Symplectic projective orbits of unimodular exponential Lie groups

For an exponential Lie group $G$ and an irreducible unitary representation $(\pi,\mathcal{H}_{\pi})$ of $G$, we consider the natural action defined by $\pi$ on the projective space of $\mathcal{H}_{\pi}$, and show that the stabilisers of this action coincide with the projective kernel of $\pi$. Using this, we prove that, if $G/\mathrm{pker}(\pi)$ is unimodular, then $\pi$ admits a symplectic projective orbit if and only if $\pi$ is square-integrable modulo its projective kernel $\mathrm{pker}(\pi)$.

A classical problem in the study of coherent state orbits is to determine the representations π and vectors η for which the associated orbit (1.1) admits an additional structure, such as a symplectic stucture [13,19,22] or even a Kähler structure [20,21,24].One motivation for studying the existence of symplectic and complex coherent state orbits is their use in geometric quantisation [16,26] and Berezin quantisation [11,29], respectively.In addition, symplectic and complex coherent state orbits are naturally related to convexity properties of moment maps of unitary representations, see, e.g., [3,24].Although groups and representations admitting complex coherent state orbits are quite well-understood [21], the picture is far less complete for symplectic coherent state orbit, see, e.g., [22,Conj. 11.1].
The aim of the present paper is to study the existence of symplectic coherent state orbits for exponential Lie groups, that is, Lie groups whose exponential map is a diffeomorphism; any such group is solvable.Representations admitting such symplectic orbits are generally referred to as coherent state representations in the literature [21,23,24], and are often assumed to have a discrete kernel.As a consequence of more general results, we obtain the following characterisation of coherent state representations of unimodular exponential Lie groups.
If one of the above conditions holds true, then G has nontrivial centre.
We provide an example (cf.Example 3.5) of a nonunimodular group that admits symplectic orbits also for non-square-integrable representations, showing that Theorem 1.1 might fail for nonunimodular groups.
For nilpotent Lie groups, Theorem 1.1 was announced and stated as [22,Thm. 4.1]; however, to the best of our knowledge, a proof has not been outlined or published.On the other hand, the class of unimodular exponential Lie groups is larger than the class of nilpotent Lie groups.Indeed, for an exponential Lie group G with Lie algebra g, the unimodular function is given by ∆ G (exp X) = e Tr (ad X ) , X ∈ g, Hence the condition that G is unimodular is equivalent with Tr (ad X ) = 0 for every X ∈ g.This is the case, for instance, for semidirect products of the form V ⋊ α D R, where V is a finite dimensional real vector space, D : V → V is a linear map with Tr (D) = 0 with any purely imaginary eigenvalue, and the action α D : R → End (V ) is given by α D (t) = e tD , t ∈ R. Another relevant example is given in Example 2.7.
Beyond unimodular groups, the relation between coherent state representations and squareintegrable representations is more delicate.Although any exponential Lie group admitting a coherent state representation also admits a square-integrable representation with the same projective kernel, and, conversely, any square-integrable representation is a coherent state representation (see Proposition 3.3 for both facts), there might exist symplectic orbits also for non-square-integrable representations, cf.Example 3.5.
Lastly, we give an application of our results to Perelomov's completeness problem [27], and show that (in the case of exponential Lie groups) necessary conditions for the completeness of coherent state subsystems can be obtained from criteria for the cyclicity of restrictions of associated projective representations obtained in [6,14,30].
The paper is organised as follows.Section 2 contains preliminary results on the projective kernel of an irreducible representation and square-integrable representations.Section 3 is devoted to the study of the existence of symplectic coherent state orbits and the squareintegrability of the representation, including, among others, a proof of Theorem 1.1.Lastly, in Section 4, we present an application to Perelomov's completeness problem for coherent state subsystems in the setting of exponential Lie groups.
Notation.Lie groups will be denoted with capital letter G, H, etc, while their respective Lie algebras are denoted with the corresponding gothic letters g, h, etc.For an irreducible representation π : G → U (H π ) we denote by the same letter π its extension to the group C * -algebra C * (G), and its unitary equivalence class in G = C * (G).
For a complex vector space H, we denote by P (H) its projective space, that is, the set of all one-dimensional subspaces of H.It can be alternatively described as the set of equivalence classes for the equivalence relation on H \ {0} defined by

