A Momentum Map for the Heisenberg Group

Abstract

We look at a momentum map associated with the Heisenberg group. We show that the cocycle associated with its momentum mapping is the value of a modulus of an associated coadjoint orbit.

1. Introduction

The symmetry of a classical Hamiltonian system is encoded in its momentum mapping, see [1]. When the momentum mapping is coadjoint equivariant and the action of the symmetry group is transitive, the geometry of the momentum mapping is determined by the geometry of an associated coadjoint orbit. In this paper, we look at a coadjoint equivariant momentum map associated with the Heisenberg group, which arises from the affine action of ${\mathbb{R}}^{2n}$ on itself. This latter action is Hamiltonian, but its momentum mapping has a cocyle because it is not coadjoint equivariant. Following Cushman and van der Kallen [2], we algebraically classify the coadjoint orbits of the Heisenberg group. We obtain one orbit with a continuous parameter called a modulus. We show that the cocycle associated with the momentum map of the Heisenberg group is the value of the modulus of the coadjoint orbit. We give a representation theoretic description of this modulus.

2. The Affine Group ${\mathbb{R}}^{\mathbf{2}\mathbf{n}}$

Let $({\mathbb{R}}^{2n},\omega )$ be a real symplectic vector space, where $\omega$ is the standard symplectic form on ${\mathbb{R}}^{2n}$ with matrix $J_{2n}=$$\left(\begin{array}{cc}0& -I_n\\ I_n& 0\end{array}\right)$ with respect to the standard basis ${\mathfrak{e}}_n=\{e_1,\dots ,e_n,e_{n+1}=f_1,\dots ,e_{2n}=f_n\}$. Consider the affine action

$$\mathsf{\Phi }:{\mathbb{R}}^{2n}\times ({\mathbb{R}}^{2n},\omega )\to ({\mathbb{R}}^{2n},\omega ):(a,v)\mapsto v+a.$$

(1)

For each x in the abelian Lie algebra $({\mathbb{R}}^{2n},[,])$ of the abelian Lie group $({\mathbb{R}}^{2n},+)$, the infinitesimal generator of the action $\mathsf{\Phi }$ in the direction x is the vector field $X^x\left(v\right)=x$.

For each $x\in ({\mathbb{R}}^{2n},[,])$, let $J^x:{\mathbb{R}}^{2n}\to \mathbb{R}:v\mapsto \omega (x,v)$. Then,

$$\mathrm{d}{J}^x\left(v\right)y=\omega (x,y)=\omega (X^x\left(v\right),y),$$

that is, $X^x=X_{J^x}$. Hence, the action $\mathsf{\Phi }$ is Hamiltonian with momentum mapping $J:{\mathbb{R}}^{2n}\to {\left({\mathbb{R}}^{2n}\right)}^{*}$, where $J\left(v\right)x=J^x\left(v\right)$. Here, ${\left({\mathbb{R}}^{2n}\right)}^{*}$ is the dual of the vector space ${\mathbb{R}}^{2n}$. For $y,z\in {\mathbb{R}}^{2n}$

$$\{J^y,J^z\}\left(v\right)=\mathrm{d}{J}^y\left(v\right)X^z\left(v\right)=\omega (X^y\left(v\right),X^z\left(v\right))=\omega (y,z)$$

Meanwhile, $J^{[y,z]}\left(v\right)=\omega ([y,z],v)=0$, since $[y,z]=0$. So,

$$\Sigma (y,z)=\left\{J^y{J}^z\right\}\left(v\right)-J^{[y,z]}\left(v\right)=\omega (y,z)$$

is the $({\mathbb{R}}^{2n},[,])$ cocycle associated with the momentum map J. We cannot use Kostant’s theorem [3] (Thm. 4.5, pp. 175–176) to determine a representation of $({\mathbb{R}}^{2n},\omega )$ because J is not coadjoint equvariant.

To continue, we find a central extension $\mathfrak{g}\subseteq {\mathbb{R}}^{2n}\times \mathbb{R}$ of the abelian Lie algebra $({\mathbb{R}}^{2n},[,])$ by the cocycle $\Sigma$, which has Lie bracket $\left[\right(x,t),(y,s\left)\right]=(0,\Sigma (x,y\left)\right)$. A Lie group G with Lie algebra $\mathfrak{g}$ has the following expression:

$$(v,r)·(w,s)=\left(v+w,r+s+{\displaystyle \frac{1}{2}}\omega (v,w)\right).$$

We now construct a model of G as a subgroup of $\mathrm{Gl}({\mathbb{R}}^{2n+2},\mathbb{R})$. Consider the map

$$\rho :G\to \mathrm{Gl}({\mathbb{R}}^{2n+2},\mathbb{R}):(v,r)\to \left(\begin{array}{ccc}1& 0& 0\\ v& I_{2n}& 0\\ r& {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(v\right)& 1\end{array}\right).$$

Here, ${\omega }^{♯}\left(v\right)\in {\left({\mathbb{R}}^{2n}\right)}^{*}$ is given by ${\omega }^{♯}\left(v\right)w=\omega (v,w)$ for every v, $w\in {\mathbb{R}}^{2n}$. Since

$$\begin{array}{cc}\hfill \rho (v+w,r+s+{\displaystyle \frac{1}{2}}\omega (v,w))& =\left(\begin{array}{ccc}1& 0& 0\\ v+w& I_{2n}& 0\\ r+s+{\displaystyle \frac{1}{2}}\omega (v,w)& {\displaystyle \frac{1}{2}}{\omega }^{♯}(v+w)& 1\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{ccc}1& 0& 0\\ v& I_{2n}& 0\\ r& {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(v\right)& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ w& I_{2n}& 0\\ s& {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(w\right)& 1\end{array}\right)=\rho (v,r)\rho (w,s),\hfill \end{array}$$

$\rho$ is a group homomorphism. However, the fact that $\rho (v,r)=I_{2n+2}$ implies $(v,r)=(0,0)$, which is the identity element of G. Hence, $\rho$ is an isomorphism onto its image. Since G is the $2n+1$ dimensional Heisenberg group ${\mathrm{H}}_{2n+1}$, $H=\rho \left(G\right)$.

