Unique maximal short root

Question

Let $\Phi$ be irreducible. Prove that $\Phi^\vee$ is also irreducible. If $\Phi$ has all roots of equal length, so does $\Phi^\vee$ (and then $\Phi^\vee$ is isomorphic to $\Phi$). On the other hand, ih $\Phi$ has two root lengths, then so does $\Phi^\vee$;but if $\alpha$ is long, then $\alpha^\vee$ is short (and vice versa). Use this fact to prove that $\Phi$ has a unique maximal short root (relative to the partial order $\prec$ defined by $\Delta$).

This is a Humphreys exercise and so far I've shown that $\Phi^\vee$ is irreducible and my thought was to solve the rest of it by the cases: only one root length ( which is ok) and two root lengths and that was the problem.

I just know that the maximal root will be obtained by the reflection of the short and simple one. But how is that unique and short/maximal at the same time?

Thank you in advance.

Answer 1

If there are two root lengths, say $\alpha,\beta\in\Phi$ have different root lengths, then we compute that $$(\alpha^\vee,\alpha^\vee)=\left(\frac{2\alpha}{(\alpha,\alpha)},\frac{2\alpha}{(\alpha,\alpha)}\right)=\frac{4}{(\alpha,\alpha)},$$ so $$(\alpha^\vee,\alpha^\vee)\ne(\beta^\vee,\beta^\vee).$$ We know that there are at least two root lengths in $\Phi^\vee$ and the shorter the length of $\alpha$, the longer the length of $\alpha^\vee$, but since you have proved that $\Phi^\vee$ is also irreducible given $\Phi$ irreducible. We conclude that $\Phi^\vee$ has exactly two root lengths if $\Phi$ has two root lengths.

To show that $\Phi$ has a unique maximal short root, suppose we have a maximal short root $\alpha\in\Phi$, then we claim that $\alpha^\vee\in\Phi^\vee$ is a maximal long root. Then by Lemma 10.4 A, $\alpha^\vee$ is unique, and hence $\alpha$ is unique. So it suffices to show that greater the height of $\alpha$, the greater the height of $\alpha^\vee$.

Write $$\alpha=\sum_i k_i\gamma_i, k_i\in\mathbb Z_{> 0},\gamma_i\in\Delta,$$ where $\Delta$ is a basis of $\Phi$. By the definition of $\alpha^\vee$, we have $$ \alpha^\vee=\frac{2\alpha}{(\alpha,\alpha)}=\sum_i\frac{2k_i}{(\alpha,\alpha)}\gamma_i=\sum_i\frac{k_i(\gamma_i,\gamma_i)}{(\alpha,\alpha)}\gamma_i^\vee, $$ where $\displaystyle\frac{k_i(\gamma_i,\gamma_i)}{(\alpha,\alpha)}>0$. Moreover, we claim that $\displaystyle\frac{k_i(\gamma_i,\gamma_i)}{(\alpha,\alpha)}\in\mathbb Z_{>0}$. Because of Exercise 10.1, we know that the elements of $\Delta^\vee$ consist of a dual basis of $\Phi^\vee$. And the expression of $\alpha^\vee$ in terms of $\gamma_i^\vee\in\Delta^\vee$ is thus unique. But we know that every positive root is a unique positive integer-valued linear combination of basis elements. Hence $$\displaystyle\frac{k_i(\gamma_i,\gamma_i)}{(\alpha,\alpha)}\in\mathbb Z_{>0}.$$ It follows easily from this that indeed the greater the $\sum_ik_i$ the greater the $\displaystyle\sum_i\frac{k_i(\gamma_i,\gamma_i)}{(\alpha,\alpha)}$ given the same root length $(\alpha,\alpha)$ and vice versa. And both the existence of maximal short root and the uniqueness follow if we first go from a maximal long coroot to a (maximal)short root to show the existence and go to the dual root system from the root system to show the uniqueness by the argument above. Hence we are done.