Timeline for Monodromy representations are "quasi-unipotent"
Current License: CC BY-SA 3.0
9 events
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| Jun 14, 2017 at 21:11 | comment | added | Venkataramana | No, I was not saying that. For example, the fundamental group of an ellptic curve is the free abelian group on two generators, and you can find (even faithful) representation, with every element unipotent. | |
| Jun 14, 2017 at 21:00 | comment | added | Randy | @Venkataramana Ok, I understand. This answers my question. I was just confused because I thought you were saying that, for all smooth projective varieties $S$, the fundamental group $\pi_1(S)$ has no unipotent elements (with respect to any faithful linear representation). But that's not true right? | |
| Jun 14, 2017 at 20:16 | comment | added | Venkataramana | @Dan Litt Thank you! | |
| Jun 14, 2017 at 17:29 | comment | added | Daniel Litt | @Venkataramana This is a really nice example! | |
| Jun 14, 2017 at 15:55 | comment | added | Venkataramana | the Shimura variety S is of the form $\Gamma\backslash G/K$ where $G$ is a Hermitian type real semi-simple group and $K$ its maximal compact subgroup. Thus, the compactness is equivalent to the compactness of $\Gamma\backslash G$. If $\Gamma $ s a co-compact arithmetic subgroup of $G$, then $\Gamma$ cannot contain unipotent elements (this is the easier part of the Godement criterion; you can find it in Raghunathan's book on discrete subgroups of Lie groups). | |
| Jun 14, 2017 at 12:02 | comment | added | Randy | @Venkataramana Thank you for your comment. I think I'm missing the argument for why "compactness of $S$ ensures $\pi_1(S)$ has no unipotent elements". How does one prove that? | |
| Jun 14, 2017 at 12:00 | comment | added | Venkataramana | The answer is no, in general. If S is a compact Shimura variety which is a closed subvariety of the moduli of ppav with "high" level structure, then you get a family of abelian variety parametrised by S. The level structure ensures that there are no torsion elements in the fundamental group of S, and compactness of S ensures there are no unipotent elements in $\pi _1(S )$ | |
| Jun 14, 2017 at 11:52 | review | First posts | |||
| Jun 14, 2017 at 12:07 | |||||
| Jun 14, 2017 at 11:50 | history | asked | Randy | CC BY-SA 3.0 |