How the modular theory of von Neumann algebras, deal with generating C*-algebras?

Question

Let $H$ be a separable infinite dimensional Hilbert space, $M \subset B(H)$ a von Neumann algebra and $A \subset M$ a separable ${\rm C}^*$-algebra such that $A''=M$. Suppose the existence of a bicyclic vector $\Omega$ for $M$ (i.e. cyclic and separating: $M\Omega$ and $M' \Omega$ dense in $H$). Let $\sigma_t^{\Omega} \in Aut(M)$ the modular action.

Let $K(A,\Omega):= \{ t \in \mathbb{R} \ \vert \ \sigma_t^{\Omega}(A)\subset A \}$ a subgroup of $\mathbb{R}$.

Question: Are they $A$ and $\Omega$ (as above) such that $K(A,\Omega) \neq \{0\}$? (If yes) such that $K(A,\Omega) = \mathbb{R}$?

Example: Let $M \subset B(H)$ be a von Neumann algebra and $\Omega \in H$ bicyclic. Suppose there is a finite set $S \subset M$ such that $S'' = M$ and let $\mathcal{A}$ be the $\star$-algebra generated by $\sigma_{\mathbb{R}}^{\Omega}(S)$.
What is $\mathcal{A}$? A separable ${\rm C}^*$-algebra or $M$ or something else?

Answer 1

All groups algebra are trivial examples because for them the modular time evolution is (can be chosen) trivial: the modular time evolution attached to a state is trivial if and only if the state is a trace and group algebra always have a trace.

The algebras of the various Bost-Connes type system have this property (that $K=\mathbb{R}$) with a non trivial time evolution (see Hecke algebras, type III factors and phase transitions with spontaneous symmetry breaking in number theory for the original one, and Paugam et Ha for the most general version as well as references to other system of this kind).

More generally, this situation (having $K=\mathbb{R}$) is equivalent to having a $C^*$-algebra with a one parameter group of automorphism and a KMS $1$ state. The KMS condition (for temperature $1$) is equivalent to the fact that in the GNS representation induced by this state the modular time evolution of the double commutant extend the initial group of automorhism. KMS state at other temperature corresponds to the same situation but with a linear change of parameter in the automorphism groups, so they are also examples.

So this is a fairly frequent situation: basically any $C^*$-algebra that have interesting KMS states produces plenty of examples like this... and finding examples of this kind is roughly the same as finding algebra with an automorphism group which have KMS states.

Edit : Due to Sebastien Palcoux precision, I'm now answering a different question:

If $M$ is a von Neuman algebra with separable predual (equivalently acting on a separable Hilbert space) and $(\sigma_t)_{t\in \mathbb{R}}$ is some weakly continuous one parameter group of automorphism of $M$ then one can always find a weakly dense separabe sub $C^*$-algebra $A$ which is stable under the action of $\sigma_t$. In fact, any separable sub-algebra of $M$ is contained into a separable sub-algebra stable by $\sigma_t$.

Indeed, start with $A_0$ some separable sub-$C^*$-algebra of $M$. and $a_1, \dots,a_n,\dots$ some countable dense subfamily.

Pick also a countable dense subset $f_0, \dots,f_m,\dots$ of $L^1(\mathbb{R})$.

Consider the set of element: $$k_{n,m} = \int_{\mathbb{R}} f_m(t) \sigma_t(a_n) dt$$ I claim that the $C^*$-algebra generated by the $k_{n,m}$ is a seprable sub-$C^*$-algebra of $M$ stable by $\sigma_t$ and containing $A_0$. I sketche the proof:

1) it is separable because the $*$-algebra generated by a countable set is countable, hence its closure (the $C^*$- algebra generated by the coutnable set) is separable.

2) it contains all the element of the form $\int_{\mathbb{R}} g(t) \sigma_t(a_n)$ for $g(t) \in L^1(\mathbb{R})$ simply by taking the limit of the $k_{n,m}$ for $f_n \rightarrow g$

3) it is stable under $\sigma_t$ : $\sigma_x(k_{n,m})$ will be an integral of the form above (with $f_m$ shifted by $x$).

4) it contains $A_0$ : using the continuity of $t \mapsto \sigma_t(a)$ at $t=0$ and a function $g \in L^1(\mathbb{R})$ with support near $0$ and of integral $1$ one can approximate $a_n$ by some $\int g(t) \sigma_t(a_n) dt$. In particular if $A_0$ was weakly dense, $A$ is also weakly dense.