The intersection of a maximal toral subalgebra with a simple ideal of a Lie algebra is a maximal toral subalgebra of the simple ideal.

I'm reading Humphreys' Introduction to Lie Algebras and Representation Theory and I have a question about Corollary 14.1, which reads:

Humphreys Corollary 14.1. Let $$L$$ be a semisimple Lie algebra, with maximal toral subalgebra $$H$$ and root system $$\Phi$$. If $$L = L_1\oplus\cdots\oplus L_t$$ is the decomposition of $$L$$ into simple ideals, then $$H_i = H\cap L_i$$ is a maximal toral subalgebra of $$L_i$$, and the corresponding (irreducible) root system $$\Phi_i$$ may be regarded canonically as a subsystem of $$\Phi$$ in such a way that $$\Phi=\Phi_1\cup\cdots\cup\Phi_t$$ is the decomposition of $$\Phi$$ into its irreducible components.

I understand most of this. What I'm struggling to understand is why the $$\Phi_i$$ must be pairwise orthogonal. That is, if $$\alpha\in\Phi_i$$, $$\beta\in\Phi_j$$, $$i\neq j$$, why is $$(\alpha, \beta)=0$$?

By definition, $$(\alpha, \beta)=\kappa(t_{\alpha}, t_{\beta})$$, where $$\kappa$$ denotes the Killing form and, for $$\gamma\in H^*$$, $$t_{\gamma}$$ is the unique element of $$H$$ such that $$\gamma(h)=\kappa(t_{\gamma}, h)$$ for all $$h\in H$$. So, we need to show that $$\alpha(t_{\beta})=0$$. Since $$\alpha(H_k)=0$$ for $$k\neq i$$, we really only need to show that $$\alpha$$ annihilates $$h_i$$, where $$t_{\beta}=h_1+\cdots +h_t$$, $$h_k\in H_k$$. From here, I'm not sure how to proceed.

• $t_\alpha\in L_i$, $t_\beta\in L_j$. Doesn't that already imply that the composition of $\operatorname{ad}(t_\alpha)$ and $\operatorname{ad}(t_\beta)$ is the zero map? Commented Jan 3, 2019 at 7:50
• @JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{\alpha}\in L_i$ and $t_{\beta}\in L_j$? Commented Jan 3, 2019 at 7:59
• Is your original question, or the one in the comment, answered by one of these: math.stackexchange.com/q/1357074/96384 and math.stackexchange.com/q/2982139/96384 ? Commented Jan 3, 2019 at 18:37
• @TorstenSchoeneberg As far as I can tell, no. Commented Jan 3, 2019 at 18:58

As @JyrkiLahtonen commented, it suffices to show that $$t_{\alpha}\in L_i$$ and $$t_{\beta}\in L_j$$, since this implies that $$\operatorname{ad}t_{\alpha}\operatorname{ad}t_{\beta}(x)=[t_{\alpha}[t_{\beta}x]]\in L_i\cap L_j=0$$ for all $$x\in L$$, i.e., $$\operatorname{ad}t_{\alpha}\operatorname{ad}t_{\beta}=0$$, so $$(\alpha, \beta)=\kappa(t_{\alpha}, t_{\beta})=\operatorname{tr}(\operatorname{ad}t_{\alpha}\operatorname{ad}t_{\beta})=\operatorname{tr}(0)=0.$$
To see that $$t_{\alpha}\in L_i$$, let $$\kappa_i$$ denote the Killing form of $$L_i$$. By Humphreys Lemma 5.1, $$\kappa_i=\kappa\vert_{L_i\times L_i}$$. By Humphreys Corollary 8.2, $$\kappa\vert_{H\times H}$$ and $$\kappa_i\vert_{H_i\times H_i}$$ are nondegenerate. This allows us to identify $$H$$ with $$H^*$$, by associating to $$\gamma\in H^*$$ the unique element $$t_{\gamma}\in H$$ such that $$\gamma(h)=\kappa(t_{\gamma}, h)$$ for all $$h\in H$$. Similarly, we can identify $$H_i$$ with $$H_i^*$$, by associating to $$\delta\in H_i^*$$ the unique element $$u_{\delta}\in H_i$$ such that $$\delta(h_i)=\kappa_i(u_{\delta}, h_i)$$ for all $$h_i\in H_i$$. With this in mind, $$u_{\alpha}\in H_i$$ and $$\alpha(h_i)=\kappa_i(u_{\alpha}, h_i)$$ for all $$h_i\in H_i$$. However, by an argument similar to that in the previous paragraph, $$\kappa(L_i, L_k)=0$$ for $$i\neq k$$, so given $$h=h_1+\cdots+h_t\in H$$, $$h_k\in H_k$$, we have $$\alpha(h)=\alpha(h_i)=\kappa_i(u_{\alpha}, h_i)=\kappa\vert_{L_i\times L_i}(u_{\alpha}, h_i)=\kappa(u_{\alpha}, h_i)=\kappa(u_{\alpha}, h).$$ Then, by uniqueness, $$t_{\alpha}=u_{\alpha}\in H_i\subseteq L_i$$. Similarly, $$t_{\beta}\in L_j$$.