Definitions: Let $G$ be a compact, connected, semi-simple (defined below) Lie group with maximal torus $T$, Weyl group $W:=N(T)/T$ and lie algebra $\mathfrak{g}$. Let $\Lambda$ be the dual of $T$, and let $\Lambda_{root}$ be the sub-lattice spanned by the weights ($\Phi$) of the adjoint representation of $T$ on $\mathfrak{g}\otimes \mathbb{C}$. Semi-simplicity means the index $[\Lambda:\Lambda_{root}]$ is finite. Give $\Lambda \otimes \mathbb{R}$ an inner product invariant under $W$ and split $\Lambda \otimes \mathbb{R}$ into two half-spaces to define the set of positive roots $\Phi^+$.
Question: The book I'm reading (Bump, Lie Groups, 2nd edition) claims that $\rho:=\frac{1}{2}\sum_{\Phi^+}\alpha$ is in $\Lambda$ (First line of the proof of prop $22.2$). How do I see this? This fact is not obvious to me and I can't find the part in the book where he's proved it.
Thanks for reading!
This follows from the fact that if $\alpha\in\Delta=\{\alpha_1,\ldots,\alpha_n\}$ is a simple root and $s_\alpha$ is the reflection $s_\alpha(\alpha)=-\alpha$ and $s_\alpha(\beta)=\beta$ for $\beta\in\alpha^\perp$, then $s_\alpha$ permutes $\Phi^+\backslash\{\alpha\}$. Therefore, $$ s_\alpha(\rho)=\rho-\alpha $$ which shows that $(\rho,\alpha)=(\alpha,\alpha)/2$.
Now, let $\omega_1,\ldots,\omega_n$ be the fundamental dominant weights (so $2(\omega_i,\alpha_j)/(\alpha_j,\alpha_j)=\delta_{ij}$). Then, writing $$\rho=\sum_ia_i\omega_i$$ we have $$(\rho,\alpha_j)=\sum_ia_i(\omega_i,\alpha_j)=a_j(\alpha_j,\alpha_j)/2.$$ Now, solving for $\alpha_j$ one easily deduces the formula $$ \rho=\sum_i\frac{2(\rho,\alpha_i)}{(\alpha_i,\alpha_i)}\omega_i=\sum_i\omega_i. $$ Therefore, $\rho$ is a sum of the fundamental weights.
As pointed out by Tobias Kildetoft in the comments, the fundamental weights belong to the weight lattice provided $G$ is simply connected.