Oscillator representation

Remarks

Such a representation exists and is unique. For each $g \in \mathrm{Sp}(W)$, the representation $(w, s) \mapsto \rho(gw, s)$ has the same central character as $\rho$. So by the Stone–von Neumann theorem there is a unitary $T_g$ with $T_g \rho(w, s) T_g^{-1} = \rho(gw, s)$, unique up to a scalar of modulus $1$. The map $g \mapsto T_g$ is a projective representation of $\mathrm{Sp}(W)$, and the scalars can be chosen so that it lifts to a genuine representation of $\mathrm{Mp}(W)$. Two lifts differ by a character of $\mathrm{Mp}(W)$, which is trivial since $\mathrm{Mp}(W)$ is connected with semisimple Lie algebra, hence equal to its commutator subgroup.

Example: for $n = 1$, the dilation $g = \operatorname{diag}(a, a^{-1})$ with $a > 0$ lifts to the operator $\psi(x) \mapsto a^{-1/2} \psi(x / a)$, up to sign. It is unitary on $L^2(\R)$. Conjugation by it turns the position operator $x$ into $x / a$ and the momentum operator $-i \, d/dx$ into $a \cdot (-i \, d/dx)$, which is the action of a diagonal symplectic map on position and momentum.

Non-example: the operators $T_g$ cannot be normalised to give a representation of $\mathrm{Sp}(W)$ itself. Only the double cover carries a genuine representation, so the condition on $\mathrm{Mp}(W)$ in the definition cannot be replaced by one on $\mathrm{Sp}(W)$.

The oscillator representation splits into two irreducible pieces, on the even and on the odd functions.