The Lie algebra $\mathfrak{h}$ of H is $\{X^{\vee }=$$\left(\begin{array}{ccc}0& 0& 0\\ x& 0& 0\\ \xi & {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(x\right)& 0\end{array}\right)$ $\xi \in \mathbb{R}$ and $x\in {\mathbb{R}}^{2n}$} with Lie bracket $[X^{\vee },Y^{\vee }]=X^{\vee }{Y}^{\vee }-Y^{\vee }{X}^{\vee }$, where $Y^{\vee }=$$\left(\begin{array}{ccc}0& 0& 0\\ y& 0& 0\\ \eta & {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(y\right)& 0\end{array}\right)$. The map

$$\sigma :\mathfrak{g}\to \mathfrak{h}:(x,\xi )\mapsto X^{\vee }=\left(\begin{array}{ccc}0& 0& 0\\ x& 0& 0\\ \xi & {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(x\right)& 0\end{array}\right)$$

is a Lie algebra isomorphism, since

$$\begin{array}{c}\hfill [\sigma (x,\xi ),\sigma (y,\eta )]=[X^{\vee },Y^{\vee }]=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \omega (x,y)& 0& 0\end{array}\right)=\sigma \left([(x,\xi ),(y,\eta )]\right).\end{array}$$

The action

$${\mathsf{\Phi }}^{\vee }:H\times ({\mathbb{R}}^{2n},\omega )\to ({\mathbb{R}}^{2n},\omega ):\left(\left(\begin{array}{ccc}1& 0& 0\\ v& 0& 0\\ r& {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(v\right)& 1\end{array}\right),w\right)\mapsto w+v$$

(2)

is Hamiltonian. For each $X^{\vee }=$$\left(\begin{array}{ccc}0& 0& 0\\ x& 0& 0\\ \xi & {\displaystyle \frac{1}{2}}{\omega }^{♯}\left(x\right)& 0\end{array}\right)$$\in \mathfrak{h}$, the infinitesimal generator of the action ${\mathsf{\Phi }}^{\vee }$ in the direction $X^{\vee }$ is the vector field $X^{X^{\vee }}\left(v\right)=x$. Let $J^{X^{\vee }}:{\mathbb{R}}^{2n}\to \mathbb{R}:v\mapsto \omega (x,v)+\xi$. Since

$$\mathrm{d}{J}^{X^{\vee }}\left(v\right)w=\omega (x,w)=\omega (X^{X^{\vee }}\left(v\right),w),$$

we obtain $X^{X^{\vee }}=X_{J^{X^{\vee }}}$. So, the action ${\mathsf{\Phi }}^{\vee }$ is Hamiltonian. The corresponding momentum mapping is $J:{\mathbb{R}}^{2n}\to {\mathfrak{h}}^{*}$, where $J\left(v\right)X^{\vee }=J^{X^{\vee }}\left(v\right)$. The mapping J is coadjoint equivariant, that is, $J\left({\mathsf{\Phi }}_g^{\vee }\left(v\right)\right)={\mathrm{Ad}}_{g^{-1}}^{T}J\left(v\right)$ for every $g\in G$ and every $v\in {\mathbb{R}}^{2n+2}$. We verify this. Since the Heisenberg group H is connected we need only show that $\{J^{X^{\vee }},J^{Y^{\vee }}\}=J^{[X^{\vee },Y^{\vee }]}$, which is the infinitesimalization of the coadjoint equivariance condition. We have

$$\begin{array}{cc}\hfill \{J^{X^{\vee }},J^{Y^{\vee }}\}\left(v\right)& =L_{X^{Y^{\vee }}}{J}^{X^{\vee }}\left(v\right)=\mathrm{d}{J}^{X^{\vee }}\left(v\right)X^{Y^{\vee }}\left(v\right)\hfill \\ \hfill & =\omega (X^{X^{\vee }}\left(v\right),X^{Y^{\vee }}\left(v\right))=\omega (x,y)=J^{[X^{\vee },Y^{\vee }]}\left(v\right).\hfill \end{array}$$

3. Coadjoint Orbits of ${\mathrm{H}}_{\mathbf{2}\mathbf{n}+\mathbf{1}}$

In this section, we classify the coadjoint orbits of the Heisenberg group ${\mathrm{H}}_{2n+1}$.

First, we find a subgroup of $\mathrm{Sp}(V,\mathcal{J})$ which has an isotropy group equal to ${\mathrm{H}}_{2n+1}$. Here, $V=\mathbb{R}\times {\mathbb{R}}^{2n}\times \mathbb{R}$. With respect to the basis $\mathfrak{e}=\{e_0;{\mathfrak{e}}_n;e_{2n+1}=f_{n+1}\}$ of V, the matrix of the symplectic form $\mathcal{J}$ is $\left(\begin{array}{ccc}0& 0& 1\\ 0& J& 0\\ -1& 0& 0\end{array}\right)$, where J is the matrix of a symplectic form on ${\mathbb{R}}^{2n}$ with respect to the basis ${\mathfrak{e}}_n$.

Let

$$\tilde{G}=\left\{\left(\begin{array}{ccc}a& 0& 0\\ d& I_{2n}& 0\\ f& g^T& h\end{array}\right)\in \mathrm{Gl}({\mathbb{R}}^{2n+2},\mathbb{R})\mid a,h\in {\mathbb{R}}^{\times };f\in \mathbb{R};d,g\in {\mathbb{R}}^{2n}\right\}.$$

Then, $\tilde{G}$ is a group of real linear mappings of ${\mathbb{R}}^{2n+2}$ into itself which sends $\left\{0\right\}\times \mathbb{R}\subseteq {\mathbb{R}}^{2n+1}\times \mathbb{R}$ into itself and is the identity map on $\left\{0\right\}\times {\mathbb{R}}^{2n}\times \left\{0\right\}$. Let $\widehat{G}=\tilde{G}\cap \mathrm{Sp}({\mathbb{R}}^{2n+2},\mathcal{J})$. Then,

$$\widehat{G}=\left\{\left(\begin{array}{ccc}a& 0& 0\\ d& I_{2n}& 0\\ f& a^{-1}{\left(Jd\right)}^T& a^{-1}\end{array}\right)\in \mathrm{Sp}({\mathbb{R}}^{2n+2},\mathcal{J})\mid a\in {\mathbb{R}}^{\times };f\in \mathbb{R};d\in {\mathbb{R}}^{2n}\right\}$$

is a Lie group with Lie algebra.

$$\widehat{\mathfrak{g}}=\left\{\left(\begin{array}{ccc}\eta & 0& 0\\ \tilde{x}& 0& 0\\ \xi & {\left(J\tilde{x}\right)}^T& -\eta \end{array}\right)\in \mathrm{sp}({\mathbb{R}}^{2n+2},\mathcal{J})\mid \eta ,\xi \in \mathbb{R};\tilde{x}\in {\mathbb{R}}^{2n}\right\}.$$

The isotropy group ${\widehat{G}}_{f_{n+1}}$ of elements of $\widehat{G}$ which leave the vector $f_{n+1}$ fixed is

$${\widehat{G}}_{f_{n+1}}=\left\{\left(\begin{array}{ccc}1& 0& 0\\ d& I_{2n}& 0\\ f& {\left(Jd\right)}^T& 1\end{array}\right)\in \mathrm{Sp}({\mathbb{R}}^{2n+2},\mathcal{J})\mid f\in \mathbb{R};d\in {\mathbb{R}}^{2n}\right\}.$$

Using the basis ${\mathfrak{e}}_n$ of ${\mathbb{R}}^{2n}$, set $J={\displaystyle \frac{1}{2}}{J}_{2n}$. We see that ${\widehat{G}}_{f_{n+1}}$ is the Heisenberg group ${\mathrm{H}}_{2n+1}$. The Lie algebra of ${\widehat{G}}_{f_{n+1}}$ is

$${\widehat{\mathfrak{g}}}_{f_{n+1}}=\left\{\left(\begin{array}{ccc}0& 0& 0\\ \tilde{x}& 0& 0\\ \xi & {\left(J\tilde{x}\right)}^T& 0\end{array}\right)\in \mathrm{sp}({\mathbb{R}}^{2n+2},\mathcal{J})\mid \xi \in \mathbb{R};\tilde{x}\in {\mathbb{R}}^{2n}\right\}.$$

We now begin the classification of the coadjoint orbits of ${\widehat{G}}_{f_{n+1}}$. Let $(V=\mathbb{R}\times {\mathbb{R}}^{2n}\times \mathbb{R},\mathcal{J})$ be a real symplectic vector space of dimension $2n+2$ with a basis $\mathfrak{e}=\{e_0;{\mathfrak{e}}_n;f_{n+1}\}$ such that the matrix of $\mathcal{J}$ with repect to the basis $\mathfrak{e}$ is $\left(\begin{array}{ccc}0& 0& 1\\ 0& J& 0\\ -1& 0& 0\end{array}\right)$, where J is one half the matrix $J_{2n}$ of the standard symplectic form $\omega$ on ${\mathbb{R}}^{2n}$. A tuple$(V,Y,f_{n+1};\mathcal{J})$ is a symplectic vector space $(V,\mathcal{J})$ with $Y\in \widehat{\mathfrak{g}}$, the Lie algebra of the Lie group $\widehat{G}$, and $f_{n+1}$ is a basis vector in $\mathfrak{e}$. Two tuples $(V,Y,f_{n+1};\mathcal{J})$ and $(V,Y^{\prime },f_{n+1};\mathcal{J})$ are equivalent if there is a bijective real linear mapping $P:V\to V$ such that (1) $P\in {\widehat{G}}_{f_{n+1}}$ and (2) there is a vector $w\in V$ such that $Y^{\prime }=P(Y+L_{w,f_{n+1}})P^{-1}$. Here, $L_{w,f_{n+1}}=w\otimes f_{n+1}^{*}+f_{n+1}\otimes w^{*}$, and $w^{*}\left(z\right)=z^T\mathcal{J}w$ for every $z\in V$.

Lemma 1.

Let $w=w_0{e}_0+\tilde{w}+w_{2n+1}{f}_{n+1}\in V$. We have $L_{w,f_{n+1}}\in \widehat{\mathfrak{g}}$.

Proof. 

We compute the matrix of $L_{w,f_{n+1}}$ with respect to the basis $\mathfrak{e}$.

$$\begin{array}{cc}\hfill L_{w,f_{n+1}}\left(e_0\right)& =(w\otimes f_{n+1}^{*})\left(e_0\right)+(f_{n+1}\otimes w^{*})\left(e_0\right)\hfill \\ \hfill & =\left(e_0^T\mathcal{J}{f}_{n+1}\right)w+\left(e_0^T\mathcal{J}w\right)f_{n+1}\hfill \\ \hfill & =w+w_{2n+1}{f}_{n+1}=w_0{e}_0+\tilde{w}+2w_{2n+1}{f}_{n+1};\hfill \\ \hfill L_{w,f_{n+1}}\left({\mathfrak{e}}_n\right)& =(w\otimes f_{n+1}^{*})\left({\mathfrak{e}}_n\right)+(f_{n+1}\otimes w^{*})\left({\mathfrak{e}}_n\right)\hfill \\ \hfill & =\left({\left({\mathfrak{e}}_n\right)}^T\mathcal{J}{f}_{n+1}\right)w+\left({\left({\mathfrak{e}}_n\right)}^T\mathcal{J}w\right)f_{n+1}\hfill \\ \hfill & ={\left(J\tilde{w}\right)}^T{f}_{n+1};\hfill \\ \hfill L_{w,f_{n+1}}{f}_{n+1}& =(w\otimes f_{n+1}^{*})\left(f_{n+1}\right)+(f_{n+1}\otimes w^{*})f_{n+1}\hfill \\ \hfill & =\left(f_{n+1}^T\mathcal{J}{f}_{n+1}\right)w+\left(f_{n+1}^T\mathcal{J}w\right)f_{n+1}=-w_0{f}_{n+1}.\hfill \end{array}$$

Thus, the matrix of $L_{w,f_{n+1}}$ with respect to the basis $\mathfrak{e}$ is $\left(\begin{array}{ccc}{w}_0& 0& 0\\ \tilde{w}& 0& 0\\ 2w_{2n+1}& {\left(J\tilde{w}\right)}^T& -w_0\end{array}\right)$, which lies in $\widehat{\mathfrak{g}}$.  □

Hence, the definition of equivalence makes sense.

Corollary 1.

$\widehat{\mathfrak{g}}=\{L_{w,f_{n+1}}\mid w\in V\}$.

Proof. 

From Lemma 1 and the definition of $\widehat{\mathfrak{g}}$, it follows that $L_{w,f_{n+1}}\in \widehat{\mathfrak{g}}$ for every $w\in V$. Suppose that $X=$$\left(\begin{array}{ccc}0& 0& 0\\ \tilde{x}& 0& 0\\ \xi & {\left(J\tilde{x}\right)}^T& 0\end{array}\right)$$\in \widehat{\mathfrak{g}}$. Let $w=w_0+\tilde{x}+{\displaystyle \frac{1}{2}}\xi f_{n+1}$. Then, $X=L_{w,f_{n+1}}$.  □

Lemma 2.

For every $P\in \mathrm{Sp}(V,\mathcal{J})$, we have $P{L}_{w,f_{n+1}}{P}^{-1}=L_{Pw,P{f}_{n+1}}$.

Proof. 

For every $z\in V$,

$$\begin{array}{cc}\hfill L_{Pw,P{f}_{n+1}}z& =(Pw\otimes {\left(P{f}_{n+1}\right)}^{*})\left(z\right)+(P{f}_{n+1}\otimes {\left(Pw\right)}^{*})\left(z\right)\hfill \\ \hfill & =\left(\left(z^T\mathcal{J}\right)P{f}_{n+1}\right)Pw+\left(z^T\mathcal{J}Pw\right)P{f}_{n+1}\hfill \\ \hfill & =\left({\left(P^{-1}z\right)}^T\mathcal{J}{f}_{n+1}\right)Pw+\left({\left(P^{-1}z\right)}^T\mathcal{J}w\right)P{f}_{n+1},\mathrm{since}P\in \mathrm{Sp}(V,\mathcal{J})\hfill \\ \hfill & =P(w\otimes f_{n+1}^{*}+f_{n+1}\otimes w^{*})P^{-1}\left(z\right)=\left(P{L}_{w,f_{n+1}}{P}^{-1}\right)z.\hfill \end{array}$$

□

Being equivalent is an equivalence relation on the set of tuples. An equivalence class ∇ of a set of tuples is a cotype. ∇ is represented by the tuple $(V,Y,f_{n+1};\mathcal{J})$ if, and only if, $(V,Y,f_{n+1};\mathcal{J})\in \nabla$. If the element Y of the tuple $(V,Y,f_{n+1};\mathcal{J})$ is nilpotent with $Y^{m+1}V=0$ but $Y^{m}V\ne 0$ for some $m\in {\mathbb{Z}}_{\ge 1}$, then the tuple is nilpotent of height m. Since equivalent nilpotent tuples have the same height, we say that the corresponding cotype is nilpotent of height m.

