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 $$\dim V(\lambda) = 1/120 (m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1+3m_2+5)?$$


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.