Definition
Explicit
Given a symplectic vector space $(V,\omega)$ and a Darboux basis $(p,q)$ where $\omega = \sum dp_i \wedge dq_i$. Then given vectors $(\vec{p},\vec{q})$ and $(\vec{p}’,\vec{q}’)$ we have $$ \omega((\vec{p},\vec{q}),(\vec{p}’,\vec{q}’)) = \vec{p}\cdot\vec{q}’ - \vec{q}\cdot\vec{p}’. $$
In QM we take the basis elements $\{P_i\}$ of the vector $\vec{p} = \sum p_i P_i$ to be operators, and same for $\vec{q}$. The commutation relations then gives $$ [\sum(p_i P_i + q_i Q_i),\sum(p’_j P_j+q’_j Q_j)] = -i\omega((\vec{p},\vec{q}),(\vec{p}’,\vec{q}’)). $$ so for $v,w \in V$ we should have $$ [v,w]=-i\omega(v,w). $$ This is similar to the relations defining Clifford algebra $\{v,w\}=-2Q(v,w)$.
A $(2n+1)$-dimensional Lie algebra is useful when studying Heisenberg algebra
The Heisenberg Lie algebra $\mathfrak{h}_n$ of $\mathbb{R}^{2n}$ is a central extension of $\mathbb{R}^{2n}$ by $\mathbb{R}$, $$ 0\to \mathbb{R}\to \mathfrak{h}_n \to \mathbb{R}^{2n} \to 0. $$
If $n=1$ this is isomorphic to a Lie algebra of upper triangular matrices with the diagonal elements being $0$ with the Lie bracket being the matrix commutator. For $n=1$, $pP + qQ + cC$ is identified with the matrix $$ \begin{pmatrix} 0 & p & c \\ 0 & 0 & q \\ 0 & 0 & 0 \end{pmatrix}. $$ Exponentialing the Heisenberg Lie algebra $\mathfrak{h}_n$ we get what is called the Heisenberg group $\mathsf{H}_n$. The Baker-Campbell-Hausdorff formula yields the group law for $\mathsf{H}_n$ immediately. Let $(\vec{p},\vec{q},c)$ represent $e^{(\vec{p},\vec{q},c)}$, we have $$ (\vec{p},\vec{q},c) \cdot (\vec{p}’,\vec{q}’,c’) = (\vec{p}+\vec{p}’,\vec{q}+\vec{q}’,c+c’+\frac{1}{2}\omega((\vec{p},\vec{q}),(\vec{p}’,\vec{q}’))). $$
Definition (geometric)
An alternative and more geometric point of view. Given a symplectic vector space $(V,\omega)$, we have the linear function $f_v = \omega(v,\cdot) \in C^\infty(V)$, the Poisson bracket is then $$ [f_v ,f_w] = \omega(v,w). $$ Thus these linear functions generate the Heisenberg Lie algebra $\mathfrak{h}(V) = V\times \mathbb{R}$ with the Lie bracket given by $[v,w]= (\omega(v,w),0)$ for $v,w\in V$.
The Heisenberg group is then $\mathsf{H}(V) = V\times U(1)$ where multiplication is defined so that the circle $U(1)$ is central. In particular, $$ (v,0)\cdot (w,0) = (v+w ,\exp(\frac{i}{2}\omega(v,w))). $$
This actually means $\mathsf{H}(V)$ is the central extension of $V$ by $U(1)$ with the 2-cocycle $\text{exp}(\pi i E)$.
The group action is $$ (v,t)\cdot(v’,t’) = (v+v’,\exp(t+t’+\frac{1}{2}\omega(v,w))) $$
We see that a quantization of $(V,\omega)$ (in the sense of geometric quantization) should yield an irreducible unitary representation of $\mathsf{H}(V)$ in which $U(1)$ acts by scalar multiplication.
Definition (general)
This definition is available for any locally compact abeliang group $K$. Recall that for a locally compact abelian group $K$ its Pontryagin dual group $\hat{K}$ is the group of continuous homomorphisms $K\to U(1)$.
