H is in little s l 2 C.

# Example: SL(2,C)

## Example: SL(2,C)

### Definition

The group S L 2 C 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 G L 2 C because it's cut out by a continuous equation det of M equals 1.

### Lie algebra

The Lie algebra of S L 2 C is little s l 2 C equals the set of matrices M = a, b, c, d such that det of exp t M equals 1 for all t To first order in t, det of exp t times the matrix a, b, c, d equals det of the matrix 1 plus t a, t b, t c, 1 plus t d plus higher order terms in t, which equals 1 plus t times (a plus d) plus higher order terms in t Differentiating with respect to t at t = 0 gives: d by dt at t = 0 of det exp t times the matrix a, b, c, d equals a plus d, which equals d by dt at t = 0 of 1, which equals 0. Therefore, if the matrix M given by a, b, c, d is in little s l 2 C, then the *trace of M* trace of M = a plus d is zero. It turns out that this is "if and only if" (we'll see this later) so that little s l 2 C is the set of matrices with trace zero The general element of little s l 2 C is therefore the matrix a, b, c, d and we see that the Lie algebra is a 3-dimensional (complex) vector space with coordinates a,b,c. By contrast, it's trickier to write down a formula for the general element of big S L 2 C.

A basis for this vector space is given by setting one coordinate at a time equal to 1 (and the others equal to zero): H is the matrix 1, 0, 0, -1; X is the matrix 0, 1, 0, 0; Y is the matrix 0, 0, 1, 0 That is, any element of little s l 2 C can be written in a unique way as a H plus b X plus c Y for some complex numbers a,b,c.

### Checking that sl(2,C) is a Lie algebra

Since little s l 2 C is a Lie algebra, it's supposed to be preserved under taking commutator bracket. Let's check: H bracket X equals H X minus X H, which equals the matrix product 1, 0, 0, -1 times 0, 1, 0, 0 minus the matrix product 0, 1, 0, 0 times 1, 0, 0, -1, which equals the matrix 0, 2, 0, 0, or 2 X H bracket Y equals the matrix product 1, 0, 0, -1 times 0, 0, 1, 0 minus the matrix product 0, 0, 1, 0 times 1, 0, 0 ,-1, which equals the matrix 0, 0, -2, 0, which equals -2Y X bracket Y equals X Y minus Y X, which equals the matrix product 0, 1, 0, 0 times 0, 0, 1, 0 minus the matrix product 0, 0, 1, 0 times 0, 1, 0, 0, which equals the matrix 1, 0, 0, -1, which is H You can see that the bracket of two things in little s l 2 C still lives in little s l 2 C. All the other bracket computations follow from antisymmetry. For example: X bracket X equals Y bracket Y equals H bracket H equals 0 and Y bracket X equals minus X bracket Y equals minus H; X bracket H equals minus H bracket X equals minus 2 X, etc This example is of such fundamental importance that I'll restate the brackets we just found: H bracket X equals 2 X, H bracket Y equals minus 2 Y and X bracket Y equals H. 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* trace of M equals 0 then M is in little s l 2 C. We have a basis (H,X,Y) for the space of trace-zero matrices. Because we know that little s l 2 C is a vector space, it suffices to show that H,X,Y are in little s l 2 C because then any linear combination of them will live in little s l 2 C.

We need to check that det ext t H equals 1 for all t. We have exp t H equals exp of the matrix t, 0, 0 ,-t, which equals the matrix 1 plus t plus a half t squared dot dot dot, 0, 0, 1 minus t plus a half t squared dot dot dot,which equals the matrix e to the t, 0, 0, e to the minus t which has determinant 1.

## Pre-class exercise

Show that X and Y are in little s l 2 C.