Timeline for Complexifications of hyperbolic manifolds
Current License: CC BY-SA 4.0
20 events
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Mar 25, 2021 at 22:12 | comment | added | Ian Agol | @MoisheKohan yes, that might be interesting. By Nash, we know any smooth manifold is a real algebraic variety. So one should be able look at it as a complex variety. The question then is whether one can resolve singularities in an equivariabt way without affecting the real locus? | |
Mar 25, 2021 at 22:07 | comment | added | Moishe Kohan | Incidentally, I think your question is interesting even without assuming that the complex manifold is hyperbolic; a natural assumption would be "projective." | |
Mar 25, 2021 at 22:01 | comment | added | Moishe Kohan | @IanAgol Ok, I will take a look. | |
Mar 25, 2021 at 20:38 | history | edited | Ian Agol | CC BY-SA 4.0 |
Added citations for the claimed results.
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Mar 25, 2021 at 20:37 | comment | added | Ian Agol | @MoisheKohan: I've added references proving the claim of the integrality result for complex hyperbolic manifolds. Apparently the Esnault-Groechenig result is applicable. | |
Mar 25, 2021 at 20:35 | history | undeleted | Ian Agol | ||
Mar 25, 2021 at 20:33 | history | edited | Ian Agol | CC BY-SA 4.0 |
Added citations for the claimed results.
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Mar 16, 2021 at 17:07 | history | deleted | Ian Agol | via Vote | |
Mar 16, 2021 at 17:04 | comment | added | Ian Agol | @MoisheKohan Thanks for this, I didn’t notice that difference. I admit that I don’t really understand their paper, and was informed of this corollary from David Fisher. I’ll delete the answer. | |
Mar 16, 2021 at 16:37 | comment | added | Moishe Kohan | I think, you are misapplying their result: They require $H^1(X, Ad\rho)=0$ where $\rho$ is a representation into $GL(n, {\mathbb C})$, while your representation is rigid in $SU(n,1)$. This is not the same. The simplest example is given by lattices with $b_1>0$: They are rigid in $SU(n,1)$ but not in $GL(n+1,C)$. Furthermore, my recollection from reading Deligne and Mostow (admittedly, long time ago) is that some of their non-arithmetic examples are non-integral. | |
Feb 24, 2019 at 0:32 | comment | added | Ian Agol | @YCor: Yes, that's right. | |
Feb 23, 2019 at 22:54 | comment | added | YCor | So if I understand correctly, you say that Esnault-Groechenig method proves that the structural representation of any $\pi_1$ of a (compact? closed?) hyperbolic complex 3-manifold has integral traces. By "structural" I mean the obvious one, in $SU(3,1)$... this certainly deserves a precise name. Second, if a compact real hyperbolic 3-manifold embeds geodesically into a compact complex one, then the restriction of the structural representation of the larger one restricts to to structural representation of the smaller one (this seems obvious but was initially unclear in my mind). | |
Feb 23, 2019 at 22:01 | comment | added | Ian Agol | @YCor : Okay, I’ve tried to clarify again, I’m only taking about the representation coming from the hyperbolic structure. | |
Feb 23, 2019 at 21:59 | history | edited | Ian Agol | CC BY-SA 4.0 |
added 48 characters in body; added 2 characters in body
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Feb 23, 2019 at 21:56 | comment | added | YCor | I still don't understand. "the discrete faithful representation" is just one representation, but you want a property of all a class of representations... all discrete faithful representations valued in $SU(n,1)$ for all $n$? | |
Feb 23, 2019 at 21:46 | vote | accept | Ian Agol | ||
Mar 16, 2021 at 17:07 | |||||
Feb 23, 2019 at 21:46 | history | edited | Ian Agol | CC BY-SA 4.0 |
More details and attributions.; Post Made Community Wiki
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Feb 23, 2019 at 21:40 | comment | added | Ian Agol | @YCor: No, just the discrete faithful rep. | |
Feb 23, 2019 at 21:11 | comment | added | YCor | Do you mean that there exist real hyperbolic 3-manifolds whose $\pi_1$ has no embedding into $\mathrm{GL}_d(\mathbf{Z})$ for any $d$? I'm not sure this is what you're saying; I'm just trying to understand. I'd be happy if you slightly expand the argument. | |
Feb 23, 2019 at 21:04 | history | answered | Ian Agol | CC BY-SA 4.0 |