Let $A$ be a nonunital C*-algebra and let $M(A)$ denote its multiplier algebra. Let $(u_t)_{t \in \mathbb{R}}$ be a strictly continuous 1-parameter group of unitary multipliers. That is, $u_t x \to x$ as $t \to 0$ for all $x \in A$. I am wondering whether there exists an unbounded operator $H : A \to A$ such that $u_t = e^{itH}$ in some sense? I would want some condititions on $H$:
- $H$ should be densely-defined and closed. That is, the domain of $H$ should be a dense subspace $A_0$ of $A$ and, whenever $x_n \in A_0$ are such that $x_n \to x \in A$ and $H x_n \to y \in A$ we should have $x \in A_0$ and $H x = y$.
- We should have $H$ symmetric with respect to the $A$-valued inner-product $\langle \cdot, \cdot \rangle : A \times A \to A$ given by $\langle x,y \rangle = x^* y$. That is, whenever $x,y \in A_0$, it should hold that $\langle Ax,y \rangle = \langle x,Ay \rangle$.
A good "core" for $H$ should be something like the set of smooth elements $x \in A$, that is elements for which $t \mapsto u_t x$ is a $C^\infty$ path $\mathbb{R} \to A$. There, we can define $H$ by $$ H(x) = \frac{1}{i} \frac{d}{dt} \big|_{t = 0} u_t x .$$
Notice that, for $x$ and $y$ smooth in the above sense, we have $$ \langle u_t x,y \rangle = (u_tx)^* y = x^* u_{-t} y = \langle x, u_{-t} y \rangle. $$ Differentiating both sides and evalutating at $t = 0$ we get gives the desired self-adjointness condition. So one, wants to do things like:
- Prove that smooth elements are a dense subalgebra
- Prove that the operator defined as above on smooth elements is closable, probably by using the symmetry condition
- Find a sense in which $e^{itH} = u_t$, probably strong convergence of the series $\sum_{n=0}^\infty \frac{(itH)^n}{n!}$ on the $\bigcap_{n=1}^\infty \operatorname{dom}(H^n)$ or something.
Hopefully someone can spell out what sorts of results are true in this context? Or point me towards references? Thanks.
There certainly exist references for this topic. Here is the one I found the most helpful.
I found it in the references of the paper
which is suggested by user "Argument" in the comments. There is also apparently a book, though I did not look at it.