Stone-von Neumann Theorem and the representation
So there is a unique irreducible representation of the Heisenberg algebra (that can be exponentiated) when $C$ is fixed to act by for example $-i\hbar$.
The representation is constructed by choosing a Lagrangian subspace. Let $L\subset V$ be Lagrangian, then $L\oplus U(1) \subset \mathsf{H}(V)$ is a maximal abelian subgroup and has a 1-dimensional representation, the projection $\pi:L\oplus U(1)\to U(1) \subset U(\mathcal{H})$. The representation $(\mathcal{H}_L,\rho_L)$ of $\mathsf{H}(V)$ is then induced from $\pi$, so it is $\text{Ind}^{\mathsf{H}(V)}_{L\oplus U(1)}\pi$.
Let $\delta_{1/2}(V/L)$ be the space of half-densities on $V/L$ (this means $\delta_{1/2}(V/L)^{\otimes 2} = \delta_1(V/L)$ where $\delta_1(V/L)$ is the space of complex-valued Haar measures on $V/L$).
- The Hilbert space $\mathcal{H}_L$ thus obtained is the completion of the space of smooth functions $\phi:V\to \delta_{1/2}(V/L)$ satisfying $\phi(x+a) = \phi(x)e^{i\omega(x,a)/2}$ for $x\in V$ and $a\in L$, finite with respect to the norm $||\phi||^2 = \int_{V/L} \bar{\phi}\phi$.
- The action of $(v,t)\in \mathsf{H}(V)$ is given by $$ \rho_L(v,t)\phi(x) = \phi(x-v)e^{i\omega(v,x)/2}t. $$
Note that $\bar{\phi}\phi:V\to \delta_{1}V/L$ is constant along $L$ and defines a function $V/L\rightarrow \delta_1(V/L)$, hence a measure on $V/L$.
The symplectic group and metaplectic representation
The symplectic group $Sp(V)$ is the group of linear transformations of $(V,\omega)$ that leaves the symplectic form $\omega$ invariant. It is the analog of the special orthogonal group $SO(V)$ in the Clifford algebra case. Since the Heisenberg algebra and group are both defined by $\omega$, the group $Sp(V)$ acts as automorphisms.
The group $Sp(V)$ is not compact. Analogous to the one used to produce the projective spinor representation of $SO(V)$ we can produce a infinite dimensional projective representation, called the metaplectic representation. The universal cover, which is a double cover, $Mp(V)$ is called the metaplectic group.
We have seen that the isomorphism class of the irreducible unitary representation of $\mathsf{H}(V)$ is independent of $L$. However there are no canonical isomorphism between the representation Hilbert space for different Lagrangian subspaces.
Consider the action of $Sp(V)$. Each $g\in Sp(V)$ determines an isomorphism $\mathcal{H}_L \to \mathcal{H}_{gL}$ of Hilbert spaces and there is an isomorphism of representations $\mathcal{H}_{gL}\to \mathcal{H}_L$. The composition of these two maps then defines an operator $\rho_L(g)$ on $\mathcal{H}_L$. In general one cannot chose these operators in such a way that $\rho_L(g) \rho_L(g’) = \rho_L(gg’)$: We only have projective representations for $Sp(V)$ on the representation space $\mathcal{H}_L$. For genuine representation $Mp(V)$ is needed.
Coadjoint orbits
The adjoint action of $H(V)$ on $\mathfrak{h}(V)$ is given by the formula $$ \operatorname{Ad}_{(v,t)}(x,s) = (x,\omega(v,x)+s). $$ Note that $t$ is central so it acts trivially on $\mathfrak{h}(V)$.
The coadjoint action $\operatorname{Ad}^\ast$ is then $$ \omega(\operatorname{Ad}^\ast_{(v,t)}(x,s),(x’,s’)) = \omega((x,s),\operatorname{Ad}_{-v,-t}(x’,st)) = \omega(x,x’) + s(s’-\omega(v,x’)) = \omega(x-sv,x’) + ss' $$ so the coadjoint action is $$ \operatorname{Ad}^\ast_{(v,t)} (x,s) = (x-sv,s). $$ Note that this action is very similar to the group action $G\times H\to H$ that defines a crossed module for $G,H$ abelian.