SO(3) as a quotient group of SU(2)?

wdlang
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we know there is a two to one homomorphism from SU(2) to SO(3)

suppose u is an element in SU(2)

then u and -u map into the same element in SO(3)

the question is, maybe SO(3) is a quotient group of SU(2)? with respect to the subgroup {I,-I}?
 
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Yes, SO(3) can be interpreted as a quotient group of SU(2).

{I,-I} is an abelian subgroup and it is a normal subgroup of SU(2). A quotient group G/H is always trivially related to a group G and a normal subgroup H of G.

Thus, indeed SO(3) equals (or is at least isomorphic to) SU(2) after dividing out by {I,-I}.

Another way to see it:

The 3-sphere (S^3) can be considered as a 'manifold' isomorphic to SU(2) and the real projective space of dimension 3, which is a quotient of S^3 under 'antipodal identification' is isomorphic to SO(3). Both are orientable manifolds.
 
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{I,-I} is an abelian subgroup and it is a normal subgroup of SU(2). A quotient group G/H is always trivially related to a group G and a normal subgroup H of G.

Thus, indeed SO(3) equals (or is at least isomorphic to) SU(2) after dividing out by {I,-I}.
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yes, you have shown that SU(2)/{I,-I} is a quotient group

but i think you did not show that it is isomorphic to SO(3).
 
Well, you didn't ask to show it.

I wrote: "The 3-sphere (S^3) can be considered as a 'manifold' isomorphic to SU(2) and the real projective space of dimension 3, which is a quotient of S^3 under 'antipodal identification' is isomorphic to SO(3). Both are orientable manifolds."

Working out the defining relations of any matrix in the fundamental representation of SU(2) will give us that SU(2) can be regarded as a 3-sphere. Doing the same for SO(3) also gives us that SO(3) can be regarded as a 3-sphere, however, an antipodal pair of points will give us the same element of SO(3).

Another way to see it: The adjoint representation of SU(2) can be identified with the fundamental representation of SO(3).
 
Hello! There is a simple line in the textbook. If ##S## is a manifold, an injectively immersed submanifold ##M## of ##S## is embedded if and only if ##M## is locally closed in ##S##. Recall the definition. M is locally closed if for each point ##x\in M## there open ##U\subset S## such that ##M\cap U## is closed in ##U##. Embedding to injective immesion is simple. The opposite direction is hard. Suppose I have ##N## as source manifold and ##f:N\rightarrow S## is the injective...

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