Timeline for Inheritance of arithmeticity properties in orbifold strata
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2023 at 16:20 | comment | added | Ian Agol | I think this paper might partially answer the intended question. arxiv.org/abs/2105.06897 | |
| Mar 14, 2023 at 17:32 | vote | accept | Ethan Dlugie | ||
| Mar 14, 2023 at 14:37 | comment | added | Ben Wieland | If a stratum has dimension zero, is it arithmetic? Is this a matter of convention? If it is an intrinsic property the locus, they must be arithmetic. A generic hyperelliptic curve is not arithmetic. Its fundamental group can be enlarged to include the hyperelliptic element, which fixes an isolated point. | |
| Mar 13, 2023 at 1:20 | answer | added | Moishe Kohan | timeline score: 1 | |
| Mar 12, 2023 at 20:59 | comment | added | Moishe Kohan | No, they will not go away. | |
| Mar 12, 2023 at 20:39 | history | edited | Ethan Dlugie | CC BY-SA 4.0 |
altering the situation to hopefully make the question more meaningful
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| Mar 12, 2023 at 20:37 | comment | added | Ethan Dlugie | Thank you. I'm also editing the question a bit to restrict to the complex hyperbolic setting (in which I think these issues may disappear?) | |
| Mar 12, 2023 at 18:39 | comment | added | Moishe Kohan | I will write an answer of sorts, correcting your question in the process, in a little while. | |
| Mar 12, 2023 at 18:17 | comment | added | Moishe Kohan | No, that will not suffice either. The orbifold locus could be a 3-valent graph. | |
| Mar 12, 2023 at 18:11 | history | edited | YCor |
edited tags
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| Mar 12, 2023 at 17:09 | history | edited | Ethan Dlugie | CC BY-SA 4.0 |
Clarifying the question
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| Mar 12, 2023 at 17:08 | comment | added | Ethan Dlugie | Ah I see. Let me modify my question to request that $M'$ be a connected component of an orbifold locus, that's what I really had in mind. | |
| Mar 12, 2023 at 0:59 | comment | added | Moishe Kohan | OK, suppose you have a lattice $\Gamma_0< PSL(2,{\mathbb R})$ uniformizing a hyperelliptic hyperbolic surface $S$. Then $S$ admits a hyperelliptic involution $\tau$. Lifting $\tau$ to the hyperbolic plane, we obtain a lattice $\Gamma< PSL(2,{\mathbb R})$ which contains $\Gamma_0$ as an index 2 subgroup. The quotient $H^2/\Gamma$ is an orbifold $O$ with at least two singular points (projections to $O$ of the fixed-points of $\tau$). Hence, the orbifold locus of $O$ is disconnected, implying that it cannot be a lattice quotient (under the most common definition). Hence, your belief is wrong. | |
| Mar 12, 2023 at 0:22 | comment | added | Ethan Dlugie | @MoisheKohan could you elaborate a bit? (Non)arithmetic lattices have come up in the things I've been working on but almost as a side effect rather than the main focus. I'm not an expert. | |
| Mar 11, 2023 at 4:19 | comment | added | Moishe Kohan | Are your symmetric spaces required to be connected? If yes, then the answer is clearly negative. | |
| Mar 11, 2023 at 2:08 | history | asked | Ethan Dlugie | CC BY-SA 4.0 |