# X and Y

## From su(2) to sl(2,C)

Last time we saw that, given a complex representation R from S U 2 to G L V there is a weight space decomposition V equals direct sum from j = 1 to n of V_j such that with respect to this splitting, R of e to the i theta, 0, 0, e to the minus i theta is the diagonal matrix with diagonal entries e to the i m_1 theta, down to e to the i m_n theta where the integers m_j are the weights of the representation.

At the level of Lie algebras, we have R of exp of theta times i, 0, 0, minus i equals exp of theta R star i, 0, 0, minus i so R star of i, 0, 0, minus i equals the diagonal matrix with diagonal entries i m_1, down to i m_n

In other words, we know where one of the matrices in little s u 2 goes under R star. But there are other matrices in little s u 2: the general element is the matrix i x, y plus i z, minus y plus i z, minus i x; where do these go?

Pick the basis sigma_1 equals the matrix i, 0, 0, minus i; sigma_2 equals the matrix 0, 1, minus 1, 0; and sigma_3 equals the matrix 0, i, i, 0 for little s u 2 (corresponding to the x, y and z directions). These (or closely related versions of them) are often called the Pauli matrices.

I've told you what R star of sigma_1 is, and since R star is linear (determined by its values on a basis) it's enough to say what R star of sigma_2 and R star of sigma_3 are to determine R star completely. It turns out to be much easier to use: X equals sigma_2 minus i sigma_3, all over 2, which equals the matrix 0, 1, 0, 0; and Y equals minus sigma_2 minus i sigma_3, all over 2, which equals the matrix 0, 0, 1, 0

The problem is that these correspond to complex values of y and z, so they don't live in little s u 2. Instead, they live in little s u 2 tensor C. In fact, X and Y live in little s l 2 C and sigma_1 is equal to i H where H equals the matrix 1, 0, 0, minus 1. So it turns out that little s u 2 tensor C equals little s l 2 C: anything inside little s l 2 C can be written as a complex linear combination of sigma_1, sigma_2, sigma_3 and anything inside little s u 2 tensor C can be written as a complex linear combination of X, Y, and H.

If R star from little s u 2 to little g l V is a representation of Lie algebras where V is a complex vector space, then we get for free a representation R star superscript C from little s u 2 tensor C to little g l C, just by setting R star superscript C of M_1 plus i M_2 equals R star M_1 plus i R star M_2.

Remark:

The point is that we don't know what i M means if M is in little s u 2 because little s u 2 is a real Lie algebra so we need to complexify it, but we do know what i R star M in little g l V means because little g l V is a complex Lie algebra and we don't to complexify it.

## How do X and Y act?

Definition:

Let R from S U 2 to G L V be a complex representation and write W_m equals direct sum of V_j where the sum runs over those j for which m_j equals m for the sum of all subspaces with weight m. In other words, we're grouping the irreducible subrepresentations for the diagonal U(1) into subspaces with a common weight. We'll call W_m the weight space with weight m.

Example:

For Sym 2 C 2, W minus 2, W_0 and W_2 are each 1-dimensional (spanned by e_2 squared, e_1 e_2 and e_1 squared respectively). We draw this as a weight diagram:

Lemma:

The following lemma tells us how H, X and Y act in the representation:

1. R star superscript C of H acts on W_m as m times the identity, i.e. if v is in W_m then R star superscript C of H applied to v equals m v.

2. R star superscript C of C sends W_m to W_{m+2}, i.e. if v is in W_m then R star superscript C of X applied to v is in W_(m+2).

3. R star superscript C of Y sends W_m to W m minus 2.

Remark:

In terms of the weight diagram earlier, we can draw arrows which show how H, X and Y act: H sends the weight spaces back to themselves, X shifts to the right, and Y shifts to the left. Note that in the example of Sym 2 C 2, X sends W_2 to W_4, but W_4=0, so X annihilates elements of W_2.

Remark:

This is what I meant when I said that the off-diagonal elements of little s u 2 mix up the weight spaces: they do so in a very precise way. Also, you can see why X and Y are more convenient than sigma_2 and sigma_3: these would mix up weight spaces in a messy way that combines the X and Y arrows.

Proof:
1. We have already proved part (1) of the lemma: we know that v in W_m implies R star sigma_1 of v equals i m v, and sigma_1 equals i H, so R star superscript C of H applied to v equals mv.

2. We'll do (2) and leave the very-similar (3) as an exercise. I'll drop the superscript C from now on. Suppose that v is in W_m. We want to show that R star X v is in W_(m plus 2). Note that v is in W_m if and only if R star H v equals m v and R star X v is in W_(m plus 2) if and only if R star H R star X v equals (m plus 2) times R star X v.

We know that R star X R star H v equals R star X m v, equals m R star X v, but we can't just commute R star H past R star X. However, we know that H bracket X equals 2 X, and since R star is a Lie algebra homomorphism we have R star H bracket R star X equals 2 R star X.

Therefore R star H R star X v equals R star X R star H v plus 2 R star X v, which equals (m plus 2) times R star X v, as required.

## Pre-class exercise

Exercise:

Suppose that R from SU(2) to G L V is a representation with weight decomposition V equals the direct sum of subspaces W_m. Show that R star superscript C of Y sends W_m to W m minus 2.