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Let $G$ be a connected Lie group and let $(\pi, \mathcal{H})$ be an irreducible unitary representation of $G$ on an infinite-dimensional Hilbert space $\mathcal{H}$. Denote by $\mathcal{H}^{\infty}$ the space of smooth vectors of $\pi$.

If $P(\mathcal{H})$ denotes the projective Hilbert space, then $G$ acts on $P(\mathcal{H})$ by $g \cdot [v] = [\pi(g) v]$, where $[v] := \mathbb{C} v$ for $v \in \mathcal{H} \setminus \{0\}$. If $v \in \mathcal{H}^{\infty} \setminus \{0\}$, then the orbit $G \cdot [v]$ is a smooth (immersed) submanifold of $P(\mathcal{H})$.

An orbit $G \cdot [v]$ of a vector $v \in \mathcal{H}^{\infty} \setminus \{0\}$ is called symplectic if it is a symplectic submanifold of $P(\mathcal{H})$.

Question: Is a symplectic orbit $G \cdot [v]$ always simply connected?

The above question stems from a claim in the paper [1], where it is claimed [1, p.106] that any symplectic orbit is always simply connected. Unfortunately, there is no hint given that justifies the claim and I have been unable to show it myself.

In [2, Proposition 3.3], a proof can be found for semi-simple Lie groups that is specific for this setting; in particular, it is used that certain coadjoint orbits are simply connected.

[1] W. Lisiecki, Symplectic and Kaehler coherent state representations of unimodular Lie groups, in Quantization and Coherent States Methods, S. T. Ali et al., eds. (World Scientific, 1993).

[2] W. Lisiecki. Kaehler coherent state orbits for representations of semisimple Lie groups. Ann. Inst. Henri Poincar´e, Phys. Th´eor., 53(2):245–258, 1990.

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    $\begingroup$ Coadjoint orbits need not be simply connected in general. I am not sure if this helps under your particular assumptions though. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Dec 11, 2021 at 7:12

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I do not have access to the article you are citing but I have made a little search and I think the answer to your question is yes (given that we are speaking about a non-compact, connected, semisimple Lie group $G$).
You can find the proof in a previous article by the same author. See: Lisiecki - Kaehler coherent state orbits for representations of semisimple Lie groups (it is proposition 3.3, p. 251).

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    $\begingroup$ Thanks for your answer. I was aware of this proposition, and maybe I should have cited it from the beginning in the question (see update). However, the claim in [1] is about general connected Lie groups. The fact that the author uses structure theory to prove this for semi-simple Lie groups makes me wonder what an argument for the general case would be. $\endgroup$
    – jvnv
    Commented Dec 13, 2021 at 11:08

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