Adjoint group of a reductive group Let $G$ be a (not necessarily quasi-split) reductive group over $\mathbb{Q}$. In order to use results from semi-simple Lie groups for the reductive group $G(\mathbb{R})$, it seems common to implicitly use the adjoint group of $G$. If I understand correclty, it is defined by the exact sequence of algebraic groups
$$
1 \rightarrow Z_G \rightarrow G \rightarrow G^{\operatorname{ad}} \rightarrow 1
$$
where $Z_G$ denotes the center of $G$. My questions are the following:

*

*What is the relation between $G^{\operatorname{ad}}(\mathbb{R})$ and $\operatorname{Ad}(G(\mathbb{R}))$, the image of $G(\mathbb{R})$ under the Adjoint representation? From what I understand, the latter is isomorphic to $G(\mathbb{R}) / C_{G(\mathbb{R})}\,(G(\mathbb{R})^\circ)$, the quotient modulo the centralizer of the connected component of $G(\mathbb{R})$. (see Wikipedia on the Adjoint representation)


*Is $G^{\operatorname{ad}}$ an adjoint semisimple group? (Def: $G$ is called adjoint, if for any connected group $H$, any isogeny $G \rightarrow H$ is an isomorphism. This implies for example that $G^{\operatorname{ad}}$ is a direct product of its simple factors).


