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Let $\mathfrak{g}$ be a complex simple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g} $. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $ \mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question : can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without much calculations, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!

Add: As pointed by Sasha, where are 2 possible length of root. But I dont see why there are only 2 possible roots in $R\cap \overline{K}$, and why both of which in the wall.

Let $\mathfrak{g}$ be a complex simple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g} $. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$. Let $W$ be the Weyl group which generator by the reflection $s_\alpha$ of simple roots.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $ \mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question : can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without much calculations, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!

Add: As pointed by Sasha, where are 2 possible length of root.

Add 2 and answer: Thank to Sasha, I can complete the proof as short as possible.

We know that $W$ acts on $R$. Fact: every two elements $\alpha,\beta$ connected by a single line in Dykin diagram is in the same $W$-orbit, since $s_\alpha s_\beta(\alpha)=\beta$.

From this, we can get that: in the (resp. non-)simple-lanced case, we have only one (resp. two) $W$-orbit. This means $R\cap \overline{K}$ has one (resp. two) element. To show the element of $\alpha \in R\cap \overline{K}$ on the Wall, by Chevalley's lemma, it is equivalent to the stabilizer $W_\alpha$ is non trivial. This is consequence of $|W|>|R|$ (resp. $|W|>|R|-1$).

Let $\mathfrak{g}$ be a complex simple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g} $. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $ \mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question (Edit): can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without calculation, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!

Let $\mathfrak{g}$ be a complex simple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g} $. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $ \mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question : can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without much calculations, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!

Add: As pointed by Sasha, where are 2 possible length of root. But I dont see why there are only 2 possible roots in $R\cap \overline{K}$, and why both of which in the wall.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g} $. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $ \mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question (Edit): can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. In fact, for non simple Lie algebra, it's trivial that $R\cap K$ is empty. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without calculation, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!

Let $\mathfrak{g}$ be a complex simple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g} $. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $ \mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question (Edit): can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without calculation, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!