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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.

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  • $\begingroup$ $t_\alpha\in L_i$, $t_\beta\in L_j$. Doesn't that already imply that the composition of $\operatorname{ad}(t_\alpha)$ and $\operatorname{ad}(t_\beta)$ is the zero map? $\endgroup$ Commented Jan 3, 2019 at 7:50
  • $\begingroup$ @JyrkiLahtonen I'm probably missing something obvious, but why must we have $t_{\alpha}\in L_i$ and $t_{\beta}\in L_j$? $\endgroup$
    – Zilliput
    Commented Jan 3, 2019 at 7:59
  • $\begingroup$ 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 ? $\endgroup$ Commented Jan 3, 2019 at 18:37
  • $\begingroup$ @TorstenSchoeneberg As far as I can tell, no. $\endgroup$
    – Zilliput
    Commented Jan 3, 2019 at 18:58

1 Answer 1

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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, $t_{\beta}\in L_j$.

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