Timeline for Birational Automorphisms and infinite divisibility
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2013 at 16:39 | comment | added | YCor | PS: an element of order 3 is infinitely 2-divisible | |
| Feb 6, 2013 at 12:59 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Feb 6, 2013 at 11:28 | comment | added | YCor | @Dmitri ok but your edit is still not correct. The argument is that a non-torsion element in the $GL$ cannot be infinitely 2-divisible. In group theory "divisible" means "divisible by any positive integer". | |
| Feb 6, 2013 at 11:18 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Feb 6, 2013 at 9:07 | comment | added | Dmitri Panov | Yves, thanks I was a bit sloppy :) . But for \mathbb Q this is ture :) | |
| Feb 6, 2013 at 0:07 | comment | added | YCor | @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). | |
| Feb 5, 2013 at 20:35 | comment | added | Daniel Loughran | 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. | |
| Feb 5, 2013 at 20:32 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Feb 5, 2013 at 20:26 | comment | added | Dmitri Panov | 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. | |
| Feb 5, 2013 at 20:03 | comment | added | Daniel Loughran | 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. | |
| Feb 5, 2013 at 19:44 | comment | added | YangMills | Lieberman's theorem was also proved independently and simultaneously by A. Fujiki here link.springer.com/article/10.1007%2FBF01403162 | |
| Feb 5, 2013 at 19:29 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Feb 5, 2013 at 19:19 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
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| Feb 5, 2013 at 18:52 | history | answered | Dmitri Panov | CC BY-SA 3.0 |