Dimension of maximal toral subalgebra and number of its roots Let $L$ be a semisimple Lie algebra, $H$ be a maximal toral subalgebra of $L$ and $\Phi$ be the set of roots relative to $H$. Then
$$L=H\bigoplus(\bigoplus\limits_{\alpha\in\Phi}L_{\alpha}).$$
We also know that if $\alpha\in\Phi$, then $-\alpha\in\Phi$ and $[L_{\alpha},L_{-\alpha}]=\mathbb{F}t_{\alpha}\in H$. Consequently, we have $$2\dim H\leq|\Phi|.$$
I want to know when the equality is attained.
 A: I assume your statement is
$$2 \dim H \le \mathrm{card}(\Phi)$$
(which is true by what you write), and you are asking when equality holds. Depending on how much theory you already have, this can be easy: One can show that the dimension of $H$ is equal to the rank of $\Phi$, which is the cardinality of any set of simple roots in $\Phi$; also, the real dimension of the Euclidean space our root system $\Phi$ lives in. That is also the index which you see in the usual naming of root systems $A_n, B_n, ..., F_4, G_2$.
Then from basic knowledge of root systems (or a case-by-case check in the classification) it is clear that $2\mathrm{rank}(\Phi)=\mathrm{card}(\Phi)$ if and only if every pair of roots $(\alpha, \beta)$ in $\Phi$ either satisfies $\alpha= \pm \beta$, or $\alpha \perp \beta$; and futher, that is the case if and only if $\Phi$ is a direct sum of copies of $A_1$, which makes your Lie algebra (assuming we are over an algebraically closed field $\mathbb F$) a direct sum of copies of $\mathfrak{sl}_2$.
In hindsight, one can probably avoid all that theory by showing: If there exist roots $\alpha, \beta \in \Phi$ with $\alpha+\beta \in \Phi$, then $t_{\alpha+\beta} \in \mathbb Ft_\alpha + \mathbb Ft_\beta$ and consequently $2\dim H < \mathrm{card} (\Phi)$. So again each pair of roots ... [see above].
PS: Note that from the decomposition you write down one also has
$$\dim L = \dim H + \mathrm{card}(\Phi)$$
and one might be able to infer from some other knowledge of Lie algebras that $\dim L = 3 \dim H$ only happens in the case of sums of copies of $\mathfrak{sl}_2$.