For each $Y\in \widehat{\mathfrak{g}}$. let ${\mathcal{\ell }}_Y:\widehat{\mathfrak{g}}\to \mathbb{R}:Z\mapsto \mathrm{tr}YZ$. Then, ${\mathcal{\ell }}_Y\in {\left(\widehat{\mathfrak{g}}\right)}^{*}$.

Theorem 1.

The correspondence

$$(V,Y,f_{n+1};\mathcal{J})\mapsto {\left({\mathcal{\ell }}_Y\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$$

(3)

between tuples and elements of ${\left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)}^{*}$ induces a bijection between cotypes and ${\widehat{G}}_{f_{n+1}}$ coadjoint orbits.

To prove the theorem, we need

Fact 1.

$$\left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°=\{X\in \widehat{\mathfrak{g}}{\mathcal{\ell }}_X\left(Y\right)=0for everyY\in {\widehat{\mathfrak{g}}}_{f_{n+1}}\}=\widehat{\mathfrak{g}}.$$

Proof. 

Suppose that $X=$$\left(\begin{array}{ccc}\zeta & 0& 0\\ d& 0& 0\\ \eta & {\left(Jd\right)}^T& -\zeta \end{array}\right)$$\in \widehat{\mathfrak{g}}$. For every $Y=$$\left(\begin{array}{ccc}0& 0& 0\\ v& 0& 0\\ \xi & {\left(Jv\right)}^T& 0\end{array}\right)$$\in {\widehat{\mathfrak{g}}}_{f_{n+1}}$, we have

$${\mathcal{\ell }}_X\left(Y\right)=\mathrm{tr}XY=\mathrm{tr}\left[\left(\begin{array}{ccc}\zeta & 0& 0\\ d& 0& 0\\ \eta & {\left(Jd\right)}^T& -\zeta \end{array}\right)\left(\begin{array}{ccc}0& 0& 0\\ v& 0& 0\\ \xi & {\left(Jv\right)}^T& 0\end{array}\right)\right]=0.$$

So $X\in \left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°$. Hence, $\widehat{\mathfrak{g}}\subseteq \left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°$. By definition, $\left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°\subseteq \widehat{\mathfrak{g}}$. Thus, $\left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°=\widehat{\mathfrak{g}}$.  □

Proof of Theorem 1.

Let $(V,Y^{\prime },f_{n+1};\mathcal{J})$ be a tuple which is equivalent to $(V,Y,f_{n+1};\mathcal{J})$. Then, for some $P\in {\widehat{G}}_{f_{n+1}}$, some $w\in V$, and every $Z\in {\widehat{\mathfrak{g}}}_{f_{n+1}}$, we have

$$\begin{array}{cc}\hfill {\mathcal{\ell }}_{Y^{\prime }}\left(Z\right)& ={\mathcal{\ell }}_{P(Y+L_{w,f_{n+1}})P^{-1}}\left(Z\right)\hfill \\ \hfill & ={\mathcal{\ell }}_{PY{P}^{-1}}\left(Z\right)+{\mathcal{\ell }}_{P{L}_{w,f_{n+1}}{P}^{-1}}\left(Z\right)\hfill \\ \hfill & ={\mathcal{\ell }}_Y\left(P^{-1}ZP\right)+{\mathcal{\ell }}_{L_{Pw,P{f}_{n+1}}}\left(Z\right),\hfill \\ \hfill & \mathrm{since} \mathrm{tr}\left(PY{P}^{-1}Z\right)=\mathrm{tr}\left(Y\left(P^{-1}ZP\right)\right)\hfill \\ \hfill & \mathrm{and}{\widehat{G}}_{f_{n+1}}\subseteq \mathrm{Sp}(V,\mathcal{J}).\hfill \\ \hfill & =\left({\mathrm{Ad}}_{P^{-1}}^T{\mathcal{\ell }}_Y\right)\left(Z\right)+{\mathcal{\ell }}_{L_{Pw,f_{n+1}}}\left(Z\right),\mathrm{since}P\in {\widehat{G}}_{f_{n+1}}\hfill \\ \hfill & =\left({\mathrm{Ad}}_{P^{-1}}^T{\mathcal{\ell }}_Y\right)\left(Z\right),\mathrm{since}{L}_{Pw,f_{n+1}}\in \widehat{\mathfrak{g}}=\left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°\mathrm{and}Z\in {\widehat{\mathfrak{g}}}_{f_{n+1}}.\hfill \end{array}$$

Thus, the map of cotypes to ${\widehat{G}}_{f_{n+1}}$ coadjoint orbits, induced by the correspondence (3), is well defined. The induced map is injective if ${\left({\mathcal{\ell }}_{Y^{\prime }}\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$ lies in the ${\widehat{G}}_{f_{n+1}}$ coadjoint orbit through ${\left({\mathcal{\ell }}_Y\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$; then, for some $P\in {\widehat{G}}_{f_{n+1}}$, we have ${\mathcal{\ell }}_{Y^{\prime }}={\mathrm{Ad}}_{P^{-1}}^T{\mathcal{\ell }}_Y={\mathcal{\ell }}_{PY{P}^{-1}}$ on ${\widehat{\mathfrak{g}}}_{f_{n+1}}$. So, $Y^{\prime }-PY{P}^{-1}\in \left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)°$. Hence, for some $w\in V$, we have $Y^{\prime }-PY{P}^{-1}=L_{w,f_{n+1}}$, that is, $Y^{\prime }=P(Y+L_{P^{-1}w,f_{n+1}})P^{-1}$. Thus, the tuples $(V,Y,f_{n+1};\mathcal{J})$ and $(V,Y^{\prime },f_{n+1};\mathcal{J})$ are equivalent and thus correspond to the same cotype. Since every element of ${\left({\widehat{\mathfrak{g}}}_{f_{n+1}}\right)}^{*}$ may be written as ${\left({\mathcal{\ell }}_Y\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$ for some $Y\in \widehat{\mathfrak{g}}$, the induced map is surjective. □

Fact 2.

$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \xi & 0& 0\end{array}\right)$$\in {\widehat{\mathfrak{g}}}_{f_{n+1}}$ is invariant under conjugation by elements of ${\widehat{G}}_{f_{n+1}}$.

Proof. 

The proof is a straightforward calculation. The details are omitted.  □

Claim 1.

The tuple $({\mathbb{R}}^{2n},Y,f_{n+1};\mathcal{J})$, where  $Y=$$\left(\begin{array}{ccc}\zeta & 0& 0\\ d& 0& 0\\ \xi & {\left(Jd\right)}^T& -\zeta \end{array}\right)$$\in \widehat{\mathfrak{g}}$, is equivalent to the tuple $({\mathbb{R}}^{2n},\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \xi & 0& 0\end{array}\right),f_{n+1};\mathcal{J})$.

