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Let $V(\lambda)$ be a highest weight module of a semi-simple Lie algebra with highest weight $\lambda$. The Weyl dimension formula is $\dim V(\lambda) = \frac{\prod_{\alpha>0} (\lambda+\rho, \alpha)}{\prod_{\alpha>0}(\rho, \alpha)}$. By multiplying $\frac{2}{(\alpha, \alpha)}$ for each $\alpha$, we have $\dim V(\lambda) = \frac{\prod_{\alpha>0} \langle \lambda+\rho, \alpha \rangle}{\prod_{\alpha>0} \langle \rho, \alpha \rangle }$. Here $\langle \cdot, \cdot \rangle$ the form such that $\langle \lambda_i, \alpha_j \rangle = \delta_{ij}$ where $\lambda_j$ are fundamental weights and $\alpha_j$ are simple roots. Let $\Delta^{\vee}$ be a base of $\Phi^{\vee}$ (dual root system). Then $\alpha^{\vee} = \sum_{i} k_i \alpha_i^{\vee} $. How to compute the coefficients $k_i$? If we can compute $k_i$, then we know how to use the Weyl dimension formula.

For type $G_2$, I computed that $\alpha_1=2\lambda_1-\lambda_2$, $\alpha_2=-3\lambda_1+2\lambda_2$, $\alpha_1+\alpha_2=-\lambda_1+\lambda_2$, $2\alpha_1+\alpha_2=\lambda_1$, $3\alpha_1+\alpha_2=3\lambda_1-\lambda_2$, $3\alpha_1+2\alpha_2=\lambda_2$. These are all positive roots. $\rho=\lambda_1+\lambda_2$.

Let $\lambda=m_1\lambda_1+m_2\lambda_2$. How can we obtain the formula $$\dim V(\lambda) = 1/120 (m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1+3m_2+5)?$$

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up vote 1 down vote accepted

If I well remember the notation in rooth systems, I think the following equality will be useful for you.

Let $\alpha$ be $\sum_{i=1}^lc_i\alpha_i$. We have

$$ \langle\lambda,\alpha\rangle = (\lambda, \alpha^{\vee})=\frac{2(\lambda,\alpha)}{(\alpha,\alpha)}=\frac{2(\lambda,\sum_{i=1}^lc_i\alpha_i)}{(\alpha,\alpha)}= \sum_{i=1}^lc_i\frac{2(\lambda,\alpha_i)}{(\alpha,\alpha)}=\sum_{i=1}^lc_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}\frac{2(\lambda,\alpha_i)}{(\alpha_i,\alpha_i)}=\sum_{i=1}^lc_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}(\lambda,\alpha_i^{\vee})=\sum_{i=1}^lc_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}\langle\lambda,\alpha_i\rangle$$

So in your notation $$k_i= c_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}$$ Now I think you could easy check the formula for $G_2$ starting from its roots system.

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The dimensionality formula for G2 modules is given explicitly in Eq.(5.7) of

R. Slansky, "Group Theory for Model Building", Physics Report vol. 79, No. 1 (1981) pp. 1-128.

Details of how it can be found from first principles (along with any other dimensionality formula) can be found therein.

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