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Let $k$ be a field; I'm going to discuss linear algebraic groups over $k$. The question I'll pose is only interesting when the characteristic is $p>0$.

1. Some motivation

A vector group is an algebraic group isomorphic (over $k$) to a product of (finitely many) copies of the additive group $\mathbf{G}_a$. Let $U$ be any connected unipotent group over $k$. If $k$ is perfect, or if $U$ is $k$-split, then $U$ has a filtration $1 = U_0 \subset U_1 \subset \cdots \subset U_n = U$ where each $U_i$ is normal in $U$ and each $U_i/U_{i-1}$ is a vector group. If $U$ is a normal subgroup of a linear group $G$, you can arrange that each $U_i$ is invariant under conjugation by $G$. This suggests that to study the group extension $$(*) \quad 1 \to U \to G \to G/U \to 1,$$ one might profitably study first the case where $U$ is a vector group.

Let $G$ be a linear group and suppose that the unipotent radical $R$ of $G$ is defined over $k$ and is $k$-split (each of these conditions can fail in general; they always hold when $k$ is perfect). Then the question of whether $(*)$ splits when $U=R$ is precisely the question of whether $G$ has a Levi factor; cf. this question of Jim Humphreys.

2. Action on a vector group

Let $U$ be a vector group and suppose that the linear group $G$ acts on $U$ by algebraic group automorphisms. The action of $G$ on $U$ determines an action of $G$ on $\mathfrak{u}=\operatorname{Lie}(U)$.

Question (first approximation): Is there a $G$-equivariant isomorphism $\mathfrak{u} \to U$ (where the vector space $\mathfrak{u}$ is viewed as a vector group in the obvious fashion)?

I'll say that the action of $G$ on $U$ is linearizable if there is such an equivariant isomorphism.

Some remarks: If the action of $G$ on $U$ is linearizable, then $G$ centralizes the action of the multiplicative group $\mathbf{G}_m$ on $U$ obtained by transport of structure from scalar multiplication on $\mathfrak{u}$. This $\mathbf{G}_m$-action determines a grading on the algebra $k[U]$ of regular functions on $U$ which is stable for the action of $G$ on $k[U]$.

3. Partial answers

Postive: If the characteristic of $k$ is $0$, the above question has always an affirmative answer. (Use the exponential mapping $\mathfrak{u} \to U$; this is $G$ equivariant e.g. because there is a faithful linear representation of the semidirect product $G \ltimes U$).

Negative: If $G = \mathbf{G}_a$, the question has a negative answer in char. $p>0$. Consider the action of $G$ on $V = {\mathbf{G}_a}^2$ given by the rule $$x.(a,b) = (a + xb^p,b).$$ The action of $G$ on $\operatorname{Lie}(V)$ is trivial, so there is no $G$-equivariant isomorphism $\operatorname{Lie}(V) \to V$.

4. What I meant to ask.

Refined question: If $G$ is reductive, is the action of $G$ on a vector group $U$ always linearizable?

An affirmative answer would mean that when $G/U$ is reductive, one can study the group extension $(*)$ using cohomology of linear representations of $G/U$.

Note that are known examples even in char. 0 (Schwarz, Kraft,...) of actions of reductive $G$ on affine space $A=\mathbf{A}^n$ which are not linear, but, at least in char. 0, these actions don't respect a vector group structure on $A$.

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2 Answers 2

I somehow just realized that although I've now know the answer for a while, I had failed to give the answer to this question...

In fact, for (any) field of characteristic $p>0$, there are (plenty of) examples of reductive groups $G$ which act by group automorphisms on a vector group $U$ for which the action is not linear -- i.e. there is no $G$-equivariant isomorphism $\operatorname{Lie}(U) \simeq U$.

For an example, see < my preprint @ gmcninch.math.tufts.edu > (especially section 5). Given a non-split extension $$(\flat) \quad 0 \to W \to E \to V \to 0$$ of finite dimensional $G$-modules, one can view the extension as being defined by a suitable sort of cocycle. One can then "twist" this cocycle by Frobenius, yielding a vector group $\tilde E$ on which $G$ acts by group automorphisms.

As a $G$-module, the Lie algebra $\operatorname{Lie}(\tilde E)$ is isomorphic to $V \oplus W^{(1)}$, where the exponent on $W$ denotes the first Frobenius twist.

By construction, there is a $G$-equivariant surjection $\pi:\tilde E \to V$, and the kernel of $\pi$ identifies with $W^{(1)}$. But there is no $G$-equivariant homomorphism of algebraic groups $V \to \tilde E$ which is a section to $\pi$, hence there can be no $G$-equivariant isomorphism between $\tilde E$ and $\operatorname{Lie}(\tilde E)$.

