.1.8 of [Th5] this means that $J\cap
A=\\{0\\}$. Let $j\in J$ and let $\delta>0$. By definition of
$A\rtimes_{\alpha,r}G$ there is a finitely supported function $F:G\to A$ such
that

$\left\|j-F(e)-\sum_{g\neq e}F(g)u_{g}\right\|\leq\delta.$ (E.0.1)

Let $P:A\rtimes_{\alpha,r}G\to A$ be the canonical conditional expectation,
cf. Section 7.3.1. Since $P$ is of norm $1$ and annihilates $F(g)u_{g}$ when
$g\neq e$, we have that $\left\|P(j)-F(e)\right\|\leq\epsilon$. Choose
$\epsilon>0$ such that

$\epsilon\sum_{g\neq e}\left\|F(g)\right\|+(n+1)\epsilon\leq\delta,$

where $n$ is the number of non-zero elements of $F(G)$. By assumption
$\alpha_{g}$ is K-outer when $g\neq e$, and it follows therefore from Lemma
E.0.5 that there is a projection $f\in A$ such that
$\left\|fF(e)f\right\|\geq\left\|F(e)\right\|-\epsilon$, $\left\|fb-
bf\right\|\leq\epsilon$ for all $b\in F(G)$ and
$\left\|f\alpha_{g}(f)\right\|\leq\epsilon$ when $g\neq e$. Let
$\pi:A\rtimes_{\alpha,r}G\to(A\rtimes_{\alpha,r}G)/J$ be the quotient map and
note that

$\left\|\pi\left(\sum_{g\in G}F(g)u_{g}\right)\right\|\leq\delta,$

thanks to (E.0.1). Since $\pi$ is injective and hence isometric on $A$ because
$A\cap J=\\{0\\}$, it follows that
$\left\|\pi(fF(e)f)\right\|=\left\|fF(e)f\right\|$. Hence

$\displaystyle\left\|F(e)\right\|\leq\left\|fF(e)f\right\|+\epsilon=\left\|\pi(fF(e)f)\right\|+\epsilon$
$\displaystyle\leq\left\|\pi(f(F(e)+\sum_{g\neq
e}F(g)u_{g})f)\right\|+\left\|\pi(f(\sum_{g\neq
e}F(g)u_{g})f)\right\|+\epsilon$
$\displaystyle\leq\|\pi(fjf)\|+\left\|\pi(f(\sum_{g\neq
e}F(g)u_{g})f)\right\|+\delta+\epsilon$ $\displaystyle\leq\left\|\sum_{g\neq
e}fF(g)u_{g}f\right\|+\delta+\epsilon$ $\displaystyle\leq\left\|\sum_{g\neq
e}F(g)fu_{g}f\right\|+\delta+\epsilon+n\epsilon$ $\displaystyle\leq\sum_{g\neq
e}\left\|F(g)\right\|\left\|f\alpha_{g}(f)u_{g}\right\|+\delta+(n+1)\epsilon$
$\displaystyle\leq\epsilon\sum_{g\neq
e}\left\|F(g)\right\|+\delta+(n+1)\epsilon\leq 2\delta.$

Since $\left\|P(j)\right\|\leq\left\|F(e)\right\|+\delta$ we conclude first
that $\left\|P(j)\right\|\leq 3\delta$, and then since $\delta>0$ was
arbitrary, that $P(j)=0$. Thus $P(J)=\\{0\\}$ and as $P$ is faithful (by Lemma
7.3.2) it follows that $J=\\{0\\}$. ∎

######  Notes and remarks E.0.6.

This section is based almost entirely on Elliotts paper [E2], but the
statement of the main result Theorem E.0.1 differs from that in [E2]. The
reason is that we here want a version where the assumptions in the theorem
only refer to K-theory data since this fits better with the way the theorem is
used in the proof of Theorem 9.3.5.

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