The adjoint representation of a Lie group

Question

Let $G$ be a Lie group and $\text{Ad}(G)$ denote its adjoint representation i.e. the adjoint action of the group $G$ on its Lie algebra $\mathfrak{g}$. The adjoint representation is a real $G$-representation.

Question 1: What are some examples of Lie groups $G$ such that $\text{Ad}(G)$ is a representation of complex type? What I mean by 'representation of complex type' is that along with being a $G$-representation of real type, there exists an endomorphism $J:\mathfrak{g}\rightarrow\mathfrak{g}$, such $J^2=-1$ and $J$ commutes with the adjoint action of $G$.

Question 2: What about examples, when the underlying group $G$ is a compact Lie group?

PS - An obvious restriction on the choice of such a $G$ would be that it would have to be even dimensional as $\text{dim}(G)=\text{dim}(\mathfrak{g})$. Any example of families of such groups or literature recommendations would be appreciated.

Answer 1

For Question 2, there are only two kinds of such examples: the case where $G$ is commutative and even-dimensional, and the case where all simple factors of $G$ come in isomorphic pairs. (In the former case, the action of $G$ is trivial, so we may arbitrarily put a complex structure on $\DeclareMathOperator\Lie{Lie}\Lie(G)$. In the latter case, for each pair of isomorphic simple factors $G_1$ and $G_2$, given any choice of $G$-equivariant isomorphism $J_{12} : \Lie(G_1) \to \Lie(G_2)$, we may (uniquely) choose $J_{21} = -J_{12}^{-1}$, and then piece these various maps together to make $J$.)

If we are not in either of those cases, then there is some simple factor of $G$ whose Lie algebra is preserved by $J$, so it suffices to consider $G$ (compact and) simple. Then the adjoint representation is absolutely irreducible, so there is at most one $G$-invariant bilinear form, up to scalars. The Killing form $\kappa$ is a non-$0$, real-valued such form, and $\kappa_J \mathrel{:=} \kappa(J\cdot, \cdot)$ is another, so $\kappa_J$ equals $\lambda\kappa$ for some real scalar $\lambda$. But then $-\kappa = \kappa_{J^2}$ equals $\lambda\kappa_J = \lambda^2\kappa$, so $\lambda^2$ equals $-1$, which is a contradiction.

In the direction of Question 1 (though not answering it), the same argument ruling out complex-type adjoint representations works for any semisimple, linear Lie group $G$, as long as its simple factors are absolutely simple, and do not come in isomorphic pairs. A simple factor may instead be (the group of $\mathbb R$-points of) the restriction of scalars of that marvellous combination of adjectives, a simple, complex algebraic group, in which case, as @MikhailBorovoi points out in a comment, the adjoint representation on the corresponding factor of the Lie algebra will be of complex type.

Answer 2

Suppose $V$ is an irreducible real representation of any group $G$, and it has an operator $J$ commuting with $G$ and satisfying $J^2=-Id$.

Suppose $V_\mathbb C=V\otimes\mathbb C$ is irreducible. Then $J_\mathbb C=J\otimes\mathbb C$ commutes with the action of $G$; so by Schur's Lemma $J_\mathbb C$ is a scalar; then $J^2=-I\Rightarrow J_\mathbb C=i*Id$. But then $J$ does not preserve the real vector space $V$, a contradiction. So $V$ cannot be "real" according to your definition. (This is more or less @LSpice's argument, emphasizing it has nothing to do with the adjoint representation.)

In the case of the adjoint representation of a simple Lie group $G$, $V_\mathbb C$ fails to be irreducible precisely when $G$ is the restriction of scalars as discussed by @MikhaelBorovoi and @LSpice. (BTW you probably should say $G$ is semisimple, for example you surely don't want $G$ to be abelian.)

I think the terminology "complex type" for a representation on a real vector space is not standard, and potentially confusing. Usually we talk about a representation on a complex vector space being real or quaternionic, having to do with an anti-holomorphic map $J$ with $J^2=-1$. If I'm wrong I'm sure someone will correct me.