is the group of 2-by-2 complex matrices which are unitary and have determinant 1 (special unitary group). Here .
The group SU(2)
I want to talk about a particular smooth homomorphism .
We can alternatively characterise as the group of matrices . To see this, note that unitarity implies , so and because we get and . The condition then becomes .
The Lie algebra su(2)
The Lie algebra is (these conditions come from differentiating and with respect to at as usual). We can write down a general element of this Lie algebra as follows: The fact that the diagonal elements are imaginary follows from . They sum to zero because the trace is zero. The bottom-left entry is determined by the top-right by .
Even though is a group of complex matrices, is a real vector space, not a complex vector space. That's because is cut out by a single real equation (rather than something holomorphic).
If , let's write . Here are some useful equations we will use.
If then for any . This means that for some . Moreover, for some rotation . In other words, we have a map such that (we will only check that is an orthogonal matrix, not that it's a rotation).
To see that , we need to show that it is tracefree and anti-Hermitian:
Because trace is conjugation invariant, we have .
We have because and .
Now define implicitly by . We want to show that is an orthogonal matrix, i.e. that it preserves dot products.
Using our formula from before, we have: which is equal to because the central cancels and the trace is unchanged by conjugation. This is then equal to by the formula from above.
We have therefore defined a map . This turns out to be a homomorphism. Let's compute . We know that , so
We have , so if then we get Differentiating with respect to gives by the product rule. Therefore , so
Therefore is the map which sends to . If we write and , note that so we have
This is a nontrivial example of a Lie algebra homomorphism . Even though it's nontrivial, it is easy to write down: it's linear in , , and .
We have this homomorphism . This is not an isomorphism. For a start, it only hits the rotations, i.e. the orthogonal transformations with determinant ( ). Even if you think of it as a map , it's still not an isomorphism: it is 2-to-1. For example, and both map to the same rotation:
At the level of Lie algebras, is an isomorphism: from we can figure out , and by dividing by , so we know which matrix to write down. That is, we have an inverse . However, it doesn't correspond to a Lie group homomorphism . This is OK because Lie's theorem about exponentiating homomorphisms only applies when the domain is simply-connected, and the group is not simply-connected.
We constructed a homomorphism . Can you see why lands in the subset of matrices with determinant 1?
Can you prove that is the set of 2-by-2 matrices such that and ?