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It is a standard and important fact that any smooth affine group scheme $G$ over a field $k$ is a closed $k$-subgroup of ${\rm{GL}}_n$ for some $n > 0$. (Smoothness can be relaxed to finite type, but assume smoothness for what follows.) The proof makes essential use of $k$ being a field, insofar as it uses freeness of finitely generated $k$-submodules of the coordinate ring of $G$. The same argument (appropriately formulated) then works when $k$ is a PID. (edit: I originally mentioned that I didn't know if this is also true over any Dedekind domain, and wasn't asking about it; nonetheless, comments from George and Kevin below give proof for Dedekind base case.)

The question is this: is the above result true for all artin local rings $k$, or even just the ring of dual numbers over a field? Or can one give a counterexample? Since monic homomorphisms between finite type groups over an artin ring are closed immersions, an equivalent formulation which may be more vivid is: does $G$ admit a (functorially) faithful linear representation on a finite free $k$-module?

(I originally thought I needed an affirmative such result over artin local rings to prove a certain general fact for smooth affine group schemes over noetherian rings, but eventually that motivation got settled in another way. So for me it is now an idle question, though I think a very natural one from the viewpoint of deformation theory of smooth linear algebraic groups.)

It sounds like the sort of thing which must have been thought about back in the 1960's when SGA3 was being written, so I mentioned the question to a couple of the SGA3 collaborators as well as some other experts in these matters. Unfortunately nobody whom I have asked knew one way or the other, even for the dual numbers. One of them suggested a couple of days ago that I should "advertise this problem; it is very provocative." Fair enough; I suppose this kind of advertising on MO is OK.

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Did someone steal the vowels in your user name?! –  Mariano Suárez-Alvarez Apr 21 '10 at 17:21
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@Mariano: Since I often drop letters to fit answers into comment boxes, I decided to drop some from my username. –  BCnrd Apr 21 '10 at 17:32
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Heh. I would expect the characters in the user name not to be counted when computing the comment length, though. –  Mariano Suárez-Alvarez Apr 21 '10 at 17:57
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Brian: stupid question. If G is smooth affine gp scheme over the dual numbers F[epsilon], F a field, does G always lift to smooth affine group scheme over F[[x]]? If so then you've reduced to the PID case. On the other hand I'm guessing that you know some crazy unipotent example that doesn't lift (multiplicative and semisimple seem to me to stand a good chance, however!) –  Kevin Buzzard Apr 21 '10 at 20:39
    
Kevin, lifting through invlim will get stuck on bounding deg of relations in coord ring presentations wrt "fixed" variables (to get polys, not restricted power series, in invlim). That's why deftion thy in affine f.type case "never" works: must bound deg of relations in presentations. And why miraculous that Groth. lifted maximal tori from special fiber in (affine!) reductive over complete local noetherian rings (even henselian local). Formal GAGA overcomes "algebraization" problem in proper case via ample line bundle with lifting, no version in affine case. –  BCnrd Apr 22 '10 at 4:45

6 Answers 6

up vote 8 down vote accepted

This is not a direct answer to the question for a general group scheme $G \to S$ and I am not an expert in this area. However, I would like to point out that the resolution property of stacks is a natural condition that appears in this context of Hilbert's 14th problem by work of R. W. Thomason:

Equivariant resolution, linearization, and Hilbert's fourteenth problem over arbitrary base schemes Advances in Mathematics 65, 16-34 (1987)

Once and for all let $ \pi \colon G \to S$ be an affine, flat, finite type group scheme over a noetherian and separated base scheme $S$.

Recall, that a noetherian algebraic stack has the resolution property if every coherent sheaf is a quotient of a vector bundle (a locally free sheaf, which will be always assumed to be of finite and constant rank). Therefore, the classifying stack $B_S G$ has the resolution property if and only if every coherent $G$-comodule on $S$ is the equivariant quotient of some locally free $G$-comodule. The latter is the definition of the $G$-equivariant resolution property of $S$.

