## Quotient of a  reductive group by a non-smooth central finite subgroup

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I need a construction in linear algebraic groups which uses taking quotient by a central finite group subscheme. My question is, whether it goes through in bad'' characteristics, when this group subscheme is not smooth. First I write this construction in a special case, and then in the general case.

Let $G$ be a connected semisimple $k$-group over a field $k$ of characteristic $p>0$. We may assume that $k$ is algebraically closed. Assume that the corresponding adjoint group $G^{ad}$ is $PGL_n$.

In general, my group $G$ is not special (recall that a $k$-group $G$ is called special if $H^1(K,G)=1$ for any field extension $K/k$). I want to construct a special $k$-group $H$ related to $G$. For this end I consider the universal covering $G^{sc}$ of $G$, then $G^{sc}=SL_n$. Let $Z$ denote the center of $G^{sc}$, then $Z=\mu_n$.

We have a canonical epimorphism $\varphi\colon SL_n \to G$. We denote by $C$ the kernel of $\varphi$. Then $C$ is a group subscheme of $Z$, defined over $k$.

Since $Z=\mu_n$, there is a canonical embedding $Z\hookrightarrow \mathbb{G}_m$ into the multiplicative group $\mathbb{G}_m$. Thus we obtain an embedding $C\hookrightarrow \mathbb{G}_m$. Consider the diagonal embedding $$C\hookrightarrow SL_n\times \mathbb{G}_m.$$ I would like to define $H:=(SL_n\times \mathbb{G}_m)/C$. Is such a quotient defined, when char($k$) divides $n$ and $C$ is not smooth?

Note that $SL_n$ embeds into $H$, and we have a short exact sequence $$1\to SL_n \to H \to \mathbb{G}_m \to 1$$ In this exact sequence both $SL_n$ and $\mathbb{G}_m$ are special, and from the Galois cohomology exact sequence we see that $H$ is special as well.

In the general case I assume that $G^{ad}$ is a product of groups $PGL_{n_i}$, $i=1,\dots s$. Then $G^{sc}$ is the product of $SL_{n_i}$. Let $C$ denote the kernel of the canonical epimorphism $\varphi\colon G^{sc}\to G$, then $C$ is contained in the center $Z$ of $G^{sc}$. We have $Z=\prod_{i=1}^s \mu_{n_i}$. Again we embed diagonally $$C\hookrightarrow (\prod_{i=1}^s SL_{n_i}) \times (\mathbb{G}_m)^s$$ and denote by $H$ the quotient. Again $H$ is special (if it is defined), and again my question is, whether this construction makes sense when char($k$) divides $n_i$ for some $i$.

Any help is welcome!

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## 3 Answers

This is an instance of what I believe is called the $z$-construction, and it is a very useful trick in the arithmetic theory of algebraic groups. (Small correction: your diagonal embedding should really be "anti-diagonal". You are really computing a "central pushout".) However, you may need to restrict your ground field to be local or global (with some caveats in case of real places) to get the cohomological vanishing.

In general, let $G$ be any split connected semisimple group over a field $k$, and $G' \rightarrow G$ the $k$-split simply connected central cover. Let $Z$ be the center of $G'$.This is a $k$-group of multiplicative type, possibly not etale. It is $k$-split since $G'$ is $k$-split, so we can choose a $k$-subgroup inclusion of $Z$ into a $k$-split $k$-torus $T$ (seen using character group). Now just form the central pushout of $G'$ along this inclusion. More precisely, the antidiagonal map $$Z \rightarrow G' \times T$$ is a central $k$-subgroup scheme, and in SGA3, VI$_{\rm{A}}$ quotients are constructed and studied for arbitrary finite type group schemes over a field modulo closed subgroup schemes of finite type (and even more generally). We can therefore form the quotient by this central subgroup scheme, and it has all of the reasonable properties one would want for a quotient. This pushout $H$ is a $k$-group containing $G'$ as a normal $k$-subgroup, and the quotient $H/G'$ is $T/Z$. Being a quotient of $k$-split torus, it is again a $k$-split torus.

Provided the ground field is a local or global function field (with some further restrictions in case of a real place), all simply connected semisimple groups have vanishing degree-1 Galois cohomology. So that does the job for such fields. One can also do a variant for split connected reductive groups, and even a variant without a split hypothesis (but then the conclusion is a little different).

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I should also mention that the appendix by T. Ono in Weil's book "Adeles and algebraic groups" also discusses the z-construction, where the point of the construction is to promote the derived group to be simply connected.  That is the reason it is so useful over general ground fields.  In my answer above, I should have noted that G' is indeed the derived group of H.  – BCnrd Feb 24 2010 at 16:12
The "quotient by a $p$-Lie algebra" construction in section 17 of Borel's book can be applied several times to also cover the quotient construction which is being used above.  It is Borel's way of handling infinitesimal kernels without directly using infinitesimal group schemes. This is due to Serre, I have been told.  – BCnrd Feb 24 2010 at 16:37
Brian, I think that the notion of a $z$-construction (there called a $z$-extension) may have been introduced by Kottwitz, in his 1982 Duke J. paper "Rational conjugacy classes in reductive groups". – L Spice Feb 24 2010 at 22:46
Loren, do you mean the terminology rather than the concept?  The trick of using central extension pushout to promote the derived group to be simply connected is there in the appendix to Weil's book.  I was told that this is what is called the z-construction (I'm not sure why; due to "central" subgroups, or a "Z" in the diagram of arrows?).  The appendix suggests it goes back at least to Weil. – BCnrd Feb 25 2010 at 1:45
Brian, I'm sorry; I misread your quote about Weil's book as discussing a later addition, rather than part of the book itself.  The Kottwitz paper was my first exposure to it, but, as you point out, it is certainly not the first occurrence of the concept.  (For what it's worth, Kottwitz cites Langlands's "stable conjugacy" paper, which I don't have to hand, but does not mention Weil.)  I'd always taken the $z$ to refer to the centre, but wondered (as I guess you do, too) why then '$z$' instead of '$Z$'. – L Spice Feb 25 2010 at 17:23

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The standard isomorphism theorems in abstract group theory all hold for group schemes of finite type over a field. This is implicit in SGA (a key point is the statement mentioned by Ekedahl) and is explicit in the notes on algebraic groups,... appearing on my website (Section 7 of Chapter I). This makes a lot of things obvious (including your questions).

[The isomorphism theorems fail when you don't allow nilpotents, which is why the standard expositions on algebraic groups are so complicated.]

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The quotient by any finite flat subgroup scheme always exists (see for example SGA 3:Exp V, Thm 7.1). In your case the subgroup scheme is of multiplicative type, the dual of a finite abelian group $A$, so an action of it on an affine $k$-scheme is just an $A$-grading of the coordinate ring on the scheme. The quotient is just the spectrum of the degree $0$-part. If $R$ is an $A$-graded commutative Hopf algebra (over $k$), it is clear that the degree $0$-part of $R$ is a commutative Hopf algebra. This gives a reasonably concrete description of the quotient in that case.

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