&=\left<\Lambda_{\psi}\left(\sigma_{-i\frac{\beta}{2}}(b)^{*}\right),T_{\phi}\Lambda_{\psi}(a)\right>=\left<T_{\phi}\Lambda_{\psi}\left(\sigma_{-i\frac{\beta}{2}}(b)^{*}\right),\Lambda_{\psi}(a)\right>\\\
&=\phi\left(a^{*}\sigma_{-i\frac{\beta}{2}}(b)^{*}\right)=\phi\left(a^{*}\sigma_{i\frac{\beta}{2}}(b^{*})\right).\end{split}$
(5.2.2)

By comparing the two expressions for
$\phi\left(a^{*}\sigma_{i\frac{\beta}{2}}(b^{*})\right)$ in (5.2.1) and
(5.2.2) it follows that $T_{\phi}$ commutes with $J_{\psi}$. By using that
$J_{\psi}^{2}=J_{\psi}$ we deduce from Theorem 4.3.6 that
$T_{\phi}\in\pi_{\psi}(A)^{\prime}\cap\pi_{\psi}(A)^{\prime\prime}$. Since

$\displaystyle\left<T_{\phi}U^{\psi}_{t}\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>=\phi(b^{*}\sigma_{t}(a))=\phi(\sigma_{-t}(b)^{*}a)$
$\displaystyle=\left<T_{\phi}\Lambda_{\psi}(a),U^{\psi}_{-t}\Lambda_{\psi}(b)\right>=\left<U_{t}^{\psi}T_{\phi}\Lambda_{\psi}(a),\Lambda_{\varphi}(b)\right>$

for $a,b\in\mathcal{M}^{\sigma}_{\psi}$, we see that $T_{\phi}$ commutes with
$U^{\psi}_{t}$. ∎

Set

$\mathcal{Z}_{\psi}:=\pi_{\psi}(A)^{\prime}\cap\pi_{\psi}(A)^{\prime\prime};$

the center of $\pi_{\varphi}(A)^{\prime\prime}$, and

$\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}:=\left\\{m\in\mathcal{Z}_{\psi}:\
\sigma^{\prime\prime}_{t}(m)=m\ \forall t\in\mathbb{R}\right\\}.$

######  Lemma 5.2.2.

$\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}=\mathcal{Z}_{\psi}$ when
$\beta\neq 0$.

###### Proof.

This follows from Theorem 4.3.1 because the modular automorphism group
associated to a faithful normal semi-finite weight on a von Neumann algebra
acts trivially on the center of the algebra. ∎

For every element
$c\in\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}\backslash\\{0\\}$, $0\leq
c\leq 1$, we can define a weight $\psi_{c}:A^{+}\to[0,\infty]$ such that

$\psi_{c}(a):=\psi^{\prime\prime}(c\pi_{\psi}(a)).$ (5.2.3)

######  Lemma 5.2.3.

$\psi_{c}$ is a $\beta$-KMS weight for $\sigma$ such that $\psi_{c}\leq\psi$.

###### Proof.

To see that $\psi_{c}$ is a weight the only non-trivial fact is that
$\psi_{c}$ is lower semi-continuous which follows easily from the fact that
$\psi^{\prime\prime}$ is lower semi-continuous with respect to the
$\sigma$-weak topology and hence also with respect to the norm-topology. To
see that $\psi_{c}$ is non-zero note that $\psi^{\prime\prime}(c)>0$ since
$\psi^{\prime\prime}$ is faithful by Lemma 4.2.4. Since $\pi_{\psi}(A)$ is
$\sigma$-weakly dense in $N$ there is a net $\\{a_{i}\\}$ in $A^{+}$ such that
$\lim_{i\to\infty}\pi_{\psi}(a_{i})=1$ in the $\sigma$-weak topology. Since
$\\{x\in N^{+}:\ \psi^{\prime\prime}(x)>\frac{\psi^{\prime\prime}(c)}{2}\\}$
is open in the $\sigma$-weak topology, there is an $i$ such that
$\psi^{\prime\prime}(c\pi_{\psi}(a_{i}))>\frac{\psi^{\prime\prime}(c)}{2}$.
Then $\psi_{c}(a_{i})>0$ , showing that $\psi_{c}$ is not zero. That
$\psi_{c}$ is $\sigma$-invariant follows from Lemma 4.2.5:

$\displaystyle\psi_{c}(\sigma_{t}(a))=\psi^{\prime\prime}(c\pi_{\psi}(\sigma_{t}(a)))=\psi^{\prime\prime}(c\sigma^{\prime\prime}_{t}(\pi_{\psi}(a)))=\psi^{\prime\prime}\left(\sigma^{\prime\prime}_{t}(\sigma^{\prime\prime}_{-t}(c)\pi_{\psi}(a))\right)$
$\displaystyle=\psi^{\prime\prime}\left(\sigma^{\prime\prime}_{-t}(c)\pi_{\psi}(a)\right)=\psi^{\prime\prime}(c\pi_{\psi}(a))=\psi_{c}(a).$

It remains to show that $\psi_{c}$ satisfies one of the four equivalent
conditions of Theorem 2.3.1. Let
$a,b\in\mathcal{N}_{\psi_{c}}\cap\mathcal{N}_{\psi_{c}}^{*}$. Then
$\sqrt{c}\pi_{\psi}(a),\sqrt{c}\pi_{\psi}(b)\in\mathcal{N}_{\psi^{\prime\prime}}\cap\mathcal{N}_{\psi^{\prime\prime}}^{*}$.
By Theorem 4.2.6 there is a continuous function
$f:{\mathcal{D}_{\beta}}\to\mathbb{C}$ which is holomorphic in the interior
$\mathcal{D}_{\beta}^{0}$ of ${\mathcal{D}_{\beta}}$ and has the property that

  * •

$f(t)=\psi^{\prime\prime}(\sqrt{c}\pi_{\psi}(b)\sigma^{\prime\prime}_{t}(\sqrt{c}\pi_{\psi}(a)))\
\ \forall t\in\mathbb{R}$, and

  * •

$f(t+i\beta)=\psi^{\prime\prime}(\sigma^{\prime\prime}_{t}(\sqrt{c}\pi_{\psi}(a))\sqrt{c}\pi_{\psi}(b))\
\ \forall t\in\mathbb{R}$.

Since
$\psi^{\prime\prime}(\sqrt{c}\pi_{\psi}(b)\sigma^{\prime\prime}_{t}(\sqrt{c}\pi_{\psi}(a)))=\psi^{\prime\prime}(c\pi_{\psi}(b\sigma_{t}(a)))=\psi_{c}(b\sigma_{t}(a))$
and similarly
$\psi^{\prime\prime}(\sigma^{\prime\prime}_{t}(\sqrt{c}\pi_{\psi}(a))\sqrt{c}\pi_{\psi}(b))=\psi_{c}(\sigma_{t}(a)b)$,
it follows that $\psi_{c}$ satisfies condition (4) of Theorem 2.3.1 and hence
is a $\beta$-KMS weight for $\sigma$. Since $c\leq 1$ it follows from Lemma
4.2.16 that
$\psi_{c}(a)=\psi^{\prime\prime}\left(\pi_{\psi}(a^{\frac{1}{2}})c\pi_{\psi}(a^{\frac{1}{2}})\right)\leq\psi^{\prime\prime}(\pi_{\psi}(a))=\psi(a)$
for all $a\in A^{+}$; i.e. $\psi_{c}\leq\psi$. ∎

######  Lemma 5.2.4.

Let $\phi_{1},\ \phi_{2}$ and $\psi$ be $\beta$-KMS weights for $\sigma$. If
$\phi_{1}(a^{*}a)=\phi_{2}(a^{*}a)$ for all $\mathcal{N}_{\psi}$, then
$\phi_{1}=\phi_{2}$.

###### Proof.

Let $\\{e_{j}\\}_{j\in J}$ be the approximate unit in $\mathcal{N}_{\psi}$
from Lemma 2.2.12. Then $e_{j}^{2}\in\mathcal{M}_{\psi}^{+}$ and

$f_{j}:=R_{1}(e_{j}^{2})\in\mathcal{M}_{\psi}^{\sigma}$

by Lemma 2.2.9. Let $a\in A$. The $C^{*}$-algebra generated by
$\left\\{\sigma_{t}(a):\ t\in\mathbb{R}\right\\}$ is separable and there is
therefore a sequence $j_{1}\leq j_{2}\leq j_{3}\leq\cdots$ in $J$ such that
$\lim_{n\to\infty}e_{j_{n}}^{2}\sigma_{t}(a)=\sigma_{t}(a)$ for all
$t\in\mathbb{R}$ and it follows then from Lemma A.2.2 in Appendix A that

$\lim_{n\to\infty}f_{j_{n}}a=\lim_{n\to\infty}\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}e^{-t^{2}}\sigma_{t}(e_{j_{n}}^{2})a\
\mathrm{d}t=a.$

Assume then that $a\in\mathcal{A}_{\sigma}$. By Lemma 4.2.1,

$\sigma_{-i\frac{\beta}{2}}(f_{j}a)=\sigma_{-i\frac{\beta}{2}}(f_{j})\sigma_{-i\frac{\beta}{2}}(a)\in\mathcal{N}_{\psi}^{*}$

and it follows therefore from the assumption that

$\phi_{1}(\sigma_{-i\frac{\beta}{2}}(f_{j}a)\sigma_{-i\frac{\beta}{2}}(f_{j}a)^{*})=\phi_{2}(\sigma_{-i\frac{\beta}{2}}(f_{j}a)\sigma_{-i\frac{\beta}{2}}(f_{j}a)^{*})$

for all $j$. Since $\phi_{1}$ and $\phi_{2}$ are both $\beta$-KMS weights for
$\sigma$ this implies that

$\phi_{1}(a^{*}f_{j}^{2}a)=\phi_{2}(a^{*}f_{j}^{2}a)$ (5.2.4)

for all $j$. Note that $\left\|f_{j}^{2}\right\|\leq 1$ for all $j$ and that
$\lim_{n\to\infty}a^{*}f_{j_{n}}^{2}a=a^{*}a$. It follows therefore from
(5.2.4) and the lower semi-continuity of $\phi_{1}$ and $\phi_{2}$ that
$\phi_{1}(a^{*}a)=\phi_{2}(a^{*}a)$. Since $a\in\mathcal{A}_{\sigma}$ was
arbitrary it follows from Lemma 3.1.5 that $\phi_{1}=\phi_{2}$. ∎

######  Corollary 5.2.5.

Let $\phi_{i},i=1,2$, be $\beta$-KMS weights for $\sigma$. If
$\phi_{1}(a^{*}a)=\phi_{2}(a^{*}a)$ for all
$a\in\mathcal{N}_{\phi_{1}}\cap\mathcal{N}_{\phi_{2}}$, then
$\phi_{1}=\phi_{2}$.

###### Proof.

By Proposition 5.1.2 the sum $\phi_{1}+\phi_{2}$ is a $\beta$-KMS weight.
Since
$\mathcal{N}_{\phi_{1}+\phi_{2}}=\mathcal{N}_{\phi_{1}}\cap\mathcal{N}_{\phi_{2}}$
it follows from Lemma 5.2.4 that $\phi_{1}=\phi_{2}$. ∎

######  Corollary 5.2.6.

Let $\phi_{i},i=1,2$, be $\beta$-KMS weights for $\sigma$. If
$S\subseteq\mathcal{N}_{\phi_{1}}$ is a subspace which is a core for
$\Lambda_{\phi_{1}}$ and $\phi_{1}(a^{*}a)=\phi_{2}(a^{*}a)$ for all $a\in S$,
then $\phi_{1}=\phi_{2}$.

###### Proof.

Let $a\in\mathcal{N}_{\phi_{1}}$. Since $S$ is a core for $\Lambda_{\phi_{1}}$
by assumption there is a sequence $\\{s_{n}\\}$ in $S$ such that
$\lim_{n\to\infty}s_{n}=a$ and
$\lim_{n\to\infty}\Lambda_{\phi_{1}}(s_{n})=\Lambda_{\phi_{1}}(a)$. Since
$\left\|\Lambda_{\phi_{2}}(s_{n})-\Lambda_{\phi_{2}}(s_{m})\right\|=\left\|\Lambda_{\phi_{1}}(s_{n})-\Lambda_{\phi_{1}}(s_{m})\right\|$
for all $n,m$, it follows from that
$\left\\{\Lambda_{\phi_{2}}(s_{n})\right\\}$ converges and hence that
$a\in\mathcal{N}_{\phi_{2}}$ since $\Lambda_{\phi_{2}}$ is closed by Lemma
2.2.1. Furthermore, $\phi_{1}(a^{*}a)=\phi_{2}(a^{*}a)$ and Corollary 5.2.5
applies. ∎

######  Theorem 5.2.7.

Let $\psi$ be a $\beta$-KMS weight for $\sigma$. The map $c\mapsto\psi_{c}$
given by (5.2.3) is a bijection from

$\left\\{c\in\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}\backslash\\{0\\}:\
0\leq c\leq 1\right\\}$

onto the set of $\beta$-KMS weights $\phi$ for $\sigma$ with the property that
$\phi\leq\psi$. Furthermore, $c\leq c^{\prime}$ in
$\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}$ if and only if
$\psi_{c}\leq\psi_{c^{\prime}}$.

###### Proof.

Let $\phi\leq\psi$. It follows from Lemma 5.2.1 that there is
$c\in\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}$ such that $0\leq c\leq 1$ and
$\phi(a^{*}a)=\left<c\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>$ for all
$a\in\mathcal{N}_{\psi}$. Let $W$ be the unitary from Lemma 4.3.4 and
$a\in\mathcal{N}_{\psi}$. By using (4.3.6) we find that

$\displaystyle\phi(a^{*}a)=\left<c\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>=\left<Wc\Lambda_{\psi}(a),W\Lambda_{\psi}(a)\right>$
$\displaystyle=\left<\pi_{\psi^{\prime\prime}}(c)W\Lambda_{\psi}(a),W\Lambda_{\psi}(a)\right>=\left<\pi_{\psi^{\prime\prime}}(c)\Lambda_{\psi^{\prime\prime}}(\pi_{\psi}(a)),\Lambda_{\psi^{\prime\prime}}(\pi_{\psi}(a))\right>$
$\displaystyle=\psi^{\prime\prime}(c\pi_{\psi}(a^{*}a))=\psi_{c}(a^{*}a).$

Thanks to Lemma 5.2.4 this shows that $\phi=\psi_{c}$, and we conclude that
the map under consideration is surjective.

Assume
$c_{i}\in\left\\{c\in\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}\backslash\\{0\\}:\
0\leq c\leq 1\right\\},\ i=1,2$, and $\psi_{c_{1}}=\psi_{c_{2}}$. For
$a\in\mathcal{N}_{\psi}$,

$\displaystyle\psi_{c_{i}}(a^{*}a)=\psi^{\prime\prime}(c_{i}\pi_{\psi}(a^{*}a))=\left<\pi_{\psi^{\prime\prime}}(c_{i})\Lambda_{\psi^{\prime\prime}}(\pi_{\psi}(a)),\Lambda_{\psi^{\prime\prime}}(\pi_{\psi}(a))\right>$
$\displaystyle=\left<\pi_{\psi^{\prime\prime}}(c_{i})W\Lambda_{\psi}(a),W\Lambda_{\psi}(a)\right>=\left<Wc_{i}\Lambda_{\psi}(a),W\Lambda_{\psi}(a)\right>$
$\displaystyle=\left<c_{i}\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>,\ \
i=1,2.$

Hence

$\left<c_{1}\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>=\psi_{c_{1}}(a^{*}a)=\psi_{c_{2}}(a^{*}a)=\left<c_{2}\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>$

for all $a\in\mathcal{N}_{\psi}$, implying that $c_{1}=c_{2}$.

Assume that $c\leq c^{\prime}$ in
$\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}$. Then $c\pi_{\psi}(a)\leq
c^{\prime}\pi_{\psi}(a)$ and hence
$\psi_{c}(a)=\psi^{\prime\prime}(c\pi_{\psi}(a))\leq\psi^{\prime\prime}(c^{\prime}\pi_{\psi}(a))=\psi_{c^{\prime}}(a)$
for all $a\in A^{+}$; i.e., $\psi_{c}\leq\psi_{c^{\prime}}$. Conversely, if
$\psi_{c}\leq\psi_{c^{\prime}}$ the calculation above shows that
$\left<c\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>\leq\left<c^{\prime}\Lambda_{\psi}(a),\Lambda_{\psi}(a)\right>$
for all $a\in\mathcal{N}_{\psi}$, implying that $c\leq c^{\prime}$. ∎

Denote by

$\operatorname{KMS}(\sigma,\beta)$

the set of $\beta$-KMS weights for $\sigma$.

######  Definition 5.2.8.

A $\beta$-KMS weight $\psi$ is _extremal_ when the only $\beta$-KMS weights
that are dominated by $\psi$ are the scalar multiplies of $\psi$.

In symbols $\psi\in\operatorname{KMS}(\sigma,\beta)$ is extremal when

$\phi\in\operatorname{KMS}(\sigma,\beta),\
\phi\leq\psi\Rightarrow\phi\in\mathbb{R}^{+}\psi.$

An immediate consequence of Theorem 5.2.7 and Lemma 5.2.2 is the following

######  Corollary 5.2.9.

Let $\sigma$ be a flow and $\beta\in\mathbb{R}\backslash\\{0\\}$. A
$\beta$-KMS weight for $\sigma$ is extremal if and only if
$\pi_{\psi}(A)^{\prime\prime}$ is a factor.

There is also a version of the last corollary for $0$-KMS weights. To
formulate it we say that a flow $\alpha$ on a von Neumann algebra $M$ is
_ergodic_ when the fixed point algebra of $\alpha$ is as small as possible;
viz. $M^{\alpha}=\mathbb{C}1$.

######  Corollary 5.2.10.

Let $\sigma$ be a flow on $A$. A $\sigma$-invariant lower semi-continuous
trace $\psi$ on $A$ is extremal as a $0$-KMS weight if and only if
$\sigma^{\prime\prime}$ acts ergodically on the center $\mathcal{Z}_{\psi}$ of
$\pi_{\psi}(A)^{\prime\prime}$.

######  Notes and remarks 5.2.11.

Theorem 5.2.7 is new. Corollary 5.2.5 is a very useful tool, and it was
obtained by Kusterman and Vaes in [KV1]. In fact, in their version it is only
required that one of the weights is a KMS weight.

###  5.3 The lattice of $\beta$-KMS weights

The set $\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$ is a partially ordered
set 111See for example Appendix A1 in [Ru2] for the definition of a partially
ordered set. when we define $\psi\leq\phi$ to mean that $\psi(a)\leq\phi(a)$
for all $a\in A^{+}$. Moreover, it follows from Proposition 5.1.2 that
$\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$ has the structure of a convex
cone. Specifically, when
$\psi,\varphi\in\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$ and
$t,s\in\mathbb{R}^{+}$, the formula

$(t\psi+s\varphi)(a):=t\psi(a)+s\varphi(a),\ \ a\in A^{+},$

defines an element $t\psi+s\varphi$ of
$\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$.

######  Speculations 5.3.1.

Usually a convex cone is a closed subset of an ambient locally convex vector
space, but in the case of KMS weights there are no obvious candidate for such
a space. This is not to say that it does not exist, and in fact it does, at
least when $A$ is separable. See Notes and Remarks 7.2.15. Under the
additional assumption that the fixed point algebra of the flow $\sigma$
contains an approximate unit for $A$ a direct construction is given in [Ch].

######  Theorem 5.3.2.

$(\operatorname{KMS}(\sigma,\beta)\cup\\{0\\},\leq)$ is a lattice in the usual
sense; every pair
$\psi_{1},\psi_{2}\in\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$ has a least
upper bound $\psi_{1}\vee\psi_{2}$ and greatest lower bound
$\psi_{1}\wedge\psi_{2}$ in $\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$.

###### Proof.

To show that $\psi_{1}$ and $\psi_{2}$ have a greatest lower bound we may
assume that $\psi_{1}\neq 0$ and $\psi_{2}\neq 0$; otherwise the greatest
lower bound is clearly $0$. Set $\varphi:=\psi_{1}+\psi_{2}$, which is a
$\beta$-KMS weight for $\sigma$ by Proposition 5.1.2. By Theorem 5.2.7 there
are elements $c_{i}\in\mathcal{Z}_{\varphi}^{\sigma^{\prime\prime}},\ 0\leq
c_{i}\leq 1$, such that $\psi_{i}=\varphi_{c_{i}},\ i=1,2$. Since
$\mathcal{Z}_{\varphi}^{\sigma^{\prime\prime}}$ is an abelian $C^{*}$-algebra
the cone ${\mathcal{Z}_{\varphi}^{\sigma^{\prime\prime}}}^{+}$ of its positive
elements is a lattice. See e.g. Example 4.2.6 in [BR]. Let
$c_{1}\wedge_{\varphi}c_{2}$ be the greatest lower bound for $c_{1}$ and
$c_{2}$ in ${\mathcal{Z}_{\varphi}^{\sigma^{\prime\prime}}}^{+}$. Set

$\psi_{1}\wedge\psi_{2}:=\varphi_{c_{1}\wedge_{\varphi}c_{2}}.$

It follows from Theorem 5.2.7 that $\psi_{1}\wedge\psi_{2}$ is the greatest
lower bound for $\psi_{1}$ and $\psi_{2}$ in the set

$\left\\{\phi\in\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}:\
\phi\leq\varphi\right\\}.$ (5.3.1)

But then $\psi_{1}\wedge\psi_{2}$ is the greatest lower bound overall; if
namely $\mu\leq\psi_{i},\ i=1,2$, it follows that $\mu\leq\varphi$ and hence
$\mu\leq\psi_{1}\wedge\psi_{2}$.

To show that $\psi_{1}$ and $\psi_{2}$ have a least upper bound we may assume
that $\psi_{1}$ and $\psi_{2}$ are not both zero; otherwise the least upper
bound is clearly $0$. Let $c_{1}\vee_{\varphi}c_{2}$ be the least upper bound
of $c_{1}$ and $c_{2}$ in $\mathcal{Z}_{\varphi}^{\sigma^{\prime\prime}}$ and
set

$\psi_{1}\vee\psi_{2}:=\varphi_{c_{1}\vee_{\varphi}c_{2}}.$

It follows from Theorem 5.2.7 that $\psi_{1}\vee\psi_{2}$ is the least upper
bound for $\psi_{1}$ and $\psi_{2}$ in the set (5.3.1). To see that it is in
fact the least upper bound overall, let
$\mu\in\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$ such that
$\psi_{i}\leq\mu,\ i=1,2$. Then $\psi_{i}\leq\mu\wedge\varphi,\ i=1,2$, and
$\mu\wedge\varphi\leq\varphi$. Hence
$\psi_{1}\vee\psi_{2}\leq\mu\wedge\varphi\leq\mu$, showing that
$\psi_{1}\vee\psi_{2}$ is the least upper bound for $\psi_{1}$ and $\psi_{2}$
in $\operatorname{KMS}(\sigma,\beta)\cup\\{0\\}$. ∎

######  Notes and remarks 5.3.3.

The main results in this chapter, Theorem 5.2.7 and Theorem 5.3.2, are new,
but Christensen obtains in Theorem 4.7 of [Ch] a version of Theorem 5.3.2 when
the fixed point algebra of the flow $\sigma$ contains an approximate unit for
$A$.

