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

1. Jul 12, 2010

### wdlang

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. Jul 12, 2010

### wdlang

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.

3. Jul 13, 2010

### DrDu

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.