Here, is a complex vector space, but is a vector space over the real numbers. When I say that is linear, the only thing that makes sense is for it to be real linear, i.e. We will later complexify to obtain a complex vector space and get an associated complex linear map .
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 , that is an assignment of a matrix to each group element such that and . The image of this representation is a group of matrices which is a quotient of . In this course, we'll focus on smooth representations of matrix groups.
Why are we focusing on representations? There are many fantastic applications:
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There are internal applications to the theory of Lie groups: by studying at the adjoint representation, you can completely classify the Lie algebras of compact Lie groups.
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There are applications to other areas of mathematics, like invariant theory.
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There are applications to particle physics. For example, by comparing tables of hadrons found in particle accelerators with weight diagrams of representations of the Lie group , Gell-Mann (and independently Ne'eman) was able to guess at the underlying internal structure of these particles and so develop the quark model of matter. This led him to predict a particle with strangeness equal to minus 3, which was discovered in 1964. We will discuss this application in detail.
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 we get a Lie algebra representation , that is a linear map such that .
The key property of was the equation This tells us that determines by differentiation: Using the formula , we see that determines for all . Does that mean determines for all ?
If is a path-connected group (i.e. any two matrices in are connected by a smooth path of matrices in ) then is determined by .
This is because, in this case, is generated as a group by . The proof of this lemma is an exercise.
Given , does give a well-defined representation ? Lie's theorem told us that this is true if is simply-connected. If is not simply-connected, we need to think. The first example we'll consider is , 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:
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study the representations of
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study the representations of
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study the representations of
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study the general theory.
Pre-class exercise
In an earlier video, we constructed a map . Check that if and only if . Given a representation , we get a representation . Show that a representation has this form if and only if for all .
Suppose we have a Lie algebra consisting of real matrices. Consider the subspace consisting of matrices of the form with and in . Show that:
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this is a Lie subalgebra, i.e. that it is preserved by Lie bracket
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if is a real-linear Lie algebra homomorphism then defined by for and in is also a Lie algebra homomorphism.