Why is the root poset is graded by height?

Question

Let $\Phi$ be a finite crytallographic root system. Let $\Phi^+$ be the positive roots and $\alpha_1$, ..., $\alpha_n$ be the simple roots. For $\beta = \sum c_i \alpha_i$ in $\Phi^+$, we define $h(\beta) = \sum c_i$. For $\beta = \sum c_i \alpha_i$ and $\gamma = \sum d_i \alpha_i$, we define $\beta \preceq \gamma$ iff $c_i \leq d_i$ for all $i$. Many sources state that $\Phi^+$ is graded by $h$. The nontrivial part of this statement is that, if $\alpha \leq \gamma$ with $h(\gamma) - h(\alpha) \geq 2$, then there is a root $\beta$ with $\alpha \leq \beta \leq \gamma$. Could someone give me a proof or reference to a proof, other than type by type check?

To show that I haven't been completely lazy: Humphreys defines $\Phi^+$ and $h$ but doesn't state that $h$ grades $\Phi^+$, Bjorner and Brenti define a different, unrelated partial order on $\Phi^+$ which they call the root poset. Cuntz and Stump cite Armstrong Section 5.4.1 but it doesn't seem to be in there.

Answer 1

Let $(\cdot,\cdot)$ be a positve definite Weyl group invariant product on $\mathbb{R}\Phi$. Let $\beta$ and $\gamma$ be positive roots with $\beta\leq \gamma$ and $h(\gamma)-h(\beta)\geq 2$. Let $v=\gamma-\beta$.

Let $\alpha_i$ be a simple root which occurs in $v$. Suppose for want of a contradiction that $\beta+\alpha_i$ and $\gamma-\alpha_i$ are not positive roots. Then $(\beta,\alpha_i)\geq 0$ and $(\gamma,\alpha_i)\leq 0$. In particular

$$(v,\alpha_i)\leq 0.$$

Since $v$ is a positive sum of the $\alpha_i$, we sum these inequalities to get $(v,v)\leq 0$. This is a contradiction since $(\cdot,\cdot)$ is positive definite. QED.