(This is a follow-up to [this question](https://mathoverflow.net/questions/333846/shimura-varieties-and-connected-components) of mine.)


Is there an example of a connected reductive algebraic group $G$ over $\mathbb{R}$ such that:


* $G$ is not isomorphic to a product $G\_1 \times G\_2$ of smaller groups (*isogenous* to a product is OK)
* $G$ is not a torus,
* the quotient of $G$ by a maximal compact-mod-centre subgroup has a complex structure,
* $Z\_G(\mathbb{R})$ is not contained in the identity component of $G(\mathbb{R})$?


The condition $Z\_G(\mathbb{R}) \subseteq G(\mathbb{R})^\circ$ is vacuously satisfied if $G$ is adjoint, because then $Z\_G = \{1\}$; but it is also vacuously satisfied if $G$ is semisimple and simply-connected, because then $G(\mathbb{R})$ is connected as a Lie group by a theorem of Cartan. So any example would have to lie somewhere in between the two (which makes me wonder if there are any examples at all).


PS: Of course $GL\_3$ is an example if the "complex structure" condition is dropped. 



Connected components of real Lie groups
There is NO such example. 


Note that any semisimple algebraic ${\mathbb{R}}$-group $H$ of Hermitian type has a *compact* (anisotropic) maximal torus. 
Indeed, by a definition of a group of Hermitian type 
(see, e.g., [Deligne, Travaux de Shimura](http://www.numdam.org/item/SB_1970-1971__13__123_0/), condition (1.5.3) on page 128), $H$ is an *inner form* of a compact algebraic $\Bbb R$-group $K$, namely, $H=\,\_\sigma K$, where $\sigma={\rm inn}(x)\in {\rm Aut}(K)$, $x^2=1$, $x\in K^{\rm ad}(\Bbb R)$,
$K^{\rm ad}=K/Z\_K$. 
Let $T\_K\subset K$ be a maximal $\Bbb R$-torus such that $T\_K^{\rm ad}(\Bbb R)$ contains $x$, where $T\_K^{\rm ad}=T\_K/Z\_K$.
Then $T\_K=\,\_\sigma T\_K\subset \,\_\sigma K=H$ is a compact maximal torus of $H$. 



> 
> **Theorem.** Let $G$ be a (connected) reductive ${\mathbb{R}}$-group. Write $G^{\rm der}=[G,G]$. Assume that $G^{\rm der}$ has a *compact* maximal torus $T^{\rm der}$.
>  If the image of $Z\_G({\mathbb{R}})$ in $\pi\_0(G({\mathbb{R}}))$ is nontrivial,
>  then there exists a nontrivial split ${\mathbb{R}}$-subtorus $T'\subset Z\_G$ and
>  a reductive ${\mathbb{R}}$-subgroup $G''\subset G$ such that $G=T'\times\_{\mathbb{R}} G''$.
> 
> 
> 


*Proof.* 
Write $T=Z\_G\cdot T^{\rm der}$; then $T$ is a maximal torus of $G$.
We have maps
$$ Z\_G({\mathbb{R}})\to \pi\_0(T({\mathbb{R}}))\to \pi\_0(G({\mathbb{R}})).$$
Since the image of $Z\_G({\mathbb{R}})$ in $\pi\_0(G({\mathbb{R}}))$ is nontrivial, we have
$ \pi\_0(T({\mathbb{R}}))\neq 1$.


Write $T=T\_0\times\_{\mathbb{R}} T\_1\times\_{\mathbb{R}} T\_2$,
where $T\_0$ is a compact ${\mathbb{R}}$-torus, $T\_1$ is a split ${\mathbb{R}}$-torus, and $T\_2\simeq (R\_{{\mathbb{C}}/{\mathbb{R}}}{\mathbb G}\_{m,{\mathbb{C}}})^{n}$.
We have $\pi\_0(T\_0({\mathbb{R}}))=1$, and $\pi\_0(T\_2({\mathbb{R}}))=1$.
Since $\pi\_0(T({\mathbb{R}}))\neq 1$, we conclude that $T\_1\neq 1$.
Note that the ${\mathbb{R}}$-torus $T/Z\_G$ is isogenous to $T^{\rm der}$, and hence, compact.
It follows that $T\_1\subset Z\_G$.


Set $T'=T\_1$ and $T''=T\_0\times\_{\mathbb{R}} T\_2$; then clearly $T=T'\times\_{\mathbb{R}} T''$.
Set $G''=T''\cdot G^{\rm der}$.
Then 
$$T'\cdot G''=T'\cdot T''\cdot G^{\rm der}=T\cdot G^{\rm der}=G\quad \text{and}\quad T'\cap G''=T'\cap T''=1.$$
Since $T'\subset Z\_G$, it commutes with $G''$.
Thus $G=T'\times\_{\mathbb{R}} G''$, as required.