Proof. 

$Y=L_{w^{\prime },f_{n+1}}$, where $w^{\prime }=\zeta e_0+d+{\displaystyle \frac{1}{2}}\xi f_{n+1}\in {\mathbb{R}}^{2n+2}$. Choose $w=-\zeta e_0-d\in {\mathbb{R}}^{2n+2}$. Then, $Y+L_{w,f_{n+1}}=$$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \xi & 0& 0\end{array}\right)$.  □

Consider the tuple $(V,Y,f_{n+1};\mathcal{J})$. If $V=V_1\oplus V_2$, where $V_1$ and $V_2$ are Y-invariant, $\mathcal{J}$-perpendicular and $\mathcal{J}$-nondegenerate subspaces of $(V,\mathcal{J})$, and $f_{n+1}\in V_1$, the cotype ∇, represented by the tuple $(V,Y,f_{n+1};\mathcal{J})$, is decomposable into a cotype ∇, represented by the tuple $(V_1,Y_{|V_1},f_{n+1};{\mathcal{J}}_{|V_1})$, and a type $\Delta$, represented by the pair $(V_2,Y_{|V_2};{\mathcal{J}}_{|V_2})$. We say that ∇ is the sum of the cotype ∇ and the type $\Delta$. If no such decomposition exists, then ∇ is said to be indecomposable. The pair $({\mathbb{R}}^{2n+2},\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \xi & 0& 0\end{array}\right);\mathcal{J})$ with $\xi =0$ is the indecomposable zero type ${\mathbf{0}}_{2n+2}$. If $\xi \ne 0$, then the cotype $\nabla ,\xi$, represented by the nipotent tuple $({\mathbb{R}}^{2n+2},\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \xi & 0& 0\end{array}\right),f_{n+1};\mathcal{J})$ of height 1, is decomposable into the sum of the zero type ${\mathbf{0}}_{2n}$ and the indecomposable cotype ${\nabla }_1\left(0\right),\xi \ne 0$ of height 1 and modulus $\xi$, represented by the nilpotent tuple $({\mathbb{R}}^2,\left(\begin{array}{cc}0& 0\\ \xi & 0\end{array}\right),f_2;{\mathcal{J}}_2=$$\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ of height 1.

We explain the geometric meaning of the modulus $\xi$ in the cotype $\nabla ,\xi$ with $\xi \ne 0$. Let $Y_1=$$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 1& 0& 0\end{array}\right)$ and $Y_{\xi }=$$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \xi & 0& 0\end{array}\right)$$=\xi Y_1$. We have

$$\nabla ,1=\left\{\begin{array}{}\\ (V,Y,f_{n+1};\mathcal{J})\mid Y=P(Y_1+L_{w,f_{n+1}})P^{-1}\hfill \\ \hfill \mathrm{for} P\in {\widehat{G}}_{f_{n+1}}\mathrm{and} w\in V\end{array}\right\}.$$

Now,

$$\begin{array}{cc}\hfill \{(V,\widehat{Y},f_{n+1};\mathcal{J})\mid \widehat{Y}& =\xi Y=\xi P(Y_1+L_{w,f_{n+1}})P^{-1}\}\hfill \\ \hfill & =\{(V,\widehat{Y},f_{n+1};\mathcal{J})\mid \widehat{Y}=P(Y_{\xi }+L_{\widehat{w},f_{n+1}})P^{-1}\}=\nabla ,\xi ,\xi \ne 0\hfill \end{array}$$

where $\widehat{w}=\xi w$. Here, $P\in {\widehat{G}}_{f_{n+1}}$ and $w\in V$ are those given in $\nabla ,1$ for the tuple $(V,Y,f_{n+1};\mathcal{J})$. The ${\widehat{G}}_{f_{n+1}}$ coadjoint orbit ${\mathcal{O}}_1$, through ${\left({\mathcal{\ell }}_{Y_1}\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$, and the ${\widehat{G}}_{f_{n+1}}$ coadjoint orbit ${\mathcal{O}}_{\xi }$, through ${\left({\mathcal{\ell }}_{Y_{\xi }}\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$, are diffeomorphic via the mapping ${\mathrm{Ad}}_{P^{-1}}^T{\left({\mathcal{\ell }}_{Y_1}\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}\mapsto {\mathrm{Ad}}_{P^{-1}}^T{\left({\mathcal{\ell }}_{Y_{\xi }}\right)}_{|{\widehat{\mathfrak{g}}}_{f_{n+1}}}$, since the set of $P\in {\widehat{G}}_{f_{n+1}}$ defining the cotypes $\nabla ,1$ and $\nabla ,\xi$ with $\xi \ne 0$ are the same. The coadjoint orbits ${\mathcal{O}}_1$ and ${\mathcal{O}}_{\xi }$ are symplectic manifolds with their natural symplectic forms being ${\omega }_{{\mathcal{O}}_1}$ and ${\omega }_{{\mathcal{O}}_{\xi }}$, respectively, see [1] (p. 288). Since $Y_{\xi }=\xi Y_1$, it follows that ${\mathcal{\ell }}_{Y_{\xi }}=\xi {\mathcal{\ell }}_{Y_1}$. Hence, ${\omega }_{{\mathcal{O}}_{\xi }}=\xi {\omega }_{{\mathcal{O}}_1}$. Thus, the symplectic manifolds $({\mathcal{O}}_{\xi },{\omega }_{{\mathcal{O}}_{\xi }})$ and $({\mathcal{O}}_1,{\omega }_{{\mathcal{O}}_1})$ are not symplectically diffeomorphic when $\xi \in \mathbb{R}\setminus \{0,1\}$. The geometric meaning of the modulus for the cotype $\nabla ,\xi$ with $\xi \ne 0$ is as follows: $\xi$ parametrizes a family $({\mathcal{O}}_{\xi },{\omega }_{{\mathcal{O}}_{\xi }})$ of symplectic manifolds, where the base manifolds ${\mathcal{O}}_{\xi }$ are diffeomorphic but are not pairwise symplectically diffeomorphic.

Appendix: Intrinsic Description

Following Wallach [4], we give an intrinsic description of the coadjoint orbits of the Heisenberg group H on ${\mathfrak{h}}^{*}$, the dual of its Lie algebra $\mathfrak{h}$.

Recall that the Heisenberg group H is a subset of $({\mathbb{R}}^{2n},\omega )\times \mathbb{R}$ with multiplication $(x,t)·(y,s)=\left(x+y,t+s+{\displaystyle \frac{1}{2}}\omega (x,y)\right)$. H is a Lie group, with Lie algebra $\mathfrak{h}$ having the bracket $\left[\right(\xi ,t),(\eta ,s\left)\right]=(0,\omega (\xi ,\eta \left)\right]$. Its exponential map $exp:\mathfrak{h}\to H:(\xi ,s)\mapsto (\xi ,s)$ is the identity map.

