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



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


