The group is defined to be the group of 2-by-2 complex matrices with determinant 1 ("S" in the name stands for "special", which means determinant 1). This is a topologically closed subgroup of because it's cut out by a continuous equation .
The Lie algebra of is To first order in , Differentiating with respect to at gives: Therefore, if is in , then the trace of is zero. It turns out that this is "if and only if" (we'll see this later) so that The general element of is therefore and we see that the Lie algebra is a 3-dimensional (complex) vector space with coordinates . By contrast, it's trickier to write down a formula for the general element of .
A basis for this vector space is given by setting one coordinate at a time equal to 1 (and the others equal to zero): That is, any element of can be written in a unique way as for some .
Checking that sl(2,C) is a Lie algebra
Since is a Lie algebra, it's supposed to be preserved under taking commutator bracket. Let's check: You can see that the bracket of two things in still lives in . All the other bracket computations follow from antisymmetry. For example: and This example is of such fundamental importance that I'll restate the brackets we just found: We will use these relations all the time later in the course.
Trace(M) = 0 implies det(exp(M)) = 1
There's one loose end to tie up, namely showing that if then . We have a basis ( ) for the space of trace-zero matrices. Because we know that is a vector space, it suffices to show that are in because then any linear combination of them will live in .
We need to check that for all . We have which has determinant 1.
Show that and are in .