Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 1 / 5

Identify the real span of the roots with its dual using the invariant inner product, and put ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$. The weight lattice isThe Weyl group is generated by the root reflections $s_{\alpha }$. Each reflection permutes the finite set of roots and preserves their lengths, hence also permutes their coroots. Thus, for $\lambda \in {\Lambda }_W$,Every generating reflection therefore maps the weight lattice to itself, and so does every product. Since the inverse product has the same property, every Weyl-group element acts bijectively on ${\Lambda }_W$. The roots span the ambient space, so the action on the finite root set is faithful; in particular the Weyl group is finite.

Choose simple roots ${\alpha }_1,\dots ,{\alpha }_r$. For the orbit statement one must use the Closed dominant Weyl chamberIts interior, defined using strict inequalities, is the open Weyl chamber. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.

Choose $\rho$ strictly inside the chamber, for example the sum of the fundamental weights. The finite Weyl group orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda ,\rho )$. If $(\lambda ,{\alpha }_i)<0$ for some simple root, thencontradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda \in \mathcal{W}$. Consequently some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber. The argument in fact works for every vector in the real weight space.

For the usual positive roots, the chamber condition is that the coefficients of $L_1,L_2,L_3$ be weakly decreasing, since their consecutive differences are the simple-coroot pairings. Adding the same constant to all three coefficients changes no weight, because $L_1+L_2+L_3=0$.

The given coefficients are $(-2,3,0)$. Let $s_1$ exchange $L_1,L_2$, and $s_2$ exchange $L_2,L_3$. With the rightmost reflection acting first,ThusIts coefficients $(3,0,-2)$ are strictly decreasing, so it lies even in the open Weyl chamber; its Dynkin labels are $(3,2)$.