Quotient of a reductive group by a non-smooth central finite subgroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:51:54Z http://mathoverflow.net/feeds/question/16261 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16261/quotient-of-a-reductive-group-by-a-non-smooth-central-finite-subgroup Quotient of a reductive group by a non-smooth central finite subgroup Mikhail Borovoi 2010-02-24T13:51:35Z 2010-02-25T02:18:15Z <p>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.</p> <p>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 $PGL_n$.</p> <p>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 $Z=\mu_n$.</p> <p>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 $k$.</p> <p>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?</p> <p>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.</p> <p>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 $i$.</p> <p>Any help is welcome!</p> http://mathoverflow.net/questions/16261/quotient-of-a-reductive-group-by-a-non-smooth-central-finite-subgroup/16272#16272 Answer by Torsten Ekedahl for Quotient of a reductive group by a non-smooth central finite subgroup Torsten Ekedahl 2010-02-24T15:58:35Z 2010-02-24T15:58:35Z <p>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.</p> http://mathoverflow.net/questions/16261/quotient-of-a-reductive-group-by-a-non-smooth-central-finite-subgroup/16273#16273 Answer by BCnrd for Quotient of a reductive group by a non-smooth central finite subgroup BCnrd 2010-02-24T16:00:45Z 2010-02-24T16:00:45Z <p>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. </p> <p>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. </p> <p>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). </p> http://mathoverflow.net/questions/16261/quotient-of-a-reductive-group-by-a-non-smooth-central-finite-subgroup/16348#16348 Answer by JS Milne for Quotient of a reductive group by a non-smooth central finite subgroup JS Milne 2010-02-25T02:18:15Z 2010-02-25T02:18:15Z <p>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).</p> <p>[The isomorphism theorems fail when you don't allow nilpotents, which is why the standard expositions on algebraic groups are so complicated.]</p>