There are no irreducible representations of dimension less than $7$ for the Lie algebra $\mathfrak g_2$.

Question

I know that any finite dimensional irreducible representation of $\mathfrak g_2$ must appear in some tensor power of the standard representation $V$ which is seven dimensional. But still there may be $3$-dimensional irreducible representation $W$ included in $V\otimes V$ which is not a subrepresentation of $V$. Can we prove that this is not possible without using the Weyl character formula?

Answer 1

If you are willing to use that $\mathfrak g_2$ is simple, then you can use the fact that any non-zero representation is injective and that $\mathfrak g_2$ has dimension $14$. So for a representation in dimension $m$ you need $m^2-1\geq 14$, so any non-trivial representation of $\mathfrak g_2$ has to have dimension at least $4$. You can also rule out dimension $4$ easily, since $\mathfrak{sl}(4,\mathbb R)$ has no simple subalgebras of codimension $1$.

Answer 2

One possibility is to use the fact that the formal character of an irreducible representation is invariant under the action of the Weyl group. In the case of $\mathfrak{g}_2$ the Weyl group has order twelve, and the stabilizer of a non-zero weight has order at most two. Either by inspection, or due to the fact that the stabilizer of a weight in the closure of the dominant Weyl chamber is generated by the simple reflections in it.

This already forces the dimension of a non-trivial representation to be at least $12/2=6$. Furthermore, for $\mathfrak{g}_2$ the weight lattice coincides with the root lattice, so $\mathfrak{sl}_2$-theory implies that the zero weight is automatically also a weight of any non-trivial representation. The minimum dimension thus goes up to $6+1=7$.