Timeline for Non- simplicity of $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2019 at 19:44 | answer | added | Jérémy Blanc | timeline score: 2 | |
| Mar 25, 2019 at 9:57 | answer | added | YCor | timeline score: 6 | |
| Mar 25, 2019 at 9:49 | comment | added | Soby | @YCor thank you very much! Meanwhile if anyone else have any insightful comments do express your views here! | |
| Mar 25, 2019 at 7:52 | comment | added | YCor | Check the Ann. Fourier paper of Anne Lonjou algebra.dmi.unibas.ch/lonjou | |
| Mar 25, 2019 at 7:47 | comment | added | Soby | @YCor Do you have the references to those papers? | |
| Mar 25, 2019 at 7:35 | comment | added | Soby | @YCor because the automorphism $g:X\rightarrow X$ is chosen on the Kummer surface $X$. Since $X$ is rational to $CP^2$ then we can say the same for $G_C$. My question is how does the criterion laid put by the authors suggests that $g$ can be chosen to be real? | |
| Mar 25, 2019 at 7:32 | comment | added | YCor | ... So one should know that in the variety $X_m$ of birational self-transformations of $P^2$ of degree $m$, there are enough real points. I guess it's a rational variety over $\mathbf{Q}$, although I'm realizing that I'm not sure about this. Anyway, it's been proved by other authors that one can even find elements $g$ chosen in $Bir(P^2_Q)$. | |
| Mar 25, 2019 at 7:30 | comment | added | YCor | "The image of $g$ in $G_C$" doesn't make sense to me. The Cantat-Lamy results looks like "for every fixed $m\ge 20$, every generic $g\in Bir(P^2)$ of degree $m$ is such that the normal subgroup generated by $g^{100}$ is a proper subgroup". (I reinvented constants.) Generic probably means, outside a countable union of subvarieties (inside those transformations of given degree). | |
| Mar 25, 2019 at 7:22 | comment | added | Soby | @YCor Thank you for your response. I think i got it Just a small doubt that I still have. How can we choose $g$ to be real? This is because $g$ is chosen in $\text{Bir}(X)$. How can we ensure the image of $g$ in $G_C$ is real as well? | |
| Mar 25, 2019 at 7:03 | comment | added | Soby | Let us continue this discussion in chat. | |
| Mar 25, 2019 at 6:46 | comment | added | YCor | Note: "is a proper subgroup, and even has a trivial intersection with $PGL_3(C)$": more precisely it's a well-known consequence of an old result of M. Noether that these are equivalent (since $PGL_3(C)$ is simple and generates $Bir(P^2_C)$ as normal subgroup). Such a fact is not used in the real case. You just use that the intersection of the normal subgroup generated by $g$ in $Bir(P^2_R)$ is contained in the normal subgroup generated by $g$ in $Bir(P^2_C)$, hence has trivial intersection with $PGL_3(C)$, hence has trivial intersection with $PGL_3(R)$. | |
| Mar 25, 2019 at 6:45 | comment | added | Soby | @YCor the fact that $N$ is proper iff it intersects trivially with $PGL_3(C)$ follows from Noether-Castelnuovo. Can we say that same for real case? Since Noether-Castelnuovo need not hold for the real plane Cremona group. (Correct me if im wrong.) | |
| Mar 25, 2019 at 6:35 | comment | added | YCor | The point is that they show that for "sufficiently many" elements $g$ of $G_C=Bir(P^2_C)$, the normal subgroup generated by $g$ in $G_C$ is a proper subgroup, and even has a trivial intersection with $PGL_3(C)$. Their criterion makes it clear that $g$ can be chosen real, and hence the normal subgroup generated by $g$ in $G_R$ is proper, since it intersects trivially $PGL_3(R)$. | |
| Mar 25, 2019 at 6:35 | comment | added | Soby | @ThiKu I don't think this is what they are suggesting. | |
| Mar 25, 2019 at 6:33 | comment | added | Soby | @YCor I have edited my post. Sorry for that | |
| Mar 25, 2019 at 6:33 | comment | added | YCor | @ThiKu no this is not what they mean | |
| Mar 25, 2019 at 6:32 | history | edited | Soby | CC BY-SA 4.0 |
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| Mar 25, 2019 at 6:32 | comment | added | ThiKu | 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. | |
| Mar 25, 2019 at 6:31 | comment | added | YCor | You're misquoting: they say (p3) "Our article directly implies...", not that "[their] result directly implies". | |
| Mar 25, 2019 at 6:28 | history | edited | ThiKu | CC BY-SA 4.0 |
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| Mar 25, 2019 at 6:26 | history | asked | Soby | CC BY-SA 4.0 |