###  5.4 Passive KMS weights

Let $\sigma$ be a flow on the $C^{*}$-algebra $A$. Let $I\subseteq A$ be a
$\sigma$-invariant closed two-sided ideal in $A$ and let $q_{I}:A\to A/I$ be
the quotient map. Let $\sigma^{A/I}$ be the flow on $A/I$ induced by $\sigma$;
viz.

$\sigma^{A/I}_{t}\circ q_{I}=q_{I}\circ\sigma_{t}.$

######  Lemma 5.4.1.

Let $\psi$ be a $\beta$-KMS weight for $\sigma^{A/I}$. Then $\psi\circ q_{I}$
is a $\beta$-KMS weight for $\sigma$.

###### Proof.

To see that $\psi$ is a weight on $A$ only the lower semi-continuity is not
trivial, but follows from Combes’ theorem, Theorem 1.1.1, which implies that

$\psi\circ q_{I}=\sup_{\omega\in\mathcal{F}_{\psi}}\omega\circ q_{I}.$

To see that $\psi\circ q_{I}$ is a $\beta$-KMS weight, note first of all that
$\psi\circ q_{I}$ is non-zero and densely defined since $\psi$ has these
properties. Let $a\in\mathcal{A}_{\sigma}$. By definition this means that
there is a sequence $\\{a_{n}\\}_{n=0}^{\infty}$ in $A$ such that
$\sigma_{t}(a)=\sum_{n=0}^{\infty}a_{n}t^{n}$ and
$\sum_{n=0}^{\infty}\|a_{n}\||t|^{n}<\infty$ for all $t\in\mathbb{R}$. It
follows that
$\sigma^{A/I}_{t}\left(q_{I}(a)\right)=\sum_{n=0}^{\infty}q_{I}(a_{n})t^{n}$
and $\sum_{n=0}^{\infty}\left\|q_{I}(a_{n})\right\||t|^{n}<\infty$ for all
$t\in\mathbb{R}$, and hence that $q_{I}(a)\in\mathcal{A}_{\sigma^{A/I}}$ and

$\sigma^{A/I}_{z}\left(q_{I}(a)\right)=q_{I}\left(\sigma_{z}(a)\right)$
(5.4.1)

for all $z\in\mathbb{C}$. Since $\psi$ is a $\beta$-KMS weight for
$\sigma^{A/I}$ it follows that

$\displaystyle\psi\circ q_{I}\left(a^{*}a\right)=\psi(q_{I}(a)^{*}q_{I}(a))$
$\displaystyle=\psi\left(\sigma^{A/I}_{-i\frac{\beta}{2}}(q_{I}(a))\sigma^{A/I}_{-i\frac{\beta}{2}}(q_{I}(a))^{*}\right)$
$\displaystyle=\psi\left(q_{I}\left(\sigma_{-i\frac{\beta}{2}}(a)\right)q_{I}\left(\sigma_{-i\frac{\beta}{2}}(a)\right)^{*}\right)$
$\displaystyle=\psi\circ
q_{I}\left(\sigma_{-i\frac{\beta}{2}}(a)\sigma_{-i\frac{\beta}{2}}(a)^{*}\right).$

Thanks to Kustermans’ theorem, Theorem 2.3.1, this shows that $\psi\circ
q_{I}$ is a $\beta$-KMS weight for $\sigma$. ∎

We say that a flow $\sigma$ is trivial when $\sigma_{t}(a)=a$ for all
$t\in\mathbb{R}$ and all $a\in A$; otherwise $\sigma$ is said to be _non-
trivial_. For a trivial flow $\sigma$ the set of $\beta$-KMS weights is the
set of all lower semi-continuous traces on $A$ for any $\beta\in\mathbb{R}$,
as one easily deduces from Kustermans theorem, Theorem 2.3.1. If follows
therefore from Lemma 5.4.1 that if, in the setting of Lemma 5.4.1, the flow
$\sigma^{A/I}$ is trivial, all lower semi-continuous traces on $A/I$ give rise
to $\beta$-KMS weights for $\sigma$ for all $\beta\in\mathbb{R}$ by
composition with the quotient map $q_{I}$. KMS weights of this sort are
relatively easy to understand and we want to be able to push them aside.

######  Lemma 5.4.2.

Let $\psi$ be a KMS weight for $\sigma$. The following conditions are
equivalent.

  * (i)

$\psi$ is a $\beta$-KMS weight for $\sigma$ for all $\beta\in\mathbb{R}$.

  * (ii)

There are real numbers $\beta\neq\beta^{\prime}$ such that $\psi$ is both a
$\beta$-KMS weight for $\sigma$ and a $\beta^{\prime}$-KMS weight for
$\sigma$.

  * (iii)

The flow $\sigma^{A/\ker_{\psi}}$ is trivial and there is a lower semi-
continuous trace $\tau$ on $A/\ker_{\psi}$ such that $\psi=\tau\circ
q_{\ker_{\psi}}$.

  * (iv)

There is a $\sigma$-invariant ideal $I$ in $A$ such that $\sigma^{A/I}$ is the
trivial flow on $A/I$ and a lower semi-continuous trace $\tau$ on $A/I$ such
that $\psi=\tau\circ q_{I}$.

###### Proof.

We remind the reader that $\ker_{\psi}$ is the $\sigma$-invariant ideal in $A$
defined in Lemma 3.1.9. The implications (i) $\Rightarrow$ (ii) and (iii)
$\Rightarrow$ (iv) are trivial and the implication (iv) $\Rightarrow$ (i)
follows from Lemma 5.4.1. It remains therefore only to show that (ii)
$\Rightarrow$ (iii). Assume therefore that (ii) holds. By (2) of Theorem
2.3.1, $\psi(b^{*}\sigma_{i\beta}(a))=\psi(b^{*}\sigma_{i\beta^{\prime}}(a))$
for all $a,b\in\mathcal{M}_{\psi}^{\sigma}$, implying that

$\left<\Lambda_{\psi}(\sigma_{i(\beta-\beta^{\prime})}(a)),\Lambda_{\psi}(b)\right>=\left<\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>$

for all $a,b\in\mathcal{M}_{\psi}^{\sigma}$. Since
$\Lambda_{\psi}\left(\mathcal{M}^{\sigma}_{\psi}\right)$ is dense in
$H_{\psi}$ by Lemma 2.2.13 this implies that
$\Lambda_{\psi}(\sigma_{i(\beta-\beta^{\prime})}(a))=\Lambda_{\psi}(a)$, and
hence that

$\psi\left((\sigma_{i(\beta-\beta^{\prime})}(a)-a)^{*}(\sigma_{i(\beta-\beta^{\prime})}(a)-a)\right)=\left\|\Lambda_{\psi}(\sigma_{i(\beta-\beta^{\prime})}(a)-a)\right\|^{2}=0.$

Thus $\sigma_{i(\beta-\beta^{\prime})}(a)-a\in\ker_{\psi}$. Using (5.4.1) we
find that

$\sigma^{A/{\ker_{\psi}}}_{it+i(\beta-\beta^{\prime})}(q_{\ker_{\psi}}(a))=\sigma^{A/{\ker_{\psi}}}_{it}(q_{\ker_{\psi}}(\sigma_{i(\beta-\beta^{\prime})}(a)))=\sigma^{A/{\ker_{\psi}}}_{it}(q_{\ker_{\psi}}(a))$

for all $t\in\mathbb{R}$. Thus the function $\mathbb{R}\ni
t\mapsto\sigma^{A/{\ker_{\psi}}}_{it}(q_{\ker_{\psi}}(a))$ is
$\beta-\beta^{\prime}$-periodic and continuous, and hence norm-bounded. Since

$\left\|\sigma^{A/{\ker_{\psi}}}_{z}(q_{\ker_{\psi}}(a))\right\|=\left\|\sigma^{A/{\ker_{\psi}}}_{i\operatorname{Im}z}(q_{\ker_{\psi}}(a))\right\|$

it follows that the function
$z\mapsto\sigma^{A/{\ker_{\psi}}}_{z}(q_{\ker_{\psi}}(a))$ is norm-bounded.
Since it is also entire holomorphic it is therefore constant by Liouville’s
theorem. This shows that
$\sigma^{A/{\ker_{\psi}}}_{t}(q_{\ker_{\psi}}(a))=q_{\ker_{\psi}}(a)$ for all
$t\in\mathbb{R}$. The set $\mathcal{M}^{\sigma}_{\psi}$ is dense in $A$ by
Lemma 2.2.10 and consequently $q_{\ker_{\psi}}(\mathcal{M}^{\sigma}_{\psi})$
is dense in $A/{\ker_{\psi}}$. It follows that $\sigma^{A/{\ker_{\psi}}}$ is
the trivial flow. We claim that we can define a lower semi-continuous weight
$\tau:(A/{\ker_{\psi}})^{+}\to[0,\infty]$ such that

$\tau\left(q_{\ker_{\psi}}(a)\right)=\psi(a)$

when $A^{+}$. To see this, let $\omega\in\mathcal{F}_{\psi}$. Then
$\omega(a^{*}a)=\psi(a^{*}a)=0$ for all ${\ker_{\psi}}$. Let
$x\in{\ker_{\psi}}^{+}$. Then $\sqrt{x}\in{\ker_{\psi}}$ and hence
$\omega(x)=\omega(\sqrt{x}\sqrt{x})=0$. Since $\ker_{\psi}$ is spanned by
${\ker_{\psi}}^{+}$, it follows that $\omega|_{\ker_{\psi}}=0$, and we can
therefore define $\tilde{\omega}:A/{\ker_{\psi}}\to\mathbb{C}$ such that
$\tilde{\omega}\circ q_{\ker_{\psi}}=\omega$. Define
$\tau:(A/{\ker_{\psi}})^{+}\to[0,\infty]$ by

$\tau(x):=\sup\left\\{\tilde{\omega}(x):\
\omega\in\mathcal{F}_{\psi}\right\\}.$

Let $a\in A^{+}$. Then

$\displaystyle\tau(q_{\ker_{\psi}}(a))=\sup\left\\{\tilde{\omega}(q_{\ker_{\psi}}(a)):\
\omega\in\mathcal{F}_{\psi}\right\\}=\sup\left\\{{\omega}(a):\
\omega\in\mathcal{F}_{\psi}\right\\}=\psi(a)$

by Combes’ theorem, implying the claim. Note that $\tau$ is non-zero and
densely defined on $A/{\ker_{\psi}}$. It remains only to show that $\tau$ is a
trace. For this purpose note that since $\sigma^{A/{\ker_{\psi}}}$ is trivial
the algebra $\mathcal{M}^{\sigma^{A/{\ker_{\psi}}}}_{\tau}$ is the set
$\mathcal{M}_{\tau}$. By Theorem 2.3.1 it suffices to show that
$\tau(x^{*}x)=\tau(xx^{*})$ when $x\in\mathcal{M}_{\tau}$. Note that we can
choose $a\in\mathcal{M}_{\psi}$ such that $q_{\ker_{\psi}}(a)=x$. Using Lemma
2.2.6 we find

$\displaystyle\tau(x^{*}x)=\lim_{n\to\infty}\tau\left(R_{n}(x)^{*}R_{n}(x)\right)=\lim_{n\to\infty}\tau\left(R_{n}(q_{\ker_{\psi}}(a))^{*}R_{n}(q_{\ker_{\psi}}(a))\right)$
$\displaystyle=\lim_{n\to\infty}\tau\circ
q_{\ker_{\psi}}\left(R_{n}(a)^{*}R_{n}(a)\right)=\lim_{n\to\infty}\psi\left(R_{n}(a)^{*}R_{n}(a)\right)$
$\displaystyle=\lim_{n\to\infty}\psi\left(\sigma_{-i\frac{\beta}{2}}(R_{n}(a))\sigma_{-i\frac{\beta}{2}}(R_{n}(a))^{*}\right)$
$\displaystyle=\lim_{n\to\infty}\tau\circ
q_{\ker_{\psi}}\left(\sigma_{-i\frac{\beta}{2}}(R_{n}(a))\sigma_{-i\frac{\beta}{2}}(R_{n}(a))^{*}\right)$
$\displaystyle=\lim_{n\to\infty}\tau\left(q_{\ker_{\psi}}(\sigma_{-i\frac{\beta}{2}}(R_{n}(a)))q_{\ker_{\psi}}\left(\sigma_{-i\frac{\beta}{2}}(R_{n}(a))^{*}\right)\right)$
$\displaystyle=\lim_{n\to\infty}\tau\left(q_{\ker_{\psi}}(R_{n}(a))q_{\ker_{\psi}}\left(R_{n}(a)^{*}\right)\right)$
$\displaystyle=\lim_{n\to\infty}\tau\left(R_{n}(x)R_{n}(x)^{*}\right)$
$\displaystyle=\tau(xx^{*}).$

∎

######  Definition 5.4.3.

A KMS weight for $\sigma$ is _passive_ when it satisfies one of the equivalent
conditions of Lemma 5.4.2. A KMS weight which is not passive is said to be
_essential_.

All passive KMS weights are traces, but not conversely; there are many
examples of flows with $0$-KMS weights that are essential. We note that all
faithful KMS weights of non-trivial flows are essential, which in particular
implies that all KMS weights of non-trivial flows are essential when $A$ is
$\sigma$-simple in the sense that the only $\sigma$-invariant ideals in $A$
are $0$ and $A$. In contrast when $\sigma$ is a flow on an abelian
$C^{*}$-algebra all $\beta$-KMS weights for $\sigma$ with $\beta\neq 0$ are
passive because they are also automatically $0$-KMS weights and hence
satisfies condition (ii) in Lemma 5.4.2. See Section 6.2 for a little more on
KMS weights for flows on abelian $C^{*}$-algebras.

We note that for a given $\beta\in\mathbb{R}$ the set of passive $\beta$-KMS
weights constitute a face in the set of all $\beta$-KMS weights:

######  Lemma 5.4.4.

Assume that $\psi$ and $\phi$ are $\beta$-KMS weights for $\sigma$, that
$\psi$ is passive and that $\psi\geq\phi$. Then $\phi$ is passive.

###### Proof.

It follows from Theorem 5.2.7 that there is an element
$c\in\mathcal{Z}_{\psi}^{\sigma^{\prime\prime}}$, $0\leq c\leq 1$, such that
$\phi(\ \cdot\ )=\psi^{\prime\prime}(c\pi_{\psi}(\ \cdot\ ))$. Since $\psi$ is
passive it follows that $\psi$ is also a $(\beta+1)$-KMS weight for $\sigma$
and then a second application of Theorem 5.2.7 shows that $\phi$ is also a
$(\beta+1)$-KMS weight for $\sigma$. By Lemma 5.4.2 this implies that $\phi$
is passive. ∎

######  Notes and remarks 5.4.5.

The material in this section is new.

##  Chapter 6 Some examples and constructions

In this chapter we present a series of examples and constructions of KMS
weights for various flows. Although some parts of the chapter will be used in
the subsequent development the reader can easily skip the chapter and return
to it on an ad hoc basis.

###  6.1 Flows on the compact operators

Let $\mathbb{K}$ denote the $C^{*}$-algebra of compact operators on an
infinite-dimensional separable Hilbert space $\mathbb{H}$, and let $H$ be a
possibly unbounded self-adjoint operator on $\mathbb{H}$.

######  Lemma 6.1.1.

Let $k\in\mathbb{K}$. The map $\mathbb{R}\ni t\mapsto e^{itH}k$ is continuous
for the norm topology.

###### Proof.

The map $t\mapsto e^{itH}$ is continuous for the strong operator topology by
Theorem 5.6.36 in [KR] and since $k$ is compact the image under $k$ of the
unit ball in $\mathbb{H}$ is compact in $\mathbb{H}$. It follows that the
continuity of $t\mapsto e^{itH}k\psi$ is uniform for $\psi$ in the unit ball
of $\mathbb{H}$; i.e. $t\mapsto e^{itH}k$ is continuous for the norm topology.
∎

Thanks to Lemma 6.1.1 we can define a flow $\sigma$ on $\mathbb{K}$ such that

$\sigma_{t}(k):=e^{itH}ke^{-itH}\ \ \forall t\in\mathbb{R},\ k\in\mathbb{K}.$
(6.1.1)

It is a well-known fact that all flows on $\mathbb{K}$ are of this form. A
proof of this can be found in Appendix B.

The aim here is to find the KMS weights for $\sigma$. Let
$\operatorname{Tr}:B(\mathbb{H})^{+}\to[0,\infty]$ denote the usual trace
given by the formula

$\operatorname{Tr}(m)=\sum_{i=1}^{\infty}\left<m\psi_{i},\psi_{i}\right>,$

where $\\{\psi_{i}\\}_{i=1}^{\infty}$ is an orthonormal basis in $\mathbb{H}$.
In the following we shall rely on properties of trace-class operators and
Hilbert-Schmidt operators on $\mathbb{H}$, and refer to Chapter VI in [RS] for
the facts we shall need. In particular, we note that
$\mathcal{N}_{\operatorname{Tr}}$ by definition is the ideal of Hilbert-
Schmidt operators and that the combination of (b) of Lemma 1.0.1 with (h) of
Theorem VI.22 in [RS] shows that $\mathcal{M}_{\operatorname{Tr}}$ is the
ideal of trace class operators.

Let $P$ be the projection valued measure on $\mathbb{R}$ representing $H$;

$H=\int_{\mathbb{R}}\lambda\ \mathrm{d}P_{\lambda},$

and set $P_{n}:=P[-n,n]$. Note that

$\displaystyle\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}aP_{n}e^{-\frac{\beta}{2}H})=\operatorname{Tr}(\sqrt{a}P_{n}e^{-\beta
H}\sqrt{a})$ $\displaystyle\leq\operatorname{Tr}(\sqrt{a}P_{n+1}e^{-\beta
H}\sqrt{a})=\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n+1}aP_{n+1}ne^{-\frac{\beta}{2}H})$

for $a\in\mathbb{K}^{+}$; that is,

$n\mapsto\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}aP_{n}e^{-\frac{\beta}{2}H})$

is non-decreasing and we can therefore define
$\psi_{\beta}:\mathbb{K}^{+}\to[0,\infty]$ by

$\psi_{\beta}(a):=\lim_{n\to\infty}\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}aP_{n}e^{-\frac{\beta}{2}H}).$
(6.1.2)

It follows easily from the properties of $\operatorname{Tr}$ that
$\psi_{\beta}:\mathbb{K}^{+}\to[0,\infty]$ is a $\sigma$-invariant weight.

######  Lemma 6.1.2.

$\psi_{\beta}$ is densely defined.

###### Proof.

By spectral theory $\lim_{n\to\infty}P_{n}=1$ in the strong operator topology,
implying that $\lim_{n\to\infty}P_{n}k=k$ in norm for all $k\in\mathbb{K}$. In
particular, $\lim_{k\to\infty}P_{k}aP_{k}=a$ when $a$ is a positive trace
class operator. Note that

$\displaystyle\psi_{\beta}(P_{k}aP_{k})=\lim_{n\to\infty}\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}P_{k}aP_{k}P_{n}e^{-\frac{\beta}{2}H})$
$\displaystyle=\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{k}aP_{k}e^{-\frac{\beta}{2}H})<\infty,$

since $e^{-\frac{\beta}{2}H}P_{k}$ is bounded and the trace class operators
form a two-sided ideal in $B(\mathbb{H})$. The lemma follows now because this
ideal is dense in $\mathbb{K}$. ∎

######  Lemma 6.1.3.

$\psi_{\beta}(a^{*}a)=\begin{cases}\operatorname{Tr}(e^{-\frac{\beta}{2}H}a^{*}ae^{-\frac{\beta}{2}H})&\
\text{when $ae^{-\frac{\beta}{2}H}$ is Hilbert-Schmidt},\\\ \infty&\ \
\text{otherwise.}\end{cases}$ (6.1.3)

###### Proof.

Assume $\psi_{\beta}(a^{*}a)<\infty$. Then
$\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H})<\infty$
and hence $aP_{n}e^{-\frac{\beta}{2}H}$ is Hilbert-Schmidt for all $n$. Let
$\left\|\ \cdot\ \right\|_{HS}$ denote the Hilbert-Schmidt norm;

$\left\|b\right\|_{HS}:=\operatorname{Tr}(b^{*}b)^{\frac{1}{2}}.$

Then, if $m\geq n$,

$\begin{split}&\left\|aP_{n}e^{-\frac{\beta}{2}H}-aP_{m}e^{-\frac{\beta}{2}H}\right\|_{HS}^{2}\\\
&=\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)+\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{m}e^{-\frac{\beta}{2}H}\right)\\\
&\ \ \ \ \ \
-\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)-\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{m}e^{-\frac{\beta}{2}H}\right)\\\
&=\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)+\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{m}e^{-\frac{\beta}{2}H}\right)\\\
&\ \ \ \ \ \
-\operatorname{Tr}\left(P_{m}e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{n}e^{-\frac{\beta}{2}H}P_{n}\right)-\operatorname{Tr}\left(P_{n}e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{m}e^{-\frac{\beta}{2}H}P_{m}\right)\\\
&=\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)+\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{m}e^{-\frac{\beta}{2}H}\right)\\\
&\ \ \ \ \ \
-\operatorname{Tr}\left(P_{n}P_{m}e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)-\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{m}e^{-\frac{\beta}{2}H}P_{m}P_{n}\right)\\\
&=\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{\frac{-\beta}{2}H}\right)+\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{m}e^{-\frac{\beta}{2}H}\right)\\\
&\ \ \ \ \ \
-\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)-\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right)\\\
&=\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{m}a^{*}aP_{m}e^{-\frac{\beta}{2}H}\right)-\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H}\right).\end{split}$
(6.1.4)

Since
$\lim_{k\to\infty}\operatorname{Tr}\left(e^{-\frac{\beta}{2}H}P_{k}a^{*}aP_{k}e^{-\frac{\beta}{2}H}\right)=\psi_{\beta}(a^{*}a)<\infty$
this equality implies that $\\{aP_{n}e^{-\frac{\beta}{2}H}\\}$ is Cauchy in
the Hilbert-Schmidt norm and converges therefore to a Hilbert-Schmidt operator
in that norm, and therefore also in the operator norm. By spectral theory
$\lim_{n\to\infty}aP_{n}e^{-\frac{\beta}{2}H}\psi=ae^{-\frac{\beta}{2}H}\psi$
for all $\psi\in D(e^{-\frac{\beta}{2}H})$ and it follows therefore that
$ae^{-\frac{\beta}{2}H}$ is not only bounded, but in fact a Hilbert-Schmidt
operator. Furthermore, since
$\lim_{n\to\infty}\left\|aP_{n}e^{-\frac{\beta}{2}H}-ae^{-\frac{\beta}{2}H}\right\|_{HS}=0$
it follows that

$\psi_{\beta}(a^{*}a)=\operatorname{Tr}(e^{-\frac{\beta}{2}H}a^{*}ae^{-\frac{\beta}{2}H}).$

Conversely, assuming that $ae^{-\frac{\beta}{2}H}$ is Hilbert-Schmidt we have
that

$\lim_{n\to\infty}P_{n}e^{-\frac{\beta}{2}H}a^{*}=e^{-\frac{\beta}{2}H}a^{*}$

in operator norm since $\lim_{n\to\infty}P_{n}=1$ in the strong operator
topology and $e^{-\frac{\beta}{2}H}a^{*}$ is compact. Hence
$\\{aP_{n}e^{-\beta H}a^{*}\\}$ converges increasingly and in norm to
$ae^{-\beta H}a^{*}$. By using the lower semi-continuity of
$\operatorname{Tr}$ we get

$\displaystyle\psi_{\beta}(a^{*}a)=\lim_{n\to\infty}\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}a^{*}aP_{n}e^{-\frac{\beta}{2}H})=\lim_{n\to\infty}\operatorname{Tr}(aP_{n}e^{-\beta
H}a^{*})$ $\displaystyle=\operatorname{Tr}(ae^{-\beta
H}a^{*})=\operatorname{Tr}(e^{-\frac{\beta}{2}H}a^{*}ae^{-\frac{\beta}{2}H}).$

∎

######  Lemma 6.1.4.

$\psi_{\beta}$ is a $\beta$-KMS weight for $\sigma$.

