# Regular embeddings of reductive groups

A regular embedding of a connected reductive linear algebraic group $G$ defined over $\mathbb{F}_q$ is a morphism $\varphi : G \rightarrow G'$ of algebraic groups which is a closed immersion where $G'$ is a connected reductive linear algebraic group defined over $\mathbb{F}_q$ with connected centre and $\varphi(G)$ contains the derived subgroup of $G'$.

For example, if $T$ is a torus and $\iota: Z(G) \rightarrow T$ an isomorphism onto its image, then mapping $G$ into the quotient of $G \times T$ by $Z = \{(z,\iota(z)^{-1})\:|\: z \in Z(G)\}$ gives a regular embedding.

I want to prove the following statement:

Given two regular embeddings, say $G \rightarrow G'$ and $G \rightarrow G''$, there exist regular embeddings $G' \rightarrow G'''$ and $G'' \rightarrow G'''$ making the resulting square of morphisms commutative.

I believe I can prove this in the special case where $G'$ and $G''$ are of the special form $(G \times T)/Z$ like above. For if $G'$ is such a quotient of $G \times T'$ and $G''$ is such a quotient of $G \times T''$, then I can take a similar quotient of $G \times T' \times T''$ for $G'''$ and this seems to work as far as I can tell.

However, I fail to understand the general case completely. Note that if $\varphi : G \rightarrow G'$ is a regular embedding, then $G' = Z(G')\varphi(G)$ and $Z(G')$ is a torus. We then have a bijective morphism $(G \times Z(G'))/Z \rightarrow G'$ where $Z$ is like above using $\varphi|_{Z(G)} : Z(G) \rightarrow Z(G')$. But this bijection is not necessarily an isomorphism (at least I could not prove that).

For reference, this is Exercise 2 in Chapter 15 of "Representation Theory of Finite Reductive Groups" by Cabanes and Enguehard (Cambridge, 2004). I found the statement also in several other papers but without any hint towards a proof. Any help is appreciated.

I had cause to think about this exercise recently so I thought I’d write an answer. I think Jim’s answer is sufficient but as you seem to want more details I’ll provide them here. I am aware that you know some of the following but I’m writing a complete solution for the benefit of future readers.

I’ll use a slight generalisation of the construction given in Cabanes—Enguehard. I’ll write $F : G \to G$ for a Frobenius endomorphism of $G$. Assume $T$ is a torus equipped with a Frobenius endomorphism, which we also denote by $F : T \to T$, and further let us assume that we have a closed embedding $\pi : Z(G) \hookrightarrow T$ which is defined over $\mathbb{F}_q$. The direct product $G \times T$ inherits a natural Frobenius endomorphism given by $F \times F$ and the subgroup $\Delta_{\pi}(Z(G)) = \{(z,\pi(z)^{-1}) \mid z \in Z(G)\}$ is an $F$-stable closed subgroup of $G \times T$. We denote by $G \times_{Z(G)} T$ the quotient group $(G\times T)/\Delta_{\pi}(Z(G))$.

Now assume $\sigma : G \hookrightarrow H$ is a regular embedding. By assumption we have $H = \sigma(G)Z(H)$ and so the morphism $G \times Z(H) \to H$ given by $(g,z) \mapsto \sigma(g)z$ is surjective. The kernel of this map is clearly $\Delta_{\sigma}(Z(G))$ so we have an induced bijective morphism of varieties $\gamma : G \times_{Z(G)} Z(H) \to H$. Your real question seems to be whether or not $\gamma$ is an isomorphism. It is worthwhile thinking about this as one wants to rule out situations like the natural projection map $\mathrm{SL}_p(K) \to \mathrm{PGL}_p(K)$ in characteristic $p$, which fails to be an isomorphism. However, this essentially doesn’t happen because our derived subgroups are assumed to be isomorphic.

By Corollary 5.3.3 of Springer’s LAG we need only check that the differential $d_e\gamma$ at the identity is bijective. However as both tangent spaces have the same dimension we need only check surjectivity, which follows from the fact that

$$T_e(H) = T_e(H_{\mathrm{der}}) + T_e(Z(H)),$$

where $H_{\mathrm{der}} \leqslant H$ is the derived subgroup of $H$. This is easily seen by noting that $T_e(H)$ is the sum of a maximal toral subalgebra together with the root spaces. Note, however, that this sum need not be direct. For instance, think of $\mathrm{GL}_p(K)$ in characteristic $p$. Although it is not direct it implies the surjectivity of our map $\gamma$ because $\sigma$ maps the derived subgroup $G_{\mathrm{der}} \leqslant G$ isomorphically onto $H_{\mathrm{der}}$ and we similarly have

$$T_e(G \times_{Z(G)} Z(H)) = T_e(G_{\mathrm{der}} \times_{Z(G)} 1) + T_e(1 \times_{Z(G)} Z(H)).$$

This is really highlighting that the problems essentially arise at the derived subgroup.