Since there are plenty of non-split extensions $(\flat)$, there are plenty of examples...

On a more positive note, let $V$ be a vector group on which $G$ acts. The main result of the preprint just mentioned shows that if $G$ is connected, if the unipotent radical of $G$ is defined over the ground field, and if $\operatorname{Lie}(V)$ is a simple $G$-module, then the action of $G$ on $V$ is linear. In particular, it follows (for $G$ as above) that if $G$ acts on a split unipotent group $U$, there is a filtration of $U$ by closed $G$-stable subgroups such that the quotients are vector groups on which $G$ acts linearly.

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[Update below: I think this answers the question now.] Let's suppose that we're working over an algebraically closed field $k$ of characteristic $p$. Let $Fr$ denote the Frobenius automorphism.

From a recent paper of Tanaka and Kaneta (Hiroshima Math. J. 2008), I found that $Aut(G_a^n)$ (the automorphism group of the vector group $G_a^n$ over $k$) can be identified with: $$\{ A \in M_n(k)[[F]]^\times : A,A^{-1} \in M_n(k)[F] \}.$$ Here, we work in a ring of noncommutative power series in the formal variable $F$, with coefficients in the matrix ring $M_n(k)$, subject to the natural relation: $$F \cdot m = Fr(m) \cdot F.$$ I.e., one may pass all F's to the right, by applying Frobenius to the entries of the matrices.

Now, there is a natural surjective homomorphism $lin$ from $Aut(G_a^n)$ to $GL_n(k)$, obtained by taking the "constant term" of the power series.

What does the kernel of $lin$ look like? Is it (representable by?) a pro-unipotent group scheme over $k$? I have no idea.

Update: Following David's comment, it appears that $Ker(lin)$ is a pro-unipotent group over $k$.

The image of a reductive group is reductive, even in the pro-affine setting (Section 3 of Mostow-Hochschild, Pro-Affine Algebraic Groups, Amer. J. of Math 1969. Happily this result is right before the assumption of char=0). Any reductive subgroup of a unipotent group is trivial, in this same reference. Hence the image of a reductive group in $Ker(lin)$ must be trivial.

Update: Let $\alpha: G \rightarrow Aut(G_a^n)$ be an algebraic action of $G$ on the vector group $G_a^n$.

It follows that $H = \alpha(G)$ is a reductive subgroup of the pro-affine group $Aut(G_a^n)$, which does not intersect the pro-unipotent subgroup $U = Ker(lin)$. It remains to prove that $H$ is conjugate to the subgroup $GL_n \subset Aut(G_a^n)$.

Let $H' = lin(H)$ be the projection of $H$ in $GL_n$ -- this projection is an isomorphism from $H$ to $H'$. We find that $H \cdot U = H' \cdot U$, and $U$ is the unipotent radical of $H \cdot U$.

Now the remaining question: are Levi factors of $H \cdot U$ (geometrically, since we work over $k = \bar k$) conjugate? To this, I think I defer back to the OP, who has recently written a paper on this topic, which will hopefully adapt to this pro-affine group setting.

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Hi Marty! You can filter $Aut(G_a^n)$ with $G^j$ being the group of such power series which are the identity modulo $F^{j}$. Then $G^0/G^1$ is $GL_n(k)$, and $G^j/G^{j+1}$ is $G_a^{n^2}$ for higher $j$. If we were dealing with finite dimensionsal group schemes, that would show $G$ is unipotent. I don't know, though, what subtleties may exist in the pro-unipotent case. –  David Speyer Jul 22 '10 at 18:06
    
@David: How can you be sure that the successive quotients are $G_a^{n^2}$? Is this written down somewhere? To me, it seems clear that the quotients are closed subgroups of $G_a^{n^2}$ -- sufficient for the pro-unipotence result -- but not that the quotients are the "whole thing". –  Marty Jul 22 '10 at 18:56
    
I have no idea whether it is written down, but isn't it clear? You just want to show that, for any $n \times n$ matrix $A$, the element $1-A F^j$ is a unit. It's inverse is $1+A F^j + A Fr^j(A) F^{2j} + A Fr^j(A) Fr^{2j}(A) F^{2j} + \cdots$. –  David Speyer Jul 22 '10 at 19:08
    
The point that I got from Tanaka-Kaneta is that one must work with power series which are polynomials in $F$, and whose inverse is also polynomial in $F$. That may imply some nilpotence condition on the coefficient matrices. –  Marty Jul 22 '10 at 19:10
    
Actually - that sounds dumb. I guess these are just invertible polynomials in the noncommutative polynomial ring. I just copied from Tanaka-Kaneta without thinking. –  Marty Jul 22 '10 at 19:12

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