Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    DrDu

    User Avatar
    Science Advisor

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook