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 from big G to big G L n C, that is an assignment of a matrix R of g to each group element such that of g_1 g_2 equals R of g_1 times R of g_2 and R of the identity equals the identity matrix. The image of this representation is a group of matrices which is a quotient of big 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 from big G to big G L n C we get a Lie algebra representation R star from little g to little g l n C, that is a linear map such that R star of X bracket Y equals R star X bracket R star Y.

Remark:

Here, little g l n C is a complex vector space, but little g is a vector space over the real numbers. When I say that R star from little g to little g l n C is linear, the only thing that makes sense is for it to be real linear, i.e. R star lambda X equals lambda R star X for all real numbers lambda We will later complexify little g to obtain a complex vector space little g tensor C and get an associated complex linear map R star superscript C from little g to little g l n C.

The key property of R star was the equation R of exp X equals exp of R star X This tells us that R determines R star by differentiation: R star X equals d by d t at t = 0 of R of exp t X Using the formula R of exp X equals exp of R star X, we see that R star determines R of g for all g in exp of little g. Does that mean R star determines R of g for all g in big G?

Lemma:

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

This is because, in this case, big G is generated as a group by the subset exp of little g inside big G. The proof of this lemma is an exercise.

Given R star from little g to little g l n C, does R of exp X equals exp of R star X give a well-defined representation R from big G to big G L n C? Lie's theorem told us that this is true if big G is simply-connected. If big 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\colon SU(2)\to SO(3). Check that R(M_1)=R(M_2) if and only if M_1=\pm M_2. Given a representation S\colon SO(3)\to GL(n,\CC), we get a representation R\circ S\colon SU(2)\to SO(3). Show that a representation T\colon SU(2)\to SO(3) has this form if and only if T(-M)=T(M) for all M\in SU(2).

Exercise:

Suppose we have a Lie algebra little g inside little g l n R consisting of real matrices. Consider the subspace little g tensor C inside little g l n C consisting of matrices of the form M + i N with M and N in little g. Show that:

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

  • if f from little g to little g l m C is a real-linear Lie algebra homomorphism then f superscript C from little g tensor C to little g l m C defined by f superscript C of M + i N equals f of M plus i f of N for M and N in little g is also a Lie algebra homomorphism.