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I am also trying to solve exercise 6.4 from "Kirillov: An Introduction to Lie Groups and Lie Algebras", just like here: Root decomposition implies semisimple, but another version.

Let $\mathfrak{g}$ be a complex Lie algebra which has a root decomposition \begin{align} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in R} \mathfrak{g}_\alpha \end{align} where $R$ is a finite subset in $\mathfrak{h}^* \setminus \{ 0 \}$, $\mathfrak{h}$ is commutative and for $h \in \mathfrak{h}$, $x \in \mathfrak{g}_\alpha$, we have $[h,x]=\langle h,\alpha \rangle x$.

And for every $\alpha\in R\cup\{0\}$, the Killing form give a non-degenerate pairing $\mathfrak{g}_{\alpha}\otimes \mathfrak{g}_{-\alpha}\rightarrow \mathbb{C}$.

Show that then $\mathfrak{g}$ is semisimple, and $\mathfrak{h}$ is a Cartan subalgebra.

The way we want to show $\mathfrak{g}$ is semisimple is to show it has the non-degenerate Killing form. To show this, assume $\kappa$ is degenerate, that is, there exists some $h\in\mathfrak{h}$ and $x_{\alpha}\in \mathfrak{g}_{\alpha}$ such that $$\kappa\left(h+\sum_{\alpha} (x_{\alpha}+x_{-\alpha}),y\right)=0$$ for every $y\in\mathfrak{g}$.

What should I do next to get contradiction? I think it does not work if we take $y=x_{\alpha}$ since we just have $\kappa$ is non-degenerate on $\mathfrak{g}_{\alpha}\otimes \mathfrak{g}_{-\alpha}$.

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Aolong Li
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1 Answers1

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Your assumptions say that $\kappa$ is non degenerate on $\mathfrak h$, and for each $\alpha\in R$, you have $-\alpha\in R$ and $\kappa$ is non--degenerate on $\mathfrak g_{\alpha}\times\mathfrak g_{-\alpha}$. The latter condition readily implies that $\kappa$ is non-degenerate as a bilinear form on the subspace $\mathfrak g_{\alpha}\oplus\mathfrak g_{-\alpha}$. Now choose $R^+\subset R$ such that $R$ is the disjoint union of $R^+$ and $\{-\alpha:\alpha\in R^+\}$ (for the current purpose, you don't even have to make the usual assumption that this is closed under addition). Then you can write the root decomposition as $\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in R^+}(\mathfrak g_{\alpha}\oplus\mathfrak g_{-\alpha})$, and from above we see that the restriction of $\kappa$ to each of the summands is non-degenerate. Thus it suffices to shows that the different summands are perpendicular with respect to $\kappa$ to conclude that $\kappa$ is non-degenerate on $\mathfrak g$.

But this is an elementary consequence of the definition of $\kappa$ as $\kappa(X,Y)=tr(ad_X\circ ad_Y)$. For $X\in\mathfrak g_{\alpha}$ and $Y\in\mathfrak g_{\beta}$ with $\alpha,\beta,\gamma \in R\cup\{0\}$ (and with $\mathfrak g_0:=\mathfrak h$), we see that $ad_Y$ maps $\mathfrak g_{\gamma}$ to $\mathfrak g_{\beta+\gamma}$, so $ad_X\circ ad_Y$ maps $\mathfrak g_{\gamma}$ to $\mathfrak g_{\alpha+\beta+\gamma}$. Unless $\alpha+\beta=0$, this means that in the matrix representation adapted to the root decompotion, only off-diagonal entries of the matrix $ ad_X\circ ad_Y$ can be non-zero and thus $\kappa(X,Y)=0$ unless $\beta=-\alpha$.

Andreas Cap
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