Let $V = \mathbb{C}^3 \otimes \mathbb{C}^3$ be a representation of $G = SL_3(\mathbb{C})$.
The weights of this representation is the set of $\varepsilon_i + \varepsilon_j$ for $i, j = 1, 2, 3$, where $\varepsilon_i$ takes $\text{diag}[h_1, h_2, h_3] \in \mathfrak{h}$ to $h_i$.
The Weyl group is $W = \{ 1, s_1, s_2, s_1s_2, s_2s_1, s_1s_2s_1 \}$.
For a simple root $\alpha_i$, the coroot $h_i$ is simply the matrix $E_{ii} – E_{i+1, i+1}$.
Then the pairing $$ \langle \varepsilon_j, h_i \rangle \alpha_i = \begin{cases} -\alpha_i & \text{ if } i = j, \\ \alpha_i & \text{ if } i = j -1, \\ 0 & \text{ else }. \end{cases}$$
By the defining equation of root reflections, $$ s_i (\beta) = \beta – \langle \beta, h_i \rangle \alpha_i$$ for $\beta \in \mathfrak{h}^*$, we have $$ s_i(\varepsilon_j) = \begin{cases} \varepsilon_j-\alpha_i & \text{ if } i = j, \\ \varepsilon_j + \alpha_i & \text{ if } i = j -1, \\ \varepsilon_j & \text{ else }. \end{cases}$$
Using this last part, the $W$-orbit of the weight $\varepsilon_1 + \varepsilon_2$ is the set $\{ \varepsilon_1 + \varepsilon_2, \varepsilon_1 + \varepsilon_3, \varepsilon_2 + \varepsilon_3 \}$.
Dominant weights are non-negative integral linear combinations of the fundamental weights, $m\varpi_1 + n \varpi_2$.
Expanding this out in terms of the $\varepsilon_i$, I get $$ m \varpi_1 + n \varpi_2 = \frac{1}{3} \left( 2(m+n) \varepsilon_1 + (2n-m) \varepsilon_2 – (m+n) \varepsilon_3 \right).$$
Equating coefficients shows that none of the set of weights $\{ \varepsilon_1 + \varepsilon_2, \varepsilon_1 + \varepsilon_3, \varepsilon_2 + \varepsilon_3 \}$ are dominant weights, contradicting what I am supposed to show.
What have I done wrong?
Best Answer
First, your coefficients for $m\varpi_1+n\varpi_2$ look wrong. As far as I know,
$$\varpi_1 = \frac13 (2\varepsilon_1 -\varepsilon_2 -\varepsilon_3), \varpi_2 = \frac13 (\varepsilon_1 +\varepsilon_2 -2\varepsilon_3)$$
and hence
$$m\varpi_1+n\varpi_2 = \frac13 \left((2m+n)\varepsilon_1 +(-m+n)\varepsilon_2 +(-m-2n)\varepsilon_3\right).$$
Now at first that looks like running into the same problem as with your coeffcients. But now remember that because of the determinant-$1$-condition in $SL$ (or the trace-$0$-condition in $\mathfrak{sl}$) we actually must have $\varepsilon_1 +\varepsilon_2 +\varepsilon_3 =0$ or in other words, we can assume $\varepsilon_3=-\varepsilon_1-\varepsilon_2$. This turns the above into
$$\varpi_1 = \varepsilon_1 , \qquad \varpi_2 = \varepsilon_1 +\varepsilon_2$$
and I'm sure you can take it from there.