Show that $\mathfrak{g}_2$ has a dimension $14$ representation

Question

This question came up in an oral exam. During the course we studied a bit of the theory of lie algebras and some representation theory.

The question: show that the lie algebra $\mathfrak{g_2}$ has a dimension $14$ representation, where dimension $14$ means the vector space $V$ where the representation is defined has dimension $14$ over $k$.

Why is this true? I think it has to do with $\mathfrak{g_2}$ having $12$ roots (and maybe the maximal toral subalgebra has dimension $2$? But why?).

I'll be glad if someone can enlighten me.

Answer 1

Hint 1: What is that Lie algebra's dimension?

Hint 2: Surely in your class the concept of the adjoint representation was mentioned?

Answer 2

Let $\mathfrak{g}$ be the simple Lie algebra of type $G_2$, with positive roots $R^+=\{\alpha,\beta,\alpha+\beta,2\alpha+\beta,3\alpha+\beta,3\alpha+2\beta \}$. With $\lambda=m_1\varpi_1+m_2\varpi_2$ the Weyl dimension formula gives \begin{align*} \dim(L(\lambda)) & = \frac{1}{120}(m_1+1)(m_2+1)(m_1+2m_2+3)(m_1+m_2+2) \\ & \hspace{1.52cm} (m_1+3m_2+4)(2m_1 + 3m_2+5) \end{align*} for the irreducible highest weight module. We easily see that the possible dimensions are $$1, 7, 14, 27, 64, 77, 182, 189, 273, 286, 378, 448, 714, 729, 748, 896, 924, \ldots $$