What we need is his Theorem 3.1: $G \to S$ can be embedded as a closed subgroup scheme of $GL(V)$ for some vector bundle $V$ on $S$ if $B_S G$ has the resolution property. If $S$ is affine, $V$ can be taken to be free.

Thomason does not say that the converse to Theorem 3.1. also holds. I guess that this is true if $S$ is affine, but as I am always getting confused while working with comodules, I cannot give a rigorous proof at the moment.

Nevertheless, it is worth to ask when $B_S G$ has the resolution property. Thomason proved this in the following cases:

  1. $S$ regular and dim $S \leq 1$,
  2. $S$ regular; dim $S \leq 2$; $\pi_* O_G$ is a locally projective $O_S$-module, e.g, if $\pi \colon G \to S$ is smooth and with connected fibres.
  3. $S$ regular or affine or has an ample family of line bundles; $G$ a reductive group scheme which is either split reductive, or semisimple, or with isotrivial radical and coradical, or over a normal base $S$.

In particular, if $S$ is the the spectrum of the ring of dual numbers, then this provides an affirmative answer to the posted question if $G \to S$ satisfies the conditions in (3).

Even for $G \to S$ arbitrary with reduction $G_0 \to S_0$, we know that the reduction $X_0=B_{S_0}G_0$ of $X= B_S G$ has the resolution property by (1). So we may reformulate the original question as follows:

(Q2) Is the resolution property preserved under the first order deformation $X_0 \to X$?

Lifting of various locally free resolutions from $X_0$ to $X$ is probably not the best approach. However, it suffices to lift a single locally free sheaf. Let us see, why this is true. A noetherian algebraic stack with affine diagonal has the resolution property if and only if there exists a vector bundle $V$ whose associated frame bundle has quasi-affine total space. The normal case was proven by Totaro in

The resolution property for schemes and stacks. J. Reine Angew. Math. 577 (2004), 1--22. 14A20 (14C35)

and in my thesis, I am currently working on, I show that this really holds for non-reduced stacks too. Therefore if we can lift $V_0$ from $X_0$ to a vector bundle $V$ on $X$, then $V$ has still quasi-affine frame bundle as its reduction is quasi-affine. The obstruction for this lies in $H^2(X_0, I \otimes V_0^\vee \otimes V_0)$ where $I$ is the coherent ideal of order two defining the deformation $X_0 \to X$. Probably, the ideal can be removed here with some tricks.

In our case this cohomology boils down to the second group cohomology of the $G_0$-representation $I \otimes V_0^\vee \otimes V_0$. In particular, if $G_0 \to S_0$ is linearly reductive, the obstruction is zero.

Therefore we have proven:

If $G \to S$ is a group scheme over an artinian base with linearly reductive special fibre, then $G \to S$ can be embedded into some $GL_{n,S}$ as a closed subgroup scheme.

Clearly this still leaves out interesting cases and probably this can be proven more directly avoiding stack theory.

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Interesting approach. However, if I understand SGA III:Exp XIII right there are is only one infinitesimal deformation of a reductive group (the reduction to the split case is easy) and hence in that case there is nothing to prove. (All linearly reductive groups are reductive so that condition does not give us anything new.) So I think that unfortunately you were right in believing that the stack techniques can be avoided and that there are lots of interesting cases left. –  Torsten Ekedahl Apr 29 '10 at 9:55
    
(cont'd) I think that if you look at Thomason's proof it seems that the split reductive proof is done by using that the group scheme is the pullback from a group scheme over $\mathbb Z$ which is another way of interpreting SGA III I believe. –  Torsten Ekedahl Apr 29 '10 at 9:55
    
Ah, tanks for pointing that out. I should read SGA more carefully... –  Philipp Gross Apr 29 '10 at 10:29
    
Although this answer doesn't provide anything new in the reductive case, its viewpoint sounds interesting from the viewpoint of the general case, and at least puts the issue into a broader setting (even if ultimately not settling any new cases at the present time). The intervention of the H^2 in these matters was already pointed out to me in a letter from Serre some months ago. So I'll vote to accept this as the "official answer". –  BCnrd Apr 30 '10 at 13:29

This doesn't answer the question posed, but maybe speaks to the one you didn't ask... In Bruhat-Tits [Groupes Reductifs sur un corps local II] 1.4.5 shows for Dedekind A that an affine A-group scheme which is flat and of finite type has a faithful linear representation.

