Representations

Representations

In the next part of the course, we're going to focus on representations of Lie groups. Remember that a (complex) representation of a group is a homomorphism R : G → G ⁢ L ⁢ ( n , 𝐂 ) , that is an assignment of a matrix R ⁢ ( g ) to each group element such that R ⁢ ( g 1 ⁢ g 2 ) = R ⁢ ( g 1 ) ⁢ R ⁢ ( g 2 ) and R ⁢ ( 1 ) = I . The image of this representation is a group of matrices which is a quotient of G . In this course, we'll focus on smooth representations of matrix groups.

Why are we focusing on representations? There are many fantastic applications:

Recap

I just want to recap the punchline of the first half of the course, as this will be the basis for everything that comes after.

Given a smooth representation R : G → G ⁢ L ⁢ ( n , 𝐂 ) we get a Lie algebra representation R * : 𝔤 → 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐂 ) , that is a linear map such that R * ⁢ [ X , Y ] = [ R * ⁢ X , R * ⁢ Y ] .

Remark:

Here, 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐂 ) is a complex vector space, but 𝔤 is a vector space over the real numbers. When I say that R * : 𝔤 → 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐂 ) is linear, the only thing that makes sense is for it to be real linear, i.e. R * ⁢ ( λ ⁢ X ) = λ ⁢ R * ⁢ ( X ) ⁢ ∀ λ ∈ 𝐑 . We will later complexify 𝔤 to obtain a complex vector space 𝔤 ⊗ 𝐂 and get an associated complex linear map R * 𝐂 : 𝔤 ⊗ 𝐂 → 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐂 ) .

The key property of R * was the equation R ⁢ ( exp ⁡ X ) = exp ⁡ ( R * ⁢ X ) . This tells us that R determines R * by differentiation: R * ⁢ ( X ) = d d ⁢ t | t = 0 ⁢ R ⁢ ( exp ⁡ ( t ⁢ X ) ) . Using the formula R ⁢ ( exp ⁡ X ) = exp ⁡ ( R * ⁢ X ) , we see that R * determines R ⁢ ( g ) for all g ∈ exp ⁡ ( 𝔤 ) . Does that mean R * determines R ⁢ ( g ) for all g ∈ G ?

Lemma:

If G is a path-connected group (i.e. any two matrices in G are connected by a smooth path of matrices in G ) then R is determined by R * .

This is because, in this case, G is generated as a group by exp ⁡ ( 𝔤 ) ⊂ G . The proof of this lemma is an exercise.

Given R * : 𝔤 → 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐂 ) , does R ⁢ ( exp ⁡ X ) = exp ⁡ ( R * ⁢ X ) give a well-defined representation R : G → G ⁢ L ⁢ ( n , 𝐂 ) ? Lie's theorem told us that this is true if G is simply-connected. If G is not simply-connected, we need to think. The first example we'll consider is U ⁢ ( 1 ) , which is not simply-connected, but all the other examples we will consider are simply-connected.

Plan

Our plan for the rest of the course is:

Pre-class exercise

Exercise:

In an earlier video, we constructed a map R : S ⁢ U ⁢ ( 2 ) → S ⁢ O ⁢ ( 3 ) . Check that R ⁢ ( M 1 ) = R ⁢ ( M 2 ) if and only if M 1 = ± M 2 . Given a representation S : S ⁢ O ⁢ ( 3 ) → G ⁢ L ⁢ ( n , 𝐂 ) , we get a representation R ∘ S : S ⁢ U ⁢ ( 2 ) → S ⁢ O ⁢ ( 3 ) . Show that a representation T : S ⁢ U ⁢ ( 2 ) → S ⁢ O ⁢ ( 3 ) has this form if and only if T ⁢ ( - M ) = T ⁢ ( M ) for all M ∈ S ⁢ U ⁢ ( 2 ) .

Exercise:

Suppose we have a Lie algebra 𝔤 ⊂ 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐑 ) consisting of real matrices. Consider the subspace 𝔤 ⊗ 𝐂 ⊂ 𝔤 ⁢ 𝔩 ⁢ ( n , 𝐂 ) consisting of matrices of the form M + i ⁢ N with M and N in 𝔤 . Show that:

  • this is a Lie subalgebra, i.e. that it is preserved by Lie bracket

  • if f : 𝔤 → 𝔤 ⁢ 𝔩 ⁢ ( m , 𝐂 ) is a real-linear Lie algebra homomorphism then f 𝐂 : 𝔤 ⊗ 𝐂 → 𝔤 ⁢ 𝔩 ⁢ ( m , 𝐂 ) defined by f 𝐂 ⁢ ( M + i ⁢ N ) = f ⁢ ( M ) + i ⁢ f ⁢ ( N ) for M and N in 𝔤 is also a Lie algebra homomorphism.