Let $G$ be a reductive group over the finite field $\mathbb{F}_q$. Then a regular embedding of $G$ is an $\mathbb{F}_q$-rational embedding $\iota \colon G \rightarrow G'$ into a second reductive group over $\mathbb{F}_q$, such that $\iota$ maps the derived group of $G$ isomorphically onto that of $G'$ and the center of $G'$ is connected. Moreover, a regular embedding is called smooth regular, if additionally the center of $G'$ is smooth. It is well-known that smooth regular embeddings always exist (see for example [1,Theorem 4.5],[2, Lemma 6.5]).

Question. Is it true that for any $G$ a smooth regular embedding $\iota \colon G \rightarrow G'$ exists, with the additional property that the center of $G'$ is an induced torus?

It seems quite plausible, but I could not find any reference (which is a bit annoying).

Note that for the somewhat dual notion of a z-extension $\widetilde G \rightarrow G$ it can be arranged -- and is even part of the definition of a z-extension (cf. [3,§1]) -- that ${\rm ker}(\widetilde{G} \rightarrow G)$ is induced.


[1] Martin, B.: Étale slices for representation varieties in characteristic p.

[2] Taylor, J.: The Structure of Root Data and Smooth Regular Embeddings of Reductive Groups.

[3] Kottwitz, R.: Rational conjugacy classes in reductive groups.

  • $\begingroup$ Let $G\hookrightarrow G'$ be a smooth regular embedding. We write $Z(G')$ for the center of $G'$. Do I understand correctly that $Z(G')$ is a torus? $\endgroup$ Jul 10, 2021 at 20:09
  • $\begingroup$ Sure. Being the center of a reductive group it's contained in a(ny) max.torus of that group. Being also connected it must a torus itself. $\endgroup$
    – AlexIvanov
    Jul 10, 2021 at 20:19
  • $\begingroup$ Good! I live in char. 0, and I feel a bit uncomfortable in positive characteristic. $\endgroup$ Jul 10, 2021 at 20:44

1 Answer 1


The answer is Yes.

Let $G\hookrightarrow G'$ be a smooth regular embedding. We write $Z(G')$ for the center of $G'$, which is an $F$-torus, where $F={\Bbb F}_q$. We construct a regular embedding $G'\hookrightarrow G''$ such that $Z'':=Z(G'')$ is an induced torus. We write $S=(G,G)=(G',G')$ for the derived group of $G$ and $G'$.

Consider the natural surjective homomorphism $$\varphi\colon Z'\times_F S\to G',\quad (z,s)\mapsto z^{-1}\cdot s.$$ Set $K=\ker\varphi=\big\{(z^{-1},z)\mid z\in Z'\cap Z(S)\big\}$. Then we have a canonical isomorphism $$(Z'\times_F S)/K\to G'.$$ Since $Z'=Z(G')$, we see that $Z'\supset Z(S)$, whence $K=\big\{(z^{-1},z)\mid z\in Z(S)\big\}$.

Write $X'$ for the character group of $Z'$. Then $X'$ is a ${\rm Gal}(\overline F/F)$-module. Clearly, there exists a surjective homomorphism $\alpha\colon X''\twoheadrightarrow X'$, where $X''$ is a permutation ${\rm Gal}(\overline F/F)$-module. Let $Z''$ denote the $F$-torus with character group $X''$, which is an induced torus. Then we have an injective homomorphism $\alpha^*\colon\, Z'\hookrightarrow Z''$.

Consider the induced injective homomorphism $$\alpha^{**}\colon\, Z'\times_F S\hookrightarrow Z''\times_F S$$ and the induced injective homomorphism $$\alpha^\vee\colon\, G'= (Z'\times_F S)/K\hookrightarrow (Z''\times_F S)/\alpha^{**}(K)=:G''.$$ Then $$\alpha^{**}(K)=\big\{(\alpha^*(z)^{-1},z)\mid z\in Z(S)\big\}.$$ It follows that the homomorphism $Z''\to Z(G'')$ is an isomorphism and that $\alpha^\vee$ is a regular embedding. The composite injective homomorphism $$ G\hookrightarrow G'\hookrightarrow G''$$ is a desired smooth regular embedding for which $Z(G'')$ is an induced torus.


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