# Classifying SU(2) representations

## Review

We have seen that for every complex representation $R\colon SU(2)\to GL(V)$ we get a Lie algebra representation $R_{*}^{\mathbf{C}}\colon\mathfrak{sl}(2,\mathbf{C})\to\mathfrak{gl}(V)$ (by complexifying $R_{*}$ ; we'll usually omit the superscript $\mathbf{C}$ and sometimes even the $R_{*}$ ). We have a basis $H=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix},\ X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\ Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ of $\mathfrak{sl}(2,\mathbf{C})$ which satisfy the commutator relations $[H,X]=2X,\quad[H,Y]=-2Y,\quad[X,Y]=H.$

We've seen that $V=\bigoplus W_{m}$ where $W_{m}=\{v\in V\ :\ Hv=mv\}$ , and that $X\colon W_{m}\to W_{m+2}$ and $Y\colon W_{m}\to W_{m-2}$ . The spaces $W_{m}$ are called weight spaces and the $m$ s are called weights.

We depict this by drawing a weight diagram with a blob at each integer point $m$ for which $W_{m}\neq 0$ (and somehow indicating the dimension of $W_{m}$ ).

## Highest weight vectors

Pick a vector $v\in W_{m}$ where $m$ is the weight that is furthest to the right in the diagram. Such a weight exists because $V$ is finite-dimensional. Act on $v$ using the matrices $R_{*}X$ , $R_{*}Y$ , and $R_{*}H$ (which we're now calling $X$ , $Y$ and $H$ for brevity). Because $V$ is a representation, all vectors we obtain in this way are again in $V$ .

$Xv=0$ because $Xv\in W_{m+2}=0$ , so we may as well start by applying $Y$ a bunch of times to obtain the following elements: $Y^{n}v,\ Y^{n-1}v,\ \ldots,Y^{2}v,\ Yv\ v.$ Here, I've ordered these elements by increasing weight: each time we apply $Y$ we decrease weight by $2$ , so $Y^{k}v\in W_{m-2k}$ . Also, let's define $n$ to be the biggest power of $Y$ we can apply before we get zero, that is: $Y^{n}v\neq 0,\qquad Y^{n+1}v=0.$ (Again this $n$ is guaranteed to exist because $V$ is finite-dimensional).

Lemma:

Take the subspace $U$ spanned by $v,Yv,Y^{2}v,\ldots,Y^{n}v$ . I claim that $U\subset V$ is a subrepresentation.

Remark:

If $V$ is irreducible then this implies $V=U$ . In other words, any irreducible representation of $SU(2)$ has a weight diagram that looks like a string of dots spaced out in twos ($W_{m-2n},\ldots,W_{m-4},W_{m-2},W_{m}$ with each weight space 1-dimensional (spanned by $Y^{k}v$ for some $k$ ).

Proof:

To check that $U$ is a subrepresentation, we need to check that if $u\in U$ and we apply $X$ , $Y$ or $H$ to $u$ then the result is still in $U$ . We will check this for $u$ running over the basis $v,Yv,Y^{2}v,\ldots,Y^{n}v$ .

If we apply $Y$ to $Y^{k}v$ then we get $Y^{k+1}v$ , and all these vectors are still in $U$ .

If we apply $H$ to $Y^{k}v$ we get $(m-2k)Y^{k}v$ because $Y^{k}v\in W_{m-2k}$ . Again, this stays in the subrepresentation $U$ .

If we apply $X$ , there is a formula which we will prove separately: $XY^{k}v=(m+1-k)kY^{k-1}v.$ This implies the lemma because it shows $XY^{k}v\in U$ .

## The formula

Lemma:

$XY^{k}v=(m+1-k)kY^{k-1}v.$

### Consequences of the formula

We'll prove the formula in a moment, but first here are some remarks.

Remark:

The formula makes sense because $X$ increases weight by $2$ and $Y^{k}v\in W_{m-2k}$ while $Y^{k-1}v\in W_{m-2k+2}$ .

Remark:

The formula tells us that the irreducible representation with highest weight $m$ is completely determined up to isomorphism by $m$ (i.e. there's a unique irrep with highest weight $m$ up to isomorphism). This is because we know how all the matrices $R_{*}X$ , $R_{*}Y$ , $R_{*}H$ act on the basis $v,Yv,Y^{2}v,\ldots,Y^{n}v$ :

• $Y(Y^{k}v)=Y^{k+1}v$

• $H(Y^{k}v)=(m-2k)Y^{k}v$

• $X(Y^{k}v)=(m+1-k)kY^{k-1}v$ .

The only choice we have is picking the vector $v$ , but different $v$ s differ only by a scale factor as $W_{m}$ is 1-dimensional, and this just amounts to changing our whole basis $Y^{k}v$ by a scale factor.

Remark:

We can now deduce that $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ is isomorphic to the representation $R(g)M_{v}=gM_{v}g^{-1}$ : we've seen both of these have the same weight diagram.

Lemma:

The formula also tells us that $n$ (the biggest power of $Y$ for which $Y^{n}v\neq 0$ but $Y^{n+1}v=0$ ) is equal to $m$ . Therefore the weights in our representations go from $m$ down to $m-2n=-m$ .

Proof:

To see why $n=m$ , note that $XY^{n+1}v=0=(m+1-(n+1))(n+1)Y^{n}v$ but $Y^{n}v\neq 0$ and $n+1\neq 0$ , so $m+1-(n+1)=0$ and hence $n=m$ .

### Proof of formula

We now prove the formula $XY^{k}v=(m+1-k)kY^{k-1}v.$ We prove it by induction on $k$ .

For $k=0$ , $Xv=0$ and $(m+1-k)kY^{k-1}v=0$ , so both sides agree.

Suppose it's true for all numbers strictly less than $k$ . We know that $XY-YX=[X,Y]=H$ . Since $R_{*}^{\mathbf{C}}$ is a representation of Lie algebras this formula is still true if we stick $R_{*}^{\mathbf{C}}$ in front of everything (which is good, since we're omitting mention of $R_{*}^{\mathbf{C}}$ from the formulae). Let's apply $XY-YX=H$ to $Y^{k-1}v$ : $XY^{k}v-YXY^{k-1}v=HY^{k-1}v.$

Therefore $XY^{k}v=YXY^{k-1}v+HY^{k-1}v$ . Using our formula to evaluate $XY^{k-1}v$ and the fact that $Y^{k-1}v\in W_{m-2(k-1)}$ , we get $XY^{k}v=(m+1-k+1)(k-1)Y^{k-1}v+(m-2k+2)Y^{k-1}v,$ which simplifies to give our formula (exercise!).

Remark:

Once again, the bracket relations in $\mathfrak{sl}(2,\mathbf{C})$ have played a crucial role in the proof.

## Summary

With this formula in hand, we have now established these nice facts:

• the weight diagrams of $SU(2)$ representations are symmetric about the origin

• there's a unique irrep with highest weight $m$ , whose weight diagram has 1-dimensional weight spaces with weights $-m,-m+2,-m+4,\ldots,m$ (spaced out by 2, starting at $-m$ and ending at $m$ ).

## Pre-class exercise

Exercise:

Check that $XY^{k}v=(m+1-k+1)(k-1)Y^{k-1}v+(m-2k+2)Y^{k-1}v,$ simplifies to give $XY^{k}v=(m+1-k)kY^{k-1}v.$