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.