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In this paper: https://perso.univ-rennes1.fr/serge.cantat/Articles/nsgc-acta-c.pdf the authors said that their article 'directly implied' that $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$ is not simple as well. How can we infer the non-simplicity of $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$ just from the fact that $\text{Bir}(\mathbb{P}_\mathbb{C}^2)$ is not simple?

I am thinking of choosing a proper normal subgroup, say $N$, of $\text{Bir}(\mathbb{P}_\mathbb{C}^2)$ and intersecting it with $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$. But there is no guarantee that $\text{Bir}(\mathbb{P}_\mathbb{R}^2)\cap N$ is non-trivial. How does one choose an $N$ s.t $\text{Bir}(\mathbb{P}_\mathbb{R}^2)\cap N$ is nontrivial?

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    $\begingroup$ You're misquoting: they say (p3) "Our article directly implies...", not that "[their] result directly implies". $\endgroup$
    – YCor
    Commented Mar 25, 2019 at 6:31
  • $\begingroup$ The logic has to be the other way: you have to start with a normal subgroup or $Bir(RP^2)$ and then prove that it is still normal in $Bir(CP^2)$ and hence trivial. $\endgroup$
    – ThiKu
    Commented Mar 25, 2019 at 6:32
  • $\begingroup$ @ThiKu no this is not what they mean $\endgroup$
    – YCor
    Commented Mar 25, 2019 at 6:33
  • $\begingroup$ @YCor I have edited my post. Sorry for that $\endgroup$
    – Soby
    Commented Mar 25, 2019 at 6:33
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    $\begingroup$ Check the Ann. Fourier paper of Anne Lonjou algebra.dmi.unibas.ch/lonjou $\endgroup$
    – YCor
    Commented Mar 25, 2019 at 7:52

2 Answers 2

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(Since the question is answered in comments, I summarize it so as to provide an acceptable answer.)

The authors do not say that the non-simplicity result of $\mathrm{Bir}_{\mathbf{C}}(\mathbb{P}^2)$ implies that of$\mathrm{Bir}_{\mathbf{R}}(\mathbb{P}^2)$. They say that (the method of) the article implies that.

Namely, they show they show that for "many" $g\in \mathrm{Bir}_{\mathbf{C}}(\mathbb{P}^2)$, the normal subgroup generated by $g$ has trivial intersection with $\mathrm{PGL}_3(\mathbf{C})$. The way they define "many" applies to showing that $g$ can be chosen to belong to $\mathrm{Bir}_{\mathbf{R}}(\mathbb{P}^2)\smallsetminus\{1\}$. Hence the normal subgroup generated by such $g$ is a nontrivial proper subgroup of $\mathrm{Bir}_{\mathbf{R}}(\mathbb{P}^2)$.

Actually, in her Ann. Fourier paper, A. Lonjou provides an example of such $g$ inside $\mathrm{Bir}_{\mathbf{Q}}(\mathbb{P}^2)$, namely some power of the birational self-transformation $(x,y)\mapsto (y,y^2-x)$.

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Let me add more information to the answer of YCor (which is of course completely correct and accurate).

In the article of S. Cantat and S. Lamy, they take an element $g\in \mathrm{Bir}(\mathbb{P}^2)$ and give a criterion to say when the normal subgroup generated by a large power of $g$ is a strict subgroup of $g\in \mathrm{Bir}(\mathbb{P}^2)$. The proof works essentially over any algebraically closed field. Then, one has to give explicit elements that satisfy this property. This is done by them in the case where the field is algebraically closed, and can also be done for any field, as it was done by A. Lonjou in "Non simplicité du groupe de Cremona sur tout corps", Annales de l'Institut Fourier, Vo 66, n°5, p.2021-2046, 2016.

The article is in french, freely available on the journal (open access free journal). The author also provides a translation on her webpage.

If you are interested in the field of real numbers, I would suggest you to keep a look at the article of S. Zimmermann "The abelianisation of the real Cremona group." Duke Math. J. vol. 167, no.2 (2018), 211--267. She shows that not only $\mathrm{Bir}_{\mathbb{R}}(\mathbb{P}^2)$ is not simple, but also that the abelianisation is a direct sum of $\mathbb{Z}/2\mathbb{Z}$ indexed by $\mathbb{R}$.

Also, for a version in higher dimension (which works over $\mathbb{C}$, $\mathbb{R}$ or $\mathbb{Q}$), see https://arxiv.org/abs/1901.04145

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