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

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 $$\pi(\mathcal{W}) \subsetneq \mathcal{B}(\mathcal{H})$$?

$$\pi(\mathcal{W})$$ 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 $$\pi(\mathcal{W})\subsetneq B(H)$$.
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.