# Decomposing into irreducibles

## Review

We have now established the following theorem.

Theorem:

Any irrep of $SU(2)$ is isomorphic to $\mathrm{Sym}^{n}(\mathbf{C}^{2})$ for some $n\in\{0,1,2,\ldots\}$ and this has weight diagram:

There is also a complete reducibility theorem:

Theorem:

Any finite-dimensional $SU(2)$ representation splits as a direct sum of irreps.

This is proved in exactly the same way as for representations of $U(1)$ once you understand how to integrate (average) over the group $SU(2)$ . Instead of talking about that, I will work through some examples, decomposing representations into irreducibles.

## Examples

### Example: C2 tensor C2

Example:

Consider the standard representation $\mathbf{C}^{2}$ tensored with itself: $\mathbf{C}^{2}\otimes\mathbf{C}^{2}$ . This will turn out to have the following weight diagram:

In other words, there are 1-dimensional weight spaces with weights $2$ and $-2$ and a 2-dimensional weight space in weight 0.

Inside this diagram, you can see the weight diagram of $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ as a subdiagram. When you remove this subdiagram what's left is the weight diagram of the trivial 1-dimensional representation.

This will allow us to deduce that $\mathbf{C}^{2}\otimes\mathbf{C}^{2}\cong\mathrm{Sym}^{2}(\mathbf{C}^{2})\oplus% \mathbf{C}$ .

A basis for $\mathbf{C}^{2}\otimes\mathbf{C}^{2}$ is given by $e_{1}\otimes e_{1},\ e_{1}\otimes e_{2},\ e_{2}\otimes e_{1},\ e_{2}\otimes e_% {2}$ (where $e_{1},e_{2}$ is the standard basis for $\mathbf{C}^{2}$ . We have $He_{1}=e_{1},\qquad He_{2}=-e_{2}.$ Because $H$ lives in the Lie algebra, we evaluate $H(x\otimes y)$ using the product rule: $H(e_{1}\otimes e_{1})=(He_{1})\otimes e_{1}+e_{1}\otimes(He_{1})=2e_{1}\otimes e% _{1}.$

Therefore $e_{1}\otimes e_{1}$ is an eigenvector for $H$ with eigenvalue $2$ , so it lives in the weight space $W_{2}$ . Similarly we get $H(e_{1}\otimes e_{2})=e_{1}\otimes e_{2}-e_{1}\otimes e_{2}=0=H(e_{2}\otimes e% _{1})$ which give us the 2-dimensional weight space $W_{0}$ and $H(e_{2}\otimes e_{2})=-2e_{2}\otimes e_{2}$ which gives the weight space $W_{-2}$ . So our weight spaces are: $W_{-2}=\mathbf{C}\cdot(e_{2}\otimes e_{2}),\quad W_{0}=\mathbf{C}\cdot(e_{1}% \otimes e_{2},e_{2}\otimes e_{1}),\quad W_{2}=\mathbf{C}\cdot(e_{1}\otimes e_{% 1}).$

Following the procedure we used in proving the classification theorem, we pick a highest weight vector $v=e_{1}\otimes e_{1}$ and use $Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ to generate a subrepresentation spanned by: $v,\ Yv,\ Y^{2}v.$ These three vectors live in the weight spaces $W_{2}$ , $W_{0}$ and $W_{-2}$ respectively, and give a subrepresentation $U$ isomorphic to $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ (because it has the right weight diagram).

There is an invariant Hermitian inner product on the representation, and we take the orthogonal complement $U^{\perp}$ to get a complementary subrepresentation. Since this complement is 1-dimensional, it must be isomorphic to the trivial 1-dimensional representation (the only irrep of this dimension).

We deduce that $\mathbf{C}^{2}\otimes\mathbf{C}^{2}\cong\mathrm{Sym}^{2}(\mathbf{C}^{2})\oplus% \mathbf{C}$ . This should not be a surprise, because $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ was defined to be a subrepresentation of $\mathbf{C}^{2}\otimes\mathbf{C}^{2}$ , in fact it was defined to be the subrepresentation spanned by the symmetric tensors $e_{1}\otimes e_{1}$ , $e_{1}\otimes e_{2}+e_{2}\otimes e_{1}$ and $e_{2}\otimes e_{2}$ .

