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Let $G$ be a semisimple algebraic group over a field $k$. Let $A_0$ be a maximal split torus of $G$, and $\mathfrak a_0 = \operatorname{Hom}(X(A_0), \mathbb R)$ the real Lie algebra of $A_0$ with dual $\mathfrak a_0^{\ast} = X(A_0) \otimes \mathbb R$.

Let $\Delta \subset \mathfrak a_0^{\ast}$ be a set of simple roots of $A_0$ in $G$. Let $\Delta^{\vee} \subset \mathfrak a_0$ be the coroots for the corresponding root system, and let $\hat{\Delta}^{\vee}$ the coweights (the dual basis to $\Delta$).

Let $$A = \{ v \in \mathfrak a_0^{\ast} : \langle v, \alpha^{\vee} \rangle > 0 \textrm{ for all $\alpha^{\vee} \in \Delta^{\vee}$}\}$$

$$B = \{ v \in \mathfrak a_0^{\ast} : \langle v, \varpi_{\alpha}^{\vee} \rangle > 0 \textrm{ for all } \varpi_{\alpha}^{\vee} \in \hat{\Delta}^{\vee} \}$$

Is it true that $A \subseteq B$?

This is claimed in line (2.2) of these notes on Langlands classification. However, I can't seem to prove this and am beginning to think it is not true.

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This statement appears to be true. The elements of $A$ are non-negative linear combinations of the fundamental weights, and elements of $B$ are non-negative linear combinations of simple roots. Formula $(2.1)$ implies that the fundamental weights are non-negative linear combinations of simple roots.

Draw a picture of $A$ and $B$ for a couple of rank 2 root systems and you'll see what is going on.

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