# SU(3) overview

## SU(3)

We now turn to the group $SU(3)$ , the group of 3-by-3 special unitary matrices. This is an instructive example: its representation theory shares many similarities to the representation theory of other semisimple groups.

When we studied $SU(2)$ , the key tool we introduced was the weight decomposition. In other words, given a representation of $SU(2)$ on a complex vector space $V$ , we decomposed $V$ as a direct sum of subspaces $W_{k}$ where $W_{k}$ was a simultaneous eigenspace (with eigenvalue $e^{ik\theta}$ ) for the action of the diagonal subgroup $T=\left\{\begin{pmatrix}e^{i\theta}&0\\ 0&e^{-i\theta}\end{pmatrix}\ :\ e^{i\theta}\in U(1)\right\}$ . This was possible because $T\cong U(1)$ .

For $SU(3)$ , the subgroup of diagonal matrices is isomorphic to $U(1)\times U(1)$ : $T=\left\{\begin{pmatrix}e^{i\theta_{1}}&0&0\\ 0&e^{i\theta_{2}}&0\\ 0&0&e^{-i(\theta_{1}+\theta_{2})}\end{pmatrix}\ :\ (e^{i\theta_{1}},e^{i\theta% _{2}})\in U(1)\times U(1)\right\}.$ $T$ here stands for torus: a 2-dimensional torus has two circular coordinates (latitude and longitude) just like $U(1)\times U(1)$ . For compact matrix groups more generally, the analogue of $T$ will be a (possibly higher-dimensional) maximal torus (more on that later).

Lemma:

If $R\colon SU(3)\to GL(V)$ is a complex representation then $V=\bigoplus W_{k,\ell}$ where $W_{k,\ell}=\left\{v\in V\ :\ R(D(\theta_{1},\theta_{2}))v=e^{i(k\theta_{1}+% \ell\theta_{2})}v\ \forall\theta_{1},\theta_{2}\right\},$ where $D(\theta_{1},\theta_{2})=\begin{pmatrix}e^{i\theta_{1}}&0&0\\ 0&e^{i\theta_{2}}&0\\ 0&0&e^{-i(\theta_{1}+\theta_{2})}\end{pmatrix}$ .

Remark:

For $SU(2)$ , our torus was just a copy of $U(1)$ , so we only needed one integer $k$ to index the weight spaces. Now we need two: one for each $U(1)$ -factor in the torus.

We will postpone the proof until we have seen some examples of weight diagrams. The weight diagrams will now be patterns of dots in the plane.

## Examples

### The standard representation

Example:

Take $V=\mathbf{C}^{3}$ , the standard representation of $SU(3)$ (i.e. considering a matrix in $SU(3)$ tautologically as an element of $GL(3,\mathbf{C})$ ). We pick the standard basis $e_{1},e_{2},e_{3}\in V$ . We have $R(D(\theta_{1},\theta_{1}))=\begin{pmatrix}e^{i\theta_{1}}&0&0\\ 0&e^{i\theta_{2}}&0\\ 0&0&e^{-i(\theta_{1}+\theta_{2})}\end{pmatrix},$

so $R(D(\theta_{1},\theta_{2}))e_{1}=e^{i\theta_{1}}e_{1},\quad R(D(\theta_{1},% \theta_{2}))e_{2}=e^{i\theta_{2}}e_{2},R(D(\theta_{1},\theta_{2}))e_{3}=e^{-i(% \theta_{1}+\theta_{2})}e_{3}$ so

• $e_{1}$ spans the weight space $W_{1,0}$ ($k=1$ , $\ell=0$ ).

• $e_{2}$ spans the weight space $W_{0,1}$ ($k=0$ , $\ell=1$ ).

• $e_{3}$ spans the weight space $W_{-1,-1}$ ($k=-1$ , $\ell=-1$ ).

The obvious way to represent these would be to simply draw the points $(k,\ell)=(1,0),\ (0,1)\ (-1,-1)$ in the plane. For reasons that will become clear when we discuss the Killing form, we will draw the $\ell$ -axis at 120 degrees to the $k$ -axis. For now, the justification is just that this restores the symmetry we broke when we chose to express the bottom-left entry of $D(\theta_{1},\theta_{2})$ in terms of the other two diagonal entries (instead of writing something equally valid like $\begin{pmatrix}e^{i\theta_{1}}&0&0\\ 0&e^{-i(\theta_{1}+\theta_{3})}&0\\ 0&0&e^{i\theta_{3}}\end{pmatrix}$ ).

We have also drawn the origin in grey in this picture: it is not a weight of the standard representation, the weights are red.

### Sym2(C3)

Example:

Consider $\mathrm{Sym}^{2}(\mathbf{C}^{3})$ . A basis for this is given by $e_{1}^{2},\ e_{1}e_{2},\ e_{1}e_{3},\ e_{2}^{2},\ e_{2}e_{3},\ e_{3}^{2}.$ How do these transform under $\mathrm{Sym}^{2}(D(\theta_{1},\theta_{2}))$ ?

