1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.