# Semisimple Lie algebra modules with $1$-dimensional weight spaces

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 $$\mathbb{Q}$$?

• 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?
– YCor
Sep 11, 2019 at 6:09
• @YCor: thanks for the edit, and yes, I mean integer up inversion, so I guess I should write an element of $\mathbb{Q}$ . . . Sep 11, 2019 at 6:16
• 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.
– YCor
Sep 11, 2019 at 6:16
• 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). Sep 11, 2019 at 6:19
• could you explain "since all irreducible representations of a Chevalley LIe algebra over the rationals are defined over the rationals" Sep 11, 2019 at 6:24

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 $$\lambda.$$
• 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 finite-dimensional representations.) Dec 29, 2021 at 22:52
• @LSpice I've tried to clarify. I don't have any experience over $\matbb{Z}$. Dec 30, 2021 at 10:14