First read about Lie algebras. Roughly a representation of a Lie algebra is a way to represent that Lie algebra by linear transformations. There is a deep and beautiful theory of such representations. I'll just scratch the surface here by giving a precise definition and then showing a few examples.

Let k be a field (like the real numbers or complex numbers). If V is a k-vector space then denote by gl(V) the collection of all linear transformations from V to itself. This is a Lie algebra if we define [f,g]=fg-gf, for two linear maps f,g:V-->V. If V is finite-dimensional then gl(V) is isomorphic as a Lie algebra to gl(n,k), which is just nxn matrices over k.

Definition A representation of a Lie algebra g is a Lie algebra homomorphism H:g-->gl(V). The dimension of the representation is the dimension of V as a vector space. A subrepresentation is given by a subspace of V that is stable under transformations from the image of H.

We are going to investigate this concept for everyone's favourite Lie algebra sl(2,k) which consists of all 2x2 matrices

 -   -
| a b |     such that a+d=0
| c d |
 -   -
Recall that the bracket for this Lie algebra is defined by [A,B]=AB-BA. (The discussion doesn't depend on the field except we do have to suppose that it has characteristic zero. Note that this is true for C.)

As a vector space sl(2,k) is three-dimensional with a basis given by

   -   -      -    -      -   -
e=| 0 1 |  h=| 1  0 |  f=| 0 0 |
  | 0 0 |    | 0 -1 |    | 1 0 |
   -   -      -    -      -   -
To understand this Lie algebra we need to know what the brackets of these basis elements are.
[e,f] = h, [h,e] = 2e, [h,f] = -2f.
So far this is the only representation of sl(2,k) that we know, a two-dimensional representation.

Here's another one the adjoint representation. For any Lie algebra g we get a representation by taking V=g and defining the representing Lie algebra homomorphism by sending x in g to the linear transformation given by y |--> [x,y]. Again one checks that this gives a Lie algebra map and so this is a representation. For sl(2,k) it is three-dimensional.

I'm am going to construct an n+1-dimensional representation of sl(2,k), for each integer n>=0. Let V be an n+1-dimensional vector space with basis e0,...,en. Then define a map H: sl(2,k)-->gl(V) as follows.

  • H(h)(ei)=(n-2i)ei
  • H(e)(ei)=(n-i+1)ei-1
  • H(f)(ei)=(i+1)ei+1
for 0<=i<=n. (We define e-1=en+1=0.) Checking that H is a Lie algebra homomorphism is easy. For example, [H(e),H(f)](ei)=H(e)H(f)(ei)-H(f)H(e)(ei) by definition. Applying the formulae above we get that the right hand side is H(e)((i+1)ei+1)-H(f)((n-i+1)ei-1). Apply the formulae once more and this is (n-i)(i+1)ei-(n-i+1)iei=(n-2i)ei. This is H([e,f]), as required. Checking the other brackets is similar.

It turns out that these representations are irreducible (they have no subrepresentations except the zero space and the whole representation) and that every representation of sl(2,k) may be contructed by direct sums of these representations.

This kind of thing is decpetively useful for dealing with angular momentum in Quantum Mechanics. If you look at the main write up there, the algebra works out exactly the same if you let:
  • h = 2J3/(h-bar)
  • e = J+/(h-bar)
  • f = J-/(h-bar)
...and then all the commutation relations come out analagously. In fact, there is another operator useful in representing sl(2,C), called the Cazimir operator; Ϊ = ef + fe + (1/2)h2. Then given this, it's not hard to show that Ϊ = J2/(h-bar)2.

Of course, all of this is no coincidence: the angular momentum theory is derived from the Lie algebra of the rotation group of 3-d space, which is sl(2,C). It's still quite cool, though.

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