###### Proof.

It follows from Lemma 6.1.3 that

$\mathcal{N}_{\psi_{\beta}}=\left\\{a\in\mathbb{K}:\ ae^{-\frac{\beta}{2}H}\
\text{ is Hilbert-Schmidt}\right\\}.$

Since the Hilbert-Schmidt operators form a two-sided ideal in $B(\mathbb{H})$
is follows that
$P_{n}e^{izH}\mathcal{N}_{\psi_{\beta}}\subseteq\mathcal{N}_{\psi_{\beta}}$
and
$\mathcal{N}_{\psi_{\beta}}P_{n}e^{izH}\subseteq\mathcal{N}_{\psi_{\beta}}$
for all $z\in\mathbb{C}$ and hence

$e^{izH}P_{n}\mathcal{M}_{\psi_{\beta}}P_{n}e^{-izH}\subseteq\mathcal{M}_{\psi_{\beta}}$
(6.1.5)

for all $z\in\mathbb{C}$ since
$\mathcal{M}_{\psi_{\beta}}=\operatorname{Span}\mathcal{N}_{\psi_{\beta}}^{*}\mathcal{N}_{\psi_{\beta}}$
by (b) of Lemma 1.0.1. Let $a\in\mathbb{K}$. Then

$z\mapsto e^{izH}P_{n}aP_{n}e^{-izH}$

is entire holomorphic and agrees with $\sigma_{t}(P_{n}aP_{n})$ for
$t\in\mathbb{R}$. It follows therefore that
$P_{n}aP_{n}\in\mathcal{A}_{\sigma}$ and

$\sigma_{z}(P_{n}aP_{n})=e^{izH}P_{n}aP_{n}e^{-izH}=P_{n}e^{izH}ae^{-izH}P_{n}$

for all $z\in\mathbb{C}$. It follows from this and (6.1.5) that

$P_{n}\mathcal{M}^{\sigma}_{\psi_{\beta}}P_{n}\subseteq\mathcal{M}^{\sigma}_{\psi_{\beta}}.$

In particular $P_{n}\mathcal{M}^{\sigma}_{\psi_{\beta}}P_{n}\subseteq
P_{n+1}\mathcal{M}^{\sigma}_{\psi_{\beta}}P_{n+1}$, implying that

$S:=\bigcup_{n}P_{n}\mathcal{M}^{\sigma}_{\psi_{\beta}}P_{n}$

is a subspace of $\mathcal{M}^{\sigma}_{\psi_{\beta}}$. We claim that $S$ has
the properties required of $S$ in Theorem 2.4.6. To see this, let
$b\in\mathcal{M}^{\sigma}_{\psi_{\beta}}$. We will show that

$\lim_{n\to\infty}P_{n}bP_{n}=b$ (6.1.6)

and

$\lim_{n\to\infty}\Lambda_{\psi_{\beta}}(P_{n}bP_{n})=\Lambda_{\psi_{\beta}}(b).$
(6.1.7)

As pointed out in the proof of Lemma 6.1.2, $\lim_{n\to\infty}P_{n}k=k$ in
norm for all $k\in\mathbb{K}$. (6.1.6) follows from this because
$\mathcal{M}_{\psi_{\beta}}$ is spanned by its positive elements. To establish
(6.1.7), note that

$\begin{split}&\left\|\Lambda_{\psi_{\beta}}(P_{n}bP_{n})-\Lambda_{\psi_{\beta}}(b)\right\|\\\
&\leq\left\|\Lambda_{\psi_{\beta}}(P_{n}bP_{n})-\Lambda_{\psi_{\beta}}(P_{n}b)\right\|+\left\|\Lambda_{\psi_{\beta}}(P_{n}b)-\Lambda_{\psi_{\beta}}(b)\right\|\\\
&=\left\|\Lambda_{\psi_{\beta}}(P_{n}b(1-P_{n}))\right\|+\left\|\Lambda_{\psi_{\beta}}((1-P_{n})b)\right\|,\end{split}$
(6.1.8)
$\begin{split}&\left\|\Lambda_{\psi_{\beta}}(P_{n}b(1-P_{n}))\right\|^{2}=\operatorname{Tr}(e^{-\frac{\beta}{2}H}(1-P_{n})b^{*}P_{n}b(1-P_{n})e^{-\frac{\beta}{2}H})\\\
&\leq\operatorname{Tr}(e^{-\frac{\beta}{2}H}(1-P_{n})b^{*}b(1-P_{n})e^{-\frac{\beta}{2}H})\\\
&=\operatorname{Tr}(be^{-\beta H}(1-P_{n})b^{*})\\\
&=\operatorname{Tr}(be^{-\beta H}b^{*})-\operatorname{Tr}(be^{-\beta
H}P_{n}b^{*})\end{split}$ (6.1.9)

and

$\begin{split}&\left\|\Lambda_{\psi_{\beta}}((1-P_{n})b)\right\|^{2}=\operatorname{Tr}(e^{-\frac{\beta}{2}H}b^{*}(1-P_{n})be^{-\frac{\beta}{2}H})\\\
&=\operatorname{Tr}(e^{-\frac{\beta}{2}H}b^{*}be^{-\frac{\beta}{2}H})-\operatorname{Tr}(e^{-\frac{\beta}{2}H}b^{*}P_{n}be^{-\frac{\beta}{2}H}).\end{split}$
(6.1.10)

The sequence $\\{be^{-\beta H}P_{n}b^{*}\\}$ increases with $n$ and converges
in norm to $be^{-\beta H}b^{*}$ because $be^{-\frac{\beta}{2}H}$ and
$e^{-\frac{\beta}{2}H}b^{*}$ are both Hilbert-Schmidt. Similarly, the sequence
$\\{e^{-\frac{\beta}{2}H}b^{*}P_{n}be^{-\frac{\beta}{2}H}\\}$ increases with
$n$ and converges in norm to
$e^{-\frac{\beta}{2}H}b^{*}be^{-\frac{\beta}{2}H}$. It follows therefore from
the lower semi-continuity of $\operatorname{Tr}$ and (6.1.9) and (6.1.10) that

$\lim_{n\to\infty}\left\|\Lambda_{\psi_{\beta}}(P_{n}b(1-P_{n}))\right\|=\lim_{n\to\infty}\left\|\Lambda_{\psi_{\beta}}((1-P_{n})b)\right\|=0.$

Inserted into (6.1.8) it follows that
$\lim_{n\to\infty}\Lambda_{\psi_{\beta}}(P_{n}bP_{n})=\Lambda_{\psi_{\beta}}(b)$
as desired.

To deduce from Theorem 2.4.6 that $\psi_{\beta}$ is a $\beta$-KMS weight for
$\sigma$ it suffices now to check that
$\psi_{\beta}(x^{*}x)=\psi_{\beta}\left(\sigma_{-i\frac{\beta}{2}}(x)\sigma_{-i\frac{\beta}{2}}(x)^{*}\right)$
for $x\in S$. Let $x=P_{n}aP_{n}$, where $a\in\mathcal{M}_{\psi_{\beta}}$.
Then

$\sigma_{-i\frac{\beta}{2}}(x)=e^{\frac{\beta}{2}P_{n}H}P_{n}aP_{n}e^{-\frac{\beta}{2}P_{n}H}$

and hence

$\displaystyle\psi_{\beta}\left(\sigma_{-i\frac{\beta}{2}}(x)\sigma_{-i\frac{\beta}{2}}(x)^{*}\right)$
$\displaystyle=\operatorname{Tr}(e^{-\frac{\beta}{2}H}e^{\frac{\beta}{2}P_{n}H}P_{n}aP_{n}e^{-\beta
P_{n}H}P_{n}a^{*}P_{n}e^{\frac{\beta}{2}P_{n}H}e^{-\frac{\beta}{2}H})$
$\displaystyle=\operatorname{Tr}(P_{n}aP_{n}e^{-\beta
H}a^{*}P_{n})=\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}a^{*}P_{n}aP_{n}e^{-\frac{\beta}{2}H})=\psi_{\beta}(x^{*}x).$

∎

######  Lemma 6.1.5.

$\psi_{\beta}$ is the only $\beta$-KMS weight for $\sigma$, up to
multiplication by scalars.

###### Proof.

Let $\psi$ be a $\beta$-KMS weight for $\sigma$. Choose $n$ so big that
$P_{n}\neq 0$. Then $P_{n}\mathbb{K}P_{n}$ is a $\sigma$-invariant hereditary
$C^{*}$-subalgebra of $\mathbb{K}$ which is full in $\mathbb{K}$ since
$\mathbb{K}$ is simple. It is easy to see that the map

$(P_{n}\mathbb{K}P_{n})^{+}\ni
T\mapsto\psi(e^{\frac{\beta}{2}H}Te^{\frac{\beta}{2}H})$ (6.1.11)

is a $\sigma$-invariant weight on $P_{n}\mathbb{K}P_{n}$ since $\psi$ is a
$\sigma$-invariant weight on $\mathbb{K}$. To see that it is a trace on
$P_{n}\mathbb{K}P_{n}$, let $T\in P_{n}\mathbb{K}P_{n}$. Then
$Te^{\frac{\beta}{2}H}\in\mathcal{A}_{\sigma}$ and
$\sigma_{-i\frac{\beta}{2}}(Te^{\frac{\beta}{2}H})=e^{\frac{\beta}{2}H}T$.
Since $\psi$ is a $\beta$-KMS weight for $\sigma$ this gives

$\displaystyle\psi(e^{\frac{\beta}{2}H}T^{*}Te^{\frac{\beta}{2}H})=\psi(\sigma_{-i\frac{\beta}{2}}(Te^{\frac{\beta}{2}H})\sigma_{-i\frac{\beta}{2}}(Te^{\frac{\beta}{2}H})^{*})=\psi(e^{\frac{\beta}{2}H}TT^{*}e^{\frac{\beta}{2}H}).$

This shows that (6.1.11) is a $\sigma$-invariant lower semi-continuous trace
on $P_{n}\mathbb{K}P_{n}$, i.e. a $0$-KMS weight for $\sigma$ on
$P_{n}\mathbb{K}P_{n}$. Applied to $0$-KMS weights, it follows from Theorem
3.0.1 and the essential uniqueness of the trace on $\mathbb{K}$ that
$\psi(e^{\frac{\beta}{2}H}\ \cdot\
e^{\frac{\beta}{2}H})=\lambda\operatorname{Tr}(\ \cdot\ )$ on
$P_{n}\mathbb{K}P_{n}$ for some $\lambda>0$. It follows that

$\displaystyle\psi(K)=\psi(e^{\frac{\beta}{2}H}e^{-\frac{\beta}{2}H}Ke^{-\frac{\beta}{2}H}\
e^{\frac{\beta}{2}H})$
$\displaystyle=\lambda\operatorname{Tr}(e^{-\frac{\beta}{2}H}Ke^{-\frac{\beta}{2}H})=\lambda\psi_{\beta}(K)$

for all $K\in(P_{n}\mathbb{K}P_{n})^{+}$. By using the uniqueness statement in
Theorem 3.0.1 again, this time applied to $\beta$-KMS weights, it follows that
$\psi=\lambda\psi_{\beta}$. ∎

To decide when $\psi_{\beta}$ is bounded we use the following lemma.

######  Lemma 6.1.6.

Let $\varphi:A^{+}\to[0,\infty]$ be a weight on the $C^{*}$-algebra $A$ and
let $\\{u_{i}\\}_{i\in I}$ an approximate unit for $A$ such that $0\leq
u_{i}\leq 1$ for all $i$. Then $\varphi$ is bounded if and only if there is
$M>0$ such that $\varphi(u_{i})\leq M$ for all $i$.

###### Proof.

Assume that such an $M$ exists, and consider $a\in A$ such that $0\leq a\leq
1$. For each $n\in\mathbb{N}$ we pick $i_{n}\in I$ such that
$\left\|u_{i_{n}}\sqrt{a}-\sqrt{a}\right\|\leq\frac{1}{n}$. Then
$\lim_{n\to\infty}u_{i_{n}}au_{i_{n}}=a$ and hence the lower semi-continuity
of $\varphi$ entails that

$\varphi(a)\leq\liminf_{n}\varphi(u_{i_{n}}au_{i_{n}})\leq\liminf_{n}\varphi(u_{i_{n}}^{2})\leq\liminf_{n}\varphi(u_{i_{n}})\leq
M.$

It follows that $\varphi$ is bounded. The reverse implication follows from
Lemma 2.4.12. ∎

Let $\\{E_{k}\\}$ be an approximate unit in $\mathbb{K}$ consisting of
projections. By Lemma 6.1.6 $\psi_{\beta}$ is bounded if and only if there is
$M>0$ such that $\psi_{\beta}(E_{k})\leq M$ for all $k$. Assume that such an
$M$ exists. Then
$\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}E_{k}P_{n}e^{-\frac{\beta}{2}H})\leq\psi_{\beta}(E_{k})\leq
M$ for all $k$ and by lower semi-continuity of $\operatorname{Tr}$ with
respect to the strong operator topology it follows that
$\operatorname{Tr}(e^{-\frac{\beta}{2}H}P_{n}e^{-\frac{\beta}{2}H})\leq M$ for
all $n$. In particular, $P_{n}e^{-\frac{\beta}{2}H}$ is Hilbert-Schmidt for
all $n$ and the calculation (6.1.4) with $a=1$ shows that
$\\{P_{n}e^{-\frac{\beta}{2}H}\\}$ converges in the Hilbert-Schmidt norm,
implying that $e^{-\frac{\beta}{2}H}$ is Hilbert-Schmidt, or alternatively,
that $e^{-\beta H}$ is trace-class. Conversely, if $e^{-\beta H}$ is trace-
class it follows that $\psi_{\beta}(E_{k})\leq\operatorname{Tr}(e^{-\beta H})$
for all $k$, and $\psi_{\beta}$ is bounded.

We summarize the results of this section in the following

######  Theorem 6.1.7.

For each $\beta\in\mathbb{R}$ the flow $\sigma$, given by (6.1.1), has up to
multiplication by scalars a unique $\beta$-KMS weight $\psi_{\beta}$ given by
(6.1.3). It is bounded if and only if $e^{-\beta H}$ is trace-class.

Set

$C_{H}=\left\\{\beta\in\mathbb{R}:\ \operatorname{Tr}(e^{-\beta
H})<\infty\right\\}.$

By Theorem 6.1.7 this is the set of real numbers $\beta$ for which the flow
$\sigma$ on $\mathbb{K}$ defined by $H$ has a bounded $\beta$-KMS weight.

######  Lemma 6.1.8.

The set $C_{H}$ is one of the following:

  * •

$C_{H}=\emptyset$,

  * •

$C_{H}=[r,\infty)$ for some $r>0$,

  * •

$C_{H}=]r,\infty[$ for some $r\geq 0$,

  * •

$C_{H}=(-\infty,r[$ for some $r\leq 0$, or

  * •

$C_{H}=(-\infty,r]$ for some $r<0$,

and all possibilities occur.

###### Proof.

Unless $H=0$ the function $\beta\mapsto\operatorname{Tr}(e^{-\beta H})$ is
strictly convex on $C_{H}$ since the exponential function is strictly convex
and $\operatorname{Tr}$ is faithful. It follows therefore that $C_{H}$ is an
interval. Since $0\notin C_{H}$, because $\operatorname{Tr}$ is unbounded, the
interval $C_{H}$ is contained in $]0,\infty)$ or $(-\infty,0[$. Since
$C_{H}=-C_{-H}$, to prove the first part of the statement, it suffices to show

$0<\beta\in C_{H},\ \beta<\beta^{\prime}\Rightarrow\beta^{\prime}\in C_{H}.$

As above we let $P$ be the projection valued measure on $\mathbb{R}$
representing $H$ and set $P_{+}=P[0,\infty)$ and $P_{-}=P(-\infty,0)$. Since
$e^{-\beta H}$ is trace class and $e^{-\beta H}=e^{-\beta H}P_{+}+e^{-\beta
H}P_{-}\geq P_{-}$, it follows that $P_{-}$ is finite dimensional and hence
$e^{-\beta H}P_{-}$ and $e^{-\beta^{\prime}H}P_{-}$ are both of trace class.
Since $e^{-\beta^{\prime}H}P_{+}\leq e^{-\beta H}P_{+}$, it follows that
$e^{-\beta^{\prime}H}=e^{-\beta^{\prime}H}P_{+}+e^{-\beta^{\prime}H}P_{-}$ is
trace class, i.e. $\beta^{\prime}\in C_{H}$.

To show that all possibilities occur observe that $C_{H}=\emptyset$ when
$H=0$. Let $\\{a_{n}\\}_{n=1}^{\infty}$ be a sequence of real numbers and
$\\{\psi_{n}\\}_{n=1}^{\infty}$ an orthonormal basis in $\mathbb{H}$. We can
then define a self-adjoint operator $H$ on $\mathbb{H}$ such that

$H\psi_{n}=a_{n}\psi_{n}.$

In this case,

$C_{H}=\left\\{\beta\in\mathbb{R}:\ \sum_{n=1}^{\infty}e^{-\beta
a_{n}}<\infty\right\\}.$

For each $r\geq 0$ it is possible to construct a sequence $\\{b_{n}\\}$ of
positive numbers such that $\sum_{n=1}^{\infty}b_{n}^{\beta}<\infty$ if and
only if $\beta>r$ so if we use $a_{n}=\log b_{n}$ to define $H$ we get
$C_{H}=(r,\infty[$. If $r>0$ it is also possible to construct a sequence
$\\{b^{\prime}_{n}\\}$ of positive numbers such that
$\sum_{n=1}^{\infty}{b^{\prime}_{n}}^{\beta}<\infty$ if and only if $\beta\geq
r$ and when we then use $a_{n}=\log b^{\prime}_{n}$ to define $H$ we get
$C_{H}=[r,\infty[$. The intervals $]-\infty,-r)$ and $]-\infty,r]$ are then
realized by using the sequences $\\{-a_{n}\\}$ instead of $\\{a_{n}\\}$. ∎

###  6.2 Flows on abelian $C^{*}$-algebras

Given a locally compact Hausdorff space $X$ we denote by $C_{0}(X)$ the
$C^{*}$-algebra of continuous functions on $X$ that vanish at infinity. A
_flow_ on $X$ is a continuous representation
$\phi=(\phi_{t})_{t\in\mathbb{R}}$ of $\mathbb{R}$ by homeomorphisms. In more
detail, each $\phi_{t}$ is a homeomorphism $\phi_{t}:X\to X$, and

  * •

$\phi_{s}(\phi_{t}(x))=\phi_{t+s}(x)$ for all $s,t\in\mathbb{R}$ and all $x\in
X$, and

  * •

the map $\mathbb{R}\times X\ni(t,x)\mapsto\phi_{t}(x)\in X$ is continuous.

A flow $\phi$ on $X$ gives rise to a flow $\sigma^{\phi}$ on $C_{0}(X)$
defined such that $\sigma^{\phi}_{t}(f)=f\circ\phi_{t}$, and all flows on
$C_{0}(X)$ arises in this way. See Section 2.1 of [Wi].

######  Lemma 6.2.1.

Let $X$ be a second countable locally compact Hausdorff space and let
$\psi:C_{0}(X)^{+}\to[0,\infty]$ be a densely defined weight on $C_{0}(X)$.
There is a regular measure $\mu$ on $X$ such that

$\psi(f)=\int_{X}f\ \mathrm{d}\mu$ (6.2.1)

for all $f\in C_{0}(X)^{+}$.

###### Proof.

Since we assume that $X$ is second countable the $C^{*}$-algebra $C_{0}(X)$ is
separable and it follows therefore from Theorem 1.2.1 that there is a sequence
$\\{\omega_{n}\\}_{n=1}^{\infty}$ of positive functionals on
$C_{0}(\mathbb{R})$ such that

$\psi(f)=\sum_{n=1}^{\infty}\omega_{n}(f)$

for all $f\in C_{0}(X)^{+}$. By the Riesz representation theorem, [Ru0], there
are bounded Borel measures $\mu_{n}$ on $\mathbb{R}$ such that

$\omega_{n}(f)=\int_{\mathbb{R}}f\ \mathrm{d}\mu_{n}\ \ \forall f\in
C_{0}(\mathbb{R}).$

Define a Borel measure $\mu$ on $\mathbb{R}$ such that

$\mu(A)=\sum_{n=1}^{\infty}\mu_{n}(A)$

for all Borel sets $A\subseteq\mathbb{R}$. For any $f\in
C_{0}(\mathbb{R})^{+}$, by approximating $f$ by an increasing sequence of
simple functions it follows that

$\int_{\mathbb{R}}f\ \mathrm{d}\mu=\sum_{n=1}^{\infty}\int_{\mathbb{R}}f\
\mathrm{d}\mu_{n}.$

Hence $\int_{\mathbb{R}}f\
\mathrm{d}\mu=\sum_{n=1}^{\infty}\omega_{n}(f)=\psi(f)$. To show that $\mu$ is
regular, let $K\subseteq X$ be a compact subset. By Urysohn’s lemma there is a
continuous function $f:X\to[0,2]$ of compact support such that $f(x)=2$ for
all $x\in K$. Since $\psi$ is densely defined there is a $g\in C_{0}(X)^{+}$
such that $\psi(g)<\infty$ and $\left\|f-g\right\|\leq\frac{1}{2}$. Then
$g(x)\geq 1$ for all $x\in K$ and hence

$\mu(K)\leq\int_{X}g\ \mathrm{d}\mu=\psi(g)<\infty.$

This shows that $\mu$ is regular. ∎

When $\phi$ is a flow on $X$ and $\sigma^{\phi}$ is the associated flow on
$C_{0}(X)$, the $\sigma^{\phi}$-invariant densely defined weights are given by
integration with respect to regular Borel measures $\mu$ on $X$ that are
$\phi$-invariant in the sense that $\mu\circ\phi_{t}=\mu$ for all
$t\in\mathbb{R}$. Hence the $0$-KMS weights for $\sigma^{\phi}$ are in
bijective correspondence with the non-zero regular $\phi$-invariant measures
on $X$ via the equation (6.2.1). As pointed out after Definition 5.4.3, when
$\beta\neq 0$ all $\beta$-KMS weights are passive. In terms of the flow $\phi$
they arise from regular measures on the set of fixed points for $\phi$. In
particular essential $\beta$-KMS weights correspond to $\phi$-invariant
regular measures that are not concentrated on the fixed points of $\phi$.

