4
$\begingroup$

I'm studying complexifications of compact Lie groups on "Representation of compact Lie groups- Dieck Brocker".

I found on internet that there is a bijection between complexifications of compact Lie groups and complex algebraic linear reductive groups.

In my book they show that given a representation $\ r:G \rightarrow \mathrm{GL}(n,\mathbb C)$ `there is a unique holomorphic representation $\ r_{\mathbb C} : G_{\mathbb C} \rightarrow \mathrm{GL}(n,\mathbb C)$ that is also injective. Then they consider $ \tilde{G}=r_{\mathbb C}(G_{\mathbb C})$ and prove that $\tilde{G}$ is homeomorphic to $ \tilde{G} \cap U(n) \times \tilde{G} \cap P(n)$ (through the map $(H,P) \rightarrow HP$). They also prove that $\tilde{G} \cap U(n)$ is a maximal compact subgroup of $\tilde{G}$ and $\tilde{G} \cap P(n)$ is homeomorphic to a Euclidean space of dimension $\dim(G)$.

My question is: I understand that $\tilde{G}$ is an algebraic linear group but I don't know how to prove that it is reductive (i.e. $G$ reductive if $\mathrm{Rad}_{\mathrm{u}}(G)$ is trivial, where $\mathrm{Rad}_{\mathrm{u}}(G)$ is the set of unipotent elements of $\mathrm{Rad}(G)$ with $\mathrm{Rad}(G)$ being the connected component of the unique normal solvable subgroup of $G$).

I would also like to know if the proof of the vice-versa needs a lot of preliminaries (at the moment I only know the definition of reductive group).

Thank you!

Improve this question
$\endgroup$
4
  • $\begingroup$ Do you assume that $G$ is compact? $\endgroup$
    – Mikhail Borovoi
    Commented May 1, 2018 at 14:24
  • 4
    $\begingroup$ replacing $G$ by its image under the representation $r$, you may assume that $r$ is faithful; since $G$ is compact, the rep is completely reducible. However, complete reducibility persists for the complexification $_{C}$. Thus the unipotent radical of $G_{C}$ acts on each irreducible component of the rep on $G_C$, and hence acts trivially: the unip radical has a fixed vector (Engel's therem), and hence the fixed vectors form an invariant subspace , hence the unip radical acts trivially on each irrep and hence on the whole space. Thus the unip radical is trivial. $\endgroup$
    – Venkataramana
    Commented May 1, 2018 at 14:37
  • $\begingroup$ @MikhailBorovoi yes $\endgroup$
    – user 123935
    Commented May 1, 2018 at 15:10
  • $\begingroup$ @Venkataramana thanks!! Not everything is clear to me yet. I will think about it $\endgroup$
    – user 123935
    Commented May 1, 2018 at 15:13

1 Answer 1

Reset to default
4
$\begingroup$

Your first question appears answered in the comments.

With regard to your second question, Maxime Bergeron has a well-written, well-referenced exposition of the characterization of complex reductive affine algebraic groups as complexifications of compact Lie groups titled Complex reductive algebraic groups (one can find a copy by searching "complex reductive algebraic groups bergeron").

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Just adding a comment that the above reference (link being dead now) goes to the article by Maxime Bergeron named "Complex reductive algebraic groups," which can probably be found at other sites of the author. $\endgroup$
    – Cheng-Chiang Tsai
    Commented Jun 3, 2021 at 16:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .