I want to talk about a particular smooth homomorphism SU(2) to SO(3).
SU(2) equals the set of invertible 2-by-2 complex matrices M such that det(M)=1 and M dagger M equals the identity is the group of 2-by-2 complex matrices which are unitary and have determinant 1 (special unitary group). Here M dagger equals M conjugate transpose.
We can alternatively characterise SU(2) as the group of matrices a,b,minus b bar, a bar such that a and b are complex numbers and absolute value of a squared + absolute value of b squared equals 1. To see this, note that unitarity implies M dagger equals M inverse, so bar a, bar c, bar b, bar d equals one over det M times d, minus b, minus c, a, and because det M equals 1 we get c equals minus b bar and d equals a bar. The condition det M equals 1 then becomes absolute value of a squared + absolute value of b squared equals 1.
The Lie algebra is little s u 2 equals the set of 2-by-2 complex matrices X such that trace X equals zero and X dagger equals minus X (these conditions come from differentiating det exp t X equals 1 and (exp t X) dagger times exp t X equals the identity with respect to t at t=0 as usual). We can write down a general element of this Lie algebra as follows:little s u 2 equals the set of matrices i x, y + i z, minus y + i z, minus i x, with x,y,z real numbers. The fact that the diagonal elements are imaginary follows from X dagger equals minus X. They sum to zero because the trace is zero. The bottom-left entry is determined by the top-right by X dagger equals minus X.
Even though SU(2) is a group of complex matrices, little s u 2 is a real vector space, not a complex vector space. That's because SU(2) is cut out by a single real equation (rather than something holomorphic).
If v equals the vector x,y,z in R 3, let's write M_v equals the 2-by-2 matrix i x, y + i z, minus y + i z, minus i x. Here are some useful equations we will use.
M_u bracket M_v equals M 2 u cross v and trace of M_u times M_v equals minus 2 u dot v.
If B is in SU(2) then B M_v B inverse is in little s u 2 for any M_v in little s u 2. This means that B M_v B inverse equals M_w for some w in R 3. Moreover, w equals R of B applied to v for some rotation R of B. In other words, we have a map R from SU(2) to SO(3) such that M subscript (R(B) of v) equals B M_v B inverse (we will only check that R of B is an orthogonal matrix, not that it's a rotation).
To see that B M_v B inverse is in little s u 2, we need to show that it is tracefree and anti-Hermitian:
Because trace is conjugation invariant, we have trace of B M_v B inverse equals the trace of M_v, which equals 0.
We have B M_v B inverse, all dagger, equals B inverse dagger M_v dagger B dagger, which equals minus B times M_v times B inverse because B is in SU(2) and M_v is in little s u 2.
Now define R of B from R 3 to R 3 implicitly by M subscript (R of B v) equals B M_v B inverse. We want to show that R of B is an orthogonal matrix, i.e. that it preserves dot products.
Using our formula from before, we have: R of B v_1 dot R of B v_2 equals minus a half trace of (M subscript (R of B v_1) times M subscript (R of B v_2)) which is equal to minus a half trace of B M v_1 B inverse B M v_2 B inverse, which equals the trace of M v_1 M v_2 because the central B inverse B cancels and the trace is unchanged by conjugation. This is then equal to v_1 dot v_2 by the formula from above.
We have therefore defined a map R from SU(2) to O(3). This turns out to be a homomorphism. Let's compute R star. We know that R of exp t M_u equals exp of t R star M_u, so R star M_u equals d by d t at t=0 of R exp t M_u.
We have M subscript (R of B v) equals B M_v B inverse, so if B equals exp t M_u then we get M subscript (R (exp t M_u) v) equals exp t M_u times M_v times exp of minus t M_u Differentiating with respect to t gives M subscript (R star M_u v) equals d by d t at t=0 of M_u M_v minus M_v M_u by the product rule. Therefore M subscript(R star M_u v) equals M_u bracket M_v, which equals M 2 u cross v, so R star M_u v equals 2 u cross v
Therefore R star M_u is the map which sends v to 2 u cross v. If we write u=(u_1,u_2,u_3) and v=(v_1,v_2,v_3), note that the 3-by-3 matrix 0, minus u_3, u_2, u_3, 0, minus u_1, minus u_2, u_1, 0 applied to the vector v equals the vector u_2 v_2 minus u_3 v_2, u_3 v_1 minus u_1 v_3, u_1 v_2 minus u_2 v_1, which equals u cross v so we have R star of the matrix ix, y + i z, minus y + i z, minus i x equals the 3-by-3 matrix 0, minus 2 z, 2 y, 2 z, 0, minus 2 x, minus 2 y, 2 x, 0.
This is a nontrivial example of a Lie algebra homomorphism little s u 2 to little o 3. Even though it's nontrivial, it is easy to write down: it's linear in x, y, and z.
We have this homomorphism R from SU(2) to O(3). This is not an isomorphism. For a start, it only hits the rotations, i.e. the orthogonal transformations with determinant 1 (SO(3) inside O(3)). Even if you think of it as a map SU(2) to SO(3), it's still not an isomorphism: it is 2-to-1. For example, B and minus B both map to the same rotation: B M_v B inverse equals minus B M_v minus B inverse.
At the level of Lie algebras, R star little s u 2 to little s o 3 is an isomorphism: from the 3-by-3 matrix 0, minus 2 z, 2 y, 2 z, 0, minus 2 x, minus 2 y, 2 x, 0. we can figure out x, y and z by dividing by 2, so we know which matrix the matrix ix, y + i z, minus y + i z, minus i x to write down. That is, we have an inverse R star inverse from little s o 3 to little s u 2. However, it doesn't correspond to a Lie group homomorphism SO(3) to SU(2). This is OK because Lie's theorem about exponentiating homomorphisms only applies when the domain is simply-connected, and the group SO(3) is not simply-connected.
We constructed a homomorphism R from SU(2) to O(3). Can you see why R lands in the subset SO(3) of matrices with determinant 1?
Can you prove that little s u 2 is the set of 2-by-2 matrices X such that X transpose equals minus X and trace of X equals 0?