An irreducible root system is isomorphic to its dual

Question

This question is regarding root systems in lie algebra. I want to prove that an irreducible root system is isomorphic to its dual. Let $\Phi$ be a root system and let $\Phi^{v}=\{\alpha^v: \alpha \in \Phi\}$ where $\alpha^v=\frac{2\alpha}{(\alpha,\alpha)}$ be its dual. Is $\alpha \to \alpha^v$ is an isomorphism between root systems? But this map will not preserve Cartan integers. Here with this map $<\alpha^v, \beta^v>=<\beta, \alpha>$. Please help me.

Answer 1

What we have is really a duality, in the sense that $\Phi^{\vee\vee} \cong \Phi$. There is no reason for $\Phi$ and $\Phi^\vee$ to be isomorphic!

This is actually false: $B_n$ and $C_n$ are dual to each other, but they are not isomorphic for $n > 2$. (For the rest of irreducible root systems, your statement is true, you can go through the classification.)

Answer 2

An irreducible root system $\Phi$ is self-dual, i.e., isomorphic to its dual root system $\Phi^{\vee}$ if and only if it is of type $A_n,D_n,B_2,G_2,F_4,E_6,E_7,E_8$. In fact, if all roots have equal lengths then $\Phi\cong \Phi^{\vee}$. If there are two roots length, then it is not always true, namely not for the two remaining types $B_n$ and $C_n$ for $n>2$, which are dual to each other. This completes the list of irreducible root systems.