I look for an example of an irreducible, but not cyclic representation of a complex semisimple Lie algebra, i.e. the irreducible representation has no highest weight.
I know that the representation has to be infinite-dimensional.
Irreducible Lie algebra representation without highest weight
lie-algebrasrepresentation-theory
Best Answer
It's not so hard to construct a representation of $\mathfrak{sl}_2$ which is not highest weight. Let $V$ be the vector space spanned by the basis $\{v_n \mid n \in 2\mathbb{Z}\}$, with the $\mathfrak{sl}_2$-action $$\begin{aligned} h \cdot v_n &= n v_n \\ e \cdot v_n &= a_n v_{n + 2}\\ f \cdot v_n &= b_n v_{n - 2} \end{aligned}$$ for some complex coefficients $a_n, b_n$. No matter then choice of $a_n, b_n$, the two relations $[h, e] = 2e$ and $[h, f] = -2f$ are already satisfied for each basis element $v_n$. We are left needing to satisfy the relation $[e, f] = h$, which on $v_n$ comes out as $$a_{n - 2} b_n - a_n b_{n + 2} = n$$ and it can be checked that setting $a_n = (\lambda + n) / 2$ and $b_n = (\lambda - n)/2$ for any $\lambda \in \mathbb{C}$ works. We can call the resulting representation $V^\lambda$.
Taking $\lambda \in \mathbb{C} \setminus \mathbb{Z}$ will ensure that the coefficients $a_n, b_n$ are never zero, and hence $V^\lambda$ is irreducible. Clearly it has no highest weight.