# Marsden–Weinstein: example of not proper action

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 $$G$$.

• 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. – Doriano Brogioli Sep 27 '20 at 11:11

(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 $$(p_1,p_2)$$.
• 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? – Doriano Brogioli Sep 27 '20 at 15:02