# Why do roots span dual space of maximal toral subalgebra?

Suppose $\Phi$ is the root system of a semi simple Lie algebra with maximal toral subalgebra $H$. I read that $\Phi$ spans $H^\ast$.

The Killing form on $H$ is nondegenerate, so $H\cong H^\ast$ by identifying $\phi\in H^\ast$ with the unique $t_\phi$ such that $\phi(t)=\kappa(t_\phi,t)$.

Every proof is by contradiction, beginning by saying if $\Phi$ does not span $H^\ast$, then there exists nonzero $h\in H$ such that $\alpha(h)=0$ for all $\alpha\in\Phi$. I understand the proof except for this line. Can anybody clarify why such $h$ must exist?

If we extend the simple roots into a basis $B=(b_i)$ of $H^{\ast}$ (possibly with additional vectors), then we can choose a dual basis $B'=(b_i')$ of $H$ such that $$b_i(b_j')=\delta_{ij}$$ If we pick $i$ so that $b_i$ is not one of the simple roots, then $\alpha(b_i')=0$ for all roots $\alpha$.
• Good answer, but the key ingredient here is that finite-dimensional vector spaces are reflexive (so that $H \cong H^{**}$), which you really should have stated explicitly. Commented Oct 1, 2016 at 5:41