Sum of two positive roots which is not a root: uniqueness of heights of the summands

Question

Consider a (finite reduced irreducible crystallographic) root system $\Phi$ and four positive roots $\alpha,\beta,\gamma,\delta$ such that $\{\alpha,\beta\} \neq \{\gamma,\delta\}$ and $\alpha+\beta=\gamma+\delta$, but these sums are not roots in $\Phi$. It turns out that $\operatorname{ht}(\alpha) \neq \operatorname{ht}(\gamma)$, where $\operatorname{ht}$ is the height of the root w.r.t. the chosen system of simple roots.

This statement is not hard to prove on a case-by-case basis (although I have not yet checked $\mathsf{E}_7$, $\mathsf{E}_8$ and $\mathsf{F}_4$), but it does look like one admitting an elegant uniform proof. Any ideas?

Answer 1

Here's a naive approach which I think works in the simply-laced cases, unless I've made a silly mistake somewhere. The restriction to simply-laced cases allows me to play fast and loose with the root-coroot pairings vs. the inner product on roots, and also assume that all roots have length 2.

Assume $\{\alpha, \beta\} \neq \{\gamma, \delta\}$, but that $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$. Since $\alpha+\beta-\gamma=\delta$ is a root, then $$ (\alpha+\beta-\gamma, \alpha+\beta-\gamma)=(\delta,\delta)=2. $$

Expanding this out, using all roots have the same length, we get $$ (\alpha,\beta)-(\alpha, \gamma)-(\beta,\gamma)=-2, $$ or that $(\alpha,\beta)-(\alpha,\gamma)=(\beta,\gamma)-2$. Since $\alpha+\beta \not \in \Phi^+$, necessarily $(\alpha, \beta) \geq 0$. Similarly, since $\alpha \neq \gamma$ and $\mathrm{ht}(\alpha)=\mathrm{ht}(\gamma)$, $\alpha-\gamma \not \in \Phi \sqcup \{0\}$ so that $(\alpha, \gamma) \leq 0$.

In total, then, $(\alpha, \beta)-(\alpha, \gamma) \geq 0$, so that $(\beta, \gamma)-2 \geq 0 \implies (\beta, \gamma) \geq 2$; since we are in the simply-laced case, necessarily $(\beta, \gamma)=2$ and thus $\beta =\gamma$, contradicting $\{\alpha, \beta\} \neq \{\gamma, \delta\}$.