Irreducible representations of exponential groups
Let G = exp(g) be an exponential Lie group and let (π, H π ) be an irreducible unitary representation of G.By the coadjoint orbit method, there exists ℓ ∈ g * and a real polarisation h of ℓ such that π is unitarily equivalent to the monomial representation π ℓ := ind G H χ ℓ , where H = exp(h) and The coadjoint orbit O ℓ := Ad * (G)ℓ associated with the equivalence class [π] ∈ G is often simply denoted by O π .See [2,15] for background on the coadjoint orbit method.
The following remark will be used repeatedly.
The group G acts continuously on the projective space P (H π ) (equipped with the quotient topology) by ( The projective stabiliser of a fixed η ∈ H π \ {0}, that is, the stabiliser with respect to the above action, is then . For exponential Lie groups, the reverse inclusion also holds. For each r > 0, a unitary representation π r of G acting on H := L 2 (R/2πZ) is given by For k ∈ Z, consider the nonzero vector η k (s) = e iks .Then This shows that for any k ∈ Z.On the other hand, note that (η k + η −k )(s) = 2 cos(ks), and hence for s, t ∈ R. Let t ∈ R be such that there exists λ ∈ T satisfying Then necessarily λ = ±1 and cos(k(s − t)) = ± cos(ks) for any s ∈ R, which implies that t ∈ π k Z.It follows therefore that Let us compute pker (π r ).For this, let ξ ∈ H be given by ξ(s) = cos(s) + cos(2s).Then Taking s = 0, and s = π, we get, respectively, This shows that cos(t) = cos(2t) = λ, so that λ = 1 and t ∈ 2πZ, and hence , it follows that pker (π r ) ⊆ {0} × 2πZ.
Note that in this case the subgroup pker (π r ) is not even connected.
2.2.Square-integrable representations.For an irreducible representation (π, H π ) of a connected Lie group G, let H be a closed normal subgroup of G that is contained in pker (π).
The representation π of G is said to be square-integrable modulo H if there exists η ∈ H π \ {0} such that the well-defined continuous function ẋ → |(η, π(x)η)| is square-integrable on G/H, with respect to the left Haar measure on G/H.
In the following proposition we gather the different characterisations of the irreducible representations of exponential Lie groups that are square-integrable modulo their projective kernel.
Proposition 2.4.Let π be an irreducible representation of an exponential Lie group G.The following assertions are equivalent: If one of the above (i)-(iv) holds, then G(ℓ) and g(ℓ) are independent of ℓ ∈ O π .
By [15,Lemma 5.3.7], the projective kernel pker (π) of π is the largest closed normal subgroup of G(ℓ).This immediately yields the following consequence.
The following lemma provides a sufficient condition on a representation and a group under which the associated coadjoint orbit is closed and affine.This result plays an essential role in proving our main theorem.
Lemma 2.6.Let G be a exponential Lie group with its Lie algebra g, and let π : G → U (H π ) be an irreducible unitary representation of G such that G/pker (π) is unimodular.Assume that π is square-integrable modulo pker (π).Then the corresponding coadjoint orbit O π is closed and affine, namely Proof.Let ker(π) ⊆ G be the kernel of the representation π : G → U (H π ), and denote by G ′ := G/ ker(π) the associated quotient group.Then G ′ is a solvable Lie group, connected since the map p : G → G ′ is continuous and G is connected.The centre of G ′ is K = pker (π)/ ker(π), which is compact (cf.Remark 2.1), and hence unimodular.On the other hand, G ′ /K = G/pker (π) is also unimodular, hence so is G ′ itself, see, e.g., [1,Rem. 6].
The following example provides an example of a unimodular group G for which Lemma 2.6 might fail without the unimodularity assumption on G/pker (π).