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@George: I had in mind closed immersions into ${\rm{GL}}_n$: faithful (as defined by Bruhat-Tits) linear repns on a finite free module. BTII uses the argument from the field case, which over non-PID Dedekind domain gives a faithful linear repn on finite projective module M. So get closed subgroup inclusion into GL(M). That's nice, but the question I didn't ask would have been whether over a Dedekind base one can do better and arrange for M to be free. There could be trick to get it, but I don't know. Thanks for posting reference to BTII (but alas the method seems hopeless over an artin ring). –  BCnrd Apr 21 '10 at 19:36
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Brian: GL(M) embeds into GL(M+N) right? (N projective). And you can arrange for M+N to be free. –  Kevin Buzzard Apr 21 '10 at 19:46
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Kevin, good point, that was dumb of me. So as I suspected, it's just some little trick. OK, back to the more interesting artin case... –  BCnrd Apr 21 '10 at 20:29
    
Kevin, in the LaTeX version of sga3, in 11.11ff, VI_B, they have improved the original field discussion to Dedekind and artin bases (the latter seems incomplete; see comments below for A. Stasinski's answer). In the Dedekind case they missed this observation with $M \oplus N$. So I'm in good company on that one. :) Anyway, I have mentioned to Gille & Polo this trick which you pointed out, so hopefully they can incorporate it into the "final" version (whatever that may mean, in view of Grothendieck's letter). –  BCnrd Apr 22 '10 at 13:32

I don't understand why the usual proof over a field base doesn't work over a (local) artinian base $R$ for a flat finite type group scheme over $R$: Let $A$ be the affine algebra of $G$. Take any basis for $A$ modulo the maximal ideal of $R$ and lift it to $A$. As $A$ is $R$-flat it is a basis $\{e_i\}$ of $A$. Now, pick $a_j$ a set of $R$-algebra generators for $A$. Applying the coproduct we get a finite expression $\Delta(a_j)=\sum_ie_i\otimes f^{\\;j}_i$. Consider now the $R$-submodule $V$ of $A$ spanned by the $e_i$ for which the $f^{\\;j}_i$ are non-zero. By coassociativity $V$ is a sub-comodule of $A$ giving a group scheme homomorphism $G\rightarrow \mathrm{GL}(V)$ which is a closed embedding as $V$ contains a set of $R$-algebra generators of $A$.

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Dear Torsten: the span of the fij's need not be a direct summand (when R is not a field), so how are you comparing coefficients in a tensor identity to see that V is a subcomodule? I think that is where the error should be (but I can't say for sure since the calculation to justify it wasn't given). –  BCnrd Apr 22 '10 at 6:52
    
I meant the first sentence literally, I didn't understand. Now I do :-) Indeed, why we should get a subcomodule seems to be where the argument breaks down. –  Torsten Ekedahl Apr 22 '10 at 7:45

The result in SGA 3, VIB Remarque 11.11.1, implies that an affine flat group scheme of finite type over a local Artinian principal ideal ring field $A$ (or over a Dedekind ring) has a closed embedding into $\mathrm{GL}(V)$, for some module $V$ locally free and of finite type over $A$. Over a Dedekind ring, V is then projective, and one can obtain the embedding into a $\mathrm{GL}(n)_A$ by Kevin Buzzard's comment to George's answer.

The above result is attributed to Raynaud, but unfortunately no proof is given. In Gille's and Polo's reissue of SGA 3, Proposition 11.11 in VIB says that an affine flat group scheme of finite type over (an arbitrary) local Artinian ring embeds in some $\mathrm{GL}(n)_A$. However, the proof given in the current version only covers the case where $A$ is a field.

