Equality on pg. 40 of Humphreys's Lie Algebras, $\kappa(t_\lambda,t_\mu)=\sum_{\alpha\in\Phi}\alpha(t_\lambda)\alpha(t_\mu)$?

Question

I don't understand part of an equality on page 40 of Humphreys's book on Lie Algebras.

Suppose $L$ is a semi-simple Lie algebra over an algebraically closed field of characteristic $0$, and $H$ a maximal toral subalgebra. Let $\kappa$ denote the Killing form, so for each $\phi\in H^\ast$, there is a unique $t_\phi\in H$ such that $\phi(h)=\kappa(t_\phi,h)$ for all $h\in H$. The equation says for any $\lambda,\mu\in H^\ast$, $$ \kappa(t_\lambda,t_\mu)=\sum\alpha(t_\lambda)\alpha(t_\mu) $$ where the sum is over the set of roots $\alpha\in\Phi$.

By definition of $t_\lambda$, $\kappa(t_\lambda,t_\mu)=\lambda(t_\mu)$, and since the roots span $H^\ast$, I can write $\lambda=\sum_{\alpha\in\Phi}c_\alpha\alpha$ for coefficients $c_\alpha$. Then $$ \lambda(t_\mu)=\sum_{\alpha\in\Phi}c_\alpha\alpha(t_\mu) $$ so this would work out if $c_\alpha=\alpha(t_\lambda)$. Is this true, or is there another way to see this?

Answer 1

By definition of Killing form $$ \kappa(t_{\lambda},t_{\mu})= \operatorname{Tr}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu}) $$ where $\operatorname{ad}\colon L\to\mathfrak{gl}(L)$, so, being $\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu}\in\mathfrak{gl}(L)$, you can split the trace over the well known decomposition $L=H\oplus\bigoplus_{\alpha\in\Phi}L_{\alpha}$: $$ \operatorname{Tr}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu})= \operatorname{Tr}_{|_{H}}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu})+ \sum_{\alpha\in\Phi}\operatorname{Tr}_{|_{_{L_{\alpha}}}}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu})\;\;. $$ At this point observe that if $h\in H$, then $\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu}(h)=[t_{\lambda},[t_{\mu},h]]=0$ (because $t_{\lambda},t_{\mu}\in H$ which is maximal toral subalgebra, thus abelian) hence $\operatorname{Tr}_{|_{H}}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu})=0.$

Consider then $x\in L_{\alpha}$: we have that $\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu}(x)=[t_{\lambda},[t_{\mu},x]]=[t_{\lambda},\alpha(t_{\mu})x]=\alpha(t_{\mu})\alpha(t_\lambda)x$, thus $\operatorname{Tr}_{|_{L_{\alpha}}}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu})=\alpha(t_{\mu})\alpha(t_\lambda)$.

Thus the last sum written turns into $$ 0+\sum_{\alpha\in\Phi}\alpha(t_{\mu})\alpha(t_\lambda) $$ ad desired.

Answer 2

I think you can do it in the following way: you know that $L$ decomposes as $L=H\oplus\bigoplus_{\alpha\in\Phi}L_\alpha$. The Killing form is $\kappa(t_\lambda,t_\mu)=\operatorname{tr}(\operatorname{ad} t_\lambda\operatorname{ad}t_\mu)$. But $H$ is abelian and nilpotent, so $t_\mu$ and $t_\lambda$ commute and are ad-nilpotent, so on $H$, $\operatorname{ad}t_\lambda$ and $\operatorname{ad}t_\mu$ are commuting, nilpotent operators, thus the trace of their restriction to $H$ is zero, so it suffices to find the trace of their composition on each root space $L_\alpha$.

However, each root space is $1$-dimensional, say with basis vector $x_\alpha\in L_\alpha$. It follows $$ \operatorname{ad} t_\lambda\operatorname{ad}t_\mu x_\alpha=[t_\lambda,[t_\mu,x_\alpha]]=[t_\lambda,\alpha(t_\mu)x_\alpha]=\alpha(t_\mu)\alpha(t_\lambda)x_\alpha. $$ So the trace on each root space $L_\alpha$ is $\alpha(t_\mu)\alpha(t_\lambda)$, and thus the trace on all of $L$ is $\sum_{\alpha\in\Phi}\alpha(t_\lambda)\alpha(t_\mu)$.

Answer 3

For any nilpotent Lie subalgebra $\mathfrak{h}\subseteq L$ we have $$ \kappa (h,h')=\sum_{\alpha\in \mathfrak{h}^{\ast}}\dim (L_{\alpha})\alpha (h)\alpha(h'), $$ which follows from Lie's theorem, giving a basis such that all $ad(h)$ on $L_{\alpha}$ are simultaneously upper triangular matrices with diagonal elements $\alpha(h)$. For $L$ semisimple, $\mathfrak{h}$ a Cartan subalgebra, and $\alpha\in \Phi$ it follows that $\dim (L_{\alpha})=1$, so that with $h=t_{\lambda}, h'=t_{\mu}$ it follows that $\kappa(t_{\lambda},t_{\mu})=\sum_{\alpha\in \Phi}\alpha(t_{\lambda})\alpha(t_{\mu})$.