Assume that $G$ is a Lie group and at the same time it admits a symplectic structure.
Does $G$ necessarily admit a symplectic structure such that the right multiplication preserves the symplectic structure?
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3$\begingroup$ Let $G$ be the universal covering of $\mathrm{SL}_2(\mathbf{R})$. Then $G\times \mathbf{R}$ is diffeomorphic to $\mathbf{R}^4$ and hence admits a symplectic structure (as a manifold). However every Lie group with a right-invariant symplectic structure is solvable (see arxiv.org/abs/1307.1629 for references), so $G\times \mathbf{R}$ has no such structure. $\endgroup$– YCorCommented Oct 23, 2017 at 9:30
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3$\begingroup$ See also mathoverflow.net/questions/71766/… and if you know the French language, see numdam.org/article/AIF_1979__29_4_17_0.pdf or this PhD thesis but in French language tel.archives-ouvertes.fr/tel-00078872/document $\endgroup$– user21574Commented Oct 23, 2017 at 10:56
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10$\begingroup$ This is the nice paper which answer to your question link.springer.com/chapter/10.1007/978-1-4613-9719-9_17 $\endgroup$– user21574Commented Oct 23, 2017 at 11:02
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9$\begingroup$ Also this is the correct definition: A Lie group $G$ admits an invariant symplectic structure if there exists on $G$ a left invariant closed 2- form whose rank is equal to the dimension of $G$. Also there is a notion of Poisson Lie group and Poisson Lie algebra(I remember I have had lecture course about it 5 years ago when I was master student in Marseille) $\endgroup$– user21574Commented Oct 23, 2017 at 11:15
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10$\begingroup$ If you are looking for Lie group which admit Kahler structure see André Lichnerowicz paper (written in French) link.springer.com/chapter/10.1007/BFb0097472 $\endgroup$– user21574Commented Oct 23, 2017 at 11:26
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1 Answer
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Let $H$ be the universal covering of $\mathrm{SL}_2(\mathbf{R})\times\mathbf{R}$. Then $H$ is diffeomorphic to $\mathbf{R}^4$ and hence has a symplectic structure (as a manifold). However, every Lie group with a right-invariant symplectic structure is solvable (see Baues-Cortès (arXiv link) for references), so $H$ has no such structure.
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2$\begingroup$ (+1) and thank you very much for your attention to my question and your very interesting answer. $\endgroup$ Commented Oct 23, 2017 at 13:15