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I understand $SO(3)$ is the right group for proper rotation of its Hamiltonian. $O(3)$ describes both proper and improper rotations and the system has symmetry even for reflection. $O(3)$ is not connected and this makes harsh to study its Lie Algebra. For this reason we study proper and improper rotation apart.

We then notice $SO(3)$ is doubly connected, so it's easier to handle the universal covering $SU(2)$. From this I wonder: why not study $U(2)$, a compact and connected group without a restriction on matrixes determinants?

Does it describe improper rotation too, or does it add simmetries not described by the Hamiltonian?

Qmechanic
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Matteo
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    Related: https://physics.stackexchange.com/q/169087/2451 – Qmechanic Jul 31 '22 at 14:43
  • It's not hard to study the Lie algebra of $O(3)$, it's actually the same as that of $SO(3)$ and $SU(2)$. On the other hand, that of $U(2)$ is different, so it's not clear why you would consider that. – fqq Aug 01 '22 at 12:11
  • O(3) is not connected, so it's easier to deal with SO(3) and study discret simmetries apart. Then we study SU(2). I thought that removing the restrictions on the determinant would make it possible to study improper rotations too, without separating the two situations. It happens that U(2) is both compact and connected, so there should be no problems as far as I understand. SO(3) and O(3) share the same algebra, so I thought SU(2) would share the same of U(2), even if it's isomorph to SU(2)×U(1) (if that's not the case, why don't they share the same one?) SU(2) has the same one of SO(3), so... – Matteo Aug 01 '22 at 14:37

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