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@A Stasinski: I knew about that part in original SGA3, where it was all over fields, so this revision is really surprising. It now seems to allow for any artin local ring (not sure where you saw the principality condition, though even for dual numbers it will be interesting to see the new idea!). But the proof at the key step near the end has not been updated at all from the field case as in the original version. In view of the SGA3 people with whom I have discussed this problem, it is astonishing to me what is now being asserted in the re-issued SGA3. I will contact Gille & Polo. Thanks. –  BCnrd Apr 22 '10 at 13:10
    
In the original (at least the version I have it says "regular locally Noetherian of dimension <= 1"). This excludes all Artinian rings except products of fields. –  Torsten Ekedahl Apr 22 '10 at 13:52
    
In the result attributed to Raynaud, one of the hypotheses is that the base be regular of $\dim\leq 1$. I see now that this actually excludes local Artinian rings which are not fields. This makes the assertion of Gille & Polo all the more interesting, if indeed they have a proof. Please keep us informed of what they have to say about this. –  A Stasinski Apr 22 '10 at 14:24
    
I don't think it is fair to say that there is a partial proof in the SGA 3 resissue. Exactly at the point where one is to deal with the implication (iv) => (v) they switch to the old version which is over a field. –  Torsten Ekedahl Apr 24 '10 at 8:53

(This is more of a comment to the question than an answer, but there are already many comments...)

For the ring of dual numbers A over k is it clear that there exist smooth affine group schemes over A which are not trivial i.e. do not come from k by base change? By smoothness we know that as schemes they are base changed and we can also assume that the identity is trivial (since we can translate). This allows one to describe any group scheme as above over A purely in terms of data on smooth affine group schemes over k.

Some computations I made suggest that there are in fact no deformations even as group schemes, but I am not not yet very confident. If someone knows a counterexample it would be useful; otherwise I will recheck my computations and post the details in a day or so.

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Deformations over $A$ are in bijection with the group cohomology $H^2(G,\mathfrak g)$, where $G$ is the smooth $k$-group to be deformed and $\mathfrak g$ is its Lie algebra. An example where this is non-trivial is when $G$ is the product of the additive group with itself. –  Torsten Ekedahl Apr 25 '10 at 13:20
    
Thanks a lot for the example. –  ulrich Apr 26 '10 at 4:33

Let's assume unipotent and characteristic zero (and eventually commutative), so that we can pass to Lie algebras. After using smoothness to choose coordinates, a deformation of $G_0$ over the dual numbers is controlled by a cocycle $\mathfrak g \otimes \mathfrak g\to \mathfrak g$, where $\mathfrak g$ is the Lie algebra of $G_0$. Adjoint to this is a map $\mathfrak g \to \mathfrak g\mathfrak l(\mathfrak g)$. If $G_0$ is commutative, then $G$ embeds in the group $G_0\rtimes GL(G_0)$, which is defined over the ground field and thus linear, as the graph of $G\to G_0\to \mathfrak{gl(g)}<GL(G_0)(F[\varepsilon])$. That is, the map takes an element of $G$, reduces mod $\varepsilon$, applies to the cocycle to get a matrix, multiplies the matrix by $\varepsilon$ and adds it to the identity. Maybe this works for unipotent groups, embedding $G$ in $\mathfrak g\rtimes Aut(G_0)$, adding the cocycle to the adjoint action, but I'm not sure why the cocycle would end up in the Lie algebra.

In positive characteristic, I think that the deformation of $G_a$ towards height 2 does not embed in $GL_2$ and thus not in $G_a\rtimes G_m$. I wouldn't be terribly surprised if all first-order deformations of $G_a$ embed in $GL_3$, though.

This was inspired by the noncommutative formal group $x+y+\varepsilon x^py$, which embeds in $G_a\rtimes G_m$ as $(x,1+\varepsilon x^p)$. Incidentally, this group doesn't lift to a domain.

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