Adjoint of the derived group

algebraic-geometryalgebraic-groupsarithmetic-geometrynumber theoryreductive-groups

Firstly, let $G$ be a nice linear algebraic group (for instance connected reductive group) over $\mathbb{Q}$. I shall first define the two other groups I require

$\textbf{Definition:}$ If $Z(G)$ is the center of the group $G$, then we define the adjoint group of $G$, denoted by $G^{\operatorname{ad}}$, to be the quotient $G/Z(G)$.

$\textbf{Definitoin:}$ The derived subgroup of $G$, denoted by $G^{\operatorname{der}}$, is the intersection of all normal subgroups $N$ of $G$ such that $G/N$ is commutative.

Is $G^{\operatorname{ad}}=(G^{\operatorname{der}})^{\operatorname{ad}}$?

I ask this because it seems to be true for $G=\operatorname{GL}_n$, where $G^{\operatorname{der}}=\operatorname{SL}_n$ and $(G^{\operatorname{der}})^{\operatorname{ad}}\cong G^{\operatorname{ad}}\cong \operatorname{PGL}_n$ and also for $G=\operatorname{GSp}_{2n}$, where $G^{\operatorname{der}}=\operatorname{Sp}_{2n}$ and $(G^{\operatorname{der}})^{\operatorname{ad}}\cong G^{\operatorname{ad}}$.

My question comes because I am trying to understand Shimura varities. If $(G,X)$ is Shimura datum, and $X^+$ is a connected component of $X$, then $(G^{\operatorname{der}},X^+)$ is a connected Shimura datum. This clearly requires the two of them to have the same adjoint group, but I am not sure why this is true.

A proof or any reference of one, will be a great help. If in fact the answer is 'no', then I might need to find some other reason why a connected component of a Shimura datum is a connected Shimura datum.

Best Answer

The statement $(G^{der})^{ad}=G^{ad}$ is equivalent to $G=Z(G).G^{der}$. This is 17.42 of Milne's book Reductive Groups.

Per Coherent Sheaf's request: $G=Z(G).G^{der}$ is equivalent to the natural map $G^{der}\to G^{ad}$ being surjective. Of course, equivalently, $(G^{der})^{ad}\to G^{ad}$ is surjective. Thus, all we need to know is that $(G^{der})^{ad}\to G^{ad}$ is injective.

The kernel is exactly $(Z(G)\cap G^{der})/Z(G^{der})$, so we need to know $Z(G^{der})=Z(G)\cap G^{der}$, i.e., that $Z(G^{der})\subset Z(G)$. But this again follows from the equality $G=Z(G).G^{der}$!

Related Solutions

Intersection of Levi subgroups is a Levi

Let $U_{\theta} = M_{\theta} \cap U_0$, which is the unipotent radical of the minimal parabolic $M_{\theta} \cap P_0$ of $M_{\theta}$. We have

$$M_{\theta \cap \Omega} \subset M_{\theta} \cap M_{\Omega} \subset P_{\theta} \cap P_{\Omega} = P_{\theta \cap \Omega} = M_{\theta \cap \Omega} \ltimes N_{\theta \cap \Omega}$$

Let $p \in M_{\theta} \cap M_{\Omega}$. We may write $p = mn$ for $m \in M_{\theta \cap \Omega}$ and $n \in N_{\theta \cap \Omega}$. The problem becomes to show that $M_{\theta} \cap M_{\Omega}$ has trivial intersection with $N_{\theta \cap \Omega}$. We have

$$M_{\theta} \cap M_{\Omega} \cap N_{\theta \cap \Omega} = U_{\theta} \cap U_{\Omega} \cap N_{\theta \cap \Omega}$$

This group is normalized by $A_0$, and this therefore connected and directly spanned by the root subgroups it contains. There are no roots in this intersection, so it must be trivial.

Adjoint group of a reductive group

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.

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