generalization of highest weight theorem for semisimple lie algebras

Let $\mathfrak g$ be a real semisimple Lie algebra (without compact factors) with Iwasawa decomposition $\mathfrak g=\mathfrak k\oplus \mathfrak a\oplus \mathfrak u$. Let $\mathfrak p$ be a parabolic subalgebra of $\mathfrak g$ containing $\mathfrak a$ and $\mathfrak u$. Let $\mathfrak u_{\mathfrak p}$ be the unipotent radical of $\mathfrak p$. Let $\Phi(\mathfrak u_{\mathfrak p})$ be the relative roots of $\Phi(\mathfrak g, \mathfrak a)$ appeared in the root space decomposition of $\mathfrak u_{\mathfrak p}$. Let $\beta\in \mathfrak a^*$ be an algebraically integral weight dominated by $\Phi(\mathfrak u_{\mathfrak p})$, i.e $\langle \beta, \alpha \rangle\ge 0$ for every $\alpha\in \Phi(\mathfrak u_{\mathfrak p})$ with respect to the usual inner product induced by the Killing form.

Is it true that there is an irreducible real representation $(\rho, V)$ of $\mathfrak g$ such that $\beta$ is a weight of $V$ and there is a nonzero vector $v\in V_{\beta}$ annihilated by $\mathfrak u_{\mathfrak p}$? I.e. $\rho(u)v=0$ for every $u \in \mathfrak u_{\mathfrak p}$.

In the case where $\mathfrak g$ is split and $\mathfrak p$ is the minimal parabolic subalgebra we get an affirmative answer from highest weight theorem. In this sense what I am asking can be considered as a generalization of highest weight theorem.

It would also be interesting to have an answer in the case where $\mathfrak g$ is split. Maybe experts in Lie algebra can answer this question immediately

• I think I should assume $\mathfrak g$ has no noncompact factors. – ronggang Mar 3 '15 at 5:47

You are asking for a "real" representation. That is not possible in general. For example, take $G=SU(2,1)$ as the group preserving the Hermitian form given by
$$\begin{pmatrix} 0& 0 & 1\cr 0 & 1 & 0 \cr 1 & 0 & 0\end{pmatrix}.$$ and take the standard representation on ${\mathbb C}^3$. The highest weight is the weight of $e_1$ (the first basis vector). However, this representation is not defined over ${\mathbb R}$, since that would mean that $SU(2,1)$ can be conjugated into $SL(3,{\mathbb R})$.
• In this case I will consider $\mathbb C^3$ as $\mathbb R^6$. As in my question, I am considering real Lie algebras and relative root systems. So the group $G=SU(2, 1)$ has rank one. – ronggang Mar 3 '15 at 5:53
• @ Venkataramana: In the question irreducible means irreducible over $\mathbb R$. – ronggang Mar 3 '15 at 17:48