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Let $G$ be a complex algebraic group embedded into $\operatorname{GL}_{n}(\mathbb{C})$. A criterion for $G$ to be reductive is the following. Let $\mathfrak{g}$ be the Lie algebra of $G$, and let $B$ be the trace form of $\mathfrak{gl}_{n}(\mathbb{C})$ restricted to $\mathfrak{g}$: $B(x,y) = \operatorname{Tr}(xy)$. Then $G$ is reductive if and only if $B$ is non-degenerate.

This criterion was used in an answer to one of my previous questions. I think I also know how to prove it. It follows because the Lie algebra of the unipotent radical, $\mathfrak{r}$, is the kernel of $B$. The reason for this is roughly as follows: using a Levi decomposition, the Lie algebra $\mathfrak{g}$ decomposes (as a vector space) into $\mathfrak{g} = \mathfrak{r} \oplus \mathfrak{z} \oplus \mathfrak{s}$, where $\mathfrak{z}$ is abelian (the Lie algebra of a torus) and $\mathfrak{s}$ is semisimple. Since $\ker(B)$ is solvable, it must be contained in $\mathfrak{r} \oplus \mathfrak{z}$, which is the radical. But since $\mathfrak{z}$ is the Lie algebra of a torus, $B$ restricts to $\mathfrak{z}$ to be positive definite. On the other hand, we can choose a basis so that $\mathfrak{r} \oplus \mathfrak{z}$ lies in the upper triangular matrices, and this implies that $\mathfrak{r} \subseteq \ker B$.

I haven't been able to find this result anywhere in the literature. I am wondering if anyone knows where I can find it?

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    $\begingroup$ Bourbaki: Lie Groups and Lie Algebras, Chapter 1, Section 4, Proposition 5, p. 56 $\endgroup$
    – F Zaldivar
    Jul 7, 2022 at 13:28
  • $\begingroup$ @FZaldivar, since this is specifically a reference request, I think that is an answer! $\endgroup$
    – LSpice
    Jul 7, 2022 at 13:31
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    $\begingroup$ @LSpice: I'll post it as an answer. $\endgroup$
    – F Zaldivar
    Jul 7, 2022 at 13:38

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Bourbaki: Lie Groups and Lie Algebras, Chapter 1, Section 4, Proposition 5, p. 56.

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  • $\begingroup$ I don't mean to be pedantic, but this doesn't have precisely the same statement. It's phrased entirely for Lie algebras, in which case the statement is false. For example, $\mathbb{C}$ is reductive, but it can be embedded in $\mathfrak{gl}_{2}$ as strictly upper triangular, so that the trace form vanishes. It's only the representations induced from group representations that are guaranteed to have the desired property. $\endgroup$ Jul 7, 2022 at 13:47
  • $\begingroup$ Do you happen to have the French-language reference for this? For me, in the French-language version, Chapitre 1, Section 4, Proposition 5 is on p. 60—although of course page numbering will vary between the languages—and is a result about nilpotent Lie algebras. But I always have a horrible time navigating the Byzantine Bourbaki numbering. $\endgroup$
    – LSpice
    Jul 7, 2022 at 13:59
  • $\begingroup$ @LSpice: I have the English version, that is why the page numbering is different. $\endgroup$
    – F Zaldivar
    Jul 7, 2022 at 14:14
  • $\begingroup$ @unknownymous: Using that in characteristic zero the connected closed subgroups of $G$ are in correspondence with the Lie subalgebras of $\text{Lie}(G)$ reduces the group statement to the Lie algebra one. But I don't have a reference for this part and I am a little unease about it. $\endgroup$
    – F Zaldivar
    Jul 7, 2022 at 14:20
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    $\begingroup$ @FZaldivar: What you say is true, but I don't think the Lie algebra alone can detect whether the group is reductive. For example, $\mathbb{C}$ is the Lie algebra of a reductive group ($\mathbb{C}^*$), and a unipotent group ($\mathbb{C}$). So I think it's key that in the criterion, $G$ is an algebraic group which is embedded into the general linear group. $\endgroup$ Jul 7, 2022 at 15:08

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