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In order to apply the Marsden–Weinstein reduction, the action of the group $G$ must be free and proper. On the other hand, if I correctly understand, the M-W reduction obtained from a given group $G$ can be used to decrease the number of degrees of freedom of a Hamiltonian $H$, provided that the Hamiltonian flow of $H$ commutes with the action of $G$.

Could you please give an example of such a Hamiltonian $H$ and of such a group $G$, whose action is not proper?

Please, try to give an example in which $G$ has the lowest possible dimension: I mean, if it is possible, provide a 1-dimensional Lie group $G$.

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  • $\begingroup$ I see. Could you please write it as an answer, so that I can accept it? Please, explicitly write an example of $H$ and the $K$ having $g$ as its Hamiltonian flux: it will help mathematicians working in different fields to understand the answer. $\endgroup$ – Doriano Brogioli Sep 27 '20 at 11:11
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(Comment $\to$ answer as requested.)

Let $G=\mathbf R$ act on the 2-torus $Z=\mathrm U(1)\times\mathrm U(1)$ by $g(z_1,z_2)=(e^{ig}z_1, e^{i\pi g}z_2)$. Lift the action to $T^*Z$ and use any $G$-invariant $H$.

Explicitly $T^*Z=\mathbf R^2\times Z\ni(p_1,p_2,z_1,z_2)$ where $G$ acts by the flow of $K=p_1+\pi p_2$ (not proper by Bourbaki, Topologie générale, Chap. III, §4, Exercice 5), and say $H=$ any function of $(p_1,p_2)$.

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  • $\begingroup$ Rigorously, this is the answer! However, the aim of my question was slightly different. I was wondering: I have an $H$, and a $K$ in involution, but the flux of $K$ is not proper so I cannot apply M-W. I believed that this meant that I could not reduce $H$. But here, instead, $H$ is already "reduced", in the sense that it does not depend on one of the variables (actually, it does not depend on two, $z_1$ and $z_2$). Is this a general property, i.e. that, whenever $K$ is not proper, there is such a reduction, in some extended sense? Do you suggest me to open a new question? $\endgroup$ – Doriano Brogioli Sep 27 '20 at 15:02
  • $\begingroup$ @DorianoBrogioli I don’t know! Your modified question seems hard to make precise. $\endgroup$ – Francois Ziegler Sep 27 '20 at 15:44

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