To summarize, for $\beta\neq 0$ the $\beta$-KMS weights all arise from
integration with respect to a regular Borel measure on $\left\\{x\in X:\
\phi_{t}(x)=x\ \forall t\in\mathbb{R}\right\\}$ while the $0$-KMS weights
arise from integration with respect to a general regular $\phi$-invariant
Borel measure. In particular, the set of $\beta$-KMS weights is the same for
all $\beta\neq 0$ while the set of $0$-KMS weights is a larger set in general.

###  6.3 Exterior equivalence

In this section we shall use the multiplier algebra $M(A)$ of a
$C^{*}$-algebra $A$ and the strict topology on $M(A)$. The reader may find the
definitions in Appendix C.

Two flows $\sigma$ and $\sigma^{\prime}$ on the same $C^{*}$-algebra $A$ are
_exterior equivalent_ when there is a unitary representation
$u=(u_{t})_{t\in\mathbb{R}}$ of $\mathbb{R}$ by multipliers of $M(A)$ such
that

  * (a)

$\mathbb{R}\ni t\mapsto u_{t}$ is strictly continuous,

  * (b)

$u_{s+t}=u_{s}\sigma_{s}(u_{t})$ for all $s,t\in\mathbb{R}$, and

  * (c)

$\sigma^{\prime}_{t}=\operatorname{Ad}u_{t}\circ\sigma_{t}$ for all
$t\in\mathbb{R}$,

cf. Section 2.2 of [Co1] or Section 8.11 in [Pe].

######  Theorem 6.3.1.

Let $\sigma$ and $\sigma^{\prime}$ be exterior equivalent flows on $A$. For
each $\beta\in\mathbb{R}$ there is an affine bijection
$\operatorname{KMS}(\sigma,\beta)\simeq\operatorname{KMS}(\sigma^{\prime},\beta)$.

###### Proof.

Let $M_{2}(A)$ denote the $C^{*}$-algebra of $2\times 2$ matrices over $A$.
Define the flow $\gamma$ on $M_{2}(A)$ such that

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

cf. Lemme 2.2.6 of [Co1] or [Pe]. The embeddings $\iota_{i}:A\to M_{2}(A),\
i=1,2$, defined by

$\iota_{1}(a)=\left(\begin{smallmatrix}a&0\\\ 0&0\end{smallmatrix}\right)$

and

$\iota_{2}(a)=\left(\begin{smallmatrix}0&0\\\ 0&a\end{smallmatrix}\right)$

identify $A$ as a $C^{*}$-subalgebra of $M_{2}(A)$ in different ways. Note
that $\iota_{1}(A)$ and $\iota_{2}(A)$ are both hereditary and full
$C^{*}$-subalgebras of $M_{2}(A)$. Since $\iota_{i}$ is equivariant and a
$*$-isomorphism onto its image it follows from Lemma 2.4.8 and Theorem 3.1.8
that $\psi\mapsto\psi\circ\iota_{1}$ is a bijection
$\operatorname{KMS}(\gamma,\beta)\simeq\operatorname{KMS}(\sigma,\beta)$ and
$\psi\mapsto\psi\circ\iota_{2}$ is a bijection
$\operatorname{KMS}(\gamma,\beta)\simeq\operatorname{KMS}(\sigma^{\prime},\beta)$.
Hence
$\operatorname{KMS}(\sigma,\beta)\simeq\operatorname{KMS}(\gamma,\beta)\simeq\operatorname{KMS}(\sigma^{\prime},\beta)$.

∎

####  6.3.1 A special case of Theorem 6.3.1

Let $\sigma$ be flow on $A$ and $u=(u_{t})_{t\in\mathbb{R}}$ a strictly
continuous unitary representation of $\mathbb{R}$ by multipliers of $M(A)$
such that $\sigma_{t}(u_{s})=u_{s}$ for all $s,t$. Then

$\sigma^{\prime}_{t}:=\operatorname{Ad}u_{t}\circ\sigma_{t}$

is a flow on $A$ exterior equivalent to $\sigma$. In this setting we can give
an explicit description of the bijection
$\operatorname{KMS}(\sigma,\beta)\simeq\operatorname{KMS}(\sigma^{\prime},\beta)$
of Theorem 6.3.1. For this we consider $A$ as a non-degenerate
$C^{*}$-subalgebra of $B(\mathbb{H})$, the $C^{*}$-algebra of bounded
operators on the Hilbert space $\mathbb{H}$. Here non-degenerate means that
$\left\\{a\xi:\ a\in A,\ \xi\in H\right\\}$ spans a dense subspace of
$\mathbb{H}$. We can then identify $M(A)$ as

$M(A)=\left\\{m\in B(\mathbb{H}):\ mA\subseteq A,\ Am\subseteq A\right\\};$

cf. Lemma C.0.5. Then $u$ is a strongly continuous one-parameter group of
unitaries on $\mathbb{H}$ and by Stone’s theorem there is a self-adjoint
operator $H$ on $\mathbb{H}$ such that

$u_{t}=e^{itH}.$

In the following we denote by $C_{0}(\mathbb{R})$ the algebra of continuous
functions on $\mathbb{R}$ that vanish at infinity and by $C_{c}(\mathbb{R})$
the sub-algebra of continuous functions on $\mathbb{R}$ with compact support,
and we extend each automorphism of $A$ to a unique automorphism of $M(A)$
which is continuous for the strict topology on norm-bounded sets of $M(A)$;
see Remark C.0.6 in Appendix C. In particular,
$\sigma_{t}\in\operatorname{Aut}M(A)$ for all $t\in\mathbb{R}$.

We shall need the Fourier transform and since there are confusingly many
different conventions regarding this transform we will be explicit: Given a
function $f\in L^{1}(\mathbb{R})$ we define the Fourier transform $\hat{f}$ of
$f$ to be the function

$\hat{f}(x):=\int_{\mathbb{R}}e^{ixt}f(t)\ \mathrm{d}t.$

With this definition we follow [KR].

######  Lemma 6.3.2.

The set $\left\\{\hat{f}:\ f\in C_{c}(\mathbb{R})\right\\}$ is a dense
$*$-subalgebra of $C_{0}(\mathbb{R})$.

###### Proof.

Observe $f\star g$ and $t\mapsto g(-t)$ are elements of $C_{c}(\mathbb{R})$
when $f,g$ are. Using this, and despite that Rudin uses other conventions, it
follows from (c) and (d) of Theorem 9.2 in [Ru0] that $\left\\{\hat{f}:\ f\in
C_{c}(\mathbb{R})\right\\}$ is a $*$-subalgebra of $C_{0}(\mathbb{R})$. (This
follows also (almost) from Corollary 3.2.28 in [KR].) Thanks to the Stone-
Weierstrass theorem it suffices therefore to show that this algebra separates
points of $\mathbb{R}$ and does not vanish at any point. Let $x\neq y$ in
$\mathbb{R}$. There is then a $t_{0}\in\mathbb{R}$ such that $e^{ixt_{0}}\neq
e^{iyt_{0}}$; say $\left|e^{ixt_{0}}-e^{iyt_{0}}\right|\geq\epsilon>0$. We can
then choose a function $f\in C_{c}(\mathbb{R})$ with a support concentrated
around $t_{0}$ such that $\int_{\mathbb{R}}f(s)\ \mathrm{d}s=1$ and

$\left|\int_{\mathbb{R}}e^{ixs}f(s)\
\mathrm{d}s-e^{it_{0}x}\right|\leq\frac{\epsilon}{3}$

and

$\left|\int_{\mathbb{R}}e^{iys}f(s)\
\mathrm{d}s-e^{it_{0}y}\right|\leq\frac{\epsilon}{3}.$

Then $\hat{f}(x)\neq\hat{f}(y)$ and $\hat{f}(x)\neq 0$ if $\epsilon<1$. ∎

######  Lemma 6.3.3.

$f(H)\in M(A)$ for all $f\in C_{0}(\mathbb{R})$ and $\sigma_{t}(f(H))=f(H)$
for all $t\in\mathbb{R}$.

###### Proof.

When $g\in C_{c}(\mathbb{R})$,

$\widehat{g}(H)=\int_{\mathbb{R}}g(t)u_{t}\ \mathrm{d}t,$

where $\widehat{g}(x)=\int_{\mathbb{R}}g(t)e^{itx}\ \mathrm{d}t$ is the
Fourier transform of $g$; cf. Theorem 5.6.36 in [KR]. For $a\in A$,

$\widehat{g}(H)a=\int_{\mathbb{R}}g(t)u_{t}a\ \mathrm{d}t,$

and

$a\widehat{g}(H)=\int_{\mathbb{R}}g(t)au_{t}\ \mathrm{d}t.$

Since $t\mapsto u_{t}a$ and $t\mapsto au_{t}$ are continuous in norm these
integral representations show that $\widehat{g}(H)a$ and $a\widehat{g}(H)$ are
elements of $A$, cf. Appendix A, and hence $\widehat{g}(H)\in M(A)$. Since
$\left\\{\widehat{g}:\ g\in C_{c}(\mathbb{R})\right\\}$ is dense in
$C_{0}(\mathbb{R})$ by Lemma 6.3.2 and $\left\|f(H)\right\|\leq\|f\|$ for all
$f\in C_{0}(\mathbb{R})$ it follows that $f(H)\in M(A)$ for all $f\in
C_{0}(\mathbb{R})$.

That $\sigma_{t}(f(H))=f(H)$ follows because $\sigma_{t}(u_{s})=u_{s}$ by
assumption: For all $a\in A$ and $g\in L^{1}(\mathbb{R})$,

$\displaystyle\sigma_{t}(\widehat{g}(H))\sigma_{t}(a)=\sigma_{t}(\widehat{g}(H)a)=\sigma_{t}\left(\int_{\mathbb{R}}g(s)u_{s}a\
\mathrm{d}s\right)$ $\displaystyle=\int_{\mathbb{R}}g(s)\sigma_{t}(u_{s}a)\
\mathrm{d}s=\left(\int_{\mathbb{R}}g(s)u_{s}\
\mathrm{d}s\right)\sigma_{t}(a)=\widehat{g}(H)\sigma_{t}(a).$

Since $\left\\{\sigma_{t}(a)\xi:\ \xi\in\mathbb{H},\ a\in A\right\\}$ is dense
in $\mathbb{H}$ this identity shows that
$\sigma_{t}(\widehat{g}(H))=\widehat{g}(H))$, and the equality
$\sigma_{t}(f(H))=f(H)$ for $f\in C_{0}(\mathbb{R})$ follows from Lemma 6.3.2
by continuity. ∎

For $z\in\mathbb{C}$ let $e^{zH}$ be the normal operator on $\mathbb{H}$
obtained from the function $t\mapsto e^{zt}$ by spectral theory applied to
$H$, e.g. Theorem 5.6.26 in [KR]. Note that $e^{zH}$ is often unbounded, but
for $f\in C_{c}(\mathbb{R})$ the operator $f(H)e^{zH}=e^{zH}f(H)$ is bounded
and an element of $M(A)$ by Lemma 6.3.3. Given a $\beta$-KMS weight $\psi$ for
$\sigma$, we can therefore define $\psi_{\beta}:A^{+}\to[0,\infty]$ such that

$\psi_{\beta}(a):=\sup\left\\{\psi\left(e^{-\frac{\beta
H}{2}}f(H)af(H)e^{-\frac{\beta H}{2}}\right):\ f\in C_{c}(\mathbb{R}),\ 0\leq
f\leq 1\right\\}.$ (6.3.1)

We aim to show that $\psi_{\beta}$ is a $\beta$-KMS weight for
$\sigma^{\prime}$. For this note first of all that $\psi_{\beta}$ is lower
semi-continuous since $\psi$ is.

######  Lemma 6.3.4.

Let $(g_{n})_{n=1}^{\infty}$ be a sequence in $C_{c}(\mathbb{R})$ such that
$0\leq g_{n}\leq g_{n+1}\leq 1$ for all $n\in\mathbb{N}$ and
$\lim_{n\to\infty}g_{n}(x)=1$ for all $x\in\mathbb{R}$. Then
$\lim_{n\to\infty}g_{n}(H)=1$ in the strict topology of $M(A)$.

###### Proof.

Let $b\in A$. It suffices to show that
$\lim_{n\to\infty}\left\|b-g_{n}(H)b\right\|=0$. Since

$\left\|b-g_{n}(H)b\right\|^{2}=\left\|b^{*}(1-g_{n}(H))^{2}b\right\|\leq\left\|b^{*}(1-g_{n}(H))b\right\|$

it suffices to show that
$\lim_{n\to\infty}\left\|b^{*}(1-g_{n}(H))b\right\|=0$. To this end let
$\omega\in B^{*}_{+},\ \|\omega\|\leq 1$. It follows from Kadison’s
representation theorem, Theorem 3.10.3 in [Pe], and Dini’s theorem that it
suffices to show that

$\lim_{n\to\infty}\omega(b^{*}(1-g_{n}(H))b)=0.$ (6.3.2)

Let $(\mathcal{H}_{\omega},\pi_{\omega},\xi_{\omega})$ be the GNS
representation of $B$ defined from $\omega$. Being non-degenerate
$\pi_{\omega}$ extends to a $*$-homomorphism $\overline{\pi}_{\omega}:M(B)\to
B(\mathcal{H}_{\omega})$ such that
$\overline{\pi}_{\omega}(m)\pi_{\omega}(b)=\pi_{\omega}(mb)$ for all $m\in
M(B)$ and $b\in B$, cf. Lemma C.0.5 in Appendix C. Hence the unitary one-
parameter group $\overline{\pi}_{\omega}(u_{t}),t\in\mathbb{R}$, is strongly
continuous. By Stone’s theorem there is a self-adjoint operator $H_{\omega}$
on $\mathcal{H}_{\omega}$ such that

$\overline{\pi}_{\omega}(u_{t})=e^{itH_{\omega}}\ \ \forall t\in\mathbb{R}.$

Since
$\widehat{f}(H_{\omega})=\int_{\mathbb{R}}f(t)\overline{\pi}_{\omega}(u_{t})\
\mathrm{d}t=\overline{\pi}_{\omega}\left(\int_{\mathbb{R}}f(t)u_{t}\
\mathrm{d}t\right)=\overline{\pi}_{\omega}(\widehat{f}(H))$ when $f\in
C_{c}(\mathbb{R})$ it follows by continuity that
$g(H_{\omega})=\overline{\pi}_{\omega}(g(H))$ for all $g\in
C_{0}(\mathbb{R})$, using Lemma 6.3.2. In particular,
$g_{n}(H_{\omega})=\overline{\pi}_{\omega}(g_{n}(H))$ and hence

$\omega\left(b^{*}(1-g_{n}(H))b\right)=\left<(1-g_{n}(H_{\omega}))\pi_{\omega}(b)\xi_{\omega},\pi_{\omega}(b)\xi_{\omega}\right>.$
(6.3.3)

It follows from spectral theory, see e.g. Theorem 5.6.26 in [KR], that

$\lim_{n\to\infty}g_{n}(H_{\omega})=1$

in the strong operator topology and hence (6.3.2) follows from (6.3.3). ∎

######  Lemma 6.3.5.

Let $\phi$ be a $\beta$-KMS weight for the flow $\sigma$ on $A$. Let $u\in
M(A)$ be a unitary such that $\sigma_{t}(u)=u$ for all $t\in\mathbb{R}$. Then
$\phi=\phi\circ\operatorname{Ad}u$.

###### Proof.

Let $a\in\mathcal{A}_{\sigma}$. Then $au^{*}\in\mathcal{A}_{\sigma}$ and
$\sigma_{z}(au^{*})=\sigma_{z}(a)u^{*}$ for all $z\in\mathbb{C}$. Hence

$\displaystyle\phi(ua^{*}au^{*})=\phi(\sigma_{-i\frac{\beta}{2}}(au^{*})\sigma_{-i\frac{\beta}{2}}(au^{*})^{*})$
$\displaystyle=\phi(\sigma_{-i\frac{\beta}{2}}(a)u^{*}u\sigma_{-i\frac{\beta}{2}}(a)^{*})=\phi(\sigma_{-i\frac{\beta}{2}}(a)\sigma_{-i\frac{\beta}{2}}(a)^{*})=\phi(a^{*}a).$

It follows therefore from Lemma 3.1.5 that $\phi=\phi\circ\operatorname{Ad}u$.
∎

######  Lemma 6.3.6.

$\psi_{\beta}$ is a $\beta$-KMS weight for $\sigma^{\prime}$.

###### Proof.

Let $0\leq f\leq g$ in $C_{c}(\mathbb{R})$ and let $b\in\mathcal{A}_{\sigma}$.
Let $g_{k}:\mathbb{R}\to[0,1]$ be a continuous function such that $g_{k}(t)=1$
when $|t|\leq k$ and $g_{k}(t)=0$ when $|t|\geq k+1$. Then

$\displaystyle\psi\left(e^{\frac{-\beta H}{2}}f(H)b^{*}bf(H)e^{-\frac{\beta
H}{2}}\right)=\lim_{k\to\infty}\psi\left(e^{-\frac{\beta
H}{2}}f(H){b^{*}}g_{k}(H)^{2}{b}f(H)e^{\frac{-\beta H}{2}}\right)$

by Lemma 6.3.4 and the lower semi-continuity of $\psi$. By Lemma 6.3.3
$g_{k}(H)$ and $f(H)e^{-\frac{\beta H}{2}}$ are fixed by $\sigma$ and since
$b$ is entire analytic for $\sigma$ by assumption, it follows that
$g_{k}(H){b}f(H)e^{\frac{-\beta H}{2}}$ is entire analytic for $\sigma$, and

$\sigma_{-i\frac{\beta}{2}}(g_{k}(H){b}f(H)e^{\frac{-\beta
H}{2}})=g_{k}(H)\sigma_{-i\frac{\beta}{2}}({b})f(H)e^{\frac{-\beta H}{2}}.$

It follows therefore that

$\displaystyle\psi\left(e^{-\frac{\beta
H}{2}}f(H){b^{*}}g_{k}(H)^{2}{b}f(H)e^{\frac{-\beta H}{2}}\right)$
$\displaystyle=\psi\left(\sigma_{-i\frac{\beta}{2}}(g_{k}(H){b}f(H)e^{\frac{-\beta
H}{2}})\sigma_{-i\frac{\beta}{2}}(g_{k}(H){b}f(H)e^{\frac{-\beta
H}{2}})^{*}\right)$
$\displaystyle=\psi\left(g_{k}(H)\sigma_{-i\frac{\beta}{2}}({b})f(H)e^{\frac{-\beta
H}{2}}e^{-\frac{\beta H}{2}}f(H)\sigma_{-i\frac{\beta}{2}}(b)g_{k}(H)\right)$
$\displaystyle\leq\psi\left(g_{k}(H)\sigma_{-i\frac{\beta}{2}}({b})g(H)e^{\frac{-\beta
H}{2}}e^{-\frac{\beta H}{2}}g(H)\sigma_{-i\frac{\beta}{2}}(b)g_{k}(H)\right)$
$\displaystyle=\psi\left(e^{-\frac{\beta
H}{2}}g(H){b^{*}}g_{k}(H)^{2}{b}g(H)e^{\frac{-\beta H}{2}}\right),$

where the last equality follows from the first two equalities applied with $f$
replaced by $g$. Letting $k$ go to infinity and by using Lemma 6.3.4 and the
lower semi-continuity of $\psi$ again, it follows that

$\psi\left(e^{\frac{-\beta H}{2}}f(H)b^{*}bf(H)e^{-\frac{\beta
H}{2}}\right)\leq\psi\left(e^{\frac{-\beta H}{2}}g(H)b^{*}bg(H)e^{-\frac{\beta
H}{2}}\right).$ (6.3.4)

Let $b\in A^{+}$ be arbitrary. Since
$\lim_{n\to\infty}R_{n}(\sqrt{b})=\sqrt{b}$ by Lemma 2.1.7 it follows from the
lower semi-continuity of $\psi$ that

$\displaystyle\psi\left(e^{\frac{-\beta H}{2}}f(H)bf(H)e^{-\frac{\beta
H}{2}}\right)$ $\displaystyle\leq\liminf_{n}\psi\left(e^{\frac{-\beta
H}{2}}f(H)R_{n}(\sqrt{b})R_{n}(\sqrt{b})f(H)e^{-\frac{\beta H}{2}}\right),$

and then from (6.3.4) that

$\begin{split}&\psi\left(e^{\frac{-\beta H}{2}}f(H)bf(H)e^{-\frac{\beta
H}{2}}\right)\\\ &\leq\liminf_{n}\psi\left(e^{\frac{-\beta
H}{2}}g(H)R_{n}(\sqrt{b})R_{n}(\sqrt{b})g(H)e^{-\frac{\beta
H}{2}}\right),\end{split}$ (6.3.5)

since $R_{n}(\sqrt{b})\in\mathcal{A}_{\sigma}$ by Lemma 2.1.7. By Kadisons
inequality, Proposition 3.2.4 in [BR],

$\displaystyle\psi\left(e^{\frac{-\beta
H}{2}}g(H)R_{n}(\sqrt{b})R_{n}(\sqrt{b})g(H)e^{-\frac{\beta H}{2}}\right)$
$\displaystyle\leq\psi\left(e^{\frac{-\beta
H}{2}}g(H)R_{n}({b})g(H)e^{-\frac{\beta H}{2}}\right),$

while the definition of $R_{n}$ implies that

$e^{\frac{-\beta H}{2}}g(H)R_{n}({b})g(H)e^{-\frac{\beta
H}{2}}=R_{n}\left(e^{\frac{-\beta H}{2}}g(H){b}g(H)e^{-\frac{\beta
H}{2}}\right)$

since $e^{\frac{-\beta H}{2}}g(H)=g(H)e^{\frac{-\beta H}{2}}$ is fixed by
$\sigma$ by Lemma 6.3.3. By use of Lemma 2.2.8 we find therefore that

$\psi\left(e^{\frac{-\beta H}{2}}g(H)R_{n}({b})g(H)e^{-\frac{\beta
H}{2}}\right)=\psi\left(e^{\frac{-\beta H}{2}}g(H){b}g(H)e^{-\frac{\beta
H}{2}}\right),$

and conclude therefore now from (6.3.5) that

$\psi\left(e^{\frac{-\beta H}{2}}f(H)bf(H)e^{-\frac{\beta
H}{2}}\right)\leq\psi\left(e^{\frac{-\beta H}{2}}g(H)bg(H)e^{-\frac{\beta
H}{2}}\right).$