Let's check that this is what we are getting. Since $Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$ we have $Ye_{1}=e_{2}$ and $Ye_{2}=0$ . If $v=e_{1}\otimes e_{1}$ then $Yv=Ye_{1}\otimes e_{1}+e_{1}\otimes Ye_{1}=e_{2}\otimes e_{1}+e_{1}\otimes e_{% 2},$ and $Y^{2}v=Ye_{2}\otimes e_{1}+e_{2}\otimes Ye_{1}+Ye_{1}\otimes e_{2}+e_{2}% \otimes Ye_{1}=2e_{2}\otimes e_{2}$ so we get precisely that $U=\mathrm{Sym}^{2}(\mathbf{C}^{2})$ on the nose.

The complementary subrepresentation turns out to be spanned by the antisymmetric tensor $e_{1}\otimes e_{2}-e_{2}\otimes e_{1}$ , which spans the exterior square $\Lambda^{2}\mathbf{C}^{2}$ . It's an exercise to check that this antisymmetric tensor does span a subrepresentation (i.e. it is fixed by the action of $SU(2)$ ).

### Example: Sym2Sym2(C2)

Example:

Let's figure out the decomposition of $\mathrm{Sym}^{2}\mathrm{Sym}^{2}(\mathbf{C}^{2})$ . We know that $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ is spanned by $\alpha=e_{1}^{2}$ , $\beta=e_{1}e_{2}$ and $\gamma=e_{2}^{2}$ . In the same way we can think of $\mathrm{Sym}^{2}(\mathbf{C}^{2})$ as quadratic polynomials in $e_{1},e_{2}$ , we can think of $\mathrm{Sym}^{2}\mathrm{Sym}^{2}(\mathbf{C}^{2})$ as quadratic polynomials in $\alpha$ , $\beta$ , and $\gamma$ . It therefore has a basis $\alpha^{2},\ \alpha\beta,\ \alpha\gamma,\ \beta^{2},\ \beta\gamma,\ \gamma^{2}.$

Let's calculate the weight space decomposition. We know that $H\alpha=2\alpha$ , $H\beta=0$ and $H\gamma=-2\gamma$ , so (using the product rule) we have $H(\alpha^{2})=(H\alpha)\alpha+\alpha(H\alpha)=2\alpha^{2}+2\alpha^{2}=4\alpha^% {2},$ $H(\alpha\beta)=(H\alpha)\beta+\alpha(H\beta)=2\alpha\beta$ $H(\alpha\gamma)=0,\quad H(\beta^{2})=0,\quad H(\beta\gamma)=-2\beta\gamma,% \quad H(\gamma^{2})=-4\gamma^{2}.$ The weight space decomposition is therefore: $W_{-4}=\mathbf{C}\cdot\gamma^{2},\ W_{-2}=\mathbf{C}\cdot\beta\gamma,\ W_{0}=% \mathbf{C}\cdot(\beta^{2},\alpha\gamma),\ W_{2}=\mathbf{C}\cdot\alpha\beta,\ W% _{4}=\mathbf{C}\cdot\alpha^{2}$ with weight diagram

From this you can see that $\mathrm{Sym}^{2}\mathrm{Sym}^{2}(\mathbf{C}^{2})=\mathrm{Sym}^{4}(\mathbf{C}^{% 2})\oplus\mathbf{C}.$ The argument is exactly as before: you pick a highest weight vector like $v=\alpha^{2}$ and you generate a highest weight subrepresentation $U$ spanned by $v,Yv,Y^{2}v,Y^{3}v,Y^{4}v$ , which is isomorphic to $\mathrm{Sym}^{4}(\mathbf{C}^{2})$ by the classification theorem. Then you take the orthogonal complement of $U$ with respect to an invariant Hermitian inner product and you get a complementary subrepresentation whose weight diagram is obtained from the weight diagram of $\mathrm{Sym}^{2}\mathrm{Sym}^{2}(\mathbf{C}^{2})$ by stripping off the weight diagram of $U$ .

The trivial complementary subrepresentation has to be spanned by some combination of $\beta^{2}$ and $\alpha\gamma$ as these span the 0-weight space. In this case, I claim that the trivial complementary subrepresentation is spanned by $\beta^{2}-\alpha\gamma$ (it will be an exercise to check this).

## Pre-class exercise

Exercise:

Compute the weight diagrams and decomposition of $\mathrm{Sym}^{k}(\mathbf{C}^{2})\otimes\mathrm{Sym}^{\ell}(\mathbf{C}^{2})$ for a few different values of $k$ and $\ell$ and see if you can formulate a conjecture as to what the answer should be in general.