Symplectic projective orbits and square-integrable representations
This section is devoted to the relation between the existence of symplectic coherent state orbits and the square-integrability of a representation.
3.1.Symplectic coherent state orbits.Let G be a connected Lie group with Lie algebra g and let (π, H π ) be an irreducible unitary representation of G. Denote by H ∞ π the space of smooth vectors of π, i.e., the family of vectors η ∈ H π such that the orbit map x → π(x)η is smooth.
The projective space P (H π ) is then a Hilbert manifold with respect to the local charts (U η , ϕ η ) at a point [η] ∈ P (H π ), η ∈ H π \ {0}, defined by The 2-form on P (H π ) defined in the chart ϕ η by ω makes P (H π ) into a symplectic Hilbert manifold.Here, for a smooth map f : M → N between Fréchet manifolds M , N , and p ∈ M , T p (f ) : T p M → T f (p) N denotes the linear tangent map.Similarly, the projective space P (H ∞ π ) is a Fréchet manifold with respect to the local charts The natural inclusion i : P (H ∞ π ) → P (H π ) is smooth, and the 2-form ω P (H ∞ π ) = i * ω P (Hπ) of P (H ∞ π ), given by ω is symplectic.The mapping i is an immersion.(See, e.g., [4,Sect. 4.3] for more details.)As mentioned before, the group G acts on the projective spaces P (H π ) and see, e.g., [4,Prop. 4.6].
For η ∈ H π \ {0}, the orbit , and since the inclusion i : The following lemma is well-known for finite-dimensional representations of connected Lie groups, see, e.g., [16,Theorem 26.8].As we are not aware of a reference in the infinitedimension case, we provide its short proof.Lemma 3.1.Let (π, H π ) be an irreducible representation of a connected Lie group G.For be the orbit map, and denote α : On the other hand, θ = α * (ω P (H ∞ π ) ), hence for X, Y ∈ g, Therefore, by (3.1), we can write where we have used that (dπ(X)η, η) is purely imaginary.Therefore, g ⊥ θ 1 = g(J π ([η])), and we have thus obtained that ω Ω is symplectic if and only if g [η] = g(J π ([η])).

Coherent state representations.
This subsection is devoted to the question which exponential Lie groups admit coherent state representations and which representations are the coherent state representations.Throughout this subsection, let G = exp(g) be an exponential Lie group and let π be an irreducible representation of G.We recall that the representation π is said to be a (symplectic) coherent state representation if there exists η ∈ H We start with the following simple consequence of Lemma 3.1.
Lemma 3.2.Let G be an exponential Lie group and π be an irreducible representation of G.
Proof.By Lemma 3.1 and since G is exponential, thus The link between the moment map and the coadjoint orbit associated to π is encoded in the fundamental identity see [3,Cor. 8,p. 274].
The next proposition uses the facts above to give the relation between the coherent state representations and the representations that are square-integrable representations modulo their projective kernel, for the case of general exponential Lie groups.Proposition 3.3.Let G be an exponential Lie group and let π : G → U (H π ) be an irreducible representation of G.
(iii) ⇒ (ii) Assume that π is square-integrable modulo pker (π).Then, by Lemma 2.6, the coadjoint orbit O π is equal to its affine hull, and hence an application of Corollary 2.5 gives Theorem 1.1 is now a consequence of Theorem 3.4.
Assume that one (and then all) of the equivalent conditions in the statement holds.If Z(G) = {0}, then π is square-integrable in the strict sense.However, since G is unimodular, this is impossible (see [10,Cor. 4.2]), therefore Z(G) must be nontrivial.
As already mentioned above, the implication (i) ⇒ (iii) of Theorem 3.4 might fail for general exponential Lie groups.This is demonstrated by the following example.
Let A * , B * , P * , Q * , R * , S * be the dual basis of {A, B, P, R, Q, S} in g * .We first note that G is not unimodular.Indeed, if ∆ G denotes the modular function of G, then ∆ G (exp X) = exp tr(ad X ) for every X ∈ g.Since tr(ad A ) = 2, the group G is not unimodular.
We next show that there exists an irreducible unitary representation of G admitting a symplectic coherent state orbit, but that fails to be square-integrable.For this, let ℓ = B * +S * and O ℓ = Ad * (G)ℓ, and denote by π = π ℓ : G → U (H π ) a realisation of an irreducible representation corresponding to O ℓ .Then g(ℓ) = RB + RQ, which is not an ideal of g, hence π ℓ is not square-integrable modulo its projective kernel by Proposition 2.4.
Let p be an ideal of g contained in g(ℓ).Assume towards a contradiction that p = {0}.For s, t ∈ R satisfying s 2 + t 2 = 0, it follows that [sB + tQ, Q] = sQ and [sB + tQ, B] = −tQ, and hence Q ∈ p.Since [P, Q] = R, it follows that R ∈ p ⊆ RB + RQ.This is a contradiction, therefore p = {0}.Since the Lie algebra p π of pker (π) is an ideal contained in g(ℓ), it follows in particular that pker (π) is trivial.Hence G/pker π is not unimodular.
We claim that there is η ∈ H ) is open in g * , so that G(J π ([η])) is trivial, hence equal to pker (π).If our claim is proved, then the action of π defines a symplectic orbit G • [η], by Lemma 3.2.

