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.

  • $\begingroup$ For the rotation group the infinite dim reps are non-unitary, of course. have not seen Sannikov, S. S. (1968): NEW REPRESENTATIONS OF LIE ALGEBRA OF ROTATION GROUP $R_3$. Sov Jou Nucl Phys, 6 (6), 939. $\endgroup$ – Cosmas Zachos Sep 17 '18 at 16:32

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.

  • $\begingroup$ Is $\mu=\lambda$ here? Why is $V^\mu$ irreducible? Sure, if a submodule contains one of the weight vectors it will contain the rest of them also, but is it immediately obvious that a submodule contains a weight vector? $\endgroup$ – Jyrki Lahtonen Sep 19 '18 at 5:24
  • 2
    $\begingroup$ Answering my own question. Any vector $x$ in $V$ is a linear combination of finitely many weight vectors. Therefore $x$ will be contained in a finite dimensional subspace $W$ of $V$ stable under the action of $h$. In a finite dimensional subspace $h$ necessarily has an eigenvector etc. $\endgroup$ – Jyrki Lahtonen Sep 19 '18 at 5:47

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