This shows that the expression $\psi\left(e^{\frac{-\beta
H}{2}}f(H)bf(H)e^{\frac{-\beta H}{2}}\right)$ increases with $f$, implying
that

$\psi_{\beta}(b)=\lim_{k\to\infty}\psi\left(e^{-\frac{\beta
H}{2}}g_{k}(H)bg_{k}(H)e^{-\frac{\beta H}{2}}\right)$

for all $b\in A^{+}$. In particular, it follows that $\psi_{\beta}$ is
additive and hence a weight. To see that $\psi_{\beta}$ is not zero, let $b\in
A^{+}$ be an element such that $\psi(b)>0$. Then $x_{k}.:=e^{\frac{\beta
H}{2}}g_{k}(H)\sqrt{b}\in A$ by Lemma 6.3.3 and

$\psi_{\beta}(x_{k}x_{k}^{*})=\lim_{l\to\infty}\psi\left(g_{l}(H)g_{k}(H)bg_{k}(H)g_{l}(H)\right)=\psi\left(g_{k}(H)bg_{k}(H)\right).$

Since $\lim_{k\to\infty}g_{k}(H)bg_{k}(H)=b$ by Lemma 6.3.4, the lower semi-
continuity of $\psi$ ensures that
$\psi_{\beta}(x_{k}x_{k}^{*})=\psi\left(g_{k}(H)bg_{k}(H)\right)>0$ for $k$
large enough. Hence $\psi_{\beta}\neq 0$. To show that $\psi_{\beta}$ is
densely defined let $b\in\mathcal{M}^{\sigma}_{\psi}$ and set
$b_{n}:=g_{n}(H){b}$. Then $g_{k}(H)b_{n}=b_{n}$ when $k>n$ and it follows
from Lemma 6.3.3 that

$\sigma_{-i\frac{\beta}{2}}(b_{n}^{*}e^{-\frac{\beta
H}{2}})=\sigma_{-i\frac{\beta}{2}}(b^{*})g_{n}(H)e^{-\frac{\beta H}{2}}.$

Hence

$\displaystyle\psi\left(e^{-\frac{\beta
H}{2}}g_{k}(H)b_{n}b_{n}^{*}g_{k}(H)e^{-\frac{\beta
H}{2}}\right)=\psi\left(e^{-\frac{\beta H}{2}}b_{n}b_{n}^{*}e^{-\frac{\beta
H}{2}}\right)$
$\displaystyle=\psi\left(\sigma_{-i\frac{\beta}{2}}(b_{n}^{*}e^{-\frac{\beta
H}{2}})\sigma_{-i\frac{\beta}{2}}(b_{n}^{*}e^{-\frac{\beta H}{2}})^{*}\right)$
$\displaystyle=\psi(\sigma_{-i\frac{\beta}{2}}(b^{*})g_{n}(H)^{2}e^{\beta
H}\sigma_{-i\frac{\beta}{2}}(b^{*})^{*})$
$\displaystyle\leq\left\|g_{n}(H)^{2}e^{-\beta
H}\right\|\psi(\sigma_{-i\frac{\beta}{2}}(b^{*})\sigma_{-i\frac{\beta}{2}}(b^{*})^{*})$

when $k>n$. It follows that $\psi_{\beta}(b_{n}b_{n}^{*})<\infty$. Since
$\lim_{n\to\infty}b_{n}=b$ by Lemma 6.3.4 and $\mathcal{M}^{\sigma}_{\psi}$ is
dense in $A$, it follows that $\psi_{\beta}$ is densely defined. That
$\psi_{\beta}$ is $\sigma^{\prime}$-invariant follows from the
$\sigma$-invariance of $\psi$ by use of Lemma 6.3.5 and Lemma 6.3.3.

Finally we chack that $\psi_{\beta}$ satifies condition (2) in Kustermans’
theorem, Theorem 2.3.1. Let $a\in\mathcal{A}_{\sigma^{\prime}}$ and set
$a_{k}=g_{k}(H)ag_{k}(H)$. To see that $a_{k}\in\mathcal{A}_{\sigma}$ note
that

$\sigma_{t}(a_{k})=g_{k}(H)\sigma_{t}(a)g_{k}(H)=g_{k}(H)u_{-t}\sigma^{\prime}_{t}(a)u_{t}g_{k}(H),$

so that

$z\mapsto
g_{k}(H)e^{-izH}\sigma^{\prime}_{z}(a)g_{k}(H)e^{izH}=e^{-izH}\sigma^{\prime}_{z}(a_{k})e^{izH}$

is an entire holomorphic extension of $t\mapsto\sigma_{t}(a)$. In particular,

$\sigma_{-i\frac{\beta}{2}}(a_{k})=e^{-\frac{\beta}{2}H}\sigma^{\prime}_{-\frac{\beta}{2}}(a_{k})e^{\frac{\beta}{2}H}.$

This is used in the following calculation:

$\displaystyle\psi_{\beta}\left(a_{k}^{*}a_{k}\right)$
$\displaystyle=\lim_{n\to\infty}\psi\left(e^{-\frac{\beta
H}{2}}g_{n}(H)a_{k}^{*}a_{k}g_{n}(H)e^{-\frac{\beta H}{2}}\right)$
$\displaystyle=\psi\left(e^{-\frac{\beta H}{2}}a_{k}^{*}a_{k}e^{-\frac{\beta
H}{2}}\right)$
$\displaystyle=\psi\left(\sigma_{-i\frac{\beta}{2}}(a_{k})e^{-{\beta
H}}\sigma_{-i\frac{\beta}{2}}(a_{k})^{*}\right)$
$\displaystyle=\psi\left((e^{-\frac{\beta
H}{2}}\sigma^{\prime}_{-i\frac{\beta}{2}}(a_{k})e^{\frac{\beta
H}{2}})e^{-\beta H}(e^{-\frac{\beta
H}{2}}\sigma^{\prime}_{-i\frac{\beta}{2}}(a_{k})e^{\frac{\beta
H}{2}})^{*}\right)$ $\displaystyle=\psi\left(e^{-\frac{\beta
H}{2}}\sigma^{\prime}_{-i\frac{\beta}{2}}(a_{k})\sigma^{\prime}_{-i\frac{\beta}{2}}(a_{k})^{*}e^{-\frac{\beta
H}{2}}\right)$ $\displaystyle=\psi\left(e^{-\frac{\beta
H}{2}}g_{k}(H)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)g_{k}(H)^{2}\sigma^{\prime}_{-i\frac{\beta}{2}}(a)^{*}g_{k}(H)e^{-\frac{\beta
H}{2}}\right)$ $\displaystyle\leq\psi\left(e^{-\frac{\beta
H}{2}}g_{k}(H)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)^{*}g_{k}(H)e^{-\frac{\beta
H}{2}}\right).$

Since $\lim_{k\to\infty}a_{k}=a$ by Lemma 6.3.4 it follows from this estimate
and the lower semi-continuity of $\psi_{\beta}$ that

$\displaystyle\psi_{\beta}(a^{*}a)\leq\liminf_{k\to\infty}\psi_{\beta}(a_{k}^{*}a_{k})$
$\displaystyle\leq\lim_{k\to\infty}\psi\left(e^{-\frac{\beta
H}{2}}g_{k}(H)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)^{*}g_{k}(H)e^{-\frac{\beta
H}{2}}\right)$
$\displaystyle=\psi_{\beta}\left(\sigma^{\prime}_{-i\frac{\beta}{2}}(a)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)^{*}\right).$

Since $a\in\mathcal{A}_{\sigma^{\prime}}$ was arbitrary, the same estimate
holds with $a$ replaced by $\sigma^{\prime}_{i\frac{\beta}{2}}(a^{*})$ and we
conclude therefore that

$\psi_{\beta}(a^{*}a)=\psi_{\beta}\left(\sigma^{\prime}_{-i\frac{\beta}{2}}(a)\sigma^{\prime}_{-i\frac{\beta}{2}}(a)^{*}\right)$

for all $a\in\mathcal{A}_{\sigma^{\prime}}$. It follows then from Kustermans’
theorem, Theorem 2.3.1, that $\psi_{\beta}$ is a $\beta$-KMS weight for
$\sigma^{\prime}$. ∎

######  Theorem 6.3.7.

Let $\sigma$ be flow on $A$ and $u=(u_{t})_{t\in\mathbb{R}}$ a strictly
continuous unitary representation of $\mathbb{R}$ by multipliers of $M(A)$
such that $\sigma_{t}(u_{s})=u_{s}$ for all $s,t$. Consider the flow
$\sigma^{\prime}$ defined by

$\sigma^{\prime}_{t}:=\operatorname{Ad}u_{t}\circ\sigma_{t}.$

The map $\psi\mapsto\psi_{\beta}$ given by (6.3.1) is a bijection from
$\operatorname{KMS}(\sigma,\beta)$ onto
$\operatorname{KMS}(\sigma^{\prime},\beta)$ for all $\beta\in\mathbb{R}$.

###### Proof.

Since $\sigma_{t}=\operatorname{Ad}u_{t}^{*}\circ\sigma^{\prime}_{t}$ and
$\sigma^{\prime}_{t}(u_{t}^{*})=u_{t}^{*}$ we get in the same way a map
$\phi\mapsto\phi^{\beta}$ from $\operatorname{KMS}(\sigma^{\prime},\beta)$ to
$\operatorname{KMS}(\sigma,\beta)$ defined such that

$\phi^{\beta}(a):=\sup\left\\{\phi\left(e^{\frac{\beta
H}{2}}f(-H)af(-H)e^{\frac{\beta H}{2}}\right):\ f\in C_{c}(\mathbb{R}),\ 0\leq
f\leq 1\right\\}.$

Let $\psi\in\operatorname{KMS}(\sigma,\beta)$ and $a\in\mathcal{A}_{\sigma}$.
For $f,g\in C_{c}(\mathbb{R})$ with $0\leq f\leq 1$ and $0\leq g\leq 1$ we
have

$\psi(f(-H)g(H)a^{*}af(-H)g(H))=\psi(\sigma_{-i\frac{\beta}{2}}(a)f^{2}(-H)g^{2}(H)\sigma_{-i\frac{\beta}{2}}(a)^{*}).$

It follows therefore from an application of Lemma 6.3.4 that

$\displaystyle\left(\psi_{\beta}\right)^{\beta}(a^{*}a)=\sup\left\\{\psi(\sigma_{-i\frac{\beta}{2}}(a)f^{2}(-H)g^{2}(H)\sigma_{-i\frac{\beta}{2}}(a)^{*}):\
0\leq f,g\leq 1\right\\}$
$\displaystyle=\psi\left(\sigma_{-i\frac{\beta}{2}}(a)\sigma_{-i\frac{\beta}{2}}(a)^{*}\right)=\psi(a^{*}a).$

Then Lemma 3.1.5 shows that $\left(\psi_{\beta}\right)^{\beta}=\psi$.
Similarly, or by symmetry, $\left(\phi^{\beta}\right)_{\beta}=\phi$ for all
$\phi\in\operatorname{KMS}(\sigma^{\prime},\beta)$. ∎

######  Notes and remarks 6.3.8.

When $A$ is unital, Theorem 6.3.1 follows from a result by Kishimoto, cf.
Proposition 2.1 in [Ki1]. For KMS weights both Theorem 6.3.1 and Theorem 6.3.7
are new.

###  6.4 Inner flows

A flow $\sigma$ on a $C^{*}$-algebra $A$ is _inner_ when there is a unitary
representation $u=(u_{t})_{t\in\mathbb{R}}$ of $\mathbb{R}$ by unitaries in
the multiplier algebra $M(A)$ of $A$ such that $\mathbb{R}\ni t\mapsto u_{t}$
is continuous with respect to the strict topology of $M(A)$ and

$\sigma_{t}(a)=u_{t}au_{-t}$

for all $a\in A$ and all $t\in\mathbb{R}$. Before studying general inner flows
we consider first a special case.

####  6.4.1 Inner flows with a bounded generator

Let $A$ be a $C^{*}$-algebra and $h=h^{*}\in M(A)$ a self-adjoint element of
the multiplier algebra $M(A)$ of $A$. In this section we consider the
corresponding inner flow

$\sigma_{t}(a)=\operatorname{Ad}e^{ith}(a)=e^{ith}ae^{-ith}.$ (6.4.1)

######  Lemma 6.4.1.

$A=\mathcal{A}_{\sigma}$, $e^{zh}\in M(A)$ and $\sigma_{z}(a)=e^{zh}ae^{-zh}$
for all $a\in A$ and $z\in\mathbb{C}$.

###### Proof.

Since $\|h\|<\infty$ the sum $e^{zh}=\sum_{n=0}^{\infty}\frac{h^{n}}{n!}z^{n}$
converges in norm and hence $e^{zh}\in M(A)$ since $M(A)$ is a
$C^{*}$-algebra. Furthermore, it follows that

$z\mapsto e^{zh}ae^{-zh}$

is entire holomorphic for all $a\in A$. ∎

Let $\beta\in\mathbb{R}$ and let $\psi$ be a $\beta$-KMS weight for $\sigma$.
It follows from Lemma 6.4.1 that

$\displaystyle\psi(e^{\frac{\beta h}{2}}a^{*}ae^{\frac{\beta
h}{2}})=\psi(\sigma_{-i\frac{\beta}{2}}(ae^{\frac{\beta
h}{2}})\sigma_{-i\frac{\beta}{2}}(ae^{\frac{\beta h}{2}})^{*})$
$\displaystyle=\psi(e^{\frac{\beta h}{2}}a(e^{\frac{\beta
h}{2}}a)^{*})=\psi(e^{\frac{\beta h}{2}}aa^{*}e^{\frac{\beta h}{2}})$

for all $a\in A$. This shows that

$\tau_{\psi}(a):=\psi(e^{\frac{\beta h}{2}}ae^{\frac{\beta h}{2}})$

defines a lower semi-continuous trace $\tau_{\psi}:A^{+}\to[0,\infty]$ such
that

$\psi(a)=\tau_{\psi}(e^{-\frac{\beta h}{2}}ae^{-\frac{\beta h}{2}})$

for all $a\in A^{+}$. Conversely, given a lower semi-continuous trace $\tau$
on $A$ we can define a $\beta$-KMS weight $\psi_{\tau}:A^{+}\to[0,\infty]$
such that

$\psi_{\tau}(a):=\tau(e^{-\frac{\beta h}{2}}ae^{-\frac{\beta h}{2}}),$ (6.4.2)

as the reader can easily verify. The two maps $\tau\mapsto\psi_{\tau}$ and
$\psi\mapsto\tau_{\psi}$ are clearly each others inverses and we get therefore
the following.

######  Proposition 6.4.2.

For each $\beta\in\mathbb{R}$ the map $\tau\mapsto\psi_{\tau}$ defined by
(6.4.2) is a bijection from the set of lower semi-continuous traces on $A$
onto the set of $\beta$-KMS weights for the flow (6.4.1).

######  Proposition 6.4.3.

For each $\beta\in\mathbb{R}$ the map $\tau\mapsto\psi_{\tau}$ defined by
(6.4.2) is a bijection from the set of bounded traces on $A$ onto the set of
bounded $\beta$-KMS weights for the flow (6.4.1).

###### Proof.

The maps $A\to A$, given by $a\mapsto e^{-\frac{\beta h}{2}}ae^{-\frac{\beta
h}{2}}$ and $a\mapsto e^{\frac{\beta h}{2}}ae^{\frac{\beta h}{2}}$, are both
bounded and hence the map $\tau\mapsto\psi_{\tau}$ and its inverse both take
bounded functionals to bounded functionals. ∎

######  Corollary 6.4.4.

Assume that $A$ is unital. For each $\beta\in\mathbb{R}$ the map
$\omega\mapsto\psi_{\omega}$ given by

$\psi_{\omega}(a):=\frac{\omega(e^{-\beta h}a)}{\omega(e^{-\beta h})},\ \
\forall a\in A,$

is a bijection from the set of tracial states $\omega$ of $A$ onto the set of
$\beta$-KMS states for the flow (6.4.1).

######  Example 6.4.5.

(Flows on a matrix algebra.) Let $M_{n}(\mathbb{C})$ be the $C^{*}$-algebra of
complex $n$ by $n$ matrices. For every self-adjoint element $h=h^{*}$ in
$M_{n}(\mathbb{C})$ there is a flow $\sigma$ on $M_{n}(\mathbb{C})$ given by

$\sigma_{t}(m)=e^{ith}me^{-ith}.$ (6.4.3)

It follows from Corollary 6.4.4 that there is a unique $\beta$-KMS state
$\omega_{\beta}$ for $\sigma$ for every $\beta\in\mathbb{R}$ and that

$\omega_{\beta}(m)=\frac{\operatorname{Tr}_{n}(e^{-\beta
h}m)}{\operatorname{Tr}_{n}(e^{-\beta h})}\ \ \ \ \forall m\in
M_{n}(\mathbb{C}),$

where $\operatorname{Tr}_{n}$ is the usual trace on $M_{n}(\mathbb{C})$.

It is straightforward to adopt the arguments from the proof of Theorem B.0.1
in Appendix B to show that every flow on $M_{n}(\mathbb{C})$ is of the form
(6.4.3) for some $h=h^{*}\in M_{n}(\mathbb{C})$.

######  Remark 6.4.6.

The map $\omega\mapsto\psi_{\omega}$ in Corollary 6.4.4 is continuous for the
weak* topology and hence a homeomorphism since a continuous bijection between
compact Hausdorff spaces is a homeomorphism. However, the map is not affine.
It fact, in general there may not be an affine homeorphism between the tracial
state space of $A$ and the compact convex set of $\beta$-KMS states in the
setting of Corollary 6.4.4. The reader can see what can go wrong in Remarks
2.3 and 2.4 of [Ki1].

####  6.4.2 General inner flows

By using Theorem 6.3.7 we can partly extend the results of the last subsection
to more general inner flows, as follows. Let $u=(u_{t})_{t\in\mathbb{R}}$ be a
strictly continuous unitary representation of $\mathbb{R}$ by unitaries in the
multiplier algebra $M(A)$ of $A$ and consider the corresponding inner flow

$\sigma^{\prime}_{t}(a)=u_{t}au_{-t}.$

Then $\sigma^{\prime}$ is related to the trivial flow
$\sigma_{t}=\operatorname{id}_{A}$ in the same way as $\sigma^{\prime}$ was
related to the $\sigma$ of Section 6.3.1. Since the $\beta$-KMS weights for
the trivial flow are the lower semi-continuous traces on $A$ we get the
following directly from Theorem 6.3.7 and its proof:

######  Theorem 6.4.7.

Let $\alpha$ be an inner flow on $A$ such that
$\alpha_{t}=\operatorname{Ad}e^{itH}$, where $u_{t}=e^{itH}$ is a strictly
continuous unitary representation of $\mathbb{R}$ by multipliers in $M(A)$.
For any $\beta\in\mathbb{R}$ the map $\tau\mapsto\tau_{\beta}$, where

$\tau_{\beta}(a):=\sup\left\\{\tau\left(e^{\frac{-\beta
H}{2}}f(H)af(H)e^{-\frac{\beta H}{2}}\right):\ f\in C_{c}(\mathbb{R}),\ 0\leq
f\leq 1\right\\},$

is a bijection from the set of lower semi-continuous traces $\tau$ on $A$ onto
the set of $\beta$-KMS weights for $\alpha$. The inverse of the map
$\tau\mapsto\tau_{\beta}$ is the map $\psi\mapsto\tau_{\psi}$ which takes a
$\beta$-KMS weight $\psi$ for $\alpha$ to the lower semi-continuous trace
$\tau_{\psi}:A^{+}\to[0,\infty]$ given by

$\tau_{\psi}(a):=\sup\left\\{\psi\left(e^{\frac{\beta
H}{2}}f(H)af(H)e^{\frac{\beta H}{2}}\right):\ f\in C_{c}(\mathbb{R}),\ 0\leq
f\leq 1\right\\}.$

######  Notes and remarks 6.4.8.

Corollary 6.4.4 is a folklore fact, but Theorem 6.4.7 seems to be new. It
generalizes Theorem 6.1.7.

####  6.4.3 KMS-states for approximately inner flows

A flow $\sigma$ on a $C^{*}$-algebra $A$ is _approximately inner_ when there
is a sequence $\\{h_{n}\\}$ of self-adjoint elements in $A$ such that for all
$a\in A$,

$\lim_{n\to\infty}e^{ith_{n}}ae^{-ith_{n}}=\sigma_{t}(a)$

uniformly for $t$ in any compact subset of $\mathbb{R}$.

######  Theorem 6.4.9.

Let $A$ be a unital $C^{*}$-algebra and $\sigma$ an approximately inner flow
on $A$. If $A$ has a trace state there is a $\beta$-KMS state for $\sigma$ for
all $\beta\in\mathbb{R}$.

###### Proof.

Let $\tau$ be a trace state on $A$ and let $\beta\in\mathbb{R}$. For each
$n\in\mathbb{N}$ set

$\omega_{n}(a):=\frac{\tau(e^{-\beta h_{n}}a)}{\tau(e^{-\beta h_{n}})}.$

Each $\omega_{n}$ is a state on $A$ and since the state space of $A$ is
compact in the weak* topology, the set $\\{\omega_{n}:n\in\mathbb{N}\\}$ has a
weak* condensation point $\omega$. We aim to show that $\omega$ is a
$\beta$-KMS state for $\sigma$. To show this let $a\in\mathcal{A}_{\sigma}$
and $b\in A$. By Theorem 2.4.14 it suffices to show that

$\omega(ab)=\omega(b\sigma_{i\beta}(a)).$ (6.4.4)

The $C^{*}$-algebra $C$ generated by

$\left\\{\sigma_{t}(x):\ x\in\cup_{n=1}^{\infty}\\{a,b,h_{n}\\}\right\\}$

is $\sigma$-invariant and separable, and there is therefore a sequence
$m_{1}<m_{2}<m_{3}<\cdots$ in $\mathbb{N}$ such that
$\lim_{n\to\infty}\omega_{m_{n}}(c)=c$ for all $c\in C$. To simplify notation,
set $\omega^{\prime}_{n}:=\omega_{m_{n}}$.