The first step toward computing the coadjoint action of H is to compute the adjoint map.

$${\mathrm{Ad}}_{(x,t)}:\mathfrak{h}\to \mathfrak{h}:(\xi ,s)\mapsto (x,t)·(\xi ,s)·{(x,t)}^{-1}$$

for every $(x,t)\in H$. We obtain

$$\begin{array}{cc}\hfill {\mathrm{Ad}}_{(x,t)}(\xi ,s)& =(x,t)·(\xi ,s)·(-x,-t)\hfill \\ \hfill & =(x,t)·\left(\xi -x,s-t+{\displaystyle \frac{1}{2}}\omega (\xi ,-x)\right)\hfill \\ \hfill & =(x+\xi -x,t+\left(s-t-{\displaystyle \frac{1}{2}}\omega (\xi ,x)+{\displaystyle \frac{1}{2}}\omega (x,\xi -x)\right)\hfill \\ \hfill & =\left(\xi ,s+\omega (x,\xi )\right).\hfill \end{array}$$

Let $f\in {\mathfrak{h}}^{*}$. For every $(\xi ,s)\in \mathfrak{h}$, we have $f(\xi ,s)={\lambda }_f\left(\xi \right)+sf(0,1)$, where ${\lambda }_f\in {\left({\mathbb{R}}^{2n}\right)}^{*}$. So, the coadjoint action • of H on ${\mathfrak{h}}^{*}$ is

$$\begin{array}{cc}\hfill \left((x,t)•f\right)(\xi ,s)& =\left({\mathrm{Ad}}_{{(x,t)}^{-1}}^{T}f\right)(\xi ,s)\hfill \\ \hfill & =f\left({\mathrm{Ad}}_{(-x,-t)}(\xi ,s)\right)=f(\xi ,s-\omega (x,\xi ))\hfill \\ \hfill & ={\lambda }_f\left(\xi \right)+\left(s-\omega (x,\xi )\right)f(0,1)\hfill \\ \hfill & =\left({\lambda }_f\left(\xi \right)+sf(0,1)\right)-\omega (x,\xi )f(0,1)\hfill \\ \hfill & =f(\xi ,s)-\left({\omega }^{♯}\left(x\right)\xi \right)f(0,1).\hfill \end{array}$$

Since ${\lambda }_f\in {\left({\mathbb{R}}^{2n}\right)}^{*}$ and $\omega$ is nondegenerate, there is $y_f\in {\mathbb{R}}^{2n}$ such that $y_f={\omega }^{♭}\left({\lambda }_f\right)$. For $\mu =f(0,1)\ne 0$, we obtain

$$\begin{array}{cc}\hfill \left(({\mu }^{-1}{y}_f,0)•f\right)(\xi ,s)& =f(\xi ,s)-{\omega }^{♯}\left({\mu }^{-1}{y}_f\right)\left(\xi \right)\mu \hfill \\ \hfill & =f(\xi ,s)-\left({\omega }^{♯}\left({\omega }^{♭}{\lambda }_f\right)\right)\left(\xi \right)\hfill \\ \hfill & ={\lambda }_f\left(\xi \right)+\mu s-{\lambda }_f\left(\xi \right)=\mu s.\hfill \end{array}$$

For $\mu \ne 0$, let $h_{\mu }:\mathfrak{h}\to \mathbb{R}:(\xi ,s)\mapsto \mu s$. Then, $h_{\mu }\in {\mathfrak{h}}^{*}$. The above calculation shows that for every, $\mu \ne 0$ we have $({\mu }^{-1}{y}_f,0)•f=h_{\mu }$. Thus, we have proved

Proposition 1.

When $\mu \ne 0$, every $f={\lambda }_f+h_{\mu }\in {\mathfrak{h}}^{*}$ lies in the H coadjoint orbit of $h_{\mu }$.

Now, we show that

Proposition 2.

The H coadjoint orbit ${\mathcal{O}}_{\mu }=H•h_{\mu }$ through $h_{\mu }\in {\mathfrak{h}}^{*}$ is the symplectic manifold $({\mathbb{R}}^{2n},\mu {\omega }_{|{\mathcal{O}}_{\mu }})$.

Proof. 

First, we compute the isotropy group $H_{h_{\mu }}=\{h\in H\mid h•h_{\mu }=h_{\mu }\}$ of the H coadjoint action at $h_{\mu }$. Since $((x,t)•h_{\mu })(\xi ,s)=h_{\mu }(\xi ,s)-\mu {\omega }^{♯}\left(x\right)\xi$, we see that $(\xi ,s)\in H_{h_{\mu }}$ implies that $0=\mu {\omega }^{♯}\left(x\right)\xi$ for every $(\xi ,s)\in \mathfrak{h}$. Hence, $x=0$, since $\mu \ne 0$ and $\omega$ is nondegenerate. Thus, $H_{h_{\mu }}$ is the center $Z=\left\{\right(0,t)\in H\mid t\in \mathbb{R}\}$ of H. By definition of the coadjoint orbit, the map $\psi :H/Z\to {\mathcal{O}}_{\mu }:hZ\mapsto h•h_{\mu }$ is smooth and bijective, as is the map $\eta :{\mathbb{R}}^{2n}\to {\mathcal{O}}_{\mu }:x\mapsto (x,0)•h_{\mu }$. Since the map $\theta :H/Z\to {\mathbb{R}}^{2n}\times \left\{0\right\}:(x,t)Z\mapsto (x,0)$ is smooth and bijective, it follows that the coadjoint orbit ${\mathcal{O}}_{\mu }$ is diffeomorphic to ${\mathbb{R}}^{2n}$. If $\alpha =(\xi ,0)$ and $\beta =(\eta ,0)$ lie in $\mathfrak{h}$, for $\pi \in {\mathcal{O}}_{\mu }$

$${\omega }_{{\mathcal{O}}_{\mu }}\left(\pi \right)(X^a\left(\pi \right),X^b\left(\pi \right))=\pi \left([a,b]\right)=\pi (0,\omega (\xi ,\eta ))=\mu {\omega }_{|{\mathcal{O}}_{\mu }}\left(\pi \right).$$

Here, $X^c\left(\pi \right)=(\frac{\mathrm{d}}{\mathrm{d}t}{\mid }_{t=0}exptc)•\pi$ for $c\in \mathfrak{h}$.  □

4. The Momentum Mapping of ${\mathrm{H}}_{\mathbf{2}\mathbf{n}+\mathbf{1}}$

In this section, we show that the ${\mathbb{R}}^{2n}$ cocycle of the momentum map $\mathsf{\Phi }$ (1) of the affine action of ${\mathbb{R}}^{2n}$ on $({\mathbb{R}}^{2n},\omega )$ becomes the value of a modulus for a coadjoint orbit of the Heisenberg group ${\mathrm{H}}_{2n+1}$ coming from the action ${\mathsf{\Phi }}^{\vee }$ (2) on $({\mathbb{R}}^{2n},\omega )$.

