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Let $\mathfrak g$ be a semisimple Lie algebra over $\mathbb{C}$ with Cartan subalgebra $\mathfrak h$. Then the Killing form on $\mathfrak g$ defined by $(X,Y) = \operatorname{Tr}(\operatorname{ad}(X) \circ \operatorname{ad}(Y))$ is nondegenerate (this can also be taken as the definition of a semisimple Lie algebra).

I'm trying to understand the following:

1 . Why is the restriction of the Killing form to $\mathfrak h$ still nondegenerate?

2 . Why do we have $(h,h) \neq 0$ for all $h \in \mathfrak h$?

3 . Let $\Phi \subseteq \mathfrak h^{\ast}$ be the roots of $\mathfrak h$ in $\mathfrak g$. Let $V$ be the $\mathbb{Q}$-span of $\Phi$, and define a bilinear form on $V$ by $(v,w) = (h_v,h_w)$, where $h_v \in \mathfrak h$ is the inverse of $v$ under the $\mathbb{C}$-vector space isomorphism $ \mathfrak h \rightarrow \mathfrak h^{\ast}, h \mapsto (-,h)$. Why does this define a symmetric, positive definite (hence nondegenerate) bilinear form on $V$?

The third statement is a claim made on page 1 of Steinberg, Lectures on Chevalley Groups.

Proof of 1: the linear transformations $\operatorname{ad}(h) : h \in \mathfrak h$ are simultaneously diagonalizable, with diagonal matrix $\operatorname{diag}(0, \ldots , 0, \alpha(h) : \alpha \in \Phi)$, where the number of zeros is the dimension of $\mathfrak h$. So the Killing form on $\mathfrak h$ is

$$(h,h') = \sum\limits_{\alpha \in \Phi} \alpha(h)\alpha(h') = 2\sum\limits_{\alpha \in \Phi^+} \alpha(h)\alpha(h')$$

where $\Phi^+$ is a given set of positive roots with base $\Delta$. Let $h_1, \ldots , h_t$ be a dual basis to simple roots $\alpha_1, \ldots , \alpha_t$. Fix $1 \leq i \leq t$, and each $n \geq 1$, let $m_n$ be the number of positive roots with the coefficient $n\alpha_i$. Then

$$(h,h_i) = 2(m_1 + 2m_2 + \cdots)\alpha_i(h)$$

So if $(h,h') = 0$ for all $h' \in \mathfrak h$, then $\alpha_i(h) = 0$ for all $i$ $\Rightarrow h = 0$.

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    $\begingroup$ When you write \textrm{Tr} instead of \operatorname{Tr} then you don't automatically get proper spacing in things like $A\operatorname{Tr} B$ and $A\operatorname{Tr}(B),$ instead seeing $A\textrm{Tr} B$ and $A\textrm{Tr}(B).$ I include both examples to show the context-dependent nature of the spacing: there's less space when $\Big($parentheses$\Big)$ are there. In LaTeX, before \begin{document} you can write \newcommand{\d}{\operatorname{diag}} and then just use \d subsequently in the document and see $\operatorname{diag} in the finished result. $\endgroup$ – Michael Hardy Dec 24 '17 at 1:03
  • $\begingroup$ This is proved, among many other places, in Carter's book on Lie algebras. Have you looked in any of the standard textbooks on the subject? $\endgroup$ – Mariano Suárez-Álvarez Dec 24 '17 at 2:28
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    $\begingroup$ Maybe I'm confused but did you just give a proof of the statement you asked about? $\endgroup$ – Tim kinsella Dec 24 '17 at 6:46
  • $\begingroup$ Only the first one $\endgroup$ – D_S Dec 24 '17 at 6:59
  • $\begingroup$ The other two statements follow from the fact the formula you have for $(h, h')$ and the fact that the roots are real valued on $V^\ast$ $\endgroup$ – Tim kinsella Dec 25 '17 at 23:53
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You have proved the first statement as part of the question. The second statement is not true and cannot be true: The Killing form is a complex bilinear form, and such a form always has isotropic vectors. you want to state this only for the subspace $\mathfrak h_0\subset \mathfrak h$ on which all roots have real values. It is part of the analysis of the root decompostion to prove that this is positive definite and this is a slightly involved sequence of arguments that you can find in standard text. The third part is a slight variation of this.

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