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 ⁡ ( M ) = 1 .

Lie algebra

The Lie algebra of S ⁢ L ⁢ ( 2 , 𝐂 ) is 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) = { ( a b c d ) : det ⁡ exp ⁡ t ⁢ ( a b c d ) = 1 ⁢ ∀ t ∈ 𝐑 } To first order in t , det ⁡ exp ⁡ t ⁢ ( a b c d ) = det ⁡ ( 1 + t ⁢ a t ⁢ b t ⁢ c 1 + t ⁢ d ) + 𝒪 ⁢ ( t 2 ) = ( 1 + t ⁢ a ) ⁢ ( 1 + t ⁢ d ) - t 2 ⁢ b ⁢ c + 𝒪 ⁢ ( t 2 ) = 1 + t ⁢ ( a + d ) + 𝒪 ⁢ ( t 2 ) . Differentiating with respect to t at t = 0 gives: d d ⁢ t | t = 0 ⁢ det ⁡ exp ⁡ t ⁢ ( a b c d ) = a + d = d d ⁢ t | t = 0 ⁢ 1 = 0 . Therefore, if M = ( a b c d ) is in 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) , then the trace of M Tr ⁢ ( M ) = a + d is zero. It turns out that this is "if and only if" (we'll see this later) so that 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) = { M : Tr ⁢ ( M ) = 0 } The general element of 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) is therefore ( a b c - a ) 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 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 = ( 1 0 0 - 1 ) , X = ( 0 1 0 0 ) , Y = ( 0 0 1 0 ) That is, any element of 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) can be written in a unique way as a ⁢ H + b ⁢ X + c ⁢ Y for some a , b , c ∈ 𝐂 .

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

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

Lemma:

H ∈ 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) .

Proof:

We need to check that det ⁡ exp ⁡ ( t ⁢ H ) = 1 for all t . We have exp ⁡ ( t ⁢ H ) = exp ⁡ ( t 0 0 - t ) = ( 1 + t + t 2 / 2 + ⋯ 0 0 1 - t + t 2 / 2 - ⋯ ) = ( e t 0 0 e - t ) which has determinant 1.

Pre-class exercise

Exercise:

Show that X and Y are in 𝔰 ⁢ 𝔩 ⁢ ( 2 , 𝐂 ) .