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Originally posted on math.stackexchange. 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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    $\begingroup$ I'm not sure whether your initial setup is relevant or whether you could ask the question for an abstract (reduced crystallographic) root system, but assuming the latter, the closure of $A$ is simply generated by the fundamental weights (it is the Weyl chamber) and the closure of $B$ is generated by the positive roots, and the inclusion $A\subseteq B$ is correct and states that fundamental weights are combinations with nonnegative coefficients of the simple roots. $\endgroup$
    – Gro-Tsen
    Commented Jul 23, 2018 at 21:07
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    $\begingroup$ If you want more details: the Cartan matrix, which expresses the simple roots in the basis of fundamental weights, is an M-matrix- en.m.wikipedia.org/wiki/M-matrix. Thus the inverse of the Cartan matrix, which expresses the fundamental weights in the basis of simple roots, has all nonnegative entries, implying exactly the inclusion of cones you want. To see that the Cartan matrix is an M-matrix, you can use the fact that the Weyl vector $\rho$ is equal to the half-sum of the positive roots as well as the sum of the fundamental weights. $\endgroup$
    – Sam Hopkins
    Commented Jul 24, 2018 at 13:14
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    $\begingroup$ (In fact an even stronger statement is true: the Cartan matrix is positive definite.) $\endgroup$
    – Sam Hopkins
    Commented Jul 24, 2018 at 13:29

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