# Complexification of compact Lie Groups and complex algebraic linear reductive groups

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!

• Do you assume that $G$ is compact? – Mikhail Borovoi May 1 '18 at 14:24
• 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. – Venkataramana May 1 '18 at 14:37
• @MikhailBorovoi yes – user 123935 May 1 '18 at 15:10
• @Venkataramana thanks!! Not everything is clear to me yet. I will think about it – user 123935 May 1 '18 at 15:13