# About certain elements in the zero weight space of an irreducible representation of the complex simple Lie algebra of type G$_2$

$$\newcommand{\fg}{\mathfrak g}\newcommand{\ee}{\varepsilon}$$Let $$\fg$$ be the complex simple Lie algebra of type G$$_2$$. We consider its root system as follows (though it is probably not necessary to state the question): $$\begin{equation*} \Delta^+(\mathfrak g,\mathfrak h) = \left\{ \begin{array}{ll} \ee_2-\ee_3,&\ee_1-2\ee_2+\ee_3,\\ \ee_1-\ee_2,&\ee_1+\ee_2-2\ee_3,\\ \ee_1-\ee_3,&2\ee_1-\ee_2-\ee_3 \end{array} \right\}, \end{equation*}$$ where $$\mathfrak h$$ is a Cartan subalgebra of $$\mathfrak g$$ with $$\mathfrak h^*=\{\sum_{i=1}^3 a_i\ee_i: a_i\in\mathbb C,\; \sum_{i=1}^3a_i=0\}$$. We will use its root decomposition $$\begin{equation*} \mathfrak g=\mathfrak h\oplus \bigoplus_{\alpha\in\Delta(\mathfrak g,\mathfrak h)} \mathfrak g_\alpha. \end{equation*}$$

Let $$(\pi,V_\pi)$$ be a non-trivial irreducible representation of $$\mathfrak g$$. Suppose $$v\in V_\pi$$ is in the weight space of weight zero (i.e. $$v\in V_\pi(0)$$) and satisfies $$\begin{equation*} \pi(X_{\ee_1-\ee_2})(v)=0 \quad\text{ and }\quad \pi(X_{\ee_1+\ee_2-2\ee_3})(v)=0, \end{equation*}$$ where $$X_{\ee_1-\ee_2}\in\mathfrak g_{\ee_1-\ee_2}$$ and $$X_{\ee_1+\ee_2-2\ee_3}\in \mathfrak g_{\ee_1+\ee_2-2\ee_3}$$ are non-trivial.

Can we ensure that $$v=0$$?

Note that the story would be very different if we replace the roots $$\ee_1-\ee_2$$ and $$\ee_1+\ee_2-2\ee_3$$ by any pair $$\alpha,\beta\in\Delta(\mathfrak g,\mathfrak h)$$ with $$\alpha$$ short, $$\beta$$ long and non-orthogonal. In that case, one easily obtains that $$\pi(X_{a\alpha+b\beta})(v)=0$$ for all $$a,b\in\mathbb Z$$, which implies that $$\mathbb C v$$ is $$\mathfrak g$$-invariant since $$\mathbb Z\alpha+\mathbb Z\beta=\operatorname{Span}_\mathbb Z\Delta(\mathfrak g,\mathfrak h)$$, contradicting the assumptions on $$\pi$$.

• $\def\e#1{\varepsilon_#1}\def\a{\alpha}\def\b{\beta}$For people, like me, more used to thinking in abstract terms: in your set-up, $\a=\e2-\e3$ is the short simple root and $\b=\e1-2\e2+\e3$ is the long simple root, so the short roots are $\a,\a+\b=\e1-\e2,2\a+\b=\e1-\e3$, and the long roots are $\b,3\a+\b=\e1+\e2-2\e3,3\a+2\b=\e1+\e2-2\e3$. As you implicitly note, what's important is that $\e1-\e2$ and $\e1+\e2-2\e3$ have different lengths and are orthogonal; since the Weyl group acts transitively on such pairs of roots, that is all that we need to know. (But I don't know the answer.) Mar 12, 2023 at 16:57

sage: G2, A1xA1 = [WeylCharacterRing(x,style="coroots") for x in ["G2","A1xA1"]]

• Nice answer. To make it understandable to people not familirized with SAGE, this code ensures that the branching of the irrep of $\mathfrak g$ with highest weight $2\omega_1+2\omega_2$ ($\omega_i$ are the fundamental weights) to the subgroup $\mathfrak{so}(4)$ generated by two orthogonal roots $\alpha,\beta$ (note they are necessarily of different length) contains a trivial representation, which gives a non-zero element $v\in V_\pi(0)$ such that $\pi(X_\alpha)\cdot v=0$ and $\pi(X_\beta)\cdot v=0$. Mar 16, 2023 at 20:02
• After playing with Sage, I obtained numerical evidences that such vector exist for an irrep with highest weight $a\omega_1+b\omega_2$ if and only if $a$ and $b$ are even. Mar 16, 2023 at 20:04