Birational Automorphisms and infinite divisibility Suppose $X$ is some algebraic variety. It can be over $\mathbb{C}$, but it doesn't have to (but char $0$ preferred).
Is it possible that the additive group $\mathbb{Q}$ acts on it birationally, without this action extending to a birational action of the additive $1$-parameter group $\mathbb{G}_a$?
What if we further assume that we do have an extension of a continuous action of $\mathbb{G}_a$, but only on some subset of the variety (the field now has a topology)?
Possible variations would be:


*

*What if we have an action of $SL_2(\mathbb Q)$?

*Instead of $\mathbb{Q}$, consider the additive group $\mathbb{Z}[1/2]$.

*The field I'm actually interested in is $\mathbb{R}$, but examples over $\mathbb{C}$ would be great, too.


Essentially, I would like to know if people have considered when can a very divisible group act on an algebraic variety?
Remark: What I mean by a "birational action" of a group might be vague, but one interpretation could be a birational map $\mathbb G_a \times X \to X $ with the compatibility conditions making it an action.
Added Feb 6, 2013 (Corrected, thanks to Jérémy) According to the paper linked at in this MO question by Francesco Polizzi, in the algebraically closed case, the birational automorphisms of any $X$ inject set-theoretically into those of $\mathbb P^n$, with $n>\dim X + 1$. However, the group structure need not be preserved.
I've added another possibility, say we have a birational action of $SL_2\mathbb Q$ (or $SL_2 (\mathbb Z[1/2])$). Need it extend to the full group (of $\mathbb C$ or $\mathbb R$ points)? 
 A: The question is about birational actions on varieties. Here is the answer for rational surfaces.
Lemma: If an element $g$ of $Bir(X)$, where $X$ is a rational surface, satisfies that for infinitely many integers $n$ there exists $h\in Bir(X)$ with $h^n=n$, then $g$ is contained in a one-parameter group of $Bir(X)$.
Proof: We can conjugate $g$ to a birational map of $\mathbb{P}^2$ and look at the sequence $\{\mathrm{deg}(g^k)\}_{k\in \mathbb{N}}$. The fact that $g$ is "divisible" by infinitely many integers implies that the degree sequence is bounded. Indeed, it is known that the sequence, if not bounded, grows linearly, quadratically or exponentially (Diller-Favre). In the first two cases, the coefficient of the quadratic or linear polynomial is bounded by below (Theorems C and D in http://arxiv.org/abs/1109.6810 ) and in the last case the eigenvalue, or dynamical degree, is more than 1.17. This implies that we cannot divide infinitely many times an element with unbounded degree sequence.
Now, the boundedness of the degree sequence implies that $g$ has finite order or that it is conjugate to an element of $GL(2)$ (see Theorem A in loc.cit.) In the first case, the element is also conjugate to an element of $GL(2)$ (see classification of elements of finite order of Bir(X), in http://arxiv.org/abs/0809.4673. )

Now, with this lemma and the classification of centralisers of linear elements in $Bir(\mathbb{P}^2)$ made in http://arxiv.org/abs/1109.6810, you get that any action of an ininfite-divisible abelian group extends to an action of an algebraic group.
A: Edit. I have not realised that this question is on Birational automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.
This answer concerns automorphisms of complex projective manifolds (and more generally Kahler ones). 
Statement. If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$ unless it factors through the action of a finite group (thanks to Yves). This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms
and deformations of Kahler manifolds, 1978).
Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.
Lieberman-Fujiki Theorem. Consider the action of $Aut(X)$ on  $H^*(X,\mathbb Z)$, 
$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$. 
Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.
Proof of the statement. Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then a finite index subgroup of it belong to $ker(\phi)$ (indeed $GL(H^*(X,\mathbb Z))$ does not have infinitely $2$-divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.
I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).
