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We know that the the roots of $\mathbb{g} = sl(3,C)$ under the adjoint action are given by $L_i - L_j$ where $L_i (diag(a_1, a_2, a_3))=a_i$ for $i = 1,2,3$. If $V$ is any irreducible representation of $sl(3,C)$, then clearly $V$ is the direct sum of its simultaneous eigenspaces given by $V_\beta$ = { $v \in V$ : $H.v$ = $\beta(H)v$ $\forall$ H $\in \mathbb{h}$ } where $\mathbb{h}$ is the subalgebra of diagonal matrices in $sl(3,C)$.In this case, $\beta$ is called the eigenvalue for the representation $V$. If $X$ $\in \mathbb{g_{\alpha}}$ and $v \in V_{\beta}$, then we know that
$\mathbb{g_{\alpha}}$ carries $V_{\beta}$ to $V_{\alpha + \beta}$ where $\mathbb{g_{\alpha}}$ = $\{ X_{\alpha} \in sl(3,C) : adH(X_{\alpha}) = \alpha(H) X_{\alpha} \forall H \in \mathbb{h}\}$. I want to now show that the eigenvalues $\alpha$ occuring in the representation $V$ differ from one another by integral linear combinations of the vectors $L_i - L_j \in \mathbb{h^*}$. Note - $adH(X) = [H,X] = HX - XH$. Thanks for any help.

Ester
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  • see Fulton, Harris book – Leox Apr 26 '16 at 06:42
  • I was reading that. Can you please elaborate the solution? – Ester Apr 26 '16 at 06:45
  • I guess if $\alpha$ and $\beta$ do not differ from one another by an integral linear combinations of the $L_i-L_j$, then they each define a nontrivial subrepresentation of $V$, say $V_{\alpha}$ and $V_{\beta}$ respectively, whose common intersection is the zero vector. This contradicts that $V$ is irreducible. – Malkoun Jul 06 '19 at 19:50
  • My previous comment was sketchy. I just found the answer to your question in Thm. 2.5 in http://www.math.uchicago.edu/~may/VIGRE/VIGRE2010/REUPapers/Patel.pdf – Malkoun Jul 06 '19 at 19:54

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