Application: Perelomov's completeness problem
This section describes an application of Proposition 2.2 to a problem considered in [27] regarding the completeness of coherent state subsystems; see [27, p. 226].More precisely, we show that necessary conditions for the completeness of coherent state subsystems of exponential Lie groups can be obtained from criteria for the cyclicity of restrictions of associated projective representations established in [6,14,30].4.1.Overcomplete coherent states.Let π be an irreducible unitary representation of a Lie group G.For a nonzero vector η ∈ H π , let G [η] be its projective stabiliser group.Denote by X = G/G [η] the associated homogeneous G-space and let s : X → G be a measurable crosssection for the projection p : G → X. Assume that X admits a G-invariant Radon measure µ X and that η is admissible, in the sense that In this situation, following [23, Sect.1.1], the vector η is said to define a π-system of coherent states based on X = G/G [η] .Given such a vector η, there exists unique d π,η > 0 such that see, e.g., [25,Thm 1.2].In many situations (i.e., when singletons in X are µ X -null sets), the relation (4.1) implies that the coherent state system {π(s(x))η} x∈G/G [η] is overcomplete, in the sense that it remains complete in H π after the removal of an arbitrary element.4.2.Coherent state subsystems.Let Γ be a discrete subgroup of G such that the factor space X/Γ has finite measure.In [27, p. 226], Perelomov considered the question of providing criteria for the completeness of a subsystem of coherent states associated with Γ ′ := p(Γ), in terms of the volume of X/Γ and the admissibility constant d π,η .As a combination of Proposition 2.2 and results in [6,30], the following theorem provides a necessary condition for the completeness of coherent state subsystems of the form (4.2) in the case of an exponential Lie group.Theorem 4.1.Let G be an exponential Lie group and let (π, H π ) be an irreducible representation of G admitting an admissible vector.Suppose that Γ is a discrete subgroup of G such that Γ ′ := p(Γ) is a uniform subgroup of X = G/G [η] .Let s : X → G be a Borel section.
Theorem 4.1 provides an extension of [32, Thm.1.2] from nilpotent Lie groups to general unimodular exponential Lie groups.In addition, Theorem 4.1 is valid for an arbitrary admissible vector η ∈ H π , whereas [32,Thm. 1.2] required the orbit G • [η] to be symplectic, which in particular implies η ∈ H ∞ π .Lastly, it is of interest to compare Theorem 4.1 to density conditions for coherent state subsystems of semisimple Lie groups.In the latter setting, the general density conditions for restricted representations [6,14,30] can be improved to depend on the projective stabiliser of an admissible vector, see, e.g., [12,Theorem 4.5].On the other hand, in the setting of exponential Lie groups, the projective stabilisers are always independent of the choice of vector, by Proposition 2.2.

Example 3 . 5 .
Let G be the exponential Lie group with Lie algebra g = span{A, B, P, R, Q, S},

. 5 )
On the other hand, the coadjoint orbit O = Ad * (G)(Q * + S * ) is open, of dimension 6 and given by