About the C*-algebra of the Schrodinger representation of the Weyl C*-algebra

Question

We start with the Weyl C*-algebra $\mathcal{W}$ for a finite dimensional symplectic space and we consider the irreducible Schrodinger representation $\pi:\mathcal{W}\rightarrow \mathcal{B}(\mathcal{H})$ where $\mathcal{H}=L^2(\mathbb{R}^n)$. Since such representation is irreducible, the von Neumann algebra generated by such represenation is $\pi(\mathcal{W})''=\mathcal{B}(\mathcal{H})$. The question now is: ¿Is the C*-algebra generated by such representation $\pi(\mathcal{W})=\mathcal{B}(\mathcal{H})$, or is it strictly smaller, i.e. $\pi(\mathcal{W}) \subsetneq \mathcal{B}(\mathcal{H})$? In such case, ¿Is there any useful characterization for such C*-algebra $\pi(\mathcal{W}) \subsetneq \mathcal{B}(\mathcal{H})$?

Answer 1

$\pi(\mathcal{W})$ is strictly smaller than $B(H)$.

Since the Weyl algebra $\mathcal{W}$ is simple, every nontrivial representation is faithful. Therefore $\pi:\mathcal{W}\rightarrow\pi(\mathcal{W})\subseteq B(H)$ is a bijective *-homomorphism (surjective to its image), hence $\pi(\mathcal{W})\cong\mathcal{W}$ as C${}^\ast$-algebras. I hope this answers your second question.

As for your first, $\pi(\mathcal{W})$ does not contain the compact operators (because it's simple), so $\pi(\mathcal{W})\subsetneq B(H)$.

For more info, see Photons In Fock Space And Beyond by Honegger and Rieckers.

Answer 2

I cannot be very explicit because I don't know what $\mathcal W$ is.

But, this can be said: if $\mathcal W$ is finite-dimensional, then so is $\mathcal H$ and $\pi(\mathcal W)=\mathcal B(\mathcal H)$. If, on the other hand, $\mathcal W$ is infinite-dimensional, it is likely that $\mathcal W$ is separable; in that case you always have $\pi(\mathcal W)\subsetneq \mathcal B(\mathcal H)$, since the latter is not separable.