Let $\sigma^{k}$ denote the flow on $C$ given by
$\sigma^{k}_{t}(c)=e^{ith_{k}}ce^{-ith_{k}}$. Choose $L_{n}>0$ such that

$\sqrt{\frac{n}{\pi}}\int_{|s|\geq L_{n}}\left|e^{-n(s-i\beta)^{2}}\right|\
\mathrm{d}s\leq\frac{1}{n},$ (6.4.5)

and

$\sqrt{\frac{n}{\pi}}\int_{|s|\geq L_{n}}e^{-ns^{2}}\
\mathrm{d}s\leq\frac{1}{n},$ (6.4.6)

and then $k_{n}\in\mathbb{N}$ such that

$\sup_{|t|\leq
L_{n}}\left\|\sigma^{k_{n}}_{t}(a)-\sigma_{t}(a)\right\|\leq\frac{M}{n},$
(6.4.7)

where

$M:=\min\left\\{\left(\sqrt{\frac{n}{\pi}}\int_{|s|\leq
L_{n}}\left|e^{-n(s-i\beta)^{2}}\right|\ \mathrm{d}s\right)^{-1},\
\left(\sqrt{\frac{n}{\pi}}\int_{|s|\leq L_{n}}e^{-ns^{2}}\
\mathrm{d}s\right)^{-1}\right\\}.$

By increasing $k_{n}$ we can arrange that

$\left|\omega^{\prime}_{k_{n}}(R_{n}(a)b)-\omega(R_{n}(a)b)\right|\leq\frac{1}{n}$
(6.4.8)

and

$\left|\omega^{\prime}_{k_{n}}(b\sigma_{i\beta}(R_{n}(a)))-\omega(b\sigma_{i\beta}(R_{n}(a)))\right|\leq\frac{1}{n}.$
(6.4.9)

Let $R^{(k_{n})}_{n}$ be the smoothing operators for flow $\sigma^{k_{n}}$.
Then (6.4.6) and (6.4.7) give the estimate

$\begin{split}&\left\|R^{(k_{n})}_{n}(a)-R_{n}(a)\right\|\leq\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-nt^{2}}\left\|\sigma^{k_{n}}_{t}(a)-\sigma_{t}(a)\right\|\
\mathrm{d}t\\\ &\\\ &\leq 2\|a\|\sqrt{\frac{n}{\pi}}\int_{|s|\geq
L_{n}}e^{-ns^{2}}\ \mathrm{d}s+\sqrt{\frac{n}{\pi}}\int_{|s|\leq
L_{n}}e^{-nt^{2}}\left\|\sigma^{k_{n}}_{t}(a)-\sigma_{t}(a)\right\|\
\mathrm{d}t\\\ &\\\
&\leq\frac{2\|a\|}{n}+\frac{M}{n}\sqrt{\frac{n}{\pi}}\int_{|s|\leq
L_{n}}e^{-nt^{2}}\ \mathrm{d}t\leq\frac{2\|a\|+1}{n}.\end{split}$ (6.4.10)

It follows from Lemma 2.1.13 that

$\displaystyle\sigma_{i\beta}(R_{n}(a))=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-n(s-i\beta)^{2}}\sigma_{s}(a)\
\mathrm{d}s$

and

$\displaystyle\sigma^{k_{n}}_{i\beta}(R^{(k_{n})}_{n}(a))=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-n(s-i\beta)^{2}}\sigma^{k_{n}}_{s}(a)\
\mathrm{d}s.$

Hence estimates similar to the preceding, using (6.4.5) and (6.4.7), show that

$\begin{split}&\left\|\sigma_{i\beta}(R_{n}(a))-\sigma^{k_{n}}_{i\beta}(R^{(k_{n})}_{n}(a))\right\|\\\
&\leq\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}\left|e^{-n(s-i\beta)^{2}}\right|\left\|\sigma_{s}(a)-\sigma^{k_{n}}_{s}(a)\right\|\
\mathrm{d}s\leq\frac{2\|a\|+1}{n}.\end{split}$ (6.4.11)

Then

$\displaystyle\left|\omega^{\prime}_{k_{n}}(b\sigma_{i\beta}(R^{(k_{n})}_{n}(a)))-\omega(b\sigma_{i\beta}(R_{n}(a)))\right|$
$\displaystyle\leq\|b\|\frac{2\|a\|+1}{n}+\left|\omega^{\prime}_{k_{n}}(b\sigma_{i\beta}(R_{n}(a)))-\omega(b\sigma_{i\beta}(R_{n}(a)))\right|\
\ \ \ \ \ \text{(by \eqref{02-10-23a})}$
$\displaystyle\leq\|b\|\frac{2\|a\|+1}{n}+\frac{1}{n}\ \ \ \ \ \ \text{(by
\eqref{01-08-23a})},$

and by using (6.4.10) and (6.4.8) in the same way we find also that

$\displaystyle\left|\omega^{\prime}_{k_{n}}(R^{(k_{n})}_{n}(a)b)-\omega(R_{n}(a)b)\right|\leq\|b\|\frac{2\|a\|+1}{n}+\frac{1}{n}.$

Since
$\omega^{\prime}_{k_{n}}(b\sigma_{i\beta}(R^{(k_{n})}_{n}(a)))=\omega^{\prime}_{k_{n}}(R^{(k_{n})}_{n}(a)b))$
this implies that

$\left|\omega(b\sigma_{i\beta}(R_{n}(a))-\omega(R_{n}(a)b)\right|\leq
2\|b\|\frac{2\|a\|+1}{n}+\frac{2}{n}.$ (6.4.12)

It follows from Lemma 2.1.13 and Lemma 2.1.7 that

$\lim_{n\to\infty}b\sigma_{i\beta}(R_{n}(a))-R_{n}(a)b=b\sigma_{i\beta}(a)-ab,$

which combined with (6.4.12) shows that (6.4.4) holds. Hence $\omega$ is a
$\beta$-KMS state. ∎

######  Notes and remarks 6.4.10.

Theorem 6.4.9 was obtained by Powers and Sakai in 1974 and published in the
very influential paper [PS] in which they conjectured that all flows on a UHF
algebra are approximately inner. The Powers-Sakai conjecture was not refuted
before the work of Matui and Sato in [MS], following a paper by Kishimoto,
[Ki2], in which he gave an example of a flow on a simple unital AF-algebra
which is not approximately inner.

We note that Theorem 6.4.9 also follows from Proposition 5.3.25 in [BR].

###  6.5 Tensor products

In this section we consider tensor products of $C^{*}$-algebras with the aim
of defining the tensor product of KMS weights. The construction we present
will work regardless of which $C^{*}$-norm we consider, as long as it is
’reasonable’ as specified on page 850 of [KR]. In particular, it works both
for the maximal and the minimal tensor product. In the construction we shall
also consider tensor products of Hilbert spaces and of operators, in addition
to certain algebraic tensor product. We use the same notation $\otimes$ for
all tensor products in the hope that the meaning will be clear from the
context.

Let $\sigma^{1}$ be a flow on the $C^{*}$-algebra $A_{1}$ and $\sigma^{2}$ a
flow on the $C^{*}$-algebra $A_{2}$, and let $\psi_{1}$ be a $\beta$-KMS wight
for $\sigma^{1}$ and $\psi_{2}$ a $\beta$-KMS weight for $\sigma^{2}$. We can
then consider the tensor product flow $\sigma$ on $A_{1}\otimes A_{2}$,
defined by

$\sigma_{t}:=\sigma^{1}_{t}\otimes\sigma^{2}_{t}.$

Thus, on simple tensors $\sigma_{t}(a_{1}\otimes
a_{2})=\sigma^{1}_{t}(a_{1})\otimes\sigma^{2}_{t}(a_{2})$.

######  Lemma 6.5.1.

$\Lambda_{\psi_{1}}\otimes\Lambda_{\psi_{2}}:\
\mathcal{N}_{\psi_{1}}\otimes\mathcal{N}_{\psi_{2}}\to H_{\psi_{1}}\otimes
H_{\psi_{2}}$ is closable.

###### Proof.

Let $(x_{i})_{i\in I}$ be a net in
$\mathcal{N}_{\psi_{1}}\otimes\mathcal{N}_{\psi_{2}}$ such that
$\lim_{i\to\infty}x_{i}=0$ in $A_{1}\otimes A_{2}$ and
$\lim_{i\to\infty}\Lambda_{\psi_{1}}\otimes\Lambda_{\psi_{2}}(x_{i})=w$ in
$H_{\psi_{1}}\otimes H_{\psi_{2}}$. Write $x_{i}$ as a finite sum

$x_{i}=\sum_{n}x_{n}^{i}\otimes y^{i}_{n},$

where $x_{n}^{i}\in\mathcal{N}_{\psi_{1}}$ and
$y_{n}^{i}\in\mathcal{N}_{\psi_{2}}$ for all $n,i$. For each of
$k\in\\{1,2\\}$ we let $(u^{k}_{j})_{j\in I_{k}}$ and $(\rho^{k}_{j})_{j\in
I_{k}}$ be the net in $\mathcal{M}^{\sigma^{k}}_{\psi_{k}}$ and the
corresponding net of operators in $B(H_{\psi_{k}})$ obtained by applying Lemma
2.4.2 to $\psi_{k}$. Then

$\displaystyle\rho^{1}_{j}\otimes\rho^{2}_{j^{\prime}}(w)=\lim_{i\to\infty}\rho^{1}_{j}\otimes\rho^{2}_{j^{\prime}}(\Lambda_{\psi_{1}}\otimes\Lambda_{\psi_{2}}(x_{i}))$
$\displaystyle=\lim_{i\to\infty}\sum_{n}\rho^{1}_{j}\otimes\rho^{2}_{j^{\prime}}(\Lambda_{\psi_{1}}(x_{n}^{i})\otimes\Lambda_{\psi_{2}}(y_{n}^{i}))$
$\displaystyle=\lim_{i\to\infty}\sum_{n}\Lambda_{\psi_{1}}(x_{n}^{i}u^{1}_{j})\otimes\Lambda_{\psi_{2}}(y_{n}^{i}u^{2}_{j^{\prime}})$
$\displaystyle=\lim_{i\to\infty}(\pi_{\psi_{1}}\otimes\pi_{\psi_{2}})(x_{i})(\Lambda_{\psi_{1}}(u^{1}_{j})\otimes\Lambda_{\psi_{2}}(u^{2}_{j^{\prime}}))=0$

for all $(j,j^{\prime})\in I_{1}\times I_{2}$. Since
$\lim_{(j,j^{\prime})\to\infty}\rho^{1}_{j}\otimes\rho^{2}_{j^{\prime}}(w)=w$
thanks to property (g) of Lemma 2.4.2, it follows that $w=0$. ∎

Let $\Lambda:D(\Lambda)\to H_{\psi_{1}}\otimes H_{\psi_{2}}$ be the closure of
$\Lambda_{\psi_{1}}\otimes\Lambda_{\psi_{2}}$. The triple

$(H_{\psi_{1}}\otimes
H_{\psi_{2}},\Lambda,\pi_{\psi_{1}}\otimes\pi_{\psi_{2}})$ (6.5.1)

is then a GNS representation of $A_{1}\otimes A_{2}$ as defined in Definition
1.2.3.

######  Lemma 6.5.2.

With respect to the flow $\sigma$ the GNS-triple (6.5.1) has the properties
(A), (B), (C), (D) and (F) from Section 2.4.1.

###### Proof.

The triple has property (B) by definition. It is straightforward to see that
the triple has property (A) because $\pi_{\psi_{k}},k=1,2$, are both non-
degenerate representations. Let $x\in D(\Lambda)$. By definition there is a
sequence $\\{x_{n}\\}$ in
$\mathcal{N}_{\psi_{1}}\otimes\mathcal{N}_{\psi_{2}}$ such that
$\lim_{n\to\infty}x_{n}=x$ and
$\lim_{n\to\infty}\Lambda(x)=\lim_{n\to\infty}(\Lambda_{\psi_{1}}\otimes\Lambda_{\psi_{2}})(x_{n})$.
Then $\sigma_{t}(x_{n})\in\mathcal{N}_{\psi_{1}}\otimes\mathcal{N}_{\psi_{2}}$
and $\lim_{n\to\infty}\sigma_{t}(x_{n})=\sigma_{t}(x)$. Furthermore,

$\displaystyle\lim_{n\to\infty}\Lambda(\sigma_{t}(x_{n}))=\lim_{n\to\infty}(U^{\psi_{1}}_{t}\otimes
U^{\psi_{2}}_{t})(\Lambda(x_{n}))=(U^{\psi_{1}}_{t}\otimes
U^{\psi_{2}}_{t})(\Lambda(x)),$

showing that $\sigma_{t}(x)\in D(\Lambda)$ and
$\Lambda(\sigma_{t}(x))=(U^{\psi_{1}}_{t}\otimes
U^{\psi_{2}}_{t})(\Lambda(x))$. The properties (C) and (D) follow from this.
To establish (F) set

$S:=\mathcal{M}^{\sigma^{1}}_{\psi_{1}}\otimes\mathcal{M}^{\sigma^{1}}_{\psi_{1}}.$

Let

$s=\sum_{n}x^{1}_{n}\otimes x^{2}_{n},$ (6.5.2)

be a finite sum with $x^{i}_{n}\in\mathcal{M}^{\sigma^{i}}_{\psi_{i}}$. For
each $n$ the map $\mathbb{C}\ni
z\mapsto\sigma^{1}_{z}(x_{n}^{1})\otimes\sigma^{2}_{z}(x_{n}^{2})$ is entire
holomorphic, implying that

$\sigma_{z}(x_{n}^{1}\otimes
x_{n}^{2})=\sigma^{1}_{z}(x^{1}_{n})\otimes\sigma^{2}_{z}(x^{2}_{n})$

for all $z\in\mathbb{C}$. It follows that
$S\subseteq\mathcal{M}^{\sigma}_{\Lambda}$. If
$a_{i}\in\mathcal{N}_{\psi^{i}}$, there is a sequence $\\{s^{i}_{n}\\}$ in
$\mathcal{M}^{\sigma^{i}}_{\psi_{i}}$ such that
$\lim_{n\to\infty}s_{n}^{i}=a_{i}$ and
$\lim_{n\to\infty}\Lambda_{\psi_{i}}(s^{i}_{n})=\Lambda_{\psi_{i}}(a_{i})$ by
Lemma 2.2.13. Then $\\{s^{1}_{n}\otimes s^{2}_{n}\\}$ is a sequence in $S$
such that $\lim_{n\to\infty}s^{1}_{n}\otimes s^{2}_{n}=a_{1}\otimes a_{2}$ and
$\lim_{n\to\infty}\Lambda(s^{1}_{n}\otimes s^{2}_{n})=\Lambda(a_{1}\otimes
a_{2})$. By linearity, and since $\Lambda$ is defined as the closure of
$\Lambda_{\psi_{1}}\otimes\Lambda_{\psi_{2}}$, it follows that for every $x\in
D(\Lambda)$ there is a sequence $\\{s_{n}\\}$ in $S$ such that
$\Lambda(x)=\lim_{n\to\infty}\Lambda(s_{n})$. In particular, the triple has
the property from the first item in condition (F) of Lemma 2.4.5. In addition,
when $s\in S$ is given as in (6.5.2),

$\sigma_{-i\frac{\beta}{2}}(s)^{*}=\sum_{n}\sigma^{1}_{-i\frac{\beta}{2}}(x^{1}_{n})^{*}\otimes\sigma^{2}_{-i\frac{\beta}{2}}(x^{2}_{n})^{*},$

and hence

$\displaystyle\left\|\Lambda\left(\sigma_{-i\frac{\beta}{2}}(s)^{*}\right)\right\|=\left\|\sum_{n}\Lambda_{\psi_{1}}\left(\sigma^{1}_{-i\frac{\beta}{2}}(x^{1}_{n})^{*}\right)\otimes\Lambda_{\psi_{2}}\left(\sigma^{2}_{-i\frac{\beta}{2}}(x^{2}_{n})^{*}\right)\right\|$
$\displaystyle=\left\|J_{\psi_{1}}\otimes
J_{\psi_{2}}\left(\sum_{n}\Lambda_{\psi_{1}}(x^{1}_{n})\otimes\Lambda_{\psi_{2}}(x^{2}_{n})\right)\right\|=\left\|J_{\psi_{1}}\otimes
J_{\psi_{2}}(\Lambda(s))\right\|=\left\|\Lambda(s)\right\|,$

since $J_{\psi_{1}}\otimes J_{\psi_{2}}$, as the tensor product of two
conjugate linear isometries is itself an isometry. This shows that the triple
(6.5.1) also has the property from the second item in condition (F) of Lemma
2.4.5, and completes the proof. ∎

######  Theorem 6.5.3.

Let $\sigma^{i}$ be a flow on the $C^{*}$-algebra $A_{i}$ and $\psi_{i}$ a
$\beta$-KMS weight for $\sigma^{i},\ i=1,2$. There is a unique $\beta$-KMS
weight $\psi_{1}\otimes\psi_{2}$ for $\sigma^{1}\otimes\sigma^{2}$ on
$A_{1}\otimes A_{2}$ such that
$\mathcal{N}_{\psi_{1}}\otimes\mathcal{N}_{\psi_{2}}$ is a core for
$\Lambda_{\psi_{1}\otimes\psi_{2}}$ and

$\psi_{1}\otimes\psi_{2}(a_{1}^{*}b_{1}\otimes
a_{2}^{*}b_{2})=\psi_{1}(a_{1}^{*}b_{1})\psi_{2}(a_{2}^{*}b_{2})$

when $a_{i},b_{i}\in\mathcal{N}_{\psi_{i}},\ i=1,2$.

###### Proof.

By Lemma 6.5.2, Lemma 2.4.5 and Proposition 2.4.3 the triple (6.5.1) is the
GNS-triple for a $\beta$-KMS weight for $\sigma^{1}\otimes\sigma^{2}$. The
resulting $\beta$-KMS weight $\psi_{1}\otimes\psi_{2}$ has the stated
properties by construction. The uniqueness is an immediate consequence of
Corollary 5.2.6. ∎

######  Notes and remarks 6.5.4.

Theorem 6.5.3 was obtained by Kustermans in Proposition 7.6 of [Ku1].

####  6.5.1 ITPFI flows

Infinite tensor products of facors of type I (abbreviated to ITPFI) were
introduced in the theory of von Neumann algebras in work by R. Powers, H.
Araki and E.J. Woods from the sixties. Here we consider the corresponding
infinite tensor product flows.

Let $n_{1},n_{2},n_{3},\cdots$ be a sequence of natural numbers. Set

$A_{k}:=M_{n_{1}}(\mathbb{C})\otimes M_{n_{2}}(\mathbb{C})\otimes
M_{n_{3}}(\mathbb{C})\otimes\cdots\otimes M_{n_{k}}(\mathbb{C}).$

A unital embedding $\phi_{k}:A_{k}\to A_{k+1}$ is given on simple tensors by

$\phi_{k}(a_{1}\otimes\cdots\otimes a_{k})=a_{1}\otimes\cdots\otimes
a_{k}\otimes 1.$

The inductive limit $C^{*}$-algebra of the sequence

$\textstyle{A_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{1}}$$\textstyle{A_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{2}}$$\textstyle{A_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{3}}$$\textstyle{\cdots}$

is a UHF algebra unless $n_{j}=1$ for all but finitely many $j$, in which case
it is a matrix algebra. We denote it by

$\otimes_{k=1}^{\infty}M_{n_{k}}(\mathbb{C}).$

Pick for each $k\in\mathbb{N}$ a selfadjoint matrix

$h_{k}\in M_{n_{k}}(\mathbb{C})$

and define a flow $\sigma^{k}$ on
$\otimes_{k=1}^{\infty}M_{n_{k}}(\mathbb{C})$ such that

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

on simple tensors $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{k}$. Define self-
adjoint elements $\hat{h}_{i}\in A_{k}$ as simple tensors where
$\hat{h}_{1}=h_{1}\otimes 1\otimes 1\otimes\cdots\otimes 1,\
\hat{h}_{2}=1\otimes h_{2}\otimes 1\otimes\cdots\otimes 1$ and so on. Then
$\sigma^{k}=\operatorname{Ad}e^{itH_{k}}$, where

$H_{k}:=\sum_{i=1}^{k}\hat{h}_{i}.$

Since $\sigma^{k+1}_{t}\circ\phi_{k}=\phi_{k}\circ\sigma^{k}_{t}$ there is a
flow $\sigma$ on $\otimes_{k=1}^{\infty}M_{n_{k}}(\mathbb{C})$ defined such
that

$\sigma_{t}\circ\phi_{k,\infty}=\phi_{k,\infty}\circ\sigma^{k}_{t},$ (6.5.3)

where $\phi_{k,\infty}:A_{k}\to\otimes_{k=1}^{\infty}M_{n_{k}}(\mathbb{C})$ is
the canonical embedding. Flows of this kind were considered in [KiR] and [Ki1]
(with $n_{j}=2$).

Let $\beta\in\mathbb{R}$. It follows from Example 6.4.5 that there is a unique
$\beta$-KMS state $\omega_{h_{j},\beta}$ for the flow
$\operatorname{Ad}e^{ith_{j}}$ on $M_{n_{j}}(\mathbb{C})$ given by

$\omega_{h_{j},\beta}(m):=\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}m)}{\operatorname{Tr}_{n_{j}}(e^{-\beta h_{j}})}$

when $\operatorname{Tr}_{n_{j}}$ denote the usual trace on
$M_{n_{j}}(\mathbb{C})$. We can then consider the tensor product state

$\otimes_{j=1}^{k}\omega_{h_{j},\beta}$

on $A_{k}$ defined such that

$\left(\otimes_{j=1}^{k}\omega_{h_{j},\beta}\right)(a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{k}):=\prod_{j=1}^{k}\omega_{h_{j},\beta}(a_{j})$

on simple tensors. Using Lemma 6.4.1 we find that

$\displaystyle\left(\otimes_{j=1}^{k}\omega_{h_{j},\beta}\right)((a_{1}\otimes\cdots\otimes
a_{k})(b_{1}\otimes\cdots\otimes
b_{k}))=\prod_{j=1}^{k}\omega_{h_{j},\beta}(a_{j}b_{j})$
$\displaystyle=\prod_{j=1}^{k}\omega_{h_{j},\beta}(b_{j}e^{-\beta
h_{j}}a_{j}e^{\beta h_{j}})$
$\displaystyle=\left(\otimes_{j=1}^{k}\omega_{h_{j},\beta}\right)(b_{1}e^{-\beta
h_{1}}a_{1}e^{\beta h_{1}}\otimes\cdots\otimes b_{k}e^{-\beta
h_{k}}a_{k}e^{\beta h_{k}})$
$\displaystyle=\left(\otimes_{j=1}^{k}\omega_{h_{j},\beta}\right)((b_{1}\otimes\cdots\otimes
b_{k})\sigma^{k}_{i\beta}(a_{1}\otimes\cdots\otimes a_{k}))$
$\displaystyle=\left(\otimes_{j=1}^{k}\omega_{h_{j},\beta}\right)((b_{1}\otimes\cdots\otimes
b_{k})e^{-\beta H_{k}}(a_{1}\otimes\cdots\otimes a_{k})e^{\beta H_{k}}),$

and we see therefore that $\otimes_{j=1}^{k}\omega_{h_{j},\beta}$ is a
$\beta$-KMS state for $\sigma^{k}$. As such it is unique by Example 6.4.5
since $A_{k}\simeq M_{m}(\mathbb{C})$ with $m=\prod_{j=1}^{k}n_{j}$. Note that

$\left(\otimes_{j=1}^{k+1}\omega_{h_{j},\beta}\right)\circ\phi_{k}=\otimes_{j=1}^{k}\omega_{h_{j},\beta}.$

It follows that there is a state $\omega_{\beta}$ on
$\otimes_{k=1}^{\infty}M_{n_{k}}(\mathbb{C})$ such that

$\omega_{\beta}\circ\phi_{k}=\otimes_{j=1}^{k}\omega_{h_{j},\beta}$ (6.5.4)

for all $k$. By Lemma 6.4.1 every element of $A_{k}$ is entire analytic for
$\sigma^{k}$ and it follows therefore from (6.5.3) that

$\bigcup_{k=1}^{\infty}\phi_{k,\infty}(A_{k})\subseteq\mathcal{A}_{\sigma}$

and that $\sigma_{z}\circ\phi_{k,\infty}=\phi_{k,\infty}\circ\sigma^{k}_{z}$
for all $k,z$. In combination with (6.5.4) this implies that

$\omega_{\beta}(ab)=\omega_{\beta}(b\sigma_{i\beta}(a))$

for all $a,b\in\bigcup_{k=1}^{\infty}\phi_{k,\infty}(A_{k})$. Since
$\Lambda_{\omega_{\beta}}:A\to H_{\omega_{\beta}}$ is bounded and
$\bigcup_{k=1}^{\infty}\phi_{k,\infty}(A_{k})$ is dense in
$\otimes_{k=1}^{\infty}M_{n_{k}}(\mathbb{C})$ it follows from Theorem 2.4.6
with $\bigcup_{k=1}^{\infty}\phi_{k,\infty}(A_{k})$ in the role as $S$, that
$\omega_{\beta}$ is a $\beta$-KMS state for $\sigma$. We denote it by

$\omega_{\beta}=\otimes_{k=1}^{\infty}\omega_{h_{k},\beta}.$

Conversely, if $\omega$ is a $\beta$-KMS state for $\sigma$ it follows from
(6.5.3) that $\omega\circ\phi_{k,\infty}$ is a $\beta$-KMS state for
$\sigma^{k}$ and and hence
$\omega\circ\phi_{k,\infty}=\otimes_{k=1}^{\infty}\omega_{h_{k},\beta}$ by
Example 6.4.5. It follows that $\omega=\omega_{\beta}$. In conclusion: The
flow $\sigma$ has exactly one $\beta$-KMS state for all $\beta\in\mathbb{R}$;
the infinite tensor product state
$\otimes_{k=1}^{\infty}\omega_{h_{k},\beta}$.