Since $e_{1}\mapsto e^{i\theta_{1}}e_{1}$ , we have $e_{1}^{2}\mapsto e^{i2\theta_{1}}e_{1}^{2},$ so lives in the weight space $W_{2,0}$ . Similarly, we get $e_{1}^{2}\in W_{1,0},\ e_{1}e_{2}\in W_{1,1},\ e_{1}e_{3}\in W_{0,-1},\ e_{2}^% {2}\in W_{0,2},\ e_{2}e_{3}\in W_{-1,0},\ e_{3}^{2}\in W_{-2,-2}.$ For example, $e_{2}e_{3}\mapsto e^{i\theta_{2}}e^{-i(\theta_{1}+\theta_{2})}e_{2}e_{3}=e^{-i% \theta_{1}}e_{2}e_{3}$ .

Plotting these weights in the plane (with respect to our tilted axes) we get:

We have also added:

• the line $k=\ell$ , to complete the symmetry of the diagram;

• the points $(0,0)$ , $(1,0)$ , $(0,1)$ , $(-1,-1)$ in grey just to give a sense of the size of the diagram. Only the red dots are weights of the representation.

The kinds of diagrams we will get for weight diagrams of $SU(3)$ representations will be triangles and hexagons.

## Proof of lemma

Consider the subgroup $T_{1}=\left\{D(\theta_{1},0)=\begin{pmatrix}e^{i\theta_{1}}&0&0\\ 0&1&0\\ 0&0&e^{-\theta_{1}}\end{pmatrix}\ :\ e^{i\theta_{1}}\in U(1)\right\}\subset T$ This is isomorphic to $U(1)$ and $V$ decomposes into weight spaces for the action of $T_{1}$ : $V=\bigoplus U_{k},\quad U_{k}=\{v\in V\ :\ R(D(\theta_{1},0))v=e^{ik\theta_{1}% }v\}.$

Define $T_{2}=\{D(0,\theta_{2})\ :\ e^{i\theta_{2}}\in U(1)\}$ .

Lemma:

The action of $T_{2}$ preserves each $U_{k}$ , i.e. $u\in U_{k}$ implies $R(D(0,\theta_{2}))u\in U_{k}$ .

This lemma will imply the result because now each $U_{k}$ decomposes under the action of $T_{2}$ : $U_{k}=\bigoplus W_{k,\ell},\quad W_{k,\ell}=\{v\in U_{k}\ :\ R(D(0,\theta_{2})% )v=e^{i\ell\theta_{2}}v\}.$ In other words, elements of $W_{k,\ell}$ transform as desired under both $T_{1}$ and $T_{2}$ and hence under $T$ : $R(D(\theta_{1},\theta_{2}))v=R(D(\theta_{1},0))R(D(0,\theta_{2}))v=e^{i(k% \theta_{1}+\ell\theta_{2})}v.$

Proof:

If $v\in U_{k}$ then $R(D(\theta_{1},0))v=e^{ik\theta_{1}}v$ . We want to show that $R(D(0,\theta_{2}))v\in U_{k}$ , i.e. that $R(D(\theta_{1},0))R(D(0,\theta_{2}))v=e^{ik\theta_{1}}R(D(0,\theta_{2}))v.$

Let's compute. Since $D(\theta_{1},0)$ and $D(0,\theta_{2})$ commute with one another, the same is true of $R(D(\theta_{1},0))$ and $R(D(0,\theta_{2}))$ . Therefore $R(D(\theta_{1},0))R(D(0,\theta_{2}))v=R(D(0,\theta_{2}))R(D(\theta_{1},0))v.$ Since $v\in U_{k}$ , we get $R(D(\theta_{1},0))R(D(0,\theta_{2}))v=R(D(0,\theta_{2}))e^{ik\theta_{1}}v=e^{% ik\theta_{1}}R(D(0,\theta_{2}))v,$ as required.

This argument works more generally for finding weight decompositions of representations of $U(1)^{n}$ : you inductively find the weight space decomposition with respect to $U(1)^{n-1}$ and show that the final factor preserves the weight spaces (because the group is abelian), then decompose these weight spaces further with respect to the action of the final $U(1)$ factor. This would give us $n$ integer weights $k_{1},\ldots,k_{n}$ and our weight diagrams would be collections of dots in $n$ -dimensional space. For $SU(4)$ we get weight diagrams in $\mathbf{R}^{3}$ , for $SU(5)$ we get weight diagrams in $\mathbf{R}^{4}$ , etc.

## Pre-class exercises

Exercise:

If $\mathbf{C}^{3}$ denotes the standard representation of $SU(3)$ , what is the weight diagram of $\mathrm{Sym}^{3}(\mathbf{C}^{3})$ ? Can you guess the weight diagram of $\mathrm{Sym}^{n}(\mathbf{C}^{3})$ ?

Exercise:

What do you think the weight diagram of the standard 4-dimensional representation of $SU(4)$ would look like?