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:

  1. What if we have an action of $SL_2(\mathbb Q)$?
  2. Instead of $\mathbb{Q}$, consider the additive group $\mathbb{Z}[1/2]$.
  3. 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)?

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    $\begingroup$ You want some conditions on the action? As $\mathbb Q$ is a direct summand of the abelian group $\mathbb R$, every action of $\mathbb Q$ (on anything) extends to an action of $\mathbb R$. $\endgroup$ – Mariano Suárez-Álvarez Feb 5 '13 at 6:17
  • $\begingroup$ Probably it can be asked by: "does not extend to an action of an algebraic group". Btw, the OP asks about birational actions, is the question irrelevant for algebraic actions? $\endgroup$ – YCor Feb 5 '13 at 9:26
  • $\begingroup$ Apologies for being vague, I've tried to fix the question a bit. Yves, unfortunately I'm ignorant even about the algebraic situation. $\endgroup$ – sfilip Feb 5 '13 at 18:30
  • $\begingroup$ @sfilip notice that my comment still applied after your remark: suppose $\mathbb Q$ acts on your variery by birational automorphisms, and let $P$ be a subgroup of $\mathbb R$ such that $\mathbb R=P\oplus\mathbb Q$. Define an action of $\mathbb R$ on the variety so that $(p,q)$ acts just as $q$ acts. Then all elements of $\mathbb R$ do act by birational automorphisms. My point was that you probably want some sort of continuity of the action on the group. $\endgroup$ – Mariano Suárez-Álvarez Feb 5 '13 at 19:27
  • $\begingroup$ @Mariano I was hoping to suggest that "birational action" means a birational map $\mathbb{G}_a\times X \to X$ with the natural compatibility conditions making it an action. This would be one way to make sense of the question, but there might be other reasonable ones. $\endgroup$ – sfilip Feb 6 '13 at 0:58

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.

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  • $\begingroup$ Jérémy, thanks a lot for the results and references! I was in fact curious if anything is known about the growth rate of the degree of a general birational automorphism (in arbitrary dimension)? I realize one can't expect a general classification, but perhaps something besides the possibility of exponential growth is known? $\endgroup$ – sfilip Feb 6 '13 at 16:21
  • $\begingroup$ This is a very good question and is for the moment not known. There are some behaviours between exponential , quadratic and linear, which should be related to some invariant fibrations but I do not think that it is really established now. It is one of the aims of future projects to determine this. $\endgroup$ – Jérémy Blanc Feb 7 '13 at 12:00
  • $\begingroup$ Thanks a lot for this answer and for all the references! $\endgroup$ – sfilip Feb 8 '13 at 15:58
  • $\begingroup$ Dear Jérémy, as a one-time dabbler in the Cremona group, I enjoy your highly informed answers a lot! $\endgroup$ – user5117 Feb 8 '13 at 21:06
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    $\begingroup$ For a general surface, just take a ample divisor D and compute the limit of $(D \cdot f^n(D))^{1/n}$. You can see that it is independent of the choice of $D$ and that it is a birational invariant (if $f\in Bir(X)$ is conjugate to $g\in Bir(Y)$ then both have the same dynamical degree). $\endgroup$ – Jérémy Blanc Aug 9 '13 at 8:40

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).

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  • $\begingroup$ Lieberman's theorem was also proved independently and simultaneously by A. Fujiki here link.springer.com/article/10.1007%2FBF01403162 $\endgroup$ – YangMills Feb 5 '13 at 19:44
  • $\begingroup$ Why is it clear that $\mathbb{Z}[1/2]$ belongs to $ker(\phi)$? I'm thinking about something like an elliptic surface which could have a copy of $\mathbb{Z} \cong \mathbb{Z}[1/2]$ in its automorphism group, given by translation by a non-torsion section. It is not clear to me that this copy of $\mathbb{Z}$ acts trivially on the cohomology of the surface. $\endgroup$ – Daniel Loughran Feb 5 '13 at 20:03
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    $\begingroup$ Daniel, the claim is that any homomorphism from $\mathbb Z[1/2]$ to $GL(n,\mathbb Z)$ sends $\mathbb Z[1/2]$ to $1$, since $1$ in $GL(n,\mathbb Z)$ is the only infinitely divisible element. $\endgroup$ – Dmitri Panov Feb 5 '13 at 20:26
  • $\begingroup$ Right I see. I got confused with the notation and thought that $\mathbb{Z}[1/2]$ meant $(1/2)\mathbb{Z}$. Thanks for clearing it up. $\endgroup$ – Daniel Loughran Feb 5 '13 at 20:35
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    $\begingroup$ @Dmitri: it is certainly not true that any homomorphism from $\mathbf{Z}[1/2]$ to $GL(n,\mathbf{Z})$ has a trivial image. For instance, $\mathbf{Z}[1/2]$ admits $\mathbf{Z}/n\mathbf{Z}$ as a quotient for any odd $n$. What survives is that any homomorphism from $\mathbf{Z}[1/2]$ to $GL(n,\mathbf{Z})$ has a finite image (indeed cyclic of odd order). $\endgroup$ – YCor Feb 6 '13 at 0:07

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