Proposition 2

The $\alpha \tilde{}$ form a root system in $V^*$, and $\alpha \tilde{} = \alpha$ for all $\alpha \in R$.

The $\alpha \tilde{}$ satisfy (RS_I) by Prop. 1. Since $s_{\alpha, \alpha \tilde{}}$ is an automorphism of the vector space $V$ equipped with the subset $R$, ${}^t(s_{\alpha, \alpha \tilde{}})^{-1}$ leaves the set $R \tilde{}$ of the $\alpha \tilde{}$ stable; but ${}^t(s_{\alpha, \alpha \tilde{}})^{-1} = s_{\alpha \tilde{}, \alpha}$, which proves that $R \tilde{}$ satisfies (RS_{II}) and that $\alpha \tilde{} = \alpha$. Finally, $\langle \alpha \tilde{}, \beta \rangle \in \mathbf{Z}$ for all $\alpha \tilde{} \in R \tilde{}$ and $\beta \in R$, so $R \tilde{}$ satisfies (RS_{III}).

The set $R \tilde{}$ is called the inverse root system of $R$. The map $\alpha \mapsto \alpha \tilde{}$ is a bijection from $R$ to $R \tilde{}$, called the canonical bijection from $R$ to $R \tilde{}$. Note that, if $\alpha, \beta$ are elements of $R$ such that $\alpha + \beta \in R$, then $(\alpha + \beta) \tilde{} \neq \alpha \tilde{} + \beta \tilde{}$ in general.

Since $s_{\alpha}(\alpha) = -\alpha$, axiom (RS_{II}) shows that $-R = R$. Evidently $(-\alpha) \tilde{} = -\alpha \tilde{}$ and $-1 \in A(R)$ (but it is not always true that $-1 \in W(R)$).

The equality ${}^t(s_{\alpha, \alpha \tilde{}})^{-1} = s_{\alpha \tilde{}, \alpha}$ shows that the map $u \mapsto {}^t u^{-1}$ is an isomorphism from the group $W(R)$ to the group $W(R \tilde{})$. We identify these two groups by means of this isomorphism; in other words, we consider $W(R)$ as acting both in $V$ and in $V^*$. Similarly for $A(R)$.