*To compare group decompositions of both groups (like Iwasawa decomposition), one needs to fix maximal compact subgroups. So let $K \subset G(\mathbb{R})$ and $\tilde{K} \subset G^{\operatorname{ad}}(\mathbb{R})$ be maximal compact subgroups. Is there any way to get a relation between $K$ and $\tilde{K}$? Since $\operatorname{Ad}$ is continuous, the image $\operatorname{Ad}(K)$ should be compact in $\operatorname{Ad}(G(\mathbb{R}))$, but I see no reason for it to be maximal.
A toy example would maybe be $GL_2$ with maximal compact subgroup $O(2) \subset GL_2(\mathbb{R})$. The Adjoint group should be $PGL_2$ with maximal compact subgroup $PO(2) = O(2) / Z(O(2))$ and $Z(O(2)) = Z_{GL_2} \, ((\mathbb{R})) \cap O(2) = \{ \pm 1 \}$.
I would be grateful for any answer. Thank you!
 A: Preamble:
Before we begin, let us give some very brief background that might be helpful.
So, let $G$ be a reductive group over $\mathbb{R}$. Then, we have defined $G^\mathrm{ad}:=G/Z(G)$ where this quotient is taken in the sense of algebraic groups (i.e. the fppf/etale sheafification of the functor on $\mathbb{R}$-algebras which takes such an algebra $R$ to $G(R)/Z(G)(R)$).
More concretely, one can think of
$$G^\mathrm{ad}(\mathbb{R})=(G(\mathbb{C})/Z(G)(\mathbb{C}))^\sigma$$
where $\sigma$ is the non-trivial element of $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ acting on $G(\mathbb{C})/Z(G)(\mathbb{C})$— so in particular, an element of $G^\mathrm{ad}(\mathbb{R})$ needn’t come from an element of $G(\mathbb{R})$ since these are elements of $G(\mathbb{C})$ which are fixed by $\sigma$, whereas elements of $G^\mathrm{ad}(\mathbb{R})$ are represented by elements of $G(\mathbb{C})$ which needn’t be fixed by $\sigma$ on the nose but, instead, fixed up to multiplication by an element of $Z(G)(\mathbb{C})$.
In particular, one has an exact sequence of groups
$$1\to Z(G)(\mathbb{R})\to G(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})\to H^1(\mathbb{R},Z(G))$$
where this latter group is $H^1(\mathbb{R},G):=H^1(\mathrm{Gal}(\mathbb{C}/\mathbb{R}),G(\mathbb{C})$ the first Galois cohomology of $Z(G)(\mathbb{C})$ with the obvious $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ action. This is particularly nice since $H^1(\mathbb{R},Z(G))$ is a finite abelian group which can be explicitly calculated (e.g. see [1, Corollary 25.29] for an explicit description). In particular, if $Z(G)$ is a split torus then Hilbert’s theorem 90 implies that $H^1(\mathbb{R},Z(G))=0$ and so $G^\mathrm{ad}(\mathbb{R})=G(\mathbb{R})/Z(G)(\mathbb{R})$.
Question 1: Ok, so let $\mathrm{Ad}:G\to \mathrm{GL}(\mathrm{Lie}(G))$ be the adjoint representation. Then, we have that $\ker(\mathrm{Ad})=Z(G)$—for example see [1, Proposition 21.7] and note that while he assumes that $G$ is split, the desired equality can be checked after base change to $\mathbb{C}$, so it’s OK.
So then, we see that
$$\ker(\mathrm{Ad}_{\mathbb{R}})=\ker(\mathrm{Ad})(\mathbb{R})=Z(G)(\mathbb{R})$$
where I am using $\mathrm{Ad}_{\mathbb{R}}$ for the $\mathbb{R}$-points of the map $\mathrm{Ad}$. In particular, we see that
$$\mathrm{Ad}_\mathbb{R}(G(\mathbb{R}))\cong G(\mathbb{R})/Z(G)(\mathbb{R})$$
and, in fact, $Z(G)(\mathbb{R})=Z(G(\mathbb{R}))$ (e.g. this follows easily from the fact that $G(\mathbb{R})$ is Zariski dense in $G$ by unirationality—e.g. see [1, Theorem 17.93]).
So, you want to see the difference between $G^\mathrm{ad}(\mathbb{R})$ and $G(\mathbb{R})/Z(G)(\mathbb{R})$. But, not much can be said beyond what we said above. Namely, from our discussion above the difference is precisely measured by $H^1(\mathbb{R},G)$. One consequence of this though, is that the map
$$\mathrm{Ad}_\mathbb{R}(G(\mathbb{R}))=G(\mathbb{R})/Z(G)(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})$$
is an isomorphism on connected components. Namely, let us denote by $H(\mathbb{R})^+$ the connected component of $H(\mathbb{R})$ in the real topology (i.e. not the Zariski topology). Then, by the above we have that
$$(G(\mathbb{R})/Z(G)(\mathbb{R}))^+\to G^\mathrm{ad}(\mathbb{R})^+$$
is an injection, which must be a surjection since it has finite cokernel (since $(G(\mathbb{R})/Z(G)(\mathbb{R}))^+$ has finite index in $G(\mathbb{R})/Z(G)(\mathbb{R})$ by [1, Corollary 25.55] and $G(\mathbb{R})/Z(G)(\mathbb{R})$ has finite cokernel in $G^\mathrm{ad}(\mathbb{R})$, namely $H^1(\mathbb{R},G)$).
One special case of this is the following. If $G(\mathbb{R})$ is connected then the map
$$(G(\mathbb{R})/Z(G)(\mathbb{R}))\to G^\mathrm{ad}(\mathbb{R})$$
is an isomorphism onto $G^\mathrm{ad}(\mathbb{R})^+$. This connectivity hypothesis is, shockingly, satisfied if $G$ is simply connected as an algebraic group (i.e. that $G(\mathbb{C})$ is a simply connected complex Lie group)—this follows form [1, Theorem 25.54]).
Question 2: Yes. This is true. In fact, it’s not hard to see that a semisimple group is $H$ adjoint (which to me means connected) if and only if it has trivial center. Indeed, if $H$ has non-trivial center then $H\to H/Z(H)$ is a non-trivial isogeny (note $Z(H)$ is finite since $H$ is semisimple). Conversely, if $f:H\to H’$ is a non-trivial isogeny then $\ker(f)$ is a non-trivial subgroup of $H$ and, in fact, it’s contained in $Z(H)$. The reason is that $\ker(f)$ is finite and so the conjugation action $H\to \mathrm{Aut}(\ker(f))$ has trivial image since $\mathrm{Aut}(\ker(f))$ is a finite group (scheme) and $H$ is connected.
So, from this, we see that your question becomes: is $Z(G^\mathrm{ad})=0$? The answer to this is yes, although the proof in complete generality is more difficult than one might first imagine. See [1, Corllary 17.62(e)].
Question 3: Let’s try to split our understanding into two parts. Namely, let us denote by $K’$ the image of $K$ under the map $G(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})$. We first try to understand the difference between $\widetilde{K}$ and $K’$, then $K’$ and $K$. For simplicitly, I’m going to assume that $G$ is semisimple. If you have a burning desire to know what happens in the general case, I can return later.
First, $K’$ is, up to conjugacy, contained in $\widetilde{K}$ (since $K’$ is contained in some maximal compact subgroup, and all such subgroups are conjugate—e.g. see [2, Theorem D.2.8]). It is also going to be finite index. Indeed, note that $\widetilde{K}^+$ is finite index in $\widetilde{K}$ and $K^+$ is finite index in $K$ and the map $G(\mathbb{R})^+\to G^\mathrm{ad}(\mathbb{R})^+$ is surjective (by prior discussion) from where the claim follows. But, in general, they won’t be equal. Indeed, note that if $G(\mathbb{R})$ is connected, then so is $K$ (e.g. see [2, Theorem D.2.8]) and thus so is $\widetilde{K}$. But, if $G^\mathrm{ad}(\mathbb{R})$ is disconnected then so is $\widetilde{K}$ (again by [2, Theorem D.2.8]) and so you’re in big trouble. This happens with $G=\mathrm{SL}_2$.
Finally, to understand the difference between $K’$ and $K$ we note that since $Z(G)(\mathbb{R})$ is compact it’s contained in every $K$ and so $K’=K/Z(G)(\mathbb{R})$.
EDIT: I wrote the below quickly, so I would double-check it for correctness.
To handle the case when $G$ is not semisimple, note that $G^\mathrm{der}\subseteq G$ is a semisimple group and the composition $G^\mathrm{der}\to G\to G^\mathrm{ad}$ realizes $G^\mathrm{ad}$ as the adjoint group of $G^\mathrm{der}$ as well (e.g. this follows from ). So then, note that $K\cap G^\mathrm{der}(\mathbb{R})$ is a compact subgroup, which must be maximal since if $L$ is a compact subgroup of $G^\mathrm{der}(\mathbb{R})$ properly containing $K\cap G^\mathrm{der}$ then $LK$ is a compact subgroup of $G(\mathbb{R})$ properly containing $K$. Note that the image of $K\cap G^\mathrm{der}(\mathbb{R})$ in $G^\mathrm{ad}(\mathbb{R})$ is contained in $K’$ up to finite index, because the product map $Z(G)(\mathbb{R})\to G^\text{der}(\mathbb{R})\to G(\mathbb{R})$ has finite cokernel and kernel (with cokernel equal to $H^1(\mathbb{R},Z(G^\text{der}))$). So, you can deduce from this that, roughly, the same argument in the semisimple case applies— $K’$ and $\widetilde{K}$ differ by finite index.
References:
[1] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
[2] Conrad, B., 2014. Reductive group schemes. Autour des schémas en groupes, 1, pp.93-444.
