Irreducible representation of sl3

Question

Please guys, I would be grateful if someone could please show me the proof of the following question

Question: Prove that $Sym^N(\mathbb{C^3})$ is an irreducible representation of $\mathcal{sl_3}$(Lie algebra) for $N \ge 1$.

Answer 1

To implement Tobias Kildetoft's suggestion, we can argue on dimensional grounds. The dimension of ${\rm Sym}^N \mathbb C^3$ is $$ \frac {3 \times 4 \times \dots \times (N+2)}{N!} = \frac 1 2 (N+1)(N+2). $$

The weight diagram of the fundamental representation (i.e. the $\mathbb C^3$ representation) contains three "dots", arranged in an equilateral triangle in weight space with sides of length $1$. If $v$ is the highest weight vector of $\mathbb C^3$, then $v \otimes \dots \otimes v$ (repeated $n$ times) is the highest weight vector of ${\rm Sym}^N \mathbb C^3$. The weight vector of $v \otimes \dots \otimes v$ lies at the corner of a larger equilateral triangle in weight space, which has sides of length $N$. Thus the irreducible representation with highest weight $v \otimes \dots \otimes v$ has dimension equal to the number of "dots" inside the convex hull of this larger equilateral triangle, which is equal to the $N$th triangular number $ \frac 1 2 (N+1)(N+2)$. This is equal to the dimension of ${\rm Sym}^N \mathbb C^3$, which implies that this irreducible representation is actually the whole of ${\rm Sym}^N \mathbb C^3$.

[I implicitly assumed that each weight in the irreducible representation with highest weight $v \otimes \dots \otimes v$ appears with multiplicity one. This is a standard fact, but if the question does not allow you to make this assumption, it doesn't matter. It suffices to know that the multiplicity of each weight within the convex hull of the equilateral triangle is at least one; this is true because every "slice" through the triangle forms a representation of some $\mathfrak{sl}_2 \mathbb C$ subalgebra of $\mathfrak{sl}_3 \mathbb C$, and you know that the weights in any $\mathfrak{sl}_2 \mathbb C$ representation have no "gaps".]