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Why so(3) is not isomorphic to su(2)?

  1. Jul 12, 2010 #1
    it is generally known that there is a two-to-one automorphism from su(2) to so(3)

    but consider the problem in this way:

    all elements in so(3) are of the form (up to a unitary transform of the basis)

    R(\alpha,\beta.\gamma)=e^{-i\alpha F_z} e^{-i\beta F_y} e^{-i \gamma F_z}

    where F_x, F_y, F_z are the 3*3 spin operators

    all elements in su(2) are of the form

    R(\alpha,\beta.\gamma)=e^{-i\alpha \sigma_z/2} e^{-i\beta \sigma_y/2} e^{-i \gamma \sigma_z/2}

    where \sigma_{x,y,z} are pauli matrices

    F_{x,y,z} and \sigma_{x,y,z}/2 are of the same lie algebra!

    so i think there should be a one-to-one correspondence between so(3) and su(2).
  2. jcsd
  3. Jul 12, 2010 #2
    i now know why

    R(\alpha,\beta.\gamma)=e^{-i\alpha \sigma_z/2} e^{-i\beta \sigma_y/2} e^{-i \gamma \sigma_z/2}

    do not cover the su(2) group completely.
  4. Jul 13, 2010 #3


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    When you write su(2) and so(3) (with minuscules), one usually understands this as a Lie algebra and not a Lie group. The algebras are in deed isomorphic. However, the very same algebra may generate different Lie groups.
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