# How the roots and weights changed under a folding?

Let $e_1,e_2,e_3,f_1,f_2,f_3$ be the generators of the Chevalley basis of the Lie algebra $sl_4$. Let $e_1' = e_1+e_3$, $e_2'=e_2$, $f_1'=f_1+f_3$, $f_2'=f_2$. Then the subalgebra generated by $e_1', e_2', f_1', f_2'$ is isomorphic to the Lie algebra $sp_4$. Let $\alpha_1, \alpha_2, \alpha_3$ be the simple roots in the root system of $sl_4$ and $\omega_1, \omega_2, \omega_3$ the fundamental weights. How to express the simple roots $\alpha_1', \alpha_2'$ and fundamental weighs $\omega_1', \omega_2'$ of $sp_4$ using the simple roots and fundamental weights of $sl_4$? I think that $\alpha_2'=\alpha_2$, $\omega_2' = \omega_2$. What about $\alpha_1'$ and $\omega_2'$? Thank you very much.

• Since there's no such thing as the generators of a Chevalley basis—there's always at least some sign issues—I think that you haven't pinned it down uniquely; but it is certainly possible to realise $\mathsf C_2$ inside $\mathsf A_3$ with basis $\{\alpha_1' = -\tfrac1 2\alpha_1 - \alpha_2 + \tfrac1 2\alpha_3, \alpha_2' = \alpha_2\}$ or with basis $\{\alpha_1' = -\tfrac3 2\alpha_1 - \alpha_2 - \tfrac1 2\alpha_3, \alpha_2' = \alpha_2\}$. In either case, $\omega_1' = \alpha_1' + \alpha_2'$ and $\omega_2' = \tfrac1 2\alpha_1' + \alpha_2'$. – LSpice Sep 20 '18 at 20:39