As you observed, with this observation the exercise becomes fairly easy. Assume $\sigma : G \to G’$ and $\tau : G \to G’’$ are closed embeddings then we denote by $T$ the torus $Z(G’) \times Z(G’’)$. This inherits a natural Frobenius endomorphism from $G’$ and $G’’$ and we have a closed embedding $Z(G) \hookrightarrow T$ given by $z \mapsto (\sigma(z),\tau(z))$ which is defined over $\mathbb{F}_q$, so we can form the group $G’’’ = G \times_{Z(G)} T$. We have regular embeddings $G’ \to G \times_{Z(G)} Z(G’) \to G \times_{Z(G)} T$ and $G’’ \to G \times_{Z(G)} Z(G’’) \to G \times_{Z(G)} T$ where the first maps are the isomorphisms constructed above. It is then easy to see that we may take your group $G’’’$ to be the group $G \times_{Z(G)} T$.

• You are right, it seems my main problem was that I was not able to show that every regular embedding is isomorphic to one of the form $G \times_Z T$. Thank you for this very detailed answer. May 8 '15 at 13:58
• It has been some time since I asked this question, I know, but lately I came to wonder whether your argument really works, because for example $SL_p \to GL_p$ is a regular embedding but $GL_p \not\cong SL_p \times Z(GL_p)$, right? Maybe you could explain to me where my misunderstanding lies? Jan 25 '16 at 10:19
• You are right that the argument above is not correct. That is the same counterexample I came to in the end. This result is not so easy to prove correctly. A proper proof of this result using root data will appear in a forthcoming joint paper that I am working on. I was going to come back and update the answer once the preprint had appeared on the arXiv. In the meantime if you want to discuss this more you should just send me an email. Jan 25 '16 at 12:43
• Thank you for your quick response. I just wanted to be sure about this and I can imagine how this is going to work on the level of root data. Jan 26 '16 at 16:29
• @JayTaylor, have you posted that preprint? May 24 '17 at 7:26

Leaving aside your minor changes in the notation and definition of Cabanes-Enguehard (who also adopt some arbitrary notation), my understanding of their exercise is that it responds to the obvious non-uniqueness in the target group. Of course, the whole problem originates in embeddings like the one of a special linear group in the corresponding general linear group when the former group has a nontrivial (but finite) center.

The target group always has a connected center, which is just a torus of some dimension (defined over $\mathbb{F}_q$, though this doesn't seem to enter directly into the question here). So the basic concern is that two different target groups may involve central tori of different dimensions. I guess the natural solution is to embed both of these tori into a possibly larger one (still defined over $\mathbb{F}_q$), which in turn allows one to construct a common target group for both of these as in the standard construction used by Cabanes-Enguehard. However you approach this you run into some arbitrary choice of central torus, so there is no universal construction. Am I oversimplifying the question? I got lost in the next-to-last paragraph of your discussion.

ADDED: To be more precise (while keeping most of the heavy notation out of the way), note that all three connected reductive groups given in the statement of the problem share a common semisimple derived group $H$ (up to isomorphism) which has a finite center $Z(H)$. Now embed the centers (both tori) of the two target groups in a large enough torus $S$ defined over $\mathbb{F}_q$ and use the Cabanes-Enguehard construction to get an ultimate target group $S \times H / Z(H)$; here $Z(H)$ acts diagonally.

• Thank you for your answer. Note that my change in definition is not really one. If we assume $[G',G'] \subseteq \varphi(G)$, then we have $G' = Z(G')\varphi(G)$ and so the commutator subgroups coincide. The idea you formulated is essentially what I wanted to do and what I actually did when only considering the special regular embeddings from the standard construction. However, I do not understand how this should work in the general case where the regular embeddings given are not of the form $(G \times T)/Z$. How would you get the desired maps into $S \times H/Z(H)$? Mar 9 '15 at 7:43
• I hope I'm not oversimplifying, but for a connected reductive group with connected center such as your $G', G''$, the group is just isomorphic to the direct product of its center (a torus) and the derived group $H \cong [G,G]$ modulo the diagonal action of $Z(H)$ (which lies in the big center). So both $G', G''$ embed naturally into $G''':=(S \times H)/Z(H)$ for a torus $S$ containing both of their centers. Mar 9 '15 at 18:24
• If this were true, all my problems would be solved indeed. I was not aware of this. Some time ago I ran into a problem (math.stackexchange.com/questions/1136933/…) where in a similar situation I got convinced that I do not necessarily have this isomorphism between e.g. $G'$ and the quotient of $[G,G] \times Z(G')$. Is this easy to see or do you have some reference for this? Mar 10 '15 at 7:23
• The basic structure of any connected reductive $G$ is laid out by Borel-Tits (1965), with treatments in books on linear algebraic groups as well as a nice summary in Chap. 0 of the text by Digne-Michel (emphasizing finite fields). To be precise, $G$ is the almost direct product of its semisimple derived group $H$ and its radical $S$, the latter being a central torus; here $Z:=Z(H) \supset H∩S$ is finite. If $Z(G)$ is connected, then it equals $S$ with $Z \subset S$; so the surjective map $S \times H \rightarrow G$ has kernel consisting of the pairs $(z,z^{−1})$ with $z \in Z$. Mar 10 '15 at 20:32
• Yes, this was known to me. But from that it seems to me that I can only deduce that I have a bijective morphism $(S \times H) / Z \rightarrow G$ of algebraic groups (which is thus an isomorphism of abstract groups). What I do not know is why this is automatically an isomorphism of algebraic groups? Mar 11 '15 at 1:26