Shifting weights by roots of a Lie algebra Let $R$ be a representation of a (finite-dimensional) semi-simple Lie group $G$.  Let $w$ be a weight of $R$ and $\alpha$ a root of $G$.  If neither $w+\alpha$ nor $w-\alpha$ are weights of $R$, is it true that $\alpha$ and $w$ are orthogonal (using the Killing form)?
 A: Recall that the Lie algebra $\mathfrak g$ contains an $\mathfrak{sl}_2 \mathbb C$ subalgebra $\langle H_{\alpha}, X_\alpha, Y_\alpha \rangle$, where $X_\alpha$ and $Y_\alpha$ are elements of the $\alpha$ and $-\alpha$ root spaces respectively. If $v$ is a vector in the $w$ weight space for the representation $R$, then $X_\alpha (v)$ and $Y_\alpha(v)$ are in the $w + \alpha$ and $w - \alpha$ weight spaces respectively of the representation $R$. Since neither $w + \alpha$ nor $w - \alpha$ are weights of $R$, it must be the case that $X_\alpha(v) = Y_\alpha (v) = 0$ for all $v$ in the $w$ weight space of $R$. In other words, the $w$ weight space of $R$ decomposes into a direct sum of trivial representations of the $\mathfrak{sl}_2 \mathbb C$ subalgebra $\langle H_{\alpha}, X_\alpha, Y_\alpha \rangle$. This implies that $H_\alpha(v) = 0$ for all $v$ in the $w$ weight space of $R$. Since $H_\alpha(v) = w(H_\alpha)v$ for all $v$ in the $w$ weight space, it follows that $w(H_\alpha) = 0$.
Now recall how the Killing form is defined. For each element $\beta$ of $\mathfrak h^\star$, there exists a unique element $T_\beta$ such that $B(T_\beta, H) = \beta(H)$ for all $H \in \mathfrak h$. (Here, $\mathfrak h$ denotes the Cartan subalgebra and $B(.,.)$ denotes the Killing form.) The inner product on $\mathfrak h^\star$ is defined by $(\beta, \gamma)=B(T_\beta, T_\gamma)$. Returning to our problem, we already know that $w(H_\alpha) = B(T_w, H_\alpha) = 0$. But in view of the well-known identity $H_\alpha = 2T_\alpha / \alpha(T_\alpha)$, this implies that $B(T_w, T_\alpha) = 0$, which is equivalent to saying that $(w,\alpha) = 0$.
