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Given a semisimple complex Lie algebra $\frak{g}$ of rank $r$, with Chevally generators $E_i,F_i,K_i$. Let $V$ be a finite dimensional representation of $\mathfrak{g}$ such that each weight space of $V$ is $1$-dimensional. Let $(i_1,\dots,i_k)$ be an ordered set of elements of $\{1,\dots,r\}$ (allowing repeats), and let $\{j_1,\dots,j_k\}$ be some permutation of $\{1,\dots,r\}$. For $v$ a highest weight of $V$, the elements $$ F_{i_1} F_{i_2} \cdots F_{i_k}(v),\quad \text { and } \quad F_{j_1} F_{j_2} \cdots F_{j_k}(v), $$ must have the same weight. Thus by our assumption they must differ by a scalar multiple. Will this scalar multiple always be an element of $\mathbb{Q}$?

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  • $\begingroup$ What do you mean by "two (vectors) differ by a scalar multiple"? multiplicatively, $1$ and $2$ differ by $2$, and also by $1/2$. That is, the scalar multiple is only defined up to inversion. Are you asking whether it is always integer up to inversion? $\endgroup$ – YCor Sep 11 '19 at 6:09
  • $\begingroup$ @YCor: thanks for the edit, and yes, I mean integer up inversion, so I guess I should write an element of $\mathbb{Q}$ . . . $\endgroup$ – Pierre Dubois Sep 11 '19 at 6:16
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    $\begingroup$ By a rational is much weaker than by an integer up to inversion. Then it's certainly true, since the Lie algebra can be defined over $\mathbf{Q}$ with the given Cartan subalgebra as split Cartan subalgebra, and the representation is then split too. $\endgroup$ – YCor Sep 11 '19 at 6:16
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    $\begingroup$ This is true with integer replaced by rational since all irreducible representations of a Chevalley LIe algebra over the rationals are defined over the rationals (JIm Humphreys book on semisimple lie algebras has a construction, I think). $\endgroup$ – Venkataramana Sep 11 '19 at 6:19
  • $\begingroup$ could you explain "since all irreducible representations of a Chevalley LIe algebra over the rationals are defined over the rationals" $\endgroup$ – Pierre Dubois Sep 11 '19 at 6:24
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Let me flesh out the answer given in the comments. Any simple Lie algebra can be defined over the field of rational numbers. Then its irreducible finite dimensional representations can be constructed as quotients of Verma modules. That is $$ V_\lambda = \mathfrak{U}(\mathfrak{g}_\mathbb{Q}) \otimes_{ \mathfrak{U}(\mathfrak{b}_\mathbb{Q})} \mathbb{Q}_\lambda / \text{maximal submodule}, $$ where $\mathbb{Q}_\lambda$ is the one dimensional representation on which the split Cartan subalgebra of $\mathfrak{g}_\mathbb{Q}$ acts by character $\lambda.$

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