I've been reading 'Introduction to Lie algebras and representation theory' of Humphreys. I have two questions.

The first question is about Coroallary 8.2 in page 37. Corollary states that "The restriction of the Killing form to the maximal toral subalgebra $H$ is nondegenerate." Then, the book says that for each $\phi \in H^*$, there exists the unique $t_\phi \in H$ such that $\phi(h)=\kappa(t_\phi, h)$ for all $h\in H$. I think the Cor. 8.2 only gives a reason for the uniqueness of $t_\phi$, not existence. How can we know that every linear functional on $H$ is represented by the Killing form?

The second question is about the proof of the Proposition 8.3.-(a). (This question is just related to the linear algebra.) The book says that "If $\Phi$ fails to span $H^*$, then (by duality) there exits nonzero $h \in H$ such that $\alpha(h)=0$ for all $\alpha \in \Phi$." Here's what I understand.

Reduce $\Phi$ to be linealry independenet, say $\Phi'$. Consider a subset of $H$ which is dual to $\Phi'$. Since $\Phi'$ cannot span $H^*$, this subset is not a basis for $H$. Extend it to be a basis for $H$. For the basis elements $h_i$'s which are added while extending, $\alpha(h_i)=0$ for all $\alpha \in \Phi'$, and hence for all $\alpha \in \Phi$.

Am I correct? Or is there any easier way to understand it?

Any help would be appreciated. Thanks.


1 Answer 1


You have $\dim H^*=\dim H$. So, since the map$$\begin{array}{ccc}H&\longrightarrow&H^*\\t&\mapsto&\kappa(\cdot,h)\end{array}$$is injective, it is also surjective.

And if $\langle\Phi\rangle\ne H^*$, take $\beta\in H^*\setminus\{0\}$ such that $(\forall\alpha\in\langle\Phi\rangle):\kappa(\alpha,\beta)=0$. Now, take $h\in H$ such that $\kappa(\cdot,h)=\beta$. Then, if $\alpha\in\langle\Phi\rangle$, $\alpha(h)=\kappa(\alpha,\beta)=0$.

  • $\begingroup$ I'm afraid your notations deviate somewhat from the OP's. Is the first map supposed to be $t \mapsto \kappa(t, \cdot)$? And after, I think your $\phi$ is OP's $\kappa$, whereas they use $\phi$ for elements of $H^\ast$. $\endgroup$ Apr 17, 2021 at 14:45
  • $\begingroup$ @TorstenSchoeneberg I've edited my answer. Thank you. $\endgroup$ Apr 17, 2021 at 15:56
  • $\begingroup$ Thanks to both of you. You solved my problems. Thanks! $\endgroup$
    – user
    Apr 18, 2021 at 6:15

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