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Simple question about SU(2) and SO(3)

  1. Nov 20, 2008 #1
    there's a surjective homomorphism from

    a : SU(2) --> SO(3)

    The kernel of this homomorphism is the center of SU(2) which is Z/2Z. Now the fundamental group of SO(3) is Z/2Z. This is a general thing.

    The simplest version of my question is how is the center of SU(2) related to the fundamental group SO(3).
     
  2. jcsd
  3. Nov 21, 2008 #2

    StatusX

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    Consider a path in SU(2) running from the identity to the other element in the center. Since SU(2) is simply connected, there is essentially one such path (ie, up to homotopy). Now project this path onto SO(3) using the surjective homomorphism. Since the center of SU(2) gets mapped to the identity of SO(3), this projected path starts and ends at the identity, so is a loop, and so defines an element of the fundamental group of SO(3).

    To see this is a non-trivial loop, ie, can't be continuously shrunk to the constant loop, note that if we deform this loop slightly, we can mirror this deformation of the original path in SU(2) so that our new loop is the projection of this new path. Now, if it was possible to iterate this procedure until our path was shrunk to the constant loop, the corresponding path in SU(2) would have to have shrunk to the constant path as well. But remember the loop in SO(3) always has both ends at the identity, so the corresponding path in SU(2) must always have each of its endpoints at one of the two points in the center. But it's clearly impossible that we could continuously deform a path connecting two distinct points to a loop ending at a single one of the points through a series of intermediate paths always ending at the two points: there would have to be a discontinuous jump at some point. Thus this is impossible, and the loop is a representative of the single non-trivial homotopy class of loops in SO(3).

    In general, if we have a simply connected space G and a surjective map of it onto another space H, the elements of the fundamental group of H are in one to one correspondence with the sheets of this covering (ie, with the elements in the preimage of a single point). In the case where G and H are groups and the map is a homomorphism, the fundamental group is isomorphic to the kernel of the map.
     
    Last edited: Nov 21, 2008
  4. Nov 22, 2008 #3
    Thank you, that was a clear explanation. Sorry, I'm not a math guy, and prefer to just ask these questions instead of really thinking about it. I guess a follow up question would be say you look at the complexifications of the groups, is the same statement is true? The center of Sl(2,C) is isomorphic to the fundamental group of PSL(2,C) under this morphism. I think that's true, but can it be extended?

    Given a surjective homorphism from a simply connected group to some other group and taking the complexifications of both the fiber group and the target group is the statement still true? That fundamental group of the complexified group in the image of the homomorphism is the same as the kernel of the homomorphism from the complexified group in the pre-image.
     
  5. Jun 23, 2010 #4
    well done explanation
     
  6. Jun 24, 2010 #5

    lavinia

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    SU(2) is simply connected - the homomorphism, SU(2) --> SO(3), is a covering projection.
     
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