A reductive group is the complexification of a compact subgroup even if not connected?

Question

The definition of a linear algebraic complex reductive group is sometimes using the connectedness hypothesis for the complex algebraic group sometimes not.

Here I use the following definition : a complex affine linear algebraic group $G$ is said to be reductive if the connected component of the identity of its unipotent radical is trivial. I am aware that the linear hypothesis is somewhat redundant.

Is the well known following theorem still true without any connectedness hypothesis (for the complex algebraic groups or the real compact ones, since being connected or irreducible is the same for the algebraic group)?

There is an equivalence of categories between the category of smooth representations of a compact real Lie group and and the category of algebraic representations of a complex linear algebraic reductive groups.

Or the following theorem?

A linear complex algebraic group is reductive if and only if it has an algebraic compact real form, which is then a maximal compact real Lie group

Or the following one?

The complexification of a compact real Lie group is a reductive linear algebraic group

The only place I found these theorems without any hypothesis on the connectedness is in Lie Groups and Algebraic Groups by Onishchik & Vinberg, but their definition of reductiveness is based on the reductiveness of the Lie algebra, which does not seem clearly to exclude the case of the $(\mathbb{C}, +)$ group.

I would also like if the hypothesis is necessary a counter-example and if not a good reference.

Edit Reposted from MathSE https://math.stackexchange.com/questions/4963922/is-connectedness-necessary-for-a-reductive-group-to-be-the-complexification-of-a

Edit Actually in the book Representations of Compact Lie Groups of Tammo tom Dieck and Theodor Bröcker, a clear proof of the last theorem mentioned above is given without any hypothesis of connectivity, in cap. III.8.

Answer 1

Let $G$ be a complex, not necessarily connected reductive group and let $K$ be a maximal compact subgroup.

There is a natural function from algebraic representations of $G$ to continuous representations of $K$. Let's check it is an equivalence.

We already know the statement for the identity component $G^0$ of $G$ and the identity component $K^0$ of $K$.

YCor points out that Mostow proved in 1955 that $K$ meets every component of $G$, i.e. that the natural map $K/ K^0 \to G/G^0$ is an isomorphism.

The function from representations of $G$ to representations of $K$ is faithful. To check it is essentially surjective, i.e. that the natural map $\operatorname{Hom}_G(V,W) \to \operatorname{Hom}_K(V,W)$ is an isomorphism, we observe that

$$\operatorname{Hom}_G(V,W) = ( \operatorname{Hom}_{G^0}(V,W))^{G/G^0} = ( \operatorname{Hom}_{G^0}(V,W))^{K/K^0}= ( \operatorname{Hom}_{K^0}(V,W))^{K/K^0}= \operatorname{Hom}_K(V,W)$$ where the first and last equalities are obvious and the middle two use the two facts we checked.

To check that it is faithful, i.e. that every representation of $K$ extends to a representation of $G$, let $V$ be a representation of $K$. As a representation of $G^0$, we know by the connected case that $V$ extends to a representation of $G^0$. Then the restriction of $\operatorname{Ind}_{G^0}^G V$ to $K$ is $\operatorname{Ind}_{K^0}^K V$ and contains $V$ as a summand over $K$. The idempotent projector to this summand is a $K$-invariant homomorphism, hence a $G$-invariant homomorphism, so this summand is a $G$-invariant subrepresentation and in particular is a representation of $G$ extending $V$, as desired.

To reverse this process, passing from a compact group to a complex group, we may simply take the Tannakian group of the category of finite-dimensional representations of $K$, which has the equivalence of categories property by definition. This is how I would define the complexification in the disconnected context, although there may be other definitions which are also helpful.