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suppose i have $O(3)$ as a group and then proceed to identify rotations on the same axis. That is, assuming an element in the simple component is written as

$$ e^{s_i I_i } $$

where $I_i$ are generators of rotations around axis X,Y and Z, basically the identification would work as

$$ s_i \sim \lambda s'_i $$

for any real $\lambda$

questions:

  • What is this quotient group that i just obtained?

  • Are infinitesimal transformations in the obtained group still generated by $I_i$?

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1 Answer

up vote 1 down vote accepted

Every rotation can be gotten as a conjugate of a rotation around, say, the $x$ axis, so the only way to kill, say, every $(\lambda-1)s_x$ with a group homomorphism is to map all of $SO(3)$ to the identity.

Note that the rotations around one axis form a subgroup, but not a normal subgroup.

If you don't care about the group structure but only the topology of SO(3), and exclude the identity before you start quotienting, you get the real projective plane.

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