The Weyl vector in fundamental coordinates

Let $\alpha \in {\mathrm{\Phi }}^{+}$ with $\alpha \mathrm{\ne }{\alpha }_i$; writing $\alpha ={\sum }_j{n}_j{\alpha }_j$ with $n_j\ge 0$ by [L2], some $n_j$ with $j\mathrm{\ne }i$ is positive, since otherwise $\alpha =n_i{\alpha }_i$ and reducedness with $\alpha$ a positive root forces $n_i=1$ and $\alpha ={\alpha }_i$; hence $s_i(\alpha )$ has the positive coefficient $n_j>0$ at position $j\mathrm{\ne }i$, and since $s_i(\alpha )$ is a root its coefficient vector has one sign by [L2], so $s_i(\alpha )\in {\mathrm{\Phi }}^{+}$; moreover $s_i(\alpha )\mathrm{\ne }{\alpha }_i$ because $s_i({\alpha }_i)=-{\alpha }_i$ and $s_i$ is an involution; thus $s_i$ maps ${\mathrm{\Phi }}^{+}\setminus \{{\alpha }_i\}$ onto itself.

Using step 1.1 and $s_i({\alpha }_i)=-{\alpha }_i$, $s_i(2\rho )={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}{s}_i(\alpha )={\sum }_{\alpha \mathrm{\ne }{\alpha }_i}\alpha -{\alpha }_i=2\rho -2{\alpha }_i$, hence $s_i(\rho )=\rho -{\alpha }_i$.

On the other hand $s_i(\rho )=\rho -(\rho ,{\alpha }_i^{\vee }){\alpha }_i$ by [L1]; comparing with step 2.1 and using that ${\alpha }_i\mathrm{\ne }0$ gives $(\rho ,{\alpha }_i^{\vee })=1$ for every $i$.

The difference $\rho -{\sum }_j{\omega }_j$ satisfies $(\rho -{\sum }_j{\omega }_j,{\alpha }_i^{\vee })=1-1=0$ for every $i$ by [L3] and step 3.1; since the simple coroots form a basis of $E$ and the inner product is nondegenerate, $\rho -{\sum }_j{\omega }_j=0$, that is, $\rho ={\sum }_j{\omega }_j$.

Both assertions are proved.