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)?

on the group. – Mariano Suárez-Alvarez♦ Feb 5 '13 at 19:27