Length of root strings is at most 4

Question

Let $\Phi$ be a root system. In his Introduction to Lie algebras and Representation Theory, J. Humphreys proves that if $$\beta-p\alpha,\dots,\beta,\dots,\beta+q\alpha$$ is the $\alpha$-root string through $\beta$, then $$\frac{2(\beta,\alpha)}{(\alpha,\alpha)}=p-q.$$ Then, he states that it follows "at once" that root strings are of length at most four.

I dont get why. Of course, we know that $-4\le 2(\beta,\alpha)/(\alpha,\alpha)\le 4$, but then length of the root string is $p+q+1$, not $p-q$...

Answer 1

By positive definiteness $0 \leq (\alpha, \beta)^2 < (\alpha, \alpha)( \beta, \beta )$ or, equivalently, $0 \leq (\alpha, \beta^\vee) (\beta, \alpha^\vee ) < 4$. Note the strict inequality.

Edit: I replaced $\langle \, , \, \rangle$ by $(\, , \,)$ to denote the inner product because of their conflicting interpretations.

Answer 2

Here is a slight refinement of Sid's argument. Using Humphreys' notation we know that $-3 \le \langle \alpha,\beta \rangle \le 3$ for all roots $\alpha,\beta$. Hence, in particular, for the $\alpha$-string through $\beta$ of length $q+r+1$ given by $\beta-r\alpha,\ldots ,\beta+q\alpha$ we find $$ \langle \beta+q\alpha,\alpha\rangle \le 3\mbox{ and} -3\le \langle \beta-r\alpha,\alpha\rangle. $$ But $\langle \alpha,\alpha\rangle=2$ therefore implies $$ 2q\le 3- \langle \beta ,\alpha\rangle \mbox{ and } 2r\le 3+\langle \beta,\alpha\rangle. $$ Thus together we find $q+r+1\le 4$.