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

# Decomposing into irreducibles

## Review

We have now established the following theorem.

There is also a complete reducibility 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: **C**^{2} tensor **C**^{2}

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 $$H{e}_{1}={e}_{1},H{e}_{2}=-{e}_{2}.$$ Because $H$ lives in the Lie algebra, we evaluate $H(x\otimes y)$ using the product rule: $$H({e}_{1}\otimes {e}_{1})=(H{e}_{1})\otimes {e}_{1}+{e}_{1}\otimes (H{e}_{1})=2{e}_{1}\otimes {e}_{1}.$$

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 $$H({e}_{2}\otimes {e}_{2})=-2{e}_{2}\otimes {e}_{2}$$ which gives the weight space ${W}_{-2}$ . So our weight spaces are: $${W}_{-2}=\mathbf{C}\cdot ({e}_{2}\otimes {e}_{2}),{W}_{0}=\mathbf{C}\cdot ({e}_{1}\otimes {e}_{2},{e}_{2}\otimes {e}_{1}),{W}_{2}=\mathbf{C}\cdot ({e}_{1}\otimes {e}_{1}).$$

Following the procedure we used in proving the classification theorem, we pick a highest weight vector $v={e}_{1}\otimes {e}_{1}$ and use $Y=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$ 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 ${U}^{\u27c2}$ 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 $Y=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$ we have $Y{e}_{1}={e}_{2}$ and $Y{e}_{2}=0$ . If $v={e}_{1}\otimes {e}_{1}$ then $$Yv=Y{e}_{1}\otimes {e}_{1}+{e}_{1}\otimes Y{e}_{1}={e}_{2}\otimes {e}_{1}+{e}_{1}\otimes {e}_{2},$$ and $${Y}^{2}v=Y{e}_{2}\otimes {e}_{1}+{e}_{2}\otimes Y{e}_{1}+Y{e}_{1}\otimes {e}_{2}+{e}_{2}\otimes Y{e}_{1}=2{e}_{2}\otimes {e}_{2}$$ 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* ${\mathrm{\Lambda}}^{2}{\mathbf{C}}^{2}$
. 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: Sym^{2}Sym^{2}(**C**^{2})

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 $$H({\alpha}^{2})=(H\alpha )\alpha +\alpha (H\alpha )=2{\alpha}^{2}+2{\alpha}^{2}=4{\alpha}^{2},$$ $$H(\alpha \beta )=(H\alpha )\beta +\alpha (H\beta )=2\alpha \beta $$ $$H(\alpha \gamma )=0,H({\beta}^{2})=0,H(\beta \gamma )=-2\beta \gamma ,H({\gamma}^{2})=-4{\gamma}^{2}.$$ 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

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