####  6.5.2 Matroid flows

In the same setting as in the last section choose non-zero projections
$p_{j}\in M_{n_{j}}(\mathbb{C}),\ j\geq 2$, and define $\kappa_{k}:A_{k}\to
A_{k+1}$ by

$\kappa_{k}(a_{1}\otimes\cdots\otimes a_{k})=a_{1}\otimes\cdots\otimes
a_{k}\otimes p_{k+1}$

for $k=1,2,3,\cdots$. Let $D$ be the $C^{*}$-algebra inductive limit of the
sequence

$\textstyle{A_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kappa_{1}}$$\textstyle{A_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kappa_{2}}$$\textstyle{A_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\kappa_{3}}$$\textstyle{\cdots}$

This kind of $C^{*}$-algebras were introduced by J. Dixmier who called them
matroid algebras. We note that $D$ is unital if and only $p_{j}$ is the unit
in $M_{n_{j}}(\mathbb{C})$ for all sufficiently large $j$, in which case $D$
is a UHF algebra.

Choose selfadjoint matrices $h_{k}\in M_{n_{k}}(\mathbb{C})$, but this time
such that $h_{k}$ commutes with $p_{k},\ k\geq 2$; i.e.

$h_{k}=h_{k}^{*}\in\\{p_{k}\\}^{\prime}\cap M_{n_{k}}(\mathbb{C}),\ k\geq 2.$

Define the flows $\sigma^{k}$ on $A_{k}$ as above using the new $h_{k}$’s and
note that $\sigma^{k+1}_{t}\circ\kappa_{k}=\kappa_{k}\circ\sigma^{k}_{t}$ so
that we have a flow $\alpha$ on $D$ defined by

$\alpha_{t}\circ\kappa_{k,\infty}=\kappa_{k,\infty}\circ\sigma^{k}_{t}$

for all $k,t$. Let $1_{n_{1}}$ be the unit in $A_{1}=M_{n_{1}}(\mathbb{C})$.
The corner

$\kappa_{1,\infty}(1_{n_{1}})D\kappa_{1,\infty}(1_{n_{1}})$

of $D$ is then $\alpha$-invariant and a full corner of $D$. Since the corner
is $*$-isomorphic to

$\otimes_{j=1}^{\infty}M_{m_{j}}(\mathbb{C}),$

where $m_{j}:=\operatorname{Tr}_{n_{j}}(p_{j})$, it follows from Section 6.5.1
and Theorem 3.1.8 that there is a $\beta$-KMS weight for $\alpha$ for every
$\beta\in\mathbb{R}$ and that it is unique up to multiplication by scalars.

The images $\kappa_{k,\infty}(1_{A_{k}})$ of the units $1_{A_{k}}$ the
$A_{k}$’s in $D$ constitute an approximate unit for $D$ and it follows
therefore from the description of the $\beta$-KMS states on
$\otimes_{j=1}^{\infty}M_{m_{j}}(\mathbb{C})$ and Lemma 6.1.6 that the
$\beta$-KMS weights for $\alpha$ will be bounded if and only if

$\prod_{j=2}^{\infty}\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}})}{\operatorname{Tr}_{n_{j}}(e^{-\beta h_{j}}p_{j})}<\infty.$ (6.5.5)

To see this let $\psi$ be a $\beta$-KMS weight for $\alpha$. Then
$\psi\circ\kappa_{k,\infty}$ is a $\beta$-KMS functional for the flow
$\sigma^{k}$ on $A_{k}$ and hence a scalar multiple of
$\otimes_{j=1}^{k}\omega_{h_{j},\beta}$ by Example 6.4.5. That is,
$\psi\circ\kappa_{k,\infty}=\lambda_{k}\otimes_{j=1}^{k}\omega_{h_{j},\beta}$
for some $\lambda_{k}>0$. Note that

$\psi\circ\kappa_{k,\infty}(1_{A_{k}})=\lambda_{k}\otimes_{j=1}^{k}\omega_{h_{j},\beta}(1_{A_{k}})=\lambda_{k}$

while

$\displaystyle\psi\circ\kappa_{1,\infty}(1_{n_{1}})=\psi\circ\kappa_{k,\infty}(1_{n_{1}}\otimes
p_{2}\otimes p_{3}\otimes\cdots\otimes p_{k-1})$
$\displaystyle=\lambda_{k}\otimes_{j=1}^{k}\omega_{h_{j},\beta}(1_{n_{1}}\otimes
p_{2}\otimes p_{3}\otimes\cdots\otimes
p_{k-1})=\lambda_{k}\prod_{k=2}^{k-1}\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}{\operatorname{Tr}_{n_{j}}(e^{-\beta h_{j}})}.$

Combined these equations imply that

$\psi\circ\kappa_{k,\infty}(1_{A_{k}})=\psi\circ\kappa_{1,\infty}(1_{n_{1}})\prod_{k=2}^{k-1}\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}})}{\operatorname{Tr}_{n_{j}}(e^{-\beta h_{j}}p_{j})}.$

Hence $\psi$ is bounded if and only if (6.5.5) holds.

We consider two special cases; the first to show that there can be bounded KMS
weights in cases where there are no bounded traces on $D$ and the second to
show that there can be unbounded KMS weights in cases where the trace(s) on
$D$ are bounded.

Consider the case where $m_{j}=n_{j}-1$ and $h_{j}=\mu_{j}e_{j}$ for some
$\mu_{j}\in\mathbb{R}$ and a minimal projection $e_{j}$ in
$M_{n_{j}}(\mathbb{C})$ with $e_{j}\leq p_{j}$. Then

$\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}})}{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}=\frac{n_{j}-1+e^{-\beta\mu_{j}}}{n_{j}-2+e^{-\beta\mu_{j}}}=1+x_{j}$

where $x_{j}=(n_{j}-2+e^{-\beta\mu_{j}})^{-1}$. Hence, in this case the
$\beta$-KMS weights for $\alpha$ are bounded if and only if111We are here
using that when $\\{a_{n}\\}_{n=1}^{\infty}$ is a sequence of positive real
numbers,
$\sum_{n=1}^{N}a_{n}\leq\prod_{n=1}^{N}(1+a_{n})\leq\exp(\sum_{n=1}^{N}a_{n})$
which leads to the well-known conclusion that
$\prod_{n=1}^{\infty}(1+a_{n})<\infty$ if and only if
$\sum_{n=1}^{\infty}a_{n}<\infty$.

$\sum_{j=1}^{\infty}(n_{j}-2+e^{-\beta\mu_{j}})^{-1}<\infty.$

Thus, if we choose $n_{j}=3$ for all $j$, the condition becomes

$\sum_{j=1}^{\infty}(1+e^{-\beta\mu_{j}})^{-1}<\infty.$

If we then set

$\mu_{j}=-(l-1)$

for $7^{l-1}\leq j<7^{l}$, $l=1,2,3,\cdots$, we see that the $\beta$-KMS
weights for $\alpha$ are bounded if and only if

$\sum_{l=1}^{\infty}\frac{7^{l-1}}{1+e^{\beta(l-1)}}<\infty;\~{A}¸$

i.e. if and only if $\beta>\log 7$. In particular, the lower semi-continuous
traces on $D$ are unbounded.

Consider then the case where $n_{j}=j!$ and
$m_{j}=\operatorname{Tr}_{n_{j}}(p_{j})=j!-1$. In this case

$\displaystyle\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}})}{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}=\frac{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}(1-p_{j}))+\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}{\operatorname{Tr}_{n_{j}}(e^{-\beta h_{j}}p_{j})}$
$\displaystyle=\frac{e^{-\beta\mu_{j}}+\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}=1+\frac{e^{-\beta\mu_{j}}}{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}$

where $\mu_{j}\in\mathbb{R}$ is the real number such that
$h_{j}(1-p_{j})=\mu_{j}(1-p_{j})$. Hence the $\beta$-KMS weight for $\alpha$
are bounded if and only if

$\sum_{j=1}^{\infty}\frac{e^{-\beta\mu_{j}}}{\operatorname{Tr}_{n_{j}}(e^{-\beta
h_{j}}p_{j})}<\infty.$

Since $\operatorname{Tr}_{n_{j}}(p_{j})=j!-1$ this condition is certainly
satisfied when $\beta=0$, regardless of the choice of $h_{j}$, implying that
$D$ has a bounded trace. On the other hand, if we take take
$h_{j}=-\log(j!)(1-p_{j})$ we get the value $\mu_{j}=-\log(j!)$ while
$e^{-\beta h_{j}}p_{j}=p_{j}$ for all $\beta$. Then

$\sum_{j=1}^{\infty}\frac{e^{-\mu_{j}}}{\operatorname{Tr}_{n_{j}}(e^{-h_{j}}p_{j})}=\sum_{j=1}^{\infty}\frac{j!}{j!-1}=\infty,$

implying that the $1$-KMS weights are unbounded. In fact, in this case the
$\beta$-KMS weights are unbounded if and only of $\beta\geq 1$.

####  6.5.3 Stabilization

Let $\sigma$ be a flow on $A$ and let $\mathbb{H}$ be a Hilbert space. Given a
flow $W=\\{W_{t}\\}_{t\in\mathbb{R}}$ of unitaries on $\mathbb{H}$ we define a
flow $\sigma\otimes\operatorname{Ad}W$ on $A\otimes\mathbb{K}(\mathbb{H})$
such that

$(\sigma\otimes\operatorname{Ad}W)_{t}(a\otimes k):=\sigma_{t}(a)\otimes
W_{t}kW_{t}^{*}$

when $a\in A,\ k\in\mathbb{K}(\mathbb{H})$. By Stone’s theorem there is a
self-adjoint operator $H$ on $\mathbb{H}$ such that

$W_{t}=e^{itH}$

for all $t\in\mathbb{R}$. For any $\beta\in\mathbb{R}$ we denote by
$\kappa_{\beta}$ the $\beta$-KMS weight for $\operatorname{Ad}W$ on
$\mathbb{K}(\mathbb{H})$ given by the formula

$\kappa_{\beta}(k)=\sup\left\\{\operatorname{Tr}\left(e^{\frac{-\beta
H}{2}}f(H)kf(H)e^{-\frac{\beta H}{2}}\right):\ f\in C_{c}(\mathbb{R}),\ 0\leq
f\leq 1\right\\},$

cf. Theorem 6.4.7.

######  Lemma 6.5.5.

For each $\beta\in\mathbb{R}$ there is a bijection

$\rho:\operatorname{KMS}(\sigma,\beta)\to\operatorname{KMS}(\sigma\otimes\operatorname{Ad}W,\beta)$

defined by

$\rho(\psi)=\psi\otimes\kappa_{\beta}.$

###### Proof.

The map $\rho$ is defined thanks to Theorem 6.5.3. We note that $\rho$ is the
composition $\rho=\rho_{2}\circ\rho_{1}$ of the map
$\rho_{1}:\operatorname{KMS}(\sigma,\beta)\to\operatorname{KMS}(\sigma\otimes\operatorname{id}_{\mathbb{K}(\mathbb{H})},\beta)$
given by $\rho_{1}(\psi):=\psi\otimes\operatorname{Tr}$ and the bijection
$\rho_{2}:\operatorname{KMS}(\sigma\otimes\operatorname{id}_{\mathbb{K}(\mathbb{H})},\beta)\to\operatorname{KMS}(\sigma\otimes\operatorname{Ad}W,\beta)$
coming from Theorem 6.3.7 by taking $u_{t}=1_{A}\otimes W_{t}$. It suffices
therefore to show that $\rho_{1}$ is a bijection. For this we pick a
1-dimensional projection $e\in\mathbb{K}(\mathbb{H})$. Then $A\otimes e$ is a
hereditary $C^{*}$-subalgebra of $A\otimes\mathbb{K}(\mathbb{H})$ which is
full and $\sigma\otimes\operatorname{id}_{\mathbb{K}(\mathbb{H})}$-invariant.
Let
$\phi\in\operatorname{KMS}(\sigma\otimes\operatorname{id}_{\mathbb{K}(\mathbb{H})},\beta)$.
It follows from Theorem 3.1.8 that there is $\beta$-KMS weight $\psi$ for
$\sigma$ such that

$\mathcal{N}_{\psi}=\left\\{a\in A:\ a\otimes e\in\mathcal{N}_{\phi}\right\\}$

and $\phi(a^{*}a\otimes e)=\psi(a^{*}a)$ for all $a\in\mathcal{N}_{\psi}$.
Then Corollary 5.2.5 implies that $\phi|_{A\otimes
e}=\psi\otimes\operatorname{Tr}|_{A\otimes e}$ and hence that
$\phi=\psi\otimes\operatorname{Tr}$ by Laca-Neshveyevs theorem or Theorem
3.1.8 again. This shows that $\rho_{1}$ is surjective. If
$\rho_{1}(\psi)=\rho_{1}(\psi^{\prime})$ it follows that

$\psi(a^{*}a)=\rho_{1}(\psi)(a^{*}a\otimes
e)=\rho_{1}(\psi^{\prime})(a^{*}a\otimes e)=\psi^{\prime}(a^{*}a)$

for all $a\in\mathcal{N}_{\psi}\cap\mathcal{N}_{\psi^{\prime}}$. Hence
$\psi=\psi^{\prime}$ by Corollary 5.2.5. ∎

######  Example 6.5.6.

The intention with this example is to illustrate Theorem 6.4.7 and in
particular the significance of the lower semi-continuity assumption in the
definition of a weight and a KMS weight. Let
$A=C_{0}(\mathbb{N},M_{2}(\mathbb{C}))$ be the $C^{*}$-algebra of sequences
$\\{a_{n}\\}_{n=1}^{\infty}$ in $M_{2}(\mathbb{C})$ with the property that
$\lim_{n\to\infty}\|a_{n}\|=0$. Let $\omega$ be the trace state of
$M_{n}(\mathbb{C})$. For any non-zero sequence $t=\\{t_{n}\\}_{n=1}^{\infty}$
of non-negative real numbers we can define a trace $\tau_{t}$ on $A$ by

$\tau_{t}(\\{a_{n}\\}_{n=1}^{\infty})=\sum_{n=1}^{\infty}t_{n}\omega(a_{n})$

when $\\{a_{n}\\}_{n=1}^{\infty}\in A^{+}$. This is a lower semi-continuous
trace on $A$, and every lower semi-continuous trace of $A$ arises in this way.
For each $h=h^{*}\in M_{2}(\mathbb{C})$ we can define an inner flow $\sigma$
on $A$ such that

$\sigma_{t}(\\{a_{n}\\}_{n=1}^{\infty}):=\\{e^{ith}a_{n}e^{-ith}\\}_{n=1}^{\infty}.$

It follows from Theorem 6.4.7 that for any $\beta\in\mathbb{R}$ the set of
$\beta$-KMS weights for $\sigma$ is in bijection with the set of lower semi-
continuous traces on $A$ and hence with the set of non-zero sequences of non-
negative real numbers.

The point we want to make here is that there are many traces on $A$ that are
not lower semi-continuous. Let $\xi\in\beta\mathbb{N}\backslash\mathbb{N}$ be
a free ultrafilter in $\mathbb{N}$ identified with an element of the Stone-
Čech compactification of $\mathbb{N}$. Define $\mu:A^{+}\to[0,\infty]$ such
that

$\displaystyle\mu((a_{n})_{n=1}^{\infty})$
$\displaystyle=\begin{cases}\lim_{N\to\xi}\frac{1}{N}\sum_{n=1}^{N}n\omega(a_{n}),&\
\text{when $\\{\frac{1}{N}\sum_{n=1}^{N}n\omega(a_{n})\\}_{N=1}^{\infty}$ is
bounded}\\\ \infty,&\ \text{when
$\\{\frac{1}{N}\sum_{n=1}^{N}n\omega(a_{n})\\}_{N=1}^{\infty}$ is
unbounded.}\end{cases}$

It is straightforward to see that this is a trace on $A$. To see that it is
not lower semi-continuous define $b(k)\in A^{+}$ such that

$b(k)_{n}=\begin{cases}\frac{1}{n},&\ n\leq k\\\ 0,&\ n>k.\end{cases}$

Then $b(k)\leq b(k+1)$ and $\lim_{k\to\infty}b(k)=b$, where $b\in A^{+}$ is
the element $b=\\{\frac{1}{n}\\}_{n=1}^{\infty}$. Since $\mu(b(k))=0$ for all
$k$ while $\mu(b)=1$ we conclude that $\omega$ is not lower semi-continuous.

For each $\beta\in\mathbb{N}$ this trace $\mu$ gives rise to a map
$\mu_{\beta}:A^{+}\to[0,\infty]$ defined such that

$\mu_{\beta}(\\{a_{n}\\}_{n=1}^{\infty})=\mu(\\{e^{-\beta
h}a_{n}\\}_{n=1}^{\infty}).$

This map $\mu_{\beta}$ has the first two of the three properties required of a
weight in Section 1.2 and it has the properties relative to $\sigma$ specified
by (1), (2) and (3) in Kustermans’ theorem, Theorem 2.3.1. But it is not a
$\beta$-KMS weight since it is not lower semi-continuous.

This observation, that there are weights with all properties required by a KMS
weight except for the lower semi-continuity assumption, may seem out of place
because all the major results on KMS weights we have presented depend on this
continuity assumption. But we note that in the first definition of a KMS
weight for flows given by F. Combes in Definition 4.1 of [C1] lower semi-
continuity is not mentioned.

##  Chapter 7 Dual KMS weights

In this chapter we describe a method to construct KMS weights for flows on a
crossed product $C^{*}$-algebra starting from KMS weight for a flow on the
core algebra.

###  7.1 Extending KMS weights to a crossed product

Let $A$ be a $C^{*}$-algebra, $G$ a locally compact group and $\alpha$ a
representation of $G$ by automorphisms $\alpha_{x},x\in G$, of $A$ such that
$x\mapsto\alpha_{x}(a)$ is continuous for all $a\in A$. Let $\sigma$ be a flow
on $A$ which commutes with $\alpha$ in the sense that

$\sigma_{t}\circ\alpha_{x}=\alpha_{x}\circ\sigma_{t}$

for all $x\in G$ and $t\in\mathbb{R}$. Let

$B:=A\rtimes_{r,\alpha}G$

be the reduced crossed product constructed from $(A,G,\alpha)$ as in Section
7.7 of [Pe] or in [Wi]. Let $C_{c}(G,A)$ denote the set of continuous
compactly supported functions $f:G\to A$. Then $C_{c}(G,A)$ is a $*$-algebra
with the convolution-like product

$(f\star g)(x):=\int_{G}f(y)\alpha_{y}(g(y^{-1}x))\ \mathrm{d}y$

and involution

$f^{\sharp}(x):=\Delta(x)^{-1}\alpha_{x}(f(x^{-1})^{*});$

see [Pe]. Here $\Delta$ is the modular function of $G$ and $dy$ denotes
integration with respect to a left-invariant Haar measure which we denote by
$\mu$ in the following. We assume that $A$ is realized as a $C^{*}$-subalgebra
of the bounded operators on a Hilbert space $\mathbb{H}$ in a
$\sigma$-covariant way; i.e. we assume that there is a strongly continuous
unitary representation $U_{t},t\in\mathbb{R}$, of $\mathbb{R}$ on $\mathbb{H}$
such that $\operatorname{Ad}U_{t}(a)=\sigma_{t}(a)$ for all $t\in\mathbb{R}$
and all $a\in A$. Such a covariant representation always exists, cf. Theorem
7.6.6 of [Pe]. We consider the corresponding injective representation

$\pi:C_{c}(G,A)\to B(L^{2}(G,\mathbb{H}))$

defined such that

$(\pi(f)\xi)(x)=\int_{G}\alpha_{x^{-1}}(f(y))\xi(y^{-1}x)\ \mathrm{d}y$

for $f\in C_{c}(G,A)$ and $\xi\in L^{2}(G,\mathbb{H})$, cf. 7.7.1 and 7.7.5 in
[Pe]. The $C^{*}$ algebra $B=A\rtimes_{r,\alpha}G$ is the closure in
$B(L^{2}(G,\mathbb{H}))$ of $\pi\left(C_{c}(G,A)\right)$. We shall work almost
exclusively with $C_{c}(G,A)$ and will therefore mostly suppress the
representation $\pi$ in the notation.

Let $\theta:G\to\mathbb{R}$ be a continuous group homomorphism. Define
unitaries $S_{t}$ on $L^{2}(G,\mathbb{H})$ by

$S_{t}\xi(x):=e^{i\theta(x)t}U_{t}\xi(x).$

Then

$\operatorname{Ad}S_{t}(\pi(f))=\pi(\sigma^{\theta}_{t}(f))$

for all $t\in\mathbb{R}$ and $f\in C_{c}(G,A)$. It follows that there is a
flow $\sigma^{\theta}$ on $B$ defined such that $\sigma^{\theta}_{t}(f)\in
C_{c}(G,A)$ and

$\sigma^{\theta}_{t}(f)(x)=e^{i\theta(x)t}\sigma_{t}(f(x))$ (7.1.1)

when $f\in C_{c}(G,A)$.

