Smooth linear algebraic groups over the dual numbers

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

(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.

Answer 5

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 $A$.

Answer 6

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.