We determine the image of the momentum map $J:{\mathbb{R}}^{2n}\to {\mathfrak{h}}^{*}$ of the Hamiltonian action ${\mathsf{\Phi }}^{\vee }$. By definition, $J\left(v\right)X^{\vee }=J^{X^{\vee }}\left(v\right)$, where $J^{X^{\vee }}:{\mathbb{R}}^{2n}\to \mathbb{R}:v\mapsto \omega (x,v)+\xi$ for $X^{\vee }=$$\left(\begin{array}{ccc}0& 0& 0\\ \tilde{x}& 0& 0\\ \xi & {\left(J\tilde{x}\right)}^T& 0\end{array}\right)$. The action ${\mathsf{\Phi }}^{\vee }$ is transitive. Thus,

$$J\left({\mathbb{R}}^{2n}\right)=J\left({\mathsf{\Phi }}_H^{\vee }\left(0\right)\right)={\mathrm{Ad}}_{H^{-1}}^T\left(J\left(0\right)\right)=H•J\left(0\right),$$

the H coadjoint orbit through $J\left(0\right)$. But $J\left(0\right)X^{\vee }=J^{X^{\vee }}\left(0\right)=\xi$. So, $J\left(0\right)=E_{0,2n+1}^{*}$, since

$$J\left(0\right)\left(\begin{array}{ccc}0& 0& 0\\ \tilde{x}& 0& 0\\ \xi & {\omega }^{♯}\left(\tilde{x}\right)& 0\end{array}\right)=\xi =E_{0,2n+1}^{*}\left(\begin{array}{ccc}0& 0& 0\\ \tilde{x}& 0& 0\\ \xi & {\omega }^{♯}\left(\tilde{x}\right)& 0\end{array}\right).$$

Thus, the H coadjoint orbit ${\mathcal{O}}_1$, through ${\mathcal{\ell }}_{E_{0,2n+1}}=E_{0,2n+1}^{*}\in {\mathfrak{h}}^{*}$, corresponds to the cotype represented by the tuple $({\mathbb{R}}^{2n+2},E_{0,2n+1},f_{n+1};\mathcal{J})$. This cotype is the sum of the indecomposable cotype ${\nabla }_1\left(0\right),1$ with modulus 1 and the zero type ${\mathbf{0}}_n$.

5. Representation of $\mathfrak{h}$ Corresponding to ${\mathcal{O}}_{\mathbf{1}}$

According to the theory of Kirillov [5], associated with the coadjoint orbit ${\mathcal{O}}_1$ of the Heisenberg group H through ${\mathcal{\ell }}_X\in {\mathfrak{h}}^{*}$, where $X=$$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 1& 0& 0\end{array}\right)$$\in \mathfrak{h}$, there is an irreducible unitary representation of $\mathfrak{h}$ by skew Hermitian differential operators on $C^{\infty }({\mathbb{R}}^n,\mathbb{C})$. We find this representation using geometric quantization, see Kostant [3] or Śniatycki [6].

Let ${\mathbb{R}}^{2n}=T^{*}{\mathbb{R}}^n$ be the cotangent bundle of ${\mathbb{R}}^n$ with coordinates $(x,y)$ and standard symplectic form $\omega \left((x,y),(z,w)\right)=\langle x,w\rangle -\langle y,z\rangle$. Here, $\langle ,\rangle$ is the Euclidean inner product on ${\mathbb{R}}^n$. Consider the trivial complex line bundle $\rho :L={\mathbb{R}}^{2n}\times \mathbb{C}\to {\mathbb{R}}^{2n}:(x,y,z)\mapsto (x,y)$. Let a, $b\in {\rho }^{-1}(x,y)$. Then, $h(x,y)(a,b)=a\overline{b}$ defines a Hermitian inner product on L. A smooth section $\tilde{\sigma }:{\mathbb{R}}^{2n}\to L:(x,y)\mapsto \left((x,y),\sigma (x,y)\right)$ of L will be identified with the smooth complex valued function $\sigma :{\mathbb{R}}^{2n}\to \mathbb{C}$. Let $s\in C^{\infty }({\mathbb{R}}^{2n},\mathbb{C})$ such that $s(x,y)=1$ for every $(x,y)\in {\mathbb{R}}^{2n}$. Then, s is a smooth section of L. We have a 1-form $\theta =\langle y,\mathrm{d}x\rangle$ on $T^{*}{\mathbb{R}}^n$, which gives rise to the symplectic form $\omega =-\mathrm{d}\theta$. For every smooth vector field X on $T^{*}{\mathbb{R}}^n$ and every smooth section $\tilde{\sigma }$ of L, let

$${\nabla }_X\tilde{\sigma }(x,y)=2\pi \mathrm{i}\left(X⅃\theta \right)(x,y)\tilde{\sigma }(x,y),\mathrm{for} \mathrm{every}(x,y)\in T^{*}{\mathbb{R}}^n.$$

For every $f\in C^{\infty }({\mathbb{R}}^{2n},\mathbb{C})$, since ${\nabla }_X\left(f\tilde{\sigma }\right)=\left(L_{X}f\right)·\tilde{\sigma }+f·\left({\nabla }_X\tilde{\sigma }\right)$, the operator ${\nabla }_X$ defines a connection on the space of smooth sections $\mathsf{\Gamma }\left(L\right)$ of the line bundle L. Using the section $s(x,y)=1$ and the definition of ${\nabla }_{X}s$, we obtain

$${\nabla }_{X}f=L_{X}f+2\pi \mathrm{i}\left(X⅃\theta \right)f\mathrm{for\; every} f\in C^{\infty }(L,\mathbb{C}).$$

Then, for $j=1,\dots ,n$,

$${\nabla }_{{\displaystyle \frac{\partial }{\partial x_j}}}f={\displaystyle \frac{\partial f}{\partial x_j}}+2\pi \mathrm{i}{y}_{j}f\mathrm{and}{\nabla }_{{\displaystyle \frac{\partial }{\partial y_j}}}f={\displaystyle \frac{\partial f}{\partial y_j}}.$$

If $F\in C^{\infty }({\mathbb{R}}^{2n},\mathbb{R})$, the corresponding Hamiltonian vector field on $(T^{*}{\mathbb{R}}^n,\omega )$ is $X_F=\langle {\displaystyle \frac{\partial F}{\partial y}},{\displaystyle \frac{\partial }{\partial x}}\rangle -\langle {\displaystyle \frac{\partial F}{\partial x}},{\displaystyle \frac{\partial }{\partial y}}\rangle$.

