Suppose $R:SU(3)\to GL(U)$ and $S:SU(3)\to GL(V)$ are irreducible representations of $SU(3)$ with highest weight $\lambda $ (with respect to some fixed choice of line of irrational slope). Then $U\cong V$ .

# Optional: Proof of uniqueness of irreps of SU(3)

## Uniqueness

This video is optional.

We have now proved that for any $SU(3)$ representation, if we pick a highest weight then we get a highest weight subrepresentation. We now need to prove that there is only one irrep with a given highest weight.

Consider the representation $R\oplus S$ on $U\oplus V$ . This is not irreducible as both $U$ and $V$ are subrepresentations. If $u\in U$ and $v\in V$ are weight vectors with weight $\lambda $ then $(u,v)\in U\oplus V$ is a weight vector with weight $\lambda $ : $${H}_{\theta}(u,v)=({H}_{\theta}u,{H}_{\theta}v)=\lambda (\theta )(u,v)$$

Take the highest weight irreducible subrepresentation $W\subset U\oplus V$ of $R\oplus S$ generated by $(u,v)$ (coming from the theorem we proved in the previous video).

The projections maps ${\mathrm{pr}}_{U}:W\to U$ and ${\mathrm{pr}}_{V}:W\to V$ are morphisms of representations (exercise!).

There's a nice fact (Schur's lemma) which says that a morphism of irreducible representations like ${\mathrm{pr}}_{U}:W\to U$ (respectively ${\mathrm{pr}}_{V}:W\to V$ ) is either an isomorphism or it's zero. To see this, note that the kernel of a morphism is a subrepresentation of $W$ , and is therefore either $0$ or $W$ (because $W$ is irreducible); the morphism is injective if the kernel is zero, and zero if the kernel is everything. Similarly, the image is a subrepresentation of $U$ (respectively $V$ ), so is either $0$ or $U$ (respectively $0$ or $V$ ). The morphism is surjective if the image is everything, and zero if the image is zero. The only possibilities are for the morphism to be an isomorphism or to be zero.

But ${\mathrm{pr}}_{U}(u,v)=u$ and ${\mathrm{pr}}_{V}(u,v)=v$ , so neither map is zero. Therefore $U\cong W\cong V$ .