Regular embeddings of reductive groups

Question

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.

Answer 1

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 $G \times_{Z(G)} T$.

Answer 2

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.