From geometric quantization, we obtain the prequantization operator $\mathcal{P}$ on $\mathsf{\Gamma }\left(L\right)$ given by

$$\begin{array}{cc}\hfill \mathcal{P}\left(F\right)& =-{\nabla }_{X_F}+2\pi \mathrm{i}F=-\langle {\displaystyle \frac{\partial F}{\partial y}},{\displaystyle \frac{\partial }{\partial x}}+2\pi \mathrm{i}y\rangle +\langle {\displaystyle \frac{\partial F}{\partial x}},{\displaystyle \frac{\partial }{\partial y}}\rangle +2\pi \mathrm{i}F\hfill \\ \hfill & =-X_F-2\pi \mathrm{i}\left(\langle y,{\displaystyle \frac{\partial F}{\partial y}}\rangle -F\right).\hfill \end{array}$$

Consider the Lie algebra mapping

$$\mathfrak{J}:(\mathfrak{h},[,])\to (C^{\infty }({\mathbb{R}}^{2n},\mathbb{C}),\{,\}):(\xi ,\eta ,s)\mapsto J^{(\xi ,\eta ,s)},$$

where

$$J^{(\xi ,\eta ,s)}(x,y)=\omega \left((\xi ,\eta ),(x,y)\right)+s=\langle \xi ,y\rangle -\langle \eta ,x\rangle +s.$$

Here, $J:{\mathbb{R}}^{2n}\to {\mathfrak{h}}^{*}$, where $J(x,y)(\xi ,\eta ,s)=J^{(\xi ,\eta ,s)}(x,y)$, is the momentum map of the Heisenberg group H acting on $({\mathbb{R}}^{2n},\omega )$. The mapping $\mathfrak{J}$ is a homomorphism of Lie algebras, namely, $J^{[(\xi ,\eta ,s),({\xi }^{\prime },{\eta }^{\prime },s^{\prime })]}=\{J^{(\xi ,\eta ,s)},J^{({\xi }^{\prime },{\eta }^{\prime },s^{\prime })}\}$. Now,

$$\begin{array}{cc}\hfill \mathcal{P}\left(J^{(\xi ,\eta ,s)}\right)& =-X_{J^{(\xi ,\eta ,s)}}-2\pi \mathrm{i}\left(\langle y,{\displaystyle \frac{\partial J^{(\xi ,\eta ,s)}}{\partial y}}\rangle -J^{(\xi ,\eta ,s)}\right)\hfill \\ \hfill & =-\langle \xi ,{\displaystyle \frac{\partial }{\partial x}}\rangle +\langle \eta ,{\displaystyle \frac{\partial }{\partial y}}\rangle -2\pi \mathrm{i}\left(\langle \eta ,x\rangle -s\right)\hfill \end{array}$$

is the prequantization operator, which gives a representation of $\mathfrak{h}$ on $C^{\infty }({\mathbb{R}}^{2n},$$\mathbb{C})$, namely, $(\xi ,\eta ,s)\mapsto \mathcal{P}\left(J^{(\xi ,\eta ,s)}\right)$. This completes the prequantization.

Using the polarization ${\displaystyle \frac{\partial }{\partial y}}=0$ of the coadjoint orbit ${\mathcal{O}}_1$, which is ${\mathbb{R}}^{2n}$, we obtain the quantization operator

$$\mathcal{Q}(\xi ,\eta ,s)=-\langle \xi , \frac{\partial }{\partial x}\rangle +2\pi \mathrm{i}(s-\langle \eta ,x\rangle )$$

(4)

on $C^{\infty }({\mathbb{R}}^n,\mathbb{C})$, where ${\mathbb{R}}^n$ has coordinate x. On $C^{\infty }({\mathbb{R}}^{2n},\mathbb{C})$, place the Hermitian inner product $\langle f,g\rangle ={\int }_{{\mathbb{R}}^n}f·g$. The quantization operator $\mathcal{Q}(\xi ,\eta ,s)$ is skew Hermitian, that is, $\langle \mathcal{Q}(\xi ,\eta ,s)f,g\rangle =-\langle f,\mathcal{Q}(\xi ,\eta ,s)g\rangle$. Using the inner product $\langle ,\rangle$, we convert $C_c^{\infty }({\mathbb{R}}^n,\mathbb{C})$, the space of smooth complex valued functions on ${\mathbb{R}}^n$ with compact support, to the Hilbert space $(\mathcal{H},\langle ,\rangle )$. Then,

$$\mathfrak{h}\to \mathrm{gl}(\mathcal{H},\langle ,\rangle ):(\xi ,\eta ,s)\mapsto \mathcal{Q}(\xi ,\eta ,s)$$

is the infinitesimalization of an irreducible unitary representation of the Heisenberg group H.

We now find the irreducible unitary representation of the Heisenberg group H, whose infinitesimalization is given by the quantum operator $\mathcal{Q}$ (4). Consider the representation ${\tilde{S}}_1=S_1\circ \psi$, where

$$S_1:H\to \mathrm{U}(\mathcal{H},\langle ,\rangle ):(x^{\prime },y^{\prime },t^{\prime })\mapsto \left(f\mapsto S_1(x^{\prime },y^{\prime },t^{\prime })f\right)$$

and $\left(S_1(x^{\prime },y^{\prime },t^{\prime })f\right)\left(z\right)={\mathrm{e}}^{2\pi \mathrm{i}[t^{\prime }+\langle x^{\prime },z+{\displaystyle \frac{1}{2}}{y}^{\prime }\rangle ]}f(z-y^{\prime })$ of H, see Wallach [4] (p. 107). Here, $\psi$ is the group homomorphism

$$\psi :H\to H:(x,y,t)\mapsto (x^{\prime },y^{\prime },t^{\prime })=(-y,x,t).$$

Let $(\xi ,\eta ,s)\in \mathfrak{h}$. Then, $exp(-u\eta ,u\xi ,us)=(-u\eta ,u\xi ,us)=(x^{\prime },y^{\prime },t^{\prime })\in H$. So, the infinitesimalization of ${\tilde{S}}_1$ is

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{du}}{\mid }_{u=0}\left({\tilde{S}}_1(u\xi ,u\eta ,us)f\right)\left(z\right)& ={\displaystyle \frac{d}{du}}{\mid }_{u=0}{\mathrm{e}}^{2\pi \mathrm{i}[us-\langle u\eta ,z+{\displaystyle \frac{1}{2}}u\xi \rangle ]}f(z-u\xi )\hfill \\ \hfill & =2\pi \mathrm{i}[s-\langle \eta ,z\rangle ]f\left(z\right)-\langle \xi ,{\displaystyle \frac{\partial f}{\partial z}}\rangle =\left(\mathcal{Q}(\xi ,\eta ,s)f\right)\left(z\right),\hfill \end{array}$$

as desired.

An irreducible unitary representation of H corresponding to the coadjoint orbit ${\mathcal{O}}_{\xi }$ is $S_{\xi }$, where

$$\left(S_{\xi }(x,y,t)f\right)\left(z\right)={\mathrm{e}}^{2\pi \mathrm{i}\xi [t+\langle -y,z+{\displaystyle \frac{1}{2}}x\rangle ]}f(z-x).$$

Here, the modulus $\xi$ of the coadjoint orbit ${\mathcal{O}}_{\xi }$ is a parameter in the irreducible unitary representation. Because every coadjoint orbit of the Heisenberg group H is of the form ${\mathcal{O}}_{\xi }$ for some value of the real nonzero parameter $\xi$, we have found all the irreducible unitary representations of the Heisenberg group up to equivalence.

6. Conclusions

This paper discusses the momentum map of the Heisenberg group and shows that it gives rise to a coadjoint orbit with a modulus. This modulus appears in the representation corresponding to this orbit given by geometric quantization.