I'm trying to grasp how Weyl's Dimension Formula works, and I'm having a bit of trouble.
As an example, I was trying to calculate the dimension of V($\varepsilon_1$) for gl(3).
First, I set the positive roots to be the set {$\varepsilon_1 - \varepsilon_2 , \varepsilon_1 - \varepsilon_3$}, and thus, I set $\rho$ (the half-sum of positive roots) to be $\frac12(2\varepsilon_1 - \varepsilon_2 - \varepsilon_3)$. So then, I go ahead, and put that in my formula; $$\prod_{\text{set of positive roots}} \frac{(\lambda + \rho, \text{positive root})}{(\rho, \text{positive root})}$$ $$ = \frac{(\varepsilon_1 + \frac12(2\varepsilon_1 - \varepsilon_2 - \varepsilon_3), (\varepsilon_1 - \varepsilon_2))}{(\frac12(2\varepsilon_1 - \varepsilon_2 - \varepsilon_3), (\varepsilon_1 - \varepsilon_2))} \frac{(\varepsilon_1 + \frac12(2\varepsilon_1 - \varepsilon_2 - \varepsilon_3), (\varepsilon_1 - \varepsilon_3))}{(\frac12(2\varepsilon_1 - \varepsilon_2 - \varepsilon_3), (\varepsilon_1 - \varepsilon_3))}$$ $$= \frac{(2\varepsilon_1 - \frac12\varepsilon_2 - \frac12\varepsilon_3), (\varepsilon_1 - \varepsilon_2))}{((\varepsilon_1 - \frac12\varepsilon_2 - \frac12\varepsilon_3), (\varepsilon_1 - \varepsilon_2))} \frac{((2\varepsilon_1 - \frac12\varepsilon_2 - \frac12\varepsilon_3), (\varepsilon_1 - \varepsilon_3))}{((\varepsilon_1 - \frac12\varepsilon_2 - \frac12\varepsilon_3), (\varepsilon_1 - \varepsilon_3))}$$
However, from here, I'm not actually sure what to do. I understand that, with inner products, $(\varepsilon_i, \varepsilon_j) = \delta_{i,j}$, but I can't work out how to use that here. I mean, if I were to do that, taking, for example, $(2\varepsilon_1 - \frac12\varepsilon_2 - \frac12\varepsilon_3), (\varepsilon_1 - \varepsilon_2))$, does that become $2 + \frac12$ ?? In the example given in class, for the dimension of V$(\varepsilon_1)$, my lecturer seemed to use that train of logic, but I'm honestly a bit uncertain.
Any help would be fantastic!!