This is an exercise 10.10 in Humphreys book on Lie algebras.
Let $\Phi$ be a root system lying in the euclidean space $E$ and let $\Delta = \{\alpha_1,\cdots,\alpha_\ell\}$ be a basis for $\Phi$. Let $\lambda = \sum_i k_i \alpha_i$ with all $k_i\geq 0$ or all $k_i\leq 0, k_i \in \mathbb Z.$ Prove that either $\lambda$ is a multiple (possibly 0) of a root, or else there exists $\sigma \in \mathscr W$ (Weyl group) such that $\sigma \lambda = \sum_i k_i'\alpha_i$ with some $k_i'>0$ and some $k_i'<0$.
He gives the following tip: If $\lambda$ is not a multiple of any root, then the hyperplane $P_\lambda$ orthogonal to $\lambda$ is not included in $\bigcup_{\alpha \in \Phi} P_\alpha$. Take $\mu \in P_\lambda \setminus \bigcup P_\alpha$ and then find $\sigma \in \mathscr W$ for which all $(\alpha_i,\sigma\mu)>0$.
I couldn't prove that $P_\lambda \not \subseteq \bigcup P_\alpha$, altough I managed to finish the exercise as follows. Taking any such a $\mu$, since every point in $E$ is $\mathscr W$-conugate to a point in the fundamental Weyl chamber, there exists $\sigma \in \mathscr W$ satisfying $(\sigma\mu, \alpha_i)>0$ as claimed. In particular, each $\sigma \alpha_i \in \Phi$, so we may write $\sigma\lambda = \sum k_i' \alpha_i$ for some (possibly new) integers $k_i'$. Now, $\mu \in P_\lambda$, so
$$ 0 = (\mu,\lambda ) = (\sigma\mu, \sigma \lambda) = \sum k_i'(\sigma\mu,\alpha_i)$$
implies that some $k_i'>0$ and some $k_i'<0$, as the terms $(\sigma\mu ,\alpha_i)$ are all positive.
The question then is: how to prove that $P_\lambda \not\subseteq \bigcup P_\alpha$? All the computations I did so far were useless, stuff like $0 = (\lambda,x) = \sum k_i (\alpha_i,x)$ cannot imply anything. I also tried to begin simple with $P_\lambda \subset P_\alpha \implies \lambda = c\alpha$ by supposig $\lambda – c\alpha\neq 0$ and $P_\lambda \subseteq P_\alpha$, but that only yelds $P_\lambda \subseteq P_{\lambda – c\alpha}$.
Any help? Thank you.
Best Answer
Lemma: If $H, H_1, ... H_r$ are hyperplanes (i.e. $(n-1)$-dimensional subspaces) in some $n$-dimensional space over an infinite field, and $H \subseteq \bigcup_{i=1}^n H_i$, then $H = H_j$ for some $1 \le j \le r$.
Proof: By assumption we have
$$H = H \cap \left( \bigcup_{i=1}^r H_i \right) = \bigcup_{i=1}^r (H \cap H_i).$$
Now the intersection of any two hyperplanes has dimension $n-2$ unless the two hyperplanes are equal. But if all the spaces in the union on the RHS are $(n-2)$- dimensional, their union cannot be the $(n-1)$-dimensional space on the LHS. QED.
To apply this to your problem: If $P_\lambda \subseteq \bigcup P_\alpha$, then by the lemma there is a root $\alpha$ such that $P_\lambda = P_\alpha$, consequently $\langle \lambda \rangle = P_\lambda^\perp = P_\alpha^\perp = \langle \alpha\rangle$, i.e. $\lambda$ is a scalar multiple of $\alpha$.