Let $G$ be a reductive complex algebraic group, $H \subset G$ a Cartan subgroup and $R^+$ a set of positive roots, and $X_+(H)$ the set of dominant weights. Let us also assume that $G$ is simply connected.
I read in these notes "Since $G$ is simply connected, $\rho := \frac{1}{2}\sum_{\alpha \in R^+} \alpha \in X_+(H)$".
For me, simply connected means $\pi_1(G,e) = \{1\}$. How to deduce the claim ?
It seems to me that it's always true. Indeed, $s_i$ preserves all positive roots but $\alpha_i$ so $s_i \cdot \rho = \rho - \alpha_i$ so it follows that $\langle \rho, \alpha_i^{\vee}\rangle =1$. In fact this equality is explicitly stated a bit after the definition of $\rho$. So how it is related to $G$ being simply connected or not ?