$\newcommand{\fg}{\mathfrak g}\newcommand{\ee}{\varepsilon}$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 $\pi$.

  • 2
    $\begingroup$ $\def\e#1{\varepsilon_#1}\def\a{\alpha}\def\b{\beta}$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.) $\endgroup$
    – LSpice
    Mar 12, 2023 at 16:57

1 Answer 1


The answer is no. Here is some SAGE code:

sage: G2, A1xA1 = [WeylCharacterRing(x,style="coroots") for x in ["G2","A1xA1"]]                         
sage: b = G2.maximal_subgroup("A1xA1")                                                                
sage: G2(2,2).branch(A1xA1,rule=b)                                                                       
A1xA1(0,0) + A1xA1(1,1) + A1xA1(1,3) + A1xA1(1,5) + 2*A1xA1(2,2) + A1xA1(2,4) + A1xA1(2,6) + 2*A1xA1(3,1) + 2*A1xA1(3,3) + A1xA1(3,5) + 2*A1xA1(4,0) + 2*A1xA1(4,2) + 2*A1xA1(4,4) + 2*A1xA1(5,1) + 2*A1xA1(5,3) + A1xA1(5,5) + A1xA1(6,0) + 3*A1xA1(6,2) + A1xA1(6,4) + 2*A1xA1(7,1) + 2*A1xA1(7,3) + A1xA1(8,0) + A1xA1(8,2) + A1xA1(8,4) + A1xA1(9,1) + A1xA1(9,3) + A1xA1(10,2) + A1xA1(0,4)

The presence of A1xA1(0,0) shows that the trivial module appears when this representation of G2 is restricted to A1xA1, giving a nonzero vector satisfying your given conditions.

  • $\begingroup$ 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 $\pi(X_\beta)\cdot v=0$. $\endgroup$
    – emiliocba
    Mar 16, 2023 at 20:02
  • $\begingroup$ 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. $\endgroup$
    – emiliocba
    Mar 16, 2023 at 20:04

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