4
$\begingroup$

Let $R$ be a Dedekind domain with fraction field $K$, and let $G$ be a smooth affine group scheme over $S = \text{Spec }R$ whose geometric fibers are connected and simple linear algebraic groups (i.e., $G$ is a "connected simple linear algebraic $S$-group scheme").

Does $G$ admit a universal cover over $S$? Here, by universal cover I mean a finite flat surjective homomorphism of $S$-group schemes $p : \tilde{G}\rightarrow G$ such that $p$ induces universal covers on every geometric fiber in the usual sense.

If not, can we always do this Zariski or etale locally on $S$?

I think this is true if $S = \text{Spec }k$ with $k$ a field, though I don't know of a reference. References are also appreciated!

EDIT: Let me elaborate on what I mean by universal cover. I believe split reductive groups all exist as smooth group schemes over $\mathbb{Z}$ (with reductive geometric fibers), and they are uniquely determined up to isomorphism by their $\mathbb{C}$-geometric fiber. Also I believe the isogeny theorem (Theorem 6.1.16 of Brian Conrad's notes) should imply that a universal cover $\tilde{\mathbf{G}}\rightarrow \mathbf{G}$ of simple algebraic groups over $\mathbb{C}$ induces an isogeny of their unique split $\mathbb{Z}$-models $\tilde{G}\rightarrow G$. I want this latter map and any of its base changes to be considered a universal cover. If $k$ is an algebraically closed field of characteristic $p$, then I think I want a universal cover of a simple linear algebraic $k$-group $G$ to be an isogeny from a simple linear algebraic $k$-group $\tilde{G}$ such that any central isogeny onto $\tilde{G}$ is trivial. Please correct me if I'm wrong!

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Group schemes like $SL_n$ in characteristic $p$ have weird finite etale covers thanks to the Lang isogeny, do we want to avoid those? $\endgroup$
    – Will Sawin
    Commented Dec 29, 2021 at 3:22
  • $\begingroup$ @WillSawin, the Lang isogeny is not a morphism of group schemes, though, so doesn't enter into this sort of theory of covers, right? $\endgroup$
    – LSpice
    Commented Dec 29, 2021 at 9:05
  • 1
    $\begingroup$ @LSpice Well, it depends what "universal covers ... in the usual sense" means. If it means that every finite etale morphism factors through, that is a problem. $\endgroup$
    – Will Sawin
    Commented Dec 29, 2021 at 14:24
  • $\begingroup$ @WillSawin Great question! I've edited the OP to address this. $\endgroup$
    – stupid_question_bot
    Commented Dec 29, 2021 at 18:12
  • 1
    $\begingroup$ Probably this is true. Over $k$, I think this follows from Exercise 1.13(2) in the linked notes. One can probably bash out the general case with the isogeny theorem, but maybe there is a better way. $\endgroup$
    – Will Sawin
    Commented Dec 29, 2021 at 18:20

1 Answer 1

Reset to default
4
$\begingroup$

I am just posting my final comment as one answer. As the OP notes, there are answers that go through the classification, and there may be a best answer that uses very little. The answer here relates the post to the theory of the automorphism group schemes of semisimple algebraic groups.

For a pinned semisimple group scheme $G$ over a base scheme $S$, there is an associated $S$-scheme $A$ over $S$ representing the function of automorphisms of $G$ as an $S$-group scheme, and $A$ is itself a linear algebraic group that is an extension of a quasi-finite group scheme by a semisimple group scheme. All of this is part of Theorem 7.1.9 of Brian Conrad's excellent notes.

Brian Conrad
Reductive group schemes.
Autour des schémas en groupes. Vol. I, 93–444, Panor. Synthèses, 42/43, Soc. Math. France, Paris, 2014.

In particular, the normal factor $G/Z_G$ of inner automorphisms is preserved by isogenies of semisimple group schemes. So an isogeny of group schemes preserves the automorphism group scheme if and only if it preserves the outer automorphism group scheme. For split, pinned semisimple group schemes, this is a group scheme of automorphisms of the associated root datum. This typically is not preserved by all isogenies, e.g., the outer automorphism group of $\textbf{SL}_n\times \textbf{PGL}_n$ is strictly smaller than the automorphism group schemes of $\textbf{SL}_n\times \textbf{SL}_n$ or $\textbf{PGL}_n\times \textbf{PGL}_n$.

As this example illustrates, for the quotient isogeny from the simply connected form to the adjoint form, the group of automorphisms of the associated root datum is preserved, as can be checked by working on field-valued points.

So now the strategy is as follows. By hypothesis, the group scheme under consideration is a form of a split, pinned, semisimple group scheme over $S$. Thus it gives a torsor for the automorphism group scheme of the split form. By forming the quotient by the center, this gives a torsor for the automorphism group scheme of the associated adjoint form. Since the map of automorphisms group schemes from the simply connected form to the associated adjoint form is an isomorphism, this torsor is equivalent to a torsor for the automorphism group scheme of the associated simply connected form of the split, semisimple group scheme. This torsor is equivalent to a form of that simply connected, semisimple group scheme. That form is the "universal cover" of the original (not necessarily split) semisimple group scheme over $S$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks to the editor! $\endgroup$
    – Jason Starr
    Commented Dec 30, 2021 at 16:30
  • $\begingroup$ Many thanks for this. If I'm understanding correctly, starting with the original group scheme $G$, you consider its split form $G_{sp}$ and its universal cover $\tilde{G}_{sp}$. Then you say that the composition $\tilde{G}_{sp}\rightarrow G_{sp}\rightarrow G_{sp}/Z(G_{sp})$ preserves automorphism group schemes. Do you assert this is always true? $\endgroup$
    – stupid_question_bot
    Commented Jan 4, 2022 at 23:32
  • $\begingroup$ Anyway, granting this, you note that $G/Z(G)$ determines an $S$-scheme $P = Isom(G/Z(G),G_{sp}/Z(G_{sp}))$ which is simultaneously a torsor for both $Aut(\tilde{G}_{sp})$ and $Aut(G_{sp}/Z(G_{sp}))$. Twisting $\tilde{G}_{sp}$ by $P$ yields a group $\tilde{G}$, and since twisting by a torsor is functorial, we get a map $\tilde{G}\rightarrow G/Z(G)$. Is it clear that this map factors through $G$? I.e., why does your “universal cover” actually map to $G$? $\endgroup$
    – stupid_question_bot
    Commented Jan 4, 2022 at 23:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .