(Preamble) In the book Representation Theory A First Course (Fulton, Harris), there is the following claim in the page 165 (written as an observation) without a proof:
The eigenvalues $\alpha$ occurring in an irreducible representation of $\mathfrak{sl}_3(\mathbb{C})$ differ from one another by integral linear combinations of the vectors $L_i – L_j \in \mathfrak{h}^*$
Prior to this claim, there is a quick derivation, where $X \in \mathfrak{g}_\alpha, v \in V_\beta, H \in \mathfrak{h}$ and $\mathfrak{g}$ is any Lie algebra, $\mathfrak{h}$ is a subspace of $\mathfrak{sl}_3(\mathbb{C})$ consisting of diagonal matrices, and $\alpha, \beta$ are eigenvalues (in this case linear functions on $H$) s.t. $\mathfrak{g}_\alpha = \{X \in \mathfrak{g}\mid \forall H \in \mathfrak{h}:[H, X] = \alpha(H)X\}$ and $V_\beta = \{v \in V\mid \forall H \in \mathfrak{h}:Hv = \beta(H)v\}$, where $V$ is to my knowledge just a vector space over some algebraically closed field.
The given derivation is: $H(X(v)) = X(H(v)) + [H, X](v) = X(\beta(H)v) + (\alpha(H)X)(v) = (\alpha(H) + \beta(H))X(v)$ by the bracket identity. The author(s) then state that
We see from this that $X(v)$ is again an eigenvector for the action of $\mathfrak{h}$ with eigenvalue $\alpha + \beta$
(Main question) It is perfectly clear to me that given any single eigenvalue $\beta$, we may jump to some other eigenvalues by the combining different roots $\alpha(H) = L_i – L_j$ for some $i, j$. What is unclear to me is that how do we know that we can jump from any eigenvalue $\beta_1$ to any other eigenvalue $\beta_2$ with the $L_i – L_j$s?
Best Answer
I expressed my problems with the other answer in comments there. But of course the basic idea is right: If the weight spaces are not "connected" in the sense that you can get from any weight to any other through a combination of roots, then the "connected components" of the weights would give non-trivial subrepresentations. I'd formalize this as follows:
Let $\mathfrak g$ be any complex semisimple Lie algebra, $\mathfrak h$ a Cartan subalgebra, $V$ a finite-dimensional representation of $\mathfrak g$. Then $V = \bigoplus_{\lambda \in P(V)} V_\lambda$ (all $V_\lambda \neq 0$) as $\mathfrak h$-modules, where $P(V)$ is a finite set of weights $\lambda: \mathfrak h \rightarrow \mathbb C$.
Let $Q$ be the $\mathbb Z$-span of all roots (or equivalently, of a set of simple roots) $\alpha$ of $\mathfrak g$ with respect to $\mathfrak h$.
We agree that if $X \in \mathfrak g_\alpha$, then $X$ induces maps $V_\lambda \rightarrow V_{\lambda + \alpha}$ for all $\lambda$.
In particular, for any given $w\in P(V)$, the subspace
$$\bigoplus_{\lambda \in P(V) \cap (w +Q)} V_\lambda \subseteq V$$
is stable under the action of each $X_\alpha$ as well as $\mathfrak h$, and hence all of $\mathfrak g$; i.e. it is a nonzero subrepresentation. So if $V$ is irreducible ...
The underlying problem with the "element-wise" approach discussed in the other answer, I think, is that to do it that way, one needs more subtle considerations about whether those maps $V_\lambda \rightarrow V_{\lambda+\alpha}$ are injective / surjective / ..., which is doable but already in the case of the adjoint representation needs more effort ($\mathfrak{sl}_2$-triples etc.) than one wants to invest at this point.