Let $\beta\in\mathbb{R}$ be a real number and let $\psi$ be a $\beta$-KMS
weight for $\sigma$ which is scaled by $\alpha$ such that

$\psi\circ\alpha_{x}=\Delta(x)^{-1}e^{-\theta(x)\beta}\psi\ \ \ \ \ \forall
x\in G.$ (7.1.2)

We fix the above setting in the rest of this section. The intention is to
construct from $\psi$ a $\beta$-KMS weight $\widehat{\psi}$ for
$\sigma^{\theta}$.

The first step is to relate $\alpha$ and $\Lambda_{\psi}$. Let
$a\in\mathcal{N}_{\psi}$. It follows from (7.1.2) that
$\Delta(x)^{\frac{1}{2}}e^{\frac{\theta(x)\beta}{2}}\alpha_{x}(a)\in\mathcal{N}_{\psi}$
and

$\displaystyle\left\|\Lambda_{\psi}\left(\Delta(x)^{\frac{1}{2}}e^{\frac{\theta(x)\beta}{2}}\alpha_{x}(a)\right)\right\|^{2}=\Delta(x)e^{\theta(x)\beta}\psi(\alpha_{x}(a^{*}a))$
$\displaystyle=\psi(a^{*}a)=\left\|\Lambda_{\psi}(a)\right\|^{2}.$

We can therefore define $W_{x}\in B(H_{\psi})$ such that

$W_{x}\Lambda_{\psi}(a):=\Delta(x)^{\frac{1}{2}}e^{\frac{\theta(x)\beta}{2}}\Lambda_{\psi}(\alpha_{x}(a))$

for all $a\in\mathcal{N}_{\psi}$.

######  Lemma 7.1.1.

$(W_{x})_{x\in G}$ is a continuous unitary representation of $G$ on
$H_{\psi}$.

###### Proof.

It is straightforward to check that $W$ is a unitary representation. To check
that it is continuous it is then sufficient to show that

$\lim_{x\to
e}\left<W_{x}\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>=\left<\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>$
(7.1.3)

for $a,b\in\mathcal{N}_{\psi}$. Let $\epsilon>0$. It follows from Combes’
theorem, Theorem 1.1.1, that there is $\omega\in\mathcal{F}_{\psi}$ such that

$\psi(a^{*}a)-\epsilon\leq\omega(a^{*}a)\leq\psi(a^{*}a).$

By Lemma 1.2.5 there is an operator $T_{\omega}$ on $H_{\psi}$ such that
$0\leq T_{\omega}\leq 1$ and

$\omega(c^{*}d)=\left<T_{\omega}\Lambda_{\psi}(d),\Lambda_{\psi}(c)\right>\ \
\ \ \ \forall c,d\in\mathcal{N}_{\psi}.$

The same calculation as in the proof of Lemma 2.2.1 shows that

$\left\|T_{\omega}\Lambda_{\psi}(a)-\Lambda_{\psi}(a)\right\|^{2}\leq
2\epsilon$

and we find therefore that

$\displaystyle\left|\left<W_{x}\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>-\left<\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>\right|$
$\displaystyle=\left|\left<\Lambda_{\psi}(a),W_{x^{-1}}\Lambda_{\psi}(b)\right>-\left<\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>\right|$
$\displaystyle\leq
2\left\|\Lambda_{\psi}(b)\right\|\sqrt{2\epsilon}+\left|\left<T_{\omega}\Lambda_{\psi}(a),W_{x^{-1}}\Lambda_{\psi}(b)\right>-\left<T_{\omega}\Lambda_{\psi}(a),\Lambda_{\psi}(b)\right>\right|$
$\displaystyle=2\left\|\Lambda_{\psi}(b)\right\|\sqrt{2\epsilon}+\left|\Delta(x)^{-\frac{1}{2}}e^{-\frac{\theta(x)\beta}{2}}\omega(\alpha_{x^{-1}}(b)^{*}a)-\omega(b^{*}a)\right|$

for all $x\in G$. Since

$\lim_{x\to
e}\left|\Delta(x)^{-\frac{1}{2}}e^{-\frac{\theta(x)\beta}{2}}\omega(\alpha_{x^{-1}}(b)^{*}a)-\omega(b^{*}a)\right|=0$

we get (7.1.3). ∎

######  Lemma 7.1.2.

Let $f\in C_{c}(G,A)$ and assume that $f(x)\in\mathcal{N}_{\psi}^{*}$ for all
$x\in G$. Then $G\ni x\to\Lambda_{\psi}(f^{\sharp}(x))$ is continuous if and
only if $G\ni x\to\Lambda_{\psi}(f(x)^{*})$ is continuous.

###### Proof.

This follows from Lemma 7.1.1 since

$\Lambda_{\psi}(f^{\sharp}(x))=\Delta(x)^{-1}\Lambda_{\psi}\left(\alpha_{x}(f(x^{-1})^{*})\right)=\Delta(x)^{-\frac{3}{2}}e^{-\frac{\theta(x)\beta}{2}}W_{x}\Lambda_{\psi}(f(x^{-1})^{*}).$

∎

Recall from Corollary 2.2.2 that $\mathcal{N}_{\psi}$ has a structure as a
Banach algebra in the norm

$|||a|||:=\|a\|+\left\|\Lambda_{\psi}(a)\right\|.$

Denote by $C_{0}(G,\mathcal{N}_{\psi})$ the space of continuous functions from
$G$ to

$(\mathcal{N}_{\psi},|||\ \cdot\ |||)$

that vanish at infinity. This is also a Banach algebra; the norm is

$|||f|||:=\sup_{x\in G}|||f(x)|||.$

There is a flow $\overline{\sigma}$ of isometries on
$C_{0}(G,\mathcal{N}_{\psi})$ defined by

$\overline{\sigma}_{t}(f)(x):=\sigma_{t}(f(x)).$

Note that $C_{0}(G,\mathcal{N}_{\psi})\subseteq C_{0}(G,A)$. We denote by
$\mathcal{A}_{\overline{\sigma}}$ the algebra of elements of
$C_{0}(G,\mathcal{N}_{\psi})$ that are entire analytic for
$\overline{\sigma}$. Set

$B_{0}:=C_{c}(G,A)\cap\mathcal{A}_{\overline{\sigma}}.$

######  Lemma 7.1.3.

$B_{0}$ is subalgebra of $B$.

###### Proof.

It is obvious that $B_{0}$ is a subspace of $B$. Let $f,g\in B_{0}$. To show
that $f\star g\in B_{0}$, note that it follows from from Lemma A.2.1 that

$f\star g(x)=\int_{G}f(y)\alpha_{y}(g(y^{-1}x))\
\mathrm{d}y\in\mathcal{N}_{\psi}$

for all $x\in G$ and

$\displaystyle\Lambda_{\psi}(f\star
g(x))=\int_{G}\Lambda_{\psi}\left(f(y)\alpha_{y}(g(y^{-1}x))\right)\
\mathrm{d}y$
$\displaystyle=\int_{G}\pi_{\psi}(f(y))\Lambda_{\psi}\left(\alpha_{y}(g(y^{-1}x))\right)\
\mathrm{d}y.$

From this expression it follows that $f\star g\in
C_{0}(C,\mathcal{N}_{\psi})$. To show that $f\star g\in B_{0}$ we must show
that $f\star g\in\mathcal{A}_{\overline{\sigma}}$. For this note that there
are sequences of functions $\\{a_{n}\\}$ and $\\{b_{n}\\}$ in
$C_{0}(G,\mathcal{N}_{\psi})$ such that

$\sigma_{t}(f(x))=\sum_{n=0}^{\infty}a_{n}(x)t^{n}$

and

$\sigma_{t}(g(x))=\sum_{n=0}^{\infty}b_{n}(x)t^{n}$

both converge in $(\mathcal{N}_{\psi},|||\ \cdot\ |||)$ uniformly in $x$ for
all $t\in\mathbb{R}$. In fact,

$\sum_{n=0}^{\infty}\sup_{x\in G}|||a_{n}(x)|||\ |z|^{n}<\infty,$
$\sum_{n=0}^{\infty}\sup_{x\in G}|||b_{n}(x)|||\ |z|^{n}<\infty,$

for all $z\in\mathbb{C}$, and there are compact sets $K$ and $L$ in $G$ such
that all $a_{n}$’s vanish outside of $K$ and all $b_{n}$’s vanish outside of
$L$. Indeed, if $h:G\to[0,1]$ is a compactly supported continuous function
such that $h(x)=1$ for all $x\in\operatorname{supp}f$, the sequence
$\\{ha_{n}\\}$ will have the same properties as $\\{a_{n}\\}$ and hence
$ha_{n}=a_{n}$ for all $n$ by the uniqueness part of Lemma 2.1.1, and we can
therefore take $K=\operatorname{supp}h$. The set $L$ is constructed in the
same way.

Set

$c_{n}(x):=\sum_{k=0}^{n}\int_{G}a_{k}(y)\alpha_{y}(b_{n-k}(y^{-1}x))\
\mathrm{d}y.$

An application of Lemma A.1.2 and (A.1.1) shows that $c_{n}\in
C_{0}(G,\mathcal{N}_{\psi})$. Note that $c_{n}$ is supported in $KL$ and

$\displaystyle\overline{\sigma}_{t}(f\star g)(x)=\sigma_{t}((f\star
g)(x))=\int_{G}\sigma_{t}(f(y))\alpha_{y}(\sigma_{t}(g(y^{-1}x)))\
\mathrm{d}y$
$\displaystyle=\int_{G}(\sum_{n=0}^{\infty}a_{n}(y)t^{n})\alpha_{y}(\sum_{n=0}^{\infty}b_{n}(y^{-1}x)t^{n})\
\mathrm{d}y$
$\displaystyle=\int_{G}\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_{k}(y)\alpha_{y}(b_{n-k}(y^{-1}x))\right)t^{n}\
\mathrm{d}y$ $\displaystyle=\sum_{n=0}^{\infty}c_{n}(x)t^{n}.$

Note also that

$\sup_{x\in G}\left\|c_{n}(x)\right\|\leq\sum_{k=0}^{n}(\sup_{x\in
G}\|a_{k}(x)\|)(\sup_{x\in G}\left\|b_{n-k}(x)\right\|)\mu(KL).$

Using Lemma A.2.1 we find that

$\displaystyle\Lambda_{\psi}(c_{n}(x))=\sum_{k=0}^{n}\int_{G}\pi_{\psi}(a_{k}(y))\Lambda_{\psi}(\alpha_{y}(b_{n-k}(y^{-1}x)))\
\mathrm{d}y$
$\displaystyle=\sum_{k=0}^{n}\int_{G}\Delta(y)^{-\frac{1}{2}}e^{-\frac{\theta(y)\beta}{2}}\pi_{\psi}(a_{k}(y))W_{y}\Lambda_{\psi}(b_{n-k}(y^{-1}x))\
\mathrm{d}y$

and hence

$\displaystyle\sup_{x\in G}\|\Lambda_{\psi}(c_{n}(x))\|\leq
M\sum_{k=0}^{n}(\sup_{x\in G}\left\|a_{k}(x)\right\|)(\sup_{x\in
G}\left\|\Lambda_{\psi}(b_{n-k}(x))\right\|)$

where $M=\int_{KL}\Delta(y)^{-\frac{1}{2}}e^{-\frac{\theta(y)\beta}{2}}\
\mathrm{d}y$. It follows that

$\sup_{x\in G}|||c_{n}(x)|||\leq\max\\{\mu(KL),M\\}\sum_{k=0}^{n}(\sup_{x\in
G}|||a_{k}(x)|||)(\sup_{x\in G}|||b_{n-k}(x)|||),$

and we conclude that

$\displaystyle\sum_{n=0}^{\infty}\sup_{x\in G}|||c_{n}(x)|||\ |z|^{n}$
$\displaystyle\leq\max\\{\mu(KL),M\\}\left(\sup_{x\in G}|||a_{n}(x)|||\
|z|^{n}\right)\left(\sup_{x\in G}|||b_{n}(x)|||\ |z|^{n}\right)<\infty$

for all $z\in\mathbb{C}$. It follows that

$\overline{\sigma}_{t}(f\star g)=\sum_{n=0}^{\infty}c_{n}t^{n}$

converges in $C_{0}(G,\mathcal{N}_{\psi})$ for all $t\in\mathbb{R}$, proving
that $f\star g\in\mathcal{A}_{\overline{\sigma}}$. Hence $f\star g\in B_{0}$.
∎

######  Lemma 7.1.4.

Let $f\in B_{0}$. Then

  * (a)

$f$ is entire analytic for $\sigma^{\theta}$ and $f(x)\in\mathcal{A}_{\sigma}$
for all $x\in G$,

  * (b)

$\sigma_{z}(f(x))\in\mathcal{N}_{\psi}$ for all $z\in\mathbb{C}$ and all $x\in
G$,

  * (c)

$G\ni x\mapsto\sigma_{z}(f(x))$ is continuous for all $z\in\mathbb{C}$,

  * (d)

$G\ni x\mapsto\Lambda_{\psi}(\sigma_{z}(f(x)))$ is continuous for all
$z\in\mathbb{C}$,

  * (e)

$\sigma_{z}^{\theta}(f)\in B_{0}$ for all $z\in\mathbb{C}$,

  * (f)

$\sigma^{\theta}_{z}(f)(x)=e^{i\theta(x)z}\sigma_{z}(f(x))$ for all $x\in G$
and all $z\in\mathbb{C}$.

###### Proof.

(a)+(e)+(f): Choose a continuous function $h:G\to[0,1]$ of compact support
such that $h(x)=1$ for $x$ in the support of $f$. Then

$\sigma^{\theta}_{t}(f)(x)=e^{i\theta(x)h(x)t}\sigma_{t}(f(x))$

for all $x\in G$. This has the effect that we may substitute $\theta h$ for
$\theta$ in the following calculations and hence work as if $\theta$ is
bounded. Since $f$ is entire analytic for $\overline{\sigma}$ there is a
sequence $\\{g_{n}\\}$ in $C_{0}(G,\mathcal{N}_{\psi})$ such that

$\sigma_{t}(f(x))=\sum_{n=0}^{\infty}g_{n}(x)t^{n}$

and

$\sum_{n=0}^{\infty}\sup_{x\in G}|||g_{n}(x)||||t|^{n}<\infty$

for all $t\in\mathbb{R}$. This implies, in particular, that
$f(x)\in\mathcal{A}_{\sigma}$ for all $x\in G$. Note that $hg_{n}=g_{n}$ by
uniqueness of $\\{g_{n}\\}$, cf. Lemma 2.1.1. Set

$h_{n}:=\sum_{k=0}^{n}\frac{(ih\theta)^{k}}{k!}g_{n-k}.$

Then

$\sigma^{\theta}_{t}(f)(x)=\sum_{n=0}^{\infty}h_{n}(x)t^{n}$

converges uniformly on $G$ with all terms supported in the compact support of
$h$. Therefore the sequence also converges in $L^{1}(G,A)$ and hence in $B$.
This shows that $f$ is entire analytic for $\sigma^{\theta}$, and

$\displaystyle\sigma_{z}^{\theta}(f)(x)=\sum_{n=0}^{\infty}h_{n}(x)z^{n}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(ih(x)\theta(x))^{k}}{k!}z^{k}g_{n-k}(x)z^{n-k}=e^{i\theta(x)z}\sigma_{z}(f(x))$

for all $z,x$. Note that

$x\mapsto
e^{i\theta(x)z}\sigma_{z}(f(x))=e^{i\theta(x)z}\sum_{n=0}^{\infty}g_{n}(x)z^{n}$

is continuous, showing that $\sigma^{\theta}_{z}(f)\in C_{c}(G,A)$. Note also
that

$\displaystyle\sigma_{t}(\sigma^{\theta}_{z}(f)(x))=e^{i\theta(x)z}\sum_{n=0}^{\infty}g_{n}(x)(t+z)^{n}$
$\displaystyle=e^{i\theta(x)z}\sum_{n=0}^{\infty}g_{n}(x)\sum_{k=0}^{n}{n\choose
k}t^{k}z^{n-k}=\sum_{n=0}^{\infty}\kappa_{k}(x)t^{k}$

where

$\kappa_{k}(x)=e^{ih(x)\theta(x)z}\sum_{n\geq k}{n\choose k}g_{n}(x)z^{n-k},$

and that

$\sum_{k=0}^{\infty}|||\kappa_{k}|||\leq K\sum_{k=0}^{\infty}\sum_{n\geq
k}{n\choose k}|||g_{n}|||\ |z|^{n-k},$

where $K=\sup_{x\in G}\left|e^{ih(x)\theta(x)z}\right|<\infty$. Since

$\sum_{k=0}^{\infty}\sum_{n\geq k}{n\choose k}|||g_{n}|||\
|z|^{n-k}\leq\sum_{n=0}^{\infty}|||g_{n}|||(1+|z|)^{n}<\infty,$

this shows that $\sigma^{\theta}_{z}(f)\in B_{0}$.

(b) + (c) + (d): Since $f$ is analytic for $\overline{\sigma}$ it follows that
$\overline{\sigma}_{z}(f)(x)=\sigma_{z}(f(x))\in\mathcal{N}_{\psi}$ for all
$z,x$. Since $\overline{\sigma}_{z}(f)\in C_{0}(G,\mathcal{N}_{\psi})$ it
follows that $G\ni x\mapsto\sigma_{z}(f(x))$ and $G\ni
x\mapsto\Lambda_{\psi}(\sigma_{z}(f(x))$ are continuous. ∎

Let $C_{c}(G,H_{\psi})$ denote the vector space of continuous compactly
supported functions from $G$ to $H_{\psi}$, and $L^{2}(G,H_{\psi})$ the
Hilbert space obtained by completing $C_{c}(G,H_{\psi})$ with respect to the
inner product

$\left<f,g\right>=\int_{G}\left<f(x),g(x)\right>\ \mathrm{d}x.$

Let $f\in B_{0}$. Then

$\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))=\Delta(x)^{\frac{1}{2}}e^{\frac{\theta(x)\beta}{2}}W_{x^{-1}}\Lambda_{\psi}(f(x)),$
(7.1.4)

and it follows from Lemma 7.1.1 and (d) of Lemma 7.1.4 that (7.1.4) depends
continuously on $x$. We can therefore define

$\Lambda_{0}:B_{0}\to L^{2}(G,H_{\psi})$

such that

$\Lambda_{0}(f)(x):=\Lambda_{\psi}(\alpha_{x^{-1}}(f(x))).$

Let $C_{c}(G,A)^{\psi}$ be the subspace of $C_{c}(G,A)$ consisting of the
functions $h\in C_{c}(G,A)$ with the properties that

  * •

$h(s)\in\mathcal{N}_{\psi}\cap\mathcal{N}_{\psi}^{*}$ for all $s\in G$, and

  * •

the functions $G\ni s\mapsto\Lambda_{\psi}(h(s))$ and $G\ni
s\mapsto\Lambda_{\psi}(h(s)^{*})$ are both continuous.

Note that $C_{c}(G,A)^{\psi}\subseteq C_{0}(G,\mathcal{N}_{\psi})$.

######  Lemma 7.1.5.

Let $f\in C_{c}(G,A)^{\psi}$ and define $\overline{R}_{n}(f)\in C_{c}(G,A)$
such that

$\overline{R}_{n}(f)(x)=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}\sigma_{s}(f(x))\
\mathrm{d}s.$

Then $\overline{R}_{n}(f)\in B_{0}\cap B_{0}^{\sharp}$,
$\lim_{n\to\infty}\overline{R}_{n}(f)=f$ in $B$ and

$\lim_{n\to\infty}\Lambda_{0}(\overline{R}_{n}(f))=\Lambda_{0}(f)$

in $L^{2}(G,H_{\psi})$.

###### Proof.

Since

$\overline{R}_{n}(f)=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}\overline{\sigma}_{s}(f)\
\mathrm{d}s$

it follows from Lemma 2.1.7 that $\overline{R}_{n}(f)\in B_{0}$ and that
$\lim_{n\to\infty}\overline{R}_{n}(f)=f$ in $C_{0}(G,\mathcal{N}_{\psi})$.
Thus $\lim_{n\to\infty}\overline{R}_{n}(f)(x)=f(x)$ uniformly and compactly
supported in $x$. Hence $\lim_{n\to\infty}\overline{R}_{n}(f)=f$ in
$L^{1}(G,A)$ and therefore also in $B$. Since $\sigma$ and $\alpha$ commute we
find that

$\overline{R}_{n}(f)^{\sharp}=\overline{R}_{n}(f^{\sharp}).$

It follows from Lemma 7.1.2 that $f^{\sharp}\in C_{c}(G,A)^{\psi}$, and we
conclude therefore that $\overline{R}_{n}(f)\in B_{0}^{\sharp}$. By using
Lemma A.2.1 we find that

$\begin{split}&\Lambda_{0}(\overline{R}_{n}(f))(x)=\Lambda_{\psi}\left(\alpha_{x^{-1}}(\overline{R}_{n}(f)(x))\right)\\\
&=\Lambda_{\psi}\left(\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}\sigma_{s}(\alpha_{x^{-1}}(f(x)))\
\mathrm{d}s\right)\\\
&=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}\Lambda_{\psi}(\sigma_{s}(\alpha_{x^{-1}}(f(x))))\
\mathrm{d}s\\\
&=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))\
\mathrm{d}s.\\\ \end{split}$ (7.1.5)

By Lemma 2.1.7

$\lim_{n\to\infty}\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))\
\mathrm{d}s=\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))$ (7.1.6)

for each $x$. Note that

$\displaystyle\left\|\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))\
\mathrm{d}s-\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x_{0}^{-1}}(f(x_{0})))\
\mathrm{d}s\right\|$
$\displaystyle\leq\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}\left\|\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))-\alpha_{x_{0}^{-1}}(f(x_{0}))))\right\|\
\mathrm{d}s$
$\displaystyle=\left\|\Lambda_{\psi}(\alpha_{x^{-1}}(f(x))-\alpha_{x_{0}^{-1}}(f(x_{0})))\right\|$
$\displaystyle=\left\|\Delta(x)^{\frac{1}{2}}e^{\frac{\theta(x)\beta}{2}}W_{x^{-1}}\Lambda_{\psi}(f(x))-\Delta(x_{0})^{\frac{1}{2}}e^{\frac{\theta(x_{0})\beta}{2}}W_{x_{0}^{-1}}\Lambda_{\psi}(f(x_{0}))\right\|.$

This estimate and an application of Lemma 7.1.1 shows that the sequence of
functions

$G\ni
x\to\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))\
\mathrm{d}s,\ n\in\mathbb{N},$

is equicontinuous in the sense that for each $x_{0}\in G$ and $\epsilon>0$
there is an open neighbourhood $\Omega_{x_{0}}$ of $x_{0}$ with the property
that

$\displaystyle\left\|\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x^{-1}}(f(x)))\
\mathrm{d}s-\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}e^{-ns^{2}}U^{\psi}_{s}\Lambda_{\psi}(\alpha_{x_{0}^{-1}}(f(x_{0})))\
\mathrm{d}s\right\|\leq\epsilon$

for all $n$ and all $x\in\Omega_{x_{0}}$. Since all the functions vanish
outside the compact support of $f$, it follows that the convergence in (7.1.6)
is in fact uniform in $x$. Using (7.1.5) and (7.1.6) we find that