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 $\mathbb{Q}$?

$\begingroup$ 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? $\endgroup$– YCorCommented Sep 11, 2019 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 DuboisCommented Sep 11, 2019 at 6:16

1$\begingroup$ 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. $\endgroup$– YCorCommented Sep 11, 2019 at 6:16

1$\begingroup$ 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). $\endgroup$– VenkataramanaCommented Sep 11, 2019 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 DuboisCommented Sep 11, 2019 at 6:24
1 Answer
Let me flesh out the answer suggested in the comments. Any complex simple Lie algebra can be obtained as a complexification of a (split) Lie algebra $\mathfrak{g}_{\mathbb{Q}}$ defined over the field of rational numbers. The irreducible finite dimensional representations of $\mathfrak{g}$ are complexifications of representations of $\mathfrak{g}_{\mathbb{Q}}$ which in turn 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 $\lambda.$
I am not sure about integrality results which were the subject of the original question.

$\begingroup$ Do you mean any split simple Lie algebra? (Otherwise, I am not sure what to make of "the split Cartan subalgebra of $\mathfrak g_{\mathbb Q}$".) If so, then doesn't Chevalley show that we can even define any such over $\mathbb Z$? (I don't know about finitedimensional representations.) $\endgroup$– LSpiceCommented Dec 29, 2021 at 22:52

1$\begingroup$ @LSpice I've tried to clarify. I don't have any experience over $\matbb{Z}$. $\endgroup$ Commented Dec 30, 2021 at 10:14